Stochastic Calculus: A practical Introduction (Probability and Stochastics Series)

-

I A Practical Introduction

PROBABILITY AND STOCHASTICS SERIES Edited by Richard Durrett and Mark Pirisky

Probability and Stochastics Series

Linear Stochastic Control Systems, Guanrong Chen, Goong Chen, and Shia-Hsun Hsu Advances in Queueing: TheonJ, Methods, and Open Problems, Jewgeni H. Dshalalow Stochastics Calculus: A Practical Introduction, Richard Durrette

A Practical Introduction

Chaos Expansion, Multiple Weiner-Ito Integrals and Applications, Christian Houdre and Victor Perez-Abreu White Noise Distribution Theory, Hui-Hsiung Kuo Topics in Contemporary Probabilitlj, J. Laurie Snell

Richard Durrett

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.

This book is the :C-e-incarnation of my fi�st: beok Bro·w�i�'!l Motion and Martingales in A nalysis, which was published by Wadsw..pr:th ip.-1984. For more than a decade I have used Chapters 1, 2, 8 , �nq-9 of tii'at'_ bo·ok to give "reading courses" to graduate students who have compJ�t.ed. t_he first year graduate prob­ ability course and were interested in learning more about processes that move continuously in space and time. Taking the advice from biology that ''form fol­ lows function" I have taken that material on stochastic integration, stochastic differential equations, Brownian motion and its relation to partial differential equations to be the core of this book {Chapters 1-5). To this I have added other practically important topics: one dimensional diffusions, semigroups and generators, Harris chains, and weak convergence. I have struggled with this material for almost twenty years. I now think that I understand most of it, so to help you master it in less time, I have tried to explain it as simply and clearly as I can. My students' motivations for learning this material have been diverse: some have wanted to apply ideas from probability to analysis or differential geometry, others have gone on to do research on diffusion processes or stochastic partial differential equations, some have been interested in applications of these ideas to finance, or to problems in operations research. My motivation for writing this book, like that for Probability Theory and Examples, was to simplify my life as a teacher by bringing together in one place useful material that is scattered ip. a variety of sources. An old joke says that "if you copy from one book that is plagiarism, but if you copy from ten books that is scholarship." From that viewpoint this is a scholarly book. Its main contributors for the various subjects are (a) stochastic integration and differential equations: Chung and Williams {1990), Ikeda and Watanabe {1981), Karatzas and Shreve {1991), Protter {1990), Revuz and Yor {1991), Rogers and Williams {1987), Stroock and Varadhan {1979); (b) partial differential equations: Folland (1976) , Friedman (1964) , {1975), Port and Stone {1978) , Chung and Zhao {1995); (c) one dimensional diffusions: Karlin and Taylor {1981); {d) semi-groups and generators: Dynkin {1965) , Revuz and Yor {1991); (e) weak convergence: Billingsley {1968) , Ethier and Kurtz {1986), Stroock and Varadhan {1979). If you bought all those books you would spend more than $1000 but for a fraction of that cost you can have this book, the intellectual equivalent of the ginzu knife.

vi

Preface

Shutting off the laugh-track and turning on the violins, the road from this book's first publication in 1984 to its rebirth in 1996 has been a long and winding one. In the second half of the 80's I accumulated an embarrassingly long list of typos from the first edition. Some time at the beginning of the 90's I talked to the editor who brought my first three books into the world, John Kimmel, about preparing a second edition. However, after the work was done, the second edition was personally killed by Bill Roberts, the President of Brooks/Cole. At the end of 1992 I entered into .a contract with Wayne Yuhasz at CRC Press to produce this book. In the first few months of 1993, June Meyerman typed most of the book into TeX. In the Fall Semester of 1993 I taught a course from this material and began to organize it into the current form. By the summer of 1994 !-thought I was almost done. At this point I had the good (and bad) fortune of having Nora Guertler, a student from Lyon, visit for two months. When she was through making an average of six corrections per page, it was clear that the book was far from finished. During the 1994-95 academic year most of my time was devoted to prepar­ ing the second edition of my first year graduate textbook Probability: Theory and Examples. After that experience my brain cells could not bear to work on another book for another several months, but toward the end of 1995 they decided "it is now or never." The delightful Cornell tradition of a long winter break, which for me stret.ched from early December to late January, provided just enough time to finally finish the book. I am grateful to my students who have read various versions of this book and also made numerous comments: Don Allers, Hassan Allouba, Robert Bat­ tig, Marty Hill, Min-jeong Kang, Susan Lee, Gang Ma, and Nikhil Shah. Earlier in the process, before I started writing, Heike Dengler, David Lando and I spent a semester reading Protter (1990) and Jacod and Shiryaev (1987), an enterprise which contributed greatly to my education. The ancient history of the revision process has unfortunately been lost. At the time of the proposed second edition, I transferred a number of lists of typos to my copy of the book, but I have no record of the people who supplied the lists. I remember getting a number of corrections from Mike Brennan and Ruth Williams, and it is impossible to forget the story of Robin Pemantle who took Brownian Motion, Martingales and Analysis as his only math book for a year­ long trek through the South Seas and later showed me his fully annotated copy. However, I must apologize to others whose contributions were recorded but whose names were lost. Flame me at [email protected] and I'll have something to say about you in the next edition. Rick Durrett

About the Author

Rick Durrett received his Ph.D. in Operations research from Stanford University in 1976. He taught in the Mathematics Department at UCLA for nine years before becoming a Professor of Mathematics at Cornell University. He was a Sloan Fellow.1981-83, Guggenheim Fellow 1988-89, and spoke at the International Congress of Math in Kyoto 1990. Durrett is the author of a graduate textbook, Probability: Theory and Examples, and an undergraduate one, The Essentials of Probability. He has written almost 100 papers with a total of 38 co-authors and seen 19 students complete their Ph.D.'s under his direction. His recent research focuses on the applications of stochastic spatial models to various problems in biology.

Stochastic Calculus: A Practical Introduction

1. Brownian Motion

1.1 1.2 1.3 1.4

Definition and Construction 1 Markov Property, Blumenthal's 0-1 Law 7 Stopping Times, Strong Markov Property 18 First Formulas 26

2. Stochastic Integration

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2- . 10

Integrands: Predictable Processes 33 Integrators: Continuous Local Martingales 37 Variance and Covariance Processes 42 Integration w .r.t. Bounded Martingales 52 The Kunita-Watanabe Inequality 59 Integration w.r.t. Local Martingales 63 Change of Variables, Ito's Formula 68 Integration w.r.t. Semimartingales 70 Associative Law 74 Functions of Several Semimartingales 76 Chapter Summary 79 2.11 Meyer-Tanaka Formula, Local Time 82 2.12 Girsanov's Formula 90

3. Brownian Motion, II

3.1 3.2 3.3 3.4 3.5 3.6

Recurrence and Transience 95 Occupation Times 100 Exit Times 105 Change of Time, Levy's Theorem 111 Burkholder Davis Gundy Inequalities 116 Martingales Adapted to Brownian Filtrations 119

4. Partial Differential Equations

A. Parabolic Equations

4.1 The Heat Equation 126 4.2 The Inhomogeneous Equation 130 4.3 The Feynman-Kac Formula 137 B. Elliptic Equations

4.4 The Dirichlet Problem 143 4.5 Poisson's Equation 151 4.6 The Schrodinger Equation 156 C. Applications to Brownian Motion

4. 7 Exit Distributions for the Ball 164 4.8 Occupation Times for the Ball 167 4.9 Laplace Transforms, Arcsine Law 170

5. Stochastic Differential Equations

5.1 5.2 5.3 5.4 5.5 5.6

Examples 177 Ito's Approach 183 Extension 190 Weak Solutions 196 Change of Measure 202 Change of Time 207

6. One Dimensional Diffusions

6.1 6.2 6.3 6.4 6.5 6.6

Construction 211 Feller's Test 214 Recurrence and Transience 219 Green's Functions 222 Boundary Behavior 229 Applications to Higher Dimensions 234

7. Diffusions as Markov Processes

7.1 7.2 7.3 7.4 7.5

Semigroups and Generators 245 Examples 250 Transition Probabilities 255 Harris Chains 258 Convergence Theorems 268

8. Weak Convergence

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

In Metric Spaces 271 Prokhorov's Theorems 276 The Space C 282 Skorohod's Existence Theorem for SDE 285 Donsker's Theorem 287 The Space D 293 Convergence to Diffusions 296 Examples 305

Solutions to Exercises 311 References Index

339

335

1

Brownian Motion

1.1. D efi nition and Construction

In this section we will define Brownian motion and construct it. This event, like the birth of a child, is messy and painful, but after a while we will be able to have fun with our new arrival. We begin by reducing the definition of a d-dimensional Brownian motion with a general starting point to that of a one dimensional Brownian motion starting at 0. The first two statements, (1.1) and (1.2), are part of our definition.

(1.1) Translation invariance. {Bt - Bo, t 2:: 0} is independent of Bo and has the same distribution as a Brownian motion with Bo = 0. (1.2) Independence of coordinates. If Bo = 0 then { B£, t 2:: 0} . . . , { Bf, t 2:: 0} are independent one dimensional Brownian motions starting at 0. Now we define a one dimensional Brownian motion starting at 0 to be a process Bt , t 2:: 0 taking values in R that has the following properties: ( a) If to < t 1 < ... < independent. (b ) If s,

t n then B(to), B(tr) - B(to), . . . B(t n ) - B(t n - 1 ) are

t 2:: 0 then P(B(s + t) - B(s) E A) =

( c) With probability one,

Bo = 0 and t

l (27rt) - 112 exp (-x2 f2t) dx ___,.

Bt is continuous.

Bt has independent increments. (b ) says that the increment B(s + t) - B(s) has a normal distribution with mean vector 0 and variance t. ( a) says that

( c) is self-explanatory. The reader should note that above we have sometimes

2

Section

Chapter 1 Brownian Motion

written B. and sometimes written B (s) , a practice we will continue in what · follows.

1.1

Definition and Construction

3

For each 0 < t 1 < . . < tn define a measure on Rn by .

An immediate consequence of the definition that will be useful many times is: (1.3) Scaling relation. If B0 = 0 then for any t > 0,

{B.t ,S � 0} 4 {t 1 1 2B. , s � 0} To be precise, the two families of random variables have the same finite dimen­ sional distributions, i.e., if s 1 < . . . < sn then (B.1t,

. . •

, B. nt ) 4 (t 1 1 2B.1 ,

• • •

, t 1 1 2B. n)

In view of {1.2) it suffices to prove this for a one dimensional Brownian motion. To check this when n = 1 , we note that t 1 1 2 times a normal with mean 0 and variance s is a normal with mean 0 and variance st. The result for n > 1 follows from independent increments. A second equivalent definition of one dimensional Brownian motion starting from Bo = 0 , which we will occasionally find useful, is that Bt , t � 0, is a real valued process satisfying· (a') Bt- is a Gaussian process (i.e., all its finite dimensional distributions are multivariate normal) , (b') EB. = 0, EB.Bt = s A t = min{s, t}, (c) With probability one, t -+ Bt is continuous.

It is easy to see that (a) and (b) imply (a'). To get (b') from (a) and (b) suppose s < t and write EB.Bt = E (B'f) + E(B. (Bt - B. )) = s + EB.E(Bt - B. ) = s

The converse is even easier. (a') and (b') specify the finite dimensional distri­ butions of Bt , which by the last calculation must agree with the ones defined in (a) and (b). The first question that must be addressed in any treatment of Brownian motion is, "Is there a process with these properties?" The answer is ''Yes," of course, or this book would not exist. For pedagogical reasons we will pursue an approach that leads to a dead end and then retreat a little to rectify the difficulty. In view of our definition we can restrict our attention to a one di­ mensional Brownian motion starting from a fix ed x E R. We could take x = 0 but will not for reasons that become clear in the remark after (1.5).

where xo = x, to = 0 ,

Pt (a , b) = (2m) - 1 / 2 exp(-(b- a) 2 f2t)

and Ai E n the Borel subsets of R. From the formula above it is easy to see that for fixed x the family J.L is a consistent set of finite dimensional distributions {f.d.d.'s), that is, if {s 1 , . . . Sn - 1 } C {t 1 , . . . tn} and ti rf {s 1 , . . . S n -d then This is clear when j = n. To check the equality when 1 :=:; j < n, it is enough to show that

By translation invariance, we can without loss of generality assume x = 0, but all this says in that case is the sum of independent normals with mean 0 and variances ti - ti - 1 and ti +l - ti has a normal distribution with mean 0 and variance ti + 1 - t1_ 1 . With the consistency of f.d.d.'s verified we get our first construction of Brownian motion: (1.4) Theorem. Let no = {functions w : [O, oo ) -+ R} and :Fa be the l1-field generated by the finite dimensional sets {w : w(ti ) E Ai for 1 $ i $ n} where A; E n. For each x E R, there is a unique probability measure llx on (no, :Fa) so that llx{w : w(O) = x} = 1 and when 0 < t 1 · · · < tn

This follows from a generalization of Kolmogorov's extension theorem. We will not bother with the details since at this point we are at the dead end referred to above. If C = {w : t -+ w(t) is continuous} then C rf :F0, that is, C is not a measurable set. The easiest way of proving C rf :Fa is to do

E :Fa if and only if there is a sequence of times t 1 , t 2 , E (O, oo) and aB E Jl{ 1 • 2 • ···l {the infinite product 0'-field R x R X · · ·) so that A = {w : (w(tt), w (t2 ), ... ) E B } . In words, all events in :Fa depend on only Exercise 1.1. A

countably many coordinates.

4

Chapter 1 Brownian Motion

Section

The above problem is easy to solve. Let Q 2 = {m2- n : m, n ;:::_ 0} be the dyadic rationals. IH 1 q = {w : Q2-+ R} and :Fq is the u-field generated by the finite dimensional sets, then enumerating the rationals ql> q2 , and applying Kolmogorov's extension theorem shows that we can construct a probability v"' on (!lq , :Fq ) so that vx {w : w (O ) = x} = 1 and ( * ) in (1.4) holds when the t; E Q2 . To extend the definition from Q2 to [0, oo) we will show: • • •

(1.5) Theorem. Let T < oo and x E R. Vx assigns probability one to paths w : Q2-+ R that are uniformly continuous on Q2 n [0, T].

�

Remark. It will take quite a bit of work to prove

{1.5). Before taking on that

task, we_ will attend to the last measure theoretic detail: we tidy things up by moving our probability measures to ( C, C) where C = {continuous w : [0, oo)-+ R} and C is the u-field generated by the coordinate maps t-+ w (t ) . To do this, we observe that the map 'if; that takes a uniformly continuous point in !lq to its extension in C is measurable, and we set

Px =

l/;r; 0

J

By (1.1) and (1.3), we can without loss of generality suppose Bo = 0 and prove the result for T = 1. In this case, part (b) of the definition and the scaling relation (1.3) imply Proof of (1.5)

where C = E I B1 1 4 < oo. From the last observation we get the desired uniform continuity by using a result due to Kolmogorov. In this proof, we do not use the independent increments property of Brownian motion; the only thing we

Definition and Construction

5

use is the moment condition. In Section 2.11 we will need this result when the Xt take values in a space S with metric p; so, we will go ahead and prove it in that generality.

(1.6) Theorem. Suppose that Ep(X. , Xt)fi � Kit - s l l+a where a, {3 > 0. If < a/{3 then with probability one there is a contant C (w) so that

'Y

p(Xq , Xr ) � C lq - ri 'Y for all q, r E Q2 n [0, 1] n Let 1 < af{3, 1J > 0, In = {( i, j) : 0 � i � j � 2n , 0 < j - i � 2 f1 } n n n and Gn = {p(X(j2 - ) , X(i2 - )) � ((j - i)2 - )'Y for all (i, j) E In } · Since afi P(IY I > a) � E!Yifi we have Proof

P(G� ) �

I: ((j - i)2-n ) -fi'Y Ep(X(j2-n ),X(i2- n ))fi

(i,i)Efn

� J(

'1/; - 1 .

Our construction guarantees that Bt (w) = Wt has the right finite dimensional distributions for t E Q2 . Continuity of paths and a simple limiting argument shows that this is true when t E [0, oo) . A'§ mentioned earlier the generalization to d > 1 is straightforward since the coordinates are independent. In this generality C = {continuous w : [0, oo ) -+ Rd} and C is the u-field generated by the coordinate maps t-+ w ( t ) . The reader should note that the result of our construction is one set of random variables Bt (w) = w (t ) , and a family of probability measures Px , x E Rd, so that under Px , Bt is a Brownian motion with Px(Bo = x) = 1. It is enough to construct the Brownian motion starting from an initial point x, since if we want a Brownian motion starting from an initial measure J.l (i.e., have PJJ(Bo E A) = J.l (A)) we simply set P11 (A) = J.l (dx)Px(A)

1.1

I: {(j i)T n ) -fi'Y+l+a

(i,j)Eln

_

2n 2nfl, so 2n 2nf} · {2nf1 2 - n ) -fi'Y +l+a = 1{ 2 - n >.

by our assumption. Now the number of {i, j) E I n is �

P(G� ) � J( where >. = ( 1 - 7J)(1 + a - {3-y) - (1 + 7J). Since 1 < af{3, we can pick 7J small enough so that >. > 0. To complete the proof now we will show (1.7) Lemma. Let A = 3 · 2(1 -'lh /{1 - 2- 'Y). On HN = n�=N Gn we have ·

p(Xq , Xr ) � Alq - r i'Y for q, r E Q2 n [0, 1] with l q - r l � 2- (1 -'I )N {1.6) follows easily from (1.7): P(H/v ) �

oo

oo

2 - N>.

I: P(G� ) � K I: 2- n>. = -1I(_2- _->.

n=N n=N This shows p(Xq , Xr ) � Alq - r i 'Y for lq - r l � 8 (w ) and implies that we have p(Xq , Xr ) � C (w )lq - r i'Y for q, r E [0, 1]. ( 1 )N Proof of (1. 7) Let q, r E Q2 n [0, 1] with 0 < r - q < 2- -'I . Pick m � N so that

a and write 6

Ch pter 1 Brownian Motion

Section

r j2 - m + 2 - r(l) + ... + 2 - r(l) q i2 - m 2- q( l ) 2 - q(k ) < <
= = r{£)andandit follows m that {1) on q(k). Now 0 < r - q < _

_ . . . _

where m < r{1) 2 - m (l -'7 ) , so (j (a) On HN it follows from the triangle inequality that (b) p(Xq , X(i2-m )) � L:k <2- q(h)F � I:=00 {2--r)h � c,. 2--rm h m h= l where C,. = 1/{1 - 2--r) 1. Repeating the last computation shows (c) Combining (a)-(c) gives · · ·

Markov Property, Blumenthal's 0-1 Law

7

We claim that An C Gn . To prove this we consider s = 1, which we claim is the worst possibility. wantcase, to conclude For this we observe thatIn jthis= 0case is theweworst but eventhat thenYn-2 ,n � 5C/n.

· · ·

Using An C Gn and the scaling relation {1.2) gives 5C ) 3 P(An ) � P(Gn ) � n P (i B 1 1n i � � { 3 3 ) = n (i Bl i � n5Cl/2 � n . n10Cl/2 . {27r)- l /2 } 2 /2) � 1. Letting n -+ shows P(An ) -+ 0. Noticing n -+ An is since exp(-xshows increasing P(An ) = 0 for all n and completes the proof. Exercise 1.2. Show by considering k increments instead of 3 that if 1/2 exponent 1/k then with with at anyprobability point of [0,1, Brownian 1]. paths are not Holder continuous The next result is more evidence that Bt - B. -/f=S. Exercise 1.3. Let .6. m, n = B(tm2 - n ) - B(t(m - 1)2 - n ). Compute p

>

oo

0

'Y >

+

p(Xq , Xr) � 3C,.2 --rm (l -l)) � 3C,.2 ( l -l)h i r - qi �' 2- m ( l -'7) � 1 -'7 i r qi.

since (1.6) and {1.5) 2 . - This completes the proof of {1.7) and hence of The scaling relation {1.2) implies EiBt - B, i 2m = Cm i t - si m where Cm = EiB1 i 2m So using {1.6) with {3 = 2m and a = m- 1 and then letting m -+ gives: {1.8) Theorem. Brownian paths are Holder continuous with exponent for any < 1/2. It is easy to show: {1.9) Theorem. tinuous (and henceWith not probability differentiable)one,atBrownian any point.paths are not Lipschitz con­ Proof Let An = {w : there is an s E [0, 1] so that i Bt - B. i � C i t - s i when i t - si � 3/n}. For 1 � k � n - 2 let Yk,n = max { IB (k�j ) - B (k+ � - 1 ) I: j = 0, 1,2} Gn = { at least one Yk,n is � 5C/n} o

oo

r

1 .2

r

r

�

and use the Borel-Cantelli lemma to conclude that Lm-< 2n .6.� .n -+ t a.s. as n -+ Remark. The last result is true if we consider a sequence of partitions II 1 C II 2 C ... with mesh - + 0. See Freedman (1970) p.42-46. The true quadratic variation, defined as the sup over all partitions, is for Brownian motion. oo.

oo

1.2. M arkov P roperty, B lumenthal's 0-1 Law

Intuitively the Markov property says "given before the present state, B. , any other information about what hap­ pened time s is irrelevant for predicting what happens after times."

8

Chapter 1 Brownian Motion

Since simply Brownian more statedmotion as: is translation invariant, the Markov property can be "ifs;:::of0what thenhappened Bt +• - B. , t 0 is a Brownian motion that is indepen­ dent before times." This shouldimplies not be that surprising: increments definition) ifs1 thes2 . independent . .::;sm =s and 0 < t1 . property . . < tn then((a) in the ;:::

::;

The major obstacle in provingtheory the necessary Markov property forandBrownian motion then is to introduce the measure to state prove the result. The fi r st step in doing this is to explain what we mean by "what happened before times." The technical name for what we are about to define is a filtra­ tion, a fancy term for an increasing collection of a--fields, :F., i. e . , ifs::; t then :F. C :Ft. Since we want B. E :F., i. e . , B. is measurable with respect to :F., the first thing that comes to mind is For technical reasons, it·is convenient to replace :F� by The fields :Fj are nicer because they are right continuous. That is, Init iswords the :Fj allow us an "infinitesimal peek at the future," i.e., A E :Ff if in :F�+e for any f 0. If Bt is a Brownian motion and f is any measurable function with f(u) 0 when u 0 the random variable limsup (Bt - B. )/f(t -s) t .l.• is measurable with respect tof(u):Fj =butfonotand:F�.f(u)Exercises 2.9 and 2.10 consider happens when we take = Juloglog(1/u). However, what as wenotwillinsee:F�.inThe(2.6two ), there are noareinteresting examples ofsets). sets that are in :Fj but a--fields the same (up to null Toa family state theof measures Markov property we need some notation. First, recall that we have Px, x E Rd, on (C, C) so that under Px, B, (w) = w(t) is a Brownian motion with Eo = In order to define "what happens after time >

>

Section

Markov Property, Blumenthal's 0-1 Law 9

s,"by it is convenient to define the shift transformations e. : c c, fors;::: 0 (e.w)(t) = w(s + t) for t;::: 0 Inso words, we cut off the part of the path before time s and then shift the path that time s becomes time 0. To prepare for the next result, we note that if R is C measurable then Yo e. is a function of the future after times. Y: C To see this, consider the simple example Y(w) = f(w(t)). In this case Yo e. = j(e. w(t)) = f(w(s + t)) = j(B. + t ) Likewise, if Y(w) = j(Wt1 , w, ,J then -lo

_...

• • •

(2.1) forTheallMarkov property. Ifs;::: 0 and Y is bounded and C measurable then x E Rd Ex(Yo e. I:F;t") = Es.Y where the right hand side is r.p(y) = EyY evaluated at y = B(s). Explanation. In words, this says that the conditional expectation of Yo e. :Fj is just the expected value of Y for a Brownian motion starting at B• . Togiven explain whyhappened this implies "given theispresent state,forBpredicting . , any other information about what before time s irrelevant what happens after time s," we begin by recalling (see (1.1) in Chapter 5 of Durrett (1995)) that if g :F and E(ZI:F) E g then E(ZI:F) = E(ZIQ). Applying this with :F = :Fj and g = u(B.) we have c

Ex(Yo e. IF;t")

>

x.

1 .2

= Ex(Yo e. I B. )

wetherecall thatof minimizing X = E(ZI:F) is our best guess at given the information in F (inDurrett sense E(Z- X) 2 over X E :F., see (1.4 ) in Chapter 4 of B. is the same (1995)) then we see our best at Yo ein. given aspredicting our bestyguesse . given :Fj, that i.e., any otherguess information :Fj is irrelevant for • Proof By the definition of conditional expectation, what we need to show is Ex(Yo e. ; A) = Ex(Es.Y;A) for all A E :r: (MP) Z

If

0

10

Chapter 1 Brownian Motion

Section

We beginwebywill proving the result>. theorem for a special classmonotone of Y's andclassa special classto ofextend A's.willThen use the and the theorem to the general case. Suppose Y(w) = l �mIT�n fm(w(tm)) where 0 < t 1 < . . . < tn and the fm Rd -+ Rare bounded and measurable. Let 0 < h < t 1 , let 0 < s1 . . . < SJ: :=:; s + h, and let A = {w : w( si) E Aj , 1 :=:; j :=:; k} where Aj E n,d for 1 :=:; j :=:; k. We will call these A's the finite dimensional :r:;+h . of Brownian motion it follows that if 0 = < < or f.d.Fromsetsthein definition Ut then the joint deQsity of (Bu 1 , Bu J is given by P:c(Bu1 = Yl > ·· .Bu t = Yt) = i=l IlPu;- u;_1 (Yi - 1> Yi ) where Yo = x. From this it follows that E:c (}] g;(Bu;)) = JdYl Pu1 - u0 (Yo, yi )gl (Yl ) J dyt Put- u t-t (Yt- 1 , Yt )Ut (Yt ) Applying resultforwiththetheg wegiven by s , . . . , SJ:, s + h, s + t1 , . . . s + tn and the obviousthischoices ; have 1 E,(Y A) = E, (Q lA;(B, ;) Ia• (B, +h ) ]/m (B, +,.)) = }At{ dxl Pst(x,x l ) ... jAk{ dXkPsk- sk-t(Xk- liXJ:) . r dyps+h - sk (x�:,y)
:

• • .

u0

• • •

l

···

u;

o

8,;

·

Using the formula for Ex TI�=l g; (Bu;) again we have

·

u1

1.2

Markov Property, Blumenthal's 0-1 Law

11

To extend the class of A's to all of :r::+h' we use (2.2) The >. theorem. Let A be a collection of subsets of nthat contains nand is closed a collection of setsunderthatintersection satisfy (i.e., if A,B E A then A n B E A). Let g be (i) if A, B E g and A B then A - B E g (ii) If An E g and An j A, then A E g. If A C g then the u-field generated by A , u(A) C g. Proof See (2.1) in the Appendix of Durrett (1995). 1r-

�

Proof Fix Y and let g be the collection of sets A for which the desired equality isconvergence true. A simple subtraction shows(ii) that (i)Ifinwe(2.2) holds, while the monotone theorem shows that holds. let A be the collection of finite dimensional +h then we have shown A C g so :r::+h = u(A) C g which is the sets desiredin :r::conclusion. Our next step is (MP-3) Ex ( Y Bs; A) = E:c(
o

>

-+

• • •

-+

-+

Using result. (MP-2) and the bounded convergence theorem now gives the desired

0

Chapter 1 Brownian Motion

12

(MP-3) shows that (MP) Toholdsextend for Yto=general Ill <m
1i

?i.

?i.

1i

1i

1i

1i

?i.

D

Take f(x) = X and f(x) = XiXj with i =f:. j in Exercise 2.1 to conclude that B; and BiB{ arei martingales if i =f:. j. The next two exercises prepare for calculations in Section 1.4. Exercise 2.3. Let To = inf{s 0 : B = 0} and let R = inf{t 1: Bt = 0}. R is for right or return. Use the Markov property at time 1 to get (2.4) Px(R 1 + t) = J Pl(x, y)Py(To t) dy Exercise 2.4. Let T0 = inf{s 0 :B. = 0} and let L = sup{t $ 1 : Bt = 0}. L is for left or last. Use the Markov property at time 0 < t < 1 to conclude

Exercise 2.2.

>

s

>

>

Po(L $ t)

or,

n

1.2

Markov Property, Blumenthal's 0-1

2::

>

>

x

s

x

will see many applicationsthatofhasthe Markov property below, soSinceweTheturnreader our attention now toother a "triviality" surprising consequences. Ex(Y !:F.+) = EB. YE :F: it follows (see, e.g., (1.1) in Chapter 5 of Durrett (1995)) that Ex (Y .j:Ff) = Ex(Y .IF:) From the last equation it is a short step to (2.6) Theorem. If ZE C is bounded then for all 2:: 0 and xE Rd, o

o

8

8s

o

8

s

>

>

(2.5)

Law 13 Exercise 2.5. Let G be an open set and letT = inf{t : B1 (j. G}. Let J( be a closed subset of G and suppose that for all xE J( we have Px (T 1, B1E K) ;::: then for all integers 1 and xE J( we have Px (T n, BtE K) ;::: arn. The next two exercises prepare for calculations in Chapter 4. Exercise 2.6. Let 0 < s < t. If h : R Rd -+ R is bounded and measurable Ex (it h(r, Br ) dr l :F. ) = i• h(r, Br) dr + EB( ) it-s h(s + u, Bu) du • Exercise 2.7. Let 0 < < t. Iff: Rd -+ R and h : R Rd -+ R are bounded and measurable then Ex ( f(Bt) exp (it h(r, Br) dr) I:F. ) = exp (it h(r, Br) dr) EB. { f(B1_3) exp (it-• h(s + u, Bu) du)} Section

= J Pt(O, y)Py (To 1 - t) dy >

Proof when

By

the monotone class theorem, (2.3), it suffices to prove the result n Z = II fm(B(tm )) m=l

14

Chapter 1 Brownian Motion

Section

and the fm are bounded and measurable. In this case Z = X(Y B3) where X E :F� and Y E C, so using a property of conditional expectation (see, e.g., (1.3) in Chapter 4 of Durrett (1995)) and the Markov property (2.1) gives E.,(Z I :FJ) = X E.,(Y B31:FJ) = X EB. Y E :F: and the proof is complete. If wesame let ZupE :Fi thensets. (2.6)Atimplies Z = E.,( Z I :F:) E :F:, so the two u-fields are the to null first The fun starts when we takes = 0 in (2.glance, 6) to getthis conclusion is not exciting. (2.7) Blumenthal's 0- 1law. If A E :Fci then for all x E Rd, P.,(A) E {0, 1} Proof Using (i) the fact that A E :Fci, (ii) (2. 6 ), (iii) :F8 = u(Bo) is trivial under P.,, and (iv) if g is trivial E(X I Q) = EX gives P., a.s. This indicator function 1A is a.s. equal to the number P.,(A) and itshows followsthatthattheP.,(A) E {0, 1}. thestudying last resultthesayslocalthatbehavior the germof field, :Fci, is trivial. This result isnotice veryInwewords, useful in Brownian paths. Until will restrict our attention to one dimensional Brownian motion.further (2.8) Theorem. If = inf{t � 0: Bt > 0} then Po(r = 0) = 1. Proof P ( :::; t) � P (Bt > 0) = 1/2 since the normal distribution is sym­ metric about 0. Letting t ! 0 we conclude Po(r = 0) = liJ;;Po(r:::; t) � 1/2 so it follows from (2.7) that Po(r = 0) = 1. must hitSince(0, t ) immediately starting from 0, it mustOnce also hitBrownian ( 0)motion immediately. Bt is continuous, this forces: (2.9) Theorem. If To = inf{t > 0: Bt = 0} then Po(To = 0) = 1. o

o

0

0

T

0 r

0

0

-oo,

oo

-+

1 .2

Markov Property, Blumenthal's 0-1 Law

15

Combining (2.8) and (2.9) with the Markov property you can prove Exercise 2. 8. If a < b then with probability one there is a local maximum of Bt maxima in (a, b).ofSince with probability local Brownian motion are one dense.this holds for all rational a< b, the Another typical application of (2.7) is Exercise 2. 9. Let f(t) be a function with f(t) > 0 for all t > 0. Use (2. 7 ) to conclude that limsupt!O B(t)ff(t) = c Po a.s. where c E (0, ] is a constant. Initerated the nextlogarithm exercise(seeweSection will see7.that c = when f(t) = t 112• The law of the 112 when 9 of Durrett (1995)) shows that = 2 1 f(t) = (t log log(l/t)) 12• Exercise 2. 10. Show that limsUP t!oB(t)jt 112 = Po a.s., so with probability one Brownian paths are not Holder continuous of order 1/2 at 0. Remark. Let 'H'Y(w) be the set of times at which the path w E C is Holder continuous ofshows orderthat1. P('H'Y (1.6) shows that P('H'Y = [0, )) = 1 for 1 < 1/2. Exercise 1. 2 = 0) = 1 for 1 > 1/2. The last exercise shows P(t E 'H1t2 ) = 0 for each t, but B. Davis (1983) has shown P('H112 f 0) = 1. Comic Relief. There is a wonderful way of expressing the complexity of Brownian paths that I learned from Wilfrid Kendall. you runit willBrownian motion in two dimensions for a positive amount of"Iftime, write your name." Ofthecourse, on Shakespeare, top of your name it willpornographic write everybody else'sandname, asof well as all works of several novels, a lot nonsense. as'rhinking follows:of the function g as our signature we can make a precise statement (2.10) Theorem. Let g : (0, 1] Rd be a continuous function with g (O) = 0, let f > 0 and let tn ! 0. Then Po almost surely, sup I B(y�tn ) - g(B) l < for infinitely many In viewthisofifBlumenthal's 0-1 law, (2.7), and the scaling relation (1.3), weProof can prove we can show that Po ( sup I B(B) - g(B) I < f) > 0 for any > 0 oo

oo

c

oo

oo

-+

0 �89

0 �89

t:

n

t:

16

Chapter 1 Brownian Motion

Section

This wasme easy forin methe todetails. believe,Thebutfirstnotstepso easy fortreatmetheto dead proveman's whensignature students asked to fill is to g(x) 0. =

Show that if e > 0 and t < then Po (sup 0sup I B! I < e) > 0 i � •9 Ingetdoing this youresult mayfrom find this Exercise 2.5 helpful. In (5.4) of Chapter 5 we wiil the general one by change of measure. Can the reader find a simple ·direct proof of (*)? With our discussion of Blumenthal's 0-1 law complete, the distinction be­ tween :Ff and :F: is no longer important, so we will make one final improvement in our u-fields and remove the superscripts. Let N:c = {A : A B with P:c(B) = 0} :F: = u(:Ff U N:c) oo

Exercise 2. 11.

c

1 .2

Markov Property, Blumenthal's 0-1 Law

17

allows usthisto relate the behavior of Bts 0-1as law t to tothea behavior as Combining idea with Blumenthal' leads very useful result. Let :Ff = u(B. : s t) = the future after time t = nt� O :F: = the tail u-field (2.12) Theorem. If A E then either P:c(A) 0 or P:c(A):::: 1. Remark. Notice that this is stronger than the conclusion of Blumenthal's 0-1 law examples : w(O) E B} show that for A in the germ u-field(2.7).:Fft The the value of P:cA(A)=may{w depend on x. Proof Since the tail u-field of B is the same as the germ u-field for X, it follows that Po(A) E {0, 1}. To improve this to the conclusion given observe that A E :Ff, so 1A can be written as 1 B 81 • Applying the Markov property, (2.1), gives t

__,.

(2.11) 0.

__,. oo

2::

T

T

=

o

P:c(A) = E:c(1 B 81 ) = E:c(E:c( 1B 81 I :F1 )) = E:c(EB1 1B ) = (2n-) - df 2 ( -ly - xl 2 /2)Py (B) dy o

o

(2.11) Theorem. If Bt is a Brownian motion starting at 0 then so is the process defined by Xo = . 0 and Xt = t B(1/t) for t > 0. Proof By (1.2) it suffices to prove the result in one dimension. We begin by observing that thepathsstrong law of large numbers implies Xt 0 as t 0, so X has continuous and we only have to check that X has the right f.d. d.'s. By0 < the second definition of Brownian motion, it suffices to show that (i) if t 1 < ... < tn then (X(t l ), ... X(tn )) has a multivariate normal distribution with mean 0 (which is obvious) and (ii) if s < t then

exp j Taking x = 0 we see that if P0(A) = 0 then Py (B) = 0 for a.e. y with respect toToLebesgue measure, and using the formula again shows P:c(A) = 0 for all x. handle the case P0(A) = 1 observe that Ac E and Po(Ac ) = 0, so the last result implies P:c(Ac ) = 0 for all x. nextofresult application of (2.12). The argument here is a closeThe relative the oneis afortypical (2.8). (2.13) Theorem. Let Bt be a one dimensional Brownian motion and let A = nn{Bt = 0 for some t n}. Then P:c(A) = 1 for all x. In"infinitely words, one dimensional Brownian motion is recurrent. It will return to 0 often, " i. e ., there is a sequence of times t n j so that Bt n = 0. We have to0, beBt careful withtothe0 infinitely interpretation the phrase since starting from will return manyoftimes by timein equotes > 0. Proof We begin by noting that under P:c, Btf.J'i has a normal distribution with mean xj.J'i and variance 1, so if we use x to denote a standard normal,

E(X.Xt) = stE(B(1/s)B(1ft)) = s

P:c(Bt < 0) = P:c(Btf-/i < 0) = P(x < -xj-/i)

:F. = nx:F:

are thethenull set and :F; are the completed u-fields for P:c. Since we do not want filtration to depend on the initialwillstatebe mentioned we take theatintersection of allthethenextcompleted u-fields. This technicality one point in section but can otherwise be ignored. (2.7) concerns the behavior of Bt as t 0. By using a trick we can use this result to get information about the behavior as t

N:c

s

__,.

__,. oo.

__,.

__,.

0

T

0

2::

oo

18

Chapter 1 Brownian Motion

Section

and limt-.oo P:r:(Bt < 0) = 1/2. If we let To = inf{t : Bt = 0} then the last result and the fact that Brownian paths are continuous implies that for all x > 0

P:r:(Bt = 0 for some t 2:: n) = E:r:(PBn (To < oo)) 2:: 1/2 --+ oo it follows that P:r:(A) >

P:r:(Bt = 0 i.o.) ::::: 1 .

1/2 but

AE

T

so (2.12) implies D

1.3. Stopping Times, Strong Markov P roperty

We call a random variable S taking values in [0, oo] a stopping time if for all t 2:: 0, { S < t} E :Ft . To bring this definition to life think of Bt as giving the price of a stock and S as the time we choose to sell it. Then the decision to sell before time t should be measurable with respect to the information known at time t. In -the last definition we have made a choice between { S < t} and { S � t} . This makes a big difference in discrete time but none in continuous time (for a right continuous filtration :.Ft) : If { S � t} E :.Ft then { S < t} = U n {S � t - 1/n} E :Ft.

If { S < t} E :.Ft then { S � t} = n n{ S < t + 1/n} E :Ft. The first conclusion requires only that t --+ :.Ft is increasing. The second relies on the fact that t --+ :.Ft is right continuous. (3.2) and (3.3) below show that when checking something is a stopping time it is nice to know that the two definitions are equivalent. (3.1) Theorem. If G is an open set and T

stopping time.

= inf{t ;:::: 0: Bt E G} then T is a

Proof Since G is open and t --+ Bt is continuous {T < t } = U q < t {Bq E G} where the union is over all rational q, so {T < t} E :Ft. Here, we need to use the rationals so we end up with a countable union. D (3.2) Theorem. If Tn is a sequence of stopping times and Tn ! T then T is a

stopping time.

Un {Tn < t}.

19 D

stopping time.

Symmetry implies that the last result holds for x < 0, while {2.9) (or the Markov property) covers the last case x = 0. Combining the fact that P:r:(To < oo) 2:: 1/2 for all x with the Markov property shows that

n

Stopping Times, Strong Markov Property

(3.3) Theorem. If Tn is a sequence of stopping times and Tn j T then T is a

P:r:(To < oo) 2:: 1/2

Letting

Proof {T < t} =

1 .3

Proof {T � t} =

n n{Tn � t}.

D

. (3 .4) Theorem. If I< is a closed set and T = inf{t;:::: 0:

stopping time.

Bt E I<} then T is a

Proof Let D(x, r) = {y : lx - Yl < r}, let Gn = U{D{x, 1/n) : x E K}, and let Tn = inf{t 2:: 0 : Bt E Gn}. Since Gn is open, it follows from (3.1) that T is a stopping time. I claim that as n j oo, Tn j T. To prove this notice that T 2:: Tn for all n, so lim Tn � T. To prove T � lim Tn we can suppose that Tn i t < oo. Since B(Tn) E Gn for all n and B(Tn ) --+ B (t), it follows that B(t ) E I< and T � t. o Remark. As the reader might guess the hitting time of a Borel set A, TA = inf{t : B, E A}, is a stopping time. However, this turns out to be a difficult result to prove and is not true unless the filtration is completed as we did at the end of the last section. Hunt was the first to prove this. The reader can find a discussion of this result in Section 10 of Chapter 1 of Blumenthal and Getoor (1968) or in Chapter 3 of Dellacherie and Meyer (1978). We will not worry about that result here since (3.1) and (3.4), or in a pinch the next result, will be adequate for all the hitting times we will consider. Exercise 3. 1. Suppose A is an Fa, i.e. , a countable union of closed sets. Show that TA = inf{t : Bt E A} is a stopping time. Exercise 3.2. Let S be a stopping time and let [x] = the largest integer � x. That is, Sn

= (m + 1)2 - n if

n m2 - �

Sn =

([2 nS] + 1)/2 n where

S < (m + 1)2 - n

In words, we stop at· the first time of the form k2- n after S (i.e., > S ) . From the verbal description it should be clear that Sn is a stopping tim�. Prove that it is. Exercise 3.3. If S and T are stopping times, then S 1\T = min{S, T} , SVT = max{S, T} , and S + T are also stopping times. In particular, if t ;:::: 0, then S 1\ t, S V t, and S + t are stopping times.

20

Section

Chapter 1 Brownian Motion

Let Tn be a sequence of stopping times. Show that infT n n, limsupT n n, limninfTn are stopping times Our nexttogoal is to state and provefrom the strong Markov property. To do this, we need generalize two definitions Section 1.2. Given a nonnegative random we define randomso that shifttime Bs which "cuts off the part ofw beforevariable S(w)S(w) and then shifts the the path S(w) becomes time 0." (Bsw)(t) = { w(S (w) + t) onon {{SS =< oo} oo} Here getsis shifted an extraaway. pointSome we addauthors to like to cover the case inconvention which thethatwholeall path to adopt the functions have f(l::!.restrict ) = 0ourto take care toof {the second case. However, we will usually explicitly attention S < oo} so that the second half of the definition will not come into play. , "the information known at time S," is a little moreThesubtle.secondWe quantity could have:Fsdefined

Exercise 3.4.

1::!.

1::!.

G

1 .3

Stopping Times, Strong Markov Property

21

LetS be a stopping time and let AE :Fs . Show that R = {Soo on on AAc is a stopping time Exercise 3.7. LetS and T be stopping times. (i)(ii) {{SS <>t}, {S = t} are in :Fs . T}, and {S = T} are in :Fs (and in :FT) · Two properties of :Fs that will be useful below are: (3.5) Theorem. IfS �T are stopping times then :Fs :FT . Proof If AE :Fs then A n { T � t} = (A n {S � t}) n {T � t}E :Ft. (3.6) Theorem. If Tn ! T are stopping times then :FT = nn:F(Tn ) · Proof (3.5) implies :F(Tn ) :FT for all n . To prove the other inclusion let AE n:F(Tn) · Since A n {Tn < t}E :Ft and Tn ! T, it follows that A n ff < t}E :Ft, so AE :FT . �he last result and Exercises 3.2 and 3.7 allow us to prove something that . obvious from the verbal definition. Exercise 3.�. B s E :Fs, � . e . , the value of Bs is measurable with respect to . for atiOn known at timeS! To prove this letSn = ([2nS ] + 1)/2n be the the m _ �times defined m_ Exercise 3.2. Show B(Sn)E :Fs,. then let n --+ oo and stoppmg use (3.6). The next result goes in the opposite direction from (3.6). Here gn l g means n --+ gn is increasing and g = ( Qn ) · Exercise 3. 9. Let S < oo and Tn be stopping times and suppose that Tn l oo as n l oo. Show that :FsAT,. l :Fs as n l oo. We are property now readyholdsto state the strong the Markov at stopping times.Markov property, which says that (3.7) Strong Markov property. Let (s,w) --+ Y(s,w) be bounded and n C measurable. If S is a stopping time then for all xE R d Ex (Ys Bsi :Fs) = EB(s)Ys on {S < oo} Exercise 3.6.

c

o

:J

o

-

IS

so by analogy we could set The definition we will now give is less transparent but easier to work with. :Fs = {A : A n {S � t}E :Ft for all t 2: 0} In{S words, this makes the reasonable demand that the part of A that lies in � t} should be measurable with respect to the information available at time t. Again we have made a choice between � t and < t but as in the case of stopping this makes no difference and it is useful to know that the two definitionstimes, are equivalent. continuous, the definition of :Fs is unchanged ifExercise we replace3.5.{SWhen � t} by:Ft {Sis right < t}. For practice with the definition of :Fs do

cr

x

o

22

Section

Chapter 1 Brownian Motion

where the right-hand side is

cp(y, t) = EyYt evaluated at y = B(S), t = S.

Remark. In most applications the function that we apply to the shifted path will not depend on s but this flexibility is important in Example 3.3. The verbal description of this equation is much like that of the ordinary Markov property:

"the conditional expectation of Y o Bs given :FJ is just the expected value of Ys for a Brownian motion starting at Bs ." Proof We first prove the result under the assumption that there is a sequence of times t n j oo, so that P:c ( S < oo ) = I: P:c( S = t n ) · In this case we simply

break things down according to the value of S, apply the Markov property and put the pieces back together. If we let Zn = Yt n (w) and A E :Fs then CXl

L

E:c(Zn Btn ; A n { S = t n }) n= l Now if A E :Fs , A n { S = t n } = (A n { S ::; t n }) - (A n { S ::; t n - 1 }) E :F(tn ), E:c(Ys Bs ; A n {S < oo}) = 0

0

so it follows from the Markov property that the above sum is =

CXl

L E:c(EB (t n ) Zn ; A n {S = t n }) = E:c(EB(S) Ys; A n { S < oo})

n= l

To prove the result in general we let Sn = ([2 n S] + 1 ) /2n where [x] = the largest integer ::; x. In Exercise 3.2 you showed that Sn is a stopping time. To be able to let n - -;. oo we restrict our attention to Y's of the form

Ys (w) = fo(s)

n

IT fm (w(tm ))

m= l

where 0 < t 1 < ... < t n and /0, , fn are real valued, bounded and continuous. If f is bounded and continuous then the dominated convergence theorem implies that x --;. dy Pt(x, y)f(y) • • .

J

is continuous. From this and induction it follows that

J

cp(x, s) = E:cYs = fo(s) dy1 Pt 1 (X, YI)f( Yl ) • • •

J

dYn Pt n - t n-l (Yn - l , Yn )f(Yn )

1.3

Stopping Times, Strong Markov Property

23

is bounded and continuous. Having assembled the necessary ingredients we can now complete the proof. Let A E :Fs. Since S ::; Sn , (3.5) implies A E :F(Sn ) · Applying the special case of (3.7) proved above to Sn and observing that { Sn < oo} = { S < oo} gives

Ex (Ysn Bsn ; A n { S < oo}) = E:c(cp (B(Sn ) , Sn ) ; A n {S < oo }) o

Now as

n--�- oo, Sn ! S, B(Sn )--�- B(S), cp(B(Sn ), Sn )--�- cp(B(S), S) and

so the bounded convergence theorem implies that (3.7) holds when Y has the form given in ( * ) . To complete the proof now we use the monotone class theorem, (2.3). Let 1l be the collection of bounded functions for which (3.7) holds. Clearly 1{. is a vector space that satisfies (ii) if Yn E 1{. are nonnegative and increase to a bounded Y, then Y E ?-{.. To check (i) now, let A be the collection of sets of the form {w : w ( ti ) E Gi } where Gi is an open set. If G is open the function 1G is a decreasing limit of the continuous functionsfk (x) = ( 1 - k dist(x, G))+, where dist( x, G) is the distance from x to G, so if A E A then 1A E ?-{.. This shows (i) holds and the desired conclusion follows from (2.3) . 0 Example 3. 1. Zeros of Brownian motion. Consider one dimensional Brow­ nian motion, let Rt = inf{u > t : Bu = 0} and let To = inf{u > 0 : Bu = 0}. Now (2.13) implies P:c(Rt < oo) = 1, so B(Rt) = 0 and the strong Markov property and ( 2.9 ) imply

P:c(To BR, o

>

Ol:FR ,) = Po(To > 0) = 0

Taking the expected value of the last equation we see that

P:c(To BR, o

>

0 for some rational t) = 0

From this it follows that with probability one, if a point u E Z(w) :: { t : Bt(w) = 0} is isolated on the left (i.e., there is a rational t < u so that (t, u ) n Z(w) = 0) then it is a decreasing limit of points in Z(w ) . This shows that the closed set Z(w) has no isolated points and hence must be uncountable. For the last step see Hewitt and Stromberg (1969), page 72. If we let I Z(w) l denote the Lebesgue measure ofZ(w) then Fubini's theorem implies

E:c( lZ(w) n [O, T] l ) =

1T P:c(Bt = O) dt= O

Chapter 1 Brownian Motion

24

So

Section 1 .3 Stopping Times, Strong Markov Property

Z(w) is a set of measure zero.

0

Example 3.2. Let G be an open set, x E G, let T = inf{t : B1 f/:. G} , and suppose P:r:(T < oo ) = 1. Let A C {)G, the boundary of G, and let. u(x) = P:r:(BT E A) . I claim that if we let 6 > 0 be chosen so that D(x, o) = {y : I Y - x l < o} c G and let s = inf{t 2: 0: Bt f/:. D(x, o)}, then

u(x) = E:r:u(Bs) Since D(x, o) C G, Bt cannot exit G without first exiting D(x, o) at B8. When Bs = y, the probability of exiting G in A is u(y) independent of how Bt got to y.

Intuition

Proof To prove the desired formula, we will apply the strong Markov prop­ erty, (3.7), to

Example 3.3. Reflection principle. Let B1 be a one dimensional Brownian a > 0 and let Ta = inf{t : B1 = a}. Then

motion, let

Po(Ta < t) = 2Po(Bt > a)

(3.8)

Intuitive proof We observe that if B, hits

a at some time s < t then the strong Markov property implies that B1 - B(Ta ) is independent of what hap­ pened before time Ta . The symmetry of the normal distribution and Po(Bt = a) = 0 then imply Po(Ta < t, Bt > a) = 21 Po(Ta < t)

(3.9)

Multiplying by 2, then using

y = 1 (BTE A)

To check that this leads to the right formula, we observe that since D(x, o) C G, we have BT o Bs = BT and 1 (BTEA) o Bs = 1 (BTEA)· In words, w and the shifted path Bsw must exit G at the same place. Since S $ T and we have supposed P:r:(T < oo ) = 1, it follows that P:r:(S < oo) = 1. Using the strong Markov property now gives

{Bt > a} C {Ta < t} we have

Po(Ta < t) = 2Po(Ta < t, Bt > a) = 2Po(Bt > a) Proof To make the intuitive proof rigorous we only have to prove (3.9). To extract this from the strong Markov property (3.7) , we let

{

if s < t , w(t - s) > a Y,(w) = 01 otherwise _ We do this so that if we let S = inf{s < t

Using the definition of u, 1 (BTEA) o Bs the previous display, we have

Exercise 3. 10.

Let G,

If we let

T, D(x, o) and S be as above, but now suppose a bounded function and let u(y) =

on

Now Bs

: B. = a} with inf 0 = oo then

{S < oo} = {Ta < t}

rp(x , s) = E:r: Y• the strong Markov property implies

Eo(Ys o Bsi :Fs) = rp (Bs , S)

0

E9T < oo for all y E G. Let g be E9(J: g (B. ) ds). Show that for x E G

that

Ys(Bsw) = 1 (B,>a)

= 1 (BTEA)> and taking expected value of

u(x) = E:r:1 (BTE A) = E:r: ( 1 (BTE A) o Bs ) = E:r:E:z: ( 1(BTEA) o Bs I Fs ) = E:r:u(Bs)

on

{S < oo} = {Ta < t}

= a on {S < oo } and rp(a, s) = 1 /2 if s < t, so Po(Ta < t, Bt > a) = Eo(1/2 ; Ta < t)

which proves (3.9). Exercise 3. 11. Generalize the proof of (3.9) to conclude that if u

then Our third application shows why we want to allow the function Y that we apply to the shifted path to depend on the stopping time S.

25

(3.10)

Po(Ta < t, u < Bt < v ) = Po(2a - v < B1 < 2a - u)


26

Section 1 .4 First Formulas

Chapter 1 Brownian Motion

(3.10), let Mt = max0$•9 B. to rewrite it as Po(Mt > a,u < Bt < v) = Po(2a - v < Bt < 2a - u) Letting the interval (u, v) shrink to x we see that 1 e - ( 2 a - x )2 / 2 t o( Mt > a, Bt - x) - o (Bt - 2a - x) - -12-if. Differentiating with respect to a now we get the joint density ) e-(2a-x)2/2t {3.11) Po(Mt = a, Bt = x) = 2� 2m3 To explain our interest in

p,

_

_

p,

_

[0, s]. dt =sf(r-s/(rs) s)2 dr, 1 100 {r + s) 2 1/2 s dr ; _. --;:-;--- {r + s)2

We have two interrelated aims in this section: the first to understand the be­ havior of the hitting times for a one dimensional Brownian motion the second to study the behavior of Brownian motion in the upper half space We begin with and observe that the E reflection principle implies

: Bt = a} Bt;H = {x Rd Ta: =>inf{t Ta {3.8) xd 0}. Po(Ta < t) = 2Po(Bt > a) = 2 100 (2m) - 112 exp(-x2 /2t) dx Here, and until further notice, we are dealing with a one dimensional Brow­ nian motion. To find the probability density of Ta, we change variables x = t1 12ajs112 , dx = -t112a/2s312 ds to get (4.1) Po(Ta < t) = 2 1° (2m) - 112 exp(-a2/2s) (-t 112a/2s312) d� t = 1 (27rs3 )- 112a exp( -a2 /2s) ds Using the last formula we can compute the distribution of L = sup{t � 1 : Bt2.3 =and0} 2.4. and R = inf{t � 1 : Bt = 0}, completing work we started in Exercises By (2.5) if 0 < s < 1 then Po(L � s) = 1: p, (O, x)Px (T0 > 1 - s) dx = 2 Jrooo {27rs) - 1 12 exp( -x2 /2s) 1r1 -oos (27rr3) - 112 x exp( -x2 /2r) dr dx

r [1 - s,

Our next step is to let t = oo ) + to convert the integral over E into one over t E so to make the calculations easier + we first rewrite the integral as

( )

_

1.4. First Formulas

27

Changing variables as indicated above and then again with

t = u2 to get

Po(L � s) = .!:_ l['o (t(1 -t)) - 1/2 dt = -2 10 .../i{1 - u2)- 1 12 du = -2 arcsin(Vs)

(4.2)

1T

1T

1T

L= 1, Po(L = t) = -1 10 ' (t(1 - t))- 112 for 0 < t < 1 is symmetric about 1/2 and blows up near 0 and 1. This is one of two arcsine laws associated with Brownian motion. We will encounter the other one in Section 4.9. The computation for R is much easier and is left to the reader. Exercise 4.1. Show that the probability density for R is given by Po(R = 1 + t) = 1/(m1 12 (1 + t)) for t � 0 Notation. In the last two displays and in what follows we will often write P(T = t) = f(t) as short hand for T has density function f(t). As our next application of (4.1) we will compute the distribution of Br where = inf{t : Bt f/:. H} and H = {z : Zd > 0} and, of course, Bt is a d dimensional Brownian motion. Since the exit time depends only on the last coordinate, it is independent of the first d- 1 coordinates and we can compute the distribution of Br by writing for x, E Rd- 1 , E R P(x,y){Br = (B , 0)) = 100 ds P(x,y) ( = s)(27rs) - (d- 1)/2e- lx- BI2/2• = {27rY) d/2 Jro oo ds 8- (d+2)/2e-
0

1T

T

{}

y

T

28 Ch apter 1 Brownian Motion

{

by 4. 1) with

Section 1.4 First Formulas 29

a = y. Changing variables s = (lx - B l2 + y2 ) f2t gives ) (d+2)/2 e-t 2t y lo -(lx - B l 2 + y2 ) dt (

(211") d/ 2

oo

By induction it follows that

l x - B l2 + y2

2t2

so we have

P(x,y) (Br = (B , 0)) = (lx l2y+ y2)d/2 f11"(d/2) d/2 00 where f(a) = f0 ya - l e- Y dy is the usual gamma function. When d = 2 and x = 0, probabilists should recognize this as a Cauchy distribution. At first glance, the fact that Br has a Cauchy distribution might be surprising, but a moment's thought reveals that this must be true. To explain this, we begin by looking at the behavior of the hitting times {Ta , a 0} as the level a varies. (4.4) Theorem. Under Po, {Ta, a 2:: 0} has stationary independent increments. Proof The first step is to notice that if 0 < < b then n eT. = n - Ta , (4.3)

_

B

2::

a

0

so if f is bounded and measurable, the strong Markov property (3.7) and trans­ lation invariance imply Eo (f(n - Ta ) E (f(n ) o B

T. 1FT. ) 1FT. ) = o E = Ea f(Tb) = of(n- a )

The desired result now follows from the next lemma which will be useful later.

(4.5) Lemma. Suppose Zt is adapted to gt and that for each bounded mea­ surable function f E(f(Zt - Zs )lgs) = Ef(Zt - s)

Z1 has stationary independent increments. Proof Let < t 1 . . . < tn , and let /;, 1 � i � n be bounded and measurable . and using the hypothesis we have Conditioning on t

Then

to

F n-l

which implies the desired conclusion. The scaling relation

D

(1.2) implies

(4.6) So consulting Section 2.7 of Durrett (1 995) we see that Ta has a stable law with index 0:' 1/2 and skewness parameter K 1 since Ta 2:: To explain the appearance of the Cauchy distribution in the hitting lo­ cations, let Ta inf t where is the second component of a two dimensional Brownian motion and observe that another application of the strong Markov property implies

=

= ( 0). = { : B'f = a}) ( B'f (4.7) Theorem. Under Po, {Cs = B(rs), s 0} has stationary independent increments. 2::

Exercise

4.2. Prove (4.7).

The scaling relation

(1 .3 ) and an obvious symmetry imply Cs d -Cs

(4.8)

=

so again consulting Section 2.7 of Durrett ( 1 995) we can conclude that Cs has the symmetric stable distribution with index a 1 and hence must be Cauchy. To give a direct derivation of the last fact let rp8 (s) E exp iBC3 . Then 4.7 and 4.8 imply

=

( ) ( )

=

r.oo(s)rpo (t) r,oo (s + t), r.oo(s)

= r.oos (1),

= ( ( )) rpo(s) = r.o- o(s)

The fact that Cs 4 -Cs implies that rpo(s) is real. Since B -+ rpo(1) is con­ tinuous, the second equation implies s -+ rp0 (s) rp0 3 (1) is continuous, and a simple argument see Exercise 4.3) shows that for each B, rp0 (s) exp ( o s) The last two equations imply that c3 sc1 and c_o = co so co = -K B for some K, so Cs has a Cauchy distribution. Since the arguments above apply equally well to where is any constant, we cannot determine K with this argument.

(

(Bf, cBl),

c

=

=

= c II

.

30

Chapter 1 Brownian Motion

Section 1.4 First Formulas

Exercise 4.3. Suppose cp(s) is real valued continuous and satisfies cp(s)cp(t) = cp(s + t), cp(O) = 1. Let .,P(s) = ln(cp(s)) to get the additive equation: .,P(s) +

.,P(t) = .,P(s + t). Use the equation to conclude that .,P(m2 - n ) = m2 - n .,P(1) for all integers m, n � 0 , and then use continuity to extend this to .,P t = t .,P 1 .

()

E(exp(i0C3)) to show Eo(exp(->.Ta)) = exp (- KVA)

Exercise 4.4. Adapt the argument for cp8 (s)

()

.

=

a

The representation of the Cauchy process given above allows us to see that its sa,mple paths a -+ and s -+ are very bad.

Ta

Exercise 4.5. If u Exercise 4.6. If u

C3

Po( -+ Ta is discontinuous in ( < v then Po(s -+ C3 is discontinuous in (

< v then

a

u,

v )) = 1 .

u,

v)) = 1 .

Hint. B y independent increments the probabilities in Exercises 4.5 and 4.6

-

only depend on v size of the interval.

u

but then scaling implies that they do not depend on the

B1

leaves H. The rest of the The discussion above has focused on how section is devoted to studying where goes before it leaves H. We begin with the case = 1 .

B1

d

(4.9) Theorem. If where

x,y > 0, then Pc(Bt = y,To > t) = Pt(x,y) - Pt(x,-y) Pt (X , Y) - (2 _.li t) - 1/2 e - (y-x)2 /2t _

The proof is a simple extension of the argument we used in Section 1.3 to prove that � a ) . Let � 0 with = 0 when :s; t) = :s; 0. Clearly

Proof

Po(Ta

2Po(B1

f(x) f Ex(f(Bt) ; To > t) = Exf(Bt) - Ex(f(Bt); To :s; t) If we let f(x) = f( -x), then it follows from the strong Markov property and symmetry of Brownian motion that Ex(f(Bt); To :s; t) = Ex[Eof(Bt -T0);To :s; t] = Ex[Eo f
/(y) = 0 for y � 0. Combining this with the first equality shows Ex(f(Bt); To > t) = Exf(Bt) - Ex/(Bt) = j (Pt(x, y) - P t(x,-y))f(y)dy The last formula generalizes easily to d � 2. (4.10) Theorem. Let r = inf{t : Bf = 0}. If x, y E H, Px (Bt = y , r > t) = Pt(x, y) - Pt(x , jj) where jj = (y1 , . . . , Yd- 1 , -ya ) . Exercise 4.7. Prove (4 . 10) .

31

since

0

2

Stochastic Integration

In this chapter we will define our stochastic integral It = J; H.dX• . To motivate the developments, think of X. as being the price of a stock at time s and H. as the number of shares we hold, which may be negative (selling short) . The integral It then represents the net profits at time t, relative to our wealth at time 0. To check this note that the infinitesimal rate of change of the integral di1 = Ht dXt = the rate of change of the stock times the number of shares we hold. In the first section we will introduce the integrands H. , the "predictable processes," a mathematical version of the notion that the number of shares held must be based on the past behavior of the stock and not on the future performance. In the second section, we will introduce our integrators X. , the "continuous local martingales." Intuitively, martingales are fair games, while the "local" refers to the fact that we reduce the integrability requirements to admit a wider class of examples. We restrict our attention to the case of martingales with continuous paths t -+ X1 to have a simpler theory. 2.1. Integrands: Predictable Proc esses

To motivate the class of integrands we consider, we will discuss integration w.r.t. discrete time martingales. Here, we will assume that the reader is familiar with the basics of martingale theory, as taught for example in Chapter 4 of Durrett ( 1995). However, we will occasionally present results whose proofs can be found there. Let Xn , n ;:::: 0 , be a martingale w.r.t. :Fn . If Hn , n ;:::: 1, is any process, we can define n

L Hm(Xm - Xm -d m=l To motivate the last formula and the restriction we are about to place on the Hm , we will consider a concrete example. Let 6 , 6 , . . . be independent with P(ei = 1) = P(ei = -1) = 1/2, and let Xn = 6 + · ·+en · Xn is the symmetric simple random walk and is a martingale with respect to :Fn = u (6 , . . . ' en ) · (H · X)n =

34

Section 2.1 Integrands: Predictable Processes

Chapter 2 Stochastic Integration

If we consider a person flipping a fair coin and betting $1 on heads each time then Xn gives their net winnings at time n. Suppose now that the person bets an amount Hm on heads at time m (with Hm < interpreted as a bet of -Hm on tails). I claim that (H · X)n gives her net winnings at time n . To check this note that if Hm > our gambler wins her bet at time m and increases her fortune by Hm if and only if Xm - Xm - 1 = 1 . The gambling interpretation of the stochastic integral suggests that it is natural to let the amount bet at time n depend on the outcomes of the first n - 1 flips but not on the flip we are betting on, or on later flips. A process Hn that has Hn E Fn - 1 for all n � 1 (here :Fa = {0, n}, the trivial a--field) is said to be predictable since its value at time n can be predicted (with certainty) at time n - 1. The next result shows that we cannot make money by gambling on a fair game.

0

0

{1 . 1) Theorem. Let Xn be a martingale. If Hn is predictable and each Hn is

bounded, then (H · X)n is a martingale.

Proof It is easy to check that (H · X)n E :Fn . The boundedness of the Hn implies EI(H · X)n l < oo for each n. With this established, we can compute conditional expectations to conclude

E((H · X)n + l i:Fn) = (H · X)n + E(Hn + l (Xn + l - Xn) I :Fn) = (HX) n + Hn +l E(Xn + l - Xn i:Fn) = (H · X)n since Hn + l E :Fn and E(Xn + l - Xn i:Fn ) =

0.

D

The last theorem can be interpreted as: you can't make money by gambling on a fair game. This conclusion does not hold if we only assume that Hn is optional, that is, Hn E :Fn, since then we can base our bet on the outcome of the coin we are betting on. Example 1. 1. If Xn is the symmetric simple random walk considered above

and Hn = en then

n

(H · X)n = since e� =

1.

I: em · em = n

m= l

D

In continuous time, we still want the metatheorem "you can't make money gambling on a fair game" to hold, i.e., we want our integrals to be martingales. However, since the present (t) and past ( < t) are not separated, the definition of the class of allowable integrands is more subtle. We will begin by considering a simple example that indicates one problem that must be dealt with.

35

0

Example 1.2. Let (n, :F, P) be a probability space on which there is defined

a random variable T with P(T ::; t) = t for ::; t ::; 1 and an independent random variable e with P(e = 1) = P(e = - 1) = 1/2. Let Xt =

{0 e

t
a:nd let :Ft = o-(X& : s ::; t). In words we wait until time T and then flip a coin. Xt is a m�rtingale with respect to :Ft . However, if we define the stochastic . mtegral It = fo X. dX. to be the ordinary Lebesgue-Stieltjes integral then

0

To check this note that the measure dX. corresponds to a mass of size e at T and hence the integral is e times the value there. Noting now that Yo = while o Y1 = 1 we see that It is not a martingale. The problem with the last example is the same as the problem with the one in discrete time our bet can depend on the outcome of the event we are betting on. Again there is a gambling interpretation that illustrates what is wrong. Consider the game of roulette. After the wheel is spun and the ball is rolled, people can bet at any time before ( <) the ball comes to rest but not after (�). One way of guaranteeing that our bet be made strictly before T, i.e., a sufficient condition, is to require that the amount of money we have bet at time t is left continuous. This implies, for instance, that we cannot react instantaneously to take advantage of a jump in the process we are b;tting on. The simplest left continuous integrand we can imagine is made by picking a < b, C E Fa , and setting

H(s, w) = C(w) 1 (a ,bj (t) !n words, we buy C(w) shares of stock at time a based on our knowledge then, I.e., C E :Fa . We hold them to time b and then sell them all. Clearly, the

examples in (* ) should be allowable integrands; one should be able to add two or more of these, and take limits. To encompass the possibilities in the previous sentence, we let II be the smallest a--field containing all sets of the form (a, b] x A where A E :Fa . The II we have just defined is called the predictable a--field. We will demand that our integrands H are measurable w.r.t. II. In the previous paragraph, we took the "bottom-up" approach to the defi­ nition of II. That is, we started with some simple examples and then extended the class of integrands by taking sums and limits. For the rest of the section ' we . will take a "top-down" approach. We will start with some natural requirements

36

Chapter 2 Stochastic Integration

Section 2.2 In tegrators: Continuous Local Martingales

and then add more until we arrive at II. The descending path is more confusing since it involves four definitions that only differ in subtle ways. However, the reader need not study this material in detail. We will only use the predictable u-fie ld. The other definitions are included only to allow the reader to make the connection with other treatments and to indirectly make the point that the can be measure theoretic questions associated with continuous time processes quite difficult. Let :Ft be a right continuous filtration. The first and most intuitive con­ cept of "depending on the past behavior of the stock and not on the future performance" is the following: H(s, w) is said to be adapted if for each t we have

Ht E :Ft.

We encountered this notion in our discussion of the Markov property in Chapter 1. In words, it says that the value at time t can be determined from the information we have at time t. The definition above, while intuitive, is not strong enough. We are dealing with a function defined on a product space [0, oo ) x n, so we need to worry about measurability as a function of the two variables. H is said to be progressively measurable if for each t the mapping (s, w) --> H(s, w) from [O, t] to R is n x :Ft measurable. This is a reasonable definition. However, the "modern" approach is to use a slightly different definition that gives us a slightly smaller u-field. Let A be the u-field of subsets of [0, 00 ) X n that is generated by the adapted processes that are right continuous and have left limits, i.e. , the smallest u-field which makes all of these processes measurable. A process H is said to be optional if H(s, w) is measurable w.r.t. A. According to Dellacherie and Meyer (1978) page 122, the optional u-field is contained in the progressive u-field and in the case of the natural filtration of Brownian motion the inclusion is strict. The subtle distinction between the last two u-fields is not important for us. The only purpose here for the last two definitions is to prepare for the next one. Let II1 be the u-field of subsets of [0 ' 00) X n that is generated by the left continuous adapted processes. A process H is said to be predictable if H( s , w) E II1• As we will now show, the new definition of predictable is the same as the old one. We have added the 1 only to make the next statement and proof possible.

37

(1.2) Theorem. II = II1 •

Proof Since all the processes used to define II are left continuous, we have II C II1• To argue the other inclusion, let H(s, w) be adapted and left continuous and let Hn (s , w ) = H(m2- n , w ) for m2 - n < s :::; (m + 1)2- n . Clearly Hn E II1• n

Further, since, H is left continuous, H (s, w) --> H(s, w) as n --> 00 .

o

. The distinction between the optional and predictable u-fields is not impor­ tant for Brownian motion since in that case the two u-fields coincide. Our last fact is that, in general, II C A. Show that if H(s, w) = 1 (a , bj (s)1A (w) where A E :Fa , then Exercise l. l. . · · of a sequence of optional processes; therefore, H is optional and the hm1t H IS

II c A.

2.2. Integrators: Continuous Lo cal Martingales

In Section 2.1 we described the class of integrands that we will consider: the predictable processes. In this section, we will describe our integrators: contin­ uous local martingales. Continuous, of course, means that for almost every w, the sample path s --> X8 (w) is continuous. To define local martingale we need some notation. If T is a nonnegative random variable and yt is any process we define on { T > 0} y:T t = YTAt on { T = 0} 0

{

(2.1) Definition. Xt is said to be a local martingale (w.r.t. {:Ft, i ;::: 0}) if there are stopping times Tn j oo so that X'{n is a martingale (w.r.t. {:FtATn : t ;::: 0}). The stopping times Tn are said to reduce X. We need to set Xf :: 0 on {T = 0} to deal with the fact that X0 need not be integrable. In most of our concrete examples Xo is a constant and we can take T1 > 0 a.s. However, the more general definition is convenient in a number of situations: for example (i) below and the definition of the variance process in

(3.1).

In the same way we can define local submartingale, locally bounded, locally of bounded variation, etc. In general, we say that a process Y is locally A if there is a sequence of stopping time Tn j oo so that the stopped process Yt has property A. (Of course, strictly speaking this means we should say locally a submartingale but we will continue to use the other term.)

38

Chapter 2 Stochastic Integration

Section 2.2 Integrators: Continuous Local Martingales

Why local martingales?

(2.2) Theorem. {XI' (t ) , .1"1' (t ) , t 2:: 0 } is a martingale.

There are several reasons for working with local martingales rather than with martingales.

First we need to show:

(i) This frees us from worrying about integrability. For example, let X1 be a martingale with continuous paths, and let cp be a convex function. Then cp (Xt) is always a local submartingale (see Exercise 2.3). However, we can conclude that cp (X1) is a submartingale only if we know Ejcp(X1) 1 < oo for each t, a fact that may be either difficult to check or false in some cases. (ii) Often we will deal with processes defined on a random time interval [0, r) . If r < oo, then the concept of martingale is meaningless, since for large t X1 is not defined on the whole space. However, it is trivial to define a local martingale: there are stopping times Tn j r so that . . .

(iii) Since most of our theorems will be proved by introducing stopping times Tn to reduce the problem to a question about nice martingales, the proofs are no harder for local martingales defined on a random time interval than for ordinary martingales. Reason (iii) is more than just a feeling. There is a construction that makes it almost a theorem. Let X1 be a local martingale defined on [0, r) and let Tn j r be a sequence of stopping times that reduces X. Let To = 0, suppose T1 > 0 a.s., and for k 2:: 1 let

,(t) =

{ �� (k - 1) rk .L i.

n - 1 + (k - 1) :::; t :::; + (k - 1) +
rk

To understand this definition it is useful to write it out for

t

T1

! (t) = tT- 1 2 t-2 T3

k = 1, 2, 3:

[0, T1]

[T1 , T1 + 1] [T1 + 1, T2 + 1] [T2 + 1, T2 + 2] [T2 + 2, T3 + 2] [T3 + 2, T3 + 3]

In words, the time change expands [0, r) onto [O, oo) by waiting one unit of time each time a Tn is encountered. Of course, strictly speaking 1 compresses [0, oo) onto [0, r ) and this is what allows Xl' (t ) to be defined for all t 2:: 0. The reason for our fascination with the time change can be explained by:

39

(2.3) The Optional Stopping Theorem. Let X be a continuous local martin­ gale. If S :::; T are stopping times and XT At is a uniformly integrable martingale then E(XT IFs ) = Xs . Proof (7 .4) in Chapter 4 of Durrett (1995) shows that if L :::; M are stopping times and YM A n is a uniformly integrable martingale w.r.t. gn then

To extend the result from discrete to continuous time let Sn = ([2n S] + 1)/2n . Applying the discrete time result to the uniformly integrable martingale Ym = XTA m2 - n with L = 2n sn and M = oo we see that Letting n -+ oo and using the dominated convergence theorem for conditional expectations ((5.9) in Chapter 4 of Durrett (1995)) the result follows. 0 Proof of (2.2) Let n = [t] + 1. Since 1(t) :::; Tn A n, using the optional stopping theorem, (2.2), gives XI' ( f) = E(XTnA n IF')'(t ) ) · Taking conditional expectation with respect to .1"1'(3 ) we get

proving the desired result.

0

The next result is an example of the simplifications that come from assum­ ing local martingales are continuous.

(2.4) Theorem. If X is a continuous local martingale, we can always take the sequence which reduces X to be Tn = inf{t : jX1 j > n} or any other sequence T� :::; Tn that has T� j oo as n j oo. Let Sn be a sequence that reduces X. If s < t, then applying the optional stopping theorem to X!n at times r = s A T:r, and t A T:r, gives Proof

40 Chapter 2 Stochastic Integration Multiplying by

Section 2.2 Integrators: Continuous Local Martingales

l (T:,.>o) E :Fo C :F(s A T:r_ A Sn ) we get

E(X(t A T:n A Sn )l (sn >o,T:,.>o) I :F(s A T:r_ A Sn )) = X(s A T:r_ A Sn )l ( sn >o,T:,.>o) As "!' j oo , :F(s A T:r_ A Sn ) j :F(s A T:r_ ), and X(r A r:r. A Sn )1 (Sn > O ,T:,.> O ) --? X(r A r:r_ )1cT:.. >o) for all r :2:: 0, and IX(r AT:r_ ASn ) I 1 (Sn> O ,T:,.>o) :::; m, so it follows from the dom­ inated convergence theorem for conditional expectations (see (5.9) in Chapter 4 of Durrett (1995)) that E(X(t A T:r_)1 (T:,. >o) l :F(s A T:r_)) = X(s A T:r_)l (T:,. >o) proving the desired result.

41

if and only if -(d - 2)p + (d - 1) > -1, i.e., p < dj(d - 2). To see that X1 is not a martingale we begin by noting that (Bt - x ) = d t 1f2(B 1 - x ) under P.., so as t --? oo , I Bd --? oo and X1 --? 0 in probability. To show that E..,X1 - 0 we observe that for any R < oo

The integral is convergent so limsup1_+<,o E..,X1 :::; R- (d- 2 ) . Since R is arbitrary E..,X1 --? 0. Since E..,X0 > 0 it follows that E..,X1 is not constant and hence X1 is not a martingale. D An even worse counterexample is provided by

D

In our definition of local martingale in (2.1), we assumed that xJn is a martingale ( w.r.t. {:FtATn , t :2:: 0}). We did this with the proof of (2.4) in mind. The next exercise shows that we get the same definition if we assume xJn is a martingale (w.r.t. {:F1 , t :2:: 0}).

S be a stopping time. Then Xl is a martingale w.r.t. :Ft i\S, t :2:: 0, if and only if it is a martingale w.r.t. :Ft, t :2:: 0.

Exercise 2. 1. Let

In the first version of this book I said "You should think of a local martingale as something that would be a martingale if it had E j X1 I < oo ." As many people have pointed out to me, there is a famous example which shows that this is WRONG.

> 3. In Section Example 2. 1. Let Bt be a Brownian motion in dimension d 3 . 1 we will see that X1 = 1/ IBtl d-2 is a local martingale. Now if and only if p < dj(d - 2) since (i ) there is no trouble with the convergence for large y and (ii) ignoring (21it) -df2 e -ly-x l 2 /2 t which is bounded near 0 and changing to polar coordinates we have

Exercise 2.2. In Section 3.1 we will learn that if B1 is a two dimensional

Brownian motion then X1 = log !Btl is a local martingale. Show for any p < oo but lim1-oo EX1 = oo so X is not a martingale.

EIX1 j P < oo

The last example may leave the reader with the impression that no amount of integrability will guarantee that a local martingale is an honest martingale, but as the next very useful result shows, this is a false impression.

(2.5) Theorem. If X1 is a local submartingale and E

(o:-:;.:-:;t IX. I) < sup

oo

for each t then X1 is a submartingale.

Proof Clearly our assumption implies E j X1 I

< oo for each t so all we have to check is that E(Xt j:F,) :2:: X, . To do this, we note that if Tn is a sequence of stopping ti_mes that reduces X

then we let n --? oo and apply the dominated convergence theorem for condi­ tional expectations. See (5.9) in Chapter 4 of Durrett (1995). D As usual, multiplying by - 1 shows that the last result holds for supermartin­ gales, and once it is true for super- and sub- it is true for martingales (by applying the two other results) . The following special case will be important:

(2.6) Corollary. A bounded local martingale is a martingale.

42

Section 2.3 Variance and Covariance Processes

Chapter 2 Stochastic Integration

Xt is bounded we mean that litya process we sayprobabi whenwith follows what in and Here � M for all t 2': 0. 1, I IXt that so M t a constan thereInissome for local mar­ theorem ence converg asufficien will need weresult tances circums t for our needs. tingales. The following simple is {2.7) Theorem. Let Xt be a local martingale on [0, r) . If E (0 sup< T I X. I) < oo $• then XT = limtTT Xt exists a.s. and EXo = EXT . gale. is aencemartin Proof Let be the time scale of {2. 2 ) and let = X-y(t ) · theorem converg ale marting the of proof the and ity inequal upincrossing The 4.2 of Durrett {1995) generalize in a straigh tforward way to Section given that e Y = limtT oo exists a. s . and conclud to us allow and time ous continu EYo = that XT = limtTT Xt exists a. s . To prove the last conclusion we note EY; then use the a.s. convergence and the dominated convergence theorem to conclude EXo = EYo = EYoo = EXT . a convex Exercise 2.3. Let Xt be a continuous local martingale and let be function. Then rp(.Xt) is a local submartingale. < be a Exercise 2.4. Let X be a continuous local martingale and let to Q, = respect with gale martin local a g time. Then Y, = XR+• is stoppin :FR+• · Exercise 2.5. Show that if X is a continuous local martingale with Xt 2': 0 and EX0 < oo then Xt is a supermartingale. Y1

'Y

00

Y1

Y1

0


R

oo

2.3 . Variance and Covariance Proce sses

The next definition may look mysterious but it is very important. em. If Xt is a continuous local martingale, then we define the 1) Theor (3.varian ing ce process (X}t to be the unique continuous predictable increas gale. process At that has Ao = 0 and makes Xl - At a local martin To warm up for the proof of (3.1) we will prove a result in discrete time. (3.2) Theorem. Suppose Xn is a martingale with EX� < for all n. Then there is a unique predictable An with Ao = 0 so that X� - An is a martingale. Furthermore, n -+process A is increasing. oo

n

Proof

43

Let Ao = 0 and define for n 1 ;:::

An = An - 1 + E(X�l:Fn - d - X�_ 1 An E Fn _ 1 , An

Fromis athesubmartingale), definition, it isandimmediate that

X�

is increasing (since

E(X� - An i :Fn - 1 ) = E(X�l:Fn - d - An = X� - 1 - An - 1 A An - Bn

A has the desired properties. To see that is unique, observe that if B ISA�o another process with desired properties, then is a martingale and n - Bn E Fn - 1 · Therefore An - Bn = E(A n - Bn I :Fn_ I ) = An - 1 - Bn - 1 An - Bn = Ao - Bo = 0.

and it follows by induction that Thetimekeymartingale to the uniqueness may" Since be summarized asmotion "any ispredictable dis­ crete is constant. Brownian a predictable . the last statement fails in continuous time and we need another martmgale, assumption. {3.3) Theorem. continuous local martingale locally of boundedAny variation is constant {in time). Xt that is predictable and �ro�f By subtracting Xo, we may suppose that X0 = 0 and prove that is Identically to check 0. Let be the variation of X on [0, t] . We leave it to the reader 3.1. If X is continuous and locally of bounded variation then t -+ isExercise continuous. e S = inf{ s : VsMt2': =I<}.X(tSince t � S implies I Xd � K, (2.4) implies thusmg t� Defi thewellnstopped process A S) is a bounded martingale. Now if s < t known properties of conditional expectation (see e.g., (1.3) and other results in Chapter 4 of Durrett {1995)) we have IF.) - 2M,E(Mt iF.) + M'f {3 .4) E((Mt - M,)2 l:F.) == E(M? E(M? IF.) - M'f = E(Mt2 - M]IF.) an equat�on that we will refer to as the orthogonality of martingale incre­ ments smce the key to its validity is the fact that E(M. (Mt - M, )j:F,) = 0 (and hence E(M, (Mt - M,)) = 0). \It

0

X

v;t

!I

44

Chapter 2 Stochastic Integration

Section 2.3 Variance and Covariance Processes

1

Letting 0 = to < t . . . < tn = t be a subdivision of [0, t] , the orthogonality of martingale increments, the inequality Ei a[ $ (L:i I a i l) supi lai I , and the fact that V1A s $ J{ imply

EMf E =

(m=tl

Mt2m - ML_1

n

=

)

E

(

V1A S S�p !Mt m

- Mt m- 1 1 $ J( E sup !Mt m - Mt m -1 1 m

$

If the reader recalls what we are trying to define (see (3.1)) then the reason for the definition is explained by (a) �emma. If X, is a bounded continuous martingale then Xl - Qf(X) is a martmgale. Proof To prove this, we first note that reasoning as in the proof of (3.4) one

E ::L: (Mt m - Mt m_J 2

m=l

45

can conclude that if r < s < t then

E ((X, - Xr) 2 !:F.)

)

If we take a sequence of partitions �n = {0 = t� < ty . . . < tk( n ) = t} in which the mesh l�n l = sup m I t� - t�_ 1 1 -+ 0 then continuity implies supm !Mt;:;, ­ M, ..m- 1 1 -+ 0. Since the supm $ 2!{ the bounded convergence theorem implies E supm !Mt;:;, - Mt::,_1 1 -+ 0 . This shows EMl = 0 so Mt = 0 a.s. In the last

conclusion t is arbitrary, so with probability one we have M, t. The desired result now follows from continuity.

= 0 for all rational

0

With (3.3) established we can do the Proof of uniqueness in (3.1 ) Suppose A, and A� are two processes with the desired properties. Then A, - A� is a continuous local martingale that is locally of bounded variation and hence must be constant. Since Ao - A� = 0 it 0 follows that A1 = A� for all t. Proof of existence in (3.1) when X, is a bounded martingale The existence of A is considerably more complicated than its uniqueness. To inspire the reader for the battle to come, we would like to point out that (i) we are about to prove the special case that we need of the celebrated Doob-Meyer decomposition and (ii) the construction will. provide some useful information, e.g. , (3.6). Though (3.1) and (3.8) are quite important, the details of their proofs are not, so the squeamish reader can skip ahead to the boldfaced "End of Existence Proof" five pages ahead. Given a partition � = {0 = to < � 1 < t 2 . . . } with limn tn = oo, we let k(t) = sup{k : t�: < t} be the index of the last point before time t. (To avoid confusion later, we emphasize now that k(t) is a number not a random variable.) We next define for a process X, an approximate quadratic variat�on by

k (t ) Qf'" (X) = ::l:(Xt k - X,k_1? + (X, - Xt k ) 2 k=l

- X. )2 !:F.) - 2(X. - Xr)E(Xt - X. !:F. ) + (X. - Xr)2 = E ((Xt - X. ) 2 I:F.) + (X. - Xr ) 2 = E ( (X,

since E(X, - X. !:F. ) = 0. Writing Qf" = Q;a + (Qf" - Q;o) and working out the difference ,

k (t ) Qf'" - Q� = ::l:CXt k - X,k_J 2 + (Xt - X,k (l ) ? kk=(• )l - ::l: (Xt k - Xt k_J 2 + (X, - Xt k<• > ) 2 k=l = (Xt k<•>+• - X, k<•> ) 2 - (X. - Xt k<•> ? k (t ) + ::L: (Xt k - Xt k_J 2 + (Xt - Xt k< t > )2 k=k(•)+2 The next step is to take conditional expectation with respect to :F. and use the first formula with = t k ( • ) and t = tk ( • )+l · To neaten up the result, define a sequence Un 1 k(s) - 1 $ n $ k(t) + 1 by letting uk ( • ) - 1 = s, u i = t i for k(s) $ i $ k(t) , and u k (t )+l = t. r

E(Xf - Qf"(X)!:F.)

46

Chapter 2 Stochastic Integration

Section

where the second equality follows from (3.4).

0

Looking at (a) and noticing that Qf(X) is increasing except for the last term, the reader can probably leap to the conclusion that we will construct the process A1 desired in (3.1) by taking a sequence of partitions with mesh going to 0 and proving that the limit exists. Lemma (c) is the heart (of darkness) of the proof. To prepare for its proof we note that using (a) , taking expected value and using (3.4) gives

E(Qf (X) - Q_;. (X)IF.) = E(Xl - X] IF.) = E((X1 - X,) 2 jF.)

(b)

To inspire you for the proof of (c) , we promise that once (c) is established the rest of the proof of (3.1) is routine. (c) Lemma. Let X1 be a bounded continuous martingale. Fix r > 0 and let �n be a sequence of partitions 0 = < . . . < = r of [0, r] with mesh = supk 0. Then Q� n (X) converges to a limit in L2.

l �n l

t� t�

It� - t�_ 1 l -+

tt

(3.5)

t -+

for any real numbers

ti :::; sk}. By the Cauchy-Schwarz inequality 1 12 E(Q��I (Q�)) :::; {EQ��I (X) 2 J ll 2 { E s p(X. k+1 + x. k - 2X1j(k) )4} � Whenever 1�1 + 1 �' 1 -+ 0 the second factor goes to 0, since X is bounded and continuous, so it is enough to prove (e) If I X1 I :::; M for all t then EQ� �� (X)2 :::; 12M4 . where j(k) = sup{j :

The value 12 is not important but it does make it clear that the bound does not depend on r, �, or �' . Proof of (e)

If r = ��I is the partition 0 = so

Q;(x )' =

2a2 + 2b2 - (a + b) 2 = (a - b) 2 � 0) we have Q��� (Y) :::; 2Q��� (Q�) + 2Q��� (Q� ' ) Combining the last two results about Q we see that it is enough to prove that

ti

+ :::; ti+1 · Recalling

To do this let Sk E �� ' and E � so that ij :::; Sk < Bk l the definition of Q� (X) and doing a little arithmetic we have

Q;:+ l - Q ;: = (X. k+1 - X1J 2 - (X, k - X1i ) 2 = (X. k+1 - X, k ) 2 + 2(X. k+1 - X, k )(X. k - X1J = (X. k+1 - X, k )(X. k+1 + X. k - 2XtJ

(f,

(x - x, _ _ , )' ••

)

'

<

81 . . . < Sn =

n = :L: <x. m - x. m_ J4 m=1n 1 + 2 L (X. m - x. m_J 2 (Q� (X) - Q!'m (X)) m=1

(e.1)

To bound the first term on the right-hand side we note that if IX1 1 (3.4) imply

t then some arithmetic and

a and b

(since

47

Q��� (Q�) :::; Q��� (X) s�p(X, k+1 + X, k - 2Xti(kl ) 2

E(Q�(X) - Q� ' (X)) 2 = EYr2 = EQ��� (Y) For reasons that will become clear in a moment, we will drop the argument from Q� when it is X and we will drop the r when we want to refer to the process Qf' . Since

Variance and Covariance Processes

Summing the squares gives

Proof If � and �' are two partitions, we call ��' the partition obtained by

taking all the points in � and in � ' . If we apply (a) twice and take differences we see that yt = Qf'(X) - Qf" (X) is a martingale. By definition Yl - Qf' � ' (Y) is a martingale, so

2.3

n m=1

r

then

:::; M for all

n m=1 m=1

E L (X. m - x. m_ J 4 :::; (2M) 2 E L (X. m - x. m-1 ? (e.2)

To bound the second term we note that and I Xr - x. Tn l :::; 2M to get (e.3)

(X, m - X, m_J2 E F, m then use (b)

E((X. m - X. m_ J2 {Q� (X) - Q!'m (X)} IF. m ) = (X. m - X, m_J2 E( {Q� (X) - Q!'m (X)} IF.m ) = (X, m - X, m_1 ) 2 E((Xr - X. m ) 2 1Fsm ) :::; (2M) 2 (X. m - X. m_ 1?

48

Section 2.3 Variance and Covariance Processes

Chapter 2 Stochastic Integration

Taking get expected value in (e.3), summing over m, and using (3.4) as before, we n . n- 1 2 x; x;m m -l E (X3 m - X3 m_ J2 (Q� (X) - Q�m (X)) :5 4M E 2.::: (e.4) m2.: m=l =l 4 -< 4M2 EXr2 -< 4M

k

_

>

k4 T k

0

k --+ --+

":..

k --+

IS

IS

IS

last detail for the case in which X is a bounded martingale is to check that The Mt2 - At is a martingale. This follows from the next result (with p = 2). (3.6)that Lemma. Suppose that for each n, Z? is a martingale with respect to :Ft; and for each t, zr Zt in LP where p 2: 1. Then Zt is a martingale. Proo� The ma�tingale property implies that if s < t then E(Zf i:F3) = Z';. The nght-hand side converges to Z3 in LP . To check that the left-hand side con­ verges to E(Zt I:F3) we note that linearity properties of and Jensen's inequality for(1995)) conditional imply expectation (see e.g., (1.1a) and (1.1d) in Chapter 4 of Durrett _,.

(e.2) into (e.1) proves (e) which completes the proof0 and bound this ng Pluggi of (d) and hence of (c). A n be the partition with Let er. togeth pieces the put to s remain only It m n n (a)),) (by gale -n2r with 0al:5inequa:5 lity2 . (seeSincee.g.,Q{>(4.3)-inQ{>Chaptis era martin k2 points (1995) t 4 of Durret L maxim the using gives (f) From this and (c) it follows that (g) There is a subsequence so that Q�n(k) --+ a limit At uniformly on [0, r]. Proof Since Q � n converges in L2, it is a Cauchy sequence in L2 and we can pick an increasing sequence n(k) so that for m 2: n(k) k EIQ� m Q� n( ) 1 2 :5 2- k Using (f) and Chebyshev's inequality it follows that (supt $_r IQ�n(k+l) - Q�n(k) I 1/k2) :5 s the implied result Canteandlli lemma Borelthe able, summ is side and thelity right-h Since desire the times on the right only fails finitely many inequa s. follow ent us,we al. argu ing aeacdiago u and uences � � ofonsubseq uences subseq � taking By ontmu c � Smc t � [O, N] for all N. � , Qt 9 not .mcreasmg. conversegence mBecau unifor get can ) X t (Xt term ( last ) the of k well. as is At Ak 2 -n r �s �ncreas�ng. Q �2 n r is increasing, so m 2: n then if er, Howev Since n is arbitrary and t At is continuous it follows that t --+ At mcreasmg. p

49

EIE(Zfi:F3) - E(Zt i:F3) 1P = EIE(Zf - Zt i:F3)1P :5 EE(IZf - Zt IP I :F. ) = EIZf - Zt IP

_,.

0

Taking limits in LP it follows that E(Zt i:F3) = Z3 and the proof is complete. Proof of existence in (3. 1) when X is a local martingale Our first step in extending theof aresult to local martingales isoneto remembers prove a resultthe about the quadratic variation stopped martingale. Once definition T given at the beginning of Section 2.2, the next result should be obvious: oftheytquadratic variation does not increase after the process stops. (3.7) Lemma. Let X be a bounded martingale, and T be a stopping time. Then (XT ) = (XV. . Proof By the optional stopping theorem (XT )2 - (X) T is a local martingale so the result follows from uniqueness. Let Tn be a sequence of stopping times increasing to oo so that y n = XTn · 1 (Tn> O) is a bounded martingale By theAn previous result there is a unique continuous predictable increasing pro­ cess so that (ytn )2 - Af is a martingale. By (3.7) Af = A� + I for t :5 Tn so wepredict can unambiguously defineThe (X} 1 = Af fort :5 Tn . Clearly, (X)t is continuous, ble, and increasing. definition implies Xfn At · 1(Tn> O) - (X)TnAt is � so Xl - (X)t is a local a martmgale martingale. o

0

0

End of Existence Proof

Finally, we have the following extension of (c) that will be useful later.

50

Section 2.3 Variance and Covariance Processes

Chapter 2 Stochastic Integration

(3.8) Theorem. Let Xt be a continuous local martingale. For every t and sequence of subdivisions .6.n of [0, t] with mesh l .6.n I --+ 0 we have

(

> c)

(

> c)

(X5) - (X 5 ). 1 � o + P sup P sup IQ� (X) - {X). I ·�t IQ� · �t Since X 5 is a bounded martingale ( c) implies that the last term goes to 0 as 0 1 .6. 1 --+ 0. -

Proof Since

sup I Q � n (X)

- {X} , I -+ 0 in probability · �t 0. We can find a stopping time S so that Xl is a bounded Proof Let o, martingale and P(S � t ) � o. It is clear that QA(X) and QA (X5) coincide on [O, S]. From the definition and (3.7) it follows that {X} and (X5) are equal on [0 , S] . So we have

c>

51

this follows immediately from the definition of {X, Y) and Our next result is useful for computing

(3.8).

0

{X, Y}t .

�3.11) T�eorem . . Suppose X� and Yt are continuous local martingales. {X, Y)t Is the umque contmuous predictable process At that is locally of bounded vari­ ation, has Ao = 0, and makes Xt Yt - At a local martingale. Proof

From the definition, it is easy to see that

(3.9) Definition. If X and Y are two continuous local martingales, we let {X, Y}t = 41 ({X + Y)t - {X - Y)t) Remark. If X and Y are random variables with mean zero,

�

EXY = (E(X + Y) 2 - E(X - Y) 2 ) 1 = 4 (var (X + Y) - var (X - Y)) so it is natural, I think, to call (X, Y}t the covariance of X and Y. cov (X, Y ) =

Given the definitions of {X)t and (X, Y}t the reader should not be surprised that if for a given partition .6. = {0 = to < t 1 < t2 . . . } with limn tn = oo, we let k (t ) = sup { k : t k < t } and define

k (t ) Qf'(x , Y) = L(Xt k - Xt k_J (Ytk - Ytk_J k =I + (Xt - Xtk<• > )(Yt - Yt k< • > ) then we have

(3.10) Theorem. Let X and Y be continuous local martingales. For every t and sequence of partitions .6.n of [0, t] with mesh l .6.n l --+ 0 we have sup I Q� n (X, Y) - {X, Y}. l -+ 0 in probability · �t

is a local martingale. To prove uniqueness, observe that if At and A' are two processes with the desired properties, then At - A� = (XtYt -A� ) - (XtYt - At) is a continuous local martingale that is locally of bounded variation and hence 0 :::: 0 by (3.3). From (3.11) we get several useful properties of the covariance. In what follows, X, Y and Z are continuous local martingales. First since Xtyt = Yt Xt ,

(X, Y)t = (Y, X}t Exercise 3.2.

{X + Y, Z)t = (X, Z)t + (Y, Z}t.

Exercise 3.3.

{X - X0 , Z} t = {X, Z) t·

a, b are real numbers then (aX, bY}t = ab(X, Y}t. Taking a = b, X = Y , and noting (X, X} t = (X) t it follows that (aX} t = a 2 (X} t . Exercise 3.5. (XT , yT ) = (X, Y}T

Exercise 3.4. If

It is also true that Since we are lazy we will wait until Section 2.5 to prove this. We invite the reader to derive this directly from the definitions. The next two exercises pre­ pare for the third which will be important in what follows:

52

Chapter 2 Stochastic Integration

Exercise 3.6. If Xt is a bounded martingale then integrable martingale.

Xl - (X}t is a uniformly

Exercise 3. 7. If X is a bounded martingale and

S is a stopping time then yt = Xs +t -Xs is a martingale with respect to :Fs+t and (Y)t = (X}S+t - (X}s. Exercise 3. 8. If S :$ T are stopping times and constant on [S, T] .

(X}s = (X} T , then X is

Exercise 3. 9. Conversely, if S :$ T are stopping times and [S, T] then (X}s = (X} T .

X is constant on

2.4. Integration w.r.t. B ounded M artingales

In this section we will explain how to integrate predictable processes w.r.t. bounded continuous martingales. Once we establish the Kunita-Watanabe in­ equality in Section 2.5, we will be able to treat a general continuous local martingale in Section 2.6. As in the case of the Lebesgue integral (i ) we will begin with the simplest functions and then take limits to get the general case, and (ii) the sequence of steps used in defining the integral will also be useful in proving results later on. St ep 1: Basic Integrands

We say H(s, w) is a basic predictable process if H(s, w) = l (a ,b](s)C(w) where C E :Fa . Let ITo = the set of basic predictable processes. If H = l (a ,b]C and X is continuous, then it is clear that we should define

J H3dX3 = C(w)(Xb(w) - Xa (w)) Here we restrict our attention to integrating over [0, oo ) , since once we know how to do that then we can define the integral over [0, t] by

(H · X)t ::

1 H3 dX3 = J H31[o,tJ (s)dX3 t

To extend our integral we will need to keep proving versions of the next three results. Under suitable assumptions on H, K, X and Y (a)

(H · X)t is a continuous local martingale

(b) ((H + K) ·X)t = (H · X)t + (K · X)t , (H · (X + Y))t = (H · X)t + (H · Y)t

Section 2.4 Integration w.r.t. Bounded Martingales 53 t (c) (H · X, I( · Y}t = fo H3K3 d(X, Y}3 In this step we will only prove a version of ( a) . The other two results will make their first appearance in Step 2. Before plunging into the details recall that at the end of Section 2.2 we defined Ht to be bounded if there was an M so that with probability one Ht :$ M for all t � 0.

(4 . 1 . a) Theorem. If X is a continuous martingale and H H is bounded } , then (H · X)t is a continuous martingale.

E bii0 = {H E II0 :

Proof

o ::; t ::; a a ::; t ::; b b :$ t < oo From this it is clear that ( H ·X) t is continuous, (H ·X) t E :Ft and E I (H ·X)t l < oo. Since (H · X)t is constant for t rf. [a, b] we can check the martingale property by considering only a :$ s < t :$ b . (See Exercise 4.1 below. ) In this case, E((H · X)ti:F3) - (H · X)3 = E((H · X)t - (H · X)3 1 :F3) = E(C(Xt - X3 ) 1 :F3) = CE(Xt - X3 ) 1 :F3) = 0 where the last two equalities follow from c E :F3 and E(Xt I:F3)

= x3 .

0

Exercise 4. 1. If yt is constant for t rf. [a, b] and E(yt I :F3) = Y3 when a :::; s < t :::; b then yt is a martingale. Step 2: Simple Integrands

We say H(s, w) is a simple predictable process and write H E II 1 if H can be written as the sum of a finite number of basic predictable processes. It is easy to see that if H E II1 then H can be written as

m H(s, w) = L_ )(t,_ 1 ,t;J (s)C;(w) i =l

where to

< t 1 < . . . < t m and C; E :Ft , _ 1 • In this case we let C;(Xt, - Xt,_J J H3dX3 = t i =l

Section 2.4 Integration w.r.t. Bounded Martingales 55

54 Chapter 2 Stochastic Integration The representation in the definition of H is not unique, since one may subdivide the intervals into two or more pieces, but it is easy to see that the right-hand side does not depend on the representation of H chosen.

(4.2.b) Theorem. Suppose X and Y are continuous martingales. If H, J( E il1 then ((H + K) · X)t = (H · X)t + (I< · X)t (H · (X + Y))t = (H · X)t + (H · Y)t H1 = H and H2 = K. By sub dividing the intervals in the defin­ tions we can suppose Hi = I:�1 1(t; _1 ,t;] (s)C/ for j = 1, 2. In this case it is

Proof Let

v�riation it defines a 0"-finite signed measure, so we fix an w and then integrate With respect to that measure. In the case under consideration, the integrand is piecewise constant so the integral is trivial. Proof To prove these results it suffices to show that

is a martingale, for then the first result follows from (3.11), the second by taking expected values, and the third formula follows from the second by taking H = J{ and X = Y. To prove Zt is a martingale we begin by noting that

((H1 + H 2 ) X)t(K Y) t = (H1 · X)t(I< · Y)1 + (H 2 X)t (I< · Y)t t t [t (Hi + H;)Ksd(X, Y}s = [ Hi I<sd(X, Y}s + H; Ksd(X, Y}.

easy to see that each side of the first identity is equal to

•

m

m

·

L Ci {(Xt; - Xt; _ J + (Yt ; - rt ; _ l )} i= l

0

Since the sum of a finite number of continuous martingales is a continuous martingale, it follows from (4.l.a) and (4.2.b) that we have (4.2.a) Theorem. If X is a continuous martingale and H E H is bounded}, then (H · X)t is a continuous martingale.

bil1 = {H E il1 :

Turning to our third result

(4.2.c) Theorem. If X and Y are bounded continuous martingales and H, K E bil1 , then t

(H · X, I< · Y}t =

1 Hsi<s d(X, Y}s

Consequently, E((H · X)t(I< · Y)t) = E I� HsKsd(X , Y}s and t E(H · X); = E H;d(X},

1

(X, Y}s are s - (X, Y}s is locally of bounded

Remark. Here and in what follows, integrals with respect to

Lebesgue-Stieltjes integrals. That is, since

•

Jo

Jo

L (C{ + C[)(Xt; - Xt;_J i= l If H3 = I:�1 1(t; _1 ,t;] (s)Ci then each side of the second identity is

·

1

The last two observations imply that if the result holds for (H l , K) and (H2 , I<), it holds for (H1 + H 2, K). Similarly, if the result holds for (H, J( 1 ) and (H, J < 2), it holds for (H, J< l + J( 2 ). In view of the results in the last paragraph, we can now prove the result by establishing it in the case H = l (a ,b]C, I< = lc c, d]D, and we can furthermore assume that (i) b :::; c or (ii) a = c, b = d.

I� HsKsd(X, Y}3 = 0, so we need to show that (H · X)t(K · Y)t is a martingale. To prove this we observe that if J = C(Xb - Xa )Dl (c , d] then (H · X)t(I< · Y)t = (J · Y)t is a martingale by {4. l.a)

Case i. In this case

Case ii. In this case

{�

s b < s -< -< b. To do this we so it suffices to check the martingale property for a . Zs =

M�

t

Zt - Zs = CD{XtYt - XsYs - Xa (Yt - Ys) - Ya (Xt - Xs ) - ((X, Y}t - (X, Y}s)} Taking expected value and noting Xa E :F3, E(yt - Y, I:Fs) = 0 we have

E(Zt - Z3 1:F.) = CDE(XtYt - (X, Y}t - {X,Y3 - (X, Y}s} I:Fs ) = 0 This completes the proof of Case ii and finishes the proof of (4.2.c) .

0

56

Chapter 2 Stochastic Integration

Section 2.4 Integration w.r.t. Bounded Martingales

1 ::; i ::; k we have Xi E M 2 and H E Il 2 (Xi ), then there E bil1 with II Hn - H llx ; -+ 0 for 1 ::; i ::; k. Proof Since Xi E M 2 , E(X i }t < oo and it follows that bil1 c Il 2 (Xi). Let 1it be the collection of predictable G that vanish on (t , oo) for which the conclusion holds. Clearly, if r < s ::; t and A E :Fr then G = 1 (r,•] 1A E 1it. Suppose now that 0 ::; Gn E 1it and Gn i G with G bounded. The dominated (4.5) Lemma. If for

Step 3: S quare Integrable Integrands, Bounded Martingales

Let Il 2 (X) be the set of all predictable processes

( . J

II H I Ix = E H';d(X} . Exercise 4.2.

57

is a sequence H n

H that have

)1/2 < oo

II H II x is a norm.

convergence theorem implies

Remark. If we define a measure on the predictable a--field by

(s,t])

= E((X}t - (X}. ; A) Jl (A x when A E :F. then Ih(X) = L 2 (Il, Jl) . Jl is called the Doleans measure after C. Doleans-Dade. For a proof that this is a measure see Chung and Williams (1990) , pages 52-53.

t

M2 be the set of all martingales adapted to {:Ft, 2:: 0} that have 1 12 IIX II 2 = (sup t EX'f) < oo In the proof of (4.6) we will see that M 2 is isomorphic to L2 (:F00 ).

Let

Exercise 4.3.

(t,

X E M2 if and only if EXfi < oo and E(X} oo < oo.

(4.6) Theorem. M 2 is complete.

The following is the key to our next extension (4.4) Isometry Property. If X is a bounded martingale and I I H · X ll 2 = I I H II x -

H E bil1 , then

Remark. We skipped over the number (4.3) since (4.3.a) and (4.3.b) will be

proved later in this step. Proof

J

Proof Standard martingale convergence theorems imply that if X E M2, then as -+ oo, Xt converges almost surely and in L2 to a limit Xoo with EX� = supt EX[, and the martingale can be recovered from Xoo by Xt = E(Xoo i:Ft) · Let :Foo = u(:Ft , i 2:: 0). Since Xoo = limXt E :F00 , the observation above shows that X -+ Xco maps M2 one-to-one into L2(:F00 ). On the other hand, if Y E L2 (:F00 ), then yt = E(YI:Ft) is a martingale with yt -+ Y as -+ oo, and Jensen's inequality shows that

t

t

Recalling the definitions and using the last formula in (4.2.c) gives

it

II H II} = E H;d(X}. = s�p E H;d(X}. = sup E(H · X)i = I I H · X I I � t

as n -+ oo, so we can pick n so that the last difference is < �:2 for 1 < i < k. Since Gn E 1it ' we can find a sequence Hn, m E bil l so that II Hn , m -anlfx ; -=-. 0 for 1 ::; i ::; k and then pick m so that II H n , m - Gn llx ; < c for 1 ::; i ::; k. The triangle inequality implies that II Hn , m - G llx ; < 2c for 1 ::; i ::; k . Since c is arbitrary it follows that G E 1it . Using the monotone class theorem now ( (2.3 ) in Chapter 1) it follows that 1it co�tains all bounded predictable processes that vanish on oo) . Now if J{ E Il2 (X') for 1 ::; i ::; k and we define J{n = K1 1 K I Sn 1[o,n] then the dominated convergence theorem implies that II Kn - Kllx ; -+ 0. Since J(n is bounded and vanishes on (n, oo), another use of the triangle inequality implies J( is a limit of H n E bil1 . D

D

The first step in extending the integral to Il 2 (X) is to prove that bil1 is dense in Il 2 (X). Later we need a slight strengthening of this result so we go ahead and prove the stronger form.

so Yi E M2. Combining this with the previous observation shows that X -+ Xco D is an isometry from M2 onto L2(:Foo ) and proves (4.6). With (4.5) and (4.6) established, we can give the Definition of the integral for Il2 (X). To define H · X when X is a bounded

continuous martingale and H

E Il2 (X), let Hn E bil1 so that I I Hn - H llx -+ 0.

58

Chapter 2 Stochastic Integration Since I IHn - H m llx -+ 0 as m, n -+ oo , (H n · X) is a Cauchy sequence in M 2 and by ( 4.6 ) must converge to a limit in M2, which we define to be the integral (H · X). To see that the limit is independent of the sequence of approximations chosen suppose II Hn - Hllx -+ 0 and I IHn - Hllx -+ 0 with Hn · X -+ Y and j[n . X -+ Y. Form a third approximating sequence by setting fin = Hn if n is odd, fin = j[ n if n is even. I I fin - HUx -+ 0 so we have fi n · X --=: Z. Looking at the even and odd subsequences of Hn it follows that Y = Z = Y so the limit

is unique. In words, the last argument shows that if the limit exists for ANY sequence of approximations, it must be unique. Note that as a corollary of the construction we get

(4.3.a) Theorem. If X is a bounded continuous martingale and H E Ih(X) then H · X E M2 and is continuous. Proof The fact that H ·X E M2 is automatic from the definition. If Hn E bil 1 have I!Hn - Hllx -+ 0 then (4.2.a) implies (H n · X)1 is continuous. Since ll (Hn · X) - (H · X) l! 2 -+ 0, using Chebyshev and the L2 maximal inequality we get that

(

)

P s�p l(Hn · X)t - (H · X)t l > c :::; c-2 E s�p l(Hn · X)t - (H · X)d 2 :::; c-2 4 1l(Hn · X) - (H · X) ll 2 -+ 0 Since this holds for any <: > 0 we say "(Hn ·X)t converges uniformly to (H · X)t •

Section 2.5 The Kunita-Watanabe Inequality

59

The proofs of the remaining two equalities

(H · (X + Y))t = (H · X)t + (H · Y)1 t (H · X, K · Y}t = H3K3 d(X, Y}3

1

will be given in the next section. To develop the results there we will need to extend the isometry property from bil 1 to Il (X). To do this it is useful to recall some of your undergraduate analysis. 2 Exercise 4.4. If 11

· 11 is a norm and ll xn - x ll -+ 0 then ll x n ll -+ ll x ll ·

Exercise 4.5. If X is a bounded martingale and

II H IIx ·

H E Il2 (X) then II H · X ll 2 = ·

2. 5 . The Kunita-Watanabe Inequality

In this section we will prove an inequality due to Kunita and Watanabe and apply this result to extend the formula given in Section 2.4 for the covariance of two stochastic integrals. (5.1) Theorem. If X and Y are local martingales and measurable processes , then almost surely

H and

!{

are two

in probability." By passing to a subsequence we can have sup l (H n · X)t -

(H · X)t l -+ 0 a.s. t and it follows that t -+ (H · X)t is continuous.

0

(4.3.b ) Theorem. If X is a bounded continuous martingale, and H, I< E Il 2 (X) then H + I< E Ih(X) and

((H + K) · X)t = (H · X)t + (K · X)t Proof The triangle inequality for the norm ll · llx implies that H +I<

E Il2 (X).

Let Hn , I< n E bil 1 with II Hn - Hll x -+ 0 and II I< n - I< llx -+ 0. The triangle inequality implies ll (Hn + I< n ) - (H + I<) llx -+ 0. (4.2.b ) implies

((Hn + I
-+ oo and use the fact that (Gn · X)t -+ (G · X)t in M2 when D G = H, I<, H + K.

Now let n

where dl(X, Y}l$ stands for dV. where on [0, s].

V. is the total variation of r -+ (X, Y}r

(i) If X = Y and d(X}3 = s then d{Y}3 = di {X, Y} l 3 = s and (5.1) reduces to the Cauchy-Schwarz inequality. (ii) Notice that H and J( are not assumed to be predictable. We assume only that H(s, w) and K(s, w) are measurable with resp ect to n X :F where n is the Borel subsets of R. The reason we can attain this level of generality is that the notion of martingale does not enter into the proof after the first step. Remarks.

Proof

If we let

Step 1: Observe that if s :::; t, (X +>.Y, X +>.Y)t � (X +>.Y, X +>.Y)3• {M, N} ! = (M, N}t - {M, N} 3 , then

0 :::; (X + >.Y, X + >.Y)t - {X + >.Y, X + >.Y}3 = {X, X} ! - 2>.(X, Y} ! + >. 2 (Y, Y} !

60

Section 2.5 The Kunita-Watanabe Inequality 61

Chapter 2 Stoclwstic Integration

s,

s t ax2 + bx + b2 ((X, Y}! )2 � (X, X} ! (Y, Y} ! Step 2: Let 0 = to < t 1 < · · · < tn be an increasing sequence of times, let k;, 1 � i � n, be random variables, and define simple measurable processes n H(s, w) = l::::::1 : h; (w)1(t ,_1 ,t ;) (s) in K(s, w) = l:::=l : k;(w )1(t ;_ 1 ,t ;) (s) i From the definition of the integral, the result of Step 1, and the Cauchy-Schwarz inequality, it follows that

for all t, and .\. If we fix and and throw away a countable number of sets of measure 0 then with probability one the last inequality will hold for all c that is nonnegative at all the rationals rational .\. Now a quadratic 4ac � 0), so we have and not identically 0 has at most one real root (i.e., that

di (X, Y}. l. J.

which is (5.1) with the absolute values outside the integral.

(X}t (Y}tH K

> M} . By or Step 3: Let M be a large number and let T = inf{t : the monotone convergence theorem, it suffices to prove (5.1) when = = 0 for s ;::: T and � M for � T. Having restricted our attention are finite measures; so using the bounded convergence to [0, T], and theorem, we see that (5.2) holds for bounded measurable processes. To improve (and complete the proof), let be a measurable process taking (5.2) to values in { 1} such that

I H.I, I K.I (X} (Y} (5.1-)1,

s

J.

1t ld(X, Y}.l 1t J,d(X, Y}, =

H, IH.I

D

With (5.1) established, we are now ready to take care of unfinished business from Section 4. (5.3) Theorem. If

h;,

Ilfo oo H.K.d(X, Y}.l � t •=1n l h; k; I I(X, Y} �;_. l 1/ � 2::.::: l h; I ( (X, X} L. (Y, Y} ;;_.) 2 l kd i:::1 112 112 � (th[ (X, X} �:-•) (t k[ (Y, Y} �;_.) •=1 •=1 proving that for simple measurable processes: (5.2) 1 100 H. K.d(X, Y}.j � (100 H;d(X}. Y'2 (100 J(;d(Y}. Y'2

K. I.Y). J.Id(X,

T? see that this exists, note that is the Radon-Nikodym derivative of and K, = = Now apply (5.2) to w1th respect to

H E IT2 (X) n IT2(Y) then H E IT2 (X + Y) and (H · (X + Y))t = (H · X)t + (H · Y)t

Proof First note that Exercise 3.2 implies

(X + Y}t = (X}t + (Y}t + 2 (X, Y}t and the Kunita-Watanabe inequality, (5.1), implies

I (X, Y}lt � ((X}t(Y}t)1 ' 2 � ((X}t + (Y}t)/2 since 2ab � a2 + b2, i.e., 0 � (a - b)2. So we have (X + Y}t � 2( (X}t + (Y} t ) remind the reader of (x + y)2 � 2 (x2 + y2) ) and it follows that To prove the formula now we note that by (4.5 ) we can find En inll2bllshould HH(which (X so+ Y). 1 that I ! Hn - Hl l z --> 0 for Z = X, Y, X + Y . (4.2.b) implies that (Hn · (X + Y))t = (Hn · X)t + (Hn · Y)t Now let n --> and use ( Hn · Z)t --> (H · Z)t for Z = X, Y, X + Y . If X , Y are bounded continuous martingales, H E IT 2 (X), K(5.4)E ITTheorem. then (Y) 2 t (H · X , K · Y}t = 1 H,K,d(X, Y} .

D

oo

Proof

By (3.11 ) , it suffices to show that

Hn and f{n be sequences of elements of bll1 that converge respectively, and let Zf be the quantity IT2H(X)n andandJ{ITn2 (Y), replace H and J{ in (t) . By ( 4 .2.c) , Zf is a

is a martingale. Let and I< in to that results when

H

62

Section 2.6 Integration w.r.t. Local Martingales

Chapter 2 Stochastic Integration

martingale. By (3.6), we can complete the proof by showing zr; -1- z. in L1. The triangle inequality and (4.3.b) imply that E sup I (Hn

t

· X)t(I
t

To bound the first term, we use (i) the sup of the product is smaller than the product of the sup's, (ii) the Cauchy-Schwarz inequality, (iii) the L 2 maximal inequality and (iv) the isometry property in Exercise 4.5: :::; E{sup I ((Hn - H) · X)t I sup I ( I
I I Hn - Hl lx

-1-

0 and I II
so we have shown (Hn · X).(I< n · Y) .

-1-

I II< I Iy <

Taking expected values and using the Cauchy-Schwarz inequality

as n -1- oo, since I I � - Hllx -1- 0 and III
in L1 . We have now shown Z'; -1- z. in L1 and the desired result follows from

(3.6).

-Jo

oo .

0 A similar

X). (I< · Y) . in L1 . -1- (H To estimate the second term in Zf - Z1 , we again begin by putting the absolute values inside, replacing the measure by its variation, and then using the triangle inequality ·

11t H: I<:d(X, Y}. - 1t H.I<.d(X, Y}. , :::; 1t I H': I<'; - H.I<. Idi(X, Y} I. :::; 1t IH: - H. I II<: I di(X, Y}l. + 1t I Hs i i i<: - I<, I di(X, Y} ls

Using the Kunita-Watanabe inequality, (5.1), the first term is

63

0

130), if M E E ll2 (M), H ·M is DEFINED to be the unique L E M 2 with L0 = 0

Remark. In some treatments (see e.g., Revuz and Yor (1991) p.

M 2 and H so that

(L , N} = H · (M, N}

Exercise 5. 1. Prove the uniqueness in the last claim.

(5.4) allows us to do easily some things that we would have had to work to prove in Section 2.3. Let X and Y be continuous local martingales, and suppose X0 = Y0 = 0. The last assumption entails no loss of generality. See Exercise

3.3.

Exercise 5.2.

(X, yT } = (X, Y} T .

Exercise 5.3.

Xf (Yt - Yl) is a local martingale.

2.6. Int egration w.r.t. Local M artingales

In this section we will extend our integral so that the integrators can be con­ tinuous local martingales and so that the integrands are in

{ 11 H'1d(X}. < oo a.s. for all t ;?: 0}

ll3 (X) = H :

64

Section 2. 6 Integration w.r.t. Local Martingales

Chapter 2 Stochastic Integration

We will argue in Section 3.4 that this is the largest possible class of integrands. To extend the integral from bounded martingales to local martingales we begin by proving an obvious fact.

X is a bounded continuous martingale, H, I< E II 2 (X) and H, = I<, for s :5 T where T is a stopping time. Then ( H · X), = (I< · X), for s :::; T. Proof Write H, = H} H'} and I<. = I<} + I<'} where H} = I<} = H31(•5T) = K.1(•5T) Clearly, ( H 1 · X) t = (K 1 · Y)t for all t. Since H'} = K'} = 0 for s :5 T, (5.4) implies (H2 • X } . = (K 2 · X }. = 0 for s :5 T and it follows from Exercise 3.8 that (H2 · X )t = (K 2 · X) t = 0 for t < T. Combining this with the first result and using (4.3.b) gives the desired (6.1)

Theorem. Suppose

+

conclusion.

D

(6 . 1) to show that if X is a bounded H,HJ(·X).-(H E II2 (X)·X)s and H. = K. for S :5 s :5 T where S, T = (K ·X). -(K ·X)s for S :::; s :::; T. (

Exercise 6. 1. Extend the proof of

continuous martingale, are stopping times then

=

In words if we set the integrand 0 after time T or stop the martingale at time T or do both, then this just stops the integral at T. Our next task is to generalize our abc's to integrands in II3 .

If X is a continuous (H · X)t is a continuous local martingale.local martingale and H E II3 (X), then Proof By stopping at Tn = inf{t : J; H'}d (X}. > n or I X11 > n}, it suffices to show that if X is a bounded martingale and H E ll 2 (X ) , then (H . X )t is a continuous martingale but this follows from (4.3.a) . (6 .4) Theorem. Let X and Y be continuous local martingales. If H, J( E II3 (X) then H + I< E II3(X) and ((H + K) · X)t = (H · X )t + (I< · X)t If H E II3 (X) n II3 (Y) then H E II3 (X + Y) and (H · (X + Y))t = (H · X)t + (H · Y)t

(6.3) Theorem.

0

Proof To prove the first formula we note that the triangle inequality for the J( II3 (X). Stopping at norm ll · llx implies

H+ E t t Tn = inf{ t : I Xt l, 1 H] d (X }., or 1 K] d (Y } . > n }

To extend the integral now, let Sn be a sequence of stopping times with

(•5Sn ) • and observe that if m < n, (6.1) implies m · X)tn = H�(Hn=· H31 X for t :5 Sm , so we can define· H · X by setting (H )t = (H · Xthe)t fordefinition X)tcomplete t :5 Sn . (H · To" we have to show that if Rn and Sn are two sequences of stopping times :5 Tn with Rn j oo and Sn j oo then we end up with the same ( H · X)t . Let H'[ be the stopped version of the process defined at the beginning of Section 2.2, let Qn = Rn 1\ Sn and note that ,6 . 1 ) implies

and Sn j oo. Let that

(H ·X)t

reduces the result to (4.3.b) For the second formula we note that the argument in (5.3) shows that II3(X Y) and by stopping we can reduce the result to (5.3) . D

HE

+

After seeing the last two proofs the reader can undoubtedly improve (5.4) to

If X, Y are continuous local martingales, H E II3 (X) and K(6.5)E IITheorem. 3 (Y) then t H.K,d X, Y}. = 1 H ·X, I< · ( Y}t ( Using Exercise 3.2, we can generalize ( 6.5) to sums of stochastic integrals.

Since Qn j oo , it follows that is independent of the sequence of stopping times chosen. The uniqueness result and ( 6 . ) imply

1

X is a continuous local martingale and H E II3 (X) then HT ·X = (H · Xf = H · XT = HT · XT

( 6 .2) Theorem. If

65

X = l:� 1 Hi · Xi i= l:'j=1 J(i i . yi where the Xi H E IT3(X ) and J(i E II3(Yi ), t (X, Y}t = � 1 H;I
(6.6) Theorem. Let and Y and yi are continuous local martingales. If then •,J

66

Section 2.6 Integration w.r.t. Local Martingales

Chapter 2 Stochastic Integration

We leave it to the reader to make the final extension of the isometry property: Exercise 6.2. If X is a continuous local martingale and H

H · X E M 2 and II H · X ll 2 = I I H I I x ·

E IT2 (X) then ·

We will conclude this section by computing some stochastic integrals. The first formula should be obvious, but for completeness must be derived from the definitions. Exercise 6.3. Let X be a continuous local martingale. Let S � T < oo be stopping times, let C (w ) be bounded with C (w) E :Fs, and define H8 = C for S < s � T and 0 otherwise. Then H E Ih(X) and

For the rest of this section, .6.n = {0 = t� < t'J. < . . . < tk( ) = t } is n a sequence of partitions of [0, t] with mesh l .6. nl = sup i l tf - tf_ d --+ 0. Our theme here is that if the integrand Ht is continuous then the integral is a limit of Riemann sums, provided we evaluate the integrand at the left endpoint of the

Proof Exercise 6.2 (or

67

4.4) implies

n II (H - H) · X II � = I I Hn - H I IJ. � ME sup ! H� - H8 1 2 --> 0 8 as n --+ oo by the bounded convergence theorem.

0

The rest of the section is devoted to investigating what happens when we evaluate at the right end point or the center of each interval. For simplicity we will only consider the special case in which H8 = 2X8 • Exercise 6.4. If X is a continuous local martingale then

Exercise 6.5. Show that if X is a continuous local martingale and we evaluate at the right end point then

intervals. (6.7) Theorem. If X is a continuous local martingale and t --+ Ht is continuous

then as n --+ oo

Comparing the last two exercises you might jump to the conclusion that if we evaluate at the midpoint we get an integral that performs like the one in calculus. However, we will not ask you to prove this in general. Exercise 6.6. Show that if Bt is a one dimensional Brownian motion starting

E ll3 (X) so the integral exists. By stopping at T = inf{s : s, {X}8 , or I H8 1 > M} and making H constant after time T to

at 0 then

Proof First note that H

n 2 -1

preserve continuity, we can suppose (X)t � M and I H8 1 � M for all s. Let H'; = Htr for s E (ti , tf+l ], H'; = Ht for s > t . Since s --+ H8 is continuous on [0, t] we have for each w that

k=O

converges in probability to

The desired result now follows from the following lemma which will be useful later.

This approach to integration, called the Stratonovich integral, is more con­ venient than Ito's approach in certain circumstances (e.g., diffusion processes on manifolds) but we will not discuss it further here. Changing topics we have one more consequence of (6.7) that we will need in Chapter 5.

(6.8) Lemma. Let Xt be a continuous local martingale with {X}t � M for all

t. If

Hn is a sequence of predictable processes with !Hn I � M for all s, w and with sup8 I H'; - H8 1 --> 0 in probability then (Hn · X) --+ (H · X) in M 2 •

L 2B((k + 1/2)2-nt){B((k + 1)2-nt) - B(k2-nt)} J� 2B8 dB8 + t = B'f .

Exercise 6. 7. Suppose h : [0, oo) --+ R is continuous. Show that a normal distribution with mean 0 and variance J� h; ds.

J� h. dB8 has

68

Section 2. 7 Change of Variables, Ito's Formula

Chapter 2 Stochastic Integration

2.7. Change of Variables, Ito's Formula

Comparing (7.4) with (7.1), it becomes clear that we want to show

This section is devoted to a proof of our first version of Ito's fromula, (7.1). In Section 2. 10, we will prove a bigger and better version, (10.2), with a slicker. proof, but the mundane proof given here has the advantage of explaining where the term with comes from.

I"

X is a continuous local martingale and I has two continuous derivatives. Then with probability 1, for all t � 0

(7.1) Theorem. Suppose

Remark. If

and

I

69

At

is a continuous function that is locally of bounded variation, has a continuous derivative then (see Exercise 7.2 below)

L I'(Xt r )(Xq+1 -Xq ) _,. it t'(X.)dx. i t (7.6) � � gi (w)(Xtf+1 - Xt r ) 2 _,. � i f" (X.)d(X}. in probability as n _,. oo. (7.5) follows from (6.8). To prove (7.6), we let G� = gf(w) = f" ( c(Xt ;, Xti+J when s E (ti,ti+l], G� = f"(Xt ) for s � t and let A� = t L +i l�· (Xti+l - Xt ; )2 (7.5)

0

'

so that

(7.2)

As the reader will see in the proof of (7.1), the second term comes from the fact that local martingale paths have quadratic variation while the 1/2 in fr�mt of it comes from expanding in a Taylor series.

I

(X}t ,

IXt I (X}t

Proof By stopping at TM = inf{t : � M} , it suffices to prove or the result when :::; M. From calculus we know that for any and and b, there is a c , b) in between and b such that

(a IXt l (X}t a 1 (7.3) l( b) - f ( a ) = ( b - a) f (a) + 2 ( b - a) 2 f" (c(a , b)) Let t be a fixed positive number. Consider a .sequence .6- n of partitions 0 = t0 < t� . . . < tt = t with mesh l .6.n l _,. 0. From (7.3) it follows that (7.4)

I(Xt) - I(Xo ) = L /(Xt;+J - I(Xti) = L f (Xt i )(Xt i+ t -Xt i ) i + � L 9i (w )(Xt i+ 1 - Xq ) 2 i

a

and what we want to show is (7.7)

To do this we begin by observing that the uniform continuity of f" implies that as n _,. oo we have uniformly in while (3.8) implies that converges in probability to • . Now by taking subsequences we can suppose that with probability 1, converges weakly to In other words, if we fix and regard as distribution functions, then and the associated measures converge weakly. Having done this, we can fix and deduce (7.7) from the following simple result:

G� -+ f"(X.At) s, (X} A A� s _,. A�AAtt s _,. (X}. At (X}. t ·

w

A�

w

f.ln [0, t] f.loo , 9n S l 9 l n 9n (s n) -+ g(s), n [0, t] _,. s

(7 .8) Lemma. If (i ) measures a finite on converge weakly to measure, and (ii) is a sequence of functions with :::; M that have the property that whenever E we have then as n -+ oo

f.l� (A) f.ln (A) /f.ln ( [O, t] ) , _,. Xn -+ Xoo

f.ln g

Proof By letting = we can assume that all the are probability measures. A standard construction (see (2. 1) in Chapter 2 of Durrett (1995)) shows that there is a sequence of random variables with so that a.s. as n oo The convergence of to distribution

f.ln

X9n n

70

Chapter 2 Stochastic Integration

implies Yn {Xn ) theorem.

�

Section 2.8 Integration w.r. t. Semimartingales

g(X00 ), so the result follows from the bounded convergence

0

(7.8) is the last piece in the proof of {7.1). Tracing back through the proof, we see that {7.8) implies {7.7), which in turn completes the proof of (7.6). So adding (7.5) and using {7.4) gives that for each t, f(Xt) - f{Xo) =

1

t

f'(Xs)dXs +

� 1 f"(Xs)d(X)s t

a.s.

Since each side of the formula is a continuous function of t, it follows that with probapility 1 the equality holds for all t ;=:: 0, the statement made in {7.1). 0

{7 .9) Remark. In Chapter 3 we will need to use Ito's formula for complex valued f. The extension is trivial. Write f = u + iv, apply Ito's formula to u and v, multiply the v formula by i and add the two. Exercise 7.1. Let At be a continuous process with total variation I Ait $ M for all t. If J(n is a sequence of measurable processes with I I< n { s, w) I $ M for all s, w and sup s II<� - I<s l ---+ 0 then supt I {Kn - K) · A)tl ---+ 0.

f(b) - f(a) = f'(c(a, b))(b - a) for some c a and b, and Exercise 7.1 to prove (7.2).

Exercise 7.2. Use the fact that

between

2.8. Int egration w.r.t. S e mimartingales

71

Exercise 8.1. Show that if Xt = Mt + At and Xf = Mf + A� are continu­ ous semimartingales then Xt + Xf = (Mt + Mf) + (At + A�) is a continuous semimartingale.

In this section we will extend the class of integrators for our stochastic in­ tegral from continuous local martingales to continuous semimartingales. There are three reasons for doing this.

{i) If X is a continuous local martingale and f is C2 then Ito's formula shows us that f(Xt) is always a semimartingale but it is not a local martingale unless f"(x) = 0 for all x. In the next section we will prove a generalization of lto's2 formula which implies that if X is a continuous semimartingale and f is C then f(Xt) is again a semimartingale. (ii) It can be argued that any "reasonable integrator" is a semimartingale. To explain this, we begin by defining an easy integrand to be a process of the form n H = L H; 1 (T;,Ti+l l i :::O · · · $ Tn +l are stopping times, and the H; E :FT, have

where 0 = To $ T1 $ I H; I < oo a.s. Let bii e ,t be the collection of bounded easy predictable processes that vanish on (t, oo ) equipped with the uniform norm

II H IIu = sup s ,w IHs(w) l

X is said to be a continuous semimartingale if Xt can be written as Mt +At , where Aft is a continuous local martingale and At is a continuous adapted pro­

For easy integrands we define the integral as

{8.1) Theorem. Let Xt be a continuous semimartingale. If the { continuous) Mt and At are chosen so that Ao = 0, then the decomposition Xt = Mt + At is unique.

Finally, let L0 be the collection of all random variables topologized by conver­ gence in probability, which comes from the metric II X IIo = E( I XI/(1 + lX I )). See Exercise 6.4 in Chapter 1 of Durrett (1995). A result proved independently by Bichteler and Dellacherie states

cess that is locally of bounded variation. A nice feature of continuous semi­ martingales that is unheard of for their more general counterparts is processes

n (H . X) = L H;(XTi+l - XT. ) i::: l

Proof

If Mf + A� is another decomposition, then At - A� = Mf - Mt is a continuous local martingale and locally of bounded variation, so by (3.3) At -A� 0 is constant and hence = 0.

(8.2) Theorem. If H ---+ (H · X) is continuous from bll e ,t ---+ L0 for all t then X is a semimartingale.

Xt = Mt + At be a continuous semimartingale" to specify that the decomposition of Xt consists of the local martingale Mt and the process At that is locally of bounded variation.

Since any extension of the integral from bii e ,t will involve taking limits, we need our integral to be a continuous function on the simple integrands (in some sense ) . We have placed a very strong topology on the domain and a weak

Remark. In what follows we will sometimes write "Let

72

Section 2.8 Integration w.r.t. Semimartingales

Chapter 2 Stochastic Integration

topology on the range, so this is a fairly weak requirement. Protter {1990) takes (8.2) as the definition of semimartingale. In his approach one gets some very deep results without much effort but then one must sweat to prove that a semimartingale {defined as a good integrator) is a semimartingale (as defined above). (iii) Last but not least, it is easy to extend the integral from local martingales to semimartingales if we replace II3 (X) by a slightly smaller class of integrands that does not depend on X. Getting back to business, let Xt = Mt +At be a continuous semimartingale. We say H E lbiT = the set of locally bounded predictable processes, if there is a sequence of stopping times Tn i oo so that I H( s, w) I ::; n for s ::; Tn. If H E lbll we can define

(H A)t(w) = ·

1 H, (w)dA, (w) t

as a Lebesgue-Stieltjes integral (which exists for a.e. respect to the local martingale Mt we note that

n

and T -+ oo. So

w). To integrate with

lbll C II3 (M), we can define (H M)t , and let ·

(H X)t = (H M)t + (H · A)t ·

·

since by the uniqueness of the decomposition this is an unambiguous definition. From the definition above, it follows immediately that we have (8.3) Theorem. If X is a continuous semimartingale and H E lbll, then (H ·X)t is a continuous semimartingale. The second of our abc's is also easy. We name it in honor of the similar rela­ tionships between addition and multiplication. (8.4) Distributive Laws. Suppose X and Y are continuous semimartingales, and H, I< E £bll . Then

73

Proo� This follows easily from the definition of the integral with respect to .

a sem1martmgale, the result for local martingales given in (6.4), and the fact that the results are true when X and Y are locally of bounded variation. In . provmg the second result we also need Exercise 8.1. o The rest of this section is devoted to defining the covariance for semimartin­ gales and proving the third of our abc's. (8.5) Definition. If X = M +A and X' = M' +A' are continuous semimartin­ gales we define the covariance (X, X')t = (M, M')t . To explain this we recall the approximate quadratic variation in Section 2.3, and note

QL!.t (X) defined

(8.6) Theorem. Suppose X = M + A and X' = M' + A' are continuous semimartingales. If .6. n is a sequence of partitions of [0, t] with mesh l .6.n l _,. 0 then Q� n (X, X') -+ (M, M')t in probability Proof Since

it suffices to show that if yt is a continuous process and Vi is continuous and locally of bounded variation then Q� n (Y, V) -+ 0 almost surely. To do this we note that Q� n (Y, V) ::; IV I t s�p I Yt;+1 - Yt; I • As n -+ oo the second term -+ 0 almost surely, while the first one has IVI t < oo and the desired result follows. o (8.7) Theor�m. �upp ?se Xi an1 yi ':re c?ntinuous semimartingales Hi , J{i E lbii, X = Li= l H' X' , Y = Lj= l [(J YJ , then t (X, Y)t = � H1K{d(X i , yi ), ·

·

• ,J

1

Ni be the local martingale parts of Xi and yi , let M = L� 1 H' M' , and N = LJ= l J(i · Ni . It follows from our definition that (X,. Y)t = (M, N)t, so using (6.6) and then (Mi , Ni )t = (Xi , Yi )t gives the Proof Let ¥i and _ ·

((H + K) X)t = (H · X)t + (Ji X)t (H · (X + Y))t = (H X)t + (H Y)t ·

·

·

·

desued result.

o

74

Chapter 2 Stochastic Integration

Section 2.9 Associative Law

2.9. Associative Law

Let Xt be a continuous semimartingale. In some computations, it is useful to write the integral relationship

To extend to more general integrands, we will take limits. First, however, we need a technical result. (9.2) Lemma. Suppose J-l is a signed measure with finite total variation ' k are bounded, and let v ([O , t]) = Jt k.J-l(ds). Then

0

as the formal equation

H

Proof of ( 9.2) The Radon-Nikodym derivative identity is just the well known fact that

(I< · X)t and observing dyt = I
d(H · Y)t = Htd Yt . Letting Yt =

d(H · (I< · X))t = Ht d(K · X)1 = Hti
D

The above proof is �ot rigorous, but the computation is useful because it tells us what the answer should be. Once we know the answer, it is routine to verify by checking that it holds for basic predictable processes and then following the extension process we used for defining the integral to conclude that it holds in general.

t

0

Proof Since the formula is linear in H and in I< separately, we can without loss of generality suppose H = l (a ,b]C and J( = l ( c, d]D and further that either (i) b :::; c or (ii) a = c, b = d. In case (i), both sides of the equation are = and hence equal. In case (ii),

o ::; t ::; a (I< · X)t = D(Xt - Xa ) a :::; t :::; b D(Xb - Xa ) b :::; t < oo o ::; t ::; a D(Xt - Xa ) a ::; t :s; b CD(Xb - Xa ) b :s; t < oo and it follows that (H · (I< · X))t = ((HI < ) · X)t .

See for example Exercise 8.7 in the Appendix of Durrett (1995).

0

To simplify the extension to II (X), we will take the limit for J( and then 2 for H. (9.3) Lemma. Let X be a continuous local martingale. If H E bii 1 and I< E II2 {X), then (*) holds. Proof Let I< n E bii 1 such that J(n --+ J( in II 2 (X) . Since H is bounded, H I
X be any process. If H, J( E II 1 , then (*) holds.

{0 {�

dvfdJ-l = k. so the desired

10 h. v(ds) = 10t h. �ddJ-1. J-l(ds)

· (K · X) = (HI<) · X

Proof using stochastic differentials

so

h and

dyt = KtdXt

where the dyt and dXt are fictitious objects known as "stochastic differentials." A good example is the derivation of the following formula, which (for obvious reasons) we call the associative law:

(9.1) Lemma. Let

75

=E

1oo 1oo H}(K. - K;')2 d(X).

:::; C I I K - Kn ll.k so as n --+ oo, .H · (I{ n

(H · (I< · X))t =

D

( 9.4) Lemma. Let

· X) --+ H · (I{ · X) in M 2 and the result follows.

0

X be a continuous local martingale. If H E II2 (K . X) and

I< E II2 (X), then (*) holds.

76

Chapter 2 Stochastic Integration Proof Let H n E bii 1 such that Hn -+ H in II 2 (J< · X). (6.4) and Exercise 6.2 imply

I IHn · (J< · X) - H · (I< · X) lb = II(Hn - H) · (I< · X)ll 2 = I I Hn - HII K·X so Hn · (I< . X) -+ H · (I< · X) in M 2 • To deal with the right side of (*), we observe that using the definition of the norm then (6.5) with (9.2)

100

I I HnI< - HI< I I} = E (H: - H.) 2 K? d(X}. = I I Hn - H l l k x -+ 0 . so applying Exercise 6.2 we have Hn I< X -+ HI< · X in M 2. ·

0

Proof Let Tn = inf{t J; K'fd(X} . or J; H'f d(I< · X} . > n}. Let H'f'" = H. 1(o5_T,. ) and R'[n = I<. 1(• 5_ K.. ) · (9.4) implies that [[Tn . (I{Tn . X) = (fiTn RTn ) . X :

·

= (HI< · X)t for t :::; Tn , and

0

(9.6 ) Associative Law. Suppose X is a continuous semimartingale. If H, I< E lbii then H · ( I< · X) = HI< · X. Proof In view of (9.5 ) and the definition of the integral it suffices to prove the result when Xt = At is continuous adapted and locally of bounded variation. However, by stopping the first time I H. I, II<. ! or the variation of A on [0, s] 0 exceeds n the statement reduces to something covered by (9.2). 2 . 1 0 . Functions of S everal S e mimartingales

In this section, we will prove a version of I�o's formula for functions of several semimartingales. The key to the proof is the following

(1 0. 1 ) Integration by Parts. then

If

[O, t] with mesh ID. n l

I)xt,+l - Xt . )(Yi ,+l - Yi , ) _,. (x, Y}t

X be a continuous local martingale. If I< E II3 (Y) and

Using- (6 . 2) now it follows that (H (I< · X))t letting n -+ oo gives the desired result.

P �oof Let D. n = {tf} ·be a sequence of subdivisions of gomg to 0. We can write

77

(8.6) implies that

(9.5) H E II 3(K · X) then (*) holds. Lemma. Let

Section 2.10 Functions of Several Semimartingales

X and Y are continuous semimartingales

i

in probability, while application of (6.7) to each of the last two sums implies that they converge to J; Y.dx. and to J; X. dY. . o When Yi is locally of bounded variation then (X, Y}t = 0 and this reduce s to the ordinary integration by parts formula of Lebesgue-Stieltjes integra tion.

1

t

Y.dX. = XtYi - XoYo -

1 X.dY. t

Note that the right hand side makes sense in a path-by-path sense. We are now ready to prove the following generalization of (7 . 1) : Rd

-+ R and 0 :::; c :::; d. If Xl , . . . , Xl are continuous semimartingales and x:+l , . . . , Xl are locally of bounded variation then (1 0 .2)

Ito's formula. Let f

provided the derivatives continuous.

:

Dj/ , 1 :::; j :::; d and Dij/ , 1 :::; i, j, :::; c exist and are

Remark. At this point saving a derivative or two may seem like pinching pennies but when we get to Chapter 4 and apply this result to u(t, Bt) we will be very happy to not have to check that the partial derivatives Utt and Ut , x ,. exist. ·

By stopping, it suffices to prove the result when I Xf j, (Xi }t :::; M for all i, t. Since any function f satisfying the hypotheses can be approximated by Proof

78

Chapter 2 Stochastic Integration

Chapter Summary

polynomials Un in such a way that Un , D;gn , 1 � i � d and D;j Un , 1 � i, j � c converge to f, Dd , and D;j f uniformly on [-M, M] d, it suffices to prove the result when f is a polynomial. For then (6.8) and Exercise 7.2 allow us to pass from the result for Un to that for f. To prove the result for a polynomial, it suffices by linearity to prove the result when f is a monomial x k 1 x k2 • • • x kn where k l > k2 , . . . , kn E {1, . . . , d}. (The reader should note that k1 , . . . , kn are superscripts, not powers, e.g., our monomial might be x 1 x4x 1 x 1 x2.) n = 1 and k 1 = k, then (10.2) says that

If

Xf - X� =

To evaluate the third term (Y, Z}t , we observe that by for the covariance of stochastic integrals, (8.7),

1 1 dX: t

y;zt - YoZo =

- Yo =

(10.3) +

+

+ (Y, Z}t

(;:.l ( g X�(m)) dX:(i)

Applying (10.2) to Y gives Y;

1t Z. dY. 1t Y. dZ.

� ;� 1 i'¢j

so using the associative law

t

(9.6),

( IT. )

x; <m ) d(Xk(i ) , x.kU ) }.

m:iJ!a ,J

t j ( rr x; <m) ) dx;
I:

-

1 +

2

i.,!,F.}l ( 1.l x;<m)) d(Xk(i) , Xk(i)}. 0

m= 1 m*i

( Jo

n +l

(Notice that for each i in the sum for (Y, Z}t , there are two terms i = i, j = n+ 1, and i = n + 1, j = i in the last sum.) The proof is complete. 0 Chapter S ummary

With the completion of the proof of the multivariate form of Ito's formula, we have enough stochastic calculus to carry us through most of the applications in the book, so we pause for a moment to recall what we've just done. In Section 2.1, we introduced the integrands H. , the predictable processes and in Section 2.2 the (first) integrators Xt , the continuous local martingales. In Section 2.3 we defined the variance process associated with a local mar­ tingale, and more generally the covariance (X, Y}t of two local martingales. Theorem (3.11) tells us that the covariance could have been defined as the unique continuous predictable process At that is locally of bounded variation, has Ao = 0, and makes Xt Y1 - At is a martingale. The next result gives a pathwise interpretation of (X, Y}t :

(3.10) Quadratic Variation. Suppose X and Y are continuous local martin­ A n is a sequence of partitions of [O, t] with mesh I .D. n l --+ 0 then kn L(Xt k - Xtk - 1 ) • (Yt k - Ytk_1 ) --+ (X, Y}t in probability

gales.

By definition,

(10.3) and the formula

Adding the last three equalities gives

which is trivially true. To prove the result for a general monomial, we use induction. Let Y1 = TI�=l x:<m) be a monomial for which (10.2) holds, and let Zt = x:
Y1 Zt - YoZo =

79

If

k=l

Taking X =:= Y we see that time t.

(X}t is the "quadratic variation" of the path up to

80

Chapter 2 Stochastic Integration

Chapter Summary

In Section 2.4-2.5, we defined the integral, beginning with bounded con­ tinuous martingales and considering a sequence of progressively more general integrands.

1 (a, bj(s)C(w) with C E :Fa finite sum of elements of IIo H with E fc� H'f d(X}s < oo H with J; H'; d(X}s < oo a.s. for each t

100 I HsKs l di(X, Y} Is ::; (100 H? d(X}s) 1/2 (100 K? d(Y}s ) 1/2 0

0

where di(X, Y} ls stands for d� where on [0, s].

Vs

is the total variation of r -->

(X, Y} r

In Section 2.6, we generalized the integrators X3 to continuous local mar­ tingales and, in Section 2.8, to continuous semimartingales which are the sum (in a unique way) of a continuous local martingale Ms and a continuous process As that is locally of bounded variation. Further, we defined the covari­ ance of two semimartingales Xt = Mt + At and x: = M: + A� to be that of their martingale parts, i.e. , (M, M'}t, and generalized the quadratic variation interpretation in (3.10), see (8.6) . To be able to integrate with respect to semimartingales we had to restrict the integrands to the class of locally bounded predictable processes, £biT, but in this generality the integral has many nice properties. (8.3) Closure under Integration. If X is a continuous semimartingale and H E lbii, then (H · X)t is a continuous semimartingale. (8.4) Distributive Laws. Let

H, J( E £bii.

X and Y be continuous semimartingales, and

((H + K) · X)t = (H X)t + (I< · X)t (H · (X + Y))t = (H X)t + (H Y)t ·

·

·

(X, Y}t = �

i H!I
(9.6) Associative Law. Let X be a continuous semimartingale. If H, J( then H · ( J{ · X) = HI< · X.

(5.1) Kunita-Watanabe Inequality. If X and Y are local martingales and H and J( are two measurable processes, then almost surely

0

(8.7) Covariance of S tochastic Integrals. Suppose Xi and yi are contin­ uous semimartingales Hi , J{i E /biT, X = L:� 1 H i . Xi , Y = L:j= 1 J(i . yi , then t 1 ,]

A key role in this development was played by the

81

E /bii

As the reader probably remembers, the first three results were only obtained after several improvements Integrands = II 2 IIa /biT Ti l a. Closure (4.3.a) (4.2.a) (6.3) (8.3) b . Distributive Laws (4.2.b) (4.3.b), (5.3) (8.4) (6.4) c. Covariance Formula (4.2.c) (5.4) (6.5), (6.6) (8.6), (8.7) In Section 2.7, we proved a change of variables formula. This led to: (10.1) Integration by Parts. Let X and Y be continuous semimartingales.

which in turn led to a bigger and better change of variables formula. {10.2) It o 's formula. Let f : Rd __. R and 0 :::; c :::; d. If Xl, . . . , Xf are continuous semimartingales and x�+l , . . . , xf are locally of bounded variation

provided the derivatives continuous.

Djf, 1 :::; j :::; d and Dii f, 1 :::; i,j, :::; exist and are c

Coming Attractions

Section 2.11 is devoted to a proof of the Meyer-Tanaka formula, which extends the version of Ito's formula given in (7.1) to f that are a difference of two convex functions. This leads in a nice way to the definition of an occupation time density (or "local time" ) L� for a continuous semimartingale that is jointly continuous in t and a.

82

Ch ap ter 2 Stoclwstic Integration

Section 2.1 1 Tl1e Meyer-Tanaka Formula, Local Time

The developments described in the previous paragraph are used only in

4

(9.3) of Chapter and in Example 2.1 of Chapter 5. The material in Section 2 . 12 concerning Girsanov's formula is not used until Section 5.5. We would like to suggest that the reader proceed now to the beginning of Chapter 3. 2.1 1 . The Meyer-Tanaka Formula, Lo cal Time

In this section we will prove an extension of Ito's formula due to Meyer (1976) that has its roots in work of Tanaka (1963). See 11 . . Loosely speaking it says that Ito's formula is valid if f is a difference of two convex functions. Since such functions are differentiable except at a countable set and have a second derivative that is a nice signed measure, this is a modest generalization. However, this is the best one can do in general: if is a standard Brownian motion and = is a semimartingale then must be the difference of two convex functions. See Cinlar, J acod, Protter, and Sharpe (1980). The developments in this section follow those in Sections 4 and 5 of Chapter IV of Protter ( 1990) but things are simpler here since we have no jumps. Our first step is

( 4)

fB1

X1 f(B1)

R be convex and let X be a continuous f(fX) isRa->semimartingale and f(Xt) - f(Xo ) fo' f'(X,) dX, + I
:

=

=

=

=

(a) Then

fn is convex and C00 • Using Ito's formula we have

(b)

->

Our strategy for the proof will be to let n oo in (b) to get the desired formula. Since a convex function is Lipschitz continuous on each bounded

83

fn (x) -> f(x) uniformly on compact sets. Since fn (Xt) -> f (Xt) uniformly in t

interval, it _ is easy to see that � N, 1t follows that

IX1I

(c)

To deal with the first term on the right, we differentiate (a) to get

f�( x) j f' (x + ;) g(y) dy j f'( x) I� f�(X, ) dM, and I1 I� J' (X,) dM, then the as n Now i! If �2 maximal me uahty for martingales and the isometry property of stochastic =

=

-. 00 · .

=

� mtegrals (Exercise 6.2) imply

E (s�p (I� - I1 ) 2) � 4 s�p E(I,n - It) 2 4E 100 (f�(X,) - /'(X,)) 2 d(M}, 0 by the bounded con�ergence theorem. (Recall (M} oo � N.) By passing to a subsequence we can Improve the last conclusion to (e) sup (I�k - /1) 2 -> 0 almost surely t Let Jf I� f�(X,) dA, and J, I� f' (X,) dA,. In this case it follows from the bounded convergence theorem that for almost every oo (f) sup I J,n - J,l � r l f�(X.) - /'(X.) I dA. - 0 t Jo To take the limit of the third and final term J{1n n ) d(X} we 2 I/o r(x note that (b) implies 1 (g) �k f J( nk (Xt) - fn J,Yo) - 1 f�JX.) dX, Combining (c), �e) an� (f) we �ee that almost surely the right hand side of (g) �onver?es to a hnut umformly m t, so the left hand side does also. Since l{�k 1s contmuous adapted and increasing the limit is also. If ":'e fix X then for a given f we call the increasing process associ­ . . ated w1th f. The mcreasmg process associated with l x - al is called the local =

=

->

=

w

=

1

•

•

=

• .

0

/{

84

Section 2. 1 1 The Meyer-Tanaka Formula, Local Time

Chapter 2 Stochastic Integration

time at a, and is denoted by

name,

Lf. To prove the first property that justifies this

(11.3), we need the following preliminary

(11.2) Lemma. The increasing process associated with (x - a) + or (x - a) - is (1/2)Lf. = (x - a) + , h (x) = (x - a) - , and let Kf be the increasing process associated with !; . We begin by observing ft + h = lx-a l, so I
the asso�iated I
:=

0.

0

(11.3) Theorem. Lf only increases when Xt = a or to be precise, if we let .ea be the measure with distribution function t -l- Lf, then .ea is supported by {t : Xt = a} . Proof Intuitively, I Xt - al is a local martingale except when Xt = a so Lf is constant when Xt =/= a. To prove (11.3), however, it is easier to use the alternative definitions in (11.2). Let S < T be stopping times so that [S, T] C {t : Xt < a} . Applying (11.1) to f(x) = (x - a)+ we have

(XT - a) + - (Xs - a) + =

T

1s 1{x.>a} dX, + � (LJ. - L']; )

The left-hand side and the integral on the right-hand side vanish so LJ. = L5 . Since this holds when S = q with q any rational and T = inf{t > S : Xt � a - 1/n} for any n, it follows that ea ({t : Xt < a}) = 0. A similar argument using (x - a) - instead of (x - a) + shows ea ({t : Xt > a}) = 0 and the proof is 0 complete. We are now ready for our main result.

(11.4) Meyer-Tanaka formula. Let X be a continuous semimartingale. Let f be the difference of two convex functions, f' be the left derivative of f and suppose f" = J1. in the sense of distribution (i.e. , J1. has distribution function f') . Then t f(Xt) - f(Xo) = J' (X,) dX, +

1

f is the difference of two convex functions and the formula is f, we can suppose without loss of generality that f is convex. By

Proof Since

linear in

� J Jl.(da)Lf

stopping we can suppose !Xt I we can let

85

$ N and (X}t :::; N for all t. Having done this

� l:

g(x) =

Jl. (da)ix - al

S �nce ( ! - g)" = 0 o� �-N, N] it follows that f(x) - g(x) = a + bx for lxl :::; N. . tnv1al for linear functions, it suffices now to prove the result Smce the result IS for f!, which is almost trivial. Differentiating the definition of g we have

g'(x) =

�1-NN

Jl. (da) sign(x - a)

Starting with the definition of the local time !Xt - ai - IXo - al = and then integrating

1t

sign(X, - a) dX, + Lf

1/2 J!!N Jl. (da) gives

t N g(Xt) - g(Xo) = 21 Jl. (da) sign(X, - a) dX, 0 -NN 1 +2 Jl. (da)Lf -N (11.5) and (11.6) below will justify interchanging the two integrals in the first

1 1

1

term on the right-hand side to give

t

N g'(X,) dX, + 21 Jl. (da) Lf o -N Since Lf = 0 for lal > N (recall !Xt I $ N and use (11.3)) this proves the result for g and completes the proof of (11.4). o g(Xt) - g(Xo) =

l

1

For the next two results let S be a set, let S be a u-field of subsets of S, and let J1. be a finite measure on (S, S). For our purposes it would be enough to take S = R, but it is just as easy to treat a general set. The first step in justifying . the Interchange of the order of integration is to deal with a measurability issue: if we have a family Hf(w) of integrands for a E S then we can define the integrals J; Hf dX, to be a measurable function of (a, t, w ).

(11.5) Lemma. Let X be a continuous semimartingale with Xo = 0 and Hf(w) = H(a,t,w) be bounded and S x II measurable. Then there is a

let

�

I

86

Chapter 2 Stochastic Integration

Z(a, t,w) E S x II such that for J.L-almost every a Z(a, t, w) is a continuous version of I; H� aX3 • Proof Let Xt = Mt + At be the decomposition of the semimartingale. By stopping we can suppose that I A it � N, !Mt l � N and (X}t = (M}t � N for all t. Let 1-l be the collection of bounded H E S x II for which the conclusion holds. We will check the assumptions of the Monotone Class Theorem, (2.3) in Chapter 1. Clearly 1-l is a vector space. To check (i) of the MCT suppose H(a, t , w) = f(a)I<(t,w) where K(t, w) E biT and f(a) E bS. In this case t t H: ax. = f(a) K. ax.

fo

fo

so fI< E 11. . Taking f and J( to be indicator functions of sets in S and II respectively, we have checked (i) of the MCT with A = {A x B : A E S, B E II} . To check (ii) of the MCT let � Hn E 1-l with Hn l H and H bounded. The L2 maximal inequality for martingales and the isometry property of the stochastic integral imply

0

)

(

E s� p !(Hn , a · M)t - (Ha · M)d 2 � 4 s�p El(Hn , a · M)t - (Ha · M)tl 2 = 4E (H;'•a - H: ? a(M}. -l-

J

0

By passing to a subsequence we have for J.L almost every sup I (H nk , a . M)t - ur . M)t 1 --l-

t

To deal with the bounded variation part we note that

a

0

I

Section 2.1 1 The Meyer-Tanaka Formula, Local Time

87

Proo� Let Xt = Mt + At be the decomp osition of the semimartingale. By stoppmg we can suppose that lAlt � N, I Mtl � N and (X) t = (M)t � N for . �ll t. As_ r:�ted In the proof of (11.5), the result in the special case X = A is Just Fubm1 s theorem from measure theory, so we will suppose for the rest of the proof that X = M. Let 1-l _ be the �ollection of bounded H E S x II for which the conclusion holds . Agam, we will check the �sumptions of the Monotone Class Theorem, . (2.3) In Chapter 1. Clearly 1-l Is a vector space. To check (i) of the MCT, suppos: H(a, t,w ! = f(a)K(t,w) where K(t, w) E biT and f(a) E bS. In this case Zt = f(a)(Ii · X)t so

is zta J.L (aa) = (I< · X)t is /(a) J.L (aa) = ( {is f(a) J.L(aa) I<} · X)t = (H . X)t

0

so fK E 11.. Taking f and K to be indicator functions of sets in S and II . respectively, we have checked (i) of the MCT with A = {A x B : A E S B E II} · n n _To check (ii) of the MCT let � H E 1-l with H l H and H bounded. L� �tmg IIJ.Lll be the total mass of J.L applying Jensen's inequality to the proba­ bility measure J.L (·)/ IIJ.Lll , then multiplying each side by IIJ.Lll we have s� p I Zf• a - Zf i J.L ( aa)f ( is �

ll II

E

� E r sup I Zf • a - Zf l 2 J.L(aa)

ls t

Usin? Fubini's theorem, the L2 maximal inequality, and the isometry property the nght-hand side

0

= r E sup IZf • a - Zfl 2 J.L(aa) t

ls

a E sup !(Hn,a · A)t - (H a · A)t l � E Jo(;0 lHf • - Hf l a!Ait -lt as n -l- oo by the bounded convergence theorem. Combining the last two results

sup E I Ztn , a _ zta i 2 J.L(aa) ls t =4 E (H;'·a - H:)2 a(X). J.L (aa) -l-

(11.6) Fubini's Theorem. Let X be a continuous semimartingale, let Hta E b(S x II), and let Zf E S x II be a continuous version of I; H� ax. for J.L-almost every a. Then yt = Is Zf J.L(aa) is a version of H · X where Ht = Is Hta J.L (aa).

by using the bounded convergence theorem three times. Putting the absolute values inside the integral we have

completes the proof.

Less formally,

D .

�4

f is (100

)

( � l is ztn,aJ.L(aa) - is ZtJ.L(aa) l) 2

0

E s p

�E

(fs s�p I Z�·a - Zf lJL(aa)f

-l-

0

Section 2. 1 1 Tl1e Meyer-Tanaka Formula, Local Time

88

Chapter 2 Stochastic Integration Taking Hf = Is Hf 'a p(da) and passing to a subsequence n�: we have that with probability one (Hn k · X)t = Is zrk• a p(da) converges unifor�ly toJs Zf p(da� . The last detail is to check that H n · X converges to H · X m M , for then 1t follows that Is Zf p(da) = (H X) t . In view of the isometry pro� e� ty, we can . complete the last detail by showing that II Hn - Hllx -+ 0 but th1s IS routme. ·

by using -the bounded convergence theorem three times.

0

We can now complete the identification of L� as the local time at a.

(11.7) Theorem. Let X be a continuous semimartingale with local time Lf . If g is a bounded Borel measurable function then t g(X,)d(X ), da = L�g(a)

1:

l

g is continuous and let f E C2 with f" = g. In this case comparing (11.4) with Ito's formula proves the identity in question. Since the identity holds for any continuous function, using the Monotone Class Theorem proves the result for a bounded measurable g. 0

Proof- Suppose first that

When X is a Brownian motion the identity in

1:

L�g(a) da =

t

(11.7) becomes

l g(X,) ds

From this, we see that Lf can be thought of as the time spe�t at .a up to time t , or to bet precise it is a density function for the occupatiOn time measure

llt(A) = Io 1 A (X.) ds.

. . . There are many ways of defining local time for Browman motwn. The a one we have given above is famous for making it easy to prove that there is hat somew is a It t. and a of n functio ous version in which L is a jointly continu remarkable that this property is true for a continuous local martingale and it is not any more difficult to prove the result in this generality. (11.8) Theorem. Let X be a continuous local martingale. There is a version of the Lf in which (a, t) -+ L� is continuous.

89

(11.8) is not true for Xt = lEd, which is a semimartingale by {11.1). Introducing subscripts to indicate the process we are referring to, we clearly have Lx (t) = L'B(t) + L1/(t) for a > 0 and LX(t) = 0 for a < 0. Since L� (t) =/=. 0 it follows that a -+ LX(t) is discontinuous at a = 0. One can complain that this example is trivial since 0 is an end point of the set of possible values. A related example with 0 in the middle is provided by skew Brownian motion. Start with lEt I and flip coins with probability p > 1/2 of + and 1 - p of - to assign signs to excursions of Et , i.e., maximal time intervals (a, b) on which lEd > 0. If we call the resulting process yt then L� (t) -+ 2pL� (t) as a l 0 and L� (t) -+ 2(1 - p)L�(t) as a j 0. For details see Walsh (1978). Example 11. 1.

Proof As nsual, by stopping we can suppose that for all t. By (11.2)

�

L� = (Xt - a)+ - (Xo - a)+ -

IXt I ::_; N and (X}t ::_; N

t

1 1{x.> a} dX.

It is clear that (a, t) -+ (Xt - a) + - (Xo - a) + is continuous, so we only have to investigate the joint continuity of the stochastic integral

Ita =

t

l 1{X. >a} dX.

Fix a time T < oo and regard a -+ Ia as a mapping from R to C([O, T] , R), the real valued functions continuous on [0, T] , which we equip with the sup norm 11 ! 11 = SUPo 0 we have

Ell r - I.b ii.B ::_; C ia - w +a Suppose without loss of generality that a < b. One of the Burkholder Davis Gundy inequalities ({5.1) in Chapter 3) implies that , T ' d(X), E II I� - I ll' "' CE l{• <X,9)

(1

To bound the right-hand side we note that using Schwarz inequality and Fubini's theorem:

(l ) '

)

(11.7) and then the Cauchy­

l(Lj.)'

LT dx S (b - a)E dx b = (b - a) { E(Lr) 2 dx ::_; (b - a) 2 sup E(Lr) 2 :r:E( a ,b) Ja �

E

90

Chapter 2 Stochastic Integration

To bound the sup, we recall the definition of LT. LT IXT - x i - IXo - xi - 1T sgn(X. - x) dX. which with implies thatthe trivial inequalities (a+b)2 :5 2a2 +2b2 and ly- x l- lz-x l :5 ly-zl T E(Lj. ) 2 ,; 2E(XT - Xo) 2 + 2E (J. •ign(X, - x) dX, ) =

'

:5 2(2N) 2 + 2E( (X} T ) :5 8( N2 + N)

by the isometry property and the boundsto wecheckhavetheimposed on IinXtlKolmogorov's and (X), by stopping. This gives us what we need inequality continuity theorem and completes the proof. TheIf proof above almost works for semimartingales X1 Mt + A,. IfRemark. we define J; 1 { X, > a } dM. and Jf J; 1 {X, > a} dA. then the argument showsassumptions that (a, t) that If isimply continuous, so we will get the desired conclusion ifabove we adopt (a, t) J,a is continuous.

0

=

=

=

-+

-+

2.12. Girsanov's Formula

Indefinition this section, westochastic will showintegral that thearecollection of semimartingales and the of the not affected by a locally equivalent change of measure. For concreteness, we will work on the canonical probability space ( C, C), with :F1 the filtration generated by the coordinate maps X1(w) w1• Two measures Q and P defined on a filtration :F1 are said to be locally equivalent if for each t their restrictions to :F1, Q, and P1 are equivalent, i. e ., mutually absolutely In this case clear we letasa1the dQt/ The reasons for our interest in thiscontinuous. quantity will become storydP1• unfolds. (12.1) Lemma. Yi is a (local) martingale/Q if and only if a1Y1 is a (local) martingale/ P. Proof The parentheses are meant to indicate that the statement is true if the locals are removed. Let Yi be a martingale/Q, s < t and E :F Now if(i),Ztwo E :F1, then (i) J Z dP J Z dP1, and (ii) J Zat dPt J Z dQ,. So using (ii), the fact that Y is a martingale/Q, (ii), and (i) we have l a1YidP l a,ytdP, · l YidQ1 L Y·dQ. l a.Y.dP. l a.Y,dP =

=

=

=

=

=

=

=

=

A

•.

Section 2. 12 Girsanov's Formula

91

!hisa sequence shows a,ytof stoppm is a :nart�ngale/ P.j Ifoo Ysoisthata local martingale/Q then there Ishence times T Yi is a martingale/Q and n A n T � P. The optional stopping theorem implies that . a,yt" n IS a m artmgalef ! . at TTonYiAprove and itthat follows(a) that a,Yt is a local martingale/ Tn IS a martmgalef the converse, observe interchanging the roles ofa (local) and Qmartmgale/ an.d applying the last result shows that if {31 dPtfdQ1, and Z1 is i P, then f3tZt a (local) martingale/Q and (b) f3t a;- 1 , so letting � Zt a,Y, we have the desned result. Since 1 is a martingale/Q we have (12.2) Corollary. a1 dQtfdP1 is a martingale/F. There is a converse of (12.2) which will be useful in constructing examples. (12.3) Lemma. Given a, a nonnegative martingale/P there is a unique locally equivalent probability measure Q so that dQtfdP1 a1• Proof The last equation in the lemma defines the restriction of Q to :F, for < tn , A ; any Tol subseeetsthatofthR�, defines aBunique measure on forC, let1 :=:;t 1i ::;< n}.t2 ...Define befimte.Bort.dimensiOnal and let {w : w(t ) E A; the ; � : distributions of a measure Q by setting Q(B) L a1dP whenevnaler t :2: tn. The martingale property of a1 implies that the finite di­ mensio (C, C). distributions are consistent so we have defined a unique measure on We are ready to prove the main result of the section. (r'12.4)1 Girsanov's formula. . X is a local martingale/P and we let At J o a.- d(a,X }., then X1 - A, IS a local martingale/Q. Although formulaIfforweAsuppose looks a little strange, it is easy to see that 1t�oroof must0, bethen theintegrating righttheanswer. that A is locally of b.v. and has by parts, i.e., using (10.1), and noting (a, A} t 0 � smce A has bounded variation gives p

i\

p

=

=

=

p.

o

=

=

=

=

o

If

=

=

92

Chapter 2 Stochastic Integration

Section 2. 12 Girsanov's Formula

At this point, we need the assumption made in Section 2.2 that our filtration only admits continuous martingales, so there is a continuous version of a to which we can apply our integration-by-parts formula. Since a3 and X3 are local martingales/ P, the first two terms on the right in (12.5) are local martingales/ P. In view of (12.1), if we want Xt - At to be a local martingale/Q, we need to choose A so that the sum of the third and fourth terms = 0, that is,

93

(12.6) Theorem. The quadratic variation (X)t and hence the covariance (X, Y)t is the same under P and Q. Proof The second conclusion follows from the first and (3.9). To prove the first we recall (8.6) shows that if .6.n = {0 = tg < t� . . . < t]:n = t} is a sequence of partitions of (0, t] with mesh l.6.nl -+ 0 then

L(Xt i+ t - Xq ? -+ (X)t i

From the last equation and the associative law (9.6), it is clear that it is neces­ sary and sufficient that The last detail remaining is to prove that the integral that defines Let Tn = inf { t : Cl't :::; n - 1 } . If t :::; Tn , then

At exists.

by the Kunita-Watanabe inequality. So if T = limn- oo Tn then At is well defined for t :::; T. The optional stopping theorem implies that Eau.Tn = Eat , so noting ll't ATn = Cl't on Tn > t we have Since

in probability for any semimartingale/ P. Since convergence in probability is not affected by locally equivalent change of measure, the desired conclusion follows. D

(12.7) Theorem. If H E £bii then (H X)t is the same under P and Q. ·

Proof The value is clearly the same for simple integrands. Let

M and N be the local martingale parts and A and B be the locally bounded variation parts of X under P and Q respectively. Note that (12.6) implies (M)t = (N)t. Let Tn = inf{t : (M)t , I Ait or IBit ;::: n} If H E £biT and Ht

= 0 for t ;::: Tn then by (4.5) we can find simple Hm so that

These conditions allow us to pass to the limit in the equality

Cl't ;::: 0 is continuous and Tn :::; T, it follows that

( Hm (M + A))t = (Hm (N + B))t ·

Letting

·

(the left-hand side being computed under P and the right under Q) to conclude that for any H E £biT the integrals under P and Q agree up to time Tn . Since n is arbitrary and Tn -+ oo the proof is complete. D

n -+ oo, we see that Cl't = 0 a.s. on {T .$ t}, so

Pt is equivalent to Qt , so 0 = Pt(T :::;. t) = P(T :::; t). Since t is arbitrary D P(T < oo) = 0 , and the proof is complete.

But

(12.4) shows that the collection of semimartingales is not affected by change of measure. Our next goal is to show that if X is a semimartingale/ P, Q is locally equi valent to P, and H E ibII , then the integral ( H X)t is the same under P and Q. The first step is ·

3

Brownian Motion , II

In this chapter we will use Ito's formula to deepen our understanding of Brow­ nian motion or, more generally, continuous local martingales. 3.1. Recurrence and Transience

If B1 is a d-dimensional Brownian motion and Bf is the ith component then Bf is a rp.artingale with (B i )t = t. If i f= j Exercise 2.2 in Chapter 1 tells us that BfBf is a martingale, so (3.11) in Chapter 2 implies (B i , Bi )t = 0. Using this information in Ito's formula we see that if f : Rd - R is C2 then

L1=l Dii f for

Writing 'Vf = (Dd, . . . , Dd f) for the gradient of f, and L'::. f = the Laplacian of f, we can write the last equation more neatly as

(1.1)

f(Bt) - f(Bo) =

11

'Vf(B3) · dB3 +

� 11 L'::.f(B3)ds

Here, the dot in the first term stands for the inner product of two vectors and the precise meaning of that is given in the previous equation. Functions with L'::. f = 0 are called harmonic. (1.1) shows that if we com­ pose a harmonic function with Brownian motion the result is a local martingale. The next result (and judicious choices of harmonic functions) is the key to de­ riving properties of Brownian motion from Ito's formula.

(1.2)

Theorem. Let G be a bounded open set and r = inf{t B1 rf. G} . If f E C2 and L'::. f = 0 in G, and f is continuous on the closure of G, G, then for :z: E G we have f(:z:) = E,::f (Br ). Proof Our first step is to prove that Pc(r < oo) = 1. Let J( = sup{ l:z: - v i : :z:, y E G} be the diameter of G. If :z: E G and I B1 - :z:l > J( then B1 rf. G and r < 1. Thus P:c(r < 1) 2:: P:c( I B1 - :z:l > K) = Po( I B1 I > K) = cK > :

0

96

Chapter 3 Brownian Motion, II

Section 3.1 Recurrence and Transience

This shows sup, P, (r 2?: k) � (1 - EK ) k holds when k = 1. To prove t)le last result by induction on k we observe that if Pt(x, y) is the transition probability for Brownian motion, the Markov property implies

P,(r 2?: k) � �

L P1 (x, y)Py (r 2?: k - 1) dy

(1 - CI( ) k- l P,(B l E G) � (1 - CI( l

where the last two inequalities follow from the induction assumption and the fact that I B1 - x l > I< implies B1 fl. G. The last result implies that P,(r < oo ) = 1 for all x E G and moreover that for all 0

(1.3)

To get the last conclusion recall that (see e.g. ,

(1995))

E,rP =

1

00

< p < oo (5.7) in Chapter 1 of Durrett

ptP - 1 P,(r > t) dt

When D.. f = 0 in G, {1.1) implies that f(Bt) is a local martingale on [0, r ) . We have assumed that f is continuous on G and G is bounded, so f is bounded and if we apply the time change -y defined in Section 2.2, Xt = f{B-y(t ) ) is a bounded martingale (with respect to Yt = F-y(t ) ) · Being a bounded martingale Xt converges almost surely to a limit Xoo which has Xt = E,(Xoo l 9t) and hence E,Xt = E,X00 • Since r < oo and f is continuous on G, Xoo = f(Br) · 0 Taking t = 0 it follows that f(x) = E,Xo = E,Xoo = E,f(Br ) . In the rest of this section we will use (1.2) to prove some results concerning the range of Brownian motion {Bt : t 2?: 0}. We start with the one-dimensional case.

(1.4) Theorem. Let a < x < b and T = inf{ t : Bt rj. (a, b)} . x-a b - x P,(B = b) = -­ P,(BT = a) = -­ T b-a b-a Proof f(x) = (b - x)J(b - a) has f" = 0 in (a, b), is continuous on [a, b], and has f(a) = 1, f(b) = 0, so (1.2) implies f(x) = E,f(BT ) = P,(BT = a). 0 Exercise 1. 1 Deduce the last result by noting that using the optional stopping theorem at time T.

Let T, = inf{t

Bt is a martingale and

: Bt = x}. From (1.4), it follows immediately that

97

(1.5) Theorem. For all x and y, P,(Ty < oo) = 1. Since P,(Ty < oo ) = P,_ Y (To < oo ), it suffices to prove the result y = 0. A little reflection (pun intended) shows we can also suppose x > 0. Now using {1.4) P,(To < TM, ) = (M - l)JM, and the right-hand side approaches 1 as M --+ oo . o

Proof

when

It is trivial to improve (1.5) to conclude that

{1.6) Theorem. For any s < oo, P,(Bt = y for some t 2?: s) = 1. Proof

By the Markov property, 0

The conclusion of {1.6) implies ( argue by contradiction) that for any y with probability 1 there is a sequence of times t n l oo ( which will depend on the outcome w ) so that Bt n = y, a conclusion we will hereafter abbreviate as "Bt = y infinitely often" or "Bt = y i.o." In the terminology of the theory of Markov processes, what we have shown is that one-dimensional Brownian motion is recurrent. Exercise 1.2

Use

{1.6) to conclude lim sup Bt

t -+ 00

= oo

liminf Bt t

-+ 00

= -oo

In order to study Brownian motion in d 2?: 2, we need to find some ap­ propriate harmonic functions. In view of the spherical symmetry of Brownian motion, an obvious way to do this is to let cp (x) = /(l x l 2) and try to pick f : R --+ R so that D.. c,o = 0. We use lxl 2 = XI + · · · x� rather than lxl since it is easier to differentiate:

Dd (lxl 2 ) = f'(lxl 2 )2x; D;d (lxl 2 ) = f"(lx l 2 )4xf + 2f'( l x l 2 ) Therefore, for D.. c,o = 0 we need

0 = l:{f"(lxl 2 )4x f + 2/'( lxl 2 )} i

Section 3.1 Recurrence and Transience

Chapter 3 Brownian Motion, II

98

y = l x l 2 , we can write the above as 4yf"(y) + 2df'(y) = 0 or, if y > 0, f"(y) = -d 2y f'(y) Taking f'(y) = cy- d/ 2 guarantees D.. cp = 0 for X i= 0, so by choosing c appro­ Letting

priately we can let

{

d=2 lxl cp(x) = log d 2 d�3 x l l We are now ready to use (1.2) in d � 2. Let Sr = inf{t : !Btl = r} and r < R. Since cp has D..cp = 0 in G = {x : r < lxl < R}, and is continuous on G , (1.2) implies

for all c: > 0, so Po(Bt = 0 for some t > So = inf{t > 0 : Bt = 0}, we have

(1.10)

99

0) = 0, and thanks to our definition of

P:c(So < oo) = 0 for all x

Thus, in d � 2 Brownian motion will not hit 0 at a positive time even if it starts there.

1.3 Use the continuity of the Brownian path and to conclude that if x f:. 0 then P:c(Sr j oo as r ! 0) = 1.

Exercise

P:c(So = oo) = 1

For d � 3, formula (1.7) says

(1.11) where

cp( r) is short for the value of cp( x) on {x lx I = r} . Solving now gives :

- cp(x) P:c(Sr < SR ) = cp(R) cp(R) _ cp(r)

(1.7) In d =

2, the last formula says log R - log lxl

P:c(Sr < SR ) = log R - log r

(1.8) If we fix r and let

R --+ oo in (1.8), the right-hand side goes to 1. So P:c(Sr < oo) = 1 for any x and any r > 0

and repeating the proof of (1.6) shows that

(1.9) Theorem. Two-dimensional Brownian motion is recurrent in the sense that if G is any open set, then P:c( Bt E G i.o.) = 1.

If we fix R, let r --+ 0 in (1.8), and let So = inf{t > 0 : Bt = 0}, then for

x f:. O

Since this holds for all R and since the continuity of Brownian paths implies SR j oo as R j oo, we have P:c (So < oo) = 0 for all x f:. 0. To extend the last result to x = 0 we note that the Markov property implies

Po(Bt = 0 for some t � ) = Eo[Pn. (To < oo)] = 0 c:

There is no point in fixing R and letting r --+ 0, here. The fact that two dimensional Brownian motion does not hit points implies that three dimensional Brownian motion does not hit points and indeed will not hit the line { x : x 1 = x 2 = 0}. we fix r and let R --+ oo in (1.11) we get

If

(1.12)

P:c(Sr < oo) = (r/ l x l ) d-2 < 1 if l x l > r

From the last result it follows easily that for d �

3, Brownian motion is tran­

sient, i.e. it does not return infinitely often to any bounded set.

(1.13) Theorem. As t --+ oo, lBt 1 --+ oo a.s. Proof Let An = { !Btl > n 1 1 2 for all t � Sn } and note that Sn < oo by (1.3). T�e strong Markov property implies

as n --+ oo. Now lim sup An

= n�=l U�=N An has

P(limsup An) � lim sup P(An)

=1

So infinitely often the Brownian path never returns to {x : time Sn and . this implies the desired result.

lxl :::; n 1 1 2 } after

D

Dvoretsky and Erdos (1951) have proved the following result about how fast Brownian motion goes to oo in d � 3.

100 Chapter 3 Brownian Motion, II (1.14) Theorem. Suppose g(t) is positive and decreasing. Then

Po( I Bt l � g(t).Ji i.o. as t i oo ) = 1 or 0 according as f'" g(t) d-2 Jt dt = oo or < oo. Here the absence of the lower limit implies that we are only concerned with the behavior of the integral "near oo." A little calculus shows that

according as a < 1 or a > 1, so Bt goes to oo faster than .Ji/(logt)'Jt/ d-2 for any a > 1. Not;that in view of the Brownian scaling relationship Bt = d t 1 12 B1 we could not sensibly expect escape at a faster rate than .Ji,. The last result shows that the escape rate is not much slower. Review. At this point, we have derived the basic facts about the recur­ rence and transience of Brownian motion. What we have found is that

(i ) P.,(jB1 I < 1 for son-te t 2:: 0) = 1 if and only if d � 2

Section 3.2 Occupation Times 101 section we will investigate the occupation time ]000 1n (Bt) dt and show that for

any x

(2.1) Pc (j000 1n (Bt) dt = oo) = 1 in d � 2 (2.2) E., ]000 1n (Bt ) dt < oo in d 2:: 3 Proof of

(2.1) Let T0 = 0 and G = B(O, 2r). For k 2:: 1, let S�o = inf{t > n - 1 : Bt E D} T�o = inf {t > S�o B1 E G} :

Writing r for T1 and using the strong Markov property, we get for k 2::

From this and

(4.5) in Chapter 1 it follows that

(ii) P.,(Bt = 0 for some t > 0) = 0 in d 2:: 2. The reader should observe that these facts can be traced to properties of what we have called cp, the (unique up to linear transformations) spherically sym­ metric function that has Llcp ( x) = 0 for all x f. 0, that is : cp ( x)

{

d=1 lxl = log l x l d = 2 lxl 2 - d d 2:: 3

and the features relevant for (i ) and (ii) above are

(i) cp (x ) -;. oo as l xl -;. oo if and only if d � 2 (ii) j cp (x) l -;. oo as x -;. 0 in d 2:: 2. 3.2. O ccupation Times

Let D = B(O, r) = {y : I Y I < r} the ball of radius r centered at 0. In Section 3. 1 , we learned that Bt will return to D i.o. in d � 2 but not in d 2:: 3. In this

1

1Tk 1n (Bt ) dt sk

are i.i.d.

Since these random variables have positive mean it follows from the strong law of large numbers that a.s. proving the desired result. Proof of

0

(2.2) If f is a nonnegative function, then Fubini's theorem implies

E.,

100 f(Bt) dt = 100 E.,f(Bt ) dt = 100 J Pt( 00 = J 1 Pt(X, y) dt f(y) dy

x, y)f(y) dy dt

where Pt(x, y) = (21li) -d/ 2 e - I"' -Y I 2 12 t is the transition density for Brownian motion. As t -;. oo, Pt(x, y) {2m)- d/2, so if d � 2 then f p1 ( x, y)dt = oo . "'

102

Chapter 3 Brownian Motion, II

3, changing variables t = l x - yl 2 /2s gives 00 1 00 Pt (x, y) dt = 0 (2m) d/ 2 e - l y - x l2 / 2 t dt 0 o d/ 2 & lx - yl 2 e- = r J 2 2s2 ds lx l Y 1l' oo (2.3) I x - y 1 2- d 00 s( d/ 2 ) -2 e -$ ds = 27l'd/ 2 0 r ( � - 1) 1 X - y 1 2- d = 27l'd/ 2 where f(a) = J000 s a - l e - & ds is the usual gamma function. If we define

When d ;:::

1

1

(

8

G(x, y) = then in d ;:::

(2.4)

1

)

(

)

100 Pt(x, y) dt

To complete the proof of (2.2) now we observe that taking B (O, r) and changing to polar coordinates

j G(x, y)f(y) dy When d =

=

loo E,f(Bt) dt

1, we let at = Pt (O, 0). With this choice, 1 G(x , y) = rn= (e - (y - x )2 / 2t - 1)r 1 12 dt y 27l' 0

1 00

and the integral converges , since the integrand is ::; 0 and ...... -(y - x ) 2 j2t3f2 as t __... oo. Changing variables t = (y - x)2 /2u gives

1oo (e-u - 1) ( (y -2ux)2 ) 1/2 - (y2u2- x)2 du - �;1 1 00 (lu e - & as) u- 312du I Y - xl loo ds e - & joo u - 3/2 0 2 du $ oo - IY - x l l ds e s - 1 / 2 = - I y - x I y 7l' 0 =

(2.5)

j

0

v'2i �

= - --

..fo

- - --c-

f = 1D with D =

--

-s

since

roo ds e - & s- 1 12 = r oo dr e - r 2 f 2 h .!.}2 . ..J<j; ,fir = lo lo 2 . The c.omputation is almost the same for d = 2. The only thing that changes Is the choice of at . If we try at = p1 (0, 0) again, then for x =f. y the integrand "". -r 1 as t 0 and the integral diverges , so we let at = p1(0, e l ) where e 1 = (1, 0). With this choice of at , we get =

To extend the last conclusion to x =f. 0 observe that applying the strong Markov property at the exit time T from B(O, lx l ) for a Brownian motion starting at 0 and using the rotational symmetry we have

We call G(x, y) the potential kernel, because G(·, y) is the electrostatic potential of a unit charge at y. See Chapter 3 of Port and Stone (1978) for more on this. In d ::; 2, J000 Pt (x, y) dt = oo so we have to take another approach to define a useful G:

G(x, y) =

100 (Pt(x, y) - at) dt

103

where the at are constants we will choose to make the integral converge (at least when x =f. y) . To see why this modified definition might be useful note that if J f(y) dy = 0 then (assuming we can use Fubini's theorem)

1 G(x, y) = -

3, G(x, y) < oo for x =f. y, and oo Ex f(Bt) dt = G(x, y)f(y) dy

1

Section 3.2 Occupation Times

__...

(2.6)

Chapter 3

104

Brownian Motion, II

df 2) · l x - yl 2-d d :;::: 3 { 1)/2tr (r{d/2 G(x,y) = {-1/tr) · log(l x - yl) d=2

To sum up, the potential kernels are given by

{2.7)

d= 1 The reader should note that in each case G(x, y) = Crp(l x - yl ) where rp is harmonic function we used in Section 3.1. This is, of course, no accident. --. spherically symmetric and, as we will see, satisfies xthe 0) formulas =0)0 isforobviously .6.G(x,TheG(x, =/= 0, so the results above imply G(x, 0) = A + Br,o(l x l ) . xabove correspond to A = 0, which is nice and simple. But what about the weird looking B 's? What is special about them? The answer is simple: They are chosen to make !.6.G(x,O) = -60 (a point mass at 0) in the distributional sense. It is easy to see that this happens in d = 1. In that case rp(x) = lxl then rp'(x) = { 11 xx <> 0O so if B = -1 then Brp"(x) = -260. More sophisticated readers can check that this is also true in d :?: 2. (See F. John {1982), pages 96-97.) To explain why we want to define the potential kernels, we return to Brow­ nian motion in the half space, first considered in Section 1.4 . Let H = {y : Yd > 0}, = inf{t : Bt f/:. H}, and let = (Yl , · · · , Yd- 1 , -yd) be the reflection of y through the plane {y E Rd : Yd = 0} (2.8) Theorem. If x E H, f :?: 0 has compact support, and {x : f(x) > 0} C H then E:c (1r f(Bt)dt) = J G(x,y)f(y)dy- J G(x, y)f(y)dy Proof Using Fubini's theorem which is justified since f :?: 0, then using (4.9) from Chapter 1 and the fact that {x : f(x) > 0} C H we have E:c 1T f(Bt) dt = 100 Ex{f(Bt); > t) dt = 100 L (Pt(x,y) - Pt(x, y))f(y)dydt = 100 L (Pt(x,y) - at)f(y)dydt - 100 L (Pt(x, y) - at)f(y)dydt = j G(x, y)f(y) dy- j G(x, y)f(y) dy T

.

-1 · lx - yl

.

Y

T

Section 3.3 Exit Times

G

105

I G(x,y)f(y)ldy The proof given above simplifies considerably in the case d :?: 3; however, part of the point of the proof above is that, with the definition we have chosen for G in the recurrent case, the formulas and proofs can be the same for all d. Let GH(x, y) = G(x, y) - G(x, y). We think of GH(x, y) as the "expected occupation time {density) at y for a Brownian motion starting at x and killed when it leaves H." The rationale for this interpretation is that (for suitable f) The compact support of f and the formulas for imply that J and J are finite, so the last two equalities are valid.

IG(x, y)f(y)l dy

D

yGH--. GH(x, y)

With this interpretation for introduced, we invite the reader to pause for a minute and imagine what looks like in one dimension. If you don't already know the answer, you will probably not guess the behavior as oo. So much for small talk. The computation is easier than guessing the answer:

y --.

G(x,y) = -lx - Y l so GH(x, y) = -lx - Y l + lx + Y l · Separating things into cases, we see that when 0 < y < x GH(x ,y) _- { -(x -(y-- x)y) ++ (x(x ++ y)y) == 2x2y when x
GH(x, y) = 2(x A y) for all x, y > 0 It is somewhat surprising that y --. GH(x, y) is constant = 2x for y :?: x, that is, all points y > x have the same expected occupation time! (2.9)

3.3. Exit Times

= t : B=t Srf/:. G}in

In this section we investigate the moments of the exit times T inf{ for various open sets. We begin with in which case T the notation of Section

G = {x : lxl < r} 3.1. {3.1) Theorem. If l x l � r then ExSr = (r2 - lxl2 )/d

Proof The key is the observation that

d {(B;)2 -t} I Bt 1 2 - dt = I: i=l

106 Chapter 3 Brownian Motion, II

Section 3.3 Exit Times 107

being the sum of d martingales is a martingale. Using the optional stopping theorem at the bounded stopping time Sr 1\ t we have .: < oo, and we have I Bsrl\t 1 2 :::; r2, so letting t -+ oo and using the dominated convergence theorem gives l x l 2 = Ex (r2 - Sr d) which 0 implies the desired result.

(1.3) tells us that E Sr Exercise 3.1.

Let

a, b > 0 and T = inf{ t : Bt ff. (-a, b)} . Show EoT = ab.

To get more formulas like (3.1) we need more martingales. Applying Ito's formula, ( 1 0. 2 ) in Chapter 2, with Xl = Xt , a continuous local martingale, and Xf = {X}t we obtain

1 Dd(X3 , {X}3)dX3 t + 1 D2 f(X3 , {X}3)d{X}3 t + � 1 Dn f(X. , {X),) d{X),

f(Xt, {X}t) - f(Xo, 0) = {3.2)

From ( 3..2) w e see that if (�D11 + D2 )f tingale. Examples of such functions are

t

=

(3.3) Theorem. Let Ta = inf{t : IBt l � a}. Then (i) Eora = a2 , (ii) EorJ = 5a4 /3. The dependence of the moments on a is easy to explain: the Brownian scaling relationship Bet =a c1 12Bt implies that Ta /a2 =a r1 .

{i) follows from (3.1), so we will only prove {ii) . To do this let X1 = B{ - 6Bft + 3t 2 and Tn :::; n be stopping times so that Tn j oo and Xti\Tn is a

Proof

martingale. Since Tn :::; n the optional stopping theorem implies

Now I Br.,I\Tn I :::; a, so using {1.3) and the dominated convergence theorem we can let n -+ oo to conclude 0 = a4 - 6a2 Eora + 3E0r� . Using (i ) and rearranging 0 gives (ii) . Find a, b, c so that Bf -aB{t+bBft 2 -ct3 is a (local) martingale and use this to compute E0r�.

Exercise 3.2

0, then f(Xt , {X}t) is a local mar­

Our next result is a special case of the Burkholder Davis Gundy inequal­ ities, {5.1), but is needed to prove (4.4) which is a key step in our proof of

(5.1).

(3.4) Theorem. If Xt is a continuous local martingale with Xo = 0 then

or to expose the pattern

fn (x, y) =

These local martingales are useful for computing expectations for one dimen­ sional Brownian motion.

O� m �S]n / 2]

cn ,m x n -2m ym

where [n /2] denotes the largest integer :::; n/ 2, cn ,o = 1 and for 0 we pick 1 2 Cn ,m (n - 2m)(n - 2m - 1) = - (m + 1)cn , m+ l so that Dx.: /2 of the mth term is cancelled by the [n /2] th term is 0.)

:::; m < [n/ 2]

Proof First suppose that IXtl and {X}t are :::; M for all t. In this case ( 2 .5) in Chapter 2 implies X{ - 6Xf {X}t + 3{X}; is a martingale, so its expectation is 0. Rearranging and using the Cauchy-Schwarz inequality

Dy of the ( m + 1 ) th. (Dxx of

The first two of our functions give us nothing new (Xt and Xf - {X}t are local martingales) , but after that we get some new local martingales:

Xl - 3Xt {X}t , X{ - 6Xf{X}t + 3{X};,

Using the L4 maximal inequality (( 4.3) in Chapter 4 of Durrett fact that (4/3)4 :::; 3.1605 < 19/6 we have

{1995)) and the

108 Chapter 3 B o nian Motion, II r

Section 3.3 Exit Times 109

w

Since I X3 1 ::; M for all s we can divide each side by E(sup3 9 Xi)1 1 2 then square to get

So if Tn is a sequence of times that reduces Z1 , the L2 maximal inequality applied to Yt ATn gives

The last inequality holds for a general martingale if we replace t by Tn = inf{t : t , I X1 I , or (X}t ;::: n } . Using that conclusion, letting n --? oo and using the monotone convergence theorem, we have the desired result. D

Letting

If we notice that f(x, y) = exp(x - y/2) satisfies (�Du + D2 ) ! = 0 , then we get another useful result.

(3.5) The Exponential Local Martingale. If X is a continuous local mar­ tingale, then E(X)1 = exp(X1 - � (X}1) is a local martingale. If we let yt = exp(X1 - t {X}1), then (3.2) says that (3.6) or, in stochastic differential notation, that dyt = ytdX3 • This property gives yt the right to be called the martingale exponential of yt . As in the case of the ordinary differential equation

f'(t) = f(t)a(t)

f(O) = 1

which for a given continuous function a(t) has unique solution f(t) = exp(At) , where At = J� a(s) ds. It is possible to prove (under suitable assumptions) that Z is the only solution of (3.6). See Doleans-Dade (1970) for details. The exponential local martingale will play an important role in Section 5.3. . Then (and now) it will be useful to know when the exponential local martingale is a martingale. The next result is not very sophisticated (see (1.14) and (1.15) in Chapter VIII of Revuz and Yor (1991) for better results) but is enough for our purposes. (3.7) Theorem. Suppose X1 is a continuous local martingale with (X}1 ::; Mt and Xo = 0. Then yt = exp(X1 - � (X}1 ) is a martingale. Proof Let Zt = exp(2X1 -

Now,

t (2X}t), which is a local martingale by (3.5).

yt2 = exp(2X1 - {X}t) = Zt exp((X}t)

n l oo and using the monotone convergence theorem we have

by Jensen's inequality. Using (2.5) in Chapter 2 now, we see that yt is a martingale. o Remark. It follows from the last proof that if X1 is a continuous local mar­

tingale with Xo = 0 and (X}t ::; M for all t then yt = exp(X1 - � (X}t ) is a martingale in M 2 .

Letting 8 E R and setting X1 = 8Btin (3.6), where B1 is a one dimensional Brownian motion, gives us a family of martingales exp(8B1 - 8 2 tj2) . These martingales are useful for computing the distribution of hitting times associated with Brownian motion. (3.8) Theorem. Let Ta = inf{t : Bt = a}. Then for a > 0 and ,\ ;::: 0 Eo exp( -.ATa ) = e - aVV: Remark. If you are good at inverting Laplace transforms, you can use (3.8)

to prove (4.1) in Chapter 1:

Proof Po (Ta < oo) = 1 by (1.5). Let X1 = exp(8B1 - 82 tj2) and Sn ::; n be stopping times so that Sn l oo and Xt A Sn is a martingale. Since Sn ::; n, the optional stopping theorem implies

If 8 2: 0 the right-hand side is ::; exp(8a) so letting n --? oo and using the bounded convergence theorem we have 1 = Eo exp(8a - 82 Ta /2). Taking 8 = J2X now gives the desired result. D

110 Chapter 3 Brownian Motion, II

If

is a martingale. B = -2J.Lfa-2 then -OJ.L - ()2 a-2 f2 = 0 and exp(-{2J.Lfa-2 )Zt) is a local martingale. Repeating the proof of (3.8) one gets

(3.9) Theorem. Let Ta = inf{t : I Bd ;:::: a}. Then for a > 0 and A 2: 0, Eo exp(-Ara ) = 2e - av'2X/(1 + e -2 av'2X) Proof Let 'lj;a (A) = Eo exp(-ATa)· Applying the strong Markov property at time Ta (and dropping the subscript a to make the formula easier to typeset)

gives

Section 3.4 Change of Time, Levy's Theorem 111

= Eo(exp( -Ar) ; Br = a) + Eo( exp( -Ar) 'lj;2 a(A) ; Br = - a) Symmetry dictates that ( r, Br) = d ( r, -Br). Since Br E {-a, a} it follows that r and Br· are independent, and we have

Exercise

3.3. Let T_a = inf{t : Zt = -a}. If a , J.L > 0 then

Eo exp( -ATa)

Using the expression for 'lj;a (A) given in

(3.8) now gives the desired result.

0

Another consequence of (3.8) is a bit of calculus that will come in handy in Section 7 .2.

(3.10) Theorem. If B > 0 then 00 -1 exp( -lzlv� 1 e - z2 f2t e - 8t dt = -2B)

1 0

.../2-ii

-128

Proof Changing variables t

= 1/2s, dt = -dsf2s2 the integral above

100 -ffs e - z 2 e -

3.4. Change o f Time , Levy's Theorem

In this section we will prove Levy's characterization of Brownian motion ( 4.1) and use it to show that every continuous local martingale is a time change of Brownian motion.

(4.1)

Theorem. If Xt is a continuous local martingale with X0 (X}t = t, then Xt is a one dimensional Brownian motion.

Proof

loo

-128 0

Using {3.8) now with a = Je and A = z2 and consulting the remark after (3.8) for the density function of Ta we see that the last expression is equal to the 0 right-hand side of {3.10). The exponential martingale can also be used to study a one dimensional Brownian motion Bt plus drift. Let Zt = a- Bt + J.Ll where a- > 0 and J.L is real. Xt = Zt - J.Lt is a martingale with (X}t = a-2t so {3.7) implies that

{4.5) in Chapter 1 it suffices to show

{4.2) Lemma. For any s and t, X3 +t - X3 is independent of :F3 and has a normal distribution with mean 0 and variance t. Proof o f ( 4.2)

ter 2, to X�

Applying the complex version oflto's formula, {7.9) in Chap­

= X3 +r - X3 and f(x) = ei8:c , we get ei8x: - 1 = iB

ds 3 - 8/23 -

2s2 o ..j'Fff 1 1 vr;;Bffe- 8 / 23 e - z2 3 ds --= -�

By

= 0 and

1 ei8X� dX' o u t

-

B 2 t i8X � e du 2 ' Jo

Let :F� = :F3 + r and let A E :F3 = :F6. The first term on the right, which we will call Yt , is a local martingale with respect to :Ff. To get rid of that term let Tn l oo be a sequence of stopping times that reduces Y;, replace t by t A Tn, and integrate over A. The definition of conditional expectation implies

since

Yo = 0. So we have

112 Chapter 3 Brownian Motion, II Since l e i0"'1 = 1, letting n -+ oo and using the bounded convergence theorem gives 82 E(e'·ox t' ; A) - P(A) = - 2 E to e'· ox ' du ; A J 2 8 i 6X � 2 E e ; A du

(

=

-

J

u

(

)

)

by Fubini's theorem. (The integrand is bounded and the two measures are finite. ) Writing j(t) = E(e;ox ; ; A), the last equality says

-1

2 t 8 j(t) - P(A) = - 2 j(u) du 0 2 Since we know that j j( s) l ::::; 1, it follows that li(t) - j ( u)l ::::; I t - u l 8 /2, so j is continuous and we can differentiate the last equation to conclude j is differentiable with

j'(t) = -282 j(t) Together with j(O) = P(A), this shows that j(t)

= P(A)e- 62tf2, or

Section 3.4 Change of Time, Levy's Theorem 113 Multiplying each side of (4.3 ) by cp ( - 8 ) and integrating we see that E(g(Xf)I:F6) is a constant and hence

E(g(X:) I :F6) = Eg(XD A monotone class argument now shows that the last conclusion is true for any bounded measurable g. Taking g = 1B and integrating the last equality over A E :F6 we have

P(A)P(x: E

B) = L E(1B (XD I:F�) dP = P(x: E B)

by the definition of conditional expectation, and we have proved the desired independence. 0 Exercise 4.1. Suppose Xf , 1 ::::; i ::::; d are continuous local martingales with Xo = 0 and i = i. (X i , Xi )t = t0 if otherwise then Xt = (Xl, . . . , Xf) is a d-dim�nsional Brownian motion.

{

An immediate consequence of (4.1 ) is: Every continuous local martingale with X0 = 0 and having (X) oo = oo is a time change of Brownian motion. To be precise if we let 1 (u) = inf{t : (X}t > u} then Bu = X-y(u ) is a Brownian motion and Xt = B(x )1 • Proof Since ! ( (X}t) = t the second equality is an immediate consequence of the first. To prove that we note Exercise 3.8 of Chapter 2 implies that u -+ Bu is continuous, so it suffices to show that Bu and B� - u are local martingales.

(4.4) Theorem.

Since this holds for all A E :F6 it follows that

(4.3 ) or in words, the conditional characteristic function of x: is that of the normal distribution with mean 0 and variance t. To get from this to (4.2 ) we first take expected values of both sides to .conclude that Xf has a normal distribution with mean 0 and variance t. The fact that the conditional characteristic function is a constant suggests1 that Xf is independent of :F6. To turn this intuition into a proof let g be a C function with compact support, and let cp ( 8 )

=

j ei6"' g(x) dx

be its Fourier transform. We have assumed more than enough to conclude that cp is integrable and hence

g(x)

=

_!_ 271"

j ei6"'cp(-8) dx

(4..5) Lemma. Bu , :F-y(u ) • u ;?: 0 is a local martingale . Proof of (4.5 ) Let Tn = inf{ t : IXtl > n}. The optional stopping theorem

implies that if u < v then

·

E(X-y(v)IITn I :F-y(u )) = X-y(u )IITn where we have used Exercise 2.1 in Chapter 2 to replace :F-y(u )IITn by :F-y(u )· To 2 let n -+ we observe that the £ maximal inequality, the fact that x;(v)IITn ­ (X)-y(v)IITn is a martingale, and the definition of 1(v) imply E sunp X;(v)IITn ::::; 4 sunp EX;(v)IITn = 4 sup E(X}-y(v)IITn ::::; 4v n 00

114 Chapter 3 Brownian Motion, II

Section 4 . 3 Change of Time, Levy's Theorem 115

The last result and the dominated convergence theorem imply that as n -+ oo, X-y(t )ATn -+ X-y(t ) in L2 for t = u, v. Since conditional expectation is a contraction in L 2 it follows that E{X-y(v ) ATJF-y(u ) ) -+ E(X-y(v) IF-r(u ) ) in L2 D and the proof is complete. To complete the proof of {4.4) now it remains to show

(4.6) Lemma. B� - u, .1"-y(u ) • u ;::: 0 is a local martingale. v

As in the proof of {4.5), the optional stopping theorem implies then

icx;(v )ATn - (X}-y(v )ATn IF-y(u )) = x;(u ) ATn - (X}-y(u )ATn To let n -+ oo we observe that using (a + b) 2 ::::; 2a2 + 2b2 , definition of 'Y( v) then

(

- (X}-y (v )ATn E sup n x�(v )ATn

)2

{3.4), and the

4 2 v) v )ATn + 2E(X}-y( ::::; 2E sup x-y(

n s; CE(X}; ( v) ::::; Cv2

The proof can now be completed as in (4.5) by using the dominated convergence theorem, 11nd the fact that conditional expectation is a contraction in L2 • D Our next goal is to extend (4.4) to X1 with P( (X} oo < case X-y(u ) is a Brownian motion run for an amount of time step in making this precise is to prove

oo) > 0. In this (X} oo . The first

{4.7) Lemma. limtroo X1 exists almost surely on {(X} oo < oo}. Tn = inf{t : (X}t ;::: n}. (3.7) in Chapter 2 implies that (XTn ) t = (X}t ATn s; n Using this with Exercise 4.3 in Chapter 2 we get Xt ATn E M 2 so limt-+ oo Xt ATn exists almost surely and in L2 • The last statement shows limt- oo Xt exists almost surely on {Tn = oo} :::> {(X} oo < n } . Letting n -+ oo now gives {4.7). D To prove the promised extension of ( 4.4) now, let 'Y( u) = inf{t : (X}t > u} when u < (X} oo , let Xoo = limt-+ oo Xt on {(X} oo < oo } , let Bt be a Brownian motion which is independent of {Xt, t ;::: 0}, and let U < (X} oo -y(u ) Y. u- X Xoo + B(u - (X} oo ) U ;::: (X} oo

Proof Let

_

{

Proof By {4.1) it suffices to show that Yu and Y; - u are local martingales with respect to the filtration cr(Yt : t $ u) . This holds on [0, (X} oo ] for reasons indicated in the proof of {4.4). It holds on [(X} oo , oo) because B is a Brownian motion independent of X. D

. The reason for our interest in

Proof of ( 4.6)

that if u <

{4.8) Theorem. Y is a Brownian motion.

{4.8) is that it leads to a converse of (4.7).

(4.9) Theorem. The following sets are equal almost surely: C = {limt-+ oo Xt exists } A = { (X} oo < oo}

B = {sup1 I Xt l < oo} B+ = {supt Xt < oo}

Proof Clearly, C C B C B+ . In Section 3.1 we showed that Brownian motion has limsup1 _., 00 Bt = oo. This and (4.8) implies that Ac C B+, or B+ C A. D Finally (4.7) shows that A C C.

The result in (4.8) can be used to justify the assertion we made at the beginning of Sections 2.6 that Ila(X) is the largest possible class of integrands. Suppose H E II and let T = sup{t : J� H; d(X}3 < oo}. (H · X)t can be defined for t < T and has t (H · X}t = H;d(X}3 so on {(H · X} T = oo} :::>

1

{T < oo} we have

limsup(H · X)t = oo

t TT

liminf(H · X)t = -oo t

TT

and there is no reasonable way to continue to define

(H · X)t for t ;::: T.

Convergence is not the only property of local martingales that can b e studied using (4.9). Almost any almost-sure property concerning the Brownian path can be translated into a corresponding result for local martingales. This immediately gives us a number of theorems about the behavior of paths of local martingales. We will state only:

(4.10) Law of the Iterated Logarithm. Let L (t) = y'2t log log t for t ;:::

Then on {(X} oo = oo } ,

limsup Xt/L( (X } t ) = 1 a.s.

t-+ oo

e.

116 Chapter 3 Brownian Motion, II This follows from (4.9) and the result for Brownian motion proved in Section 7.9 of Durrett (1995). D

(b) P(r 1 12

Proof

Finally we have a distributional result that can be derived by time change.

-+ R is measurable and locally bounded. Use (4.4) to generalize Exercise 6.7 in Chapter 2 and conclude that 1 1 X1 = h. dB. is normal with mean 0 and variance h� ds Exercise 4.2. Suppose h : [0, oo)

1

3.5.

1

> {J>., B ; � 6>.) � 13g� 1 P(r1 12 > >.)

Remark. The inequalities above are called "good )." inequalities, although the reason for the name is obscured by our formulation (which is from Burkholder (1973)). The name "good >." comes from the fact that early versions of this and similar inequalities (see Theorems 3.1 and 4.1 in Burkholder, Gundy, and Silverstein (1971)) were formulated as P(f > >.) � Cp,K P(g > >.) for all >. that satisfy P(g > >.) � KP(g > [3>.). Here {J, I<. > 1. Proof It is enough to prove the result for bounded r A n for all n, it also holds for r. Let

for

(5.1) Theorem. For any 0 < p < oo there are constants 0 < c, C < oo so that

Remark. This result should be contrasted with the martingales that only holds for 1 < p < oo

LP maximal inequality for

Levy's Theorem, (4.4), tells us that any continuous local martingale is a time change of Brownian motion Bt , so it suffices to let B; = sup•
(5.2) Theorem. For any 0 < p < oo there are constants 0 < c, C < oo so that for any stopping time r

Proof of {5.2)

The key is the following pair of odd looking inequalities.

(5.3) Lemma. Let {3 > 1 and 6 > 0. Then for any >. > 0 (a) P(B; > {3>., r1 12 � 6>.) � (/3�21 p P(B; > >.)

r for if the result holds

S1 = inf{t : IB(t A r) l > >.} s2 inf{t : IB(t A r) l > {3>.} T inf{t : (t A r) 1 12 > 6>.}

B urkholder Davis Gundy Inequalities

Let X1 be a local martingale with Xo = 0 and let x; = sup•
·

Section 3.5 Burkholder Davis Gundy Inequalities 117

=

=

Since B;

> [3>. implies sl < s2 < T and r1 12 � 6>. implies T

= 00

we have

P(B; > [3>., r1 ' 2 � 6>.) � P( I B(r A S2 A T) - B(r A S1 A T) I 2:: ({3 - 1)>.) � ({3 - 1) -2 >. - 2 E{(B(r A S2 A T) - B(r A St A T)) 2 } where the second inequality is due to Chebyshev. Now if R1 stopping times then

� R2 are bounded

So .we have

E(B(R2 ) - B(R1 )) 2

EB(R2 ) 2 - 2EB(R1 )B(R2 ) + EB(R1 ) 2 E(R2 - Rt) = EB(R2 ) 2 - EB(R1 ) 2 =

=

since Bl - t is a martingale. Resuming our first computation we find

({3 - 1) - 2 >. - 2 E{(r A S2 A T) - (r A S1 A T)} � ({3 - 1) - 2 >. -2 (6>.) 2 P(S1 < oo) = ({3 - 1) -2 62 P(B; > >.) =

since T A r � (6>.)2 proving (a) .

118

Chapter 3 Brownian Motion, II

Section 3.6 Martingales Adapted to Brownian Filtrations

1 To prove (b) we interchange the roles of B ( t 1\ r) and (t 1\ r) 12 in the first set of definitions and let sl = inf{t : (r 1\ t ? l 2 > ..\ } s2 = inf{t : (r 1\ t) 1 12 > ,8..\ } T = inf{t : I B (r 1\ t)l > 6..\ } 1 Reversing the roles of B3 and s 12 in the proof, it is easy to check that P (r

Replacing h by a nonnegative random variable Z, taking expectations and using Fubini's theorem gives

Ecp (Z) =

(5.5)

1oo P(Z

>

..\) dcp (..\)

From our assumption it follows that

2 2 1'2 ,B ; > ..\, B :::; 6..\) :::; P (( r A s2 A T) - (r A s1 A T) � (,8 - 1)..\ ) :::; (,82 - 1)- 1 ..\- 2 E{( r 1\ S2 1\ T) - (r 1\ S1 1\ T) }

and using the stopping time result mentioned in the proof of (a) it follows that the above is

(,82 - 1)- 1 ..\ - 2 E{ B (r A s2 A T) 2 - B ( r A s1 A T) 2 } :::; (,82 - 1)- 1 ..\- 2 ( 6..\) 2 P (Sl < oo) :::; (,82 _ 1)- 1 62 P (r1 12 > ..\) 2 since I B (r 1\ T) l � ( 6..\) proving (b).

119

=

0

It will take one more lemma to extract ( 5.2) from (5.3) . First, we need a definition. A function cp is said to be moderately increasing if cp is a , nondecreasing function with cp ( O) = 0 and if there is a constant J( so that cp (2..\ ) :::; J( cp (,\). It is easy to see that cp(x) = xP is moderately increasing (I< = 2P) but cp ( x) = e a:r: - 1 is not for any a > 0. To complete the proof of (5.2) now it suffices to show (take ,8 = 2 in (5.3) and note 1/3 :::; 1)

2

(5.4) Lemma. If X, Y � 0 satisfy P (X > 2..\, Y :::; 6..\ ) :::; 6 P (X > ..\ ) for all 6 � 0 and cp is a moderately increasing function, then there is a constant C that only depends on the growth rate J{ so that

Ecp(X ) :::; CEcp (Y) Proof It is enough to prove the result for bounded cp for if the result holds for cp 1\ n for all n � 1, it also holds for cp. Now cp is the distribution function of a measure on [0, oo ) that has

P (X > 2..\) = P (X > 2..\, Y :::; 6..\) + P (X > 2..\, Y > 6..\) :::; 62 P (X > ..\) + P (Y > 6..\) Integrating dcp( ..\) and using (5.5) with Z = X/ 2, X, Y/6 Ecp(X/2) :::; 62Ecp (X) + Ecp (Y/ 6) N

2

1

Pick 6 so that J( 6 < 1 and then pick N � 0 so that 2 > 6- • From the growth condition and the monotonicity of cp, it follows that Combining this with the previous inequality and using the growth condition again gives

Solving for Ecp(X) now gives 0

p = 4. In this case J( = 24 = 2 1 / 8 to make 1 - J( 6 = 3/4 and N = 3, we get a constant which

381 Revisited. To compare with (3.4), suppose

16. Taking 6

is

=

164 . 4/3 = 8 7 , 3 81 .333 . . .

Of course we really don't care about the value, just that positive finite constants 0 < c, C < oo exist in (5.1).

3.6.

Martingales Adapted t o B rownian Filtrations

Let {Bt , t � 0} be the filtration generated by a d-dimensional Brownian motion Bt with Bo = 0, defined on some probability space (n, :F, P). In this section we will show (i) all local martingales adapted to {Bt , t � 0} are continuous and

120

Chapter 3 Brownian Motion, II

(ii) every random variable integral.

X E L 2 (n, Boo , P) can be written as a stochastic

(6.1) Theorem. All local martingales adapted to {Bt , t ;::._ 0 } are continuous. Proof Let Xt be a local martingale adapted to {Bt , t ;::._ 0 } and let Tn ::; n be a sequence of stopping times that reduces X. It suffices to show that for each n, X(t A Tn ) is continuous, or in other words, it is enough to show that the result holds for martingales of the form yt = E(Y!Bt) where Y E Bn . We

Section 3.6 Martingales Adapted to Brownian Filtrations 12 1 and it follows from step 1 that yt is continuous on (0, t 1 ]. Repeating the ar­ gument above and using induction, it follows that the result holds if Y = fl (Bt 1 ) • • • fk (Bt�o) where t 1 < t 2 < < t k ::; n and fl , . . . , fk are bounded · · ·

continuous functions.

Step 3. Let Y E Bn with ElY I < oo. It follows from a standard application of the monotone class theorem that for any E > 0, there is a random variable XE of the form considered in step 2 that has E!XE - Yl < E. Now

build up to this level of generality in three steps.

S tep 1. - Let Y = f(Bn ) where f is a bounded continuous function. If t 2':: n, then yt = f(Bn ), so t __. yt is trivially continuous for t > n. If t < n, the Markov property implies yt = E(Y!Bt) = h( n - t, Bt) where

and if we let Zt = E(!XE - YI !Bt), it follows from Doob's inequality (see e.g., (4.2) in Chapter 4 of Durrett ( 1995)) that

(t :::;n

>.P sup Zt It is easy to see that h( s, x) is a continuous function on (0, oo ) x R, so yt is continuous for t < n. To check continuity at t = n, observe that changing variables y = x + z-jS, dy = sdf 2 dz gives

) J (21r)1 d/ 2 e-lzl2 /2 j(x + zvt:s) dz

h(s, x =

so the dominated convergence theorem implies that as

f(Bn ) ·

S tep 2. Let

t i n, h (n - t, Bt)

__.

Y = fl (BtJh(Bt 2 ) where t 1 < t 2 ::; n and fl , h are bounded and

continuous. If t ;::._ t 1 , then

so the argument from step 1 implies that yt is continuous on other hand, if t ::; t 1 , then yt = E(Yt1 !Bt) and where

g(x) =

[t 1 , oo )

j (27r(t2 -1 t l) ) d/2 e- ly-xj2/2(t 2-t l) j?(y) - dy

is a bounded continuous function, so

.

On the

)

> >. $ EZn = E!X E - Yl < E

Now XE(t) = E(XE!Bt) is continuous, so letting E 0 we see that for a.e. w, Yt (w) is a uniform limit of continuous functions, so yt is continuous. 0 __.

Remark. I would like to thank Michael Sharpe for telling me about the proof given above.

We now turn to our second goal. Let Bt = (Bf , . . . , Bf) with B0 = 0 and {� , t ;::._ 0 } be the filtrations generated by the coordinates. (6.2) Theorem. For any X E L 2 (r!, B00 , P) there are unique Hi E II2 (B i )

let

with

Proof We follow Section V.3 of Revuz and Yor ( 1991 ) . The uniqueness is immediate since the isometry property, Exercise 6.2 of Chapter 2, implies there is only one way to write the zero random variable, i.e., using Hi ( s, w) := 0. To prove the existence of H3, the first step is to reduce to one dimension by proving:

(6.3) Lemma. Let X E L 2 (r!, Boo , P ) with EX = 0, and let X; = E(X!Bi00). = X1 + + Xd .

Then X

· · ·

Proof Geometrically, see ( 1.4) in Chapter 4 of Durrett ( 1995 ) , X; is the pro­ jection of X onto L[ the subspace of mean zero random variables measurable

Chapter 3 Brownian Motion, II with respect to 800 • If Y E L1 and Z E L} then Y and Z are independent so EY Z = EY EZ = 0. The last result shows that the spaces L[ and L} are orthogonal. Since Lr, . . . , L� together span all the elements in L2 with mean 0, 122

·

(6.7) follows.

0

Proof in one dimension Let I be the set of integrands which can be written as L,j=1 Aj 1 (3j_1,3j] where the >.i and si are (nonrandom!) real numbers and 0 = so < s 1 < . . . < Sn . For any integrand H E I C II2 (B), we can define the stochastic integral Yt = J; H3 dB3 , and use the isometry property, Exercise 6.2 of Chapter 2, to conclude Yt E M 2 • Using (3.7) and the remark after its proof, we can further define the martingale exponential of the integral, Zt = &(Y)t and conclude that Zt E M 2 • Ito's formula, see (3.6), implies that Zt - Z0 = J; Z3 dY3• Recalling Yt = (H B)t and using the associative law, (9.6) in Chapter 2, we have ·

Zt - Zo =

1 Z3H3 dB3 t

Yo = 0 and (Y) o = 0, so Zo = 1. Since Zt E M2 we have EZt = 1 = Zo and it

follows that

{ oo

Zt = EZt + J z. H3 1[o ,tJ (s) dB. . o

i.e., the _ desired representation holds for each of the random variables in the set :1 = {&(H · B)t : H E I, t ? 0} . To complete the proof of (6.2) now, it suffices to show

(6.4) Lemma. If W E L 2 has E(ZW) = 0 for all Z E :1 then W = 0. (6.5) Lemma. Let g be the collection of L2 random variables X that can be written as EX + J000 H3 dB Then g is a closed subspace of L2 • Proof of { 6.4) We will show that the measure W dP (i.e., the measure J.L with dJ.L/dP = W) is the zero measure. To do this, it suffices to show that W dP is the zero measure on the O"-field ( Bt1 , • • • , Bt,.) for any finite sequence 0 = to < t 1 < t2 < . . . < t n . Let Aj , 1 � j � n be real numbers and z be a •.

C7

complex number. The function

is easily seen to be analytic in C, i.e., cp ( z) is represented by an absolutely convergent power series. By the assumed orthogonality, we have cp ( x ) = 0 for

Section 3.6 Martingales Adapted to Brownian Filtrations

123

all real x , so complex variable theory tells us that cp must vanish identically. In particular cp (i) = 0, when i = A. The last equality implies that the image of W dP under the map is the zero measure since its Fourier transform vanishes. This shows that W dP vanishes on C7 ( Bt1 - Bt0 , • • • , Bt,. - B1,_ J = C7( Bt1 , • • • , B1 J and the proof of 0 (6.4) is complete. Proof of

(6.5) It is clear that g is a subspace. Let Xn E g with

(6.6) and suppose that Xn __... X in L2 • Using an elementary fact about the variance, then (6.6), and the isometry property of the stochastic integral, Exercise 6.2 in Chapter 2, we have

E(Xn - Xm ) 2 - (EXn - EXm ) 2 = E{(Xn - Xm ) - (EXn - EXm )} 2 = E (H� - H';')2 ds

1oo

Since Xn __... X in L2 implies E(Xn - Xm ) 2 __... 0 and EXn __... EX, it follows that II H m - Hn iiB __... 0. The completeness of II 2 (B), see the remark at the beginning of Step 3 in Section 2.4, implies that there is a predictable H so that II Hn - H II B � 0. Another use of the isometry property implies that H m B converges to H . B in M2 • Taking limits in (6.6) gives the representation for X. This completes the proof of (6.5) and thus of (6.2) . o ·

4

Partial D ifferential Equations

A. Parab olic Equations

In the first third of this chapter, we will show how Brownian motion can be used to construct (classical) solutions of the following equations:

Ut Ut Ut

( )

1

=

2 .6.u

=

2 .6. u + g

=

2 .6. u + cu

1

1

in (0, oo) x Rd subject to the boundary condition: u is continuous at each point of {0} X Rd and u O , x = f(x) for x E Rd . Here,

D. u

82 u l

2

8 ud + . . · + -axi

= -?

OXJ

and by a classical solution, we mean one that has enough derivatives for the equation to make sense. That is, u E C1 • 2 , the functions that have one contin­ The uous derivative with respect to t and two with respect to each of the continuity in u in the boundary condition is needed to establish a connection between the equation which holds in (O, oo) x Rd and u ( O , x ) = f(x) which holds on {0} x Rd . Note that the boundary condition cannot possibly hold unless Rd -+ R is continuous. We will see that the solutions to these equations are (under suitable as­ sumptions) given by

Xi ·

f:

f(Bt) Ex (t(B ) + 1t g(t - s, B.) ds) t Ex (t(B ) (1t c(t - s, B.) ds)) t

Ex

exp

126

Section 4.1 The Heat Equation

Chapter 4 Partial Differential Equations

In words, the solutions may be described as follows:

(i) To solve the heat equation, run a Brownian motion and let u(t, x) = Exf(Bt ) · (ii) To introduce a term g(x) add the integral of g along the path. (iii) To introduce cu, multiply f(B1) by m1 = exp { f; c(t - s, B3)ds) before taking expected values. Here, we think of the Brownian particle as having mass 1 at time 0 and changing mass according to m� = c(t - s, B3)m3 , and when we

take expected values, we take the particle's mass into account. An alternative interpretation when c ::; 0 is that exp { f; c(t - s, B3) ds) is the probability the particle survives until time t, or - c(r, x) gives the killing rate when the particle is at x at time r. In the first three sections of this chapter, we will say more about why the expressions we have written above solve the indicated equations. In order to bring out the similarities and differences between these equations and their el­ liptic counterparts discussed in Sections 4.4 to 4.6, we have adopted a rather robotic style. Formulas (m.2 ) through (m.6 ) and their proofs have been devel­ oped in parallel in the first six sections, and at the end of most sections we discuss what happens when something becomes unbounded. 4.1. The Heat Equation

127

X� = t - s and X! = B! for 1 ::; i ::; d gives 3 u(t - s, B3) - u(t, Bo) = -ut (t - r, Br) dr 3 + V'u(t - r, Br) · dBr 3 + -1 6.u(t - r, Br) dr 2 0

1 1 1

To check this note that dX� = -dr and X� has bounded variation, while the x: with 1 ::; i ::; d are independent Brownian motions, so

( 1.2 ) follows easily from the Ito's formula equation since - u1 + the second term on the right-hand side is a local martingale.

!6.u = 0 and

D

Our next step is to prove a uniqueness theorem.

( 1.3) Theorem. If there is a solution of ( 1.1 ) that is bounded then it must be

v(t, x) = Exf(Bt)

In this section, we will consider the following equation:

( 1.1a) u E C1 • 2 and u1 = �6.u in (0, oo ) x Rd . ( 1 .1b ) u is continuous at each point of {0 } x Rd and u(O, x) = f(x).

Here = means that the last equation defines v. We will always use generic solution of the equation and v for our special solution. Proof If we now assume that u is bounded, then M3 , 0 ::; s < �artingale. The martingale convergence theorem implies that

u for a

t, is a bounded

Mt = lim 3 Tt M3 exists a.s.

This equation derives its name, the heat equation, from the fact that if the units of measurement are chosen suitably then the solution u(t, x) gives the temperature at the point x E Rd at time t when the temperature profile at time 0 is given by f(x) . The first step in solving ( 1.1 ) , as it will be six times below, is to find a local martingale.

If u satisfies ( 1.1b ) , this limit must be it follows that

( 1.2) Theorem. If u satisfies ( 1.1a) , then M3 = u(t-s, B3) is a local martingale on [O, t).

Now that ( 1.3 ) has told us what the solution must be, the next logical step is to find conditions under which v is a solution. It is ( and always will be ) easy to show that if v is smooth enough then it is a classical solution.

Proof

Applying Ito's formula, ( 10.2 ) in Chapter 2, to

u(xo, . . . , xd ) with

f(B1 ). Since M3 is uniformly integrable,

u(t , x) ExMo = ExMt = v(t, x) =

( 1.4) Theorem. Suppose f is bounded. If v E C1 • 2 then it satisfies ( 1.1a) .

D

128

Chapter 4 Partial Differential Equations

Proof

The Markov property implies, see Exercise

Section 4. 1 The IIeat Equation where p1(x, y) = (2 m ) - df2e - lx- yl2/ 2 1• Writing little calculus gives

2.1 in Chapter 1, that

The left-hand side is a martingale, so the right-hand side is also. If v E C1 • 2 , then repeating the calculation in the proof of (1.2) shows that

1 - r, Br )dr v(t - s , B3) - v(t, Bo) = 0 (-Vt + -Av)(t 2 + a local martingale

13

It is easy to give conditions that imply that v satisfies (l.lb). In order to keep the exposition simple, we first consider the situation when f is bounded.

(1.5) Theorem. If f is bounded and continuous, then v satisfies (l.lb ) . (Bt - Bo) 4. t 1 1 2N , where N has a normal distribution with mean 0 and variance 1, so if t n -+ 0 and X n -+ x, the bounded convergence theorem Proof

D; = 8f8x; and Dt = 8/8t, a

D;pt(x, y) = -(x; - y;) Pt(x, y) t (x; y;) 2 - t Pt(x, y) D;;pt(x, y) = t2 (x y;)(xi - Yi ) (X y) ; -1- J· ; Dij Pt ( X , y) = Pt , • T t2 -d/2 l x - Yl 2 DtPt(x, y) = t - + 2t 2 Pt(x , y) If f is bounded, then it is easy to see that for = i, ij , or t

)

(

The left-hand side is a local martingale, so the integral on the right-hand side is also. However, the integral is continuous and locally of bounded variation, so by (3.3) in Chapter 2 it must be = 0 almost surely. Since Vt and Av are continuous, it follows that - Vt + �Av = 0. For if it were f. 0 at some point (t, x), then it would be f. 0 on an op en neighborhood of that point, and, hence, with positive probability the integral would not be ::: 0, a contradiction. 0

129

a

j IDaPt (x, Y) f(y)l dy < oo and is continuous in Rd , so (1.6) follows from the next result on differentiating under the integral sign. This result is lengthy to state but short to prove since we assume everything we need for the proof to work. Nonetheless we will see that this result is useful.

(1.7) Lemma. Let (S, S, m) be a a--finite measure space, and g : S -+ R be measurable. Suppose that for x E G an open subset of Rd and some h0 > 0 we have: (a) u(x) = fs I< (x, y)g(y)m(dy) where I< and 8I
h ��

The final step in showing that v is a solution is to find conditions that guarantee that it is smooth. In this case, the computations are not very difficult.

I<(x* + he; , y) - I<(x * , y) = fo (x * + Be ; , y) dB for I h i :::; ho and y E S (c) u; (X ) = fs (X , y)g(y) m( dy) is continUOUS at x * and (d) fs J��o (x * + Be ;, y)g(y) dB m(dy) < oo. Then 8uf8x; exists at x * and equals u;(x * ).

(1.6) Theorem. If f is bounded, then v E �Y1 · 2 and hence satisfies (l.la) .

Proof Using the definition of u i n (a) , then (b) and Fubini's theorem, which is justified for I hi :::; h0 by ( d) , we have

implies that

0

Proof

B y definition,

(b )

�� I ��

l

is (K(x* + he;, y) - K(x* , y))g(y) m(dy) = fh fs 88x1� (x* + Be ;, y)g(y) m(dy) dB lo l ,

u(x*+ he;) - u(x * ) =

j

v(t, x) = Exf(Bt ) = Pt(x, y)f(y) dy

130

Section 4.2 The Inhomogeneous Equation

Chapter 4 Partial Differential Equations

Dividing by

h and letting h -+ 0 the desired result follows from (c) .

0

Later we will need a result about differentiating sums. Taking S = Z with all subsets of S, and is counting measure in ( 1.7) , then setting g = 1 = and fn(x) = K(x, n ) gives the following. Note that in ( a) and ( c) of ( 1.7) it is implicit that the integrals exist, so here in (a) and ( c) we assume that the sums

S

J.L

converge absolutely.

x

( 1 .8) Lemma. Suppose that for E G an open subset of Rd and some

we have:

ho

>

0

u(x) = I:n fn(x) where fn and fJfn/fJxi G -+ R, n E Z are measurable functions with (b ) fn(x* + h ei ) - fn(x*) = foh �(x* + B ei ) dB for lhl :::; ho and n E Z (c) Ui(x) = I:n �(x) is continuous at x* and ( d) I:n J��o l*:-(x* + B ei )l dB < oo. Then fJujfJxi exists at x* and equals ui (x*). Unbounded f. For some applications, the assumption that f is bounded is too restrictive. To see what type of unbounded f we can allow, we observe that, at the bare minimum, we need Ex l f (Bt )l < oo for all t. Since ( a)

:

131

u Ut :- �D.u =contmuous.

u(O,x) = f(x). uf C1 • 2 g(t, x) Our first step in treating the new equation is to observe that if u 1 is a solution of the equation with f = fo and g = 0 which we studied in the last se.ction, and u 2 is a solution of the equation with f = 0 and g = g0 then u 1 + u 2 is a solution of the equation with f = fo and g = g0 so we can restrict our attention to the case f = 0. (2.1b) is continuous at each point of {0} x Rd and We observed i n Section 4 . 1 that ( 2.1b ) cannot hold unless is continuous. Here unless so the equation in ( 2.1a) cannot hold with E ¥

IS

Having made this simplification, we will now study the equation above by following' the procedure used in the last section. The first step is to find an associated local martingale.

u 3 M3 = u(t- s,B3) + 1 g(t- r,Br)dr is a local martingale on [0, t). (2.2) Theorem. If satisfies (2.1a) , then

Proof

Applying Ito's formula as in the proof of ( 1.2 ) gives

3 u(t- s,B3) - u(t,Bo) = 10 (-ut + 21 D.u)(t- r, Br) dr 3 + 1 Vu(t- r,Br) dBr which proves { 2.2 ) , since - Ut + �D.u = -g and the second term on the right­ hand side is a local martingale. ·

a condition that guarantees this for locally bounded

f is

o

·

Replacing the bounded convergence theorem in ( 1.5 ) and ( 1.6 ) by the dominated convergence theorem, it is not hard to show:

( 1.9 ) Theorem. If f is continuous and satisfies

( ), then v satisfies (1.1 ). *

4.2. The Inhomogeneous Equation

In this section, we will consider what happens when we add a function to the equation we considered in the last section. That is, we will study

u C1 •2 and Ut = �D.u + g in (O,oo) x Rd

(2.1a) E

g(t, x)

Again the next step is a uniqueness result.

g

( 2.3 ) Theorem. Suppose is bounded. If there is a solution of (2.1 ) that is bounded on [0, T] x Rd for any T it must be

< oo, v(t,x) := Ex (1t g(t- s, B3) ds) Proof Under the assumptions on g and u, M3, 0 :::; s < t, defined in ( 2.2 ) is a bounded martingale and u(O, x) := 0 so Mt := lim 3 Tt M3 = }t0 g(t- s, Bs) ds

132

Section 4.2 The Inhomogeneous Equation

Chapter 4 Partial Differential Equations

Since Ms is uniformly integrable,

u(t, x) = E:cMo = E:cMt = v(t, x)

0

Again, it is easy to show that if v is smooth enough it is a solution. (2.4) Theorem. Suppose g is bounded and continuous. If v E satisfies (2.1a) in (0, oo) x Rd . Proof

C1 • 2 , then it

( 1t g(t - r, Br) dr I:Fs) t • = 1s g(t - r, Br) dr + EB, ( 1 g(t - s - u, Bu) du) = 1s g(t - r, Br) dr + v(t - s, Bs)

_

v(t , x) = C1 • 2 ,

1s g(t - r, Br ) dr = 10 (-vt + 2 Av + g)(t - r, Br) dr $

+ a local martingale

The left-hand side is a local martingale, so the integral on the right-hand side is also. Since the integral is continuous and locally of bounded variation, (3.3) in Chapter 2 implies it must be = 0 almost surely. Again this implies that . ( -Vt + !Av+ g) := 0, for our assumptions imply that this quantity is continuous 0 and if it were =f:. 0 at some point (t, x) we would have a contradiction. The next step is to give a condition that guarantees that v satisfies (2.1b ) . As in the last section, we will begin by considering what happens when every­ thing is bounded.

--1-

lv(t, x) l :::; Ex

0

1t lg(t - s, B.)l ds :::; Mt

where p. (x, y) = (211"s ) -df 2e - l u - :c l 2 1 2 • . The expression we have just written for v above is what Friedman (1964) would call a volume potential and would write as

V(x,t) = ( Z(x, t ; � , r)g(� , r)d� dr }To Jn To translate between notations, set To = O, D = Rd , Z(x, t;�, r) = Pt - r (x, �) and change variables s = t - r, y = � - Because of their importance for the

parametrix method, the differentiability properties of volume potentials are well known. The results we will state are just Theorems 2 to 5 in Chapter 1 of. Friedman (1964), so the reader who is interested in knowing the whole story can find the missing details there. (2.6a) Theorem. If g is a bounded measurable function, then v(t, x) is contin­ uous on (0, oo) x Rd . Proof

Use the bounded convergence theorem.

D

(2.6b) Theorem. There is a constant C so that if I Y I :::; M is measurable, then the partial derivatives D;v = 8vj8x; have I D;vl :::; CMt 1 12 , are continuous, and are given by

( 2. 5) Theorem. If g is bounded, then v satisfies (2.1b ) . If I Y I :::; M, then as t

1t ds J p. (x , y)g(t - s, y) dy {

v(t - s, B8 ) - v(t , Bo) +

1

Proof

Take home message is: it is not enough to assume g is continuous to have v E C1 • 2• We must assume g to be Holder continuous locally in t. That is, for any N < oo there are constants C, E (0, oo), which may depend on N, such that lg(t, x) - g(t, y)l :::; Clx - Y l a whenever t :::; N.

The reason for this assumption can be found in the proof of (2.6c) and again in (2.6d). The first step in showing v E C1 • 2 is to assume g is bounded and use Fubini's theorem to conclude

-

The left-hand side is a martingale, so the right-hand side is also. If v E then repeating the calculation in the proof of (2.2) shows that

The final step in showing that v is a solution is to check that v E C1 • 2• Since the calculations necessary t o establish these properties are quite tedious, we will content ourselves to state what the results are and do enough of the proofs to indicate why they are true. If you get dazed and confused and bail out before the end, the

a:

Using the Markov property, see Exercise 2.6 in Chapter 1, gives

Ex

133

--1-

0

0

D;v =

lt j D;p.(x, y)g(t - s, y) dy ds

134

Section 4.2 The Inhomogeneous Equation

Chapter 4 Partial Differential Equations

Proof

Using the formula for D;p, from (1.6), the right-hand side is

Although the last formula looks suspicious because we are integrating s - 1 near 0, everything is really all right. If I Y I :::; M, then

E, l (x; - B!)g(t - s, B,)l :::; M E,lx; - B!l = CMs 1 12 so we have

11 -ds E, I(x; - B!)g(t - s, B.)l :::; CMt 112 0

2

s

We can overcome this problem if g is Holder continuous at x, because the fact that E,(x; - B!) 2 = s allows us to write

Using the Holder continuity of g one can show the above is :::; Cs- l+ a/ 2, so its integral from s = 0 to t converges absolutely, and with a little work (2.6c) 0 follows. (See Friedman (1964), pages 10-12, for more details.) The last detail now is:

< oo

Using our result on differentiating under the integral sign, (1.7), it follows that the partial derivatives D;v exist, are continuous, and have the indicated form.

0

(2.6d) Theorem. Let g be as in (2.6c). Then ovfot exists, and

ov (t, x) = g(t, x) r dr +J ot o Proof

v(t, x) =

(2.6c) Theorem. Suppose that g is bounded and Holder continuous locally in t. Then the partial derivatives D;i v = EPvjox ;f)xi are continuous, and

Suppose for simplicity that i = j. Consulting (1.6) for the formula for

D;;p3 , the right-hand side is

J ota Pt - r(x, y)g(r, y) dy

t, we rewrite v as

To take the derivative w.r.t.

Things get even worse when we take second derivatives.

Proof

135

11 j Pt-r(x, y)g(r, y) dy dr

Differentiating the right-hand side w.r.t. t gives two terms. Differentiating the upper limit of the integral gives g(t, x). Differentiating the integrand and using a formula from (1.6) gives

11 j :t Pt-r(x, y)g(r, y) dy dr = 11 J (27r) d/ 2(���) (d+ )/ (��� = ;r) g(r, y) dy dr 2 2 -lx - Yl 2 g(r, y) dy dr l x - Yl 2 + J( ! (27r(t - r)) - d/ 2 2(t ( 2(t - r) ) - r)2 1 o 1 1 dr -d 1 dr exp

exp

=2

This time, however, E,l(x;

- B!) 2 - si = sEo i(Bf) 2 - I I so

0

-E,g(r, Bt - r) + t-r

0

2( t - r)2

E, (Ix - B t - rl 2 g(r, Bt - r))

In the second integral, we can use the fact that E,( lx - Bt - r l 2 ) = C(t - r) to cancel one of the t - r's and make the second expression like the first, but even if we do this,

1t -dr t-r 0

- oo

136

Chapter 4 Partial Differential Equations

Section

This is the difficulty that we experienced in the proof of (2.6c), and the remedy is the same: We can save the day if g is Holder continuous locally in t. For further details, see pages 12-13 of Friedman (1964). 0 To see what type of unbounded g 's can be allowed, we will restrict our attention to the temporally homogeneous case g (t, x) = h(x ) . A t the bare minimum, w e need Unbounded

g.

4.3

The Feynman-Kac Formula

137

source can be consulted for a wealth of information about these spaces. We will content ourselves here with an Example 2.1. Let h(z) We will now show that

(2.7) Theorem.

= k( lzl ) where

k(r) =,r-P for r :::; 1 and 0 for r

>

1.

h E J(d if p < 2.

The special case p = 1 is the Coulomb potential which is important in physics. and if we want (2.1b) to hold, we need to know that if tn ! 0 and Xn -+ x, then

Proof If a

< 1 then changing to polar coordinates and noticing the integral x = 0 we have

is largest when

x 1jx- yj5_a cp( lx - yl) l h(y)l dy :::; cd Jor cp(r)k(r)rd- l dr 2 If we replace k( r) by r- P and cp(r) by r- ( d- ) , which holds in d =/: 2, then the above a = cd 1 rl - p dr 0 sup

If we put absolute values inside the integral and strengthen the last result to uniform convergence for x E R then we get a definition that is essentially due to Kato (1973). A function h is said to be in if

d,

J{d

t.l.O x Ex (}0t l h(B,)jds) = 0

-+

when p < 2. In

lim sup

d = 2 when p < 2 we get

0

By Fub1ni's theorem, we can write the above as

t.l.O x J kt(x, y)jh(y)jdy = 0 t kt(x, y) = 1 (2 -;rs) -df 2e - l x - y j 2 f 2• ds lim sup

where

By considering the asymptotic behavior of kt(x, y) as t -+ 0 and we can cast this condition in a more analytical form as

a.l.O x Jflx- yj 5_ a cp(lx - yl)jh(y) j dy = O ( d- ) cp(r) = ;r-log r2 dd 2::= 23 d= 1

lim sup where

Remark. We will have more to say about these spaces at the end of the next

section. For the developments there, we will also need the space J(�ac, which is defined in the obvious way: f E J(�c iffor every R < oo, the spatially truncated function fl (l xi
d·

l x - yj 2jt -+ c,

{

The equivalence of (* ) and ( ** ) is Theorem 4.5 of Aizenman and Simon (1982) or Theorem 3.6 of Chung and Zhao (1995) . Section 3.1 of the latter

4 . 3 . The Feynman-Kac Formula

In this section, we will consider what happens when we add cu to the right-hand side of the heat equation. That is, we will study

= tdu + cu in (0, oo) X Rd . (3.1b) u is continuous at each point of {0} x Rd and u(O, x) = f(x). 12

(3.1a) u E C • and U t

:::;

If c (t , x) 0, then this equation describes heat flow with cooling. That is, u (t, x) gives the temperature at the point x E R at time t, when the heat at x at time t dissipates at the rate - c(t, x) .

d

138

Chapter 4 Partial Differential Equations

Section 4.3 The Feynman-Kac Formula

The first step, as usual, is to find a local martingale.

As before, it is easy to show that if v is smooth enough it is a solution.

(3.2) Theorem. Let c; = J; c(t - r, Br) dr. If u satisfies (3.1a), then

M.

=

u(t - s, B. ) exp(c! )

(3.4) Theorem. Suppose f is bounded and that c is bounded and continuous. If v E C1 • 2 , then it satisfies (3.1a). Proof

is a local martingale on [0, t).

The Markov property implies, see Exercise 2.7 in Chapter 1 and take

h (r, x) = c(t - r, x), that

Proof Applying Ito's formula with X� = t - s , X! xf+l = c; gives that

=

139

.

Ex(f(Bt) exp(c!}IF.) = exp(c! ) EB ( f(Bt - s ) exp(c!:�)) = exp(c�)v(t - s, B. )

B! for 1 :::; i :::; d, and

The left-hand side is a martingale, so the right-hand side is also. If v E C1 • 2 , then repeating the calculation in the proof of (3.2) shows that

1

{

since we have

(Xi, xi ) t = t if 1 :::; i_ = i :::; d 0 otherwise Using de� = c(t - r, Br ) dr and rearranging, the right-hand side is =

1• (-Ut + cu + � .6.u) (t - r, Br) exp(c; ) dr • + 1 exp(c;) 'vu(t - r, Br) dBr ·

which proves (3.2); since -Ut + cu + �.6.u = 0 and the second term is a local martingale. D The next step, again, is a uniqueness result. (3.3) Theorem. Suppose that c is bounded. If there is a solution of (3.1) that is bounded on [O, T] x Rd for any T < oo, it must be v(t, x) = Ex {f(Bt) exp(c! )}

-

Under our assumptions on c and u, M. , 0 :::; s < t, is a bounded mar­ tingale and Aft = lims t M. = f(B1 ) exp(cD. Since Ms is uniformly integrable it follows that D u(t , x ) = ExMo = ExMt = v(t, x)

Proof

v(t - s, B. ) exp(c! ) - v(t, Bo) 1 (-vt + cv + -2 .6.v)(t - r, Br ) exp(c� ) dr = 0 + a local martingale

.

The left-hand side is a local martingale, so the integral on the right-hand side is also. Since the integral is continuous and locally of bounded variation, (3.3) in Chapter 2 implies that it must be = 0 almost surely. Again this implies that ( - Vt + cv + �.6.v)(t - r, Br) exp(c;) = 0 for our assumptions imply this quantity is continuous and if it were =f 0 at some point we would have a contradiction.

D

The next step is to give a condition that guarantees that v satisfies (3.1b ). As before, we begin by considering what happens when everything is bounded. (3.5) Theorem. If c is bounded and f is bounded and continuous, then v satisfies (3.1b). If l ei :::; M, then e - Mt :::; exp(cD :::; eMt , so exp(cD -r 1 as t Letting ll !ll oo = supx lf(x) l this result implies Proof

-r

0.

lEx exp(c! )f(Bt) - Exf(Bt )l :::; llfll oo Ex I exp(c!} - ll -r 0 (1.5) implies that (t, x) -r Exf (Bt) is continuous at each point of {0 } x Rd and D the desired result follows. This brings us to the problem of determining when v is smooth enough to be a solution.

140

Section 4.3 The Feynman-Kac Formula

Chapter 4 Partial Differential Equations

(3.6) Theorem. Suppose that f is bounded and Holder continuous. If c is bounded and Holder continuous locally in t, then v E CI • 2 and, hence, satisfies (3.1a) . To solve the problem in this case, we use a trick to reduce our result to the previous case. We begin by observing that

Proof

!=

c

1" c(t - r, Br) dr

is continuous and locally of bounded variation. So Ito's formula, (10.2) in Chapter 2, ·implies that if h E C1 then

Taking h(x) = e - x we have exp( - c�) - 1 =

- 11 exp(-c�)c(t - s, B, ) ds

Multiplying by - exp(ci) gives exp(c�) - 1 =

11 c(t - s, B, ) exp(c� - c!) ds

Plugging in the definitions of c� and c; we have exp

(11 c(t - r, Br) dr) = 1 + 11 c(t - s, B, ) exp (it c(t - r, Br) dr) ds

Multiplying by f (B1 ) , taking expected values, and using Fubini's theorem, which is justified since everything is bounded, gives

Taking the expected value of the last equation and plugging into the previous one, we have

v(t, x) = Exf(Bt) +

(

Ex c(t - s, B,) exp

(it c(t - r, Br) dr) f(Bt) ' :F,) = c(t - s, B, )v(t - s, B,)

1t Ex{c(t - s, B,)v(t - s, B,)}ds

VI ( t, X ) + V2 ( t, X ) The first term on the right, vi (t, x), is CI • 2 by (1.6). The second term, v2 (t, x), is of the form considered in the last section with g( r, x) = c(r, x)v( r, x ). If we start with the trivial observation that if c and f are bounded, then v is bounded on [O, N] x Rd , and apply (2.6b), we see that l v2 (t, x) - v2 (t, y) l :::; CN lx - v i whenever t :::; N To get a similar estimate for VI let Bt be a Brownian motion starting at 0 and observe that since f is Holder continuous l vi(t, x) - vi(t, Y)l = IE{f(x + Bt) - f(y + Bt)}l :::; Elf(x + Bt ) - f(y + Bt )l :::; Clx - v i a Combining the last two estimates we see that v(r, x) is Holder continuous locally in t. Since c and v are bounded and we have supposed that c(r, x ) is Holder continuous locally in t the triangle inequality implies g(r, x) is Holder continuous locally in t. Using (2.6c) and (2.6d) now, we have v2 (t, x) E CI •2 • D Unbounded c. As in the last section, we can generalize the results above to unbounded c's, but for simplicity we restrict our attention to the temporally homogeneous case c(t, x) = h(x). Given the formula (*) above, which expresses v as a volume potential, it is perhaps not too surprising that the appropriate assumption is c E J(d · The key to working in this generality is what Simon (1982) calls :=

(3.7) Khasminskii's

lemma.

Let g � 0 be a function on Rd with

a := s�p Ex Then

Conditioning on :F, , noticing c(t - s, B, ) E :F, , and using the Markov property as in Exercise 2.7 of Chapter 1,

141

Proof

(1t g(B, ) ds)

<

1

The Markov property and the nonnegativity of g imply that sup Ex

x

f · · · Jfo <•1<···<•n
142

Section 4.4 The Dirichlet Problem

Chapter 4 Partial Ditferen tial Equations

Bt rj. G t. (t - s,B. )

(s1 , . . . , sn) we have 1 ··· 1O<• l<···<• n<1 ds1 . . . dsng(B.1) g(B.J 1 1 . . . 1 ds . .. ds g(B.J . . . g(B.J n =r n. 1o1 =0 1•n =O 1 n = �! (1� / (B. ) ds)

where T = inf{t > 0 : } . To see the similarities to the solutions given for the equations in Part A of this chapter think of those solutions in terms of space-time Brownian motions iJ. = run until time f = inf{s : iJ. (0, oo) x Rd }. Of course, f :::

Since the integrand is symmetric in

•• •

Summing on n now gives the desired formula.

From _the last result, it should be clear why assuming this context. This condition guarantees that and, hence, that

0

h E J(d is natural in

(fo1 l h(B. )lds) -+ 0 t s�p Ex exp (l l h (B. )lds) -+ 1

With these two results in hand, we can proceed with developing the theory much as we did in the case of bounded coefficients. B . Elliptic Equations

In the next three sections of this chapter, we show how Brownian motion can be used to construct classical solutions of the following equations: 1 0= 2 1 0= 0= 1

G u = f.

8G, u

in an open set subject to the boundary condition: at each point of is continuous and We will see that the solutions to these equations are (under suitable as­ sumptions) given by

Exf(Br) r Ex (f(Br) + l g(B. )ds) r Ex (f(Br) exp (fo c(B. )ds))

rj.

4.4. The D irichlet P roble1n

s�p Ex

-b.u 2 .6.u+ g 2 .6.u + cu

143

In this section, we will consider the Dirichlet problem. That is, we will study (4.1a) C2 and = 0 in

uE

b.u

G.

8G, u is continuous and u = f. To see what this means, note that if we let h(t,x) = u(x), then h satisfies the heat equation 8h = �b.h 8t 2 Thus, u is an equilibrium temperature distribution when 8G is held at a fixed temperature profile f. As in the first three sections, the first step in solving (4.1) is to find a local (4.1b) At each point of

martingale.

(4.2) Theorem. Let T = inf{t > 0 B rj. G}. If u satisfies (4.1a), then M1 = u(Bt) is a local martingale on [0, ) 1 :

r .

Proof

Applying Ito's formula gives

u(Bt) - u(Bo) = Jto 'Vu(B. ) · dB. + 2"1 Jto b.u(B.)ds for t < T. This proves (4.2), since .6. u 0 in G and the first term is a local martingale on [0, ) 0 =

r .

The second step is a uniqueness theorem. For this we introduce (t) Px r <; oo) = 1 for all which 4. 7b) below will show is necessary for uniqueness. Note that 1.3) in Chapter 3 implies bounded sets satisfy (t).

( (

(4.3) Theorem. Suppose bounded, it must be

x

(

G

G satisfies (t). If there is a solution of (4.1) that is v(x) Exf(Br) =:

144

Section 4.4 The Dirichlet Problem

Chapter 4 Partial Differential Equations

If u is bounded and satisfies (4.1) then M. , 0 � s < T, is a bounded local martingale. Using (2.7) in Chapter 2, (t), and (4. 1b), we have

Mr := limr M. = f(Br) •t u(x) = E:cMo = E:cMr = v(x)

Xn -+

D

Proof

As in the first three sections, it is easy to show:

� t - c)

y -+ Py (r � t - c) is bounded and measurable and p , (x , y) = (27rc) - df 2 e - l :c - yl2 / 2 £

dy

-+

is continuous for each c > 0. Letting f ! 0 shows that x P:c(r � t) is an increasing limit of continuous functions and, hence, lower semicontinuous. 0

yXn + y , -

If is regular for G and t > 0, then Py( T � t) = 1, so it follows from (4.5a) that if then

The lefl-hand side is a local martingale on (0, r), so v(B. ) is also. If v. E C2 , then repeating the calculation in the proof of (4.2) shows that, for s E [0, r), . 1 ' v(B. ) - v(B0) = -2 0 .6.v(Br) + a local martmgale

dr

The left-hand side is a local martingale on [0, r), so the integral on the right­ hand side is also. However, the integral is continuous and locally of bounded variation, so by (3.3) in Chapter 2 it must be = 0. Since .6.v is continuous in G, it follows that .6.v = 0 in G. For if .6.v f 0 at some point we would have a contradiction. D Up to this point, everything has been the same as in Section 4.1. Differences appear when we consider the boundary condition (4.1b ) , since it is no longer sufficient for f to be bounded and continuous. The open set G must satisfy a regularity condition. A point y E 8G is said to be a regular point if Py (r = 0) = 1.

(4.5) Theorem. Let G be any open set. Suppose f is bounded and continuous and y is a regular point of 8G. If E G and then v(xn ) -+ f ( ) .

Xn -+ y,

r

X -+ P:c(B. E ac for some s E (c, t])

The Markov property implies that on { T > s}

Xn

J p,(x, y)Py (

it follows from the dominated convergence theorem that

(4.4) Theorem. Let G be any open set and suppose f is bounded. If v E C2, then it satisfies (4.1a).

Definition.

By the Markov property

P:c(B. E Gc for some s E (c, t]) = Since

1

-+

(4.5a) Lemma. If t > 0, then x P:c( T � t) is lower semicontinuous. That is, if x, then l ��i�f P:c n (r � t) 2: P:c(T � t)

(t) is not needed for the existence of solutions, but to drop (t) we need to modify the definition to take the possibility of T = oo into account. Let

Proof

The first step is to show

Proof

Proof

145

y

l��i�f P:c n (T � t) 2: 1 for all t > 0

With this established, it is easy to complete the proof. Since f is bounded and continuous, it suffices to show that if = {x : l x - Y l < 6}, then .

D(y, li) (4.5b) Lemma. If y is regular for G and Xn -+ y , then for all P:cJr < oo, Br E D(y, li)) -+ 1

Proof

Let c > 0 and pick t so small that

(

{j

)

Po sup lB. I > -26 < f O �s�t

>0

-+ 1 as Xn -+ y, it follows from the choices above that liminf sup lB. - xn l � i2 ) n -oo P:cn (r < oo, Br E D(y, li)) 2: liminf n -oo P:cn (r � t, O�s� t 2: liminf I Bd > i2 ) > 1 n- oo P:cJr � t) - Po (osup �s�t

Since P:c n (T � t)

<

.! !

146

Chapter 4 Partial Differential Equations

(

Section 4.4 The Dirichlet Problem

(

0

Since f was arbitrary, this proves 4.5b) and, hence, 4.5).

(4.5) shows that if every point of 8G is regular, then v(x) will satisfy the boundary condition (4.1b) for any bounded continuous The next two exercises develop a converse to this result. The first identifies a trivial case.

f.

Exercise 4.1.

If G C R is open then each point of 8G is regular.

Exercise 4.2. Let G be an open set in Rd with d 2::: 2 and let y E 8G have Py(T = 0) < 1. Let f be a continuous function with f(y) = 1 and f(z) < 1 for all z :f y . Show that there is a sequence of points Xn --;. y such that liminfn .... v(xn) < 1. From the discussion above, we see that for v to satisfy (4.1b) it is sufficient (and almost necessary) that each point of 8G is a regular point. This situation oo

raises two questions: Do irregular points exist? What are sufficient conditions for a point to be regular? In order to answer the first question we will give two examples.

> B2 1 l x l < 3, To 3 3 Example 4.2. Lebesgue's Thorn. Let d 2::: 3 and let

DPo(To D

Let d 2::: and let G = {0}, where = oo) = 1 D = {x : 1}. If we let = inf{t 0 : = 0}, then by (1.10) in Chapter so 0 is not a regular point of 8D. One can get an example with a bigger boundary in d 2::: by looking at G = - J{ where J( = {x : x 1 = x 2 = 0}. In d = this is a ball minus a line through the origin. Example 4.1. Punctured Disc.

where the n = oo term is the single point {0}.

an ! 0 sufficiently fast, then 0 is not a regular point of 8G. Proof P0 ((B£, B�) = (0, 0) for some t > 0) = 0, so with probability 1, a Brownian motion B1 starting at 0 will not hit In = { X 1 E [2- n , 2- n +l] , X2 - - - Xd - O} Since B1 is transient in d 2::: 3 it follows that for a.e. the distance between {B. 0 � s < oo} and In is positive. From the last observation, it follows Claim. If

X

:

:

_

Xa

_

• • •

-

w

-

147

n , 2- n+l] [an n , an]d- l } and pick [2 B Tn 1 n - -1 1/2, >Po(TnB1 103: B1=rJ.3{-1,(3/2)1)d},= then 00 Po(T < ) � nI:=l Po(Tn < oo) � 21

immediately that if we let = inf{t E small enough, then Now oo) � so if we let T = inf{t 0 : G} and = inf{t we have :

an

X

O"

O"

Thus

Po( T > 0) 2::: Po(T =

O"

) 2:::

1/2 and 0 is an irregular point.

0

y, y y (4.5c) Cone Condition. Ifthen there is a cone V having vertex y and an r > 0 such that V D(y, r) y is a regular point. Proof The first thing to do is to define a cone with vertex y, pointing in direction v , with opening a as follows: V(y, v, a) = {x : x = y+ B(v + ) where 0 E (O, oo), z ..l v , and l z l < a}

The last two examples show that if ac is too small near y, then may be irregular. The next result shows that if a c is not too small near then is regular. c ac ,

n

z

Now that we have defined a cone, the rest is easy. Since the normal distribution is spherically symmetric,

Py(Bt E V(y, v, a)) = fa > 0

fa is a constant that depends only on the opening a. Let r > 0 be such V(y, v, a) D(y, r) C The continuity of Brownian paths implies tlim-o Py (sup • ::S t jB. - yj > r) = 0

where that

ac .

n

Combining the last two results with a trivial inequality we have

-< liminf t !O Py(Bt E Gc) -< limPy t !O {T -< t) -< Py(T = 0) and it follows from Blumenthal's zero-one law that Py( T = 0) = 1. fa

0

(4.5c) is sufficient for most cases.

{

<

Let G = x : g( x) 0} where g is a C1 function with \l for each E 8G. Show that each point of 8G is regular. Exercise 4.3.

y

g(y) :f 0

148

Chapter 4 Partial Differential Equations

Section 4.4 The Dirichlet Problem

!Y - xl� = l::;(Yi - x;)2

V(y,

d-

{y + V(y,E R}.)

d=

:

v, a

g(x) = 1D(0,6) 'r/J( Izl2)h(x + z) dz = c 1o 6 drrd- 1 'rf; (r2 ) JaD(O,r) r h(x + z) 7r(dz) = C J.' dr r'- 1 .P(r2 ) h(x) So h is a constant times g and hence coo . To conclude now that Llh = 0 we note that the multivariate version of Taylor's theorem implies that if I Y - xl :::; r then h(y) = h(x) + L(Y; - x;)D;h(x) i + Lij (Y; - x; )(Yi - Xj)D;jh(x) + <.(y, x) where l c (y! x)l :::; C3r3. Integrating over D(x, r) w.r.t. dyf i D (O, r)l and using the averagmg property we have h(x) = h(x) + 0 + C2Llh(x)r2 + O(r3) Subtracting h( x) from each side, dividing by r2 , and letting r 0 it follows that Llh(:i: ) = 0. 0

z

When 2 this says that if we can find a line segment ending at that lies in the complement then is regular. For example this implies that each point of 2: the boundary of the slit disc G 0 } is regular.

z

= D - {x : x 1 0, x2 =

(

)

E coo and,

x E G and pick 6 > 0 so that D(x, 6) C G. If we let = inf{ t : Bt rf. D(x, 6)}, then the strong Markov property implies that (for more details see Example 3.2 in Chapter 1 Let

Proof

u

)

= Ex(ii(B17)) = JaD(x,6) { v(y)1r(dy) where 1r is surface measure on D(x, 6) normalized to be a probability measure. The last result is the "averaging property" of harmonic functions. Now a simple analytical argument takes over to show E coo. (4.6a) Lemma. Let G be an open set and h be a bounded function which has the averaging property - ii( x) = Ex (f( Br) 1 (r< oo ))

Unbounded G. As in the previous three sections, our last topic is to . . discuss what happens when something becomes unbounded. This time we will focus on G and ignore f. Combining (4.3), (4.5), and (4.6) we have:

r h(y) 7r (dy) h(x) = JaD(x,li) when x E G and 6 < 6x where 6x > 0. Then h E C00 and Llh = 0 in G. (62 ,

on that

)

--+

ii

Proof

( )

(

Having completed our discussion of the boundary condition, we now turn our attention to determining when ii is smooth. As in Section 4.1, this is true under minimal assumptions on f. 4.6) Theorem. Let G be any open set. If f is bounded, then ii hence, satisfies (4. 1 a .

I Y - xl

Note that we use rather than is not smooth at ? · Makmg a �1mple change of variables, then looking which at things in polar coordmates, and usmg the averaging property we have

However, in a pinch the following generalization can be useful. Exercise 4.4. Define a flat cone v, a ) to be the intersection of with a 1 dimensional hyperplane that contains the line Bv (} Show that (4.5c) remains true if "cone" is replaced by "flat cone."

149

)

(4. 7 a Theorem. Suppose that f is bounded and continuous and that each point of aG is regular. If for all G, Px( T < 00 ) 1, then v is the unique bounded solution of (4.1).

XE

=

To see that there might be other unbounded solutions consider G f(O ) is a solution. Conversely, we have and note that

= 0,

Let 'ljJ be a nonnegative infinitely differentiable function that vanishes oo ) but is not = By repeatedly applying (1.7) it is routine to show

0. g(x) = 1D(x,6) 'rfJ( iY - x i2)h(y) dy E coo

u(x) = ex

= (0, ) , 00

(4.7b ) Theorem. Suppose that f is bounded and continuous and that each point of aG is regular. If for some G, Px ( T < 00 ) < 1, then the solution of (4.1 ) is not unique.

XE

1

150

Chapter 4 Partial Differential Equations

Section 4.5 Poisson 's Equation

h(x).t1h P:c(T 0

Proof Since = = oo ) has the averaging property given in (4.6a), it is and has = in G. Since each p oint y E 8G is regular, a trivial comparison and (4.5a) implies

coo

Py(r > 1) = 0 The last two observations show that h is a solution of (4.1) with f 0, which 0 completes the proof. By working a little harder, we can show that adding aP:c(T = oo ) to v(x) is the only way to produce new bounded solutions. limsup P:z: (r = oo )

:::; limsup P:z:(r > 1) :::;

=

(4.7c) Theorem. Suppose that f is bounded and continuous and that each point of oG is regular. If is bounded and satisfies ( 4.1) in G, then there is a constant such that

u

a

We will warm up for this by proving the following sp ecial case in which G = Rd .

u is constant. Proof (4.2) above and (2.6) in Chapter 2 imply that u(Bt ) is a bounded oo, martingale. So the martingale convergence theorem implies that as t is measurable with respect to the tail CT-field, it follows ) -;. U u(B • Since U oo t 00 from (2.12) in Chapter 1 that P:c(a < Uoo < b) is either 0 or 1 for any a < b. The last result implies that there is a constant c independent of x so that P:c(U = c) oo :: 1. Taking expected values it follows that u(x) = E:cUoo c. 0 Proof of ( 4. 7 ) From the proof of (4. 7d) we see that u( Bt ) is a bounded local martingale on [0, r) so U,. = limtt-r u(Bt ) · On {r < oo } , we have U,. = f(B,.) so what we need to show is that there is a constant independent of the starting point Bo so that U,. = a on { T = oo } . Intuitively, this is a consequence of the triviality of the asymptotic CT-field, but the fact that 0 < P:c( T = oo ) < 1 makes it difficult to extract this from (2. 12) in Chapter 1. To get around the difficulty identified in the previous paragraph, we will extend u to the whole space. The two steps in doing this are to (a) Let h(x) = u(x) - E:c( f (B,.) ; r < oo) . (4.6) and (4.5) imply that h is bounded and satisfies (4.1) with boundary function f 0. (b) Let M = l h l oo and look at w(x) = { �(x) + MP:c(T = oo) xX EE GGc (4.7d) Theorem. If u is bounded and harmonic in Rd then

_,.

=

=

=

c

a

=

151

f3P:c( r = oo) , (i) When restricted to G, w satisfies (4. 1) with boundary function f :: 0. The proof of (4.7b) implies that M P:c( T < oo ) satisfies (4.1) with boundary function f 0. Combining this with (a) the desired result follows. (ii) w ;::: 0. To do this we use the optional stopping theorem on the martingale h(Bt ) at time T 1\ t and note h(B,.) = 0 to get h(x) = E:c(h(Bt) ;r > t) ;?: -MP:c(r > t) Letting t -;. oo proves h(x) ;?: -MP:c(T = oo ) . (iii) w(Bt ) is a submartingale. Because of the Markov property it is enough to show that for all x and t we have w(x) :::; E:cw(Bt ) . Since w ;?: 0 this is trivial if x rJ. G. To prove this for x E G, we note that (i) implies = w(Bt) is a bounded local martingale on [O,r) and W,. = 0 on {r < oo} , so using the optional stopping theorem To complete the proof now, we will show in four steps that w(x) = from which the desired result follows immediately.

=

Wi

f3 so that w(x) = f3P:c(r = oo) . Since w is a bounded submartingale it follows that as t oo, w(Bt ) converges to a limit Woo . The argument in (4.7d) implies there is a constant (3 so that P:c(Woo = (3) = 1 for all x. Letting t oo in w(x) = E:c(w(Bt );r > t) (iv) There is a constant

-

_,.

arid using the bounded convergence gives (iv).

0

4. 5. Poisso n's Equation

In this section, we will see what happens when we add a function of equation considered in the last section. That is, we will study: (5.1a)

u E C2 and �.t1u =

-g

(5. 1b) At each point of oG,

x to the

in G.

u is continuous and = 0. u

0

As in Section 4.2, we can add a solution of (4.1) to replace u = in (5.1b) by = . As always, the first step in solving (5.1) is to find a local martingale.

u f

152

T

Chapter 4 Partial Differential Equations

(5.2) Theorem. Let

Proof

[0, T).

153

The left-hand 2 side is a local martingale on [0, T), so the right-hand side is also. If v E C , then repeating the calculation in the proof of (5.2) shows that for

T = inf{t > 0 : B1 rf. G} . If u satisfies (5.1a) , then t Mt = u(Bt) + g(B.) ds

is a local martingale on

Section 4.5 Poisson 's Equation

i

s E [O, T),

v(B.) - v(Bo) +

Applying Ito's formula as we did in the last section gives

u(Bt) - u(Bo) = Jt \lu(B.) · dB. + 21 Jt 6.u(B.)ds o o for t < T: This proves (5.2) , since t6.u = -g and the first term on the right­ hand side is a local martingale on (0, T). 0 The next step is to prove a uniqueness result. (5.3) Theorem. Suppose that G and g are bounded. If there is a solution of (5.1) that is bounded , it must be

1• g(Br) dr = 1• (� 6.v + g) (Br) dr

+ a local martingale The left-hand side is a local martingale on (0, T), so the integral on the right­

hand side is also. However, the integral is continuous and locally of bounded variation, so by (3.3) in Chapter 2 it must be = 0. Since �6.v + g is continuous in G, it follows that it is = 0 in G, for if it were :f:. at some point then we would have a contradiction. 0

0

After the extensive discussion in the last section, the conditions needed to guarantee that the boundary conditions hold should come as no surprise. (5.5) Theorem. Suppose that G and g are bounded. Let of 8G. If X n E G and X n ..__. y, then v(x n ) ..__. 0.

y

be a regular point

We begin by observing: (i) It follows from {4.5a) that if f > 0, then Px n ( T > f ) ..__. 0. (ii) If G is bounded, then (1 .3) in Chapter 3 implies C = supx Ex T < oo and, hence, ll v ll oo :s:; C II Y II oo < oo. Let e: > 0. Beginning with some elementary inequalities then using the Markov property we have rA £ Proof

u satisfies (5. 1a) then M1 defined in (5.2) is a local martingale on (0, T). If G is bounded, then (1.3) in Chapter 3 implies Ex T < oo for all x E G. If u and g are bounded then for t < T Proof - If

Since the right-hand side is integrable, (2.7) in Chapter 2 and (5.1b) imply r = Mr := lim Mt g(Bt) dt t Tr 0

1

u(x) = ExMo = Ex (M;) = v(x)

i v(x n ) I :s:; Ex n

0

As usual , it is easy to show (5.4) Theorem. Suppose that G is bounded and then it satisfies (5.1a). Proof

g

is continuous. If v

The Markov property implies that on { T > s } ,

E C2,

(i

)

(l iT

l )

lg(B. ) I ds + Ex n g(B.) ds ; T > e: :s:; t: I I Y II oo + Ex n (lv(Bf )l ; T > e:) :s:; e:II Y II oo + ll v ll oo Px n (T > e:)

Letting n ..__. oo, and using (i) and (ii) proves (5.5) since f is arbitrary.

0

Last, but not least, we come to the question of smoothness. For these developments we will assume that g is defined on Rd not just in G and has compact support. Recall we are supposing G is bounded and notice that the values of g on ac are irrelevant for (5.1), so there will be no loss of generality if we later want to suppose that J g(x) dx = 0. We begin with the case d ;;?: 3, because in this case (2.2) in Chapter 3 implies

w(x) = Ex

loa ig(Bt)i

dt < oo

154

Section 4.5 Poisson's Equation

Chapter 4 Partial Differential Equations

As in Section 4.2, trouble starts when we consider second derivatives. If

and, moreover, is a bounded function of x. This means we can define

w(x) = Ex

155

i =!= j, then

100 g(Bt) dt

use the strong Markov property to conclude

In this case, the estimate used above leads to

and change notation to get

which is (just barely) not locally integrable. As in Section 4.2, if g is Holder continuous of order a, we can get an extra lx - Y l a to save the day. The details are tedious, so we will content ourselves to state the result.

x) = w( x) - Ex w( BT ) (4.6) tells us that the second term is C00 in G, so to verify that v E C2 we need only prove that w is, a task that is made simple by the fact that (2.4) and (*)

v(

(2.3) in Chapter 3 gives the following explicit formula

j

w(x) = Ca l x - yf - dg(y) dy The first derivative is easy.

J{g;tO} I

dy yl d- 1

X -

w is C1 and

w(x) =

( ) (�(xi - Yi f)-d/2 2(x; - y;)

X

dy < oo - Y l d- 1

{

-� log ( l x - yl) d = 2 G(x, y) = -l x - yl d=1

where the see that

-y)

J I(I X - y li·d g(y) I dy $ llu lloo J{g;tO} I

D

j G(x, y)g(y) dy

G is the potential kernel defined in {2.7) of Chapter 3 , that is,

J

The integral on the right-hand side is convergent since .

where

G was defined as

So differentiating under the integral sign

I

>

The last result settles the question of smoothness in d 3. To extend the result to d $ 2, we need to find a substitute for ( * ) . To do this, we let

< 00

\Ve will content ourselves to show that the expression we get by differ­ entiating under the integral sign converges and leave it to the reader to apply (1.7) to make the argument rigorous. Now

X

The reader can find a proof either in Port and Stone {1978), pages 116-1 17, or in Gilbarg and Trudinger { 1977) , pages 53-55. Combining (*) with (5 .6b) gives {5.6) Theorem. Suppose that G is bounded. If g is Holder continuous, then E C2 and hence satisfies (5.1a).

Proof

2-d D; lx - Y l 2- d = 2-

w is C2 •

v

(5.6a) Theorem. If g is bounded and has compact support, then there is a constant C which only depends on d so that

I D;w(x)l $ Cll u ll oo

(5.6b) Theorem. If g is Holder continuous, then

100 {Pt(x, y) - at} dt

at were chosen to make the integral converge. So if J g dx = 0, we

J G(x, y)g(y)dy

{T

lim Ex g(Bt)dt T-+oo Jo Using this interpretation of w, we can easily show that ( *) holds, so again our problem is reduced to proving that w is C2, which is a problem in calculus. Once all the computations are done, we find that (5.6) holds in d $ 2 and that in d = 1 , it is sufficient to assume that g is continuous. The reader can find =

156

Chapter 4 Partial Differential Equations

details for d = 1 in (4.5) of Chapter 6, and for cited above.

d � 2 in either of the sources

Section 4.6 The Schrodinger Equation

B

The general solution is A cos bx + sin bx, where b = V'Fj. So if we want the boundary condition to be satisfied we must have

B

1 = A cos ba + sin ba 1 = A cos ( -ba) + sin ( -ba ) = A cos ba -

4 . 6 . The S chrodinger Equation

cu

In this section, we will consider what happens when we add to the left-hand side of the equation considered in Section 4.4. That is, we will study

( 6.1a) u E C2 and k.D.. u + cu = 0 in G. {6.1b ) At each point of 8G, u is continuous and u = f. As always, the first step in solving { 6.1 ) is to find a local martingale. ( 6.2) Theorem. Let r = inf{t > 0 : B1 rf. G}. If u satisfies (6.1a) then

B sin ba

Adding the two equations and then subtracting them it follows that

From this we see that solve for A.

B

=

0 = 2B sin ba

0 always works and we may or may not be able to

If cos ba = 0 then there is no solution. If cos ba =f. 0 then x = cos bx / cos ba is a solution.

u( )

We will see later (in Example 9.1 ) that if ab < rr/2 then v ( x ) = cos bx f cos ba

(0, r) .

c1 = J; c(B,)ds. Applying Ito's formula gives u(B1-) exp (c1) - u(Bo) = 11 exp (c,)V'u(B,) · dB, + 11 u(B,) exp (c, ) de, t + -1 1 .D.. u (B,)exp (c,)ds 2 for t r. This proves ( 6.2) , since dc5 c(Bs ) ds, k.D.. u + cu 0, and the first term on the right-hand side is a local martingale on (0, r) . Proof

B

2 = 2A cos ba

M1 = u(B1 ) exp (11 c(B.)ds)

is a local martingale on

157

Let

However this cannot possibly hold for ab > rr/ 2 since v ( x ) � hand side is < 0 for some values of x.

0 while the right­

0

We will see below ( again in Example 9.1 ) that the trouble with the last example is that if ab > rr /2 then = 'Y is too large, or to be precise, if we let

c

0

=

<

=

D

At this point, the reader might expect that the next step, as it has been five times before, is to assume that everything is bounded and conclude that if there is a solution of ( 6.1 ) that is bounded, it must be v ( x)

=

Ex(f(BT) exp(cT ))

co in ( -a, a ) . The rest of this section is devoted to showing that if then "everything is fine." The development will require several stages. The first step is to show

then

w =f:.

w

co,

=

( 6.3a) Lemma. Let () > 0. There is a f1 > 0 so that if H is an open set with Lebesgue measure !H I :::; f1 and TH = inf{t > 0 : tJ. H} then sup

We will not do this, however, because the following simple example shows that this result is false. Example 6.1. Let

considering is

d = 1, G = (-a, a) , c and f 1. The equation we are 1 -u" + -yu = 0 u(a) = u(-a) = 1 2 = 'Y,

=

X

Proof Pick

'Y

Ex (exp ( )) B rH

> 0 so that e6'Y :::; 4/ 3. Clearly,

B1

:::;

2

158

Chapter 4 Partial Differential Equations

Section 4.6 The Schrodinger Equation

if we pick p so that pf(27r'Y ) df 2 � 1/ 4. Using the Markov property as in the proof of (1.2) in Chapter 3 we can conclude that

> >

>

So it follows by induction that for all integers k � 0 we have

00

I:: exp(B'Yk)Pz ((k - 1)1 TH � k1) 00

<

( 4 )k

00

4 1 4 1 =2 = 3 3 4k - 1 = 3 1 k 1 13 k= 1 3 Careful readers may have noticed that we left TH = 0 out of the expected value. However, by the Blumenthal 0-1 law either Pz(TH = 0) = 0 in which case our computation is valid, or Pz( TH = 0) = 1 in which case Ez exp (B rH ) = 1. 0 �

L k= 1

1

L

·

_

Let c* = supz l c(x) l . By (6.3a) , we can pick ro so small that if Tr = inf{t : ! Bt - Bol r} and r � ro , then Ez exp (c*Tr) � 2 for all x.

>

(6.3b) Lemma. Let

26 � r0 • If D(x, 26) C G and y E D(x, 6), then w(y) � 2d+2 w(x)

The reason for our interest in this is that it shows w(x) for y E D(x, 6). Proof

If D(y, r) C G, and r �

< oo implies w(y) < oo

o then the strong Markov property implies

r

w(y ) = Ey[exp (cTr) w(B (Tr))] � Ey[exp (c*Tr) w(B (Tr))] = Ey[exp ( c*Tr)] w(z ) 1r(dz ) w(z ) 1r(dz ) � 2 laD(y,r) laD(y,r) where is surface measure on 8D ( y, r) normalized to be a probability measure, since the exit time, Tr , and exit location, B (Tr ) , are independent.

f

1r

f

rra is the surface area of {x : l x l = 1}. Rearranging we have od w(z ) dz � 2 - 1 c w(y) jD(y,6) o where Co = d/ua is a constant that depends only on d. Repeating the first argument in the proof with y = x and using the fact that CTr � -c*Tr gives w(x) = E:�:[exp(cTr) w( B (Tr))] � E:�:[exp(-c*Tr) w( B (Tr))] w(z ) 1r(dz ) = E:�:[exp(-c*Tr)] JaD(x,r) Since 1 / x is convex, Jensen 's inequality implies where

r

Since ez is increasing, and e 8'Y � 4/3 by assumption, we have

k=1

rd- l and

>

Pz(TH k1) = E:�:(Pn.,(TH ( k - 1)1) ; TH 1) 1 � sup Py (TH ( k - 1) 1) 4 y

E:�: exp (BrH ) �

If 6 � ro and D ( y, 6) C G, multiplying the last inequality by integrating from 0 to 6 gives

159

f

E:�:[exp(-c*Tr)] � 1/ Ez [exp (c*Tr)] � 1 / 2 Combining the last two displays, multiplying by rd- l , and integrating from 0 to 26 we get the lower bound 1 (26) d -d-w(x) � 2 - 1 a z w(z ) dz rr D( ,26) Rearranging and using D ( x, 26) ::) D(y, 6), w � 0 we have ·

-

j

� r

w(x) � 2 - 1 ( o) d w(z ) dz }D(y,6) where again C0 = dfua. Combining (**) and (*) it follows that

� r

w(x) � 2 - 1 ( o) d w(z ) dz jD(y,6) co_ . 2 - 1 sa w(y) = 2- (d+2) w(y) 1_ 2 - (2fi) d Co

>

(6.3b) and a simple covering argument lead to

0

160

Chapter 4 Partial Differential Equations

(6.3c) Theorem. Let

Section 4.6 The Schrodinger Equation

G be a connected open set. If w :j. oo then

(A3) w :j. oo .

w(x) < oo for all x E G

(6.3) Theorem.

Proof From (6.3b ), we see that if w(x) < oo, 2o :::; ro, and D(x, 2o) C G, the:n w < oo on D(x, o). From this result, it follows that G0 = {x : w(x) < oo } 1s

an open subset of G. To argue now that Go is also closed (when considered as a subset of G) we observe that if 2o < ro , D(y, 3o) C G, and we have Xn E Go with Xn ---> y E G then for n sufficiently large, y E D(xn , 5) and D(xn , 2o) C G, 0 so w(y) < oo. Before we proceed to the uniqueness result, we want to strengthen the last conclusion. (6.3d) Theorem. Let G be a connected open set with finite Lebesgue measure, I G I < oo. If w :j. oo then sup w(x) < oo X

C G be compact so that I G - K l < f1. the constant in (6.3a) for c* . For each x E [( we can pick a Ox so that 2ox :::; ro and D(x, 2ox) C G. The open sets D(x , ox) cover I< so there is a finite subcover D(x;, ox.), 1 :::; i :::; I. ClearlY-, sup w(x;) < oo Proof Let I< 0

=

If there is a solution of (6.1) that is bounded , it must be

Proof (6.2) implies that M3 = u(B, ) exp(c3 ) is a local martingale on [O, r) . Since J, c, and u are bounded, letting s j r A t and using the bounded conver­ gence theorem gives

u(x) = Ex (f(Br ) exp(cr ); r :::; t) + Ex ( u(Bt) exp(ct); r > t) Since f is bounded and w(x ) = Ex exp(cr ) < oo, the dominated convergence theorem implies that as t ---> oo, the first term converges to Ex (f(Br ) exp(cr )). To show that the second term ---> 0, we begin with the observation that since {r. > t} E :F1 , the definition of conditional expectation and the Markov property imply Ex(u(Bt) exp(cr ) ; r >

l :::=; i $1

w(y) 2: exp(-c* )Py (r :::; 1) 2: f > 0 The first inequality is trivial. The last two follow easily from (A1). See the first display in the proof of (1.2) in Chapter 3. Replacing w(B1 ) by f,

M = sup w(y) < oo yEK

w(y)

[( ,

then Ey (exp(c* rH)) :::; 2 by (6.3a) so using the strong

Ey (exp(crH ); TH = r) + Ey (exp(crH )w(BrH); BrH :::; 2 + MEy (exp(crH ) ; BrH E K) :::; 2 + 2.1\1

=

E K) 0

With (6.3d) established, we are now m?re than ready to prove our �nique­ ness result. To simplify the statements of the results that follow we w1ll now list the assumptions that we will make for the rest of the section. (A1)

G is a bounded connected open set.

(A2) f and c are bounded and continuous.

t) = Ex (Ex( u(Bt) exp(cr )1Ft); r > t) = Ex (u(Bt) exp(ct)w(Bt) ; r > t)

Now we claim that for all y E G

(6.3b) implies w(y) :::; 2 d+ 2 w(x;) for y E D(x;, ox; ) , so

If y E H = G Markov property

161

Ex (iu(Bt)l exp(ct ); r > t) :::; f - l Ex (iu(Bt) l exp(cr ); T > t) :::; f - 1 llullooEx (exp(cr ) ; r > t) ---> 0 as t ---> oo, by the dominated convergence theorem since w( x) = Ex exp(Cr ) < oo and Px( r < oo ) = 1. Going back to the first equation in the proof, we have shown u(x) = v ( x ) and the proof is complete. 0 This completes our consideration of uniqueness. The next stage in our program, fortunately, is as easy as it always has been. Recall that here and in what follows we are assuming (A1)-(A3). (6.4) Theorem. If v

E C2 , then it satisfies (6. 1a) in G.

162 Proof

T

Chapter 4 Partial Differential Equations The Markov property implies that on

{ r > s},

Ex(exp(cr )f(Br )!F.) = exp(c,)EB. (exp(cr )f(Br )) = exp(c, )v(B, ) The left-hand side is a local martingale on [0, r), so the right-hand side is also. If v E C2 , then repeating the calculation in the proof of ( 6. 2) shows that for

s E [O, r),

v(B,) exp(c,) - v(Bo) =

1' (�

) (Br ) exp (cr )dr

�v + cv

163

to the previous case. We begin with the identity established there, which holds for all t and Brownian paths w and, hence, holds when t = r(w)

Multiplying by f(Br ) and taking expected values gives

Conditioning on F, and using the Markov property, we can write the above as

+ a local martingale

The left-hand side is a local martingale on [0, r), so the integral on the right­ hand side is also. However, the integral is continuous and locally of bounded variation, so by ( 3.3 ) in Chapter 2 it must be = 0. Since v E C2 and c is continuous, it follows that t �v + cv = 0, for if it were ::f. 0 at some point then we would have a contradiction. D Having proved ( 6.4 ) , the next step is to consider the boundary condition. As in the last two sections, we need the boundary to be regular.

( 6.5) Theorem. v satisfies (6.1b ) at each regular point of aG.

Proof Let y be a regular point of aG. We showed in (4.5a) and (4.5b) that if Xn -+ y, then PxJr � 6) -+ 1 and Px n (Br E D ( y , 6 )) -+ 1 for all 6 > 0. Since c is bounded and f is bounded and continuous, the bounded convergence theorem implies that

To control the contribution from the rest of the space, we observe that if lei � M then using the Markov property and the boundedness of w established in ( 6.3d ) we have

Ex n (exp(cr )f(Br ); r > 1 ) � e M IIfll oo Ex n (w(Bi ); r > 1 ) � e M IIfll oo llwll oo PxJr > 1 ) -+ 0

Section 4.6 The SclJrodinger Equation

D

This brings us finally to the problem of determining when v is smooth enough to be a solution. We use the same trick used in Section 4.3 to reduce

100

Ex(c(B, )v(B,); T > s) ds v(x) = Exf(Br ) + = v 1 (x) + v (x) 2 The first term, v 1 (x), is coo by (4.6 ) . The second term is so if we let g( x) = c( x ) v ( x) then we can apply results from the last section. If c and f are bounded and w ;j:. oo, then v is bounded by ( 6.3d) , so it follows from results in the last section that v2 is C 1 and has a bounded derivative. Since v1 E coo and G is bounded, it follows that v is C1 and has a bounded derivative. If c is Holder continuous, then g(x) = c (x) v (x) is Holder continuous, and we can use ( 5.6b ) from the last section to conclude v2 E C2 and hence

(6.6 ) Theorem. If in addition to (A1 ) -(A3 ) , c is Holder continuous, then v E C2 and, hence, satisfies (6.1a) . C. Applicat ions t o Brownian Mot ion

In the next three sections we will use the p.d.e. results proved in the last three to derive some formulas for Brownian motion. The first two sections are closely related but the third can be read independently.

164

Chapter 4 Partial Differential Equations

T

4. 7. Exit D istributions for the B all

In4.4this section, we will use resultsforforD the= {xDirichlet provedresultin Section to find the exit distributions : lxl < 1}.problem Our main is (7.1) Theorem. Iff is bounded and measurable, then Exf(Br) = lanf �x--lxY:: f(y) 1r(dy) where 7r is surface measure on an normalized to be a probability measure. Anoo itapplication ofbounded the monotone class f.theorem shows thatweifcan(*) prove holds forProof f E c is valid for measurable In view of ( 4.3), (*) for f E C00 by showing that if ky(x) = (1 - l x l2)/lx - Y ld and v(x) = { I<�ty(x)f(y)7r(dy) � � �D then v solves the Dirichlet problem (4.1): (7.2a) In D, v E C2 and .6.v = 0. (7.2b) At each point of aD, v is continuous and v(x) = f(x). The first, somewhat painful, step in doing this is to show (7.3) Lemma. If y E aD then .6.ky = 0 in D. Proof To warm up for this, we observe so taking = -d we have D;ky(x) = (-2x; ) · I x -1 y l d + (1 - lx l 2) · -Ixd(x;- y-l d+y?-; ) Differentiating again and using our fact with = -(d + 2) gives D;;ky(x) = (-2) · lx -1 y ld + 2 · (-2x;) · -Ixd(x;- y-l d+y-?; ) - y;)2 - d d } + (1 - lx l2) { d(d +lx 2)(x; d - Yl +4 lx - Y l +2 p

p

Section 4. 7 Exit Distributions for the Ball

165

Summing the last expression on i gives .6.ky(x) = Ix--2dy ld + 4d . lIxxl 2--ylxd+· 2y + (1 - lxl2). { lxd2-+Yl2dd+2 - lx -d2Yld+2 } = -lx2dl-xY-l d+yl22 + 4d(llxx-l 2 Y-ldx+2· y) + �-( 1.,.-l x-_-!.l xY.L,-.,.1 2l!::-d)+·-:- 22_:- d If we replace the 1 by IY I 2 , the expression collapses .6.ky(x) = l x 2dY ld+2 (-l x - Yl2 + 2l x l 2 - 2x · Y + I YI2 - lxl2) = 0 since lx - Y l2 = (x - y) · (x - y) = l x l2 - 2x · y + I YI 2 . , v is a linear combination of the ky . So bringing the open settheDintegral Inside the under (and leaving it to the reader to justify this rentiation e diff using (1.7)) gives .6.v(x) = lanr 7r(dy) f(y) .6.ky (x) = 0 Thus, v satisfies (7.2a). To check (7.2b), the first step is to show _

I

approach toI(O)use=a1,softI isnoncomputational but we prefer "observing is justWe"calculus, This rotations, under invariant that by begin instead. 0 in D. (For the last conclusion apply the result for .6.v with f = 1. ) = .6.I and : IBt l > r}. inf{t = let <now1, and r = lxl now, let x3 EwithD with ToApplying conclude(1.I2)=of1 Chapter shows r) G = D(O, I(O) = Eoi(Br) = I( x) whereTotheshowsecond followsas from X --+ yinvariance E an, weunder observerotations. that if z =f:. y, that equality v(x) --+ f(y) kz (x) = 1l x--l xz lld2 --+ 1 y -0 z l d as x -+ y r

166

Chapter 4 Partial Differential Equations

From the last calculation it is clear that if 8 > 0, then the convergence is uniform for z E B0( 8) = 8D - D(y, 8 ). Thus, if we let B1 (8) = 8D - B0(8), then

r

Jno(o) since I(x) =

kz (x)?T(dz) -+ 0 and

1. To prove that v(x) -+ f(y) now we observe

lv(x) - f(y) l = �

I f kz (x)f(z)?T(dz) - f(y) J kz (x)1r(dz) l

2 llfll oo

r

Jn0 ( 5 )

kz (x)?T( dz) + sup lf(z) - f(y)l zE B1(o)

T

I

I

Section 4.8 Occupation Times for the Ball 4.8. Occupation Times for the B all

In the last section we considered how B1 leaves D = {x : l x l < 1 } . In this section we will investigate how it spends its time before it leaves. Let T = inf {t : Bt fl. D} and let G(x, y) be the potential kernel defined in (2.7) of Chapter 3. That is,

cd = f(d/2 - 1}/21Td/ 2 • (8.1) Theorem. If g is bounded and measurable then

where

The first term -+ 0 as x -+ y, while the sup in the second is small if 8 is. This shows that (7 .2b) holds and completes the proof of (7 .1 ) . 0 The derivation given above is a little unsatisfying, since it starts with the answer and then verifies it, but it is simpler than messing around with Kelvin's transformations (see pages 100-103 in Port and Stone (1978)). It also has the merit of explaining why ky (x) is the probability density of exiting at y: ky is a nonnegative harmonic function that has ky(O) = 1 and ky(x) -+ 0 when x -+ z E 8D and z :f. y. Exercise 7.1. The point of this exercise is to apply the reasoning of this section to T = inf{ t : Bt fl. H} where H = {(x, y) : X E nd- I , y > 0 } . For (} E nd- 1 let

ho(x, y) = ( x - 8C2d+y y2) d/ 2 l 1 where

Ex where

(a) Show that 6.ho = 0 in H and use (1.7) to conclude 6.u = 0 in H. (b ) Show I(x, y) = J d(} ho(x, y) = 1. ( c) Show that if X n -+ x, Yn -+ 0 then u(x n , Yn ) -+ f(x, 0). ( d) Conclude that E( y) f(B ) = J d(} ho(x, y)f(8, 0). :c ,

r

GD(x, y) = G(x, y) -

J 1lx- l xz lld2 G(z, y)1r(dz) _

{7.1).

D

We think of GD(x, y) as the expected amount of time a Brownian motion starting at x spends at y before exiting G. To be precise, if A C G then the expected amount of time B1 spends in A before exiting G is fA GD(x, y) dy. Our task in this section is to compute GD (x, y). In d = 1, where D = (-1, 1) , (8.1) tells us that

x + 1 G{1, y) - 1 - x G{-1, y) GD(x, y) = G(x, y) - 22x+1 1 -x = -lx - Y l + -- {1 - y) + -- (y + 1) 2 2 = -lx - Y l + 1 - xy

J d(} ho(x, y)f(8, 0)

where f is bounded and continuous.

1r g(Bt) dt = J GD(x, y)g(y) dy

Proof Combine ( * ) in Section 4.5 with

cd is chosen so that I d(} ho(O, 1} = 1 and let u(x, y) =

167

Considering the two cases

(8.2)

x ;::: y and x � y leads to

{

- x)(1 + y) -1 � y � x � 1 GD (x, y) = (1 (1 - y)(1 + x) -1 � x � y � 1

Geometrically, if we fix y then x -+ GD(x, y) is determined by the conditions that it is linear on (O, y] and on (y, l] with GD(O, y) = GD(l, y) = 0, and GD(y, y) = 1 - y2•

168

Section 4.9 Laplace Transforms, Arcsine Law

Chapter 4 Partial Differential Equations

In d ;::: 2, (8.1) works well when y = 0 for then G(z, O) is constant on { l zl = 1} and the expression in (8.1) reduces to

Proof To check (A) and (B), we observe that if we let 11x denote the Laplacian

acting in the

x variable for fixed y then

11x G(x, y) = 0 for x =/= y

(8.3) To get an explicit formula for Gn (x, y) in d ;::: 2 for y =/= 0 we will cheat and look up the answer. This weakness of character has the advantage of making the point that Gn (x, y) is nothing more than the Green's function for D with Dirichlet boundary conditions. Folland {1976), page 109, defines the Green's function for- D to be the function J{ ( x, y) on D x D determined by the following properties:

(a) Since yjjyj 2 cf_ D , (* ) implies that the second term is harmonic for x E D. (b ) Let x E 8D. Clearly, the right-hand side is continuous at x. To see that it vanishes, we note

0

(A) For each y E D, K(· , y) - G( · , y) is C2 and harmonic in D. (B) For each y E D, if X n -+ x E 8D, K(xn , y) -+ 0.

The last equality follows from a fact useful for the next proof:

Remark. For convenience, we have changed Folland's notation to conform

(8.6) Lemma. If l x l = 1 then jxjyj - YIYI - 1 1 = !x yj.

to ours and interchanged the roles of x and y. This interchange makes no difference, since the Green's function is symmetric, that is, K(x, y) = K(y, x). (See Folland {1976), page 110.)

Proof Using

·

defined above.

Proof Using

Again turning to page

(8.1) and (7.1) we have

Gn(x, z) - G(x, z) = -

1an I1x--lxlyjd2 G(y, z)1r(dy) = -Ex G(Br , y)

This is harmonic in D by (4.6).

++

0

123 of Folland {1976) we "guess"

(8.7) Theorem. In d = 2 if 0 < I Y I < 1 then -1 Gn (x, y) = - (ln l x - yj - ln lxlyl - YI Y I - 1 1 ) 1f

(4.5) implies if X n -+ x E 8D, then D

Having made the connection in {8.4), we can now find Gn by "guessing and verifying" functions with properties ( A) and (B). To "guess," we turn to page 123 in Folland {1976) and find

(8.5) Theorem. In d ;::: 3, if 0 < I Y I < 1 then

lzl 2 = z · z and then lxl2 = 1 we have l x lyl - Y IY I - 1 1 2 = l x i 2 I Y I 2 - 2x · Y 1 = I Y I 2 - 2x Y l x l2 = lx - Y l 2

{8.4) Lemma. The occupation time density Gn is equal to the Green's function

f(

169

Proof Again we need to check ( A) and (B).

(a)

Gn(x, y) - G(x, y) =

; (ln l x - ��2 � + ln I YI )

Again the first term is harmonic by ( * ) since yjjyj 2 rJ. D. The second term does not depend on x and hence is trivially harmonic. (b ) Let x E 8D. Clearly, the right-hand side is continuous at x. The fact that it vanishes follows immediately from {8.6). o

170

Section 4.9 Laplace Transforms, Arcsine Law

Chapter 4 Partial Differential Equations

4.9. Laplace Transfor1ns , Arcsine Law

In this section we will apply results from Section 4.6 to do three things: (a) Complete the discussion of Example 6.1. (b) Prove a remarkable observation of Ciesielski and Taylor (1962) that for Brownian motion starting at 0, the distribution of the exit time from D = {x : lxl < 1} in dimension d is the same as that of the total occupation time of D in dimension d + 2. (c) Prove Levy's arcsine law for the occupation time of (0, oo ). The third topic is independent of the first two. Example 9.1. Take d = 1, G = (-a, a), c(x) = problem considered in Section 4.6.

(9.1)

Theorem.

Let

> 0, and f =

1

r = inf{t : B1 ¢ (-a, a)}. If 0 <

1

<

1 in the

1r2 /8a2 , then

cos(x.J2"7)

Ex e-yr - cos(a.J2"7) _

u(x) = cos(x..j'i:y)j cos(a..j'i:y) is a nonnegative solution of

1 -u" 2 + 1u = 0

u(-a) = u(a) = 1

To check that w :;j oo, we observe that (6.2) implies that Mt = u(B1 )e"Y 1 is a If we let Tn be a sequence of stopping times that local martingale on (0, reduces !111 , then u(x) = Ex u(BTn A n )e-y(Tn J\ n)

r).

(

Letting n -+ oo and noting Tn A n j

computations in Example 6.1 imply that in this case there is no nonnegative solution. o Exercise 9.1. Show that if{3 � 0 then

Ex exp( Since cos(z)

J2:8} -f3r) = cosh(x cosh(aJ2:8}

= (eiz + e - iz ) /2 this is what we get if we set 1 = -{3 in {9.1).

Example 9.2. Our second topic is the observation of Ciesielski and Taylor

{1962). The proof of the general result, see Getoor and Sharpe {1979), sections 5 and 8, or Knight (1981), pages 88-89, requires more than a little familiarity with Bessel functions, so we will only show that the distribution of the exit time from ( -1, 1) in one dimension starting from 0 is the same as that of the total occupation time of D = {x : lxl < 1} in three dimensions starting from the origin. Exercise 9.1 gives the Laplace transform of the exit time from (-1, 1) so we need only compute v(x) = Ex exp

Proof - By Example 6.1,

)

r, it follows from Fatou's lemma that

u (x) � Ex e -yr

The last equation implies w :;j oo, so (6.3) implies the result for 1 < 1r2 /8a2• To see that w(x) = oo when 1 � 1r2 /8a2 , suppose not. In this case (6.3) implies v(x) = Ex ( f( Br ) exp( cr )) is the unique solution of our equation but

171

(- 100 1D (Bt) dt) /3

Of course, we only have to compute v(O) but as in Example 9.1, and Exercise 9.1, we will do this by using a differential equation to compute v(x) for all x . Several properties of v are immediately obvious {in any dimension d � 3): {i) Spherical symmetry implies v(x) = f(lx l). (ii) Using the strong Markov property at the hitting time of D and {1.12) from Chapter 3 gives for r > 1 (a:)

f(r)

= r2- d !{1) + (1 - r2- d ) · 1 = 1 + r2 -d {! (1) - 1) ·

{iii) The strong Markov property and (6.3) imply that v is the unique solution of 1 -.6-v - f3v = 0 for x E D

2

v(x) = f(1) for x E 8D To express the last equation in terms of f we note that if v(x) = f(lxl) then

: nii V(x) = !' ( lx n { 1! 1 - ,:13} + f"(lxD ,:12 Di v(x)

= f' (lxl) l il

172

Ch apter 4 Partial Differential Equations

Summing over i gives

d-

1 ! .Q.v = ! f"( l x l ) + 2 2 l x l f'(lx l ) 2

d

!' (1-) = C(cz) cosh( a) - C(a)a- 1 sinh( a) !'(1+) = -{ ! {1) - 1} = 1 - C(a)a - 1 sinh(a)

d-

To tie equations (a) and (b) together, we note that by applying the reasoning in (iii) to D(O, q) with q > 1 and using the proof of (6.6) (which refers us to (5 .6a) ) we can conclude (iv)

v is C1 and hence f'(r) is continuous at r = 1.

Facts (ii) -(iv) give us the information we need to solve for f at least in principle: (b) is a second order ordinary differential equation, so if we specify f(O) = C and f'(O) = 0, the latter from (i), there is a unique solution fc on [0, 1]. Given fc (1), (ii) gives us the solution on [1 , oo). We then pick C so that (iv) holds. To carry out our plan, We begin with the "well known" fact that the only so­ lutions to_ (b) which stay bounded near 0 have the form cr1 -( d/2 ) I( d/2 ) - 1 ( r,ffJJ) where oo

Iv(z) =

z

+m

( j2 y 2 'L: m!f(v + m + 1)

m= O

is one of the happy families of Bessel functions. Letting a = ,ffJJ and recalling r(x) = (x - 1)f(x - 1), we see that when = 3 and hence v = 1/2 the solution has a simple form

d

00

(ar)2 m = C(a) C(a) """" " I - sinh{ar) = O ( 2m + 1 ) . m� m·

173

which shows that (b) holds. To complete the solution, we have to pick C(a) to make f E C1 . Using formulas for f'(r) for r < 1 from the previous display, for r > 1 from (a), and recalling = 3, we have

Multiplying by 2 we see that (b)

f satisfies 1 f"(r) + r f'(r) - 2{Jf(r) = 0 for r < 1

Section 4.9 Lap/ace Transforms, Arcsine Law

Solving gives C(a) = 1/ cosh{ a). Since Exercise 9.1 when x = 0.

a=

../2l3 this matches the result in

0

Example 9.3. Our third and final topic is Kac's (1951) derivation of

(9.2) L evy' s arcsine law. Let

Ht = l{s E [O, t] : Bs ;:::: 0}1. If O ::; (J ::; 1 then

1 dr

8 . ..fB) Po( Ht ::; Bt) = -7r1 = -2 arcsm( o J7'(1 - r) 7r Remark. The reader should note that by scaling the distribution of Htft does not depend on t, so the fraction of time in [0, t] that a Brownian gambler is ahead does not converge to 1/2 in probability as t --> oo. Indeed r = 1/2 is the value for which the probability density 1/7rJr(1 - r) is smallest. Proof For the last equality see the other arcsine law, prove the first we start with

(9.3) Lemma. Let c(x) = bo1:1nded, C1 , and satisfies for all x :f.:

0. Then

(4.2) in Chapter 1. To

-a - f3 1[o, oo) (x) with a, {J ;:::: 0. Suppose v

IS

1 -.Q.v 2 + cv = -1

Having "found" the solution we want, we can now forget about its origins and simply check that it works. Differentiating we have

C(a) co h(ar) - C(a) smh . (ar) J'(r) = -arr s 2C(a) smh C(a)a sinh(ar) - 2C(a) co h(ar) + � . (ar) j"(1· ) = s r rar 2-f'(r) = C( a)a . h(ar) = a-j (r) j"(1· ) + r sm -?

--

r

?

-

--

?

Proof Our assumptions about v imply that v"(x) = -2(1 + c(x)v(x)) dx in the sense of distribution. So two results from Chapter 2, the Meyer-Tanaka formula, (11.4), and (11.7) imply

174

Chapter 4 Partial Differential Equations

Section

Letting Ct = I� c(B3 ) ds and using the integration by parts formula, (10.1) in Chapter 2, with Xt = v(Bt) and Yi = exp(ct ), which is locally of bounded variation, we have

i exp(c3 )v' (B. ) dB3 t - i exp(c3){l + c(B3)v ( B.)} ds t + i v(B. ) exp(c3)dc3

So Mt = v(Bt) exp(ct ) + I� exp(c3) ds is a local martingale. Since v is bounded and Ct :5 0, left is bounded. As t -+ oo, exp(ct) :5 e - at -+ 0, so using the martingale and b ounded convergence theorems gives

v(x) = ExMo = ExMeo = Ex

A = va+lJ - fo (a + [J)fo

1

(a + (J) v = 2 v + 1 1/

x>O

1 -v" + 1

x
1

v(O) = A + a +1 - = --;==:=1 =.= fJ Ja(a + [J)

(9.4)

-

so we have v0(x) = Ce±x,;:Fi_ Picking the signs to keep v bounded, we have _ - x J2 (a +.O) + _1 a +.O X > 0 v(x) = Ae x..f2a x<0 Be + 1.a To find A and B we note that we want v to be C1 so 1

A + a +1 = B + -a (J -AV2(a + fJ) = B�

=

To go backwards from here to the answer written in (9.2) , we warm up by changing variables x = .,j'Fji, dx = (1/2) J2ijt dt to get

e- -yt eo l0eo dt = l e - x 1 2..f2Jr dx Vi 2

0

=

· -21 · ..,fi; = �

'Y

This identity and Fubini's theorem imply 1 eo e - (a +.O) s eo e - a (t - s ) = Ja +. (J fo 0 Js t - s dt ds t e - .0• = eo e - a t _l ds dt o o Js(t - s) 1

1

--=== --

1r

l

l

11"

i

i •

From (9.3), (9.4), and the uniqueness of Laplace transforms it follows that

Eo exp( -f3Ht) = -1 11"

1/

-yvo = 2 vo

--

gives a solution. What we are interested in is

leo exp(c. ) ds

Plugging in the definition of c(x) now and using Fubini's theorem leads to the 0 formula given above. (9.3) tells us that we want to find a bounded C1 function v with

{

It is a little tedious to solve these equations but it is easy to check that

t

v(Bt) exp(ct) - v(Bo) =

Laplace Transforms, Arcsine Law 175

4.9

io t Js(te-{3-3 s) ds dt

and invoking the uniqueness again we have the desired result.

0

5

Stochastic D ifferential Equations

5.1. Examples

In this chapter we will be interested in solving stochastic differential equations (or SDE's) that are written informally as

dX3 = b(X.) ds + u(X.) dB. or more formally, as an integral equation:

We will give a precise formulation of (*) at the beginning of Section 5.2. In (*) , Bt is an m dimensional Brownian motion, u is a d x m matrix, and to make the matrix operations work out right we will think of X, b and B as column vectors. That is, they are matrices of size d x 1, d x 1, and m x 1 respectively. Writing out the coordinates (*) says that for 1 :::; i :::; d

Note that the number m of Brownian motions used to "drive" the equation may be more or less than d. This generalization does not cause any additional difficulty, is useful in some situations (see Example 1.5 ) , and will help us dis­ tinguish between uuT which is x d and uT u which is m x m. Here uT denotes the transpose. of the matrix u. To explain what we have in mind when we write down (*) and why we want to solve it, we will now consider some examples. In some of the later examples we will use the term diffusion process. This is customarily defined to b e a "strong Markov process with continuous paths." Ilowever, i n this section,

d

178

Chapter 5 Stochastic Differential Equations

the term will be used to mean "a family of solutions of the SDE, one for each starting point." (4.6) will show that under suitable conditions the family of solutions defines a diffusion process.

l

Section 5.1 Examples

179

using the associative law, and taking expected values

t

1 X�bi (X3 ) ds t t + E 1 X!b; (X3 ) ds + E 1 ai; (X3 ) ds

E(X1X{ ) = xi xi + E

Example 1.1. Exponential Brownian Motion. Let Xt = Xo exp(J.Lt + uBt )

where Xo is a real number and Bt is a standard one dimensional Brownian motion. Using Ito's formula

If a i; is also continuous, differentiating gives

(1.1) so Xt solves (*) with b(x) = (J.L + ( u 2/2))x and u( x) = ux. Exponential Brownian motion is often used as a simple model for stock prices because it stays nonnegative and fluctuations in the price are proportional to the price of the stock. To explain the last phrase, we return to the general equation (*) and sup­ pose that b and u are continuous. If Xo = x then stopping at t A Tn for suitable stopping times Tn j oo, taking expected values in (*) and letting n --1- oo we have

Using EXI = xi + E J; bi(X3 ) ds and differentiating again we get

Subtracting the last two equations gives

So if bi is continuous

d EX;. I t = o = bi (x) dt Because of the last equation, b is called the infinitesimal drift. To give the meaning of the other coefficient we note that if

then the formula for the covariance of two stochastic integrals, 2, implies

(8.7) in Chapter

justifying the name infinitesimal covariance. When d = 1, we call a(x) the infinitesimal variance and u(x) = .Ja'W the infinitesimal standard deviation. In Example 1.1, u(x) = ux is pro­ portional to the stock price x. The infinitesimal drift b( x) = (J.L + ( u2 /2) )x is also proportional to x. Note, however, that in addition to the drift J.L built in to the exponential there is a drift u2 x /2 that comes from the Brownian motion. More precisely it comes from the fact that e"" is convex and hence exp(uBt) is a submartingale. Example

1.2.

The Bessel Processes. Let

motion with d > 1 , and Xt = Using the integration by parts formula x:xf = xi xi + Substituting

t

d- 1 .6.lxl = Txf

t

1 X� dX! + 1 X! dX� + (Xi , Xi ) t

dx: = bk (X3 ) ds + I::>k; (X3 ) dBt j

Wt be a d-dimensional Brownian

I Wt I· Differentiating gives

So using Ito's formula

180

Chapter 5 Stochastic Differential Equations

Section 5.1 Examples

We will use Bt to denote the first term on the right-hand side, since Bt is a local martingale and (B} 1 = t , i.e., Bt is a Brownian motion by Levy's theorem, in Chapter Changing notation now we have

(4.1) (1.2)

3.

so (*) holds with

b(x)

- 1 ds Xt - Xo = Bt + Jot d2X3 and = = (d -

Example 1.3. The Ornstein Uhlenbeck Process. Let

sional Brownian motion and consider

(1.3)

Bt b e a one dimen­

dXt = -cxXt dt + u dBt

which describes one component of the velocity of a Brownian particle which is slowed by friction (i.e., experiences an acceleration -a times its velocity Xt) · This is one case where we can find the solution explicitly.

( i

Xt = e - at Xo +

(1.4)

t 3 e a o- dB3

)

To check this informally, we note that differentiating the first term in the prod­ uct gives -cxXt dt and differentiating the second gives u dBt. To check this formally, we use Ito's formula with

Ut = t to conclude

Xt - Xo =

(1.3)

(1987),

(1992),

Example 1.5. Brownian Motion on the Circle. Let Bt be a one dimen­ sional Brownian motion, let Xt = cos Bt and yt = sin Bt . Since Xf + Y? = (Xt, Yt) always stays o n the unit circle. Ito's formula implies that

1,

(1.7)

1 Xt dBt - 21 Yt dt

dXt = -yt dBt - 2 Xt dt dyt =

To write this in the form (*) we take

b(x, y) = ( -x/2 -y/2 )

Note that u is always a unit vector tangent to the circle but we need the drift which is perpendicular to the circle and points inward, to keep (Xt, Yt ) from flying off.

b(x, y),

Example 1.6. Feller's Branching Diffusion. This

i -cxX3 ds + i

t

t

e - a 3 (ea 3 u) dB3

which is in integral form. From the representation in

(1.4) we get the following useful fact (1.5a) tTheorem. When�X0 =ar x, the distribution of Xt is normal with mean xe - and variance o-2 J e -2 dr.

Proof To see that the integral has a normal distribution with mean 0 and the indicated variance we use Exercise in Chapter Adding e - at Xo = e - at x now we have the desired result. D

6.7

2.

Example 1.4. The Kalman-Bucy Filter. Consider

(1.6)

Here vt is an Ornstein-Uhlenbeck velocity process, and yt is an observation of the position process subject to observation error described by ex dWt , where W is a Brownian motion independent of B. The fundamental problem here is: how does one best estimate the present velocity vt from the observation {Y6 : s � t}? This is a problem of important practical significance but not one that we will treat here. See for example, Rogers and Williams p. or 0ksendal Chapter VI.

327-329,

1)/2l x l u(x) 1.

dyt = vt dt + dWt dvt = -cvt dt + u dBt ex

181

(1.8)

Xt

2: 0 and satisfies

dXt = f3Xt dt + u..jX; dBt

To explain where this equation comes from, we recall that in a branching process, each individual in generation m has an independent and identically dist1;ibuted number of offspring in generation m + Consider a sequence of branching processes { z:;. I m 2: 0} in which the probability of k children is Pk and suppose the mean of Pk is + with -+ {3 the variance of Pk is o-� with o-� -+ o-2 > 0 for any o > 0, P pk -+ 0 .

1.

{A1) (A2) (A3)

1 (f3n /n) Lk> on

f3n

Since the individuals in generation 0 have an independent and identically dis­ tributed number of children

nx) = nx ( 1 + �) var (Zf IZ� = nx) = nx · u� E (Zf IZ� =

·

Section 5.2 Ito's Approach 183

182 Chapter 5 Stochastic Differential Equations So if we let Xf = Z{:.t] fn then

( Xftn - X I xn (o) = X) = xf3n . ; var ( Xft n I x n (0) = x) = xu; · ; E

The last two equations say that the infinitesimal mean and variance of the rescaled process Xf are xf3n and xu� . These quantities obviously converge to those of the process X1 in (1.8). In Example 8.2 of Chapter 8, we will show that under (A1)-(A3) the processes {Xf , t ;:::: 0} converge weakly to {Xt , t ;:::: 0}. Example

(1.9)

1.7.

Wright-Fisher Diffusion. This Xt

E [0, 1] and satisfies

dXt = ( -aXt + {3(1 - Xt)) dt + y'Xt(1 - Xt) dBt

The motivation for this model comes from genetics, but to avoid the details of that subject we will suppose instead we have an urn with n balls in it that may be labelled A or a. To build up the urn at time m + 1 we sample with replacement from the urn at time m . However, with probability afn we ignore the draw and place an a in, and with probability {3/n we ignore the draw and place an A in. In genetics terms the last two events correspond to mutations. Let Z� -be the number of A's in the urn at time m. Now when the fraction of A's in the urn at time 0 is :r:, the probability of drawing an A on a given trial is

Pn = X

•

(1 - ;) + ( 1 - X)

•

�

Since we are sampling with replacement,

E (Zf I Z� = nx ) = npn var (Zf I Z� = n x ) = npn (1 - Pn )

arise from genetics. See Karlin and Taylor (1981), vol. II, pages 176-191 for some of the others. 5 .2. Ito's Approach

We begin this section by giving some essential definitions and introducing a coup.terexample that will explain the need for some of the formalities. We will then proceed to our main business: the first existence and uniqueness result for SDE, which was proved by K. Ito long before the subject became entangled in the complications of the general theory of processes. To finally become precise, a solution to our SDE (*) is a triple (Xt, Bt , :Ft) where Xt and Bt are continuous processes adapted to a filtration :Ft so that (i) Bt is a Brownian motion with respect to :Ft, i.e. , for each s and t, Bt + 3 - Bt is independent of :Ft and is normally distributed with mean 0 and variance s. (ii)

Xt satisfies

b

We will always assume that 11 and are measurable and locally bounded so the integral exists. It is easy to see that if the SDE holds for some :Ft it will hold for :Ff•B the filtration generated by (Xt, Bt). When Xt is adapted to :Ff the filtration generated by the Brownian motion, then we can take :Ft = :Ff and Xt is called a strong solution. It is difficult to imagine how Xt could satisfy (*) without being adapted to :Ff but there is a famous example due to H. Tanaka showing that this can happen. Example

and let

2.1. Let Wt be a one dimensional Brownian motion with Wo

Bt =

If we let Xf = Z(:, t/n then

(

I ( l

E xr, n - X x n (o) = X var Xft n Xn (O) = X

) = {-ax + {3(1 - :r:)} . ; ) = Pn (1 - Pn) · ;

Now Pn -+ x as n -+ oo so again the infinitesimal mean and variance of Xf converge to those of Xt in (1.9) but the rigorous connection has to wait until Example 8.3 of Chapter 8. This is just one of many of the diffusions that can

1t sgn(W3 ) dW3

=

0

where sgn(:r:) = 1 if x ;:::: 0, sgn(:r:) = -1 if x < 0. Since Bt is a local martingale with (B}t = t, (4.2) in Chapter 3 implies that B1 is a Brownian motion with re­ spect to :Ftw , the filtration generated by Wt. Since sgn(:r:) 2 = 1, the associative law implies that

t

t

1 sgn(W3 ) dB3 = 1 dWs = Wt So W is a solution of (*) with b = 0, u(:r:) = sgn(:r:) , and :Ft = :Ftw = :Ftw,B .

184 Ch apter 5 Stochastic Differential Equations

W ( 4)

Section 5.2 Ito's Approach 185

To see that is a not a strong solution we apply the Meyer-Tanaka for­ mula, 1 1 . in Chapter 2, with = and f ( x ) = to get

Xt Wt

lxl

(*)

!Wt!

where L� is the local time of at 0. The occupation time density formula in ( 1 1.7) of Chapter 2, and the continuity of the local time established in ( 1 1 .8) of Chapter 2, imply that

t

1 10 1{1W. I ::; €} ds = 2e 1- € L� da L� So L0 and hence Bt = ! Wt I - L� is measurable with respect to the filtration gene:ated by IWtl, t ;:::: 0. However, Wt is clearly not adapted to the filtration generated by !Wtl· 1 2e

€

-

_.

D

As usual when we have an equation, we are interested in having a unique solution. For SDE there are several notions of uniqueness. Our first, pathwise uniqueness holds if whenever and Xi are solutions of *) with =x = driven by the same Brownian motion = Xi then with probability one for all t ;::::- 0. Here, the filtrations and :Fi are allowed to be different.

Xt

(

:Ft Bt,

Xo XXt �

-W

W was a solution. To show that ) = -sgn (x ) except when x = 0, in

is also a solution, we note that sgn ( -x which case the left side is 1 and the right side is -1. So

-Wt)

To prove that the second integral is 0 ( and hence the right.:.hand side is = recall that is a Brownian motion. Example 3 . 1 in Chapter 2 implies that { s : = 0} has measure 0 , and hence

w3 Wt

Turning to positive results:

has a strong solution and pathwise uniqueness holds. Proof We construct the solution by a successive approximation scheme that reduces to the classical Picard iteration method for solving ordinary differential equations in the special case 0. We begin by introducing the notation

!T :::t Xt = Xo + i !7(X3) dB3 + it b(X3) ds to describe one iteration. In words, we use the process {X3, t ;:::: 0} as input for the coefficients !T and b, and the output i� the process {Xt , t ;:::: 0}. To solve ) we begin with the initial guess X� = x and define the successive guess by (*

iteration:

3

x n+ l

=

3

-

x n for n > 0

The rest of this section is devoted to proving that this procedure constructs a strong solution and that pathwise uniqueness holds. To help the reader digest the argument we have divided it into sections. One of the claims is easy. It is clear by induction that is adapted to :Ff .

Xf

The basic estimate which is the key to the proof is:

{ 2.1 ) Theorem. Pathwise uniqueness does not hold in Example 2.1 Proof We observed in Example 2.1 that

(Y)l � Kl x - Yl ;j (x) - !T;jequation l b;(x) - b;(y)l � Klx - yl, then theystochastic I!Tdifferential t t Xt = x + i !7(X3) dB3 + i b(X3) ds

(2.2) Theorem. If for all i, j, x, and we have

and

( 2.3 ) Lemma. Let r be a stopping time with respect to :F[, let T < oo and let

B = (4Td + 16d2)K2• Then

Proof The inequality is trivial if the right-hand side is infinite so we can suppose that each term on the right is finite. To begin we recall that � 2 Using this twice

2a2 + b2

•

(a + b + c)2 � 2a2 + 2(b + c) 2 � 2a2 + 4b2 + 4c2 So the left-hand side of ( 2.3) is ti\ T 2 . � 2EI Yo - Zo l 2 + 4E sup ' 3) ) !7( dB$ !7(Z ¥:, 1 t 1 (a) tilT o::; ::;T o 2 + 4E o9::; sup I { b(Y3) - b( Z3) ds 1 T lo

(a+ b)2

s

Section 5.2 Ito Approach 187 '

186 Chapter 5 Stochastic Differential Equations To bound the third term in (a), we observe that the Cauchy-Schwarz inequality implies that t l\ r t l\ t AT ·

Combining inequalities (a) , (c) , and (d) proves

(l b;(Y.) - b;(Z.) dsy :5 1 r (b;(Y.) - b;(Z.))2 ds l 1 ds l v(t)l2 :5 I:d

sup v[(t) sup O�t �T i= 1 O�t �T with the Lipschitz continuity assumption gives t T - 4E sup f i\ O�t �T lo t i\ T d 4E I: sup i:1 O� t �T 0 (c) d [TAT 4T i= 1 ° TAT [ 4Tdi< 2

I b(Y.) - b(Z.) ds 1 2 (1 b;(Y.) - b;(Z.) ds)2 :5

:5 E I: Jo (b;(Y.) - b;(Z.))2 ds E Jo IY. - Z.l2 ds :5 To bound the second term in (a), let cr; be the ith row of u , let Mf = Jto i\T (u;(Y, ) - cr;(Z.)) · dB. and let Tn = inf{t : IMfl ;:=: n} /\ r. The L2 maximal inequality and the formula for the variance of a stochastic integral imply that E O�tsup (Mf )2 :5 4E (M� J 2 :5 4E l{1\T o iu;(Y.) - u;(Z.W ds �TI\Tn r " oo, then using (b) and Lipschitz continuity we have Letting 2 4E sup I t" r (u(Y.) - u(Z.))dB., 0 9�T lo 2 sup I t" r (u;(Y.) - u;(Z.)) · dB., :5 4E t i = 1 09 �T lo (d) d TI\T :5 16E2: l lu;(Y.) - u;(Z.)I2 ds i=1 ·

and observe that since X�

°

= X�_ 1 , (2.3) implies that for n ;::: 1

(2.4) To get started we have to estimate l.6. 1 (s) l . Our earlier inequality for squares implies that for norms + W 2 l l + 2 l • So we have

Ia

:5 a 2 bl 2

Using the fact that sup O� • � t

•

n �

D

Convergence of the sequence of approximations. Let

Taking sup 0�t �T and using (b)

(2.3).

iu(x)B.I 4 t11 2 0�sup• �1 l u(x)B.I

and that the right-hand side has finite expected value we have

Combining this with

(2.4) and using induction we have

(2.5) From (2.5) we easily get the existence of a limit. Chebyshev's inequality shows that

(2.5) implies the right-hand side is summable, so the Borel-Cantelli gives p

(

sup I Xf - Xf - 1 1 > O� t �T

)

2 - n i.o. = 0

Since Ln 2 - n < 00 it follows that with probability uniformly on [0, 11 ·

1, Xf

�

a limit xr

188 Chapter 5 Stochastic Differential Equations

Section 5.2 Ito's Approach 189

The limit is a solution. We begin by showing that convergence also

occurs uniformly in L 2 •

(2.6) Lemma.

For all 0 :::; m < n :::; oo,

E

(O�s�T IX;' - x:1 2) :::; (k=tm+ 1 Ak (T)112)2

To prove pathwise uniqueness now, let (Yi, Bt , :F'l ) and ( Zt , Bt , :Fl) be two solutions of (*) . Replacing the :Ff by :Ft = :Fl V :Ff we can suppose :Fl = :Fl. To prepare for developments in the next section, we will use a more general formulation than we need now.

sup

Let II Z II 2 = (EZ2 ) 1 12• If n < oo, then it follows from monotonicity of expected value and the triangle inequality for the norm ll · l b that Proof

I O�s�up�T I X;' - x: 1 1 2 :::; I k=tm+ 1 O�sup· �T 1x: - x: - 1 1 1 2 sup IX: - x: - 1 1 1 ? = L A k (T) 1 12 :::; =L+1 I O �s�T - k=m+1 k m

Letting n --+ oo and using Fatou's lemma proves the result for n = oo.

Let Yi and Zt be continuous processes adapted to :Ft with and let r(R) = inf{t : IYi l or !Ztl > R}. Suppose that Bt is Brownian motion w.r.t. :Ft and that Yi and Zt satisfy (*) on [O, r(R)] . Then Yi = Zt on [0, r(R)] .

(2.8) Lemma.

Yo = Zo =

0

Since Yi and Zt are solutions,

To see that Xt" is a solution, we let Yi = Xf and Zt = Xt00 in (2.3) to get T E 0 sup 1 Xf +1 - Xt" 1 2 :::; BE I X: - xr: 1 2 ds 9�T

(

by

)

(2.6), so Xt"

1

::=; B T · E

= limXf +l = Xt" .

(O� • �T IX: - X';" l2) sup

--+ 0

Uniqueness. The key to the proof is ·

(2.7) Gronwall's Inequality. Suppose cp(t) ::=; A + B f� cp(s) ds for all t � 0 and cp(t) is continuous. Then cp(t) :::; AeBt . Proof If we let .,P(t) = (A + e)eB t then .,P'(t) = B.,P(t) so

.,P(t) = A + e + B

1

t

.,P(s) ds

Let r = inf{t : cp(t) � .,P(t)}. Since '1/J and

for .,P(s) ds > A + B for cp(s) ds � cp(t)

x

Proof Let

n

n

The last contradiction implies that cp(t) :::; (A + e)eBt , but e > 0 is arbitrary, so the desired result follows. 0

cp(t) ::=; B E ::=; B E so

(2.3) implies

10 IYs - Zs l2 1 IY.AT - Zsi\T I 2 t i\T(R)

t

ds

. ds ::=; B

(2.7) holds with A = 0 and it follows that

1

t

cp(s) ds 0

The last result implies that two solutions must agree until the first time one of them leaves the ball of radius R. (Of course this implies that the two solutions must exit at the same time.) Since solutions are by definition contin­ uous functions from [O, oo) --+ Rd , we must have r(R) j oo as R j oo and the desired result follows. 0 Temporal Inhomogeneity. In some applications it is important to allow the coefficients and cr to depend on time, that is, to consider

b

In many cases it is straightforward to generalize proofs from (*) to (**) · For example, by repeating the argument above with some minor modifications one can show

Section 5.3 Extensions 191

Chapter 5 Stochastic Differential Equations

190

(2 .9) Theorem. If for all i, j, x, y, and T < oo there is a constant for t $ T, !cr;j (t, O)l $ KT , lb;(t, 0) 1 $ J(T ,

KT so that

lcrii (t, x) - CTij {t, y)l $ KT lx - Y l and lb;(t, x) - b;(t, y)l $ KT lx - Y l

then (ii) in (3.1) holds with A = B and cp(x) We begin with the easier proof.

Using Ito's formula with X1° = t and

Proof of (3.2)

then the stochastic differential equation (**) has a strong solution and pathwise uniqueness holds. Comparing with (2.2) one sees that the new result simply assumes Lipschitz continuity locally uniformly in t but needs new conditions: lcrii (t, O)l $ ]{T and lb;(t, 0) 1 $ KT that are automatic in the temporally homogeneous case. The dependence of the coefficients on t usually does not introduce any new difficulties but it does make the statements and proofs uglier. ( Here it would be impolite to point to Karatzas and Shreve (1991) and Stroock and Varadhan (1979).) So with the exception of (5.1) below, where we need to prove the result for time-dependent coefficients, we will restrict our attention to the temporally homogeneous case. 5.3. Extensions

In this section we will ( a) extend Ito's result to coefficients that are only locally Lipschitz, (b ) prove a result for one dimensional SDE due to Yamada and Watanabe, and ( c) introduce examples to show that the results in (2.2) and (3.3) are fairly sharp. a.

Locally Lipschitz Coefficients

In this subsection we will extend .

(3 . 1 )

(2.2) in the following way.

Theorem. Suppose (i ) for any n

< oo we have

when lxl, I Y I $ n and (ii) there is a constant A < oo and a function cp(x) 2::: 0 so that if X1 is a solution of (*) then e - At cp(Xt) is a local supermartingale. Then (*) has a strong solution and pathwise uniqueness holds.

(3.2) Theorem. Let a = crcrT and suppose d

L {2x;b;(x) + a;;(x)} $ B( 1 + lxl 2 ) i=l

= 1 + lxl 2 •

f(xo,

· · · ,

Xd)

we have

= e - Bx a cp(xt , . . . , X d)

t

i e-B3 cp(X3 ) ds d d t + � r e - B3 2x; dX; + � � 1 e - B3 2 d(Xi ) 3 lo

e - At cp(Xt ) - cp(Xo) = -B

f.' ,-B• (-B�(X,) t, {2X; b;(X, )

= local martingale +

+

Our assumption implies that the last term is supermartingale.

$

0 so

+

)

au (X, ) ) d•

e - At cp(Xt) is a local

o

R < oo, and introduce erR and bR with (a) crR (x) = cr(x) and bR (x) = b(x) for lxl $ R (b ) crR (x) = 0 and bR (x) = 0 for lxl 2: 2R ( c) l crfj (x) - cr� (y) l $ K'lx - Y l, lbfl (x) - b[l (y) l $ K'lx - Y l

Proof of (3.1 )

Let

For an explicit construction do

lh(x) - h(y) l $ Clx - Y l when lxl $ R. Let h(x) = 2R R- Ixl h(Rxf!xl) for R $ lxl $ 2R

Exercise 3.1. Suppose

·

and h(x) = 0 for lxl 2::: 2R. Show that constant G = 2Gt + R- 1 lh(O)I.

h is Lipschitz continuous on Rd with

Fix a Brownian motion and let Xf be the unique strong solution of

192 Chapter 5 Stochastic Differential Equations

Section 5.3 Extensions 193

Let Tn = inf{t : IXt l � n}. (2.8) implies that if m < n then xr = Xf for t ::; Tm . From this it follows that we can define a process Xf' so that and Xf' will be a solution of (*) for t < T00 = limn f oo Tn . To complete the proof now it suffices to show that T00 = oo a.s. If X0 = x then using the optional stopping theorem at time t A Tn , which is valid since the local supermartingale is bounded before that time, we have

cp(x) � E (e-A (t ATn ) cp(Xt ATn )) � e-A t P(Tn ::; t) y :linf y l =n cp(y) Rearranging gives

I

cp(y) P(Tn ::; t) ::; eAt cp( x) y inf : l yl =n Since we have supposed cp(y) -+ oo as y -+ oo it follows that P(Tn ::; t) -+ 0 as n -+ oo which proves the desired result. 0 In words, what we have shown is that for locally Lipschitz coefficients the solution is unique up to Tn the exit time from the ball of radius n. If Tn i oo we get a -solution defined for all time. If not then the solution reaches infinity in finite time and we say that an explosion has occurred. Intuitively (3.2) says there is no explosion if, when l x l is large, the part of the drift that points out, b(x) · xf l x l , is smaller than C l x l and the variance of each component is smaller than C l x l 2• We do not need conditions on the off diagonal elements of a;i since the Kunita-Watanabe inequality implies · Note that since our condition concerns a sum, a drift toward the origin can compensate for a larger variance.

3.1. Consider d = 1 and suppose b(x) = -x3 • Then (3.2) holds if a(x) ::; B(1 + x 2 ) + 2x4 •

Example

The next two examples show that when considered separately the conditions on b and a are close to the best possible. Example

3.2. Consider d = 1 and let

O"(x) = 0,

with 6 >

1

Suppose Xo = 0. b > 0 so t -+ Xt is strictly increasing. Let Tn = inf{t : Xt = n}. While Xt E (n - 1, n) its velocity b(Xt ) � n6 so Tn - Tn _ 1 ::; n- 6 • Letting Too = lim Tn and summing we have T00 ::; 2:::::'= 1 n-6 < oo if 6 > 1. We like the last proof since it on'ly depends on the asymptotic behavior of the coefficients. One can also solve the equation explicitly, which has the advantage of giving the exact explosion time. Let yt = 1 + Xt . dYt = Y/ dt so if we guess yt = (1 - at)-P then

ap (1 - at)P+l Setting a = 1/p and then p = 1/(6 - 1) to have p + 1 = 6p we get a solution that explodes at time 1/a = p = 1/(6 - 1). dytjdt =

Example 3.3. Suppose b = 0; and O" = (1 + l xi) 6 J. (2.2) implies there is no explosion when 6 ::; 1. Later we will show (see Example 6.2 and (6.3)) that in d � 3 explosion occurs for 6 > 1 in d ::; 2 no explosion for any 6 < oo b. Yamada and Watanabe in d

=

1

In one dimension one can get by with less smoothness in

(3.3) Theorem.

tion p with

O".

Suppose that (i) there is a strictly increasing continuous func­

I O"(x) - O"(Y ) I ::; p( l x - yl) where p(O) = 0 and for all c > 0

1E p-2 (u) du = oo

(ii) there is a strictly increasing and concave function

x:(lx - yl) where x:(O) = 0 and for all f > 0

x: with lb(x) - b(y) l ::;

1 f x:- 1 (u) du = oo

Then pathwise uniqueness holds.

Remark. Note that there is no mention here of existence of solutions. That is taken care of by a result in Section 8.4. Proof The first step is to define a sequence cpn of smooth approximations of

l x l . Let an ! 0 be defined by ao = 1 and

Section 5.3 Extensions 195

194 Chapter 5 Stochastic Differential Equations Let 0 � t/Jn (u) � and

2p- 2 (u)/n be a continuous function with support in (an , an - 1 )

Since the upper bound in the definition of t/Jn integrates to such a function exists. Let


{

2 over (an , an - 1 )

1'xl dy 1Y dz t/Jn (z)

By definition 0, so

= 0 for lxl � an � 1 for an � lxl � an - 1 = 1 for an - 1 � lxl and it follows that
�t = t { l1(X} ) - l1 (X])} dB. + lot {b(X} ) - b(X])} ds lo . so Ito's foimula gives

1t

1 + 2 lot
=

Io(t) + I1 (t) + I2 (t)

Io(t) is a local martingale so there are stopping times Tm j oo with Eio(t A Tm ) = 0 (b ) To deal with 12 in ( a) we note that
For I1 in ( a) we observe have

l
�

1, so using (ii) and the concavity of K we

1t Elb(X} ) - b(X]) i ds � 1t EK( I�3 1 ) ds � 1t K(E I�3 1 ) ds

Ei h (t) i �

( d)

Combining ( a)- ( d) we have

Letting m -+ oo and using Fatou's lemma, then letting n -+ oo and using the monotone convergence theorem (recall
To finish the proof now we prove a slight improvement of Gronwall's inequality.

(3.4) Lemma. Suppose f(t) � J; K(j(s)) ds where f is continuous, K(x) > 0 is increasing for x > 0 , and has foe K - 1 (x) dx = oo for any f > 0. Then f(t) = 0 . Proof Let g(t) be the solution of g'(t) = K(g(t)) with g(O) = c. It follows from the proof of (2.7) that f(t) � g(t) . To show that g is small when c is small note that g is strictly increasing and let h be its inverse. Since g(h(x)) = x, we have h'(x) = 1/g'(h(x)) = 1/K(g(h(x)) = 1/K(x) and hence if a < b

h(b) - h(a) =

1b 1/K(x) dx

In words the right-hand side gives the amount time it takes g to climb from a to b. Taking a = f and b = S > 0, the assumed divergence of the integral implies that the amount of time to climb to S approaches oo as c -+ 0 and the desired result follows. D c.

Examples

We will now give examples to show that for bounded coefficients the com­ bination of (2.2) and (3.3) is fairly sharp. In the first case the counterexamples will have to wait until Section 5.6.

Section 5.4 Weak Solutions 197

196 Chapter 5 Stochastic Differential Equations Example

that

3.4. Consider b(x) = 0 and (T(x) = lxl6 /1. 1. (2.2) and (3.3) imply 2': 1/2 in d = 1 pathwise uniqueness holds for 2': 1 in d > 1

{ 88

Example 6.1 will show . . . pathwiSe umqueness fa1.ls lf

{ 88 << 1/2 1

in d = 1 in d > 1

·

88

3.5. Consider (T(x) = 0, b(x) = lxl6 /1. 1, and Xo = 0 . If 2': 1 then . < 1 there (2.2) implies that there is a unique solution: X. = 0. However, 1f are others. To find one, we guess Xt = CtP. Then dX. = CpsP - 1 ds Example

so to make

8

dX. = b(X. ) ds we set

8)

If < 1 then p > 0 and we have created a second solution starting at 0. Once we have tyvo solutions we can construct infinitely many. Let a > 0 and let 0

Exercise

_

t
chitz continuity of (T. However, the condition on b is also an improvement, and further sharpens the line between theorems and counterexamples. (a) Show that (ii) in (3.3) holds when C > 0 and ·

(x

)-

_

{ C(e-2

p

> 0 and show that g'(t)

solution. However that proof is more complicated and the conclusion is not important for our purposes so we refer the reader to Revuz and Yor (1991), p. 340-341. Proof Suppose

(X 1 , B 1 ) and (X2, B2) are two solutions with XJ = X6 = x.

(As remarked above the solutions are specified by giving the joint distributions

Cx log(1/x) for x < e-2 + x) for x � e -2

(b) Consider g (t) = exp(-1/tP ) with where 1/Jp ( Y) = py{log(1/y)} (p+l )/p .

(4.1) Theorem. If pathwise uniqueness holds then there is uniqueness in dis­ tribution. Remark. Pathwise uniqueness also implies that every solution is a strong

3.2. The main reason for interest in (3.3) is the weakening of Lips­

n.

4.1. Since sgn(x)2 = 1, any solution to the equation in Example 2.1 is a local martingale with (X)t = t. So Levy's theorem, (4.1) in Chapter 3, implies that Xt is a Brownian motion. The last result asserts that unique­ ness in distribution holds, but (2.1) shows that pathwise uniqueness fails. The next result, also due to Yamada and Watanabe (1971), shows that the other implication is true. Example

(p - 1) = p8 i.e. , p = 1/(1 Cp = C6 i.e., C = p- 1 /( 1 - o) Xt = { C(t

the Brownian motion .and the solution at the same time. In signal processing applications one must deal with the noisy signal that one is given, so strong solutions are required. However, if the aim is to construct diffusion processes then there is nothing wrong with a weak solution. Now if we do not insist on staying on the original space ( n, :F, P) then we can use the map w - (Xt (w), Bt (w)) to move from the original space (r2, :F) to (C X C, C X C) where we can use the filtration generated by the coordinates w1 (s), w2 (s), s � t. Thus a weak solution is completely specified by giving the joint distribution (Xt , Bt) · Reflecting our interest in constructing the process Xt and sneaking in some notation we will need in a minute, we say that there is uniqueness in distri­ bution if whenever (X, B) and (X', B') are solutions of SDE(b, (T) with X0 = X� = x then X and X' have the same distribution, i.e., P(X E A) = P(X' E A) for all A E C. Here a solution of SDE(b, (T) is something that satisfies (*) in the sense defined in Section 5.2.

=

1/Jp ( g (t))

5 . 4. Weak S olutions

Intuitively, a strong solution corresponds to solving the SDE for a given Brow­ nian motion, while in producing a weak solution we are allowed to construct

(Xi , B i) .) Since (C, C) is a complete separable metric space, we can find regular conditional distributions Q i (w , A) defined for w E C and A E C (see Section 4.1

in Durrett (1995)) so that

P(X i E A I B i ) = Qi (B i ,A) Let Po be the measure on ( C, C) for the standard Brownian motion and define a measure i on (C3 , C3) by

1r(Ao X A1 X A2 ) =

{

}A0

dPo(wo)Q l (wo, A1 )Q 2 (wo, A2 )

198 Chapter 5 Stochastic Differential Equations If we let

Section 5.4 Weak Solutions 199

�i (wo, w2) = w;(t) then it is clear that Wt ,

A2 = C A 1 = C Y1 = Y 2

=

y2

4

x2

a(:z:) = uuT (:z:) and recall from Section 5.1 that (Xi , Xi ) t = 11 a;i (X.) ds We say that X is a solution to the martingale problem for simply X solves MP(b,a), if for each i and j, 1 1 xf - 1 b; (X.) ds and x;x{ - 1 a;j (X.) ds

0

b

and

a, or

:F1

To be precise in the definition above we need to specify a filtration for the local martingales. In posing the martingale problem we are only interested in the process so we can assume without loss of generality that the underlying space is and the filtration is the one generated by C) with In the formulation above and are the basic data for the mar­ tingale problem. This presents us with the problem of obtaining from . To solve this problem we begin by introducing some properties of Since then Also, if z E is symmetric, that is,

(Xi ,Xi )1 = (Xi ,Xi) 1 , a

a;j = aii ·

see Stroock and

T ;::: al 2 and lla(:z:) - a(y)l l :::; Cl :z: - Y l for all :z:, y al 1 12(:z:) - a1 12e(y)lla(:z::::;)O(C/2a81 112 )l:z: - Yl for all :z:,y. When a can degenerate we need to suppose more smoothness to get a Lipschitz continuous square root. (4.4) Theorem. Suppose aij E C2 and 1rgft-d l eT D;;a(:z:)B.I :::; C IBI 2 then ll a 1 12 (:z: ) - a 1 12 (y)l l :::; d(2C) 1 12 I :z: - Y l for all :z:, y. To see why we need a E C2 to get a Lipschitz a 1 12 , consider d = 1 and a(x) = lxl>- . In this case u(x) = l :z: l>./2 is Lipschitz continuous at 0 if and only if >. ;::: 2. (4.3) Theorem. If then

are local martingales. The second condition is, of course, equivalent to

a = uuT

0"

Varadhan (1979), Section 5.2.

Let

X (C, 1 , Xt (w) = w1b

;:::

• • •

Take or respectively in the definition. Thus, we have two solu­ tions of the equation driven by the same Brownian motion. Pathwise uniqueness implies with probability one. (i) follows since

xt 4 yt

and A with diagonal entries >.1 ;::: >.2 ;::: >.d 0 so that a = UaTdiagonal AU. a ismatrix invertible if and only if >. d > 0. (4.2 }_tells us a = uT AU, so if we want a = O"O"T we can take = uT .;A, where VA is the diagonal matrix with entries .;>:i. (4.2) tells us how to find the square root of a given a. The usual algorithms for finding U are such that if we apply them to a measurable a( :z:) then the resulting square root u( :z:) will also. be measurable. It takes more sophistication to start with a smooth a(:z: ) and construct a smooth u(:z: ) . As in the case of (4.2), we will content ourselves to just state the result. For detailed proofs of (4.3) and (4.4)

X1 . ua. a. Rd

Returning to probability theory, our first step is to make the connection between solutions to the martingale problem and solutions of the SDE.

a

(4.2) Lemma. For any symmetric nonnegative definite matrix a, we can find an orthogonal matrix U (i.e., its rows are perpendicular vectors of length 1)

MP(b,a) u Bt , u). (X, B) If u is invertible at each X this is easy. Let �i = Xf - J� b; (X.) ds

a.

Proof

and let So is nonnegative definite and linear algebra provides us with the following useful information.

X

{4.1)) Theorem. Let be a solution to and a measurable square root of Then there is a Brownian motion possibly defined on an enlarge­ ment of the original probability space, so that solves SDE(b,

(a)

From the last two definitions and the associative law it is immediate that (b )

1t u(X.) dB. = yt - Yo = X1 - Xo - 11 b(X.)ds

200 Chapter 5 Stochastic Differential Equations so (*) holds. The B; are local martingales with (B i , Bi )t = O;jt so Exercise 4.1 in Chapter 3 implies that B is a Brownian motion. To deal with the general case in one dimension, we let

let J3 = 1 - 13 , let W3 be an independent Brownian motion (to define this we may need to enlarge the probability space) and let (a') The two integrals are well defined since their variance processes are :::; t. Bt is a local martingale with (B}t = t so (4.1) in Chapter 3 implies that Bt is a Brownian motion. To extract the SDE now, we observe that the associative law and the fact that J3 u(X3 ) = 0 imply

it u(X3) dB3 = it 13 dY3 = yt - Yo -

it J3 dY3

In view of (b) the proof will be complete when we show this we note that

Section 5.4 Weak Solutions 201 Our result for one dimension implies that the zf may be realized as stochastic integrals with respect to independent Brownian motions. Tracing back through the definitions gives the desired representation. For more details see Ikeda and Watanabe (1981), p. 89-91, Revuz and Yor (1991), p. 190-191, or Karatzas and Shreve (1991), p. 315-317. 0

(4.5) shows that there is a 1-1 correspondence between distributions of so­ lutions of the SDE and solutions of the martingale problem, which by definition are distributions on (C, C). Thus there is uniqueness in distribution for the SDE if and only if the martingale problem has a unique solution. Our final topic is an important reason to be interested in uniqueness. (4.6) Theorem. Suppose a and b are locally bounded functions and MP (b , a) has a unique solution. Then the strong Markov property holds. Proof For each x E Rd let Px be the probability measure on ( C, C) that gives the distribution of the unique solution. starting from X0 = x. If you talk fast the proof is easy. If T is a stopping time then the conditional distribution of {XT+3 , s � 0} given :FT is a solution of the martingale problem starting from XT and so by uniqueness it must be Px(T) · Thus, if we let BT be the random shift defined in Section 1.3, then for any bounded measurable Y : C -+ R we have the strong Markov property

(4.7) J� J3 dY3 = 0. To do

E (Y o BT i:FT ) = Ex(T) y

To begin to turn the last paragraph into a proof, fix a starting point Xo = x0• To simplify this rather complicated argument, we will follow the standard

practice (see Stroock and Varadhan (1979), p. 145-146, Karatzas and Shreve (1991), p . 321-322, or Rogers and Williams (1987), p. 162-163) of giving the details only for the case in which the coefficients are bounded and hence

it b;(X3 ) ds t Mfi = XfX{ - i a;i (X. ) ds

Mf 0 = Xf To handle higher dimensions we let structed in (4.2) and let

Zt = Introducing Kronecker's

U(x) be the orthogonal matrices con­

t

i UT (X3 ) dY3

O;j = 1 if i = j and 0 otherwise, we have

are martingales. We have added an extra index in the first case so we can refer to all the martingales at once by saying "the Mfi ." Consider a bounded stopping time T and let Q(w , A) be a regular condi­ tional distribution for {XT+t , t � 0} given :FT . We want to claim that (4.8) Lemma. For Px0 a. e. w , all the

ij t - 1�Lrii = Mtij 1�u·T+ T

0

e

T

202 Chapter 5 Stochastic Differential Equations

Section 5.5 Change of Measure 203

Q(w, ·).

are martingales under To prove this we consider rational < and B E stopping theorem implies that if A E then

s :Ft :F, and note that the optional T E.,0 ({ (Mfi - M;i )1B) ()T } 1A) = 0 Letting B run through a countable collection that is closed under intersection and generates :F, it follows that P.,0 a.s., the expected value of (Mfi -M!i )1B is 0, which implies (4.8). Using uniqueness now gives (4.7) and completes the o

0

proo(

5.5.

2.12

Our next step is to use Girsanov's formula from Section to solve martingale problems or more precisely to change the drift in an existing solution. To accomodate Example below we must consider the case in which the added drift depends on time as well as space.

(5.1)

5.2

a), Xt a- 1 (x) P)x, :Ft b(s, x) Xt(w)bT=aW- t1 · l P, b(s, x)l :::; M. b, a). Let Xt = Xt - J� ,B(X. )ds, let c(s, x) = a - 1 (x)b(s, x) and let t Yt = lo c(s,X,) · dX,

is a solution of MP(,B, which, for concreteness Theorem. Suppose and without loss of generality, we suppose is defined on the canonical probability space (C, C, with the filtration generated by Suppose exists for_all and that is measurable and has We can define a probability measure Q locally equivalent to so that under Q , is a solution t o MP(,B +

Xt

Proof

at 2),

So by the formula for the covariance of stochastic integrals

t (a,Xi )t = :Li i0 c; (s,X,) d(Xi , Xi ), t = lo a,(cT a)i (s, X,) ds = lot a, bj(s, X,) ds since c(s,x) = a - 1 (x)b(s,x). It follows that a,

Change of Measure

b

(Y)t(12.3) :::; Mt, (3.7) 2 3 ((12 ) = a;-1 d(a,Xi ), .

Since in Chapter implies that is a martingale, and invoking in Chapter we have defined Q. To apply Girsanov's formula .4 from Chapter we have to compute A{ J� Ito's formula and the associative law imply that

Using Girsanov's formula now, it follows that

is a local martingalefQ. This is half of the desired conclusion but the other half is easy. We note that under

P

.which exists since

t (Y)t = lo � c; (s,X. )ci (s,X, )d(Xi , J(i ), t t = lo CT ac(s, x. ) ds = lo bT a- 1 b(s, x. ) ds :::; Mt by our assumption that l bT a - 1 b(s, x)l M. Letting Q t and Pt be the restric­ tions of Q and P to :Ft we define our Radon-Nikodym derivative to be I)

::=:;

(12.6)

and in Chapter under Q.

t (Xi ,Xi )t = lo a;i (X. )ds

2 implies that the covariance of X; and Xi is the same

0

Exercise 5.1. To check your understanding of the formulas, consider the case in which does not depend on start with the Q constructed above, change to a measure R to remove the added drift and show that so

R = P.

b

s,

Xt

b,

dRtfdQt = !fat

Example 5.1. Suppose is a solution of MP(O, I) , i.e., a standard Brownian motion and consider the special situation in which In this case

b(x) = 'V U(x).

204 Ch apter 5 Stochastic Differential Equations

Section 5.5 Change of Measure 205

recalling the definition of yt in the proof of (5.1) and applying Ito's formula to U(X1 ) we have yt

=

1 0

t

t \lU(X.) · X. = U(Xt ) - U(Xo) - -2l 0 � U(X. ) ds

i

Thus, we can get rid of the stochastic integral in the definition of the Radon­ Nikodym derivative:

��: = (U(X1 ) - U(Xo) - � 1t �U(X. ) + IVU(X. ) 12 ds) exp

In the special case of the Ornstein-Uhlenbeck process, b(x) is a positive real number, we have U(x) = -alxl2 � U(x) expression above simplifies to

= - 2ax, where a = - 2da, and the

��: = (-a1Xtl2 + a1Xo l2 + dat - 1t 2a2 1Xs l2 ds)

= Jl · X

dQt = exp . Jl · Xt - Jl · Xo - -21 lf.ll 2t d� It is comforting to note that when Xo = x, the one dimensional density functions satisfy

(

� )

Qt (Xt = y) = exp Jl · Y - Jl · x - 1 Jltl 2 Pt(Xt = y) = (211i) - d/ 2 exp( -ly - x - Jlil 2 /2t) as it should be since under Q, Xt has the same distribution as Bt + Jli.

D

(5.1) deals with existence of solutions, our next result with uniqueness. (5.2)

Theorem. Under the hypotheses of (5.1) there is a

between solutions of MP(,B, a) and MP(,B + b, a).

(5.3) Theorem. Suppose X is a solution of MP(O, a) constructed on (C, C). Suppose that a - 1 (x) exists, bT a - 1 b (x) is locally bounded and measurable, and the quadratic growth condition from (3.2) holds. That is, d

i=l

)

(

One can considerably relax the boundedness condition in (5.1). The next result will cover many examples. If it does not cover yours, note that the main ingredients of the proof are that the conditions of (5.1) hold locally, and we know that solutions of MP(b, a) do not explode.

2:: {2x;b; (x) + aii (x)} � A(l + lxl2 )

exp

In the trivial case of constant drift, i.e., b(x) = Jl = \lU(x) we have U(x) and � U(x) = 0 so

CASE 2. Suppose P1 and P are locally equivalent. In the definition of at, {Y}t 2 and yt are independent of P; by (12.6) and (12.7) in Chapter 2, so dQrfdP1 = dQ 2/ dP2 and Q 1 f:. Q 2 . D

1-1 corresp ondence

P1 and P2 be solutions of MP(,B, a). By change of measure we can produce solutions Q 1 and Q 2 of MP(,B + b, a). We claim that if P1 f:. P2 then Q 1 f:. Q 2 . To prove this we consider two cases:

Then there is a 1-1 correpondence between solutions to MP(b, a ) and MP(O, a). Remark. Combining (5.3) with

(3.1) and (4.1) we see that if in addition to the conditions in (5.3) we have a = u(F where u is locally Lipschitz, then MP(b, a) has a unique solution. By using (3.3) instead of (3.1) in the previous sentence we can get a better result in one dimension. Proof Let Z and a be as in the proof of (5.1) and let Tn = inf{t : I Xt l > n}. From the proof of (5.1) it follows that a(t 1\ Tn ) is a martingale. So if we let P? be the restriction of P to :Fv.Tn and let dQf fdP? = a(t 1\ Tn ) then under Q n , Xt is a solution of MP(b, a) up to time Tn . The quadratic growth condition implies that solutions of MP(b, a) do not explode. We will now show that in the absence of explosions, at is a martingale. First, to show at is integrable, we note that Fatou's lemma implies

(a) where

Proof Let

CASE 1 . Suppose P1 and P are not locally equivalent measures. Then since 2 Q; is locally equivalent to P;, Q 1 and Q 2 are not locally equivalent and cannot

be equal.

(b)

Eat � lim n -inf oo E(a(t 1\ Tn )) = 1 E denotes expected value with respect to P. Next we note that Elat - at i\TJ = E(lat - a(Tn ) I ; Tn � t) � E(a(Tn );Tn � t) + Eat - E(at; Tn > t)

The absence of explosions implies that as n --+ oo (c)

Section 5.6 Change of Time 201

206 Chapter 5 Stochastic Differential Equations This and the fact that

a(t

1\

2b3 dB3 - (1/2) l2b3 12 ds) 1.

Tn) is a martingale implies

( d)

a(t at in L1 . It follows (see a(t Tn ) = E (at I:FtATn ) and from this that a . , 0 s t is a martingale. The rest of the proof is the same as that of (5.1) and (5.2). as n -+ oo . (b ) , ( c) , and ( d) imply that 1\ Tn ) -+ e.g. , (5.6) in Chapter 4 of Durrett (1995)) that 1\

::=;

::=;

D

Our la.St chore in this section is to complete the proof of (2.10) in Chapter

1 by showing (5.4) Theorem.

Brownian motion starting at 0, let g be a contin­ g(O)Bt =be0,a and let f > 0. Then P (0sup 99 !Bt -g(t)l < f) > 0 Proof We can find a C1 .function h with h(O) = 0 and l h (s) - g(s)l < E/2 < for s so we can suppose without loss of generality that g E C1 and let t b3 = g'(s). If we let yt = J0 b3 dB3 and define Qt by dQ: = exp (i0t b3 dB3 - 2l i0t l b3l 2 dB3 ) dP then it follows from ( 5.1 ) that under Q, Bt -g(t) is a standard Brownian motion. Let G = { I B3 - g(s)l < f for s t}. Exercise 2.11 in Chapter 1 implies that Qt ( G) > 0. It follows from the Cauchy-Schwarz inequality that 1/2 Qt (G) = j ��: · la dPt :::; j ( ��:) dPt Pt (G) 112 To complete the proof now we let E denote expectation with respect to Pt and observe that if l b3l :::; M for s :::; t then j (��:Y dPt = E exp (it 2b. dB3 - it l b. l 2 ds) = exp (it l b$ 12 ds) :::; eM2 t

Let uous function with

since exp (I� J� is a martingale and hence has ex­ pected value For this step it is important to remember that is a non-random vector. D

5.6.

b3 = g'(s)

C hange of Time

b

In the last section we learned that we could alter the in the SDE by changing the measure. However, we could not alter with this technique because the quadratic variation is not affected by absolutely continuous change of measure. In this section we will introduce a new technique, change of time, that will allow us to alter the diffusion coefficient. To prepare for developments in Section 6.1 we will allow for the possibility that our processes are defined on a random time interval.

u

Xt be a solution of MP(b, a) for t < (. That is, for each t t x: - i bi (X. ) ds and x:xf - i aii (X3) ds are local martingales on [0,(). Let g be a positive function and suppose that

( 6.1 ) Theorem. Let

and j,

i

t,

.

::=;

(

2

)

u "'{ or t � ( } and let = X('Y3) for : Ut > s ajg) for s < .; . Y. 3 = MP(bjg, Proof We begin by noting .that Xt = Y(ut ), then in the second step change variables r = u. , dr = g(X. ) ds, X. = X-y(r) = Yr, to get

Define the inverse of by inf{t < e U( . Then is a solution of

s =

Changing variables

'Yu

Y.

t = 'Yu gives MP(b/g, a/[0,g)..;)

Since the are stopping times, the right-hand side is a local martingale on and we have checked one of the two conditions to be a solution of

210 Chapter 5 Stochastic Differential Equations Then MP(O, a) is well posed, i.e., uniqueness in distribution holds and there is no explosion.

Xt = Bt , a Brownian motion, which solves MP(O, I) and g(x) = h(x)- 1 • Since g(Bt ) ::; E_R1 for t ::; TR = inf{t : ! Btl > R} we have J; g(B,) ds < oo for any t. On the other hand Proof We start with

take

6

One D imensional D iffusions

since Brownian motion is recurrent, so applying (6.1) we have a solution of

MP(O, a). Uniqueness follows from (6.2) so the proof is complete.

D

Combining (6.3) with (5.2) we can construct solutions of MP(b, a) in one dimension for very general b and a.

d

6 . 1 . Construction

In this section we will take an approach to constructing solutions of

(6.4) Theorem. Consider dimension = 1 and suppose (i)

0 < fR ::; a(y) ::; CR < oo when IYI ::; R

b(x) is locally bounded (iii) x b(x) + a(x) ::; A(1 + x2) Then MP(b, a) is well posed. (ii)

·

that is special to one dimension, but that will allow us to obtain a very detailed understanding of this case. To obtain results with a minimum of fuss and in a generality that encompasses most applications we will assume: ( 1D) b and u are continuous and

>

a(x) = u2 (x) 0 for all x

The purpose of this section is to answer the following questions: Does a solution exist when (1D) holds? Is it unique? Does it explode? Our approach may be outlined as follows:

Xt

(i) We define a function cp so that if is a solution of the SDE on is a local martingale on [O, e). yt =

cp(Xt )

[o, e) then

(ii) Our yt has (Y)t = J; h(Y, ) ds so construct yt to be a solution of MP(O, h) by time changing a Brownian motion. is a solution of the SDE. = cp- 1 (Yt) and check that (iii) We define

Xt

Xt

Xt f t '(X.) dX, t J"(X.)d(X . X ) f(Xo) = JC t 1! �1 } t = local mart. + 1 Lf(X,) ds

To begin to carry out our plan, suppose that is a solution of MP(b, a) 2 for t < e. If E C , Ito's formula implies that for t < e

(1.1)

+

Section 6.1 Construction 213

212 Chapter 6 One Dimensional Diffw?ions where

Lf(x) = 21 a(x) t'(x) + b(x)f'(x)

(1.2)

and a(x) = u2 ( x). From this we see that f (Xt) is a local martingale on [0, ( ) if and only if Lf(x) = 0. Setting Lf = 0 and noticing ( 1D) implies a(x) > 0 gives a first order differential equation for f'

' ( !' )' = -2b a !

(1.3) Solving this equation we find,

f'(y) = B exp and it follows that

f(x) = A +

(1Y-�(�} dz)

1x B (1Y-2:g; dz) dy exp

Any of these functions ca� be called the natural scale. We will usually take A = 0 and B = 1 to get (1.4)

so(x) =

1x (1Y-2:g; dz) dy exp

However, in some situations it will be convenient to replace the O 's at the lower limits by some other point. Note that our assumption (1D) implies so is C2 so our use of Ito's formula in (1.1) is justified. Since yt =
1tt so'(X3)2 d(X}3 t = 1 so' (X3 ) 2 a(X3) ds = 1 h(Y3) ds

(Y}t = (1.5) where

To construct solutions of MP(b, a) we will reverse the calculations above: we will use a time change of Brownian motion to construct a yt which solves MP(O, h) and then let Xt = so - 1 (1'1). With Examples 1.6 and 1.7 of Chapter 5 in mind we generalize our set-up to allow the coefficients b and u to be defined on an open interval ( ex , {3) with ex < 0 < {3. Since so' ( x) > 0 for all x, the image of ( ex , /3) under so is an open interval (£, r) with - oo :::; l < 0 < r :::; oo. Letting Wt be a Brownian motion, ( = inf{t : Wt rf:. (£, r)}, g = 1 /h ,

h(y) = {so'(so - 1 (y)) } 2 a(so - 1 (y)) > 0 is continuous So if we let Tt = inf{s : (Y}3 > t } then Wt = YT, is a Brownian motion run for an amount of time (Y} e .

lTt =

1 g(W3) ds t

for t < ( and

Using (6.1) in Chapter 5, we see that

s < e = O'( ·

{3 = inf{t : O't > s or t ;::: ( }

Ys = W(!3 ) is a solution of MP(O, h) for

To define X1 now, we let '1/J b e the inverse of so and let X1 = 'l/J(Y1). To check that X solves MP(b, a) until it exits from ( a , {3) at time e, we differentiate so('l/J(x)) = x and rearrange, then differentiate again to get

:

'1/J'(x) = so' ( (x)) '1/J"(x) = { so' ( ( ))}2 so" (.,P(x )).,P' (x) Using the first equality and S011(y) = -(2b(y)/a(y))so'(y), which follows from ( 1 3 ) , we have 1 2b(.,P(x)) '1/J11(x) - {so'('l/J(x))}2 a(.,P(x))

;�

.

_

These calculations show for t < e

'1/J E C2 so Ito's formula, (1.1), and (1.5) imply that

'l/J(Y1 ) - .,P(Yo) =

1t .,P'(¥3) d¥3 + � 1t 'l/J11 (Y3 )h(Y3 ) ds

For the second term we note that combining the formulas for (recall '1/J = so - 1 ) 1 II

'l/J11 and h gives

2 '1/J (y)h(y) = b(.,P(y)) To deal with the first term observe that Y1 is a solution of MP(O, h) up to time e so by (4.5) in Chapter 5 there is a Brownian motion B3 with d¥3 = .jh{'YJ dB3. Since .,P'(y),jh(ilj = u(.,P(y)), letting X1 = .,P(yt) we have for t < e t t Xt - Xo = u(X3 ) dB3 + b(X3 ) ds

1

1

Section 6.2 Feller's Test 215

214 Chapter 6 One Dimensional Diffusions The last computation shows that if yt is a solution of MP(O, h) then X1 = ,P(yt) is a solution of MP(b, a), while (1.1) shows that if X1 is a solution of MP(b, a) then yt = so (X1) is a solution of MP(O, h). This establishes there is a 1-1 correspondence between solutions of MP(O, h) and MP(b, a ) . Using (6.2) of Chapter 5 now, we see that (1.6) Theorem. Consider (C, C) and let Xt (w) = w1• Let a < {3 and T( cr,,B) = inf{X1(w ) fJ. (a, {3)} . Under (1D), uniqueness in distribution holds for MP(b, a) on [0, T( cr,,B))· Using the uniqueness result with (4.6) of Chapter 5 we get (1.7) Theorem. For each x E (a, /3) let Px be the law of Xt , t < T( a ,,B) from (1.6). If we set X1 = A. for t ;:::: T(a,.B)• where A. is the cemetery state of Section 1.3, then the resulting process has the strong Markov property. 6.2. Feller's Test

In the previous section we showed that under (1D) there were unique solutions to MP(b, a) on [O, T(cr ,,B)) where T(a ,,B) = inf{t : X1 fJ. (a, {3)} . The next result gives necessary and sufficient conditions for no explosions, i.e., T(a,,B) = oo a.s. Let

Ty = inf{t : X1 = y} for y E ( a , {3) Ta = lim y! cr Ty and Tp = lim YT.B Ty

In stating (2.1) we have without loss of generality supposed 0 E ( a , {3) . If you are confronted b y an (a, {3 ) that does not have this property pick your favorite "' in the interval and translate the system by -"! . Changing variables in the integrals we see that (2.1) holds when all of the O 's are replaced by 1 's. (2.1) Feller's test. Let

m(x) = 1/(so ( x)a(x )). '

(a)

so(x) be the natural scale defined by (1.4) and let

Px (Tp < To) is positive for some (all) x E (0, {3) if and only if

1,8 dx m(

(b)

X) ( SO (f3) - SO( X ))

<

00

Px (Ta < To) is positive for some (all) x E (a, 0) if and only if

0 1 dx m(X) (SO(X) - SO(a)) <

00

(c) If both integrals are infinite then Px (T(a ,,B) = oo) = 1 for all x E (a, {3). Remarks. If so(f3) = oo then the first integrand is :: oo and the integral is oo. Similarly the second integral is oo if so(a) = oo The statement in (a) means that the following are equivalent (a1) Px (Tp < To) > 0 for some x E (0, {3) -

. (a2) (a3)

.

It dx m(x) (so(f3) - so(x)) < oo

Px (Tp < To) > 0 for all x E ( 0 , {3)

Proof The key to the proof of our sufficient condition for no explosions given in (3.1) and (3.2) of Chapter 5 was the fact that if A is sufficiently large S1 = ( 1 + l Xt l 2 )e - At is a supermartingale

To get the optimal result on explosions we have to replace 1 + lxl 2 by a function that is tailor-made for the process. A natural choice that gives up nothing is a function g ;:::: 0 so that e - 1 g (X1) is a local martingale To find such a g it is convenient to look at things on the natural scale, i.e. , let Yi = so (Xt) · Let £ = limx!cr so(x) and let r = limx r.B so(x). We will find a C2 function f(x ) so that f is decreasing on (£, 0), f is increasing on (0, r) and e - t f(Yi ) is a local martingale Denouement. Once we have f the conclusion of the argument is easy, so to explain our motivations we begin with the end. Let

f(£) = lim y f(y)

!l

/(r) = lim f(y)

YTr

(2.6) and the calculations at the end of the proof will show that (2.2a) The integral in (a) is finite if and only if f(r) < oo (2.2b) The integral in (b) is finite if and only if f(£) < oo

.

.

Once these are established the conclusions of (2.1) follow easily. Let

Ty = inf{t : yt = y} for y E (£, r) Tr = lim yTr Ty and Tt = lim y!l Ty T(t, r) = Tt A Tr

216 Chapter 6 One Dimensional Diffusions

Section 6.2 Feller's Test 217

Proof of {c) Suppose 1(1!.) = l(r) = oo. Let an < 0 < bn be chosen so that l ( a n ) = l ( bn ) = n and let Tn = Ta n A Tb n · Since e- • I(Y, ) is bounded before time Tn At we can apply the optional stopping theorem at that time to conclude that if y E (an , bn ), then et l(y) 0 as n -jo oo Py (Tn < t) ::; --jo

n

Letting t -jo oo we conclude 0 = Py(f(t,r ) < oo) = Prp-l (y) (T( a ,,8) < oo).

Let 0 < y < r, let bn j r with b 1 > y, and let Tn = To A Tb n · If l ( r) = oo , applying the optional stopping theorem at time Tn A t, we conclude that et l(y) --!- 0 as n -l- oo Py(Tb n < To A t) ::; Proof of {a)

l(bn )

Letting t -jo oo we conclude 0 = Py(Tr < To) = Prp-l(y ) (Tp < T0). This shows that if (a2) is false then (al) is false, i.e., (a1) implies (a2). If l(r) < oo, applying the optional stopping theorem at time Tn (which is justified since e- t I (Yt ) ::; l(bn ) for t ::; Tn ) we conclude that 1 < l(y) = Ey (e-Tn I(YrJ)

(

::; 1 + l ( r) Ey e-1\n ; Tb n < To

Rearranging gives

)

)

(

l(y) - 1 >0 ;.::::: Ey e-Tb n ; T-b n < To l ( r)

Noting Tr = T0 < oo is impossible, and letting n -jo oo we have Ey(e-Tb n ; Tb n < To) ! Ey (e-Tr ; Tr < To)

which is a contradiction, unless 0 < Py(Tr < To) = Prp-l(y) (Tp < To). This shows that (a2) implies (a3). (a3) implies (a1) is trivial so the proof of (a) is complete. Proof of (b) is identical to that of (a) .

I so that e-t I(Yt ) is a local mar­ tingale, we apply Ito's formula to l(x l , x 2 ) = e-"'1 l ( x 2 ) with Xl = t, Xl = Yt, and recall Yt is a local martingale with variance process given b y (1.5), so The search for f. To find a function

e-t I(Yt ) - f (Yo) =

t

i {- e- • I(Y, ) + � e- • f" (Y, )h(Y, )} ds

+ local mart.

so we want

�h(x)/"(x) - l(x) =

(2.3)

0

To solve this equation we let lo = 1 and define for n ;.::::: 1

ln (x) = r dy r dz 2ln - l ( z ) h(z) lo lo

(2.4)

(1 .8) in Chapter 4 implies that if we set l(x) = I:�= O ln (x) and (2.5)

00

I: sup lln (x)l < oo for any R < oo n =o i':_I � R

then f"(x) I:�= O l::(x). Using l::(x) = 2ln - l ( x ) fh (x ) for n > 1 and l�'(x) = 0 it follows that

f" (x) =

x l::(x) = t 21n - l (x) = 2l( ) t h(x) n=O n= l h(x)

i.e., I satisfies (2.3). To prove (2.5), we begin with some simple properties of the In ·

In ;.::::: 0 (2.6b) In is convex with 1�(0) = 0 (2.6c) In is increasing on (0, r) and decreasing on (£, 0) (2.6a)

h(z) > 0 (2.6a) follows easily by induction from the definition in (2.4). Using (2.6a) in the definition, (2.6b) follows. (2.6c) is an immediate consequence of {2.6b ). 0

Proof Since

Our next step is to use induction to show {2.7) Lemma.

ln (x) ::; (fi(x)) n /n!, (2.5) holds, and 1 + !I

::; I ::; exp( fi )

Proof Using (2.6c) and (2.6a) the second and third conclusions are immediate consquences of the first one. The inequality ln (x) ::; (fi(x)) n /n! is obvious if

Section 6.3 Recurrence and Transience 219

218 Chapter 6 One Dimensional Diffusions

n = 1. Using (i) the definition of fn in (2.4}, (ii} (2.6c}, (iii) the definition of /1 and /0 , (iv) the result for n - 1, and (v) doing a little calculus, we have 2 z) fn (x) = dy dz

x 1 1Y /�{;} dy f ( ) r dz h(z) � Jr o x n - 1 Y Jo = 1 dy fn - 1 (Y )f{(y) 2

(fl (y)) n - 1 ' (y) = (fl (x)) n r d Y (n - 1}! !1 n! - Jo <

Letting

{3

1u

du
To see what ( 2. 1) says in a concrete case we consider

2.1. Let u(x) = 1, b(x) = (1 + l x l ) 6 / 2 where o > 0. When o � 1 the

coefficients are Lipschitz continuous, so there is no explosion. We will now use (2.1) to reprove that result and show that explosion occurs when o > 1. When y < 0,
a(y) = 1 so m(y) = 1/
)

(

( oo 1oo loo . (

_

_ _ _ _

)

)

)

where in the last step we have changed variables x = u + 1, y = v + 1 to get rid of the 1 's. The next lemma estimates the last expression and shows that it is < 00 if = 00 if

(2.8) Lemma. If o > 0,

0> 1 0< 0� 1

(

)

oo 1+ 0 l+ " y6 }r dx exp y (1 xo) -+ 1 as y -+ oo � y

x = y + zy-6 we have y y- 6 l+ o l+ o -6 J dz exp - }r +z w6 dw dx exp y 1 - x0 = y + o y

Proof Changing variables

}ry oo

dv m(v) (
A similar argument shows that the second expression in (2.1) is (except for a factor of 2) h (f) and the proof is complete. D

Example

(

____

{J {J r r 2 dv fl (r) = Jo
1

exp

l+o 1+ dv exp (1 + 1v) - 1 roo du exp -(1 + u) 0 + 1 Jr>O o +o 1+o Jv oo o l+ l+" r = J dv r du exp (1 + v) - (1 + u) o 1+0 Jv 1+ 6 l+ o = dy dx exp .::.Y . 1 x +0 y 1

x j r and using Fubini's theorem

=2

(1Y -(1 + z)" dz) = ( -(1 +1�1;6 + 1 )

so
D

2 r fl (x) = Jr o dy Jo dz
= o


00 =

To complete the proof now we only have to prove (2.2a) and (2.2b ) . In view of (2.7} it suffices to prove these results with f replaced by fl . Changing variables z =
11/J(x)

To evaluate the first integral we note that if y > 0

(

-6

)

roo

(

Since z � g+z y w6 dw � zy-6 (y + zy-6 ) 6 -+ z as y -+ follows from the dominated convergence theorem.

6.3.

)

oo, the desired result

o

Recurrence and Transience

The natural scale, which was used in Section 6.1 to construct one dimensional diffusions, can also be used to study their recurrence and transience. Let Xt be a solution of MP(b, a) and suppose

Section 6.3 Recurrence and Transience 221

220 Chapter 6 One Dimensional Diffusions

(1D) b and u are continuous with u(x) > 0 for all x

(3.4) Corollary. X is recurrent if and only if t.p{R) = R.

Let a(x) = u2 (x), and let t.p be the natural scale defined by � dz dy t.p(x) = exp

!�is co.nclusion should not be surprising. yt = t.p{X1) is a local martingale so It IS a time change of a Brownian motion run for a random amount of time and if t.p{R) =f. R that random amount of time must be finite. To understand the dividing line between recurrence and transience we con­ sider some examples. We begin by noting that the natural scale only depends on b and u through the ratio 2bfa.

1 (1Y -2:(�] )

Let Ty = inf{t > 0 : X1 = y} and let r = Ta 1\ Tb . The first detail is to show

(3.1) Lemma.

If a

< x < b then P�(r < oo) = 1.

3.1. (i) Show that if b(y) � 0 for y � y0 then P�(To < oo) = 1 for all x > 0. (ii) Show that if b(y)fa(y) � > 0 for y � y0 then P�(To < oo) < 1 for all x > 0. Exercise

We saw in Section 6.1 that Yi = t.p(X1) is a solution of MP(O, h) where h given by (1.5) is positive and continuous. Thus (6.1) in Chapter 5 implies that Y can be constructed as a time change of Brownian motion, B1 :

Proof

Since Brownian motion will exit (t.p(a), t.p(b)) with probability one, Yi will exit (t.p(a), t.p(b)) with probability one, and X1 will exit (a, b) with probability one.

0

With (3.1) established we can study the recurrence and transience as we did for Brownian motion. t.p(Xti\T) is a uniformly bounded martingale, so the optional stopping theorem implies

3.1. The exercise identifies the interesting case as being b(y) -> 0 with b(y)fa(y) ....,. 0 as y ....,. oo. Suppose that for x � 0 2b(x)fa(x) = C(1 + x) - r where C, r > 0 Example

If r >

t.p(b) - t.p( X) P�(Ta < n) = t.p(b) t.p(a) _

t.p(x) - t.p(a) P�( n < Ta ) = t.p(b) t.p(a)

100 -C(1

+ z) - r dz = -J( > -oo

l

t.p( 00) = r oo exp lo

(-_2_ 1 - r { (1

+ y) l - r - 1

l) dy < 00

In the borderline case r = 1 the outcome depends on the value of C.

_

Letting t.p{oo) = limb..... t.p(b) and t.p(-oo) = lima..... - t.p(a) (the limits exist oo we have oo since t.p is strictly increasing)

(3.3) Theorem. Suppose a < x < b. P�(Ta < oo) = 1 it' and only if t.p{oo) = oo. P�(Tb < oo) = 1 if and only if t.p( -oo) = -oo.

1 then

so t.p'(y) � e - K for all y > 0 and t.p(oo) = oo. If r < 1 y c 1 -r - r dz = - 1 - r { (1 + y) - 1 } o -C(l + z) and it follows that

and solving we have

(3.2)

€

In one dimension we say that X is recurrent if P�(Ty < oo) = 1 for all y. From (3.3) it follows that

so

{ =< oooo

if C > 1 0 if C � l To check the last calculation note that Example 1.2 in Chapter 5 shows that the radial part of d dimensional Brownian motion has b(r)fa(r) = (d - 1)/2r, while Section 3.1 shows that Brownian motion is recurrent in d < 2 and transient in d > 2, which corresponds to C � 1 and to C > 1 respectively. Sticklers for

222 Chapter 6 One Dimensional Diffusions detail may complain that we do not have any Brownian motions defined for dimensions 2 < d < 3. However, this analysis suggests that if we did then they would be transient. Exercise 3.2. Suppose a and b satisfy {1D) and are periodic with period 1, i.e., a(y+ 1) = a(y) and b(y+ 1) = b(y) for all y. Find a necessary and sufficient condition for Xt to be recurrent. 6.4. Green's Functions

Suppose !Ct is a solution of MP(b, a), where b and a satisfy {1D) . Let a < b be real numbers, which you should not confuse with the coefficients, let D = (a, b), and let T = inf{t : Xt fl. (a, b)} . Our goals in this section, which are accomplished in {4.5) and {4.4) below, are to show that (i) if g is bounded and measurable then

E:c

1T g(Xs) ds = j GD(X, y)g(y) dy

and {ii) to give a formula for the Green's function GD (x, y). As in the discussion of {8.1) of Chapter 4, we think of GD(x, y) as the expected amount of time spent at y before the process exits D when it starts at x. The rigorous meaning of the lasf sentence is given by {i) . The analysis here is similar to that in Section 4.5 but instead of the Lapla­ cian, D., we are concerned with 1

Lf(x) = 2 a(x) J" (x) + b(x)f'(x) The first step is to generalize {1.3) from Chapter 3. {4.1) Lemma. sup:c e( a , b ) E:cT <

oo.

Proof Let

v = ( 0, f(x) = (£2 - (x - v)2)lp,

Section 6.4 Green 's Functions 223 and note that f ;::: 0 in (
-21p. Ito's formula implies f(Yt ) - f(Yo) = local mart. - t h(Ys) ds Jo P Our choice of p implies h(y) 2:: p, so f(Yt) + t is a local supermartingale on [0, r) . Using the optional stopping theorem at time Tn A n where Tn = inf{t : Yt fl. (
n -+ oo and use the monotone convergence theorem.

D

(4 . 2) Theorem. Suppose g is bounded. If there is a function v with (i) v E C2 , Lv = -g in (a, b) {ii) v is continuous at a and b with v(a) = v(b) = 0 then T v(x) = E:c g(Xs) ds

1

Proof Let

Mt = v(Xt) + J; g(Xs) ds. (i) and {1.1) imply that for t < T t v(Xt) - v(Xo) = local mart. - g(Xs) ds

1

so Mt is a local martingale on

[0, r) . If v and g are bounded then for t < T

(4.1) implies that the right-hand side is an integrable random variable so (2.7) in Chapter 2 and (ii) imply

MT = lim t jT Mt = lor g(Bt) dt T v(x) = E:cMo = E:cMT = E:c g(Bt) dt

1

D

Solving ( 4.2). The next two pages will be devoted to deriving the solution to the equation in (4.2). Readers who are content to guess and verify the answer can skip to (4.4) now. To solve the equation it is convenient to introduce 1 m(x) =

Section 6.4 Green 's Functions 225

224 Chapter 6 One Dimensional Diffusions

cp( x) is the natural scale and note that (4.3) _1_ .!!_ -1- df - a(x) d2f + a(x) -cp11(x) df - Lf ( x ) 2m(x) dx cp' (x) dx 2 dx2 2 cp'(x) dx since the definition of the natural scale implies cp"(x)fcp'(x) = - 2b(x)fa(x). m where

(

)

(

_

)

(

)

d 1 dv = -2m(x)g(x) dx s(x) dx - -

·

and integrate to conclude that for some constant f3

1Y

1 dv s(y) dy = f3 - 2 a dz m(z)g(z) Multiplying by s(y) on each side, integrating y from a to x, and recalling that v(a) = Q and s = cp' we have ( a)

v(x) = f3( cp(x) - cp(a)) - 2

1 dy s(y) 1Y dz m(z)g(z)

(c)

+

u(x)

1b dy s(y) :��� = ;�:� (cp(b) - cp(x))

_

1

ly

u(x) = cp (x) _- cp(a) = Px (Tb < Ta) cp(b) cp(a) we have (b )

1b dy s(y) 1Y dz m(z)g(z) x - 2 1 dy s(y) 1Y dz m(z)g(z)

v(x) = 2u(x)

=

·

= (1 - u(x))

Multiplying the last identity by 2 J: dz m(z)g(z), we have

2u(x)

1b dy s(y) 1x dz m(z)g(z)

= 2(1 - u(x))

1x dy s(y)

1x dy s(y) 1x dz m(z)g(z)

Using this in (c) we can break off part of the second term and cancel with the first one to end up with (d)

lx dy s(y) lx dz m(z)g(z) b 2u(x) 1 dy s(y) 1Y dz m(z)g(z) x 2(1 - u(x)) 1 dz m(z)g(z) 1z dy s(y) 2u(x) 1 dz m(z)g(z) 1 dy s(y)

v(x) = 2(1 - u(x)) +

x

2 f3 = cp(b) cp (a) dy s(y) dz m(z)g(z) a a Plugging the formula for f3 into ( a) and writing

1x dy s(y) 1Y dz m(z)g(z) b 2u(x) 1 dy s(y) 1Y dz m(z)g(z)

Recalling s(y) = cp'(y) we have

Using Fubini's theorem now gives

In order to have v(b) = 0 we must have

b

J: dy into an integral over [a, x] and one over [x, b] we have

v(x) = 2(u(x) - 1)

_

is called the (density function of the) speed measure, though as we will see in Example 4.1 the rate of movement of the process near x is inversely proportional to m(x)! To simplify notation and to facilitate comparison with our source for this material, Karlin and Taylor (1981), Vol. II, we let s(x) = cp'(x) be the derivative of the natural scale. To solve equation Lf = -g now, we use (4.3) to write -

Breaking the first

(e)

v(x) =

+

b

>

b

If we define G(x, z) to be

(4.4)

- cp (a) . (cp(b) - cp (z)) m(z) 2 cp(x) cp(b) - cp(a) (b) - cp (x) . ( (z) - (a)) m(z) 2 cp (b) cp cp - cp(a) cp

then we have (f)

v(x) =

1b G(x, z)g(z) dz

when

z 2:: x

when z �

x

Section 6.4 Green 's Functions 227

226 Chapter 6 One Dimensional Diffusions

4.1. Speed measure? If X has no drift (e.g., Brownian motion or any diffusion on its natural scale) then cp (x) = x and (4.4) becomes

Combining the calculations above with (4.2) shows

Example

{

{4.5) Theorem. If g is bounded and measurable then

E:r:

1r g(X8 ) ds = 1b G(x, z)g(z) dz

Proof By the monotone class theorem it suffices to prove the result when g is

continuous. To do this it suffices to show that the v defined in (f) satisfies the hypotheses of (4.2) . To check (ii) , we note that as x --. a or x --. b, G(x, z) --. 0, and G is bounded so the bounded convergence theorem implies v(x) --. 0. To check that v E C2 we note that :r: cp(b) cp(a)

1 (cp(z) - cp(a)) m(z)g(z) dz b + (cp(x) - cp(a)) 1 ( cp(b) - cp(z)) m(z)g(z) dz

;

v(x) = (cp(b) - cp(x))

Differentiating and noting that the terms from the limits of the integrals cancel we have :r:

cp(b) ; cp(a) v'(x) = -cp' (x) + cp'(x)

Differentiating again we have

1 (cp(z) - cp(a)) m(z)g(z) dz

1b (cp(b) -
+ cp"(x)

J +

1� h (z - x + h) m(z) dz "' m(x)h2

as h __. 0. Thus m(x) gives us the time that Xt takes to exit a small interval centered at x, or to be precise , the ratio of the time for Xt to that for Brownian motion. For another interpretation of m note that {11.7) in Chapter 2 implies that if Lf is the local time at x up to time t then for a bounded measurable g

J Lfg(x) dx = 1 g(X.) d(X}t = 1 g(X.)a(X.) ds t

t

= 1 so m(x) = 1/a(x) and taking g(x) = f(x)m(x) t

Thus multiplying the local times by m(x) converts them into occupation times. This and the previous interpretation suggest that it would be natural to call m{ x) dx the occupation measure. However, . it is too late to try to change the original name, speed measure. D

1b (cp(b) - cp(z)) m(z)g(z) dz

- cp'(x)( cp (b) - cp(a))m(x)g(x)

a(x)/2,

1 cp(b) - cp(a) Lv(x) = - a(x ) · cp'(x) (cp(b) - cp (a)) g(x) 2 2 cp' (x)a(x) so Lv =

{4.6)

+h E:r: T(:r: - h,:r:+h) = r :r: (x + h - z) m(z) dz

J m(x)Lf f(x) dx = 1 f(X.) ds

1a

This shows v E C2 • Multiplying the last two equations by b(x) and adding, then using Lcp = 0 and the definition of m(x) we have

{

When Xt has no drift cp'(x) we have

cp (b) - cp(a) v"(x) = -cp" (x) :r: (cp(z) - cp (a))m(z )g(z ) dz 2

2m(z)(b - z)(x - a)f(b - a) when z ;::: x G(x ' z) = 2m(z)(b - x)(z - a)f(b - a) when z :::; x Taking g = 1 , a = x - h, and b = x + h the last expression becomes m(z)(x + h - z) when z ;::: x G(x ' z) = m(z)(z - x + h) when z :::; x Letting T(a , b ) = inf{t : Xt ¢ (a, b)}, and using (4.5) with g = 1 we have

-g. This shows that v satisfies (i) in (4.2) and the result follows.

4.2. Second Proof of Feller's test. Let T:r: = inf{t : Xt = x}, Ta = lim:r:J.a T:r: and T(a , b ) = Ta 1\ n. We content ourselves to establish a variant of {b) in {2.1). The first of two steps is Example

{4.7) Lemma. Let D

0 < b < f3. Then

1° (cp(z) - cp(a)) m(z) dz < oo

228 Chapter 6 One Dimensional Diffusions if and only if infcx< a < O Po(Ta

Section 6.5 Boundary Behavior 229

< n) > 0 and SUPcx
Letting T = Ta 1\ Tb and repeating the proof of (1.2) in Chapter 3 now, we have

00 .

Px (T > kM) = Ex (PxM (r > (k - 1)M); T > M) � (1 - f ) sup Py (r > (k - 1)M) cx
Proof (3.2) implies

- so(O) Po(Ta < n) = so(b) so(b) _ so(a) so infcx < a < O Po(Ta < n ) and using (4.4) we have

It follows by induction that for all integers

> 0 if and only if so( a) > -oo. Taking g = 1 in (4. 5)

- so( a) 1b ( - so(z)) m(z) dz Ex T(a ,b) = 2 so(x) so(b) _ so(b) so(a) (4.8) - so(x) lx ( + 2 so(b) so(b) _ so( a) a so(z) - so( a)) m(z) dz The first integral always stays bounded as a ! a. So EoT(a ,b) stays bounded as

x

a --.. a if and only if so( a) > -oo and

1°(so(z) - so(a))m(z) dz < oo The- two conditions in ( 4. 7) are equivalent to Po(Tcx n ) < oo . To complete the proof now we will show (4.9) Lemma. If Po(Ta

D

<

Tb) > 0 and Eo(Tcx l\

P(r > kM) � (1 - f)k

Using (5.7) in Chapter 1 of Durrett the moments of the exit time r.

6.5.

(1995) now, we get an upper bound on all

D

B oundary B ehavior

In Section 6.1 when we constructed a diffusion on a half line or a b ounded interval, we gave up when the process reached the boundary and we said that it had exploded. In this section, we will identify situations in which we can extend the life of the process. To explain how we might continue we begin with a trivial example. Example 5.1. Reflecting boundary at 0. Suppose b(x) = 0, and u(x) is positive and continuous on [0, oo ). To define a solution to dXt = u(Xt) dBt with a "reflecting boundary at 0," we extend u to R by setting u(-x) = u(x), let yt be a solution of dyt = u(yt ) dBt on R and then let Xt IYt I ·

=

< n) > 0 then Eo(Ta 1\ n) < oo.

P0(Ta < n) > 0 implies Po(Ta < oo) > 0. Pick M large enough so that P0(Ta � M) � f > 0 and that P0(Tb � M) � f > 0. By considering Proof

the first time the process starting from 0 hits property, it follows that if 0 < x < b then .

sup

xe( cx,b)

k�1

To use the last recipe in general we start with Xt a solution of MP(b, a) and let so be its natural scale. yt = so(Xt) solves MP(O, h) where

x and using the strong Markov

Po(Tb � t) = Eo(Px (Tb � t - Tx ) i Tx � t) � Px (n � t)Po(Tx � t) � Px (Tb � t) A similar argument shows

To see if we can start the process yt at 0 we extend h to the negative half-line by setting h( -y) = h(y) and let Zt be a solution of MP(O, h) on R. If we let m(y) = 1/h(lyl) be the speed measure for Zt , which is on its natural scale, we can use (4. 6) and the symmetry m( -y) = m(y) to conclude

1

for a

<x<0

Combining the last two results shows that for all a < x


2 Eor( - f, f) = Jro { f - y)m( y) dy Changing variables y = so ( X ) ' =

dy = so' ( X ) dx ' f = so( 6) the above

1 dx 106 (so(6) - so(x)) so'(x)a(x)

l

230 Chapter 6 One Dimensional Diffusions Now, (i) recalling that the speed measure for X1 is m(x) = 1/rp'(x)a(x), and changing notation rp'(z) = s(z), and (ii) using Fubini's theorem, the above

16 (16 s(z) dz) m(x) dx = 16 lz m(x) dx s(z) dz

=

Introducing M as the antiderivative of m to bring out the analogy with the condition in Feller's test (2.1), we have

EoT( - E,£) = 2

(5.1)

16 (M(z) - M(O)) s(z) dz

To see that this means that the process cannot escape from 0, we note that repeating the proof of (4.9) shows

(5.2) Lemma.

If Po(T(- E,£) < oo) > 0 then Eor( - E,£) < oo.

M so that Po(T( - E, £) :::; M) = o > 0. Using the strong Markov property as in the proof of (4.9) we first get Proof Pick

sup Px ( 1"(- £,£)

xE ( - £,£)

and then conclude that for each integer sup Px (T(- E,£)

xE ( - £,£)

> M) :::; 1 - {j

> kM) :::; (1 - o)k

Consider a diffusion on (0, r) where r :::;

0

oo, let q E (0, r) , and let

lg (rp(z) - rp(O)) m(z) dz J = fo q (M(z) - M(O)) s(z) dz I=

Feller's test implies that when I < oo we can get IN to the boundary point. The analysis above shows that when J < oo we can get OUT from the boundary point.

This leads to four possible combinations, which were named by Feller as follows I name J < oo < oo regular < oo = 00 absorbing = 00 < oo entrance = oo = 00 natural The second case is called absorbing because we can get in to 0 but cannot get out. The third is called an entrance boundary because we cannot get to 0 but we can start the process there. Finally, in the fourth case the process can neither get to nor start at 0, so it is reasonable to exclude 0 from the state space. We will now give examples of the various possibilities. Example

5.2.

Feller's branching diffusion. Let

dX1 = (JX1 dt + ujX;dBt Of course we want to suppose u > 0 but we will also suppose (J > 0 since the calculations are somewhat different in the cases (J = 0 and (J < 0. See Exercise 5.1 below. Using the formula in (1.4), the natural scale is rp(x) =

k ;?: 1

from which the desired result follows.

Section 6.5 Boundary Behavior 231

lx

exp

(lay -:�z dz) dy

0"2 ( 1 - e - 2{3xf u 2 ) = r e - 2{3y/u2 dy = 2(3 Jo which maps [0, oo) onto [0, u2 /2(3). The speed measure is 1 = e2f3xf u2 j(u 2x) m(x) = (x)a(x) rp' To investigate the boundary at 0, we note that

since the integrand converges to 1/u2 as x � 0. To calculate J we note that m(x) ,.,. 1/(u2 x) as x � 0 so M(O) = - oo and J = oo. The combination I < oo and J = oo says that the process can get into 0 but not get out, so 0 is an absorbing boundary. To investigate the boundary at oo, we note that

1100 m(x)(rp(oo) - rp(x)) dx = 100 2(Jx1 dx = oo 1

Section 6.5 Boundary Behavior 233

232 Chapter 6 One Dimensional Diffusions To calculate J we note that when {3 2:: 0, m{x) 2:: 1/{cr2x) so M(oo) = oo and J = oo. The combination I = oo and J = oo says that the process cannot get into or out of oo, so

The first conclusion is reasonable since in d > 2 we can start Brownian motion at 0 but then it does not hit 0 at positive ti�es. The second conclusion can be thought of as saying that in dimensions d < 2 Brownian motion will hit 0.

oo is a natural boundary.

Exercise 5.2. Show that 0 is an absorbing boundary if 1 :::; -1. We leave it to the reader to ponder the meaning of Brownian motion in dimension d :::; 0.

Exercise 5.1. Show that 0 is an absorbing boundary and boundary when (a) {3 = 0, (b) {3 < 0. Example

5.3.

oo is a natural

Example

5.4.

Power noise. Consider

Bessel process. Consider

dXt = Xf dBt

dXt = 2 t dt + dBt

�

Here 1 > -1 is the index of the Bessel process. To explain the restriction on 1 and to prepare for the results we will derive, note that Example 1.2 in Chapter 5 shows that the radial part of d dimensional Brownian motion is a Bessel process with 1 = (d - 1). The natural scale is

cp ( x) =

lx exp (- 1Y 1/z dz) dy

= }r y- '"'( dy =

1

{ (x1 -"' - 1)/{1 - 1) ln x

if / = 1 if / # 1

From the last computation we see that if 1 2:: 1 then cp {O) = -oo and I = oo. To handle - 1 < 1 < 1 we observe that the speed measure

1 = z"� m(z) = (z)a(z) cp' So taking q = 1 in the definition of I I=

11 1z1--"t'Y z"� dz < oo 0

--

To compute J we observe that for any 1 > -1, J=

M(z) = z"1 +1 /(1 + 1) and

1o 1 1z"1++11 z-"1 dz < oo --

0 is an entrance boundary if 1 E [1, oo) if r E {-1, 1) regular b oundary

Combining the last two conclusions we see that

{0, oo ). The natural scale is cp(x) = x and the speed measure is m{ x) 1/(cp'(x)a(x)) = x -26 so 1 1 oo �f 5 < 1 I= x - 26 dx = =< 00 If 5 2:: 1 0

on

1

When 5 2::

{

1/2, M{O) = -oo and hence J = oo. When 5 < 1/2 1 z 1 -26 J= -- dz < oo 0 1 - 25

1

Combining the last two conclusions we see that the boundary point 0 is if 5 E [1, oo) if 5 E [1/2, 1) if 5 E {0, 1/2) The fact that we can start at 0 if and only if 5 < 1/2 is suggested by (3.3) and Example 6.1 in Chapter 5. The new information here is that the solution can reach 0 if and only if 5 < 1. For another proof of that result, recall the time change recipe in Example 6.1 of Chapter 5 for constructing solutions of the equation above and do natural absorbing regular

Exercise

5.3. Consider

Show that (a) for any 5 > 0, P1 (H� {c) P1 (H6 = oo) = 1 when 5 ;:::) .

< oo) = 1. (b) E1 H6 < oo when 5 < 1.

Exercise 5.4. Show that the boundary at oo in Example 5.4 is natural if 5 :::; and entrance if 5 > 1.

1

234 Chapter 6 One Dimensional Diffusions Remark. The fact that we cannot reach oo is expected. (5.3) in Chapter 5 implies that these processes do not explode for any 6. The second conclusion is surprising at first glance since the process starting at oo is a time change of a Brownian motion starting from oo. However the function we use in (5.1) of Chapter 5 is g(x) = x - 26 so (2.9) in Chapter 3 implies

EM

1T1 g(B. )1(B. s;M) 1M -26 ds =

which stays bounded as M

x

•

Section 6.6 Applications to Higher Dimensions 235 for all x with lxl :::; r. Let h (x) =

Noting that cosh(y) �

5.5. Wright-Fisher diffusion. Recall the definition given in Ex­ 1.7 in Chapter 5. Show that the boundary point 0 is absorbing if {3 = 0 regular if {3 E (0, 1 /2)

t

10 I:d a sinh(aX! )b;(X.) ds + local mart. i=l 1 11 d a 2 cosh(aX.i )a;;(X. ) ds +2 0 I: i=l

h(X1 ) - h(Xo) =

2x dx

� oo if and only if 6 > 1.

E1= 1 cosh( ax;) . Using Ito's formula we have

1 and

Exercise

ample

entrance if {3 � 1/2 Hint: simplify calculations by first considering the case a = 0 and then arguing that the value of a is not important.

6.6.

Applications to Higher Dimensions

In this section we will use the technique of comparing a multidimensional dif­ fusion with a one dimensional one to obtain sufficient conditions (a) for no explosion, (b) for recurrence and transience, and (c) for diffusions to hit poin:s or not. Let Sr = inf{t : I X1 1 = r} and S00 = limr-+ oo Sr . Throughout this section we will suppose that X1 is a solution to MP(b, a) on [0, Soo ) , and that the coefficients satisfy (i) b is measurable and locally bounded (ii) a is continuous and nondegenerate for each x, i.e., ; Y a;i (x)yi > 0 when y f. 0

Li,i

C < oo

Proof Assumptions (i) and (ii) imply that we can pick an a large so that if

lxl :::; r

(

eY e -Y 2' 2

)

� l sinh(y) l

our choice of a implies

�2 cosh(aX! )a;;(X.) + a sinh(aX! )b; (X. ) � a cosh(aX!) ( � a;;(X. ) - lb;(X. ) I ) � a cosh(aX!) � a

so h(Xt ) - tad is a local submartingale on theorem at time Sr 1\ t we have

[0, Sr ). Using the optional stopping

0 :5 h (x) :5 Ex{h(XsrAt) - ad(Sr 1\ t)} Since

h(XsrAt) :::; d cosh(ar) it follows that Ex(Sr 1\ t) ::::; cosh(ar) /a

Letting t � oo gives the desired result.

0

Remark. Tracing back through the proof shows

chosen so that

The first detail is to show that X1 will eventually exit from any ball.

(6.1) Theorem. Let Sr = inf{t : IXd = r} . Then there is a constant so that ExSr :::; C for all x with lxl :5 r.

cosh(y) � max

C = cosh(ar)/a where a is

a 2a;;(x) � 1 + lb;(x) l To see that the last estimate is fairly sharp consider a special case Example 6.1. Suppose d = The natural scale has

so (x) =

1, a(x)

=

1, and b(x)

= -{3 sgn(x) with {3

10 :z; e2f3Y dy = -2{31 (e 2f3x - 1)

> 0.

236 Chapter 6 One Dimensional Diffusions

Section 6.6 Applications to Higher Dimensions 237

for x � 0 and cp (-x) = -
Our plan is to compare Rt =

x

_ = e - 2.6l l m(x) = _1(x) cp' Using (4.8) now with b = r, a = -r, x = 0 and noting that (r) - cp (O) = 2 c,o (O) - <,o (-r) = 1 2 cp cp(r) - <,o (-r) cp(r) - <,o (-r) we have

dZt = p(Zt ) dt + B(Zt ) dBt on

s(x) = exp cp(x) =

z z

+

1° � (-e2.61z l

- r 2/3 The two integrals are equal so r e 2.6 (r Eor( - r ,r ) =

� fo (

(

-

-

"'

)

+ e2,6 r e -2,6lzl dz

- !.. ) l r

t

d

1"' s(y) dy

In view of Feller's test, (2.1), the next result says that if Z1 can't reach finite time then X1 does not explode.

oo in

(6.2) Theorem. There is no explosion if

1 1 e 2,6(r - z) = __ ( e 2,6r - 1) - !:_ = f3 2� f3 0 2� To compare with (6.1) now, note that when a(x) = 1 and b(x) = -{Jsgn(x), the remark after the proof says we can pick a = 2/3 + 2 so the inequality in (6.1) is -

J.L (y) y) (-!"' 2B2(y)

m(x) for Zt :

m(x) = 1/(s(x)B2 (x))

z - 1) dz )

(0, oo)

So we introduce the natural scale cp (x) and speed measure

1 e ""'r o Eor( - r,r) = r ( - " - e "-" ) e -"" dz J 2/3 "'

1Xtl 2 with

.

100 dx m(x)(cp(oo) - cp(x))

= oo

0

Proof To prove the result we will let Rt = 1 Xtl 2 and find a function g with g(r) -;. oo as r -;. oo so that e- t g (Rt) is a supermartingale while Rt � 1. To see that this is enough, let Sr = inf{t : Rt = r} and use the optional stopping theorem at time St A Sn A t to conclude that for l x l > 1

With the little detail of the finiteness of E.,Sr out of the way we now intro­ duce a comparison between a multidimensional diffusion and a one dimensional one, which is due to Khasminskii (1960). We begin by introducing three pairs of definitions that may look strange but are tailor made for computations in (d) , (e) , and (a) of the proof of (6.2) .

L�tting Soo = limr joo Sr , then n -;. oo and t -;. oo we have P ( S < St ) = 0. To define the function g , we follow the proof of (2.1). Let yt = cp(Zt), use the construction there t o produce a function f so that e - t f(Yt ) is a local martingale, then take g = f o

Eor( - r, r) a.

+ 2)r) � cosh((2/3 2/3 + 2

Explosions

d

a(x) = 2xT a(x)x and f3(x) = I ) x;b;(x) + a;;(x) i=l : l x l2 = r and p(r) = sup { a(x) : l x l2 = r } v(r) = sup

{ !�:�

}

p(r) = p(r)v(r) and

B(r) = �

x 00

e - 1 g(Z1) - g (Zo) = + +

-11 1t t �1

e -• g (Z, ) ds

e -• g ' (Z, )p(Z, ) ds + local mart. e -• g" ( Z. )B 2 ( Z. ) ds

Section 6.6 Applications to Higher Dimensions 239

238 Chapter 6 One Dimensional Diffusions Since e - t g (Zt ) is a local martingale it follows that 1 2 B2 (z)g11(z) + J-L (z)g'(z) = g (z) Using the definition of B and J-L now we can rewrite this as p(z)g"(z) + v(z)p(z)g'(z) = g (z) (a) Before plunging into the proof we need one more property which follows from (2.6c) and the fact that our natural scale is an increasing function with 1p(1) = 0: (b) g ' ( r) ;?: 0 for r ;?: 1 A little calculus gives (c)

D; g ( lxl2 ) = g' (lxl 2 )2x; D;j g ( lx l2 ) = g" (lxl 2 )4x;Xj i :j:. j D;;g( lx l2 ) = g" (lxl 2 )4x; + g' (lxl2 )2

Using Ito's formula now we have e - t g (Rt) - g (Ro) =

1t

-e -• g (R. ) ds t + L: e -• g' (R. )2X; b;(X. ) ds + 1ocal mart. i

l0

� � 1t + � 2;::: 1t e -• g ' (R. )2aii (X,) ds

+

• ,J

Using the functions a(r) and (3(r) introduced above we can write the last equa­ tion compactly as e - t g (Rt) - g (Ro) = local mart. t (d) + e -• { - g (R. ) + (3(Xt) g' (R. ) + a(X, ) g" (R. )} ds

1

To bound the integral when R. ;?: 1 we use (i) the definition of 11 and (b) , then (ii) the equation in (a) , the fact that g ;?: 0, and the definition of p

(e)

b. Recurrence and Transience

The next two results give sufficient conditions for transience and recurrence in d > 1. In this part we use the formulation of Meyers and Serrin (1960). Let a(x) =

{ !��:>

}

' (R. ) + g" (R. ) a(X. )

:::; - g (R. ) + { v (R.)g ' (R. ) + g" (R.) } a(X. ) g (R,) p(R. ) :::; 0 :::; -g (R. ) + p(R. )

{ }

1:1

. a(x)

1:1

de ( X ) =

2x · b(x) + tr (a(x)) a(x)

where tr (a(x)) = 2:::; a;;(x) is the trace of a. We call de(x) the effective dimension at x because (6.3) and (6.4) will show that there will be recurrence or transience if the dimension is < 2 or > 2 in a neighborhood of infinity. To formulate a result that allows de(x) to approach 2 as lxl --+ oo we need a definition: c5(t) is said to be a Dini function if and only if

100 c5(t)t dt < 1

00

In the next two results we will apply the definition to

( j c�s) ds)

e -• g " (R. )4X!X�a;i (X. ) ds

•

- g (R. ) +

Combining (d) and (e) shows that e - t g (Rt) is a local supermartingale while o Rt ;?: 1 and completes the proof.

c5(t) = exp -

t

To see what the conditions say note that c5(t) = (log t) -P is a Dini function if p > 1 but not if p :::; 1 and this corresponds to c(s) = pf log s . (6_.3) Theorem. Suppose de(x) ;?: 2(1 + c:(lxl)) for lxl ;?: R and c5(t) is a Dini function, then Xt is transient. To be precise, if lxl > R then P.,(SR < oo) < 1. Here Sr = inf{t : IXt l = r} , and being transient includes the possibility of explosion. (6.4) Theorem. Suppose de(x) :::; 2(1 +c(lx l)) for lxl ;?: R and c5(t) is not a Dini function, then Xt is recurrent. To be precise, if lxl > R then P.,(SR < oo) = 1. Proof of { 6.3)

The first step is to let

240

Chapter 6 One Dimensional Diffusions

Section 6.6 Applications to Higher Dimensions

(-lr � ) ( ], d )

which has

The upper bound is independent of s and

c t) dt ::; 0
and hence satisfies

Proof of (6.4)

r
Letting Rt = 1 Xtl 2 and using Ito's formula we get the following which we will use four times below:

(6.5) Lemma. If rg11(r) + 'Y(r)g'(r) = 0 for r E (u2, v2) and g'(lxl 2 ) (de (x) 2'Y( I xl2)) ::; 0 when lxl E ( u, v), then g(Rt) is a local supermartingale while u < !Xt l < v. •

Proof Using Ito's formula with ( c) from the proof of (6.2)

g(Rt ) - g(Ro) = L ( g 1(R,)2X!b;(X, ) ds + local mart. ; lo + L l' (R. )4x;x{a;i (X.) ds

This time we let

Since 1/;'(r) = -
r'l/;"(r) + (1 + c(r))'l/;'(r) = 0 but this time 1/;'(r) 2:: 0. Since we have assumed de (x) ::; 2(1+c(lxl)) for lxl 2:: R it follows from (6.5) that 1/;(Rt) is a local supermartingale while Rt ;::: R2 • If R ::; r < s < oo using the optional stopping theorem at time T = Sr A S, ((6.1) implies that Px( T < oo ) = 1) we see that

s2) - 1/J (I xl2) Px(Sr < S,) 2:: 1/;( 1/;(s2) - 1/;(r2) oo and noting that 1/;(s2) oo since t5(t) is not a Dini function

Letting s --+ gives the desired result.

t

1 2ii(X, ) (R.g''(R,) + g'(R.) de(�.) ) ds

Using our two assumptions on g, we get for u <

0

Rearranging gives

� ij + � L l(R,)2a;;(X,) ds i

= local mart. +

(6.3) follows.

241

c.

--+

0

Hitting Points

The proofs of (6.3) and (6.4) generalize easily to investigate whether mul­ tidimensional diffusions hit 0. Let T0 = inf{t > 0 : Xt = 0}.

!Xt I < v

(6.6) Theorem. If de (x) 2:: 2(1 - c(lxl)) for lxl ::; 7J and dy y

0

which proves the result.

(6.5) implies that
r

- = oo then

Px (To < oo ) = 0.

(6.7) Theorem. If de (x) ::; 2(1 - c(lxl)) for lxl ::; 7J and dy y

- < oo

Rearranging gives then

Px(To < oo ) > 0.

242 Chapter 6 One Dimensional Diffusions

Section 6.6 Applications to Higher Dimensions 243

We have changed the sign of e to make the proofs more closely parallel the previous pair. Another explanation of the change is that Brownian motion

Using ( 3.2) and the Markov property now extends the conclusion to

is recurrent in d :::; 2 hits points in d < 2 doesn't hit points in d 2: 2 is transient in d > 2 so the boundary is now a little below 2 dimensions rather than a little above.

Proof of (6.7)

Proof of (6.6}

Sin-ce '1/J'(r) =

Let

-so'(r) we have '1/J'(r) 2: 0 and

�1 exp (- 11 e�t) dt) :::; 0 1 - e(r) exp (- 11 e(t) dt) so"( r) =

and hence satisfies

r

t

-

Rearranging gives

(1 -

rso"(r) + e(r))so'(r) = 0 Letting Rt = I Xtl 2 and using ( 6.5 ) shows so(Rt) is a local supermartingale while Rt E (0, rP). If 0 < r < s :::; TJ using the optional stopping theorem at time T = Sr /i. s8 we see that Rearranging gives

Px ( Sr < S8 ) :::; so(lxl 2 )/ so(r2 ) Letting r -+ 0 and noting so (r2) -+ oo gives Px (To < S11 ) = 0 for 0 < lxl < TJ

To replace sf) by 00 in the last equation let Uo = 0 and for Tn = inf{ t > Un - 1 : I Xt l Un = inf{ t > Tn : I Xt l =

n

2: 1 let

= TJ}

TJ/2}

The strong Markov property implies that the Un - Tn are i.i.d. so Un -+ oo as n -+ oo. It is impossible for the process to hit 0 in [rn, un] and hitting 0 in [un , Tn+1 ] has probability 0 so we have

Px (To < oo ) = 0 for 0 < lxl < TJ

(1 - e(r)),P'(r) = 0

I Xtl 2 and using (6.5 ) shows ,P(Rt) is a local supermartingale while Rt E (0, TJ2 ). If 0 < r < s :::; TJ using the optional stopping theorem at time T = Sr /1. s8 we see that

Letting Rt =

so'(r) =

r2

0

This time we let

r,P"(r) + which has

lxl 2: TJ ·

2 ) - 'I/J( I xl2 ) Px ( Sr < S. ) 2: ,P(s ,P(s2) _ ,P(r2) Letting r -+ 0 and noting ,P(r2) 0 gives Px (To < oo) > 0 for lxl < TJ -+

Using ( 3.2 ) and the Markov property now extends the conclusion to

lxl 2: 7J·

0

(6.8) Corollary. If (i) d 2: 3 or (ii) d = 2 and a is Holder continuous then Px (To < oo ) = 0 for all x. Proof Referring to (4.2) in Chapter 5, we can write a(O) = UT AU where A is diagonal matrix with entries A i . Let r be the diagonal matrix with entries /A and let v = uT r. The formula for the covariance of stochastic integrals implies that Xt = VXt has ii(O) = VT a(O)V = I

1

a

So we can without loss of generality suppose that a(O) = I. In d 2: 3, the continuity of a implies de (x) -+ d as x -+ 0 so we can take e(r) = 0 and the desired result follows from (6.6). 6 If d = 2 and a is Holder continuous with exponent 6 we can take e(r) = Cr To extract the desired result from (6.6) we note that •

c - y6 )) ydy = oo ( 11 Cz6-1 dz) ydy = J{o1 (-8(1

[1 Jo exp -

Y

exp

244 Chapter 6 One Dimensional Diffusic;ms since the exponential in the integrand converges to a positive limit as y �

0.

D

It follows from (6.7) that a two dimensional diffusion can hit 0. Example

6.1. Suppose b(x) = 0, a(O) = I, and for 0 < l x l < 1/e let a(x) be

7

Diffusions as Markov Processes

the matrix with eigenvectors

and associated eigenvalues

A2 = 1 - 2pf log l x l These definitions are chosen so that de (x ) = 2 - 2pflog(lxl) and hence we can take c(r) = pf log(r). Plugging into the integral

1 0

1 /e

exp

(- 1 y

1 / e p dz 1

)

dy z og z y

-

1 11/ e =

=

1/ e

0

o

dy exp (p log l log yl) y dy y I Iog y I P

<

00

if p >

In this chapter we will assume that the martingale problem MP(b, a) has a unique solution and hence gives rise to a strong Markov process. As the title says, we will be interested in diffusions as Markov processes, in particular, in their asymptotic behavior as t � oo. 7.1. S e migroups and Generators

The results in this section hold for any Markov process, measurable /, and t 2:: 0, let

1

Xt . For any bounded

T't f(x) = E,z:f (Xt) The Markov property implies that

Remark. The problem of whether or not diffusions can hit points has been investigated in a different guise by Gilbarg and Serrin (1956), see page 315.

So taking expected values gives the semigroup property T3 +d (x) = T3 (Td) (x) (1.1) for. s, t 2:: 0. Introducing the norm II/II = sup lf(x) l and L00 = {! : II/II < oo } we see that Tt is a contraction semigroup on L00 , i.e., (1.2) IITt / 11 ::; II III Following Dynkin (1965), but using slightly different notation, we will de­ fine the domain of T to be 'D(T)

= {! E L00 IITt f - fll � 0 as i � 0} :

As the next result shows, this is a reasonable choice.

(1.3) Theorem. 'D(T) is a vector space, i.e., if fl , 12 E 'D(T) and c1 , c E R, cd1 + c2 !2 E 'D(T) . 'D(T) is closed. If f E 'D(T) then T3 f E 'D(T) and s 2� T3 f

is continuous.

246 Chapter 7 Diffusions

as

Markov Processes

Section

Remark. Here and in what follows, limits are always taken in, and hence

11 · 11· Proof Since T't (cd1 + c2!2) = c1Tt ft + c2T't h , the first conclusion follows from the triangle inequality. To prove 1J(T) is closed, let fn E 1J(T) with we pick n large llfn - fl l < f/3. fn E 1J(T) so l fnfn- -fnll fll 0 . f/3Letforf t><0.to.IfUsing the triangle inequality and the contraction IIproperty Tt (1.2)< now, it follows that for t < to IIT't f - fll :5 IIT't f - T't /nll + IITt fn - fnl l + l fn - fl l :5 l f - fnll + f/3 + f/3 < f continuity is defined for the uniform norm

-4

oo is an entrance boundary.

coo

fk Af�:(O) =

(k )k, Af = f Yk(x) = x2 x

Yk

Returning to properties of actual generators, we have

f E 1J(A) then Af E 1J(T), d =A f= dt Ttl Tt TtAf Tt f - f = 1t T. AJ ds

{1.5) Theorem. 1J(A) is dense in 1J{T). If

Tt f E 1J (A), and

Remark. Here we are differentiating and integrating functions that take values in a Banach space, i.e., that map s E [0, oo ) into L 00 • Most of the proofs from calculus apply if you replace the absolute value by the norm. Readers who want help justifying our computations can consult Dynkin {1965), Volume 1, pages 19-22, or Ethier and Kurtz (1986) , pages 8-10.

E 1J(T) and Da = foa T.J ds. To explain the motivation for the f last definition, we note that Proof Let

r: - til :5 � 1a I T.J - ! I ds

The infinitesimal generator of a semigroup is defined by

T f-f Af = lim h!O h h Its domain 1J(A) is the set of f for which the limit exists, that is, the f for which there is a g so that

Semigroups and Generators 247

Example 1.1. The property in {1.4) looks innocent but it eliminates a number of operators. For example, when k 2:: 3 is an integer we cannot have the kth derivative. To prove this, we begin by noting that 1- + has a local maximum at 0. From this it follows that we can pick a small 8 > 0 and define a function 2:: 0 that agrees with on ( - 8, 8) and has a strict 0 global maximum there. Since k! > 0, this contradicts (1.4).

To prove the last claim in the theorem we note that the semigroup and co_n­ traction properties, (1.1) and (1.2), imply that if s < t 0 IITt f -T.JII = IIT. (Tt -.J- !) I :5 1171 - .! - ! II Remark. While 1J(T) is a reasonable choice it is certainly not the only one. A common alternative is define the domain to be Co, the continous functions which converge to 0 at oo equipped with the sup norm. When the semi-group Co into itself, it is called a Feller semi-group. We will not follow Tthist maps approach since this assumption does not hold, for example, in d = 1 when

7. 1

as

a

-4

-4

o

f E 1J(T) . Now a Th Ya = Jro T• +h f ds = Ya + 1a +h T.J ds - Jro h T. f ds

0 since

so we have Before delving into properties of generators, we pause to identify some operators that cannot be generators.

f E 1J(A) and f(xo) 2:: f(y) for all y then Af(xo) :5 0. Since xo is a maximum, Tt f (xo) - f(xo) :5 0 so Af(xo) :5 0.

{1.4) Theorem. If Proof

0

Ya

Aga = Taf - f. Since the first calculation f as a 0 the first claim follows.

This shows that E 1J(A) and shows E 1J(T) and converges to

Yafa

-4

248 Chapter 7 Diffusions

as

Markov Processes

Section 7.1 Semigroups and Generators 249

7i I Th1i fh- 1i f - 1i At l = 1 7i (Th fh- Af) I Tf A/ I -- 0 $ 1 h h-

The fact that Af E 1J(T) follows from (1.3) which implies (Th f - f)fh E V(T) and V(T) is closed. To prove that / E 1J(A) we note that the contrac­ tion property and the fact that f E 1J(A) imply

But for any y , it follows from the definition of Tr and Fubini's theorem that

I-

I-

0

which is by

The last equality shows that A1i f = 1i Af and that the right derivative of 7i / is 1i Af. To see that the left derivative exists and has the same value we note that ·

1 1 ( $1

)l

Th f - f - A + ll (Af - T Af)ll 1i f - - h f - Ii A/ $ 1i 1i - h t h -h h Th f - I - A t + II Af - Th All i h

1 ; -+ 0

I

as h since f E 1J(A) and Af E 'D(T). The final equation in obtained by integrating the one above it.

(1.5) is

D

(1.6) Theorem. If f E 1J(A) then for any probability measure J.L 1 M/ := /(X1 ) - f(Xo) - Af(Xs ) is a PJ.l martingale with resp ect to the filtration Proof Since

:F8 measurable

ds

0

t.

By the Markov property the second term on the right-hand side is

1

oo

dt

e - >.1 1i f(:c) = Ex

1oo e->.1 /(X1 ) dt

II U>. /11 $ 11/fl/ A. A more subtle fact is that

{1.7) Theorem. U>. maps V(T) 1-1 onto V(A). Its inverse is (A - A). Proof Let A h f = (nf - f)/h. Plugging in the definition of U>. and changing variables = h and = we have

s t+

To conclude that

s t

U>. f E 1J(A) and AU>. / = AU>. / - f we note that h AU>. / - f = A e - >.8 T8 f f

100

:F1 generated by X1 .

f and Af are bounded, M/ is integrable for each Since Mf is

0

Our final topic is an important concept for some developments but will play a minor role here. For each A > define the resolvent operator for bounded measurable f by

Clearly

To make the connection between generators and martingale problems we will now prove

1

(1.5)

U:>J(:c) =

-+ 0

11-8 Af(Xr) dr) 1-8 = I't - s f( y) - f( y) - 1 TrAf(y) dr

(

Ey t (X1 - 8 ) - /{Xo) -

ds - X 1 ds

ll1i/ll $ 11/11 ) that oo h >. e - >.s ll /11 e - 8 11/11

and compare the last two displays to conclude (using

e >.h II Ah U>. f - (AU>./ - 1) 11 $ h- 1 - A h h (1 - e - >. • ) 11/11 II Td - /II

+X1

0.

1

ll ds + X 1

as h -+ The formula for AU>. / implies that and its inverse is A - A.

ds + A 1 ds -+ 0

ds

(A - A)U>. f = f. Thus U>. is 1-1

250 Ch apter 7 Diffusions To prove that

as

Section

·Markov Processes

7.2

Examples 251

Xt defines our pure jump Markov process. To compute its generator we note that Nt = sup { n : Tn ::::; t} has a Poisson distribution with mean >.t so if f is bounded and measurable

U>. is onto, let f E V(A) and note (1.5) implies

1oo e - >.t Tt (>.f - A!) dt = 100 >.e - >.t Tt f dt - 100 e - >.t ! Ttf dt

U>. (>.f - Af) =

Doing a little arithmetic it follows that

Integrating by parts

Tt f(x) - f(x) - >. P(x, dy)(f(y) - f(x)) t = C 1 (e - >.t - 1 + >.t)f(x) + >.(e- >.t - 1) P(x, dy)f(y) + O (t)

J

Combining the last two displays shows complete.

U>. (>.f - A!)

f and the proof is

0

7.2. Examples

In this section we will consider two families of Markov processes and compute their generators.

2.1. Pure jump Markov processes. Let (S, S) be a measurable space a_nd let Q(x , A) S X S -+ [0, oo) be a transition kernel. That is, (i) for each A E S, x -+ Q(x, A) is measurable (ii) for each x E S, A -+ Q(x , A) is a finite measure (iii) Q(x, { x}) = 0 Intuitively Q(x, A) gives the rate at which our Markov chains makes jumps from x into A and we exclude jumps from x to x since they are invisible. To be able to construct a process Xt with a minimum of fuss, we will suppose that >. = SUPze s Q(x, S) < oo Define a transition probability P by Example

:

Since r 1 (e - >. t - 1 + >-.t) -+,0, e - >.t it follows that

as t -+ 0.

-+ 1, and P(x, dy) is a transition probability

I Ttf(x)t- f(x) - ).. J P(x, dy)(f(y) - f(x)) l -+ 0

Recalling the definition of P(x, dy) it follows that

(2.1 ) Theorem. V(A) = V(T) = Af(x) = ·

j

L00 and

j Q(x, dy)(f(y) - f(x))

We are now ready for the main event.

.

±

P(x, A) = Q(x, A) if x rf. A 1

P(x, { x}) = X (>. - Q(x, S) ) Let t 1 . t 2 , . . . be i.i.d. exponential with parameter >., that is, P(ti > t) = e->-t . Let Tn = t 1 + + tn for n � 1 and let To = 0. Let Yn be a Markov chain with transition probability P(x, dy) (see e.g. , Section 5.1 of Durrett (1995) for a defintion), and let · · ·

Example 2.2. Diffusion processes. Suppose we have a family of measures Pz on (C, C) so that under Pz the coordinate maps Xt (w) = w(t) give the unique solution to the MP(b, a) starting from x, and :Ft is the filtration generated by the Xt . Let C_k be the C2 functions with compact support.

(2.2) Theorem. Suppose a and b are continuous and MP(b, a) is well posed. Then C_k C 'P(A) and for all f E C_k

252 Chapter 7 Diffusions

as

Markov Processes

Section 7.2 Examples 253

C2 then Ito's formula implies t f(Xt) - f(Xo) D;J(X. )b;(X.) ds + local mart.

Proof If f E

h(y) L; (Yi - x;) 2 • Using Ito's formula we have t 2(X! - x;)b; (X.) ds + local mart. h(Xt) - h(Xo)

Proof Let

�1 + 21 � Jto D;j/(X. )a;j(X.) ds

=

=

=

I

+ 21

1 ,}

If we let Tn be a sequence of times that reduce the local martingale, stopping at time t A Tn and taking expected value gives AT,. Exf(XtAT,. ) - f (x) Ex Jt D;f(X.)b;(X.) ds o I tAT,. + -21 :l: Ex D;j/(X. )a;i (X. ) ds . . 0 =

2;:

1 ,}

�1

l

If f E C_k and the coefficients are continuous then the integrands are bounded

I

t � Jo 2aii(X.) ds I

Letting Tn be a sequence of. times that reduce the local martingale, stopping at t A Sr A Tn , taking expected value, then letting n -+ oo and invoking the bounded convergence theorem we have

t i\Sr Ex h(XtAs. ) = Ex J{ 2: { 2(X! - x;)b;(X.) + a;;(X.) } ds � 2t(rB + A) 0 n i Since

h 2: 0 and h(Xs. ) r2 it follows that r2 Px (Sr � t) � 2t(rB + A) =

and the bounded convergence theorem implies

which is the desired result.

(2.3)

(2.2) Pick R so that H {x : lxl � R - 1} contains the support of f and let I< {x : l x l � R}. Since f E C_k, and the coefficients are continuous, Af is continuous and has compact support. This implies Af is uniformly continuous, that is, given c > 0 we can pick 8 E (0, 1] so that if l x - Y l < 8 then I AJ(x) - Af(y)j < c. If we let C = supz I Af(z) l and use (2.3) it follows that for x E I< Exf(Xt - f(x ) - Af(x) � Ex t AJ(X. - Af( x)l ds ) I Proof of

]

(2.4) Lemma. Let r > 0, ]{ C Rd be compact, and Kr {y : lx - yj � r for some x E K}. Suppose lb(x) l � B and Li a;;(x) � A for all x E I
=

( �)

B sup P (S � t) � t · 2 r +r x EK x r

l � i

� c + 2C sup Px (So < t) x EK For x rf_ K, f(x) 0 and Af 0. So using (2.3) again gives that for x rf_ f{ Ex f(Xt - f(x) - Af(x) Exf (Xt) � CP (TH < t) x t � C sup Py(S1 < t) y E8K =

The first step is to bound the movement of the diffusion process.

=

=

I

To prove the conclusion now we have to show that

0

I

]

=

l I =

I

where the second equation comes from using the strong Markov property at time TK . (2.4) shows that sup

x E8K

Px(Sl < t) � sup Px(So < t) -+ 0 as t -+ 0 x EK

254 Chapter 7 Diffusions as :Markov Processes Since f > 0 is arbitrary, the desired result follows.

Section 0

For general b and a it is a hopelessly difficult problem to compute V(A). To make this p oint we will now use (1.7) to compute the exact domains of the semi-group and of the generator for one dimensional Brownian motion. Let Cu be the functions f on R that are bounded and uniformly continuous. Let C� be the functions f on R so that f, f', !" E Cu .

7.3

Transition Probabilities 255

Differentiating again gives

It is easy to use the dominated convergence theorem to justify differenti­ ating the integrals. From the formulas it is easy to see that g, g1, g11 E Cu so g·E C� and

(2.5) Theorem. For Brownian motion in d = 1 , V(T) = Cu, V(A) = C� , and Af(:x) = f"(:x)/2 for all f E V(A). Remarl!:. For Brownian motion in d > 1, V(A) is larger than C� and consists of the functions so that D.. f exists in the sense of distribution and lies in Cu (see Revuz and Yor (1991), p. 226).

g 11 (x) = 2Ag(x) - 2f(x) Since Ag(:x) = Ag(x) - f(x) by the proof of (1.7), it follows that Ag(x) = g11(x)f2. To complete the proof now we need to show V(A) :::> C� . Let h E C� and define a function f E Cu by

Proof If t > 0 and f is bounded then writing p1( :x , z) for the Brownian transition probability and using the triangle inequality we have

f(x) = Ah(x) - �h"(x)

I Ttf(:x ) - Tt f(y)l :::; II/II

j IPt(x, z) - Pt (Y, z)l dz

The function y = h - U>. f satisfies

The right-hand side only depends on l x - Y l and converges to 0 as l x - Y l -+ 0 so Ttf E Cu . If f E V(T) then II Tt f - !II -+ 0. We claim that this implies To check this, pick f > 0, let t > 0 so that I I Tt! - !II < c/3 then pick 6 fE so that if l x - Y l < 6 then I Tt f(x) - Tt f(y)l < t:/3. Using the triangle inequality now it follows that if l x - Y l < 6 then

c�.

lf(x) - f(y) l :::; lf(x) - Tt f(x)l + lTt f(x) - Tt f(y) l + lTt f(y) - f(y)l < f To prove that V(A) = C� we begin by observing that (3. 10) in Chapter 3

implies

Our first task is to show that if f E Cu then U>..f E C� . Letting g(x) = we have

U>. f(x)

All solutions of this differential equation have the form Aex V2"X + Be - x V2X . Since h - U>. f is bounded it follows that h = U>. f and the proof is complete. 0 7.3. Transition Probabilities

In the last section we saw that if a and b are continuous and MP(b, a ) is well posed then

d dt Tt f(x) = ATt f(x) for f E C_k , or changing notation v(t, x) = Tt f(x), we have dv = Av(t x) ' dt

(3.1) Differentiating once with respect to x and noting that the terms from differen­ tiating the limits cancel, we have

where A acts in the x variable. In the previous discussion we have gone from the process to the p.d.e. We will now turn around and go the other way. Consider

(3.2)

(a) (b)

u E C 1 • 2 and u1 = Lu in (O, oo) x Rd u is continuous on [O, oo) x Rd and u(O, x) = f(x)

256 Chapter 7 Diffusions as Markov Processes Here we have returned to our old notation

L=

i � aij (x)D;i f(x) + 2;: b; (x)Dd(x) J

J)

To explain the dual notation note that (3.1) says that v(t, · ) E V(A) and satisfies the equality, while (a) of (3.2) asks for u E C 1 • 2. As for the boundary condition (b) note that f E V(A) C V(T) implies that 11 11 ! - !II -+ 0 as t -+ 0. To prove the existence of solutions to ( 3.2 ) we turn to Friedman (1975 ) , Section 6.4. (3.3 ) Theorem. Suppose a and b are bounded and Holder continuous. That is, there are constants 0 < o, C < oo so that

l b; (x) - b;(Y) I :::; C l x - Y l 5 Suppose also that a is uniformly elliptic, that is, there is a constant >. so that for all x, y (UE) L Y;a;j(x)yi 2: ..\ lyl 2 i ,j Then there is a function Pt(x, y) > 0 jointly continuous in t > 0, x, y , and C2 as a function of x which satisfies dpjdt = Lp (with L acting on the x variable) and so _that if f is bounded and continuous (HC)

u(t, x) =

1 Pt (x, y)f(y) dy

satisfies ( 3.2) . The maximum principle implies

l u(t, x) l :::; llfll .

To make the connection with the SDE we prove (3.4) Theorem. Suppose Xt is a nonexplosive solution of MP(b, a). If u is a bounded solution of (3.2) then

Section 7.3 Transition Probabilities 257 so (a) tells us that u(t - s, X3 ) is a bounded martingale on [O, t). (b) implies that u(t - s, X3) -+ f(Xt) as s -+ t, so the martingale convergence theorem implies

u(t, x) = Exf(Xt)

Pt (x, y) is called a fundamental solution of the parabolic equation Ut = LiL since it can be used to produce solutions for any bounded continuous initial data f. Its importance for us is that (3.4) implies that

i.e., Pt(x, y) is the transition density for our diffusion process. Let D = l x l < R}. To obtain result for unbounded coefficients, we will consider ( 3.5 )

(a) (b)

1s - �� (t - r, Xr) dr 3 + :2;::: 1 D;u(Xr )b;(Xr) dr + local mart.

u(t - s, X3) - u(O, Xo) =

J

{x :

u E C1 •2 and dujdt = Lu in (O, oo) X D u is continuol!s on (O, oo) x iJ with u(O, x) = f(x), and u(t, y) = 0 when t > O, y E aD

To prove existence of solutions to ( 3.5 ) we turn to Dynkin ( 1965 ) , Vol. II, pages 230-231. (3.6) Theorem. Let D = {x : l xl :::; R} and suppose that (HC) and (UE) hold in D. 2Then there is a function p{l(x, y) > 0 jointly continuous in t > 0, x, y E D , C as a function of x which satisfies dpRjdt = Lp (with L acting on the x variable) and so that if f is bounded and continuous

u(t, x) =

u(t, x) = Exf(Xt)

Proof Ito's formula implies

0

1 p{l(x, y)f(y) dy

satisfies (3.5 ) . To make the connection with the SDE we prove (3.7) Theorem. Suppose Xt is any solution of MP(b, a) and let r = inf{t : Xt f/:. D} . If u is any solution of ( 3.5 ) then

u(t, x) = Ex(f(Xt)i r > t)

258 Chapter 7 Diffusions as Marko v Proce ses

Section 7.4 Harris Chains 259

s

u is continuous in (0, t] x jj implies it is bounded there. Ito's formula implies that for 0 � s < t 1\ r 3 u(t - s, X3 ) - u(O, Xo) = - (t - r, Xr) dr 3 + 2;:: D;u(Xr)b;(Xr) dr + local mart. Proof The fact that

1 �� 1 •

1

+ 2 � Jor D;j u(Xr)a;j (Xr) dr IJ

so (a) tells us that u(t - s, Xs) is a bounded local martingale on [0, t 1\ r). (b) implies that u(t - s, X3 ) --;. 0 as s t T and u(t - s, Xs) --;. f(X1) as s --;. t on { r > t}. Using (2.7) from Chapter 2 now, we have

u(t, x) = E:r:(f(X1 ) ; r > t)

D

Combining (3.6) and (3.7) we see that

Letting R t oo and Pt(x, y) = limR-. oo pf(x, y), which exists since (3.7) implies R --;. pf(x, y) is increasing, we have (3.8) Theorem. Suppose that the martingale problem for MP(b, a) is well posed and that a and b satisfy (HC) and (UE) hold locally. That is, they hold in {x : l x l � R} for any R < oo. Then for each t > 0 there is a lower semicontinuous function p(t, x, y) > 0 so that if f is bounded and continuous then

E:r: f(Xt) =

J Pt(x , y) dy

Remark. The energetic reader can probably show that Pt ( x, y) is continuous.

the next, we will assume that the reader is familiar with the basic theory of Markov chains on a countable state space as explained for example in the first five sections of Chapter 5 of Durrett (1995). This section is a close relative of Section 5.6 there. We will formulate the results here for a transition probability defined on a measurable space (S, S). Intuitively, P(Xn +l E AIXn = x) = p(x, A). Formally, it is a function p : S x S --;. [0, 1] that satisfies for fixed A E S, x --;. p(x, A) is measurable for fixed x E S, A --;. p( x, A) is a probability measure In our applications to diffusions we will take S = Rd , and let

p(x, A) =

1 Pl (x, y) dy

where p1(x, y) is the transition probability introduced in the previous section. By taking t = 1 we will be able to investigate the asymptotic behavior of Xn as n --;. oo through the integers but this and the Markov property will allow us to get results for t --;. oo through the real numbers. We say that a Markov chain Xn with transition probability p is a Harris chain if we can find sets A, B E S, a function q with q( x, y) � E > 0 for x E A, y E B, and a probability measure p concentrated on B so that: (i) If TA = inf{n � 0 : Xn E A}, then Pz (TA < oo) > 0 for all z E S (ii) If x E A and C C B then p(x, C) � fc q(x, y)p(dy) In the diffusions we consider we can take A = B to be a ball with radius r, p to be a constant c times Lebesgue measure restricted to B, and q(x, y) = p1 (x, y)jc. See (5.2) below. It is interesting to note that the new theory still contains most of the old one as a special case.

4.1. Countable State Space. Suppose Xn is a Markov chain on a countable state space S. In order for Xn to be a Harris chain it is necessary and sufficient that there be a state u with Px(Xn = u for some n � 0) > 0 for all X E S.

Example

7.4. Harris C hains

Proof To prove sufficiency, pick v so that p( u, v) > 0. If we let A = { u} and B = { v} then (i) and (ii) hold. To prove necessity, let { u} be a point with p( { u}) > 0 and note that for all x D . P:r:(Xn = u for some n � 0) � Ex{q(XrA )p({u})} > 0

In this section we will give a quick treatment of the theory of Harris chains to prepare for applications to diffusions in the next section. In this section and

The developments in this section are based on two simple ideas. (i) To make the theory of Markov chains on a countable state space work , all we need

However, lower semicontinuity implies that p1(x, y) is bounded away from 0 on compact sets and this will be enough for our results in Section 7.5.

260 Chapter 7 Diffusions

as

Markov Processes

is to have one point in the state space which is hit. (ii) In a Harris chain we can manufacture such a point (called a below) which corresponds to "being in B with distribution p." The notation needed to carry out this plan is occasionally obnoxious but we hope these words of wisdom will help the reader stay focused on the simple ideas that underlie these developments. Given a Harris chain on (S, S), we will construct a Markov chain Xn with transition probability p on (S, S) where S = S U {a} and S = { B , B U {a} : B E S}. Thinking of a as corresp onding to being on B with distribution p, we define the modified transition probability as follows: If x E S - A, p( x, C) = p( x, C) for C E S p(X' {a}) = f If X E A, p(x, C) = p(x, C) - e:p(C) for C E S If x = a, p(a, D) = J p(dx)p(x, D) for D E S Here and in what follows, we will reserve A and B for the special sets that occur in the definition and use C and D for generic elements of S. We will often simplify notation by writing p(x, a) instead of p(x, {a}), p(a) instead of p({a}), etc. Our first step is to prove three technical lemmas that will help us carry out the proofs below. Define a transition probability v by

v( X, {X}) = 1 if X E S v( a, C) = p ( C)

In words, v leaves mass in S alone but returns the mass at a to S and distributes it according to p.

(4.1) Lemma. (a) vp = p and (b) pv = p. Proof Before giving the proof we would like to remind the reader that mea­

sures multiply the transition probability on the left, i.e., in the first case we want to show pvp = pp. If we first make a transition according to v and then one according to p, this amounts to one transition according to p, since only mass at a is affected by v and

p(a, D) =

J p(dx)p(x, D).

The second equality also follows easily from the definition. In words, if p acts first and then v, then this is the same as one transition according to p since v returns the mass at a to where it came from. 0 From (4.1) it follows easily that we have:

Section

Harris Chains 261

7.4

(4.2) Lemma. L:_t Yn be an inhomogeneous Markov chain with p2 �: = v and = fi. Then Xn = Y2 n is a Markov chain with transition probability p and Xn = Y2 n + 1 is a Markov chain with transition probability p.

P2 k+l

(� -2) shows that the�e is an intimate relationship between the asymptotic behavwr of Xn and of Xn. To quantif;r this we need a definition. If f is a bounded measurable function on S, let f = vf, i.e., /(x) = f(x) for x E S /(a) = f dp

j

(4.3) Lemma. If J.l is a probability measure on (S, S) then Proof Observe tha� if Xn and

S) = 1 then Xo according to v.

= Xo and Xn

are constructe� as in (4.2), and P(Xo E is obtained from Xn by making a transition

Xn

o

Before developing the theory we give one example to explain why some of the statements to come will be messy.

4.2. Perverted Brownian motion. For x that is not an integer 2: 2, let p(x, ) be the transition probability of a one dimensional Brownian motion. When x 2: 2 is an integer let p( X, {X + 1}) = 1 - X - 2 p(x, C) = x - 2 IC n [0, 1] 1 if X + 1 ¢.

Example

·

c

is the transition probability of a Harris chain (take A Lebesgue measure on B) but .

p

= B = (0, 1),

p

=

P2 (Xn = n + 2 for all n) > 0 I can sympathize with the reader who thinks that such crazy chains will not arise "in applications," but it seems easier (and better) to adapt the theory to include them than to modify the assumptions to exclude them. a. Recurrence and transie.nce

We begin with the dichotomy between recurrence and transience. Let inf{n 2: 1 : Xn = a}. If Po:(R < oo ) = 1 then we call the chain

R=

262 Chapter 7 Diffusions

as

Markov Processes

R1 = R and for k � 2, let R1: = inf{ n > R�:- 1 : Xn = a } be the time of the kth return to a. The strong Markov property implies Pa(RI: < oo) = Pa(R < oo) k so Pa(Xn = a i.o. ) = 1 recurrent, otherwise we call it transient. Let

in the recurrent case and is follows easily.

0 in the transient case. From this the next result

(4.4) Theorem. Let A( C) = I:n 2 - n _pn (a, C) . In the recurrent case if A( C) > 0 then Pa(Xn E C i.o. ) = 1. For A a.e. x, P,(R < oo) = 1. Remark. Here and in what follows ?(x, C) = P,(Xn E C) is the nth iterate of the transition probability defined inductively by p1 = p and for n � 1

Proof The first conclusion follows from the following fact (see e.g., (2.3) in Chapter 5 of Durrett (1995)). Let Xn be a Markov chain and suppose

on

1 '

Section 7.4 Harris Chains 263

= 0 then the definition of A implies Pa(Xn E C) = 0, so Pa(Xn E C, R > n) = 0 for all n, and ji.(C) = 0. We next check that j1 is a-finite. Let G1:, o = {x : _pk (x, a) � b" } . Let To = 0 and let Tn = inf{ m � Tn - 1 + k : Xm E G�:, o }. The definition of G1:, o implies P(Tn < Ta i Tn - 1 < Ta) :::; (1 - b") Pro?f If A ( C )

N = inf{ n : Tn � Ta} then EN < 1/b". Since we can only have with R > m when Tn :::; m < Tn + k for some 0 :::; n < N it follows that ji. (G�:, o ) :::; kj b". Part (i) of the definition of a Harris chain implies S C U�:,m;:: 1 Gk , 1 / m and a-finiteness follows. Next we show that jlp = fl. so · if we let

Xm E G1:, o

CASE 1 . Let C be a set that does not contain Fubini's theorem.

j Ji.(dy)p(y, C) = n=Of j Pa(Xn E dy, R > n)p(y, C)

{Xn E An}

00

P({Xn E An i.o.}) - P({Xn E Bn i.o.}) = 0. Taking An = {a} and Bn = C gives the desired result. To prove the second conclusion, let D = { x : P,(R < oo) < 1 } and observe that if pn (a, D) > 0 for some n then

= L Pa(Xn + l E C, R > n + 1) = ji.(C)

Then

Pa(Xm = a i.o. ) :::; J pn (a, dx )P (R < oo) < 1 ,

Remark. Example 4.2 shows that we cannot expect to have

0

b. Stationary measures

R = inf{ n � 1 : Xn = a}.

n=O

a (/. C and Pa(Xo = a) = 1 . CASE 2. To complete the proof now i t suffices t o consider C = {a}.

since

f J Pa(Xn E dy, R > n)p(y, a) J Jl(dy)p(y, a) = n=O

P,(R < oo) = 1

for all x. To see that this can occur even when the state space is countable, consider a branching process in which the offspring distribution has Po = 0 and I: kp1: > 1. If we take A = B = {0 } this is a Harris chain by Example 4.1. Since pn (o, 0) = 1 it is recurrent.

(4.5) Theorem. Let

a. Using the definition j1 and

In the recurrent case,

00

= L Pa(R = n + 1) = 1 = ji.(a) n=O

where in the last two equalities we have used recurrence and the fact that when C = {a} only the n = 0 term contributes in the definition. Turning to the properties of J.l., we note that J.1. = jlv and Ji. ( a) = 1 so J.1. is a-finite. To check J.l.P = J.l., we note that (i) using the definition J.1. = jlv and ( b ) in (3.1), (ii) using ( a) in (4.1), (iii) using jlp = jl, and (iv) using the definition J.1.

defines a a-finite stationary measure for p, with j1 << A, the measure defined in (4.4). J.1. = jlv is a a-finite stationary measure for p.

= jlv:

J.l.P = (Jl v)(pv)

= jlpv = jlv = J.1.

To investigate uniqueness of the stationary measure we begin with:

0

264 Chapter 7 Diffusions

as

Section

Markov Processes

(4.6) Lemma. If v is a a--finite stationary measure for p, then v (A )

v = vp is a a--finite stationary measure for p with v( a) < oo.

< oo and

< oo. If v(A) = oo then part (ii) of the . definition of a Harris chain implies v ( C ) = oo for all sets C with p(C) > 0. If B = U;B; with v (B;) < oo then p(B;) = 0 by the last observation and p(B) = 0 by countable subadditivity, a contradiction. So v(A) < oo and v(a) = vp(a) = ev(A) < oo. Using the fact that vp = v, we find v(C) = vp(C) = v(C) f v(A)p(B n C) the last subtraction being well defined since v ( A) < oo. From the last equality it is clear that v is a--finite. Since v(a) = w (A ) , it also follows that vv = v. To check vp = v, we observe that ( a) of (4. 1 ) the last result, and the definition of v imply 0 vp = vvp = vp = v Proof We will first show that v (A )

-

,

(4 . 7) Theorem. Suppose p is recurrent. If v is a a--finite stationary measure then v = v(a)p where Jl is the measure constructed in the proof of (4.5). Proof By (4.6) it suffices to prove that if v is a a--finite stationary measure for p with v( a) < oo then v = v( a )fl. Our first step is to observe that

v(C) = v(a)p(a, C) + Using the last identity to replace

+

{

ls -{a}

v(a)p(a, dy)p(y, C)

1S-{a} v(dx) J{s-{a} p(x, dy)p(y, C)

= v(a)Pa(Xl E C) + v(a)Pa(Xl =f. a, X2 E C) + P;;(Xo =f. a, X1 =f. a,X2 E C) Continuing in the obvious way we have

n v(C) = v(a) L Pa (XJ: =f. for 1 � k < m, Xm E C) m=l Pv(X�: =f. for 0 � k < n + 1,Xn + l E C) + Q'

a

Harris Chains 265

The last term is nonnegative, so letting n � oo we have

v(C) ;:::: v(a)p(C) To turn the ;:::: into an = we observe that

v(a) = J v(dx)pn (x, a) ;:::: v(a) J fl(dx)pn (x, a) = v(a)jl(a) = v(a)

fl(a) = 1. Let Sn = {x : pn (x, a) > 0}. By assumption UnSn = S. If v(D) > v(a)jl(D) for some D, then v(Dn Sn) > v(a)jl(D n Sn) for some n and it follows that v(a) > v(a), a contradiction. o since

(4.5) and { 4. 7) show that a recurrent Harris chain has a a--finite stationary measure that is unique up to constant multiples. The next result, which goes in the other direction, will be useful in the proof of the convergence theorem and in checking recurrence of concrete examples. (4.8) Theorem.

is recurrent.

If there is a stationary probability distribution then the chain

Proof Let 1f = 1rp, which is a stationary distribution by (4.6), and note that part (i) of the definition of a Harris chain implies if{a) > 0. Suppose that the chain is transient, i.e., Pa (R < oo) = q < 1. If R1: is the time of the kth return then Pa(R�: < oo) = qJ: so if x =f. a

v(dy) on the right-hand side we have

v(C) = v(a)p(a, C) + .

1S-{a} v(dy)p(y, C)

7.4

The last upper bound is also valid when

x = a since

Integrating with respect to 1f and using the fact that 1f is a stationary distribu­ tion we have a contradiction that proves the result 0

266 Chapter 7 Diffusions

as

Markov Processes

Section 7.4 Harris Chains 267

c. Convergence theorem

If I is a set of positive integers, we let g.c.d. (I) be the greatest common divisor of the elements of I. We say that a recurrent Harris chain Xn is aperi­ odic if g.c.d. ( {n � 1 : pn (a, a) > 0}) = This occurs, for example, if we can take A = B in the definition for then p(a, a) > 0. A well known consequence of aperiodicity is

1.

(4.9) Lemma. There is an

mo < oo so that j1"'(a, a) > 0 for m � mo.

For a proof see (5.4) of Chapter 5 in Durrett (1995) . We are now ready to prove (4.10) Theorem. Let Xn be an aperiodic recurrent Harris chain with stationary distribution 1r. If P:r:(R < oo) = 1 then as n ___. oo,

Remark. Here 11 · 11 denotes the total variation distance between the measures. (4.4) guarantees that .\ a.e. x satisfies the hypothesis, while (4.5), (4.7), and (4.8) imply 1r is absolutely continuous with respect to .\. Proof In view of (4.3) �nd (4.6) it suffices to show that if if = 1rp then

llpn (x, · ) - if(·) ll

___.

We claim that the product chain is a Harris chain with A = {(a, a)} and B = B X B. Clearly {ii) is satisfi e d. To check {i) we need to show that for all (x 1 , x 2 ) there is an N so that pN((x 1 , x 2 ), (a, a)) > 0. To prove this let I< and L be such that pK (x 1 , a) > 0 and pL(x 2 , a) > 0, let M � mo, the constant in {4.9) , and take N = J( + L + M. From the definitions it follows that

pK+L+M (x l , a) � pK (x l , a)pL+M (a, a) > 0 pK+L +M (x 2 , a) � pL (x l , a)pK+M (a, a) > 0 and hence pN((x 1 , x 2 ), (a, a)) > 0.

The next step is show that the product chain is recurrent. To do this, note that (4.6) implies 7f = 1rp is a stationary distribution for p, and the two coordinates move independently, so

defines a stationary probability distribution for p, and the desired conclusion follows from (4.8). To prove the convergence theorem now we will write Zn = (Xn, Yn) and consider Zn with initial distribution lia x if. That is X0 = a with probability 1 and Yo has the stationary distribution if. Let T = inf{ n : Zn = (a, a)} = R. (4.8) , (4.7), and (4.5) imply if x if « 5. so (4.4) implies �

0

Po, x'it(T <

Our second reduction is to note that if x =f. a then

n pn (x, · ) = L P:r: (R = m)pn - m (a, · ) m=l

so we have

n n IIP (x, ·) - if(·) ll :::; L P:r: (R = m) IIPn - m (a, ·) - if( · ) II m=l and it suffices to prove the result when x = a. Let S2 = S x S. Define a transition probability p on S x S by That is, each coordinate moves independently. We will call this the "product chain" ( think of product measure) and denote it by Zn. �

Dropping the subscript lia P (Xn

x

_

_

oo) = 1

if from P and considering the time T,

n

E C, T :::; n) = L P(T = m)pn - m (a, C) = P{Yn E C, T :::; n)

m=l since on {T = m}, Xm = Ym = a. Now P (Xn E C) = P (Xn E C, T :::; n) + P ( Xn E C, T > n) = P (Yn E C, T :::; n) + P (Xn E C, T > n) :::; P(Yn E C) + P(T > n) Interchanging the roles of X and Y we have P (Yn

and it follows that

E C) :::; P (Xn E C) + P(T > n)

Section

268 Chapter 7 Diffusions as Markov Processes

llvn (a, · ) - 7r( - ) ll = II P (Xn E ·) - P(Yn E ·) II

= sup !P(Xn E C) - P(Yn E C ) I c

as n __. oo.

7.5.

l x l2 fc: gives the following (5.4) Corollary. Suppose tr a (x ) + 2x · b(x) :::; then ExTK :::; l x l 2 /c . Taking u(x) =

:::; P(T > n) __. 0 0

-€

for

x E gc = {x : l x l > r}

Though ( 5.4) was easy to prove, it is remarkably sharp.

Convergence Theorems

Assumption.

Convergence Theorems 269

Letting n __. oo and using the monotone convergence theorem the desired result follows. o

In this section we will apply the theory of Harris chains developed in the previ­ ous section to diffusion processes. To be able to use that theory we will suppose throughout this section that

(5.1)

7.5

MP(b, a ) is well posed and the coefficients a and b satisfy

Example 5.1. Consider d = 1. Suppose a(x) = 1 and b(x) = - cfx for l x l � 1 where c � 0 and set I< = [-1 , 1] . If c > 1/ 2 then ( 5.4) holds with c = 2c- l. To compute ExTK , suppose c -:/= 1 / 2, let € = 2c - 1, and let u(x) = x2fc . Lu = - 1 in � 1, b) so u(�t) + t is a local martingale until time T( l , b) = inf{t : Xt rf. ( 1, b)}. Usmg the optiOnal stopping theorem at time T(l , b) A t we have

(HC) and ( UE) locally.

(5.2) Theorem. If A = B = {x : l x l :::; r} and p is Lebesgue measure on B normalized to be a probability measure then Xn is an aperiodic Harris chain. Proof By

( 3.8) there is a lower semicontinuous function Pt(x, y) > 0 so that Px ( Xt E A) =

1 Pt(x, y) dy

E X2 X2 Ex T.( l b) = - - X T(l,b) '

From this we see that Px(R = 1) > 0 for each x so (i) holds. Lower semi­ continuity implies P l (x, y) � c > 0 when x, y E A so (ii) holds and we have

p(a, a) > 0.

Rearranging, then letting t __. oo and using the monotone and bounded conver­ gence theorems, we have

0

To finish checking the hypotheses of the convergence theorem (4.10), we have to show that a stationary distribution 1r exists. Our approach will be to show that Ea R < oo, so the construction in (4.5 ) produces a stationary measure with finite total mass. To check Ea R < oo We will use a slight generalization of Lemma 5.1 of Khasminskii ( 1960).

(5.3) Theorem. Suppose u � 0 is C2 and has Lu :::; -1 for all x rf. K where K is a compact set. Let TK = inf {t > 0 : Xt. E K}. Then u(x) � ExTK. Proof Ito's formula implies that Vt = u(Xt) + t is a local supermartingale on [0, TK ) . Letting Tn l TK be a sequence of times that reduce vt, we have

€

€

To evaluate the right-hand side, we need the natural scale and doing this it is convenient to let start from 1 rather than 0: x c �(x) = exp x =C Y 2 c dy = C'(y 2c+l - 1)

l (ly � dz) l

Using ( 3.2) in Chapter 6 with a = 1 it follows that

x 2 - b2 c+l - x2 c+l . 1 x 2 c+l - 1 b2 Ex r(l,b) = 7 b2 c+l - 1 � - b2c+l - 1 The second term on the right always converges to - 1 / c as b __. oo. The third term on the right converges to 0 when c > 1 /2 and to oo when c < 1/2 (recall that € < 0 in this case) . Exercise

5.1. Use (4.8) in Chapter 6 to show that ExTK = oo when c :::; 1 / 2.

Ch apter 7 Diffusions Markov Processes Exercise 5.2. Consider d = 1. Suppose a(x) = 1 and b(x) � - f.j x 6 for x � 1 where f. > 0 and 0 < 6 < 1. Let T1 = inf{t : Xt = 1} and show that for x � 1, Ex T! � (x 1 +6 - 1)/c. Example 5.2. When d > 1 and a(x) = I the condition in (5.4) becomes x · b(x) � -(d + c)/2 To see this is sharp, let yt = ! Xt l . and use Ito's formula with f(x) = lxl which has Dd = x;fl x l to get t X · b(X3)ds 1 X · d 3 + 1 11 d - 1 ds 1 Yo = + 1o IX: I W 2 o IX3 1 yt o IX: I We have written W here for a d dimensional Brownian motion, so that we can let Bt = 1t IXX33 1 . dW3 be a one dimensional Brownian motion B (it is a local martingale with (B)t = t) and write d --1 ) dt + dBt dyt = I.X1t l (Xt . b(Xt ) + 2 If x · b(x) = -(d - 1 + c)/2 this reduces to the previous example and shows that the condition x b(x) � -(d + c)/2 is sharp. The last detail is to make the transition from Ex TK < oo to Ea R < oo. {5.5) Theorem. Let I<. = {x : lxl � r} and H = {x : lxl � r + 1}. If SUPxeH Ex TK < oo then Ea R < oo. Proof Let Uo = 0, and for m � 1 let Vm = inf{ t > Um - 1 : Xt E K }, Um = inf{t > V�m : IX1 1 (j. H or t E Z}, and M = inf{m � 1 : Um E Z}. Since X(Um) E A, R UM . To estimate Ea R, we note that X(Um - d E A so E (Vm - Um- 1 I Fum-l ) � Co = xEsupA ExTK To estimate M , we observe that if TH is the exit time from H then (3.6) implies inf p� + 1 (x,y) = co > 0 xEinfK Px (TA > 1) � I I< ! x ,yE K 1 + where p� (x, y) is the transition probability for the process killed when it leaves the ball of radius r + 1. From this it follows that P(M > m) < (1 - co) m . Since Um - Vm � 1 we have EUM � {1 + Co)/co and the proof is complete. 0 as

270

0

·

8

Weak Convergence

8.1. In M etric Spaces

We begin with a tre'!tment of weak convergence on a general space S with a metric and {iii) i.e., a function with {i ) 0 , (ii) z) z) . Open balls are defined by < r} with r > 0 and the B orel sets S are the u-field generated by the open balls. Throughout this section we will suppose S is a metric space and S is the collection of Borel sets, even though several results are true in much greater generality. For later use, recall that S is said to be separable if there is a countable dense set, and complete if every Cauchy sequence converges. A sequence of probability measures on (S, S) is said to converge weakly if for each bounded continuous function I _,. I In this case we write In many situations it will be convenient to deal directly with random variables rather than with the associated distribu­ tions converges weakly to and E We say that write _,. if for any bounded continuous function Our first result, sometimes called the Portmanteau Theorem, gives five equivalent definitions of weak convergence.

p, � p(:i, p(x, y)+p(y,

·

{yp(x,y) : p(x,=y)p(y ,x),

p(x,x) =

J.Ln

J.Ln => f.L· X J.Ln (A) = P(Xn A). n Xn => X

f(x)J.Ln(dx)

f(x)J.L(dx). X f, Ef(Xn) Ef(X).

Xn

(1.1) Theorem. The following statements are equivalent. (i ) Ef(Xn) Ef(X) for any bounded continuous function f. (ii) For all closed sets I<. , limsup n P(Xn E K) � P(X E K). (iii) For all open sets G, liminfn P(Xn E G) � P(X E G). (iv) For all sets A with P(X E 8A) = 0, limsupn P(Xn E A) = P(X E A). (v ) Let D1 = the set of discontinuities of f . For all bounded functions with _,.

.....

.....

oo

oo

.....

P(X E DJ ) = 0, Ef(Xn) Ef(X). _,.

oo

272 Chapter 8 Weak Convergence Remark. To help remember (ii) and (iii) think about what can happen when P(Xn = Xn) = 1, X n -+ x , and x lies on the boundary of the set. If I< is closed

we can have Xn (/. f{ for all n but all n but x (/. G.

x E I<. If G is open we can have Xn E G for

Proof We will go around the loop in roughly the order indicated. The last step, (v) implies {i), is trivial so we have four things to show.

Let p(x, I<) = inf{p(x, y) : y E K } , 1/Jj (r) = (1 - jr) + where + z = max{z, 0} is the positive part, and let fi (x) = 1/Jj (p(x , I<)). fi is bounded, (i ) implies (ii)

continuous, and ;:::: 1K so

limsup P(Xn E I<) � nlimoo Efi (Xn) � n � oo Now /j ! 1 K as j t oo, so letting j -+ oo gives (ii).

(ii) is equivalent to (iii) This follows immediately from (a) P(A)+P(Ac) 1 , and (b) A is open if and only if A c is closed.

P(X E

=

=

the

K) = P(X E A) = P(X E G)

Using (ii) and {iii) now with the last equality we have limsup P(Xn E A) � limsup P(Xn E K) � P(X E I<) = P(X E A) n -> oo n -> oo lim inf P(Xn E A) 2: liminf n --+oo n -+ oo P(Xn E G) � P(X E G) = P(X E A) At this point we have shown liminf sup P(Xn E A) n -+ oo P(Xn E A) 2: P(X E A) 2: lim n --+ oo which with the triviality liminf � limsup gives the desired result. ( iv) implies (v ) Suppose lf(x) l �

I< and pick ao < a 1 < . . . < at with a0 < - I< and at > I< so that P(f(X) = a; ) = 0 for 0 � i � £ and aj -aj - 1 < f for 1 � j � £. This is always possible since {a : P(X = a) > 0} is a countable set. Let A; = {x : a; - 1 < f ( x ) � a;}, 8A; C { x : f( x ) E {a; - 1 , a;}} U Dj, so P(X E 8A;) = 0, and it follows from (iv) that l

l

L: a;P(Xn E A; ) -+ L: a;P(X E A;)

I

Section 8. 1 In Metric Spaces 273 The definition of the a; implies 0�

l

L: a;P(Xn E A;) - Ef(Xn ) � c

i =1

and this inequality holds with X in place of Xn . Combining our conclusions we have limsup IEJ(Xn) - Ef(X) I � 2c but f is arbitrary so the proof is complete.

0

An important consequence of the {1.1) is

= Efi (X)

(ii) and (iii) imply ( iv) Let [{ = A = the closure of A, G = A0 interior of A, and note that 8A = J( - G, so our assumptions imply

1

{1.2) Continuous mapping theorem. Suppose J.ln on ( S, S) converge weakly to J.l, and let t.p : S -+ S' have a discontinuity set D'P with J.l(D'P) = 0 . Then f.ln o t.p- 1 :::} j.l o t.p- 1 . 'Proof Let f : S' -+ R be b ounded and continuous. Since follows from (v) in (1.1) that

Djo

it

1 f.ln(dx)f(t.p(x)) -+ 1 J.l(dx)f(t.p(x))

Changing variables gives

Since this holds for any bounded continuous f, the desired result follows.

0

In checking convergence in distribution the following lemma is sometimes useful. {1.3) Converging together lemma. Suppose Xn in probability then Yn :::} X.

:::?

X and p(Xn , Yn)

-+ 0

Proof Let J( be a closed set and I 8 so that B(z,-y ) n I< = 0, so (KaY is open and Ko is closed. With this little detail out of the way the rest is easy.

P(Yn E K) � P(Xn E I
> 8)

274 Chapter 8 Weak Convergence

-+

0 lim sup P(Yn E I<) :$ lim sup P(Xn E I<6) :$ P(X E I<6) n-+oo n-+oo Measure theory tells us that P(X E K6 ) ! P(X E I<) as 6 ! 0 so we have shown limsup P(Yn E I<) :$ P(X E I<) n -oo

Letting n oo, noticing the second term on the right goes to by assumption, and using (ii) in (1.1) we have

The desired conclusion now follows from (1.1).

D

The next result, due to Skorokhod, is useful for proving results about a sequence that converges weakly, because it converts weak convergence into almost sure convergence. (1.4) Skorokhod's representation theorem. Suppose S is a complete sep­ arable metric space. If => then we can define random variables on 1] (equipped with the Borel sets and Lebesgue measure .:\) with distribution so that almost surely.

Jl[0,n

Jln Jloo

Yn -+ Y00

Yn

Fn(x) = Jln(-oo, x] be the distribution F;; 1 (y) Fn(x) ;:::: y} let U (w) = w for w E [0, 1] be a random variable that is uniform on (0, 1), and let Yn = F;; 1 (U). Then Yn -+ Y almost surely. (For a proof, see (2.1) of 00 the reader to contemplate the question: Chapter 2 of Durrett (1995).) We invite "Is there an easy way to do this when S = R2 (or even S = [0, 1] 2 )?" while we give the general proof. The details are somewhat messy and not important for

Remark. When S = R this is easy. Let function, let = inf{x :

the developments that follow, so if the reader finds herself asking "Who cares?" , she can skip to the beginning of the next section. Proof We will warm up for the real proof by treating the special case

Jln Jl· =

Jl there is a random variable Proof For each k construct a decomposition Ak = {Ak , l ,Ak , 2 , . . . } of S into disjoint sets of diameter less than 1/k and so that Ak refines Ak - 1 > i.e., each set A k - l ; is a union of sets in A k· For each k construct a corresponding decompo�ition Ik of the unit interval so that .:\ ( Ik,i ) = Jl(Ak ,j ) and arrange things so that A k - l ,i Ak ,i if and only if Ik - l , i Ik,j . (1.5) Theorem. For each probability measure defined on 1] with

[0,

Y

P(Y E A) = Jl(A). ::::>

::::>

1 l

Section 8. 1 In Metric Spaces

275

x ,j be some pointin insomeAk,element for w E Ik ,i · Since i and letofXAkk,i,(w)which= Xk,byidefinition has diam­ Xeterk ,XLet k:::; +1/k,l >k it. isiscontained a Cauchy sequence. So X(w) = limk Xk(w) exists and satisfies p(Xk,X) :::; 1/k. To show that X has distribution Jl we let J( be a closed set, let S(k, I<) = {j : A k,j I< ::f. 0}, and let I< = {x : p(x, I<) :::; 6} as in the proof of (1.3). .h(Xk E I<) :$ .h(Xk(w) E UjeS(k,K)Ak,j ) < 2::: .h(Xk (w) E Ak,i ) jES(k,K) jES(2:::k,K ) .h(Ik,i ) = j ES(2:::k,K) Jl(Ak,i ) :$ Jl(I
n

a

=>

c

D

c

'

" LJ

'

" �

'

276 Chapter 8 Weak Convergence

f3n

as n oo. IJ::i Xn IJ:,io f3nIJ:,i forf3oolarge n, and lim sup p(Xn, X ) $ 2/k n- oo 00 Letting k oo we see that if is not one of the countable set of points that is the end point of some IJ::i we have Xn X , which proves (1.4 ). 00 __...

w

__...

__...

Similarly if we let be the right endpoint of then Hence if w is in the interior of it lies in the interior of it follows from the definition of the that

__...

1

Since f is arbitrary, (iii) in

0

( Jl.(K) 1 -

Jl.n Jl.,

e:

J1. (2. 1) Theorem. If II is tight then it is relatively compact. (2.2) Theorem. Suppose

pact, it is tight.

countable , asomewhat 00 is approach This S. of compact sets, and finally, a general

union lengthy but to inspire the reader for the effort, we note that the first two special cases are important examples, and then the last two steps and the converse are remarkably easy. Before we begin we will prove a lemma that will be useful for the first two examples. (2.3) Lemma. Let A be a subclass of S that is closed under intersection, and __... so that each open set is a countable union of elements of A. If => for all E A then

A

Jl.n(A) Jl.(A)

Jl.n Jl. ·

Proof The inclusion-exclusion formula says m

P(U�1 A;) = L i
· · ·

+ (-1)

m

+ l P(n� 1 A;)

0

G(q) and define the proposed limiting distribution function by P(x) = inf{G(q) : q E Q d and q > x}

where q > x means q; > x; for each i. To check that is a distribution function we note that it is an immediate consequence of the definition that

P

S is complete and separable. If II is relatively com­

Proof of ( 2.1) We shall prove the result successively for Rd , R

(1.1) follows and the proof is(complete.

Jl.( Yi By enumerating the points in Q d and then using a diagonal argument, we can find a subsequence Pnk so that Pnk (q) converges for each q E Q d . Call the limit

e:

The first result is the one we want for applications, but the second one is comforting since it says that, in nice spaces , the two notions are equivalent.

-

Proof for Rd This is a fairly straightforward generalization of the argument for d = · 1. (See e.g., (2.5) in Chapter 2 of Durrett (1995).) In this case we can rely on the distribution functions P( x) = {y $ x}), where for two vectors y $ x means $ x; for each i. Let Q d be the points in Rd with rational coordinates.

II be a family of probability measures on a metric space S, p). We call II relatively compact if every sequence of elements of II contains a subse­ quence that converges weakly to a limit which may be rf; II. We call II tight if for each > 0 there is a compact set ]{ so that 2: for all E II. This section is devoted to the proofs of

Jl.nk

Combining this with the fact that A is closed under intersection it is easy to see that __... Given an open set G and an f > 0, choose sets so that > Jl. ( G) e:. Using the convergence just proved it follows that

A1 , . Jl.. .nA(Ufk =1 A; ) Jl.(J1Uf.(Uf=l=A;)1 A;).

8.2. Prokhorov's Theorems

Let

Section 8.2 Prokhorov's Theorems 277

.j

{i) if x $ y then P(x) $ P(y) {ii) limy.tx P(y) = P(x)

Yi

where y ! x means ! x; for each i. The third condition for being a distribution function is easier to prove than it is to state:

A = (a1 ,

b 1] x X (ad , bd] be a rectangle with vertices V = { a1 , b1 } x where n a , v is the number of a's {a d , bd } , and let sgn ( v ) =

(iii) let

n (a , v ) ( ) (-1) that appear in v. The inclusion-exclusion formula implies that the probability of A is I:vev sgn ( v ) P( v ) 2: 0. The fact that (iii) holds for the Pn 's implies that it holds for G and by taking limits we see that it holds for P. Conditions (i)-(iii) imply that P is the distribution function of a measure on Rd , which in our case will always have total mass < 1. Define the ith marginal distribution function of P by pi ( x) = lim P(y) where the limit is taken through vectors y with Yi = x and the other coordinates Yi oo. pi is a one dimensional distribution function and hence its discontinuities Di are a · · ·

_x

· · ·

__...

countable set. Call x a good point if x; rf; Di for each i. We claim that P is continuous at good points. To see this, let w < x < y, let z; be the vector which uses the first

278

Ch ap ter 8

Section

Weak Convergence

i components from y and the last d - i from w and note that the monotonicity properties that come from the interpretation of F as a measure imply d

d

F(y) - F(w) = L F(z;) - F(zi - 1 ) � L F;(y;) - F;(w;) i=1 i=1 The continuity of F now follows from the monotonicity and the continuity of the F;. Our next claim is that for good x, Fnk (x) -;. F(x). To prove this pick p, q, r E Q d with p < q < x < r and F(x) - c: � F(p) � F(q) � F(x) � F(r) � F(x) + c: Since u

< v implies F(u) � G(v) � F(v) it follows that F(x) - � G(q) � F(x) � G(r) � F(x) + c:

c:

Using Fn (q) � Fn (x) � Fn (r) and the convergence Fnk (q) -;. G(q) for q E Q d the claim follows easily. Up to this point we have not used the tightness assumption. It enters into the proof only to guarantee that F is the distribution function of a probability measure. Given c: choose a compact set I<. so that Fn (I<.) 2: 1 - c: for all n, where Fn (I<.) is short for the probability of I<. under the measure associated with the distribution function Fn . Now pick a good rectangle A, i.e., one with a; , b; ¢ Di for all i, so that I<. C A. The formula in (iii) for the probability of A and the previous claim about convergence at good points imply

F(A) = lim F (A) 2: liminf n oo Fn(K) 2: 1 k_..oo nk Finally, we have to prove that Fnk => F. For this we use the following -

c:

result which is of interest in its own right.

(2.4) Theorem. In Rd probability measure J.Ln => J.L if and only if the associated distribution functions have Fn (x) -;. F(x) for all x that are good for F. x is good for F, then A = {y : y � x} has J.L(8A) = 0, so if J.ln ::;. J.L then (iv) of (1.1) implies Fn (x) -;. F(x). To prove the converse we will apply (2.3) to the collection A of finite disjoint unions of good rectangles.

Proof If

(i.e., rectangles, all of whose vertices are good) . We have already observed that J.L n (A) -;. J.L(A) for all A E A. To complete the proof we note that A is closed under intersection, and any open set is a countable disjoint union of good 0 rectangles.

8.2

Prokhorov's Theorems 279

Proof for R00 Let R00 be the collection of all sequences (xi> x 2 , ) of real numbers. To define a metric on R00 we introduce the bounded metric p0(x, y) = lx - Y l/(1 + lx - yl) on R and let • • •

00

p(x, y) = L 2 - i p0(x;, y; ) i=1 It is easy to see that the induced topology is that of coordinatewise convergence, (i.e., x n -;. x if and only if xf -;. x;) for each i and hence this metric makes R00 complete and separable. A countable dense set is the collection of points with finitely many nonzero coordinates, all of which are rational numbers.

S = noo , the product u-field gnerated by the finite dimensional sets { x : x; E A; for 1 � i � k}. Proof To argue that s ::) noo observe that if G; are open then {x : X; E G; for 1 � i � k} is open. For the other inclusion note that (2.5)

Lemma. The Borel sets

{

{y : p(x, y) � 8} = n�= 1 y : so

t= 2- m Po (xm , Ym ) � 8

m 1

{y : p(x, y) < 'Y } = U�= 1 {y : p(x, y) � 'Y - 1 /n } E noo .

}

E noo 0

Let 1f'd : R00 -;. Rd be the projection 7rd (x) = (xi> . . . , x d ) · Our first claim is that if II is a tight family on (R00 , noo ) then {J.L o 7rJ 1 : J.L E II} is a tight family on (ltd , n d ). This follows from the following general result.

(2.6) Lemma. ' If II is a tight family on (S, S) and if h is a continuous map from S to S then {J.L o h - 1 : J.L E II} is a tight family on (S', S'). Proof Given c: , choose in S a compact set K so that J.L( K) 2: 1- c: for all J.L E II. If I<' = h(K) then K' is compact and h - 1 (I<.') ::) I<. so J.L o h - 1 (I<.') 2: 1 - c: for all J.L E II. 0

Using (2.6), the result for Rd , and a diagonal argument we can, for a given sequence J.Ln , pick a subsequence J.Lnk so that for all d, J.lnk o 1f'd converges to a limit Vd . Since the measures Vd obviously satisfy the consistency conditions of Kolmogorov's existence theorem, there is a probability measure v on (R00 , noo ) so that v o 1r;J 1 = Vd . We now want to check that J.Lnk => v. To do this we will apply (2.3) with A = the finite dimensional sets A with v(8A) = 0. Convergence of finite dimensional distributions and (1.1) imply that J.Ln k (A) -;. v(A) for all A E A.

280

Chapter 8 Weak Convergence

The intersection of two finite dimensional sets is a finite dimensional set and o(A n B) c 8A u 8B so A is closed under intersection. To prove that any open set G is a countable union of sets in A, we observe that if y E G there is a 8 > 0 so that B(y, 8) = {x : p(x, y) < 8} C G. Pick 8 so large that B(y, 28) n Gc =j:. 0. (G = R00 E A so we can suppose Gc =j:. 0.) If we pick N so that 2 - N < 8/2 then it follows from the definition of the metric that for r < 8/2 the finite dimensional set

A(y, r, N) = { y : lx; - y; j < r for 1 :::; i :::; N} C B(y, 8) C G The boundaries of these sets are disjoint for different values of r so we can pick r > 8/4 so that the boundary of A(y, r, N) has v measure 0. As noted above R00 is separable. Let y1 , Y2 , . . . be an enumeration of the members of the countable dense set that lie in G and let A; = A(y;, r;, N;) be the sets chosen in the last paragraph. To prove that U;A; = G suppose x E G - U;A;, let r > 0 so that B(x, r) C G, and pick a point Yi so that B(x, y; ) < r/9. We claim that x E A; . To see this, note that the triangle inequality implies B(y;, Br/9) C G, so 8 2:: 4r/9 and r > 8/4 = r/9, which completes the proof of J.lnk => v. 0 Before passing to the next case we would like to observe that we have shown

Section 8.2 Prokhorov's Theorems 281 If the points Xn � x in S then limp(xn , q; ) = p(x, q;) and hence h (x n ) � h ( x ) . Suppose x =j:. x' and let c = p(x, x') . If p(x, q;) < c/2, which must be true for some i, then p(x' , q;) > c/2 or the triangle inequality would lead to the contradiction f < p(x, x'). This shows that h is 1-1. Our final task is to show that h - 1 is continuous. If X n does not converge to x then limsup p(xn , x) = f > 0. If p(x , q;) < c/2 which must be true for some q; then lim sup p (x n , q;) 2:: c/2, or again the triangle inequality would give a· contradicition, and hence h( xn) does not converge to h(x) . This shows that 0 if h ( xn ) � h(x) then Xn � x so h - 1 is continuous.

It follows from (2.6) that if II is tight on S then {J.L o h - 1 : J.l E II} is a tight family of measures on R00 • Using the result for R00 we see that if J.ln is a sequence of measures in II and we let Vn = J.ln o h - 1 then there is a convergent subsequence Vnk · Applying the continuous mapping theorem, (1.2), now to the function cp = h - 1 it follows that J.lnk = Vn k o h converges weakly. The general case Whatever S is, if

II is tight, and we let K; be

so that

J.L (K;) 2:: 1 - 1/i for all J.l E II then all the measures are supported on the u-compact set So = U;K; and this case reduces to the previous one. 0 Proof of (2.2) We begin by introducing the intermediate statement:

(H) For each f, 8 > 0 there is a finite collection A1 , . . . , An of balls of radius so that J.L (Ui �n Ai ) 2:: 1 - f for all J.l E II.

8

(2.7) Theorem. In R00 , weak convergence is equivalent to convergence offinite

We will show that (i) if (H) holds then II is tight and (ii) if {H) fails then II is not relativley compact. Combining (ii) and (i) gives (2.2).

Proof for the u-compact case We will prove the result in this case by

Proof of (i) Fix f and choose for each k finitely many balls Af , . . . , A� k of radius 1/k so that J.L (U; �nk Af) 2:: 1 - c/2k for all J.l E II. Let J( be the closure of nf= 1 U;�n k A� . Clearly J.L (K) 2:: 1 - c. K is totally bounded since if Bj is a ball with the same center as AJ and radius 2/k then Bj, 1 :::; j :::; n k covers J(. Being a closed and totally bounded subset of a complete space, J( is compact and the proof is complete. 0

dimensional distributions.

reducing it to the previous one. We start by observing

(2.8) Lemma.

If S is u-compact then S is separable.

Proof Since a countable union of countable sets is countable, it suffices to prove this if S is compact. To do this cover S by balls of radius 1/n, let x�, m < mn be the centers of the balls of a finite sub cover, and check that the x� ar;-a countable dense set. 0

(2.9) Lemma. If S is a separable metric space then it can be emb edded home­ omorphically into R00 • Proof Let

q1 , q2 ,

• • •

from S into R00 by

be a sequence of points dense in S and define a mapping h(x) = (p(x, q1 ), p(x, q2 ), . . . )

Proof of (ii) Suppose (H) fails for some c and 8. Enumerate the members of the countable dense set q 1 , q2 , . . . , let A; = B ( q; , 8), and Gn = Uf= 1 A;. For each n there is a measure J.ln E II so that J.l n ( Gn ) < 1 - f. We claim that J.ln has no convergent subsequence. To prove this, suppose J.lnk => v. (iii) of {1. 1) implies that for each m v(Gm ) :::; lim sup J.lnk (Gm )

k-+oo :::; lim sup J.lnk (Gnk ) :::; 1 - f k-+oo

Section

282 Chapter 8 Weak Convergence However, Gm j S as m j

oo so we have (S) � 1 - e, a contradiction.

0

v

{w : Xt; (w) E A; for 1 � i � k} where 0 � t 1 < t 2 < . . . < tk � 1 and A; E nd , the Borel subsets of Rd . (3. 1 ) Lemma. B is the same as the u-field C generated by the finite dimensional

The Space C 283

As n -+ oo, fn (t) -+ foo = 0 but not uniformly. To see that Jl.n does not converge weakly to J.l oo , note that h(w) = sup 05 t 9 w(t) is a continuous function but

j h(w)Jl.n (dw)

8.3. The Space C

d Let C = C([O, 1], Rd ) be the space of continuous functions from [0, 1] to R equipped with the norm llwll = sup t 9 !w(t) l Let B be the collection of Borel subsets of C. Introduce the coordinate random variables Xt (w) = w(t) and define the finite dimensional sets by

8.3

= 1 -f> 0 =

j h(w)Jl.oo (dw)

� n � oo be probability measures on C. If the finite dimensional distributions of Jl.n converge to those of p.00 and if the Jl.n are tight then J.l n ::} p.00

{3 .2 ) Theorem. Let f1.n , 1 •

Proof If fl.n is tight then by (2.1) it is relatively compact and hence each subsequence Jl.n m has a further subsequence Jl.n ',. that converges to a limit v . If f : Rk -+ R is bounded and continuous then f(Xt u · . . Xt k ) is bounded and continuous from C to R and hence

sets.

Proof Observe that if e is a given continuous function

{w : llw - ell � e - 1/n} = nq {w : lw(q) - e(q) l � E - 1/n} where the intersection is over all rationals q E [0, 1]. Letting n -+ oo shows {w : ilw - ell < e} E C and B C C. To prove the reverse incl�sion observe that if the A; are open the finite dimensional set {w : w(t; ) E A; } open so the A 0 theorem implies C C B. Let 0 � t 1 < t2 < . . . < tn � 1 and 1rt : C -+ (Rdt be defined by 11" -

IS

1rt (w) = (w(t1 ), . . . , w(tn )) I Given a measure fl. on (C, C), the measures f1. o 1r;- , which give the dis�ribu­ the finite dimensxoz;tal called are tion of the vectors (Xt1 , , X1 J under Jl., . 1 ) one might hope that 3 ( of distributions or f.d.d.'s for short. On the basis convergence of the fl.n · weak for convergence of the f.d.d.'s might be enough • • •

However, a simple example shows this is false.

Example 3.1. Let an = mass on the function

{

1/2 - 1/2n, bn = 1/2 - 1/4n, and let fl.n be the point

fn (t) =

·

0

4n(x - an ) �n(1/2 - x)

x E [O, an] X E [an , bn] X E (bn , 1/2] x E (1/2, 1)

The last conclusion implies that Jl.n � o 1r;- 1 ::} v o 1r;- 1 , so from the assumed convergence of the f. d. d.'s we see that the f.d.d's of v are determined and hence there is only one subsequential limit. To see that this implies that the whole sequence converges to v, we use the following result, which as the proof shows, is valid for weak convergence on any space. To prepare for the proof we ask the reader to do Exercise 3.1. Let rn be a sequence of real numbers. If each subsequence of rn has a further subsequence that converges to r then rn -+ r.

(3.3) Lemma. If each subsequence of J.l n has a further subsequence that con­ verges to v then Jl.n ::} v . Proof Note that if f is a bounded continuous function, the sequence of real numbers I f(w)Jl.n (dw) have the property that every subsequence has a further subsequence that converges to I f(w)v(dw). Exercise 3.1 implies that the whole sequence of real numbers converges to the indicated limit. Since this holds for 0 any bounded continuous f the desired result follows.

As Example 3.1 suggests, Jl.n will not be tight if it concentrates on paths that oscillate too much. To find conditions that guarantee tightness, we intro­ duce the modulus of continuity

wa(w) = sup{ lw(s) - w(t) l : I s - t! � 8}

284 Chapter 8 Weak Convergence

{are3 .4n) Theorem. The sequence J..ln is tight if and only if for each e > 0 there 0 , M and 6 so that {i) J..ln (lw (O)l > M) � e for all n ;::: no (ii) J..ln (w 6 > e) � e for all n ;::: no Remark. Of course by increasing M and decreasing 6 we can always check the condition with n0 = 1 but the formulation in ( 3.4) eliminates the need for that final adjustment. Also by taking e = TJ A ( it follows that if J..ln is tight then there is a 6 and an no so that J..l n (w6 > TJ) � ( for n ;::: no. Proof We begin by recalling (see e.g. , Royden (1988), page 169) (3.5) The Arzela-Ascoli Theorem. A subset of C has compact closure if and only if supw E A l w (O)l < oo and lim6.... o SUPw e A w 6(w) = 0. To prove the necessity of (i) and (ii), we note that if J..ln is tight and e > 0 we can choose a compact set (I<) ;::: 1 {w- e �fore}allfornsmall . By (3.5), I< C {X{O) � M} for large M andsoifthat e > 0J..lnthen 6. I< 6 To prove the sufficiency of (i) and (ii) , choose M so that J.l. n(I X (O) I > M) � e/2 for all n and choose 6k so that J..ln (w6k > 1/k) � ej2k+1 for all n. If we let I< be the closure of {X(O) � M, W6k � 1/k for all k} then ( 3.5 ) implies I< is compact- and J..l n (I<) ;::: 1 - e for all n. Condition (i) is usually easy to check. For example it is trivial when Xn(O) = Xn and Xn x as n ---+- oo The next result, (3.6), will be useful in checking condition (ii). Here, we formulate the result in terms of random variables Xn taking values in C, that is, to say random processes {Xn(t),O � tJ..ln�(A)1}.=However, easily rewrite the result in terms of their distributions P(Xn E one A). can Here, A E C. (3.6) Theorem. Suppose for some a, (:J > 0 A

[{

C

D

---+-

.

then (ii) in ( 3.4) holds. Remark. The condition should remind the reader of Kolmogorov's continuity

(1.6) (1.6) (1968).

1.

criterion, in Chapter The key to our proof is the observation that the proof of gives quantitative estimates on the modulus of continuity. For a much different approach to a slightly more general result, see Theorem in Billingsley

12.3

Section 8.4 Skorokhod 's Existence Theorem for SDE 285

Proof Let

r

< af(:J, and pick TJ > 0 small enough so that

(1 - 1})(1 + a - (:J-y) - (1 + TJ) > 0 From (1.7 ) in Chapter 1 it follows that if A = 3 2(1 - 'lh /(1 - 2 - 'Y) then with probability ;::: 1 - I<2 - N>.j( l - 2 - >.) we have IXn(q) - Xn(r)l � A l q - rl'Y for q, r E Q2 n [0, 1] with 2- (l - !J)N �ick N so that I<2 - N>./(1 - 2 - >.) � e then 6 � 2 - (l - !J)N so that A6'Y < e and - 0 It follows that P(w6 > e) � e. ..\

=

·

8.4. S korokhod 's Existence Theorem for SDE

In this sect�on, we wil! describ : Skorokhod's approach to constructing solutions . of stochastic differential equatiOns. We will consider the special case

(X ) dt

lT

where is bounded and continuous, since we can introduce the term b t by change of measure. The new feature here is that u is only assumed to be cont!nuous, not Lipschitz or even Holder continuous. Examples at the end of SectiOn 5.3 show �hat we cannot hope to have uniqueness in this generality. Sko okhod's Idea for solving stochastic differential equations was to dis­ . : �re�Ize time to get an equation that is trivial to solve, and then pass to the hm1t and extract subsequential limits to solve the original equation. For each n, define by setting = and, for < +

Xn(t)

Xn(O) x

m2- n t � (m 1)2- n ,

Xn

Since is a stochastic integral with respect to Brownian motion ' the formula for covariance of stochastic integrals implies

ns]j2n))ds Xn([ i\lT 2 klT j k}( i k i a;j (Xn.([2ns]/2n ))ds where as usual a = uuT . If we suppose that a;j(x) � M for all i, j and x, it follows that if s < t, then (X� , X� ) t = L 0 t =

I

' I !

286 Chapter 8 Weak Con vergence so

Section

(5.1) in Chapter 3 implies sup IX�(u) - X�(s)IP � Cp EI(X� ) t - (X� ) . IP/ 2 � Cp {M(t - s) }P/ 2 E ue[•,t]

Taking p = 4, we see that

E ue[sup• ,t] IX�(u) - X�(s)l4 � CM2 (t - s)2 Using (3.6) to check (ii) in ( 3 .4) and then noting that Xn(O) = x so (i) in 3.4) is trivial, it follows that the sequence Xn is tight. Invoking Prohorov's (Theorem (2 .1 ) now, we can conclude that there is a subsequence Xn (k) that converges weakly to a limit X. We claim that X satisfies MP (O, a) . To prove this, we will show (4.1) Lemma. If f E C2 and Lf = � L:ii aii Dii f then f(Xt ) - f(Xo) - 1t Lf(X. ) ds is a local martingale Once this is done the desired conclusion follows by applying the result to = and

Xi f(x) XiXj .

f, Dd, Dij f f(Xn(t)) - f(Xn(s)) = 2;:::: 1t Dd(Xn(r))dX�(r) . . 1 "' t + 2 � i Dij /(Xn(r))d(X� ,X� )r • So it follows from the definition of Xn that

f(x) =

are bounded, then the Proof It suffices to show that if and process above is a martingale. Ito's formula implies that I

I]

8.5

Donsker's Theorem

287

Skorokhod's representation theorem, (1.4), implies that we can construct processes with the same distributions as the on some probability space in such a way that with probability 1 as k -+ oo, t converges to t uniformly on T] for any T < oo. If < t and : C -+ R is a bounded continuous function that is measurable with respect to :F. , then

Yk

Xn(k) Yk ( ) g s

[0,

Y( )

E (g(Y)· {!(Yi ) - f("Y, ) - it Lf(Yr) dr}) t = lim E (u(Y k ) · { f(Yl) - f(Y.k ) - i Lnf(Y/.') dr }) = 0

Since this holds for any continuous g , an application of the monotone class theorem shows

which proves (4.1).

D

8 . 5 . Donsker's Theorem

El;.i 0 E(f Sm , 0Sn m n · · · x. S(nt] /Vn [x] 0 t = m/n Sm /Vn ift Btn - { linear if t E [m/n, (m + 1 )/n] (5.1) Donsker's Theorem. As n oo, B n B, where B is a standard

= 1. Let = 6 + + /;.n be the = and Let 6, 6, . . . be i.i.d. with nth partial sum. The most natural way to turn � � into a process is the largest integer � where indexed by � � 1 let il'( = To have a continuous trajectory we will instead let

=>

-+

Brownian motion.

(3 2) . Thus there are two things to do: Convergence of finite dimensional distributions. Since S(nt] is the sum of [n t] independent random variables, it follows from the central limit theorem that if 0 < t1 < . . tn � 1 then Proof We will prove this result using

.

.

so if we let

Lnf(r) = � L:ij aij {Xn { [2nr]2- n ))Dij /(X� ) then f(Xn(t)) - f(Xn(s)) - J.t Lnf(r) dr is a local martingale.

To extend the last conclusion from = nt - nt then

rn

[]

iJn to Bn , we begin by observing that if

288 Chapter 8 Weak Convergence

rn

Section

rn�[ntJ+dVn--+

Since 0 ::; < 1 , we have 0 in probability and the converging together lemma, (1.3), implies that the individual random variables Bf =? Bt . To treat the vector we observe that

( Bf, - Bf,_J - ( Bf, - Bf,_ J = ( Bf, - Bf, ) - ( Bf,_ 1 - Bf,_J - o in probability, so it follows from (1.3) that

(x1 , x2 ,

8.5

Donsker's Theorem

289

C1 = 32 {fl� + iiM } and C2 = 24 {ji._L + iiM }. Remark. We give the explicit values of C1 and C2 only to make it clear that they only depend on fi.M and iiM . Proof Only (iii) needs to be proved. Since E(ei - fi.M ) = 0 and the ei are independent where

( --+) (xi> X 1 x 2 , ... , x 1

Using the fact that . . . , Xn) + · · · + X m ) is a + continuous mapping and invoking 1. 2 it follows that the finite dimensional distributions of B n converge to those of B .

L2

Tightness. The maximal inequality for martingales (see e.g., Chapter 4 of Durrett (1995)) implies

(4.3) in

2 -< 4ES'f s E (0max ) i 5,i5.l

Taking £ =

If p is an even integer (we only care about p =

E(ej - fi.M )P ::; 2P(ji.�{ + Eef )

From this, (ii) , and our assumptions it follows that

njm and using Chebyshev's inequality it follows that P (05,j5,maxn/m lSi I > ivfn) ::; 4/me2

C2£2 . To estimate the fourth moment we notice that E(ej) = 1M 4x3P(I{t l > x)dx ::; 2M2 1M 2xP( Iei 1 > x) dx = 2M2 iiM

This gives the term

or writing things in terms of Bf

m

So using our inequality for the pth power again, we have

1/m

Since we need intervals of length to cover [0, 1] the last estimate is not . good enough to prove tightness . To improve this, we will truncate and compute fourth moments. To isolate these details we will formulate a general result.

.6,=. . . be i.i.d. with mean Let {i = �i l ( I€15: M ) , _let � 1 2 , let > St el +et o:(M) E(l�d ; l�d M), fi.M = E�i , and iiM = E(([) . P(�i ei) ::; M- 2E(I�i l 2 ; 1�1 > M) = o:(M)/M2 (ii) I J.l - fi.M I ::; E(l�d; 1�1 > M) ::; o: (M ) /M (iii) If M � 1 and o: (M) ::; 1 then (5.2) Lemma. Let = +. let Then we have ( i) :f::

2, 4) then

f.l ·

M � 1, this gives the term C1£M2 and the proof is We will apply ( 5 2 ) with M = 8 vfn. Part (i) implies (5.3) nP(�i :f:: �i)- ::; n o:(882nvfn) --+ 0 Since we have supposed complete.

o

.

·

by the dominated convergence theorem. Since J.l = 0 part (ii) implies

(5.4)

Section

290 Chapter 8 Weak Con vergence

n oo. Using the L4 maximal inequality for martingales (again see e.g., (4.3) in Chapter 4 of Durrett (1995)) and part (ii9 , it follows tha� if i = .nfm and n is large then (here and in what follows C Will change from hne to hne) E sup I Sk - kiiMI 4 � GE I St - liiMI 4

as

--.

O�k9

- P

2 Sk - kiiMI � c Vn/9) � Cc-4 ( 6m + � (osup I m) �k�l

Let Bf = S[n1] /Vn and Ik ,m = probabilities in (5.5), it follows that

[kjm, (k + 1)/m].

(

maxm max I B� - Bl:; m I lim sup P O�k< •Elk,m n-oo

Adding up

)

(

m of the

> c/9 � Cc -4 o2 + _!_ m

)

To turn this into an estimate of the modulus of continuity, we note that maxm max lf( s) - f (k/m) l w1 ; m (f) � 3 O�k< •Eh,m

� s � (k + 1)/m � t � (k + 2)/m lf(t) - f (s) l � lf (t) - f ((k + 1)/m) l + lf((k + 1)/m) - f (kjm) l + if(kjm) - f (s) i If we pick m and o so that Cc4 ( o 2 + 1/m) � f. it follows that limsup P(wl fm(f:J n ) > c/3) � f. (5.6) n-oo

since for example if k/m

To convert this into an estimate the modulus of continuity of iJn we begin by observing that (5.3) and (5.4) imply

P as

n

--.

(0�sup1 9 IB� - B� l > c/3) -o 0

oo so the triangle inequality implies limsup P(wt ; m ( iJn ) n-oo

> c) � f.

Donsker's Theorem

291

Now the maximum oscillation of B n over [s, t] is smaller than the maximum oscillation of iJ n over [[ns]jn, ([nt] + 1)/n] so

(5.7) and it follows that

limsup P(w1/ 2m (B n ) > c) � f. n-oo This verifies (ii) in (3.4) and completes the proof of Donsker's theorem.

0

The main motivation for proving Donsker's theorem is that it gives as corollaries a number of interesting facts about random walks. The key to the vault is the continuous mapping theorem, (1.2), which with (5.1) implies:

Using Chebyshev's inequality now it follows that

(5.5)

8.5

.,P : C[O, 1] R has the property that it is continuous .,P(Bn ) => .,P(B) . Example 5.1. Let .,P(w) = max {w(t) : 0 � t � 1}. It is easy to see that I .,P(w) - .,P(e) l � llw - .,P II s o .,P : C[O, 1) --. R is continuous, and (5.8) implies max = max Bt O� m � n Sm /Vn => Mt 0� 1 9 (5.8)

Theorem. If

--.

Po-a.s. then

To complete the picture, we observe that by (3.8) in Chapter 1 the distribution of the right-hand side is

Po(Mt � a) = Po(Ta � 1) = 2Po(Bt � a) 5.2. Let .,P(w) :::d sup {t � 1 : w(t) = 0}. This time .,P is not continuous, for if w, has we(O) = 0, we(1/3) = 1, we(2/3) = f., w(1) = 2, and linear on each interval (j, (j + 1)/3) then .,P(wo) = 2/3 but .,P(we) = 0 for f. > 0. It is easy to see that if .,P(w) < 1 and w(t ) has positive and negative values in each interval (.,P(w) - o, .,P(w)) then .,P is continuous at w. By arguments in Example 3.1 of Chapter 1, the last set has Po measure 1. (If the zero at .,P(w) was isolated on the left, it would not be isolated on the right. ) Using (5.8) now Example

n : Sm-1 Sm � 0}/n => L = sup {t � 1 : B1 = 0} The distribution of L, given in (4.2) in Chapter 1, is an arcsine law. Example 5.3. Let .,P(w) = l {t E [0, 1] : w(t) > a } ! . The point w = a shows that .,P is not continuous but it is easy to see that .,P is continuous at paths w with l {t E (0, 1] : w(t) = a} l = 0. Fubini's theorem implies that sup { m �

·

Eo l {t E (0, 1] : Bt = a} l =

11 Po(Bt = a) dt = 0

292 Chapter 8 Weak Con vergence

Section

so 7/J is continuous Po-a.s. With a little work (5.8) implies

l {m :::; n : Sm > a vfn}l/n l{t E [0, 1] : Bt > a} l Before doing that work we would like to observe that (9.2) in Chapter 4 shows that l {t E [0, 1] : Bt > O}l has an arcsine law under P0 • Proof Application of (5.8) gives that for any a, l {t E [0, 1] : Bf > avfn}l l {t E [0, 1] : Bt > a}l To convert this into a result about l { m :::; n : Sm > afo}l we note that if e > 0 then =>

=>

(5.9)

{ m n IXml :::; ey'Ti}, we have l {t E [0, 1] : Bf > (a + e) vfn}l :::; -n1 l {m :::; n : Sm > avfn}l :::; l {t E [0, 1] : Bf (a - e)vfn}l Combining this with the first conclusion of the proof and using the fact that b -+ l{t E [0, 1] : Bt > b} I is continuous at b = a with probability one, we arrive easily at the desired conclusion. 0 Example 5.4. Let 1/J(w) = Jr0 , 11 w(t) k dt where k > 0 is an integer. 7/J is

by dominated convergence, and on max

.::;

>

continuous so applying (5.8) gives

IJ.' <s(nt)/Vn)' dt - n- l-(k/') t. s� I

we have

x<

lx - Yl :::; e lxl, I YI :::; M eMk+1 l z lk+ 1 -< -lx k k I < 1Y --dz k+1 k+ 1 -y

From this it follows that on

-

:z:

The Space D 293

(k +\)M For fixed M we have P(m m I Xml :::; M - ( k+ 2 ) Vn) -+ 0 by (5.9) so using Example 5.1, and (1.1) it follows that ax

,;

.::; n

IBtl < M) n- oo P(Gn(M)) � P (omax .::;t _::; 1 The right-hand side is close to 1 if M is large so n {1 k k ) ( / 1 2 / n dt(Bf ) L vn l0o m=1 s� -+ 0 liminf

in probability and it follows from the converging together lemma

(1.3) that

n n- 1 - (k/2) m=L s� 11 Bz dt 1 It is remarkable that the last·result holds under the assumption that EXi = 0 and EX[ = 1, i.e., we do not need to assume that EI Xl l < oo. =>

8.6.

°

The Space D

In the previous section, forming the piecewise linear approximation was an an­ noying bookkeeping detail. In the next sectioZ: when we consider processes that jump at random times, making them piecewise linear will be a genuine nui­ sance. To deal with that problem and to educate the reader, we will introduce the space Rd ) of functions from into Rd that are right continuous and have left limits. Since we only need two simple results, (6.4) and 6.5 be­ low, and one can find this material in Chapter of Billingsley 68 Chapter of Ethier and Kurtz 86 or Chapter VI of Jacod and Shiryaev 87 we will content ourselves to simply state the results. We begin by defining the Skorokhod topology on D. To motivate this consider

D([O, 1],

To convert this into a result about the original sequence, we begin by observing that if y with and then

8. 6

3

Example

[0, 1]

(19 ),

6.1. For 1

3

:::; n :::; oo let 1)/2n) fn(t) = { 01 tt EE [O,(n [(n + +1)/2n, 1]

(19 ), ( ) (19 ),

294

Chapter 8 Weak Convergence

where (n+I)/2n = I/2 for n = oo . We certainly want fn -+ foo but l fn-foo ll = I for all n. Let A be the class of strictly increasing continuous mappings of [0, I] onto itself. Such functions necessarily have = 0 and (I ) = 1. For J, E D define to be the infimum of those positive c for which there is a >. E A so that ::; c sup ::; c and sup t t It is easy to see that d is a metric. If we consider = and = in Example 6.I then for c < I we must take >.((n + I)/2n) = (m + I)/2m so

>.(0)

d(f,g)

>.

g

l f(t) -g(>.(t))l f fn g fm

i >.(t) -t i

I l I I When m = we have d(fn, Joo ) = I/2n so fn -+ foo in the metric d. We will see in (6.2) that d defines the correct topology on D. However, in n+I - m+I = I - I d(fn, fm ) = � � 2n 2m

oo

view of (2.2), it is unfortunate that the metric d is not complete. Example 6.2. For

{

= I/2)

� � The pointwise limit of Un is g 00 0 but d(gn,Uoo ) = 1. We leave it to the reader to show =

> 0.

To fix the problem with completeness we require that >. be close to the identity in a more stringent sense: the slopes of all of its chords are close to 1. If >. E A let = ��� log

11>-1 1

I ( >.(t� = �(s) ) I

The Space D 295

For J, E D define do(x , y) to be the infimum of those positive c for which there is a >. E A so that

11>-11 ::; c

l f(t) - g(>.(t))l ::; c It is easy to see that do is a metric. The functions Un in Example 6.2 have do(Un , Um ) = min { I, pog (n/ m)l} and

sup t

so they no longer form a Cauchy sequence. In fact, there are no more problems.

(6.I) Theorem. The space D is complete under the metric d0• For the reader who is curious why we discussed the simpler metric d we note:

(6.2) Theorem. The metrics d and do are equivalent, i.e., they give rise to the same topology on D .

S C [O, I]. For 0 < 8 < I put w�(f) = {inft i} O<maxi :$; r W[t ". t -+1) (!) where the infimum extends over the finite sets {t ; } with 0 = to < t 1 < < tr and t; - ti - l > o for I i ::;

for each

d(gn , Um ) = 2� - 2�

Exercise 6.1. If h E D then lim infn- oo d(gn, h)

l

g

8.6

ws(f) = sup{lf(t) - f(s).l : s, t E S}

0 t E [O, I/2) I t E [I/2, (n + I)/2n) 0 t E [(n + I)/2n, I]

In order to have c < I in the definition of d(gn, Um ) we must have >.(I/2) and >.((n + I)/2n) = (m + I)/2m so

If

Section

Our first step in studying weak convergence in D is to characterize tight­ ness. Generalizing a definition from Section 7.3 we let

I ::; n < oo let

Un (t) =

I

'

::;

···

r

The analogue of (3 .4) in the current setting is

(6.3 ) Theorem. The sequence are no, M, and 8 so that

Jl.n is tight if and only if for each c > 0 there

fl.n(lw (O)I > M) ::; c for all n ;:=: no (ii) fl. n (w� > c) ::; c for all n ;:=: n 0 (i )

Note that the only difference from (3 .4) is the 1 in (ii). If we remove the 1 we get the main result we need.

296

Chapter 8 Weak Con vergence

Section

(6.4) Theorem. If for each E > 0 there are n0 , M, and 6 so that (i) J.Ln( lw(O) I > M) � f for all n � no (ii) then

(6.5) Theorem. If each subsequence of converges to J.L then J.L n => J.L ·

{

dy)jh h (x, dy.) -_ ITQhh (x, (x, dy)

r.(

(6.1)-(6.4) are Theorems 14.2, 14.1, 15.2, and 15.5 i n Billingsley (1968) . We will also need the following, which is a consequence of the proof of (3.3) .

8. 7.

in the limit then we have weak convergence. To make a precise statement we need some notation. To incorporate both cases we introduce a kernel

J.Ln(w6 > c) � f for all n � no

J.Ln is tight and every subsequential limit J.L has J.L(C) = 1 .

J.Ln

�'

and define

afj (x) = bf (x) =

has a further subsequence that

Convergence t o Diffusions

In this section we will prove a result, due to Stroock and Varadhan, about the convergence of Markov chains to diffusion processes. Our treatment here follows that in Chapter 11 of Stroock and Varadhan (1979) but we will discuss both discrete and continuous time in parallel. Our duplicity lengthens the proof somewhat, but is preferable we believe to the usual (somewhat dangerous) cop­ out: "the same argument with minor changes handles the other case." In discrete time, the basic data is a sequence of transition probabilities I h (x, ay) for a Markov chain Y� h • m = 0, 1 , 2, . . . , taking values in Sh C Rd , I.e., P (Yc�+l) h E A I Y� h = x) = I h (x, A) for X E sh , A E nd In this case we defin e X1h = Yh11 J h] • i.e., we make Xf constant on intervals [mh , (m + 1)h). In continuous time, the basic data is a sequence of transition rates Qh (x, dy) for a Markov chain Xf , t � 0, taking values in Sh c Rd , i.e.,

Note that although x E A is excluded from the interpretation, we will allow Q( x, { x}) > 0 in which case jumps from x to x (which are invisible) occur at a positive rate. To have a well behaved Markov chain we will suppose

8. 7 Con vergence to Diffusions 297

in discrete time . . . m contmuous time

1ly-xl $1 (Yi - x;)(Yi - Xj )Kh(x, dy) 1IY-XI$1 (Yi - x;)Kh (x, dy)

.6-� (x) = Kh (x, B(x, en where B(x , ) = {y : IY - x l < c}. Suppose (A) aii and bi are continuous coefficients for which the martingale problem is well posed, i.e., for each x there is a unique measure Px on (C, C) so that the coordinate maps X1 (w) = w(t) satisfy Px ( Xo = x) = 1 and E

Xf

- 11 b;(X, ) ds

and XfXf

- 11 a;i (X, ) ds

are local martingales. (7.1) Theorem. Suppose in continuous time that (B) holds, and in either case that (A) holds and for each i, j, R < oo, and c > 0 (i) limh!O sup l xi$R i a�(x) - a;j (x) l = 0 (ii) limh .t.o SUP 1xi$R l bf {x) - b; (x) l = 0 (iii) limh!O SUP 1xi$R .6- �(x) = 0

If X8 = Xh - X then we have Xf with Xo = x.

=>

x1 the solution of the martingale problem

K, SUPxe K Q h (x, Rd ) < oo.

Remarks. Here => denotes convergence in D([O, 1] , Rd ) but the result can be trivially generalized to give convergence D([O, T] , Rd ) for all T < oo. In (i), (ii) , and (iii) the sup is taken only over points x E Sh . Condition (iii) cannot hold unless the points in Sh are getting closer together. However, there is no implicit assumption that Sh is becoming dense in all of Rd .

In words, our convergence theorem states that if the infinitesimal mean and covariance of the Markov chain converge to those of a diffusion for which the martingale problem is well posed, and we have a condition that rules out jumps

Proof The rest of the section is devoted to the proof of this result. We will first prove the result under the stronger assumptions that for all i, j and E > 0 we have

(B) For any compact set

298

Chapter 8 Weak Convergence

Section

l a�(x) - a;j (x) l = 0 (iv) suph,:c la�(x) l < l bf<x) - b;(x) l = 0 (v) SUPh,:c l bf{x) l < (vi) SUP h, :c Ll�(x) < Ll� ( x) = 0 and in continuous time that (B') sup, Q(x, Rd ) < (i') limh!o sup, (ii') limh!o sup, (iii') limh!O sup,

oo

00

00

oo .

a. Tightness

f

If is bounded and measurable, let

Lh f(x) = J I
As the notation may suggest, we expect this to converge to

Lf(x) = � Li, aij (x)D;j /(x) + Li b;(x)Dd(x) i To explain the reason for this definition we note that if f is bounded and measurable then (7.2a) In discrete time f(Xfh ) - 2:::7,: : � h Lh f(Xjh ) is a martingale. (7.2b)_In continuous time f(Xf') - J� Lh ! (X:) is a martingale. Proof (7.2a) follows easily from the definition of Lh and induction. Since we have (B'), (7.2b) follows from ( 2 .1) and (1.6) in Chapter 7. 0 The first step in the tightness proof is to see what happens when we apply Lh to f,,y (x) = !E(x - y) where f,(x) = g( l x i2 /E2), and g(x) = { (10 - x2)(1 - x)3 0X �;:::: x1 � 1 The exact formula for g is not important. All we care about is that 0 g(x) � 1 for X ;:::: 0, g(O) = 1, g(x) = 0 for X ;:::: 1, and g E C2 (since g'(O) = g" (O) = 0). (7.3) Lemma. There is a which only depends on E so that I Lh /y, , l � CE. Proof Applying Taylor's theorem to the function h(t) = f(x t(y - x)) for 0 � t � 1 we have for some c:c, y E [0, 1] f(y) - f(x) = h(1) - h(O) = h' (O) + h"(c,,y)/2! (7.4) = L ( Yi - x;)Dd (x) + L ( Yi - Xi )( Yi - Xj)Dij / (z:c ,y ) i i ,j �

G,

+

8. 7

Convergence to Diffusions 299

C:c,y (Yx). Integrating with respect to I
{

Noticing that the Cauchy-Schwarz inequality implies

(7.5)

I'Li u;v;l � l u llvl

llm;i l gives ��(y; - x;)(yi - Xj)D;j /(z) l � jy - x i2 11 Dii f(z) i

and the definition of

(7.6) it follows that

N.ow

}

• ,J

I

l

Lh f(x) � AE 1ly-:cl9 (y - x)I
Li a�; (x) = 1ly-:cl 9 I Y - x l 2 I
o

To estimate the oscillations of the sample paths, we will let r0 = 0, E/4} Tn = inf{t > 1"n - 1 : = inf{n ;:::: 0 : Tn > u = min rn Tn - 1 : 1 n () = 0<

IXf - x;,._l l ;:::: N 1} { � � N} max{!Xh (t) - Xh (t-) 1 : t � 1}

Section

300 Chapter 8 Weak Con vergence

u

>

o and 8 $

E/4 then w0(Xh ) $ f..

Proof To check this we note that if

cases to consider 1.

CASE

Con vergence to Diffusions

u > o and 0 < t - s < o there are two

Tn $ < t < Tn+1 lf(t) - f(s)l $ lf(t) - f(rn)l + lf(rn) - f(s)l $ 2E/4 S

l f(t) - f(s)l :5 l f(t) - f(rn)l + l f(rn) - f(rn-) 1 + l f (Tn-) - f(Tn - 1 ) 1 + l f(Tn - 1 ) - f(s)l $ f.

Combining (7.10) and (7. 11) we have

Py(u $ o) $ k sup Px (r $ o) + Py(N X

>

k) $ C•14k0 + e.Xk

If We pick k SO that e,Xk $ E/3 and then pick 0 SO that C E) :::; f.. This verifies (ii) in (6.4). Since X8 = (i) is trivial and the proof of tightness is complete in discrete time. 0

Py(w6(Xh )

Xh ,

0

Tightness proof, discrete time. One of the probabilities in (7.8) is

Tightness proof, continuous time. One of the probabilities in (7.8)

E/4 occur at a rate smaller than llh (x, B(x, (7.9b) Py(O ;::: E/4) $ 1 - exp(- sup Q h (x, B(x, Ej4t)) $ sup ��14 (x) --+ 0 by (iii'). The first step in estimating Py(u > o) is to estimate Py(r1 $ o). To do this we begin by observing that (7.2b) and (7.3) imply

is trivial to estimate. Since jumps of size supx E/4Y) and e-z ;::: 1 - z

>

X

(7.9a)

Py(u

_xn

(7.11)

(7.12)

trivial to estimate

by (iii'). The first step in estimating > o) is to estimate do this we begin by observing that (7.2a) and (7.3) imply

301

Iterating and using the strong Markov property it follows that Ey( e- r" ) :::; and

To relate these definitions to tightness, we will now prove (7.8) Lemma. If

8. 7

Py(r1 $ o). To

fy •f4 (Xfh ) + C.t4kh, k = 0, 1, 2, . . . is a submartingale ,

!y,
X

t

;:::

0 is a submartingale

Using the optional stopping theorem at time T A o it follows that

Using the optional stopping theorem at time T A 6 where 6 = [ojh]h $ o and noticing Xr 11 6 = Xr 11 o i t follows that The remainder of the proof is identical to the one in discrete time.

0

or rearranging and using T A o $ o

b. Convergence

(7.10)

Since we have assumed that the martingale problem has a unique solution, it suffices by (6.5) to prove that the limit of any convergent subsequence solves the martingale problem to conclude that the whole sequence converges. The first step is to show

Py(N k). To do this, we begin by observing Ey (e- T1) :::; Py(T1 :::; o) + e -o Py (T1 o) $ C./48 + e - o (1 - C•/48) .X < 1

Our next task is to estimate

>

>

=:

(7.13) Lemma. If f E C'k then

I Lh f - Lfl loo --+ 0.

Proof Using (7.4) then integrating with respect to

IY - xl � 1 we have Lh f (x) = I )f (x)Dd(x)

Kh(x, dy)

over the set

+ 1l y- xl9 .L:( Yi - x;)(Yj - Xj )D;jf (zx,y ) Il {f(y) - f(x)} I
(7.14)

Recalling the definition of .D.� and using (vi) i t follows that the third term goes to 0 uniformly in x . To treat the first term we note that (7.5) and (v ) imply

l 4 )f (x)Dd(x) - � b;(x)Dd(x) l � s�p lbj (x) - bj (x) l � !Dd(x)l -+ 0 I

I

I

uniformly in x . Now the difference between the second term in (7. 14) and is smaller than

(7.15)

):;j a;j D;j/( x)

where

Convergence to Diffusions

303

II D;j/ (zx ,y ) - D;j/ (x) ll r (t:) = ysup l - x l::; e

Now Zx ,y is on the line segment connecting x and y, D;j/ is continuous and has compact support, so f (t:) --+ 0 as t: --+ 0. Using (7.7) and (iv) now the quantity in (7.16) converges to 0 and the proof of (7.13) is complete. 0 Convergence proof, discrete time. Let hn weakly to a limit X0 • (7 .2a) implies k- 1

f (X�i:J - .L: hn L h ,. f (Xj;,.), j=O

--+ 0 so that X h ,. converges

k = 0, 1, 2, . . . is a martingale

To pass to the limit, we rewrite the martingale property by introducing a bounded continuous function g : D --+ R which is measurable with respect to :F3 and observe that if kn = [s/hn] 1 and .en = [tfhn] 1 the martingale property implies

+

+

� � a;j (x)D;j/(x) - � afj (x)D;j/(x) l IJ

IJ

- x;)(Yj - Xj) {D;j/ (zx, y) - D;j f (x)} I
By (7 .5) the first term is smaller than

which goes to 0 uniformly in x by (iv) . To estimate the second term in (7.15) we note that for any t: the integral over {y : t: < ! Y - xI � 1 } goes to 0 uniformly in x by ( vi) and the fact that the integrand is uniformly bounded. Using (7.6), the last piece

(7.16)

8. 7

Section

302 Chapter 8 Weak Convergence

Let h 00 = 0 . Skorokhod's representation theorem (1 .4) implies that we can construct processes yn ' 1 � n � 00 with the same distribution as the x h ,. so that yn --+ y oo almost surely. Combining this with (7.13) and using the bounded convergence theorem it follows that if f E C'k then

Since this holds for all continuous g : C --+ R measurable with respect to :F3 it follows that if f E Ck: then

f (X? ) -

it Lf(X�) ds

is a martingale

Applying the last result to smoothly truncated versions of f ( x) = x; and f (x) = x;xj it follows that X0 is a solution of the martingale problem. Since the martingale problem has a unique solution, the desired conclusion now follows from (6.5). 0

304 Chapter 8 Weak Convergence

Section

Convergence proof, continuous time. Let

verges weakly to a limit

X0 • (7.2b) implies

and repeating the discrete time proof shows that tingale problem and the desired result follows.

hn -�o 0 so that Xhn con­

X0 is a solution of the mar­0

be a coo func­ cp k 1 1) I
martingale problem with coefficients

a b (6.1) h� X 0 • h� X k h n , x h � , k xo , k . h� X 1}. Gk (2.1)

X0• k

since up until the exit from B(O, k) , is a solution of the martingale problem for and and hence its distribution agrees with that of Now up to the first exit from B(O, k) so we has the same distribution as that of have

X0 • X h � , k liminf P(X h� E H) :2:: liminf n -co P(Xh� E Gk n H) -> P(X0 E Gk n H) ...... Since there is no explosion the right-hand side increases to P(X 0 E H) as k j oo and the proof is complete. b a n

oo

X h�

305

(7.1) beginning with the following

8.1. Ehrenfest Chain. Here, the physical system is a box filled with air and divided in half by a plane with a small hole in it. We model this mat4ematically by two urns that contain a total of balls, which we think of as the air molecules. At each time we pick a ball from the in the two urns at random and move it to the other urn, which we think of as a molecule going through the hole. We expect the number of balls in the left urn to be about + C..jii, so we let z:;. be the number of balls at time m in the left urn minus and let Y��� z:;,j..jii. The state space ::; k ::; { kf..jii : while the transition probability is + II +

Example

2n

2n

n n,

= S1/n = n} -n Ill/n (x ' x n- 1/2) = n -2xn..jii 1/n (x , X - - 1/2) - n 2nx..jii When > n - 1 12 , 6.! / n (x) = 0 for all x so (iii) holds. To check conditions (i) and (ii) we note that n - x..jii __!__ . n - x ..jii } = -x . blfn (x) = n { __!__ 2n ..jii 2n Vn a1/n ( x) - n { -n1 · n -2nx ..jii + -n1 · n -2nx..jii } = 1 The limiting coefficients b(x) = -x and a(x) = 1 are Lipschitz continuous, so n n the martingale problem is well posed and (A) holds. Letting x1l/ = y,[nl/1]/ n . (7.1) now it follows that and applymg n (8.1) Theorem. As n -�o oo, X11/ converges weakly to an Ornstein-Uhlenbeck process x1 , i.e. , the solution of n

-

t:

_

We have supposed that the martingale problem for and has a unique nonexplosive solution In view of (3.3), to prove it suffices to show that given any sequence =? To -�o 0 there is a subsequence so that prove this we note that since each sequence is tight, a diagonal argument =? shows that we can select a subsequence so that for each k, Let is an open set so if w : w(t) E B(O, k) , 0 ::; t ::; { , =? and H c C is open then implies that

G Xhn ,k Xk0·k=

Examples

Examples

In this section we will give applications of very simple situation.

c. Localization

0• X hn

8 .8.

8.8

_

Next we state a lemma that will help us check the hypotheses in the next two examples. The point here is to replace the truncated moments by ordinary moments that are easier to compute.

lifi (x) = j(y; - x;)(yj - Xj)I
306 Chapter 8 Weak Convergence

Section

(8.2) Lemma. If p ;:::: 2 and for all R < oo we have (a) limh!o sup l xi � R la?j (x) - a ij (x) l = 0 (b) limh! o sup l xi � R l bA ;h (x) - b;(x) l = 0

Proof We first prove the result under the assumption

ll.�(x) � cP ,;(x)

Using the Cauchy-Schwarz inequality with the triviality ( Yk - x k ?

�

f

}l y -x l >1

{

}l y -x j >1

� IY - xl 2

i (y; - x;)(Yi - Xj) I I
The desired conclusion follows easily from the last three observations. Our first two examples are the last two in Section the beginnings of the subject, see Feller (1951). Example

8.2.

0

5.1, and date back to

Branching Processes. Consider a sequence of branching

Our goal is to show

n

If (A1), (A2), and (A3) hold then Xt1 / converges weakly to Feller's branching diffusion Xt, i.e., the solution of

(8.3) Theorem.

dXt = f3Xt dt + IT.j)(; dBt

(A3') for any 6 > 0, Lk >on k2 pA: = 0 for large n Then we will argue that if n is large then with high probability we never see any families of size larger than n6. The limiting coefficients satisfy (3.3) in Chapter 5, so using (4.1) there, we see that the martingale problem is well posed. Calculations in Example 1.6 of Chapter 5 show that (a) and (b) hold. To check (c) with p = 4, we begin by noting that when zr; = f., zr is the sum of f. independent random variables �r with mean 1 + f3n/n ;:::: 0 and variance IT�. Let 6 > 0. Under (A3') we have in addition l�f I � n6 for large n. Using (a + b )4 � 24a4 + 24b4 we have �� / n (x) = nE(n-4 {Z1 - nx} 4 I Z� = nx) � 16n -3(xf3n)4 + 16n -3E( {Zf - nx (1 + f3n/n)}4 I Z� = nx) To bound the second term we note that

{z:;. , m ;:::: 0 } in which the probability of k children Pk has mean 1 + (f3n/n) and variance IT� . Suppose that (A1) f3n -+ (3 E ( -oo, oo), (A2) ITn -+ IT E (O, oo), (A3) for any 6 > 0, Lk>o n k2 pA: -+ 0 n Following the motivation in Example 1.6 of Chapter 5, we let Xt1 / = Z�t] fn.

processes

307

the result, see Sections 3.3-3.4 of Jagers (1975). For an approach based on semigroups and generators see Section 9.1 of Ethier and Kurtz (1986).

then (i) , (ii), and (iii) of (7.1) hold.

lat (x) - afj (x) l �

Examples

References. For a classical generating function approach and the history of

(c) limh!O SUP j xj� R /; (x) = 0

Proof Taking the conditions in reverse order, we note

8. 8

E({Zf - nx(1 + f3nfn)} 4 I Z� = nx ) = nxE(�i - (1 + f3n/n))4 + 6 n2x E(�i - (1 + f3n/n))2

()

E(ei - (1 + f3n fn))4, we note that if 1 + f3nfn � n6 no 4x3 P( l�f - (1 + f3n/n) i > x ) dx E(�i - (1 + f3n/n))4 = no � 2(n6) 2 2x P( l�i - (1 + f3n/n) l > x) dx � 262 1T�n2

To bound

1

1

::; R then �� /n (x) � 3262 1T�R + 961T;R2 n - 1 + 16( xf3n/n)4

Combining the last three estimates we see that if lxl

Since 6 > 0 is arbitrary we have established (c) and the desired conclusion follows from (8.2) and (7.1). To replace (A3') by (A3) now, we observe that the convergence for the special case . and use of the continuous mapping theorem as in Example 5.4 imply

n-1 1 2 n 2.:::: z:;. => 1 X. ds m=O

0

Section

308 Chapter 8 Weak Con vergence

(A3) implies that the probability of a family of size larger than n6 is o(n- 2). Thus if w e trunca�e the original sequence b y not allowing more than bn n children where bn -> 0 slowly, then {A1), {A2) and {A3') will hold and the probability of a difference between the two systems will converge to 0.

0

Example 8.3. Wright Fisher Diffusion. Recall the setup of Example 1.7 in Chapter 5. We have an urn with n letters in it which may be A or a. To build up the urn at time m + 1 we sample with replacement from the urn at time n but with probability a/n we ignore the draw and place an a in, and with probability f3/n we ignore the draw and place an A in. Let z::, be the number n of A's in the urn at time n, and let xfl = Z{:.t /n. Our goal is to show

n

{8.4) Theorem. As n -> oo, Xtl / converges weakly to the Wright-Fisher diffusion Xt , i.e., the solution of

{1951). Chapter 10 of {1986) treats this problem and generalizations allowing more

References. Again the first results are due to Feller

Ethier and Kurtz than 2 alleles.

Proof The limiting coefficients satisfy (3.3) in Chapter 5 so using (4.1) there we see that the martingale problem is well posed. Calculations in Example 1.7 in Chapter 5 show that (a) and (b) hold. To check condition (c) with p = 4 now, we note that if z� = nx then the distribution of zr is the same as that of Sn = 6 + · · · + �n where 6 , . . . , �n E {0, 1} are i.i.d. with P(�i = 1) =

x ( l - afn) + {1 - x)f3/n.

E(Sn - np)4 = E (t �i - p) ] =1 = nE(�i - p)4 + 6 (;) E(�i - p) 2 � Cn2

E(€i - p) m � 1 for all m ;::: 0. 24a4 + 24b4 it follows that

From this and the inequality

Proof The new {a) and (b) imply the ones in {8.2). To check (c) of (8.2) with

p = 2 we note that

� af; (x) = j l y - xi 2I
0

a

Example 8.4. Epidemics. Consider a population with a fixed size of n individuals and think of a disease like measles so that recovered individuals are �mmune to :u�ther inf�ction. Let Sf and If be the number of susceptible and _ �nf�c�ed mdlVlduals at time n, so that n - (Sf + If) is the number of recovered mdrvrduals. We suppose that (Sf, If) makes transitions at the following rates:

{s, i) -> (s + 1, i) (s, i) -> (s - 1, i + 1) ( i) ( i - 1) 81

->

81

a(n - s - i) f3sifn Ji rate a and is

In wor�s, each immune individual dies at replaced by a new susc�ptrble. Each susceptible individual becomes infected at rate f3 times the fractiOn of the population that is infected, ifn. Finally, each infected individual recovers at rate 1 and enters the immune class. n Let Xtl / = (Sf!n, Ifln). To check (a) in (8.5) we note that

1 {an(1 a111/ n (x1 , x 2 ) = ? x1 - x 2 ) + f3nx1x 2 } n1 f3n a 11 /2n (x1, x 2 ) = ? n- x1x 2 1 {f3nx1x + a 221 / n (x1 , x 2 ) = 2 2 1nx1x 2 } n · and all three terms converge to 0 uniformly in r = {(xr, x 2 ) : xr, x 2 ;::: 0, x1 + x 2 � 1}. To check (b) in (8.5) we observe b 11 / n (xr, x 2 ) = an(1 - x1.- x 2 ) n1 f3nx1x 2 1 n = a(1 - xr - x 2) - f3x1x 2 = br(x) b21/ n (xr, x 2 ) = f3n x1x 2 -n1 - 1nx 2 n1 = f3xrx 2 - 'YX 2 = b2 (x) ·

(a + b)4

�

0

Our final example has a deterministic limit. In such situations the following lemma is useful.

309

(8.5) Lemma. If for all R < oo we have (a) limh!o sup l xi � R lafi (x) l = 0 (b) limh!o sup l xi � R lbfex ) - b;(x ) l = 0 then (i), (ii) , and (iii) of (7.1) hold with a;j { x) := 0.

· - -

uniformly for x E [0, 1) and the proof is complete.

Examples

·

4

since

·

8.8

·

·

-

·

­

310 Ch apter 8 Weak Con vergen e c

The limiting coefficients are Lipschitz continuous on f. Their values outside f are irrelevant so we extend them to be Lipschitz continuous. It follows that the martingale problem is well posed and we have

n (8.6) Theorem. As n --1- oo, X:l / converges weakly to X: , the solution of the

Solutions to Exercises

ordinary differential equation

dX: = b(Xt) dt = 0 then our epidemic model reduces to one considered in Sections 11.1-11.3 of Ethier and Kurtz (1986). In addition to (8.6) they prove n a ce�tral limit theorem for JTi(Xi 1 - X:). Remark. If we set

a

C hapter 1

1.1. Let A = {A = {w : (w(t 1 ) , w(t 2 ), . . . ) E B} : B E n{ 1 • 2 ····1 }. Clearly, any

A E A is in the cr-field generated by the finite dimensional sets.

To complete the proof, we only have to check that A is a cr-field. The first and easier step is to note if A = {w : (w(t l ), w(t 2 ), . . . ) E B} then Ae = {w : (w(t 1 ), w(t 2 ), . . . ) E Be} E A. To check that A is closed under countable unions, let An = {w : (w(t�), w(t�), . . . ) E Bn}, let t 1 , t 2 , . . . be an ordering of {t� : n, m 2:: 1 } and note that we can write An = {w : (w(t l ),w(t 2 ), . . . ) E En } so Un An = {w :

(w(t1 ), w(t2 ), . . .) E UnEn} E A. 1.2. Let An = {w : there is an s E [0, 1] so that I B: - B. I $ Ci t - s i""Y when it - s l $ kfn}. For 1 $ $ n - k + 1 let Yi,n = max B -B �- 1 = 0, 1, . . . k - 1 Bn = { at least one Yi ,n is $ (2k - 1)C/n1'} Again An C Bn but this time if 'Y > 1/2 1/k, i.e. , k(1/2 - 'Y) < - 1, then P(Bn) $ nP( I B(1/n) l $ (2k - 1)C/n-y ) k ::; nP( I B(1 ) 1 ::; (2k - 1)Cn 11 2 --y ) k $ C' n k(l /2 --y) +l 0 1.3. The first step is to observe that the scaling relationship (1.3) implies D. m,n .!!- 2- n /2 D. 1, 0 while the definition of Brownian motion shows E D.. r '0 = t, and E(D.. i 0 - t) 2 = C < oo. Using (*) and the definition of Brownian motion, it follo�s that if k -::/= m then. D..� ,n - t2- n and D..� ,n - t2 - n are independent and have mean 0 so

i { I (i : j) (i + ) I : j + --�-

}

312 Solutions to Exercises

Answers for Chapter

where in the last equality we have used (*) again. The last result and Cheby­ shev's inequality imply

{T n,

1

313

> Bn E I<}, recalling the definition of Integrating both sides over An = conditional expectation, and using our assumption shows The right-hand side is smaller than P(An+l ) and the desired result follows.

The right-hand side is summable so the Borel-Cantelli lemma implies

P

( m�2"

L �;;, , n - t

2:

1/n infinitely often

)

=0

Ex

2.1. Take Y(w) = f(wt - 3 ) ·

2.1

2.2. When f(x) = Xi , becomes Ex(Bf !J=·3 ) = EB(3)BL 3 = B! since under becomes Px, BL 3 is normal with mean X i . When f(x) = Xi Xj with i =f. j

Ex(B: B{ I:F3 ) = EB, (BL 3 B{_ J = B!B�

2.1

since under Px , BL 3 and B{_3 are independent normals with means X i and Xj .

1 (To>t) and note that To o 81 = R- 1 so Y o 81 = 1 (R> l+t )· Using the_ Markov property gives Px(R > 1 + t i :Ft) = PB1 (To > t) Taking expected value now and recalling Px( B1 = y) = Pt (x, y) gives Px(R > 1 + t) = J P1 (x , y) Py (To > t) dy 2.4. Let Y = 1 (To>1 - t ) and note that Y o 8t = 1 (L � t ) · Using the Markov property gives Po(L :::; t i :Ft) = PB1 (To > 1 - t) Taking expected value now and recalling Po (Bt = y) = Pt (O, y) gives Po( L $ t) = J p1 (0, y)Py (To > 1 - t) dy 2.5. We will prove the result by induction on n. By assumption it holds for n = 1. Suppose now it is true for n. Let Y(w) = 1(T>t,w1eK) · Applying the Markov property at time n gives 2.3. Let Y =

2.6. Since J; h(r, Br ) dr E :F3 , we have

(it h(r, Br ) drl :Fs ) = 13 h(r, Br ) dr t + Ex (i h(r, Br ) dr i :F3 )

To evaluate the second term we let Y(w)

Y o 8s = so the Markov property implies

= J; - s h(s + u,wu) du and note

t

10 -s h(s + u, Bs+u) du

2. 7. Since J; c(Br ) dr E :F3 , we have

3 To evaluate the second term we let Y(w) = f(Bt - s ) exp(J; - c(wu ) du) and 3 note Y o 83 = f(B ) exp(J; - c(B3 + u ) du) so the Markov property implies

t

(2.8) , (2.9s) , t 1 1

and the Markov property imply that with probability one 2.8. there are times > > s2 > that converge to a so that B(sn ) = B(a) and B(tn ) > B(a). In each interval (sn +l l sn ) there is a local maximum.

t2

• • •

(2.7)

implies Po(Z 2.9. Z = limsup J.o B(t)ff (t) E :Fit so each c, which implies that Z is constant almost surely.

t

> c)

E

{0,1} for

Answers for Chapter 1

314 Solutions to Exercises 0, AN = {B(tn) 2: C,;t;; for some n 2: A = nNAN. A trivial inequality and the scaling relation (1.3) imply

2.10. Let C <

oo ,

tn !

Po(AN) 2: Po(B (tN)

2:

N}

and

CVtN) = Po(B(1) 2: C) > 0

Letting N -+- oo and noting AN ! A we have Po (A) 2: P0 (B1 2: C) > 0. Since A E Ffi it follows from (2.7) that Po(A) = 1, that is, limsupt..... o B (t)/Vt 2: C with probability one. Since C is arbitrary the proof is complete.

2.11. Since coordinates are independent it suffices to prove the result in one dimension. Continuity implies that if 8 is small Po

(

sup I B. I < E/2 0� ·�6

)

=

2a > 0

Pz

2: 0 and the bottom one for

IBt I < (0sup 99

€,

T m ;:: n m

x

and liminf n Tn = sup infn Tm

n m ;::

3.5. First if A E Fs then

(0�sup· � 6 I B. I < Ej2, B6 < o) = Po ( sup I B. I < E/2, B6 > 0) = a > O 0� ·9 x

3.4. Define Rn by R1 = T1 , Rn = Rn - 1 V Tn . Repeated use of Exercise 3.3 shows that Rn is a stopping time. As n j oo Rn j supn Tn so the first conclusion , follows from (3.3). Define Sn by S1 = T1 , Sn = Sn - l A Tn. Repeated use of Exercise 3.3 shows that Sn is a stopping time. As n j oo, Sn ! infn Tn so the second conclusion follows from (3.2). The last two conclusions follow from the first two since

n

A n {S <

Po

Using the top result for X E [-E/2, E/2)

3.3. Since constant times are stopping times the last three statements follow from the first three. {S A T � t} = {S � t} U {T � t} E F1• {S V T � t} = {S � t} n {T � t} E Ft {S + T < t} = Uq,r E Q:q+r
limsup Tn = inf n sup

. so it follows from symmetry that

315

t} = Un (A n {S � t - 1/n} ) E Ft

On the other hand if A n {S < t} E Ft and the filtration is right continuous then A n {S � t} = n n (A n {S < t + 1/n} ) E nn Ft+l/n = F1 •

� 0 shows that if

3.7. (i) Let r = s A t.

)

B6 E [-E/2, E/2] 2: a > 0

{S < t} n {S < s} = {S < r} E Fr {S � t} n { S � s} = { S � r} E Fr

Iterating and using Exercise 2.5 gives that if x E [-E/2, E/2)

c

c

F. F.

This shows {S < t} and {S � t} are in Fs. Taking complements and intersec­ tions we get {S 2: t}, {S > t}, and {S = t} are in Fs .

3.1. Suppose A = U�=1 I
3.2. If m2 - n < t � ( m + 1)2 - n then {Sn < t} = {S < m2 - n } E Fm 2

3.6. {R � t} = {S � t} n A E F1 since A E Fs

-n

C

Ft .

(ii) {S < T} n {S < t} = Uq q} E Ft by (i) , so {S < T} E Fs. {S < T} n {T < t} = Uq T} in Fs n FT. Taking complements and intersections we get {S 2: T}, {S � T}, and {S = T} are in Fs n FT .

3.8. If A E 1l then

...

...

316 Solutions to Exercises by (i ) in Exercise 3.7 . This shows {B(Sn ) E A } E :Fs so B(Sn) E :Fs . Letting n oo we have Bs = lilTln B(Sn ) E nn:Fs = :Fs . 3.9. (3.5) implies :FsAT,. C :Fs. To argue the other inclusion let A E :Fs . Since A = Un(A n {Tn > S}) suffices to show that A n {Tn > S} E :FsAT To do this we observe A n {Tn > S} n {Tn S < t} = A n {Tn > S} n {S < t} = Uq q} E :Fq .

...

�

...

it

1\

·

>

3.10. Since f05 g(B3 ) ds E :Fs we have s s Ex 1 g(B3 ) ds :Fs = 1 g(B3) ds T + Ex fs g(B3 ) ds :Fs

(

l)

(

Applying the strong Markov property to Y(w)

·'

--

r

The rest of the argument is identical to that for (4.4).

= J:(w) g(w3) ds we have

v,

4.3. As s -+ 0, cp(s) -+ 1, so for some o > 0 we have cp9(s) > 0 for all s 6. Using the equation cp(s)cp(t) = cp(s + t) we can now conclude by induction that for all n, cp(s) > 0 for s � n o , and we can take logarithms without feeling guilty. The rest is easy: 1/1(2- n ) + 1/1(2- n ) = 1/1(2-(n -l)) and induction tells us that 1/1{2-n) = 2- n?/1(1), while 1/l((m - 1)2- n) + 1/1(2- n) = 1/l(m2- n) and induction implies 1/l(m2- n ) = m2- n ?/1(1). Since cp(s) is continuous and does not vanish, 1/1(s) = cp(s) is continuous and the desired result follows. :::;

4.4. Let

. (a) = Ea(exp( ->.Ta )). (4.4) and (4.5) imply

u

{S S

Symmetry of the normal distribution implies Ea Y3 = Ea Y3 , so if we let = inf{s < t : B3 = a} and apply the strong Markov property then on < oo}

Taking expected values now gives the desired result.

4.1. We begin by noting symmetry and (2.5) imply Po(R � 1 + t) = 21 00 Pl (O, y) 1t Py(To = s) ds dy t = 1 21 00 Pl (O, y) 1t Py (To = s) dy ds by Fubini's theorem, so the integrand gives the density P0(R = 1 + t). Since Py (To = t) = Po(Ty = t), (4.1) gives 1 -Y2,.,- 1 ye-Y2/2t dy Po(R = 1 + t) = 2 100 --e 0 -../2i � 1 2 1__t_ _ _ -- 2rrt3/ 2 }r0 oo Ye -Y ( l +t)f 2 t dy -- _ 2rrt3/2 (1 + t) 4.2. Translation invariance implies that if < s then using the last identity we have

l)

v

1

E(x, r)f (Br. - Bo) = Eco,o)f (Br._r ) Applying the strong Markov property to Y(w) = f (w(r3 ) - w0) at time Tr, and

3.11. Let Y3 (w) = 1 if s < t and < w(t - s) < 0 otherwise. Let y3 (w) = { 1 if s < t., 2a - < w(t - s) < 2a 0 otherwise u

Answers for Chapter 317

.(a)

. (b) =

.(a + b) As before the first equation implies

. ( a ) = exp(c>.a ) while the second with >. = and a = gives = and this implies C>. = -1\,J>..

1

v'b

cb v'bc1 4.5.> LetB1 and M1 hence = maxo< 3 < 1 B3 • Exercise 3.10 implies that with probability 1, M1 a :_.-Ta is discontinuous at a = M1. This shows Po( a -+ Ta is discontinuous in [0, n]) -+ 1 as n -+ oo and the result follows from the hint.

318 Solu tions to Exercises

Answers for Chapter 2

> Ml }. Since B'f < M1 the strong Markov property and (4.3) imply that with probability one Bj. =/= B11 so s -+ C3 is discontinuous at s = M1 . This shows P0(s -+ C3 is discontinuous in [0, n]) -+ 1 as n -+ oo and the result follows from the hint.

319

4.6. Let M1 = maxo� 3:9 B} and T = inf{t > 1 : B}

2.2. To check integrability we note that

4.7. If we let /(x) = f(x), then it follows from the strong Markov property

since pog IYII is integrable near 0 and e -l y - :c l 2 f 2 t takes care of the behavior for large y. To show that E:r:Xt -+ oo we observe that for any R < oo

and symmetry of Brownian motion that

E:r:Cf (Bt); T � t) = E:r:[Eo f (Bt - r ) ; T � t] = E:r:[EofCBt - r); r � t] = E:r:[/(Bt); r � t] = E:c(fCBt)) The last equality holds since follows as in (4.8). Chapter

/(y) = 0 when Yd ;::: 0, and the desired formula

2

1.1. If A E converges to

:Fa and a < b then Hn (s, w) = 1[a + l / n ,b +l / n) 1A 1 (a ,bj(s)1A(w) as n -+ oo. 2.1. The martingale property w.r.t. a filtration 9t is that

is optional and

E(X/ ; A) = E(X:; A) for all A E 93. (3.5) in Chapter 1 implies that :Fs ,.. 3 C :F3 so if Xf is a martingale with respect to :F3 it is a martingale with respect to :FSA 3 . To argue the other direction let A E :F3. We claim that A n {S > s} E :Fs,.. 3 . To prove this we have to check that for all r

(A n { S > s}) n { S 1\ s � r} E :Fr but this is 0 for r < s and true for r ;::: s. If Xf is a martingale with respect to :Fs,..3 it follows that if A E :F3 then

E(X/;A n {S > s}) = E(X:;A n {S > s}) On the other hand

E:r:Xt 2:: (2-rrt) -d/2 1

IYI : 9

log IYI dy + (log R)P:r:( I Bt I

> R)

The integral is convergent and as we showed in Example 2.1, P:r:( I Bt l > R) -+ 1 as t -+ oo so the desired result follows from the fact that R is arbitrary.

2.3. Let Tn = inf{t : I Xt l > n}. By (2.3) this sequence will reduce Xt. Note that x'[n � n. Jensen's inequality for conditional expectation (see e.g., {l.l.d) in Chapter 4 of Durrett (1995)) implies

E(cp(X[n ) I :FTn A 3) 2:: cp(E(X[n i :FTn A 3)) = cp(X'[n ) 2.4. Let Tn � n be a sequence which reduces X, and Sn = {Tn - R) + . If s < t then defi n itions involved {note S'n = Tn - R on {Sn > 0} = {Tn > R}), the optional stopping theorem, and the fact 1 (Tn>R) E :FR C :FR+3 imply E(Y1ASn 1 (Sn >O}IQ3) = E(X(R+t }ATn 1(Tn >R}I :FR+ 3) = 1(Tn >R} E{X(R+t)ATn I :FR+3) = x(R+3)ATn 1 (Tn >R) = Y.,.. sn 1 csn >O) 2.5. Let Tn be a sequence that reduces X. The martingale property shows that if A E :Fo then E(Xt ATn 1(Tn > 0) ; A) = E{Xo1(Tn >O) ; A) Letting n -+ oo and using Fatou's lemma and the dominated convergence theorem gives ; A) 1 E(Xt ; A) � liminf n-+oo E(Xt ATn (Tn>O) � nlim -+oo E{Xo1 (Tn >O} ; A) = E(Xo; A) Taking A = n we see that EXt � EXo < oo. Since this holds for all A E :Fo we get E(Xt I :Fo) � Xo. To replace 0 by s apply the last conclusion to Y1 = x. +t ' which by Exercise 2.4 is a local martingale.

Adding the last two equations gives the desired conclusion.

3.1. Since t -+ Vt is increasing, we only have to rule out jump discontinuities. Suppose there is a t and an

E

> 0 so that s < t < u then V(ut - V(s) > 3�: for

Answers for Chapter 2

320 Solu tions to Exercises

321

= (Xs+t - Xs) 2 - {(X}s+t - (X}s}. Using the last equality we get E(Zt lFs+3) = E(X�+t - X�+3 - {(X} s+t - (X}s} IFs+3 ) + (Xs+3 - X3)2 = E(X�+t - (X}s+t lFs+ 3) - X�+3 + (X}s + (Xs+3 - X3 ) 2 = -(X}s +3 + (X}s + (Xs+3 - X3) 2 = Z3

all n. It is easy to see that V(u) - V(s) is the variation of X over [s, u]. Let s0 < t < u0 • We will now derive a contradiction that the variation over [s0 , uo] is infinite. Pick 6 > 0 so that if l r - t l < 6 then I Xt -Xrl < c. Assuming sn and Un have been defined, pick a partition of [sn , un] not containing t with mesh < 6 and variation > 2c. Let Sn + l be the largest point in the partition < t and Un + l be the smallest point in the partition > t. Our construction shows that the variation of X over [sn , un] - [sn + l l t n + l ] is always > c, so the total variation over [so, u 0] is infinite, a contradiction which implies t -l- Vt is continuous.

Let Zt

3.2. (X + Y)tZt - (X, Z}t - (Y, Z}t is a local martingale.

3.8. By stopping at Tn = inf{t : I Xt l > n} and using Exercise 3.5 it suffices to prove the result when X is a �ounded martingal_e. Using Exerc�se 3. 7, we c_an _ �ahty suppose without loss of generahty that S = 0. Usmg the L2 max1mal meq on the martingale Xt AT • and the optional stopping theorem on the martmgale X{ - (X}t it follows that

3.3. -XoZt is a local martingale, so if Yt desired result follows from Exercise 3.2.

= -Xo

then

(Y, Z}t

=

0 and the

3.4. (aXt)(bYt) - ab(X, Y}t = ab (XtYt - (X, Y}t) is a local martingale.

where the last equality follows from the optional stopping theorem and Exercise 3�6. This shows that Zt is a martingale, so (Y}t = (X} s+t - (X} s .

E (sup t�n XfAT ) � 4E(Xf An ) = E((X }t AT = 0

3.5. If T is such that XT is a bounded martingale it follows from (3.7) and the definition of (X} that (XT } = (X}T . For a general stopping time T, let

Tn = inf{t : I Xt l > n}, note that the last result implies (XTATn ) = (X) T ATn and then let n -l- oo The result for the covariance process follows immediately

Letting

3.6. Since X{ - (X}t is a martingale and I Xt I � M for all t

4.1. It is simply a matter of patiently checking all the cases.

.

from the definition and the result for the variance process.

E(X}t = EXt2 - EXJ � M2 so letting t -l- oo we have E(X}co � M2 • This shows that Xt2 - (X}t is dominated by an integrable random variable and hence is uniformly integrable.

3. 7. The optional stopping theorem implies and Xs E Fs C Fs+3 so

n -l- oo now gives the desired conclusion. 3.9. This is an immediate consequence of (3.8).

s < t � a. Yt = Yo E Fo so E(Yt lF3 ) = E(YolF3) = Yo = Ys 2. s < a � t � b E(Yt 1 .1"3 ) = E(E(Yt I Fa)F3) = E(Ya 1 .1"3 ) = E(YolF3 ) = Yo = Y3 3. s < b, t > b. Yt = Yb so this reduces to case 2 if s < a or to our assumption if a � s < b. 4. b � s < t. E(Yt lFs) = E(YblF3) = Yb = Ys .

1.

w then t -l- (X}t defines a measure. The triangle inequality for L2 of that measure implies

4.2. If we fix i.e., Y3 is a martingale. To prepare for the proof of the second result we note that imitating the proof of (2.4)

E((Xs+t - Xs) 2 1Fs +3) = E((Xs+t - Xs+3 ) 2 1Fs + 3) + (Xs+3 - X3)2 = E(Xj+t - xj+3 1Fs+3) + (Xs+3 - x3 ) 2

Now take expected value.

Answers for Chapter 2

322 Solu tions to Exercises 4.3. X E M2 implies E supt X? < oo and hence EX� < oo. To check the other

condition let Tn be a sequence of stopping times l 00 so that xTn is bounded and {X) Tn ::::; n. The optional stopping theorem implies

Clearly, (H 1 X)1 = (K 1 Y)t for all t. Since H'f = K'f = 0 for S ::::; s (5.4) implies (H2 · X). = {K2 · X } . = 0 for S ::::; s ::::; T •

•

5

T,

and it follows from Exercise 3.8 that (H 2 X)t and (K2 X)t are constant on [S, T]. Combining this with the first result and using ( 4.3.b ) gives the desired conclusion. •

•

Letting n --+ oo it follows that E(X! - (X)oo = EX6 . To prove the converse rearrange the displayed equation to conclude

323

6.2. Stop X at Tn = inf{t : ! Xt l > n or J; H; d(X}. } to get a bounded martingale and an integrand in II2 (X) . Exercise 4.5 implies

so X is L2 bounded.

4.4. The triangle inequality implies ll x ll ::::; ll xn ll + ll x - xn ll so

ll x ll ::::; liminf ll xn ll For the other inequality note that ll x nll ::::; ll x ll + ll xn - x ll so ll x ll ;:::: limsup ll x nll 4.5. Let Hn E bii 1 with II Hn - H llx --+ 0. Since II (Hn · X) - (H · X) ll 2 --+ 0, using Ex(!rcise 4.4, the isometry property for bii 1 , and Exercise 4.4 again II H · X l b = lim II Hn · X ll 2 = lim II Hn llx = II H II x 5.1. If L, L' E M2 have the desired properties then taking N = L - L' we have {L, L - L'} = {L', L - L') so {L - L') = 0. Using Exercise 3.8 now it follows that L - L' is constant and hence must be :: 0.

5.2. I! we let H. = 1 and K. = lc. 5T) then X = H . X and yT = [{ . Y. (It is

for this reason we have assumed Xo = Yo = 0.) Using (5.4) now it follows that

5.3. If we let H. = lc•5T) and K. = lc• >T) then xT = H . X and y - yT = K · Y. Using (5.4) now it follows that (XT, Y - YT)t :: 0 so the desired result follows from (3.11). 6.1. Write H. = H} + H; and I<. = K} + K; where

Using the L2 maximal inequality it follows that

)

(

Tn E sup (H · X); ::::; 4E J{ H? d(X}t t 5Tn o

Letting n --+ oo now we can conclude

With this in hand we can start with

E(H · X)}n = E and let n --+ oo to get the desired result.

1Tn Hl d(X}t 0

6.3. By stopping we can reduce to the case in which X is a bounded continuous martingale and (X}t ::::; N for all t, which implies H E II2 (X). If we replace S and T by Sn and Tn which stop at the next dyadic rational and let H'; = C for Sn < s ::::; Tn then H'; E II 1 and it follows easily from the definition of the integral in Step 2 in Section 2.4 that

To complete the proof now we observe that if I C (w ) I

::::; K then

II Hn - H llx ::::; K2 { E((X}Tn - (X } T ) + E((X} sn - (X } s ) } --+ 0

324 Solutions to Exercises

) · X)l1 2 -+ ·X) (H H'; dX -+ ( Hn 3 li C(XT.. - XsJ -+ C(XT - Xs) 6.4. x; - x5 = LX2 (t i+ 1 ) -X2 (ti) = L {X(t i+d - X(t f )} 2 + L 2X(tf){X(t i+ l ) -X(tf )}

as n -+ oo by the bounded convergence theorem. Using Exercise 4.4 and {4.3.b 0 and hence J J H3 dX3 . it follows that Clearly, and the desired result follows.

i

i

{3.8) implies that the first term converges in probability to (X)t. The previous exercise shows the second term converges in probability to J� 2X3 d 3 .

X

6.5. The difference between evaluating at the right and the left end points is

2 L{X(ti+l ) -X(tf )}2 -+ 2{X)t 6.6. Let Ck, n = B((k + 1/2)2 - n t) - B(k2 - n t) and Dk, n = B((k + 1)2 - nt) ­ B((k + 1/2)2- nt). In view of {6 .7) it suffices to show that as n -+ oo Sn = .Lk Ck,n(Ck,n + Dk,n ) -+ t/2 in probability. E C'f, n = 2 - n t/2 and ECk,nDk, n = 0 so ESn = t/2. To complete the proof now we note that the terms in the sum are independent so 2 2 2 t/2 + D , , n = !/2) , . E (� Gi,n " G, ) E(Sn 2 n - 2- n t/2)2 + ck,: n Dk,2 n -< 2n ct2- 2n -+ 0 = E �(Ck, k i

L.J

and then use Chebyshev's inequality.

6. 7.

as n

that if t f is a sequence of partitions of [0, t] with mesh -+ 0 -+(6.oo7) implies we have

Answers for Chapter 3

-+

as n oo by the convergence of the Riemann approximating sums for the integral. Convergence of the means and variances for a sequence of normal random variables is sufficent for them to converge in distribution {consider the characteristic functions) and the desired result follows.

A ;J At )) then as in {7.4) we get t (A ! t ) - f(Ao) = 1 G! dA3 As o -+ 0, G� -+ f'(A3) uniformly on [0, t] so the desired conclusion from Exercise 7 .1. 8.1. Clearly Mt + Mf' is a continuous local martingale and A + A� is continuous adapted, locally of bounded variation and has A 0 + A� = 0. 7.2. By stopping it suffices to prove the result when I it :::; M. If we let

G� = f'(c(At

,+ 1

t

Chapter

3

1.1. The optional stopping theorem implies that

x = E:cBT = aP:c(BT = a) + b(1 - P:c(BT = a)) Solving gives P:c(BT = a) = (b - x)f(b- a). 1.2. From { 1 6) it follows that for any s and P(supt?:3 Bt 2:: ) = 1. This implies P(sup t?:3 Bt = oo ) = 1 for all s and hence limsup 3 _ 00 B3 = oo a.s. 1.3. If Sr(w) j t(w) which is < oo with positive probability then Bt (w) (w) = 0 and we have a contradiction. 3.1. Using the optional stopping theorem at time T t we have EoBfAt = t). Letting t -+ oo and using the bounded and monotone convergence Eo(T theorems with { 1 .4) we have EaT = EoBf = a2 -b+b a + b2 -b+a a = ab 3.2. Let f (x, t) = x6 - ax 4t + bx 2t2 - ct3 • Differentiating gives Dd = -ax4 + 2bx2t - 3ct2 (1/2)D:c:cf = 15x4 - 6ax2t + bt2 .

n

� hir (ti+l -tf ) -+ fo h; ds '

.

n

1\

1\

·

in probability. The left-hand side has a normal distribution with mean 0 and vanance t

325

-

•

-

Answers for Chapter 4

326 Solutions to Exercises Setting a = 15, 2b = 6 a , and b = 3c, that is a = 15, b = 45, c = 15, we have Dd + ( 1 /2)D l = 0, so f (B , t) is a local martingale. Using the optional stopping theorem at time Ta A t we have Eo{ BL\t - 1 5B;a At (Ta A t) + 45 B;a At (Ta A t) 2 } = 15Eo(Ta A t)3 Letting t --+ oo and using the bounded and monotone convergence theorems ,

:r::r:

with the results in (3.3) we have

As n --+ oo the first term on the right --+ 0, so

and it follows that for large n, infx e lW (y , 1 / n ) v(x) :::; 1 - f .

4.3. Let y E aG and consider U = V(y, \lg(y), 1 ) . Calculus gives us

15Eor: = a6 - 15 a4Era + 45a2Er; = a6 ( 1 - 15 + 75)

g(z) - g(y) =

so Eo r; = 61/15.

3.3. Using the optional stopping theorem at time T_a A n we have

= Eo exp( -(2J.L/u2 )ZT- a An) Letting n --+ oo and noting that on {T- a = oo } we have Zt/t = u Bt/t + J.L --+ J.L almost surely and hence Zt --+ oo a.s., the desired result follows. 4.1. Let 0 : [0, oo ) --+ Rd be piecewise constant and have 1 03 1 = 1 for all s. Let yt = I:; J; O! dX! . The formula for the covariance of stochastic integrals shows yt is a local martingale with (Y)t = t 1 so (4.1) implies that yt is a Brownian motion. Letting 0 = t 0 < t 1 < . . . < t n and taking 03 = Vj for s E (ti _ 1 , tj ] for Xt n - Xt n- are independent multivariate 1 :::; j :::; n_ shows that x,l - Xt normals with mean 0 and covariance matrices (t 1 - t0)I1 , (tn - tn - 1 )!, and 1

oI . . . I

1

• • •

it follows that Xt is a d-dimensional Brownian motion.

4.2. X3 is a local martingale with (X) 3 = J; h; dr. By modifying h3 after

time t we can suppose without loss of generality that (X) oo = oo Using (4.4) now we see that if 1(u) = inf{ s : (X) 3 > u} then X..,. (u ) is a Brownian motion. Since (X) 3 and hence the time change ! ( u) are deterministic, the desired result follows. .

4.1. Let r = inf{t > 0 : Bt f/. G}1 let y E aG1 and let Ty = inf{t > 0 : Bt = y}.

(2.9) in Chapter 1 implies regular.

Py (Ty = 0)

= 1. Since r

:::; Ty

it follows that

y is

4.2. Py(r = 0) < 1 and Blumenthal's 0-1 law imply Py( r = 0) = 0. Since we are in d 2:: 21 it follows that Py (Br = y) = 0 and hence f = 1 - Ey/(Br ) > 0. Let Un = inf{t > 0 : Bt f/. D(y 1 1/n) } . Un ! 0 as n j oo so Py (un < r) --+ 1. 1 - f = Ey/ (Br)

11 \lg(y + O(z - y))

= Ey { f (Br ) ; T :::; Un) + Ey (v(Bun ) ; T > Un)

·

O( z - y) dO

Continuity of \lg implies that if l z - Y l < r and z E U then \lg(y + O(z ­ y)) · O(z - y) 2:: 0 so g(z) 2:: 0. This shows that for small r the truncated cone U n D(y, r) c ac so the regularity of y follows from (4.5c).

4.4. Let r = inf{t > 0 : Bt E V(y, v, a) } . By translation and rotation we can suppose y = 0, v = (1, 0, . . . , 0) , and the d-1 dimensional hyperplane is Zd = 0. Let To = inf{t > 0 : .Bf = 0}. (2.9) in Chapter 1 implies that P0(T0 = 0) = 1. I_! a > 0 then for some k < oo the hyperplane can be covered by k rotations of V so Po(r = 0) 2:: 1/k and it follows from the 0-1 law that P0(r = 0) = 1. 7.1. (a) Ignoring Cd and differentiating gives

; - O; ) Y D h 8 _- - �2 . ( x _2(x 2 0 1 + y2 ) (d+2 )/2 l + 2)(x; - 0) 2 y D h 8 - - d · ( l x _ 0 1 2 +y y2 ) (d+2 )/ 2 + ( ld(d x 0 1 2 + y2 ) (d+4)/2 2 Dyh 8 = ( x - 0 1 21+ y2 ) d/2 - d . ( x - 0 1 2 +y y2 ) (d+2 )/2 l l 2y y + 3 DYY h8 - -d · ( x 0 1 2 + y2 ) (d+2 )/2 + ( l x d(0 1 2d ++ y2)y 2 ) ( d+4)/2 l :r:;

:r: ; :r: ;

_

_

_

Chapter 4

327

Adding up we see that

d- 1

- 1) +" (-d) · 3}y "\;"" D {;;:_ h 8 + Dyy h 8 _- {(-d)(d ( l x - Oj 2 + y2 ) (d+2 )/2 :r: ; :r: ;

+ 2)y( l x - 0 1 2 + y2 ) = 0 + d(d ( l x 0 1 2 + y2 ) (d+4)/2 _

The fact that .6. u( x)

= 0 follows from ( 1. 7)'as in the proof of (7 .1).

Answers for Chapter 5

328 Solutions to Exercises (b) Clearly J dO h 9(x, y) is independent of x. To show that it is independent of y, let x = 0 and change variables 0; = yep ; for 1 ::; i ::; d - 1 to get

(c) Changing variables 0; = x ; + r; y and using dominated convergence

{

JD(:c ,e)c (d) Since P:c(r

< oo) = 1 for all x E H, this follows from (4.3).

=

1 2

-u" - {Ju

::; 1

u(-a)

=0

and it follows from (6.3) that

= u(a) = 1

Guessing u (x) = B cosh ( bx) with b > 0 we find

1

2u" - {Ju =

if b = -12!3. Then we take B = Chapter

1/ cosh( a.J2/3) to satisfy the boundary condition.

5

3.1. We first prove that h is Lipschitz continuous with constant C2 = 2C1 + Let

f (x) = (2R - l x i)/R and g(x)

=

l f(x) - f (y) l = I IYI � l x l l ::; l x - Yl

�

= { l x l ::; R}, and Rxf l x l is the projection

Since h is Lipschitz continuous on D1 of x onto D1 we have

lu(x) - g(y)i = ih(Rxfl x l ) - h(Ryfiyl) l Rx Ry ::; cl � - IYT ::; C1l x - Yl

I

l

lh(x) - h(y)l ::; ll f lloo , 2 I Y (x) - g(y) l + l f(x) - f (y) lll u lloo , 2 ::; 1 · C1 l x - Yl + R1 llhll oo , l ::; (2Cl + lh (O) I /R) l x - Y l = C2 l x - Yl since for x E D1 , lh(x) l ::; lh(O) I + C1 R.

lh(x) - h(y) l ::; lh(x) - h(z) l + lh (z) - h(y) l ::; C1l x - z l + C2 lz - Yl ::; C2 l x - Yl since C1 ::; C2 and l x - zl + lz - Yl = l x - Yl · This shows that h is Lipschitz continuous with constant C2 on lzl ::; 2R. Repeating the last argument taking l xl ::; 2R and y > 2R completes the proof. 3.2. (a) Differentiating we have

) (2 B b2 - {JB cosh ( bx ) = 0

·R- 1 I h(O) I on D2 = {R ::; l x l ::; 2R}. h(Rxf l x l). If x, y E D2 then

: x E D;}, and

To extend the last result to Lipschitz continuity on Rd we begin by ob­ serving that if x E D1 , y E D2 , and z is the point of 8D1 on the line segment between x and y then

dO h6(x, y) = J

-{3 < 0 so w(x) = E:c e - f3r f3 e r E:c - is the unique solution of

9.1. c(x)

v(x) =

{D(O,efy)c dr hr(O, 1) -+ 0

Combining the last two results, introducing ll k lloo , i = sup { l k(x) l using the triangle inequality:

329

��:(x) = -Cx log x ��:'(x) = -C log x - C > 0 if O < x < e - 1 ��:"(x) = - Cfx < 0 if x > 0 so is strictly increasing and concave on [0, e - 2 ]. Since ��:(e - 2 ) = 2Ce - 2 and ��:'(e- 2 ) = C, is strictly increasing and concave on [O, oo ). When f ::; e - 2 - 1 dx = - c- 1 loglog x = oo I' C x Iog x (b) If g(t) = exp( - 1/tP ) with p > 0 then g' (t) = t �l exp(- lftP) = pg(t){log(1/g(� ))} (P+l )/p P 5.1. Under Q the coordinate maps Xt (w) satisfy MP ([J + b, ) Let 11:

11:

1,

0

0

u .

it [J(X. ) + b(X. ) ds t = Xt - i b (X. ) ds

Xt = Xt -

Answers for Chapter

330 Solutions to Exercises Since we are interested in adding drift

-b we let c = a- 1 b as before, but let

-1t c(X.) dX. t = -Yt + 1 c(X.) b(X.) ds t 1 = -yt + 1 ba- b(X.) ds

Yt =

exp

·

(- 1Y 2b(z)fa(z) dz) ::; e-2
so
3;2. The co�1it:on is [01 -2b(y)fa(y) dy = 0. Let I = J01 - 2b(y)fa(y) dy.

-oo and if I < 0 then so(oo) < 00 .

+

Since yt and Yt differ by a process of bounded variation we have

e

If I = 0 then

&t = exp (Ayt - 21 (YA }t) = exp ( -yt + �(Y) ) = 1/ o:t t

·

1ju2 x so 1 1 m(x)(so(x) - so(O)) dx = 11 u-2 dx

-n } , Rn +I = +1)

= has a geometric distribution with E(N where the number of terms, N n n 2 = and E(Sm Rm) Sm - Rm are i.i.d. with E(Sm - Rm) = 2 2 , and the n that noticing (5/3)2 -4 . Using the formula for the variance of a random sum or as n --+ oo one concludes easily that limsup ;::: C > 0 a.s. P(N ;::: 2n ) --+

e- I

6 3.1. (i) If b(z) :5 0 then

oo

M(O) = -oo so J = oo and it follows that 0 is absorbing. To deal with oo note so(oo) = M(oo) = oo so I = J = oo. (b) When (3 < 0, so'(x) = e2I .B i x fu2 , so so(x) =

-2

<

that

6.1. It suffices to show that Po a.s.

+1,

so'(y) < c < oo, and it follows that

5.1. (a) If f3 = 0 then so(x) = x and m(x) =

The change of measure in this case is given by

Bt

331

so
·

To prove the last result, let Ro = 0, Sn = inf{t > Rn : I Btl = 2 inf{t > Sn : = 0} for n 2::: 0, and N = sup{n : Rn < Tr} . Now

6

2�;l (e2I.Bixfu2 _ 1)

m(x) = e-2I.Bixfu2 ju2x

To evaluate I for the boundary at 0, we note

{I m (x)(so(x) - so(O)) dx = {1 1 - e-2I.Bixfu2 dx < oo Jo Jo 2 lf3lx sine� the integrand c�nverges to 1/u2 as x --+ 0. Again M(O) = -oo so J = 00 and It follows that 0 Is absorbing. To deal with oo we note that so(oo) = oo so I = oo. To evaluate J we begin by noting that

Chapter

exp

(- lay 2b(z)/a(z) dz) ;::: 1

Since most of the integral comes from x ::; y ::; x + Vx it follows easily that

332

Solutions to Exercises

Answers for Chapter 8 333

Since
oo is a natural boundary. 5.2. The computations in Example 5.3 show that (a) I < oo when 1 < 1 and (b ) for 1 < -1 we have M(O) = -oo and hence J = oo. Consulting the table of definitions we see that 0 is an absorbing boundary for 1 < -1. 5.3. (a) H� ::; To < oo. (b ) (2.9) in Chapter 3 implies E1Ho =

11 2yy-26 dy < oo

does not depend on x. Letting S.: = (T.:/ 2 A T2x ) o Markov property it follows that the distribution of

BT,

and using the strong

under P1 does not depend on x. Pick 8 > 0 so that P1 (Ix 2:: 8) > 0. The I2 -n are inde­ pendent so the second Borel-Cantelli lemma implies that P1 {I2 -n 2:: 8 i.o.) = 1 but this is inconsistent with P1(H1 < oo) > 0. 5.4.

1/2 z1 - 26 - 1 dz < oo if. 8 > 1 J= <1 = 00 1f u 28 - 1 1

hence J =

100

{

c

Combining the results for I and J and consulting the table we have the indicated result. 5.5. When a = 0

(

)

dz
5.1. Letting

b -+ oo in (4.7) of Chapter 6 we have

E.: T(1 ,ooj' = 2(
100 m(z) dz + 2 1.: (
The result now follows from m(z) =

under P.:

0

7

Chapter

since 8 < 1 implies 1 - 28 > -1. (c) If 8 > 1 then H6 > H1 so it suffices to prove the result when 8 = 1. To do this, we note that if Bo = x then x - 1 B3.:2 is a Brownian motion starting at 1, so the distribution of

1T,/21\T2, B-;2 ds

Here and in what follows C is a constant which will change from line to line. If f3 2:: 1/2 then 0, M(z) - M(O) ....., Cz2� as z -+ 0 so J < oo if f3 > 0. Comparing the possibilities for I and J gives the desired result.

= Cy- 2�

1/
5.2. We apply (5.3) to v(x) = x 1 +6 - 1 and note that

x6- 1 + b(x)(1 + 8)x6 < -1 Lu = (1 +2 8)8 · f f since b(x) ::; -t:jx6 , (1 + 8)/2 ::; 1, and x6- 1 ::; 1. Chapter

8

3.1. Suppose rn -f+ r. Then there is an f > 0 and a subsequence rn k so that l rn·k - r l > f for all k. However it is impossible for a subsequence of rn k ' so we have a contradiction which proves rn -+ r. •

6.1. If d(gn, h ) -+ 0 then h can only take the values 0 and 1. In Example 6.2 we showed that h cannot be = 0 so it must take the value 1 at some point a. Since h is right continuous at a there must be some point b f:. 1/2 where h (b) = 1 but in this case it is impossible for d(gn, h) -+ 0.

References

M. Aizenman and B . Simon (1982) Brownian motion and a Harnack inequality for Schrodinger operators. Comm. Pure Appl. Math. 35, 209-273 P. Billingsley (1968) Convergence of Probability Measures. John Wiley and Sons, New York R.M. Blumenthal and R.K. Getoor (1968) Markov Processes and their Potential Theory. Academic Press, New York D. Burkholder (1973) Distribution function inequalities for martingales. A nn. Probab. 1, 19-42 D. Burkholder, R. Gundy, and M. Silverstein (1971) A maximal function char­ acterization of the class HP . Trans A.M.S. 157, 137-153 K.L. Chung and R.J. Williams (1990) Introduction to Stochastic Integration. Second edition. Birkhauser, Boston K.L. Chung and Z. Zhao (1995) From Brownian Motion to Schrodinger's Equa­ tion. Springer, New York Z. Ciesielski and S.J. Taylor (1962) First passage times and sojurn times for Brownian motion in space and exact Hausdorff measure. Trans. A.M.S. 103, 434-450 E . Cinlar, J . Jacod, P. Protter, and M. Sharpe (1980) Semimartingales and Markov processes. Z. fur. Wahr. 54, 161-220 .

B. Davis (1983) On Brownian slow points. Z. fur Wahr. 64, 359-367 C. Dellacherie and P.A. Meyer (1978) Probabilities and Potentials. North Hol­ land, Amsterdam C. Dellacherie and P.A. Meyer (1982) Probabilities and Potentials B. Theory of Martingales. North Holland, Amsterdam C. Doleans-Dade (1970) Quelques applications de la formule de changement de variables pour les semimartingales. Z. fur Wahr. 16, 181-194 R. Durrett (1995) Probability: Theory and Examples. Second edition. Duxbury Press , Belmont, CA

336 References

References 337

A. Dvoretsky and P. Erdos (1951) Some problems on random walk in space. Pages 353-367 in Proc. 2nd Berkeley Symp., U. of California Press

M. Kac ( 1951) On some connections between probability theory and differential and integral equations. Pages 189-215 in Proc. 2nd Berkeley Symposium, U. of California Press

E.B. Dynkin (1965) Markov Processes. Springer, New York

I. Karatzas and S. Shreve ( 1991) Brownian Motion and Stochastic Calculus. Second edition. Springer, New York

P. Echeverria (1982) A criterion for invariant measures of Markov processes. Z. fur Wahr. 61, 1-16

S. Karlin and H. Taylor (1981) A Second Course in Stochastic Processes. Aca­ demic Press , New York

S. Ethier and T. Kurtz (1986) Markov Processes: Characterization and Con­ vergence. John Wiley and Sons, New York

T. Kato (1973) Schrodinger operators with singular potentials. Israel J. Math. 13, 135-148

W. Feller ( 1951) Diffusion processes in genetics. Rages 227-246 in Proc. 2nd Berkeley Symp., U. of California Press

R.Z. Khasminskii (1960) Ergodic properties of recurrent diffusion processes and stabilization of the solution to the Cauchy problem for parabolic equations. Theor. Prob. Appl. 5, 179-196

G.B. Folland (1976) An Introduction to Partial Differential Equations. Prince­ ton U. Press, Princeton, NJ D . Freedman (1970) Brownian Motion and Diffusion. Holden-Day, San Fran­ cisco, CA

F. Knight (1981) Essentials of Brownian Motion and Diffusion. American Math. Society, Providence, RI

A. Friedman (1964) Partial Differential Equations of Parabolic Type. Prentice­ Hall, Englewood Cliffs, NJ

H.P. McKean (1969) Stochastic Integrals. Academic Press, New York P.A. Meyer (1976) Un cours sur les integrals stochastiques. Pages 246-400 in Lecture Notes in Math. 511, Springer, New York

A. Friedman (1975) Stochastic Differential Equations and Applications, Volume 1. Academic Press , New York

N. Meyers and J . Serrin (1960) The exterior Dirichlet problem for second order elliptic partial differential equations. J. Math. Mech. 9, 513-538

R.K. Getoor and M. Sharpe (1979) Excursions of Brownian motion and Bessel processes . Z. fur. Wahr. 47, 83-106

B. 0ksendal (1992) Stochastic Differential Equations. Third edition. Springer, New York

D. Gilbarg and J. Serrin (1956) On isolated singularities of second order elliptic differential equations. J. d 'Analyse Math. 4, 309-340 D. Gilbarg and N.S. Trudinger (1977) Elliptic Partial Differential Equations of Second Order. Springer, New York

·

.E. Hewitt and K. Stromberg (1969) Real and Abstract Analysis. Springer, New York N. Ikeda and S. Watanabe (1981) Stochastic Differential Equations and Diffu­ sion Processes. North Holland, Amsterdam J . J acod and A.N. Shiryaev (1987) Limit Theorems for Stochastic Processes. Springer, New York P. Jagers (1975) Branching Processes with Biological Applications. John Wiley and Sons, New York F. John (1982) Partial Differential Equations. Fourth edition. Springer-Verlag, New York ·

S. Port and C. Stone (1978) Brownian Motion and Classical Potential Theory. Academic Press, New York P: Protter (1990) Stochastic Integration and Differential Equations. Springer, New York D. Revuz and M. Yor (1991) Continuous Martingales and Brownian Motion. Springer, New York L.C.G. Rogers and D. Williams {1987) Diffusions, Markov Processes, and Martingales. Volume 2: Ito Calculus. John Wiley and Sons, New York H. Royden (1988) Real Analysis. Third edition. MacMillan, New York B. Simon (1982) Schrodinger semigroups. Bulletin A.M.S. 7, 447-526 D.W. Stroock and S.R.S. Varadhan (1979) Multidimensional Diffusion Pro­ cesses. Springer, New York

338 References H. ·Tanaka ( 1963 } Note on continuous additive functionals of one-dimensional Brownian motion. Z. fur Wahr. 1, 251-257 J. Walsh ( 1978} A diffusion with a discontinuous local time. Asterique 52-53, 37-45 T. Yamada and S. Watanabe ( 1971 } On the uniqueness of solutions to stochastic differential equations. J. Math. Kyoto 11, I, 155-167; II, 553-563

Index

adapted 36 arcsine law 173, 291 Arzela-Ascoli theorem 284 associative law 76 averaging property 148 basic predictable process 52 Bessel process 179, 232 Bichteler-Dellacherie theorem 71 Blumenthal's 0-1 law 14 boundary behavior of diffusions 231 Brownian motion continuity of paths 4 definition 1 Holder continuity 6 Levy's characterization of 111 nondifferentiability 6 quadratic variation 7 writes your name 15 zeros of 23 Burkholder Davi& Gundy theorem 116 C, the space 4, 282 Cauchy process 29 change of measure in SDE 202 change of time in SDE 207 Ciesielski-Taylor theorem 171 cone condition 147 continuous mapping theorem 273 contraction semigroup 245 converging together lemma 273 covariance process 50

D, the space 293 differentiating under an integral 129 diffusion process 177 Dini function 239 Dirichlet problem 143 distributive laws 72 Doleans measure 56 Donsker's theorem 287 Dvoretsky-Erdos theorem 99 effective dimension 239 exit distributions ball 164 half space 166 exponential Brownian motion 178 exponential martingale 108 Feller semigroup 246 Feller's branching diffusion 181, 231 Feller's test 214, 227 Feynman-Kac formula 137 filtration 8 finite dimensional sets 3, 10 Fubini's theorem 86 fundamental solution 251 generator of a semigroup 246 Girsanov's formula 91 Green's functions Brownian motion 104 one dimensional diffusions 222 Gronwall's inequality 188

Index 341

340 Index harmonic function 95 averaging property 148 Dirichlet problem 143 Harris chains 259 heat equation 12Q inhomogeneous 130 hitting times 19, 26

half space 31 optional process 36 in discrete time 34 optional stopping theorem 39 Ornstein-Uhlenbeck process 180

stopping time 18 strong Markov property for Brownian motion 2 1 for diffusions 2 0 1 strong solution to SDE 183

pathwise uniqueness for SDE 184 1r >. theorem 1 1 Poisson's equation 151 potential kernel 102 predictable process 36 basic 52 in discrete time 34 simple 53 predictable u-field 35 progressivley measurable process 36 Prokhorov's theorems 276 punctured disc 146

tail u-field 17 tight sequence of measures 276 transience, see recurrence transition density 257 transition kernels 250

quadratic variation 50

Yamada-Watanabe theorem 193

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infinitesimal drift 178 infinitesimal generator 246 infinitesimal variance 179 integration by parts 76 isometry property 56 Ito's formula 68 multivariate 77 Kalman-Bucy filter 180 Khasminskii's lemma 141 Khasminskii's test 237 Kolmogorov's continuity theorem 5 Kunita-Watanabe inequality 59 law of the iterated logarithm 115 Lebesgue's thorn 146 Levy's theorem 111 local martingale 37, 113 local time 83 continuity of 88 locally equivalent measures 90 Markov property 9 martingale problem 198 well posed 210 Meyer-Tanaka formula 84 modulus of continuity 283 monotone class theorem 12 natural scale 212 nonnegative definite matrices 198 occupation times ball 167

recurrence and transience Brownian motion 17, 98 one dimensional diffusions 220 Harris chains 262 reflecting boundary 229 reflection principle 25 regular point 144 resolvent operator 249 relatively compact 276 right continuous filtration scaling relation 2 Schrodinger equation 156 semigroup property 245 semimartingale 70 shift transformations 9 simple predictable process 53 skew Brownian motion 89 Skorokhod representation 274 Skorokhod topology on D 293 speed measure 224

uniqueness in distribution for SDE 197 variance process 42 weak convergence 271 Wright-Fisher diffusion 182, 234, 308