Probability and Random Processes for Electrical and Computer Engineers Solution Guide

Solutions Manual for

Probability and Random Processes for Electrical and Computer Engineers John A. Gubner University of Wisconsin–Madison

File Generated July 13, 2007

CHAPTER 1

Problem Solutions 1. Ω = {1, 2, 3, 4, 5, 6}. 2. Ω = {0, 1, 2, . . . , 24, 25}. 3. Ω = [0, ∞). RTT > 10 ms is given by the event (10, ∞). 4.

(a) Ω = {(x, y) ∈ IR2 : x2 + y2 ≤ 100}.

(b) {(x, y) ∈ IR2 : 4 ≤ x2 + y2 ≤ 25}. 5.

(a) [2, 3] c = (−∞, 2) ∪ (3, ∞). (b) (1, 3) ∪ (2, 4) = (1, 4). (c) (1, 3) ∩ [2, 4) = [2, 3). (d) (3, 6] \ (5, 7) = (3, 5].

6. Sketches:

y

y

y

1 1

x

x

−1

x

−1

B0

B1

B−1

y

y

y

3 x

C1

3

H3

1

x

x

J3

2

Chapter 1 Problem Solutions y

y

3 3 3

U H3

x

x

3

J3 = M 3

H3 U J3 = N 3 y

y

4

2

3

2

M2

7.

U

N 3 = M2

x

x

3 4

M4

U

N3

      (a) [1, 4] ∩ [0, 2] ∪ [3, 5] = [1, 4] ∩ [0, 2] ∪ [1, 4] ∩ [3, 5] = [1, 2] ∪ [3, 4].

(b)



c = [0, 1] c ∩ [2, 3] c [0, 1] ∪ [2, 3] h i h i = (−∞, 0) ∪ (1, ∞) ∩ (−∞, 2) ∪ (3, ∞)  h i = (−∞, 0) ∩ (−∞, 2) ∪ (3, ∞)  h i ∪ (1, ∞) ∩ (−∞, 2) ∪ (3, ∞) = (−∞, 0) ∪ (1, 2) ∪ (3, ∞).

(c) (d) (e) (f)

∞ \

(− 1n , n1 ) = {0}.

n=1 ∞ \

1 ) = [0, 3]. [0, 3 + 2n

n=1 ∞ [

1 ] = [5, 7). [5, 7 − 3n

n=1 ∞ [

[0, n] = [0, ∞).

n=1

Chapter 1 Problem Solutions

3

8. We first let C ⊂ A and show that for all B, (A ∩ B) ∪C = A ∩ (B ∪C). Write A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C), = (A ∩ B) ∪C,

by the distributive law,

since C ⊂ A ⇒ A ∩C = C.

For the second part of the problem, suppose (A ∩ B) ∪C = A ∩ (B ∪C). We must show that C ⊂ A. Let ω ∈ C. Then ω ∈ (A ∩ B) ∪ C. But then ω ∈ A ∩ (B ∪ C), which implies ω ∈ A. 9. Let I := {ω ∈ Ω : ω ∈ A ⇒ ω ∈ B}. We must show that A ∩ I = A ∩ B. ⊂: Let ω ∈ A ∩ I. Then ω ∈ A and ω ∈ I. Therefore, ω ∈ B, and then ω ∈ A ∩ B.

⊃: Let ω ∈ A ∩ B. Then ω ∈ A and ω ∈ B. We must show that ω ∈ I too. In other words, we must show that ω ∈ A ⇒ ω ∈ B. But we already have ω ∈ B. 10. The function f : (−∞, ∞) → [0, ∞) with f (x) = x3 is not well defined because not all values of f (x) lie in the claimed co-domain [0, ∞). 11.

(a) The function will be invertible if Y = [−1, 1]. (b) {x : f (x) ≤ 1/2} = [−π /2, π /6]. (c) {x : f (x) < 0} = [−π /2, 0).

12.

(a) Since f is not one-to-one, no choice of co-domain Y can make f : [0, π ] → Y invertible. (b) {x : f (x) ≤ 1/2} = [0, π /6] ∪ [5π /6, π ]. (c) {x : f (x) < 0} = ∅.

13. For B ⊂ IR,

 X,    A, −1 f (B) = c, A    ∅,

0 ∈ B and 1 ∈ B, 1 ∈ B but 0 ∈ / B, 0 ∈ B but 1 ∈ / B, 0∈ / B and 1 ∈ / B.

14. Let f : X → Y be a function such that f takes only n distinct values, say y1 , . . . , yn . Let B ⊂ Y be such that f −1 (B) is nonempty. By definition, each x ∈ f −1 (B) has the property that f (x) ∈ B. But f (x) must be one of the values y1 , . . . , yn , say yi . Now f (x) = yi if and only if x ∈ Ai := f −1 ({yi }). Hence, f −1 (B) =

[

Ai .

i:yi ∈B

15.

(a) f (x) ∈ B c ⇔ f (x) ∈ /B⇔x∈ / f −1 (B) ⇔ x ∈ f −1 (B) c . (b) f (x) ∈

∞ [

n=1

Bn if and only if f (x) ∈ Bn for some n; i.e., if and only if x ∈ f −1 (Bn )

for some n. But this says that x ∈

∞ [

n=1

f −1 (Bn ).

4

Chapter 1 Problem Solutions

(c) f (x) ∈

∞ \

n=1

Bn if and only if f (x) ∈ Bn for all n; i.e., if and only if x ∈ f −1 (Bn )

for all n. But this says that x ∈ S

16. If B = i {bi } and C = is countable.

S

i {ci },

∞ \

f −1 (Bn ).

n=1

put a2i := bi and a2i−1 := ci . Then A =

17. Since each Ci is countable, we can write Ci = B :=

∞ [

Ci =

i=1

S

j ci j .

∞ [ ∞ [

S

= B ∪C

i ai

It then follows that

{ci j }

i=1 j=1

is a doubly indexed sequence and is therefore countable as shown in the text. S

18. Let A = m {am } be a countable set, and let B ⊂ A. We must show that B is countable. If B = ∅, we’re done by definition. Otherwise, there is at least one element of B in S A, say ak . Then put bn := an if an ∈ B, and put bn := ak if an ∈ / B. Then n {bn } = B and we see that B is countable. 19. Let A ⊂ B where A is uncountable. We must show that B is uncountable. We prove this by contradiction. Suppose that B is countable. Then by the previous problem, A is countable, contradicting the assumption that A is uncountable. 20. Suppose A is countable and B is uncountable. We must show that A∪B is uncountable. We prove this by contradiction. Suppose that A ∪ B is countable. Then since B ⊂ A ∪ B, we would have B countable as well, contradicting the assumption that B is uncountable. 21. MATLAB. OMITTED. 22. MATLAB. Intuitive explanation: Using only the numbers 1, 2, 3, 4, 5, 6, consider how many ways there are to write the following numbers: 2 3 4 5 6 7 8 9 10 11 12

= = = = = = = = = = =

1+1 1+2 = 2+1 1+3 = 2+2 = 3+1 1+4 = 2+3 = 3+2 = 4+1 1+5 = 2+4 = 3+3 = 4+2 = 5+1 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 2+6 = 3+5 = 4+4 = 5+3 = 6+2 3+6 = 4+5 = 5+4 = 6+3 4+6 = 5+5 = 6+4 5+6 = 6+5 6+6

23. Take Ω := {1, . . . , 26} and put P(A) :=

|A| |A| = . |Ω| 26

1 way, 2 ways, 3 ways, 4 ways, 5 ways, 6 ways, 5 ways, 4 ways, 3 ways, 2 ways, 1 way, 36 ways,

1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 36/36

= = = = = = = = = = = =

0.0278 0.0556 0.0833 0.1111 0.1389 0.1667 0.1389 0.1111 0.0833 0.0556 0.0278 1

Chapter 1 Problem Solutions

5

The event that a vowel is chosen is V = {1, 5, 9, 15, 21}, and P(V ) = |V |/26 = 5/26. 24. Let Ω := {(i, j) : 1 ≤ i, j ≤ 26 and i 6= j}. For A ⊂ Ω, put P(A) := |A|/|Ω|. The event that a vowel is chosen followed by a consonant is  Bvc = (i, j) ∈ Ω : i = 1, 5, 9, 15, or 21 and j ∈ {1, . . . , 26} \ {1, 5, 9, 15, 21} . Similarly, the event that a consonant is followed by a vowel is  Bcv = (i, j) ∈ Ω : i ∈ {1, . . . , 26} \ {1, 5, 9, 15, 21} and j = 1, 5, 9, 15, or 21 .

We need to compute

P(Bvc ∪ Bcv ) =

|Bvc | + |Bcv | 5 · (26 − 5) + (26 − 5) · 5 21 = = ≈ 0.323. |Ω| 650 65

The event that two vowels are chosen is  Bvv = (i, j) ∈ Ω : i, j ∈ {1, 5, 9, 15, 21} with i 6= j ,

and P(Bvv ) = |Bvv |/|Ω| = 20/650 = 2/65 ≈ .031.

25. MATLAB. The code for simulating the drawing of a face card is % Simulation of Drawing a Face Card % n = 10000; % Number of draws. X = ceil(52*rand(1,n)); faces = (41 <= X & X <= 52); nfaces = sum(faces); fprintf(’There were %g face cards in %g draws.\n’,nfaces,n)

26. Since 9 pm to 7 am is 10 hours, take Ω := [0, 10]. The probability that the baby wakes up during a time interval 0 ≤ t1 < t2 ≤ 10 is P([t1 ,t2 ]) := Hence, P([2, 10] c ) = P([0, 2]) = 27. Starting with the equations

R2 0

Z t2 1 t1

10

dω .

1/10 d ω = 1/5.

SN = 1 + z + z2 + · · · + zN−2 + zN−1

zSN =

z + z2 + · · · + zN−2 + zN−1 + zN ,

subtract the second line from the first. Canceling common terms leaves SN − zSN = 1 − zN ,

or

SN (1 − z) = 1 − zN .

If z 6= 1, we can divide both sides by 1 − z to get SN = (1 − zN )/(1 − z).

6

Chapter 1 Problem Solutions 28. Let x = p(1). Then p(2) = 2p(1) = 2x, p(3) = 2p(2) = 22 x, p(4) = 2p(3) = 23 x, p(5) = 24 x, and p(6) = 25 x. In general, p(ω ) = 2ω −1 x and we can write 6

1 =

∑ p(ω )

ω =1

5

6

=

∑ 2ω −1 x

= x

ω =1

∑ 2ω

=

ω =0

1 − 26 x = 63x. 1−2

Hence, x = 1/63, and p(ω ) = 2ω −1 /63 for ω = 1, . . . , 6. 29.

(a) By inclusion–exclusion, P(A ∪ B) = P(A) + P(B) − P(A ∩ B), which can be rearranged as P(A ∩ B) = P(A) + P(B) − P(A ∪ B).

(b) Since P(A) = P(A ∩ B) + P(A ∩ B c ),

P(A ∩ B c ) = P(A) − P(A ∩ B) = P(A ∪ B) − P(B),

by part (a).

(c) Since B and A ∩ B c are disjoint, P(B ∪ (A ∩ B c )) = P(B) + P(A ∩ B c ) = P(A ∪ B),

by part (b).

(d) By De Morgan’s law, P(A c ∩ B c ) = P([A ∪ B] c ) = 1 − P(A ∪ B). 30. We must check the four axioms of a probability measure. First, P(∅) = λ P1 (∅) + (1 − λ )P2 (∅) = λ · 0 + (1 − λ ) · 0 = 0. Second, P(A) = λ P1 (A) + (1 − λ )P2 (A) ≥ λ · 0 + (1 − λ ) · 0 = 0. Third,  ∞   ∞   ∞  [ [ [ An = λ P1 P An + (1 − λ )P2 An n=1

n=1

∞

n=1 ∞

n=1

= λ =

n=1

∞

∑ P1 (An ) + (1 − λ ) ∑ P2 (An )

∑ [λ P1 (An ) + (1 − λ )P2 (An )]

n=1 ∞

=

∑ P(An ).

n=1

Fourth, P(Ω) = λ P1 (Ω) + (1 − λ )P2 (Ω) = λ + (1 − λ ) = 1.

/ ∅S, µ (∅) = 0. Second, by definition, µ (A) ≥ 0. Third, for disjoint 31. First, since ω0 ∈ / An for n 6= m. Then An , suppose ω0 ∈ n An . Then ω0 ∈ Am for some S m,and ω0 ∈ µ (Am ) = 1 and µ (An ) = 0 for n 6= m. Hence, µ n An = 1 and ∑n µ (An ) = µ (Am ) = S  S 1. A similar analysis shows that if ω0 ∈ / n An then µ n An and ∑n µ (An ) are both zero. Finally, since ω0 ∈ Ω, µ (Ω) = 1.

Chapter 1 Problem Solutions

7

32. Starting with the assumption that for any two disjoint events A and B, P(A ∪ B) = P(A) + P(B), we have that for N = 2,  N  N [ An = ∑ P(An ). (∗) P n=1

n=1

Now we must show that if (∗) holds for any N ≥ 2, then (∗) holds for N + 1. Write  N+1   N   [ [ An = P An ∪ AN+1 P n=1

n=1

 N  [ = P An + P(AN+1 ),

additivity for two events,

n=1

N

=

∑ P(An ) + P(AN+1 ),

by (∗),

n=1 N+1

=

∑ P(An ).

n=1

c ∩ · · · ∩ F c ⊂ F , it is easy to see that 33. Since An := Fn ∩ Fn−1 n 1 N [

n=1

An ⊂

N [

Fn .

n=1

S

The hard part is to show the reverse inclusion ⊃. Suppose ω ∈ Nn=1 Fn . Then ω ∈ Fn for some n in the range 1, . . . , N. However, ω may belong to Fn for several values of n since the Fn may not be disjoint. Let k := min{n : ω ∈ Fn and 1 ≤ n ≤ N}. / Fn for n < k; in symbols, In other words, 1 ≤ k ≤ N and ω ∈ Fk , but ω ∈ S

c ω ∈ Fk ∩ Fk−1 ∩ · · · ∩ F1c =: Ak .

Hence, ω ∈ Ak ⊂ Nn=1 An . The proof that k := min{n : ω ∈ Fn and n ≥ 1}.

S∞

n=1 An

⊂

S∞

n=1 Fn

is similar except that

34. For arbitrary events Fn , let An be as in the preceding problem. We can then write  ∞   ∞  ∞ [ [ P Fn = P An = ∑ P(An ), since the An are disjoint, n=1

n=1

n=1

N

∑ P(An ), N→∞

= lim

n=1

 N  [ = lim P An N→∞

n=1

 N  [ = lim P Fn . N→∞

n=1

by def. of infinite sum,

8

Chapter 1 Problem Solutions 35. For arbitrary events Gn , put Fn := Gnc . Then  ∞   ∞  \ [ P Gn = 1 − P Fn , by De Morgan’s law, n=1

n=1

 N  [ Fn , = 1 − lim P N→∞

by the preceding problem,

n=1

  N  \ = 1 − lim 1 − P Gn , N→∞

by De Morgan’s law,

n=1

 N  \ Gn . = lim P N→∞

n=1

36. By the inclusion–exclusion formula, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ≤ P(A) + P(B). This establishes the union bound for N = 2. Now suppose the union bound holds for some N ≥ 2. We must show it holds for N + 1. Write  N+1   N   [ [ Fn = P Fn ∪ FN+1 P n=1

n=1

 N  [ ≤ P Fn + P(FN+1 ),

by the union bound for two events,

n=1

N

≤ =

∑ P(Fn ) + P(FN+1 ),

by the union bound for N events,

n=1 N+1

∑ P(Fn ).

n=1

37. To establish the union bound for a countable sequence of events, we proceed as folc ∩ · · · ∩ F c ⊂ F be disjoint with S∞ A = S∞ F . Then lows. Let An := Fn ∩ Fn−1 n n=1 n n=1 n 1  ∞   ∞  [ [ P Fn = P An n=1

n=1

∞

=

∑ P(An ),

since the An are disjoint,

∑ P(Fn ),

since An ⊂ Fn .

n=1 ∞

≤

n=1

38. Following the hint, we put Gn :=

S∞

k=n Bk

so that we can write

 ∞ ∞   ∞   N  \ [ \ \ P Bk = P Gn = lim P Gn , n=1 k=n

n=1

N→∞

n=1

limit property of P,

Chapter 1 Problem Solutions

9

= lim P(GN ), since Gn ⊃ Gn+1 , N→∞  ∞  [ = lim P Bk , definition of GN , N→∞

k=N

∞

≤ lim

N→∞

∑ P(Bk ),

union bound.

k=N

This last limit must be zero since ∑∞ k=1 P(Bk ) < ∞. 39. In this problem, the probability of an interval is its length. (a) P(A0 ) = 1, P(A1 ) = 2/3, P(A2 ) = 4/9 = (2/3)2 , and P(A3 ) = 8/27 = (2/3)3 . (b) P(An ) = (2/3)n . (c) Write    N  ∞ \ \ An , An = lim P P(A) = P N→∞

n=1

limit property of P,

n=1

= lim P(AN ), N→∞

since An ⊃ An+1 ,

= lim (2/3)N = 0. N→∞

40. Consider the collection consisting of the empty set along with all unions of the form S i Aki for some finite subsequence of distinct elements from {1, . . . , n}. We first show that this collection is a σ -field. First, it contains ∅ by definition. Second, since A1 , . . . , An is a partition,   [

c

Aki

i

=

[

Ami ,

i

where mi is the subsequence {1, . . . , n} \ {ki }. Hence, the collection is closed under complementation. Third,  ∞ [ [ [ Akn,i = Am j , i

n=1

j

where an integer l ∈ {1, . . . , n} is in {m j } if and only if kn,i = l for some n and some i. This shows that the collection is a σ -field. Finally, since every element in our collection must be contained in every σ -field that contains A1 , . . . , An , our collection must be σ (A1 , . . . , An ). 41. We claim that A is not a σ -field. Our proof is by contradiction: We assume A is a σ -field and derive a contradiction. Consider the set ∞ \

[0, 1/2n ) = {0}.

n=1

Since [0, 1/2n ) ∈ Cn ⊂ An ⊂ A , the intersection must be in A since we are assuming A is a σ -field. Hence, {0} ∈ A . Now, any set in A must belong to some An . By the

10

Chapter 1 Problem Solutions preceding problem, every set in An must be a finite union of sets from Cn . However, the singleton set {0} cannot be expressed as a finite union of sets from any Cn . Hence, {0} ∈ /A.

42. Let Ai := X −1 ({xi }) for i = 1, . . . , n. By the problems mentioned in the hint, for any subset B, if X −1 (B) 6= ∅, then X −1 (B) =

[

i:xi ∈B

Ai ∈ σ (A1 , . . . , An ).

It follows that the smallest σ -field containing all the X −1 (B) is σ (A1 , . . . , An ).  43. (a) F = ∅, A, B, {3}, {1, 2}, {4, 5}, {1, 2, 4, 5}, Ω (b) The corresponding probabilities are 0, 5/8, 7/8, 1/2, 1/8, 3/8, 1/2, 1. (c) Since {1} ∈ / F , P({1}) is not defined. 44. Suppose that a σ -field A contains an infinite sequence Fn of sets. If the sequence is not disjoint, we can construct a new sequence An that is disjoint with each An ∈ A . Let a = a1 , a2 , . . . be an infinite sequence of zeros and ones. Then A contains each union of the form [ Ai . i:ai =1

Furthermore, since the Ai are disjoint, each sequence a gives a different union, and we know from the text that the number of infinite sequences a is uncountably infinite. 45.

T

T

A ∈ α Aα , then A ∈ Aα (a) First, since ∅ is in each Aα , ∅ ∈ α Aα . Second, if T for each α , and so A c ∈ Aα for each α . Hence, A c ∈ α Aα . Third, if An ∈ A S for all n, then for each n and each α , An ∈ Aα . Then n An ∈ Aα for each α , S T and so n An ∈ α Aα .

(b) We first note that

A1 = {∅, {1}, {2}, {3, 4}, {2, 3, 4}, {1, 3, 4}, {1, 2}, Ω} and A2 = {∅, {1}, {3}, {2, 4}, {2, 3, 4}, {1, 2, 4}, {1, 3}, Ω}. It is then easy to see that A1 ∩ A2 = {∅, {1}, {2, 3, 4}, Ω}. T

(c) First note that by part (a), A :C ⊂A A is a σ -field, and since C ⊂ A for each T A , the σ -field A :C ⊂A A contains C . Finally, if D is any σ -field that contains C , then D is one of the A s in the intersection. Hence, C ⊂ Thus

T

A :C ⊂A

\

A :C ⊂A

A ⊂ D.

A is the smallest σ -field that contains C .

Chapter 1 Problem Solutions

11

46. The union of two σ -fields is not always a σ -field. Here is an example. Let Ω := {1, 2, 3, 4}, and put   F := ∅, {1, 2}, {3, 4}, Ω and G := ∅, {1, 3}, {2, 4}, Ω . Then

F ∪G =





∅, {1, 2}, {3, 4}, {1, 3}, {2, 4}, Ω

is not a σ -field since it does not contain {1, 2} ∩ {1, 3} = {1}.

47. Let Ω denote the positive integers, and let A denote the collection of subsets A such that either A or A c is finite. (a) Let E denote the subset of even integers. Then E does not belong to A since neither E nor E c (the odd integers) is a finite set. (b) To show that A is closed under finite unions, we consider two cases. First suppose that A1 , . . . , An are all finite. Then n n [ Ai ≤ ∑ |Ai | < ∞, i=1

i=1

and so

Sn

i=1 Ai

∈ A . In the second case, suppose that some A cj is finite. Then 

Hence, the complement of

n [

i=1

Ai

c

Sn

i=1 Ai

=

n \

i=1

Aic ⊂ A cj .

is finite, and so the union belongs to A .

(c) A is not a σ -field. To see this, put Ai := {2i} for i = 1, 2, . . . . Then E∈ / A by part (a).

S∞

i=1 Ai

=

48. Let Ω be an uncountable set. Let A denote the collection of all subsets A such that either A is countable or A c is countable. We show that A is a σ -field. First, the empty set is countable. Second, if A ∈ A , we must show that A c ∈ A . There are two cases. If A is countable, then the complement of A c is A, and so A c ∈ A . If A c is countable, then A c ∈ A . Third, let A1S, A2 , . . . belong to A . There are two cases to consider. If all An are countable, then n An is also countable by an earlier problem. Otherwise, if some Amc is countable, then write 

49.

∞ [

n=1

An

c

=

∞ \

n=1

Anc ⊂ Amc .

Since the subset of a countable set is countable, we see that the complement of S∞ A n=1 n is countable, and thus the union belongs to A . (b)

T∞

1 1 n=1 (a, b + n ), and since each (a, b + n ) ∈ B, (a, b] ∈ B. T 1 1 1 1 Since {a} = ∞ n=1 (a− n , a+ n ), and since each (a− n , a+ n ) ∈ B, the singleton

(a) Since (a, b] = {a} ∈ B.

12

Chapter 1 Problem Solutions (c) Since by part (b), singleton sets are Borel sets, and since A is a countable union of Borel sets, A ∈ B; i.e., A is a Borel set. (d) Using part (a), write
λ (a, b] = λ



∞ \

(a, b +

n=1

= lim λ N→∞



N \



1 n)

 (a, b + 1n ) ,

limit property of probability,

n=1


= lim λ (a, b + N1 ) ,

decreasing sets,

N→∞

= lim (b + N1 ) − a, N→∞

characterization of λ ,

= b − a. Similarly, using part (b), we can write
λ {a} = λ



∞ \



1 1 n,a+ n)

(a −

n=1

= lim λ N→∞



N \

(a −

n=1


= lim λ (a − N1 , a + N1 ) , N→∞

= lim 2/N, N→∞

 ,

1 1 n,a+ n)

limit property of probability,

decreasing sets,

characterization of λ ,

= 0. 50. Let I denote the collection of open intervals, and let O denote the collection of open sets. We need to show that σ (I ) = σ (O). Since I ⊂ O, every σ -field containing O also contains I . Hence, the smallest σ -field containing O contains I ; i.e., I ⊂ σ (O). By the definition of the smallest σ -field containing I , it follows that σ (I ) ⊂ σ (O). Now, if we can show that O ⊂ σ (I ), then it will similarly follow that σ (O) ⊂ σ (I ). Recall that in the problem statement, it was shown that every open set U can be written as a countable union of open intervals. This means U ∈ σ (I ). This proves that O ⊂ σ (I ) as required. 51. MATLAB. Chips from S1 are 80% reliable; chips from S2 are 70% reliable. 52. Observe that  N(Od,S1 )  N(Ow,S1 ) = N(OS1 ) − N(Od,S1 ) = N(OS1 ) 1 − N(OS1 ) and

 N(Od,S2 )  . N(Ow,S2 ) = N(OS2 ) − N(Od,S2 ) = N(OS2 ) 1 − N(OS2 )

Chapter 1 Problem Solutions

13

53. First write P(A|B ∩C) P(B|C) =

P(A ∩ [B ∩C]) P(B ∩C) P([A ∩ B] ∩C) · = = P(A ∩ B|C). P(B ∩C) P(C) P(C)

From this formula, we can isolate the equation P(A|B ∩C) P(B|C) =

P([A ∩ B] ∩C) . P(C)

Multiplying through by P(C) yields P(A|B ∩C) P(B|C) P(C) = P(A ∩ B ∩C). 54.

(a) P(MM) = 140/(140 + 60) = 140/200 = 14/20 = 7/10 = 0.7. Then P(HT) = 1 − P(MM) = 0.3.

(b) Let D denote the event that a workstation is defective. Then

P(D) = P(D|MM)P(MM) + P(D|HT)P(HT) = (.1)(.7) + (.2)(.3) = .07 + .06 = 0.13. (c) Write P(MM|D) =

P(D|MM)P(MM) .07 7 = = . P(D) .13 13

55. Let O denote the event that a cell is overloaded, and let B denote the event that a call is blocked. The problem statement tells us that P(O) = 1/3,

P(B|O) = 3/10,

and

P(B|O c ) = 1/10.

To find P(O|B), first write P(O|B) =

P(B|O)P(O) 3/10 · 1/3 = . P(B) P(B)

Next compute P(B) = P(B|O)P(O) + P(B|O c )P(O c ) = (3/10)(1/3) + (1/10)(2/3) = 5/30 = 1/6. We conclude that P(O|B) =

1/10 6 3 = = = 0.6. 1/6 10 5

56. The problem statement tells us that P(R1 |T0 ) = ε and P(R0 |T1 ) = δ . We also know that 1 = P(Ω) = P(T0 ∪ T1 ) = P(T0 ) + P(T1 ). The problem statement tells us that these last two probabilities are the same; hence they are both equal to 1/2. To find P(T1 |R1 ), we begin by writing P(T1 |R1 ) =

P(R1 |T1 )P(T1 ) . P(R1 )

14

Chapter 1 Problem Solutions Next, we note that P(R1 |T1 ) = 1 − P(R0 |T1 ) = 1 − δ . By the law of total probability, P(R1 ) = P(R1 |T1 )P(T1 ) + P(R1 |T0 )P(T0 )

= (1 − δ )(1/2) + ε (1/2) = (1 − δ + ε )/2.

So, P(T1 |R1 ) =

(1 − δ )(1/2) 1−δ = . (1 − δ + ε )/2 1−δ +ε

57. Let H denote the event that a student does the homework, and let E denote the event that a student passes the exam. Then the problem statement tells us that P(E|H c ) = .1,

P(E|H) = .8,

and

P(H) = .6.

We need to compute P(E) and P(H|E). To begin, write P(E) = P(E|H)P(H) + P(E|H c )P(H c ) = (.8)(.6) + (.1)(1 − .6) = .48 + .04 = .52. Next, P(H|E) =

.48 12 P(E|H)P(H) = = . P(E) .52 13

58. The problem statement tells us that P(AF |CF ) = 1/3,

P(AF |CFc ) = 1/10,

and

P(CF ) = 1/4.

We must compute P(CF |AF ) =

P(AF |CF )P(CF ) (1/3)(1/4) 1/12 = = . P(AF ) P(AF ) P(AF )

To compute the denominator, write P(AF ) = P(AF |CF )P(CF ) + P(AF |CFc )P(CFc )

= (1/3)(1/4) + (1/10)(1 − 1/4) = 1/12 + 3/40 = 19/120.

It then follows that P(CF |AF ) =

10 1 120 · = . 12 19 19

59. Let F denote the event that a patient receives a flu shot. Let S, M, and R denote the events that Sue, Minnie, or Robin sees the patient. The problem tells us that P(S) = .2, P(M) = .4, P(R) = .4, P(F|S) = .6, P(F|M) = .3, and P(F|R) = .1. We must compute P(S|F) =

P(F|S)P(S) (.6)(.2) .12 = = . P(F) P(F) P(F)

Chapter 1 Problem Solutions

15

Next, P(F) = P(F|S)P(S) + P(F|M)P(M) + P(F|R)P(R) = (.6)(.2) + (.3)(.4) + (.1)(.4) = .12 + .12 + .04 = 0.28. Thus, P(S|F) = 60.

12 100 3 · = . 100 28 7

(a) Let Ω = {1, 2, 3, 4, 5} with P(A) := |A|/|Ω|. Without loss of generality, let 1 and 2 correspond to the two defective chips. Then D := {1, 2} is the event that a defective chip is tested. Hence, P(D) = |D|/5 = 2/5.

(b) Your friend’s information tells you that of the three chips you may test, one is defective and two are not. Hence, the conditional probability that the chip you test is defective is 1/3. (c) Yes, your intuition is correct. To prove this, we construct a sample space and probability measure and compute the desired conditional probability. Let Ω := {(i, j, k) : i < j and k 6= i, k 6= j}, where i, j, k ∈ {1, 2, 3, 4, 5}. Here i and j are the chips taken by the friend, and k is the chip that you test. We again take 1 and 2 to be the defective chips. The 10 possibilities for i and j are 12 13 14 15 23 24 25 34 35 45 For each pair in the above table, there are three possible values of k: 345 245 235 235 145 135 134 125 124 123 Hence, there are 30 triples in Ω. For the probability measure we take P(A) := |A|/|Ω|. Now let Fi j denote the event that the friend takes chips i and j with i < j. For example, if the friend takes chips 1 and 2, then from the second table, k has to be 3 or 4 or 5; i.e., F12 = {(1, 2, 3), (1, 2, 4), (1, 2, 5)}. The event that the friend takes two chips is then T := F12 ∪ F13 ∪ F14 ∪ F15 ∪ F23 ∪ F24 ∪ F25 ∪ F34 ∪ F35 ∪ F45 . Now the event that you test a defective chip is D := {(i, j, k) : k = 1 or 2 and i < j with i, j 6= k}.

16

Chapter 1 Problem Solutions We can now compute P(D|T ) =

P(D ∩ T ) . P(T )

Since the Fi j that make up T are disjoint, |T | = 10·3 = 30 and P(T ) = |T |/|Ω| = 1. We next observe that D ∩ T = ∅ ∪ [D ∩ F13 ] ∪ [D ∩ F14 ] ∪ [D ∩ F15 ] ∪ [D ∩ F23 ] ∪ [D ∩ F24 ] ∪ [D ∩ F25 ] ∪ [D ∩ F34 ] ∪ [D ∩ F35 ] ∪ [D ∩ F45 ].

Of the above intersections, the first six intersections are singleton sets, and the last three are pairs. Hence, |D∩T | = 6·1+3·2 and so P(D∩T ) = 12/30 = 2/5. We conclude that P(D|T ) = P(D ∩ T )/P(T ) = (2/5)/1 = 2/5, which is the answer in part (a). Remark. The model in part (c) can be used to solve part (b) by observing that the probability in part (b) is P(D|F12 ∪ F13 ∪ F14 ∪ F15 ∪ F23 ∪ F24 ∪ F25 ), which can be similarly evaluated. 61.

(a) If two sets A and B are disjoint, then by definition, A ∩ B = ∅.

(b) If two events A and B are independent, then by definition, P(A∩B) = P(A)P(B). (c) If two events A and B are disjoint, then P(A ∩ B) = P(∅) = 0. In order for them to be independent, we must have P(A)P(B) = 0; i.e., at least one of the two events must have zero probability. If two disjoint events both have positive probability, then they cannot be independent. 62. Let W denote the event that the decoder outputs the wrong message. Of course, W c is the event that the decoder outputs the correct message. We must find P(W ) = 1 − P(W c ). Now, W c occurs if only the first bit is flipped, or only the second bit is flipped, or only the third bit is flipped, or if no bits are flipped. Denote these disjoint events by F100 , F010 , F001 , and F000 , respectively. Then P(W c ) = P(F100 ∪ F010 ∪ F001 ∪ F000 ) = P(F100 ) + P(F010 ) + P(F001 ) + P(F000 ) = p(1 − p)2 + (1 − p)p(1 − p) + (1 − p)2 p + (1 − p)3 = 3p(1 − p)2 + (1 − p)3 . Hence, P(W ) = 1 − 3p(1 − p)2 − (1 − p)3 = 3p2 − 2p3 . If p = 0.1, then P(W ) = 0.03 − 0.002 = 0.028. 63. Let Ai denote the event that your phone selects channel i, i = 1, . . . , 10. Let B j denote the event that your neighbor’s phone selects channel j, j = 1, . . . , 10. Let P(Ai ) =

Chapter 1 Problem Solutions

17

P(B j ) = 1/10, and assume Ai and B j are independent. Then  10  10 [ P [Ai ∪ B j ] = ∑ P(Ai ∩ B j ) = i=1

i=1

10

10

∑ P(Ai )P(B j )

i=1

=

1

1

∑ 10 · 10

= 0.1.

i=1

64. Let L denote the event that the left airbag works properly, and let R denote the event that the right airbag works properly. Assume L and R are independent with P(L c ) = P(R c ) = p. The probability that at least one airbag works properly is P(L ∪ R) = 1 − P(L c ∩ R c ) = 1 − P(L c )P(R c ) = 1 − p2 . 65. The probability that the dart never lands within 2 cm of the center is  ∞   N  \ \ c c P An = lim P An , N→∞

n=1

limit property of P,

n=1

N

= lim

N→∞

= lim

N→∞

∏ P(Anc ),

independence,

n=1 N

∏ (1 − p)

n=1

= lim (1 − p)N = 0. N→∞

66. Let Wi denote the event that you win on your ith play of the lottery. The probability that you win at least once in n plays is   n  n  n \ [ c = 1 − ∏ P(Wic ), by independence, Wi Wi = 1 − P P i=1

i=1

i=1

n

= 1 − ∏(1 − p) = 1 − (1 − p)n . i=1

We need to choose n so that 1 − (1 − p)n > 1/2, which happens if and only if 1/2 > (1 − p)n

or

− ln 2 > n ln(1 − p) or

− ln 2 < n, ln(1 − p)

where the last step uses the fact that ln(1 − p) is negative. For p = 10−6 , we need n > 693147. 67. Let A denote the event that Anne catches no fish, and let B denote the event that Betty catches no fish. Assume A and B are independent with P(A) = P(B) = p. We must compute P(A ∩ [A ∪ B]) P(A) P(A|A ∪ B) = = , P(A ∪ B) P(A ∪ B) where the last step uses the fact that A ⊂ A ∪ B. To compute the denominator, write

P(A ∪ B) = 1 − P(A c ∩ B c ) = 1 − P(A c )P(B c ) = 1 − (1 − p)2 = 2p − p2 = p(2 − p).

18

Chapter 1 Problem Solutions Then P(A|A ∪ B) =

p 1 = . p(2 − p) 2− p

68. We show that A and B \C are independent as follows. First, since C ⊂ B, P(B) = P(C) + P(B \C). Next, since A and B are independent and since A and C are independent, P(A ∩ B) = P(A)P(B) = P(A)[P(C) + P(B \C)] = P(A ∩C) + P(A)P(B \C). Again using the fact that C ⊂ B, we now write P(A ∩ B) = P(A ∩ [C ∪ B \C]) = P(A ∩C) + P(A ∩ B \C). It follows that P(A ∩ B \C) = P(A)P(B \C), which establishes the claimed independence. 69. We show that A, B, and C are mutually independent. To begin, note that P(A) = P(B) = P(C) = 1/2. Next, we need to identify the events A ∩ B = [0, 1/4)

A ∩C = [0, 1/8) ∪ [1/4, 3/8) B ∩C = [0, 1/8) ∪ [1/2, 5/8)

A ∩ B ∩C = [0, 1/8) so that we can compute

P(A ∩ B) = P(A ∩C) = P(B ∩C) = 1/4

and

P(A ∩ B ∩C) = 1/8.

We find that P(A ∩ B) = P(A)P(B),

P(A ∩C) = P(A)P(C),

P(B ∩C) = P(B)P(C),

and P(A ∩ B ∩C) = P(A)P(B)P(C). 70. From a previous problem we have that P(A ∩ C|B) = P(A|B ∩ C)P(C|B). Hence, P(A ∩C|B) = P(A|B)P(C|B) if and only if P(A|B ∩C) = P(A|B). 71. We show that the probability of the complementary event is zero. By the union bound,    ∞  ∞ ∞ ∞ \ [ \ c c Bk ≤ ∑ P Bk . P n=1 k=n

n=1

k=n

We show that every term on the right is zero. Write  ∞   N  \ \ c c P Bk = lim P Bk , k=n

N→∞

k=n

limit property of P,

Chapter 1 Problem Solutions

19

N

∏ P(Bkc ), N→∞

= lim

independence,

k=n N

= lim

N→∞

≤ lim

N→∞

∏ [1 − P(Bk )]

k=n N

∏ exp[−P(Bk )],

the hint,

k=n

 N  = lim exp − ∑ P(Bk ) N→∞

 = exp − lim

N→∞

k=n N

 P(B ) ∑ k ,

since exp is continuous,

k=n

  ∞ = exp − ∑ P(Bk ) k=n

−∞

= e

= 0,

∞ where the second-to-last step uses the fact that ∑∞ k=1 P(Bk ) = ∞ ⇒ ∑k=n P(Bk ) = ∞.

72. There are 3 · 5 · 7 = 105 possible systems.

73. There are 2n n-bit numbers.

74. There are 100! different orderings of the 100 message packets. In order that the first header packet to be received is the 10th packet to arrive, the first 9 packets to be received must come from the 96 data packets, the 10th packet must come from the 4 header packets, and the remaining 90 packets can be in any order. More specifically, there are 96 possibilities for the first packet, 95 for the second, . . . , 88 for the ninth, 4 for the tenth, and 90! for the remaining 90 packets. Hence, the desired probability is

75.

5
2

96 · · · 88 · 4 · 90! 96 · · · 88 · 4 90 · 89 · 88 · 4 = = = 0.02996. 100! 100 · · · 91 100 · 99 · 98 · 97 = 10 pictures are needed.

76. Suppose the player chooses distinct digits wxyz. The player wins if any of the 4! = 24 permutations of wxyz occurs. Since each permutation has probability 1/10 000 of occurring, the probability of winning is 24/10 000 = 0.0024.
77. There are 83 = 56 8-bit words with 3 ones (and 5 zeros).
78. The probability that a random byte has 4 ones and 4 zeros is 84 /28 = 70/256 = 0.2734. 79. In the first case, since the prizes are different, order is important. Hence, there are 41 · 40 · 39 = 63 960 outcomes. In the
second case, since the prizes are the same, order is not important. Hence, there are 41 3 = 10 660 outcomes.
80. There are 52 the deck contains 13 spades, 13 hearts, 13 14 possible hands. Since
13
13
13
diamonds, and 13 clubs, there are 13 5 hands with 2 spades, 3 hearts, 4 4 3 2

20

Chapter 1 Problem Solutions diamonds, and 5 clubs. The probability of such a hand is 13
13
13
13
2

3

4 52
14

5

= 0.0116.

81. All five cards are of the same suit if and only if they are all spades or all hearts or all diamonds or all clubs. These are four disjoint events. Hence, the answer is four times the probability of getting all spades: 13
1287 5 4 52 = 0.00198.
= 4 2 598 960 5

82. There are

n
k1 ,...,km

such partitions.

83. The general result is



n k0 , . . . , km−1



mn .

4
/10 000 = When n = 4 and m = 10 and a player chooses xxyz, we compute 2,1,1 4
0.0012. For xxyy, we compute 2,2 /10 000 = 0.0006. For xxxy, we compute 4
3,1 /10 000 = 0.0004.

84. Two apples and three carrots corresponds to (0, 0, 1, 1, 0, 0, 0). Five apples corresponds to (0, 0, 0, 0, 0, 1, 1).

CHAPTER 2

Problem Solutions 1.

(a) {ω : X(ω ) ≤ 3} = {1, 2, 3}.

(b) {ω : X(ω ) > 4} = {5, 6}.

(c) P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) = 3(2/15) = 2/5, and P(X > 4) = P(X = 5) + P(X = 6) = 2/15 + 1/3 = 7/15.

2.

(a) {ω : X(ω ) = 2} = {1, 2, 3, 4}.

(b) {ω : X(ω ) = 1} = {41, 42, . . . , 52}.

(c) P(X = 1 or X = 2) = P({1, 2, 3, 4} ∪ {41, 42, . . . , 52}). Since these are disjoint events, the probability of their union is 4/52 + 12/52 = 16/52 = 4/13.

3.

(a) {ω ∈ [0, ∞) : X(ω ) ≤ 1} = [0, 1].

(b) {ω ∈ [0, ∞) : X(ω ) ≤ 3} = [0, 3]. R

(c) P(X ≤ 1) = 01 e−ω d ω = 1 − e−1 . P(X ≤ 3) = 1 − e−3 , P(1 < X ≤ 3) = P(X ≤ 3) − P(X ≤ 1) = e−1 − e−3 .

4. First, since X −1 (∅) = ∅, µ (∅) = P(X −1 (∅)) = P(∅) = 0. Second, µ (B) = P(X −1 (B)) ≥ 0. Third, for disjoint Bn ,    ∞    ∞  ∞ ∞ ∞ [ [ [ −1 −1 X (Bn ) = ∑ P(X −1 (Bn )) = ∑ µ (Bn ). =P Bn Bn = P X µ n=1

n=1

n=1

n=1

n=1

Fourth, µ (IR) = P(X −1 (IR)) = P(Ω) = 1. 5. Since P(Y > n − 1) =

∞

∞

k=n

k=n+1

∑ P(Y = k) = P(Y = n) + ∑

P(Y = k) = P(Y = n) + P(Y > n),

it follows that P(Y = n) = P(Y > n − 1) − P(Y > n). 6. P(Y = 0) = P({TTT,THH,HTH,HHT}) = 4/8 = 1/2, and P(Y = 1) = P({TTH,THT,HTT,HHH}) = 4/8 = 1/2. 7. P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = 2/15 and P(X = 6) = 1/3. 8. P(X = 2) = P({1, 2, 3, 4}) = 4/52 = 1/13. P(X = 1) = P({41, 42, . . . , 52}) = 12/52 = 3/13. P(X = 0) = 1 − P(X = 2) − P(X = 1) = 9/13. 9. The possible values of X are 0, 1, 4, 9, 16. We have P(X = 0) = P({0}) = 1/7, P(X = 1) = P({−1, 1}) = 2/7, P(X = 4) = P({−2, 2}) = 2/7, P(X = 9) = P({3}) = 1/7, and P(X = 16) = P({4}) = 1/7. 21

22

Chapter 2 Problem Solutions

10. We have P(X > 1) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)] = 1 − [e−λ + λ e−λ ] = 1 − e−λ (1 + λ ).

When λ = 1, P(X > 1) = 1 − e−2 (2) = 1 − 2/e = 0.264. 11. The probability that the sensor fails to activate is P(X < 4) = P(X ≤ 3) = P(X = 0) + · · · + P(X = 3) = e−λ (1 + λ + λ 2 /2! + λ 3 /3!). If λ = 2, P(X < 4) = e−2 (1 + 2 + 2 + 4/3) = e−2 (19/3) = 0.857. The probability that the sensor activates is 1 − P(X < 4) = 0.143. 12. Let {Xk = 1} correspond to the event that the kth student gets an A. This event has probability P(Xk = 1) = p. Now, the event that only the kth student gets an A is {Xk = 1 and Xl = 0 for l 6= k}. Hence, the probability that exactly one student gets an A is   15 [ {Xk = 1 and Xl = 0 for l 6= k} = P k=1

=

15

∑ P({Xk = 1 and Xl = 0 for l 6= k})

k=1 15

∑ p(1 − p)14

k=1

= 15(.1)(.9)14 = 0.3432. 13. Let X1 , X2 , X3 be the random digits of the drawing. Then P(Xi = k) = 1/10 for k = 0, . . . , 9 since each digit has probability 1/10 of being chosen. Then if the player chooses d1 d2 d3 , the probability of winning is  P {X1 = d1 , X2 = d2 , X3 = d3 } ∪ {X1 = d1 , X2 = d2 , X3 6= d3 }  ∪{X1 = d1 , X2 6= d2 , X3 = d3 } ∪ {X1 6= d1 , X2 = d2 , X3 = d3 } ,

which is equal to .13 +3[.12 (.9)] = 0.028 since the union is disjoint and since X1 , X2 , X3 are independent.

14.  m   m  [ \ P {Xk < 2} = 1 − P {Xk ≥ 2} k=1

k=1

m

m

= 1 − ∏ P(Xk ≥ 2) = 1 − ∏ [1 − P(Xk ≤ 1)] k=1

k=1

−λ

= 1 − [1 − {e

−λ

+λe

m

}]

= 1 − [1 − e−λ (1 + λ )]m .

Chapter 2 Problem Solutions 15.

23

(a)  n   n  n [ \ P {Xi ≥ 2} = 1 − P {Xi ≤ 1} = 1 − ∏ P(Xi ≤ 1) i=1

i=1

i=1

n

= 1 − ∏[P(Xi = 0) + P(Xi = 1)] i=1 −λ

= 1 − [e  n  \ (b) P {Xi ≥ 1} = i=1

  n \ {Xi = 1} = (c) P i=1

n

∏ P(Xi ≥ 1)

n

=

i=1 n

∏ P(Xi = 1)

+ λ e−λ ]n = 1 − e−nλ (1 + λ )n .

∏[1 − P(Xi = 0)] i=1

= (1 − e−λ )n .

= (λ e−λ )n = λ n e−nλ .

i=1

16. For the geometric0 pmf, write ∞

∞

k=0

k=0

∑ (1 − p)pk = (1 − p) ∑ pk = (1 − p) ·

1 = 1. 1− p

For the geometric1 pmf, write ∞

∑ (1 − p)pk−1

k=1

∞

∞

k=1

n=0

= (1 − p) ∑ pk−1 = (1 − p) ∑ pn = (1 − p) ·

1 = 1. 1− p

17. Let Xi be the price of stock i, which is a geometric0 (p) random variable. Then    29  29 29 \ [ {Xi ≤ 10} = 1 − ∏[(1 − p)(1 + p + · · · + p10 )] {Xi > 10} = 1 − P P i=1

i=1

i=1

 1 − p11 29 = 1 − (1 − p) = 1 − (1 − p11 )29 . 1− p

Substituting p = .7, we have 1 − (1 − .711 )29 = 1 − (.98)29 = 1 − .560 = 0.44. 18. For the first problem, we have  n  \ P(min(X1 , . . . , Xn ) > `) = P {Xk > `} = k=1

n

n

∏ P(Xk > `)

k=1

=

∏ p`

= pn` .

k=1

Similarly,   n n n \ {Xk ≤ `} = ∏ P(Xk ≤ `) = ∏ (1− p` ) = (1− p` )n . P(max(X1 , . . . , Xn ) ≤ `) = P k=1

k=1

k=1

19. Let Xk denote the number of coins in the pocket of the kth student. Then the Xk are independent and is uniformly distributed from 0 to 20; i.e., P(Xk = i) = 1/21.

24

Chapter 2 Problem Solutions  25  25 \ {Xk ≥ 5} = ∏ 16/21 = (16/21)25 = 1.12 × 10−3 . (a) P k=1

k=1

   25  25 \ [ {Xk ≤ 18} = 1 − (1 − 2/21)25 = 0.918. {Xk ≥ 19} = 1 − P (b) P k=1

k=1

(c) The probability that only student k has 19 coins in his or her pocket is   \ P {Xk = 10} ∩ {Xl 6= 19} = (1/21)(20/21)24 = 0.01477. l6=k

Hence, the probability that exactly one student has 19 coins is  25  [ \ {Xk = 10} ∩ {Xl 6= 19} = 25(0.01477) = 0.369. P l6=k

k=1

20. Let Xi = 1 if block i is good. Then P(Xi = 1) = p and
P(Y = k) = P {X1 = 1}∩· · ·∩{Xk−1 = 1}∩{Xk = 0} = pk−1 (1− p),

k = 1, 2, . . . .

Hence, Y ∼ geometric1 (p).

21.

(a) Write ∞

P(X > n) =

∑

(1 − p)pk−1 =

k=n+1

∞

∞

∑ (1 − p)p`+n

`=0

= (1 − p)pn ∑ p` = (1 − p)pn `=0

1 = pn . 1− p

(b) Write P(X > n + k|X > n) =

pn+k P(X > n + k, X > n) P(X > n + k) = = n = pk . P(X > n) P(X > n) p

22. Since P(Y > k) = P(Y > n + k|Y > n), we can write P(Y > k) = P(Y > n + k|Y > n) =

P(Y > n + k,Y > n) P(Y > n + k) = . P(Y > n) P(Y > n)

Let p := P(Y > 1). Taking k = 1 above yields P(Y > n + 1) = P(Y > n)p. Then with n = 1 we have P(Y > 2) = P(Y > 1)p = p2 . With n = 2 we have P(Y > 3) = P(Y > 2)P(Y > 1) = p3 . In general then P(Y > n) = pn . Finally, P(Y = n) = P(Y > n − 1) − P(Y > n) = pn−1 − pn = pn−1 (1 − p), which is the geometric1 (p) pmf. 23.

(a) To compute pX (i), we sum row i of the matrix. This yields pX (1) = pX (3) = 1/4 and pX (2) = 1/2. To compute pY ( j), we sum column j to get pY (1) = pY (3) = 1/4 and pY (2) = 1/2.

Chapter 2 Problem Solutions

25

(b) To compute P(X < Y ), we sum pXY (i, j) over i and j such that i < j. We have P(X < Y ) = pXY (1, 2) + pXY (1, 3) + pXY (2, 3) = 0 + 1/8 + 0 = 1/8. (c) We claim that X and Y are not independent. For example, pXY (1, 2) = 0 is not equal to pX (1)pY (2) = 1/8. 24.

(a) To compute pX (i), we sum row i of the matrix. This yields pX (1) = pX (3) = 1/4 and pX (2) = 1/2. To compute pY ( j), we sum column j to get pY (1) = pY (3) = 1/6 and pY (2) = 2/3. (b) To compute P(X < Y ), we sum pXY (i, j) over i and j such that i < j. We have P(X < Y ) = pXY (1, 2) + pXY (1, 3) + pXY (2, 3) = 1/6 + 1/24 + 1/12 = 7/24. (c) Using the results of part (a), it is easy to verify that pX (i)pY ( j) = pXY (i, j) for i, j = 1, 2, 3. Hence, X and Y are independent.

25. To compute the marginal of X, write pX (1) =

∞

e−3 3

3j e−3 3 = e = 1/3. 3 j=0 j!

4e−6 6

4e−6 6 6j = e = 2/3. 6 j=0 j!

∑

Similarly, pX (2) =

∞

∑

Alternatively, pX (2) = 1 − pX (1) = 1 − 1/3 = 2/3. Of course pX (i) = 0 for i 6= 1, 2. We clearly have pY ( j) = 0 for j < 0 and pY ( j) =

6 j−1 e−6 3 j−1 e−3 +4 , j! j!

j ≥ 0.

Since pX (1)pY ( j) 6= pXY (1, j), X and Y are not independent. 26.

(a) For k ≥ 1, ∞

pX (k) =

(1 − p)pk−1 kn e−k n! n=0

∑

= (1 − p)pk−1 e−k

∞

kn

∑ n!

n=0

= (1 − p)pk−1 e−k ek = (1 − p)pk−1 ,

which we recognize as the geometric1 (p) pmf. (b) Next, ∞

pY (0) =

∑ (1 − p)pk−1 e−k =

k=1

=

1− p ∞ ∑ (p/e)k−1 e k=1

1− p ∞ 1 1− p 1− p ∑ (p/e)m = e · 1 − p/e = e − p . e m=0

26

Chapter 2 Problem Solutions (c) Since pX (1)pY (0) = (1 − p)2 /(e − p) is not equal to pXY (1, 0) = (1 − p)/e, X and Y are not independent.

27. MATLAB. Here is a script: p = ones(1,51)/51; k=[0:50]; i = find(g(k) >= -16); fprintf(’The answer is %g\n’,sum(p(i)))

where function y = g(x) y = 5*x.*(x-10).*(x-20).*(x-30).*(x-40).*(x-50)/1e6;

28. MATLAB. If you modified your program for the preceding problem only by the way you compute P(X = k), then you may get only 0.5001 = P(g(X) ≥ −16 and X ≤ 50). Note that g(x) > 0 for x > 50. Hence, you also have to add P(X ≥ 51) = p51 = 0.0731 to 0.5001 to get 0.5732. 29. MATLAB. OMITTED. 30. MATLAB. OMITTED. 31. MATLAB. OMITTED. 32. E[X] = 2(1/3) + 5(2/3) = 12/3 = 4. 33. E[I(2,6) (X)] = ∑5k=3 P(X = k) = (1− p)[p3 + p4 + p5 ]. For p = 1/2, we get E[I(2,6) (X)] = 7/64 = 0.109375. 34. Write ∞

λ n e−λ e−λ ∞ λ n+1 e−λ ∞ λ n = = ∑ ∑ n! n! λ n=0 (n + 1)! λ n=1 n=0   e−λ λ 1 − e−λ e−λ ∞ λ n −1 = [e − 1] = . = ∑ λ n=0 n! λ λ

E[1/(X + 1)] =

1

∑ n+1

35. Since var(X) = E[X 2 ] − (E[X])2 , E[X 2 ] = var(X) + (E[X])2 . Hence, E[X 2 ] = 7 + 22 = 7 + 4 = 11. 36. Since Y = cX, E[Y ] = E[cX] = cm. Hence, var(Y ) = E[(Y − cm)2 ] = E[(cX − cm)2 ] = E[c2 (X − m)2 ] = c2 E[(X − m)2 ] = c2 var(X) = c2 σ 2 . 37. We begin with E[(X +Y )3 ] = E[X 3 + 3X 2Y + 3XY 2 +Y 3 ] = E[X 3 ] + 3E[X 2Y ] + 3E[XY 2 ] + E[Y 3 ] = E[X 3 ] + 3E[X 2 ]E[Y ] + 3E[X]E[Y 2 ] + E[Y 3 ],

by independence.

Chapter 2 Problem Solutions

27

Now, as noted in the text, for a Bernoulli(p) random variable, X n = X, and so E[X n ] = E[X] = p. Similarly E[Y n ] = q. Thus, E[(X +Y )3 ] = p + 3pq + 3pq + q = p + 6pq + q. 38. The straightforward approach is to put f (c) := E[(X − c)2 ] = E[X 2 ] − 2mc + c2 and differentiate with respect to c to get f 0 (c) = −2m + 2c. Solving f 0 (c) = 0 results in c = m. An alternative approach is to write E[(X − c)2 ] = E[{(X − m) + (m − c)}2 ] = σ 2 + (m − c)2 . From this expression, it is obvious that c = m minimizes the expectation. 39. The two sketches are:

x /a 1/2

I[ a , )( x )

1

8

8

( x /a ) I[ a , )( x )

1

a

a

40. The right-hand side is easy: E[X]/2 = (3/4)/2 = 3/8 = 0.375. The left-hand side is more work: P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)] = 1 − e−λ (1 + λ ). For λ = 3/4, P(X ≥ 2) = 0.1734. So the bound is a little more than twice the value of the probability. 41. The Chebyshev bound is (λ + λ 2 )/4. For λ = 3/4, the bound is 0.3281, which is a little better than the Markov inequality bound in the preceding problem. The true probability is 0.1734. 42. Comparing the definitions of ρXY and cov(X,Y ), we find ρXY = cov(X,Y )/(σX σY ). Hence, cov(X,Y ) = σX σY ρXY . Since cov(X,Y ) := E[(X − mX )(Y − mY ), if Y = X, we see that cov(X, X) = E[(X − mX )2 ] =: var(X). 43. Put f (a) := E[(X − aY )2 ] = E[X 2 ] − 2aE[XY ] + a2 E[Y 2 ] = σX2 − 2aρσX σY + a2 σY2 . Then f 0 (a) = −2ρσX σY + 2aσY2 . Setting this equal to zero and solving for a yields a = ρ (σX /σY ).

28

Chapter 2 Problem Solutions

44. Since P(X = ±1) = P(X = ±2) = 1/4, E[X] = 0. Similarly, since P(XY = ±1) = 1/4 and P(XY = ±4) = 1/4, E[XY ] = 0. Thus, E[XY ] = 0 = E[X]E[Y ] and we see that X and Y are uncorrelated. Next, since X = 1 implies Y = 1, P(X = 1,Y = 1) = P(X = 1) = 1/4 while P(Y = 1) = P(X = 1 or X = −1) = 1/2. Thus, P(X = 1)P(Y = 1) = (1/4)(1/2) = 1/8,

P(X = 1,Y = 1) = 1/4.

but

45. As discussed in the text, for uncorrelated random variables, the variance of the sum is the sum of the variances. Since independent random variables are uncorrelated, the same results holds for them too. Hence, for Y = X1 + · · · + XM , M

∑ var(Xk ).

var(Y ) =

k=1

We also have E[Y ] = ∑M k=1 E[Xk ]. Next, since the Xk are i.i.d. geometric1 (p), E[Xk ] = 1/(1− p) and var(Xk ) = p/(1− p)2 . It follows that var(Y ) = M p/(1− p)2 and E[Y ] = M/(1 − p). We conclude by writing E[Y 2 ] = var(Y ) + (E[Y ])2 =

Mp M2 M(p + M) + = . 2 (1 − p) (1 − p)2 (1 − p)2

46. From E[Y ] = E[dX − s(1 − X)] = d p − s(1 − p) = 0, we find that d/s = (1 − p)/p. 47.

(a) p = 1/1000. (b) Since (1− p)/p = (999/1000)/(1/1000) = 999, the fair odds against are 999 :1. (c) Since the fair odds of 999 :1 are not equal to the offered odds of 500 :1, the game is not fair. To make the game fair, the lottery should pay $900 instead of $500.

48. First note that

  

1 1 ∞ · p−1 , p 6= 1, 1− p t 1 dt = ∞  tp  p = 1. lnt ,

Z ∞ 1 1

1

For p > 1, the integral is equal to 1/(p − 1). For p ≤ 1, the integral is infinite. For 0 < p ≤ 1, write ∞

1 ∑ kp ≥ k=1

∞

∑

Z k+1 1

k=1 k

t

dt = p

Z ∞ 1 1

tp

dt = ∞.

p For p > 1, it suffices to show that ∑∞ k=2 1/k < ∞. To this end, write ∞

1 ∑ kp = k=2

∞

∞ 1 ∑ (k + 1) p ≤ ∑ k=1 k=1

49. First write E[X n ] =

∞

∑ kn

k=1

Z k+1 1 k

t

dt = p

Z ∞ 1 1

tp

∞ C−1 1 p −1 = C . ∑ p p p−n k k=1 k

dt < ∞.

Chapter 2 Problem Solutions

29

By the preceding problem this last sum is finite for p − n > 1, or equivalently, n < p − 1. Otherwise the sum is infinite; the case 1 ≥ p − n > 0 being handled by the preceding problem, and the case 0 ≥ p − n being obvious. 50. If all outcomes are equally likely, n

H(X) =

1

∑ pi log pi

=

i=1

1 n ∑ log n = log n. n i=1

If X is a constant random variable with pi = 0 for i 6= j, then n

H(X) =

1

∑ pi log pi

= p j log

i=1

1 = 1 log 1 = 0. pj

51. Let P(X = xi ) = pi for i = 1, . . . , n. Then n

E[g(X)] =

∑ g(xi )pi

and

i=1

 n  g(E[X]) = g ∑ xi pi . i=1

For n = 2, Jensen’s inequality says that p1 g(x1 ) + p2 g(x2 ) ≥ g(p1 x1 + p2 x2 ). If we put λ = p1 , then 1 − λ = p2 and the above inequality becomes

λ g(x1 ) + (1 − λ )g(x2 ) ≥ g(λ x1 + (1 − λ )x2 ), which is just the definition of a convex function. Hence, if g is convex, Jensen’s inequality holds for n = 2. Now suppose Jensen’s inequality holds for some n ≥ 2. We must show it holds for n + 1. The case of n is  n  n g(x )p ≥ g x p , if p1 + · · · + pn = 1. i i i i ∑ ∑ i=1

i=1

Now suppose that p1 + · · · + pn+1 = 1, and write  n ∑ g(xi )pi = (1 − pn+1 ) ∑ g(xi )

n+1

i=1

i=1

 pi + pn+1 g(xn+1 ). 1 − pn+1

Let us focus on the quantity in brackets. Since n

pi

∑ 1 − pn+1

i=1

=

p1 + · · · + pn 1 − pn+1 = = 1, pn+1 1 − pn+1

Jensen’s inequality for n terms yields  n  pi pi ∑ g(xi ) 1 − pn+1 ≥ g ∑ xi 1 − pn+1 . i=1 i=1 n

30

Chapter 2 Problem Solutions Hence,  n g(x )p ≥ (1 − p )g i i n+1 ∑ ∑ xi

n+1 i=1

i=1

pi 1 − pn+1



+ pn+1 g(xn+1 ).

Now apply the two-term Jensen inequality to get   n ∑ g(xi )pi ≥ g (1 − pn+1 ) ∑ xi

n+1 i=1

  n+1 = g ∑ pi xi .

i=1

  pi + pn+1 xn+1 1 − pn+1

i=1

52. With X = |Z|α and g(x) = xβ /α , we have E[g(X)] = E[X β /α ] = E[(|Z|α )β /α ] = E[|Z|β ] and

E[X] = E[|Z|α ].

Then Jensen’s inequality tells us that E[|Z|β ] ≥ (E[|Z|α ])β /α . Raising both sides the 1/β power yields Lyapunov’s inequality. 53.

(a) For all discrete random variables, we have ∑i P(X = xi ) = 1. For a nonnegative random variable, if xk < 0, we have 1 =

∑ P(X = xi ) i

≥

∑ I[0,∞) (xi )P(X = xi ) + P(X = xk )

= 1 + P(X = xk ).

i

From this it follows that 0 ≥ P(X = xk ) ≥ 0, and so P(X = xk ) = 0.

(b) Write

E[X] =

∑ xi P(X = xi ) i

=

∑

i:xi ≥0

xi P(X = xi ) +

∑

xi P(X = xk ).

k:xk <0

By part (a), the last sum is zero. The remaining sum is obviously nonnegative. (c) By part (b), 0 ≤ E[X −Y ] = E[X] − E[Y ]. Hence, E[Y ] ≤ E[X].

CHAPTER 3

Problem Solutions 1. First, E[X] = G0X (1) =



1 6

 + 43 z

z=1

1 6

=

+

4 3

=

9 6

= 32 . Second, E[X(X − 1)] =

G00X (1) = 4/3. Third, from E[X(X −1)] = E[X 2 ]−E[X], we have E[X 2 ] = 4/3+3/2 = 17/6. Finally, var(X) = E[X 2 ] − (E[X])2 = 17/6 − 9/4 = (34 − 27)/12 = 7/12.     2. pX (0) = GX (0) = 16 + 61 z + 23 z2 = 1/6, pX (1) = G0X (0) = 16 + 43 z = 1/6, z=0

z=0

and pX (2) = G00X (0)/2 = (4/3)/2 = 2/3.

3. To begin, note that G0X (z) = 5((2 + z)/3)4 /3, and G00X (z) = 20((2 + z)/3)3 /9. Hence, E[X] = G0X (1) = 5/3 and E[X(X − 1)] = 20/9. Since E[X(X − 1)] = E[X 2 ] − E[X], E[X 2 ] = 20/9 + 5/3 = 35/9. Finally, var(X) = E[X 2 ] − (E[X])2 = 35/9 − 25/9 = 10/9.

4. For X ∼ geometric0 (p), ∞

∞

∞

n=0

n=0

n=0

1

∑ zn P(X = n) = ∑ zn (1 − p)pn = (1 − p) ∑ (zp)n = (1 − p) 1 − pz .

GX (z) = Then

E[X] = G0X (1) = Next,

1− p 1 − p p = p p= . (1 − pz)2 z=1 (1 − p)2 1− p G00X (1)

2

E[X ] − E[X] = E[X(X − 1)] = =

(1 − p)p · 2p(1 − pz) = (1 − pz)4 z=1

(1 − p)p · 2p 2p2 = . 3 (1 − p) (1 − p)2

This implies that E[X 2 ] =

2p2 p 2p2 + p(1 − p) p + p2 + = = . (1 − p)2 1 − p (1 − p)2 (1 − p)2

Finally, p + p2 p2 p − = . 2 2 (1 − p) (1 − p) (1 − p)2

var(X) = E[X 2 ] − (E[X])2 = For X ∼ geometric1 (p), ∞

GX (z) =

∑ zn P(X = n) =

n=1

∞

∑ zn (1 − p)pn−1 =

n=1

(1 − p)z 1 − p pz = . = p 1 − pz 1 − pz 31

1− p ∞ ∑ (zp)n p n=1

32

Chapter 3 Problem Solutions Now G0X (z) = Hence,

1− p (1 − pz)(1 − p) + (1 − p)pz = . (1 − pz)2 (1 − pz)2 E[X] = G0X (1) =

Next, G00X (z) = We then have

1 . 1− p

(1 − p) 2p(1 − p) · 2(1 − pz)p = . (1 − pz)4 (1 − pz)3

E[X 2 ] − E[X] = E[X(X − 1)] = G00X (1) = and E[X 2 ] = Finally,

2p(1 − p) 2p = , (1 − p)3 (1 − p)2

1 2p + 1 − p 1+ p 2p + = = . 2 2 (1 − p) 1− p (1 − p) (1 − p)2

var(X) = E[X 2 ] − (E[X])2 =

1 p 1+ p − = . (1 − p)2 (1 − p)2 (1 − p)2

5. Since the Xi are independent Poisson(λi ), we use probability generating functions to find GY (z), which turns out to be Poisson(λ ) with λ := λ1 + · · · + λn . It then follows that P(Y = 2) = λ 2 e−λ /2. It remains to write GY (z) = E[zY ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ] =  = exp ∑ λi (z − 1) = eλ (z−1) , 

n

∏ E[zXi ] i=1

n

=

∏ eλi (z−1) i=1

n

i=1

which is the Poisson(λ ) pgf. Hence, Y ∼ Poisson(λ ) as claimed.

∞ k 6. If GX (z) = ∑∞ k=0 z P(X = k), then GX (1) = ∑k=0 P(X = k) = 1. For the particular formula in the problem, we must have 1 = GX (1) = (a0 + a1 + a2 + · · · + an )m /D, or D = (a0 + a1 + a2 + · · · + an )m .

7. From the table on the inside of the front cover of the text, E[Xi ] = 1/(1 − p). Thus, n

E[Y ] =

∑ E[Xi ]

=

i=1

n . 1− p

Second, since the variance of the sum of uncorrelated random variables is the sum of the variances, and since var(Xi ) = p/(1 − p)2 , we have n

var(Y ) =

∑ var(Xi )

i=1

=

np . (1 − p)2

Chapter 3 Problem Solutions

33

It then easily follows that np n2 n(p + n) + = . 2 (1 − p) (1 − p)2 (1 − p)2

E[Y 2 ] = var(Y ) + (E[Y ])2 =

Since Y is the sum of i.i.d. geometric1 (p) random variables, the pgf of Y is the product of the individual pgfs. Thus,   n (1 − p)z (1 − p)z n = . GY (z) = ∏ 1 − pz i=1 1 − pz 8. Starting with GY (z) = [(1 − p) + pz]n , we have

E[Y ] = GY0 (1) = n[(1 − p) + pz]n−1 p

Next,

= np.

z=1

E[Y (Y − 1)] = GY00 (1) = n(n − 1)[(1 − p) + pz]n−2 p2

z=1

= n(n − 1)p2 .

It then follows that

E[Y 2 ] = E[Y (Y − 1)] + E[Y ] = n(n − 1)p2 + np = n2 p2 − np2 + np. Finally, var(Y ) = E[Y 2 ] − (E[Y ])2 = n2 p2 − np2 + np − (np)2 = np(1 − p). 9. For the binomial(n, p) random variable, n

GY (z) =

∑ P(Y = k)zk =

k=0

Hence,

  n ∑ k pk (1 − p)n−k . k=0 n

1 = GY (1) = 10. Starting from

  n ∑ k pk (1 − p)n−k zk . k=0 n

  n ∑ k pk (1 − p)n−k = 1, k=0 n

we follow the hint and replace p with a/(a + b). Note that 1 − p = b/(a + b). With these substitutions, the above equation becomes k  n−k n   n   n a b n ak bn−k = 1 or = 1. ∑ ∑ k n−k a+b a+b k=0 k k=0 k (a + b) (a + b) Multiplying through by (a + b)n we have n   n ∑ k ak bn−k = (a + b)n . k=0

34

Chapter 3 Problem Solutions

11. Let Xi = 1 if bit i is in error, Xi = 0 otherwise. Then Yn := X1 + · · · + Xn is the number of errors in n bits. Assume the Xi are independent. Then GYn (z) = E[zYn ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ]

= E[zX1 ] · · · E[zXn ], by independence, = [(1 − p) + pz]n , since the Xi are i.i.d. Bernoulli(p).

We recognize this last expression as the binomial(n, p) probability generating func
tion. Thus, P(Yn = k) = nk pk (1 − p)n−k .

12. Let Xi ∼ binomial(ni , p) denote the number of students in the ith room. Then Y = X1 + · · · + XM is the total number of students in the school. Next, GY (z) = E[zY ] = E[zX1 +···+XM ] = E[zX1 ] · · · E[zXM ],

by independence,

M

=

∏[(1 − p) + pz]ni ,

since Xi ∼ binomial(ni , p),

i=1

= [(1 − p) + pz]n1 +···+nM . Setting n := n1 + · · · + nM , we see that Y ∼ binomial(n, p). Hence, P(Y = k) = n
k p (1 − p)n−k . k

13. Let Y = X1 +· · ·+Xn , where the Xi are i.i.d. with P(Xi = 1) = 1− p and P(Xi = 2) = p. Observe that Xi − 1 ∼ Bernoulli(p). Hence, n

Y = n + ∑ (Xi − 1) . i=1

|

{z

=: Z

}

Since Z is the sum of i.i.d. Bernoulli(p) random variables, Z ∼ binomial(n, p). Hence, P(Y = k) = P(n + Z = k) = P(Z = k − n)   n = pk−n (1 − p)2n−k , k = n, . . . , 2n. k−n 14. Let Xi be i.i.d. Bernoulli(p), where Xi = 1 means bit i is flipped. Then Y := X1 + · · · + Xn (n = 10) is the number of bits flipped. A codeword cannot be decoded if Y > 2. We need to find P(Y > 2). Observe that GY (z) = E[zY ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ] =

n

∏ E[zXi ] i=1

= [(1 − p) + pz]n .

This is the pgf of a binomial(n, p) random variable. Hence, P(Y > 2) = 1 − P(Y ≤ 2) = 1 − [P(Y = 0) + P(Y = 1) + P(Y = 2)]        10 10 10 2 = 1− (1 − p)10 + p(1 − p)9 + p (1 − p)8 0 1 2 = 1 − (1 − p)8 [(1 − p)2 + 10p(1 − p) + 45p2 ].

Chapter 3 Problem Solutions

35

15. For n = 150 and p = 1/100, we have     k P Binomial(n, p) = k P Poisson(np) = k 0 1 2 3 4 5

0.2215 0.3355 0.2525 0.1258 0.0467 0.0138

0.2231 0.3347 0.2510 0.1255 0.0471 0.0141

16. If the Xi are i.i.d. with mean m, then  n  1 n nm 1 n 1 X E[X ] = m = = = m. E[Mn ] = E i i ∑ ∑ ∑ n i=1 n i=1 n i=1 n If X is any random variable with mean m, then E[cX] = cE[X] = cm, and var(cX) = E[(cX − cm)2 ] = E[c2 (X − m)2 ] = c2 E[(X − m)2 ] = c2 var(X). 17. In general, we have P(|Mn − m| < ε ) ≥ 0.9

⇔

P(|Mn − m| ≥ ε ) < 0.1.

By Chebyshev’s inequality, P(|Mn − m| ≥ ε ) ≤

σ2 < 0.1 nε 2

if n > σ 2 /(.1)ε 2 . For σ 2 = 1 and ε = 0.25, we require n > 1/(.1)(.25)2 = 1/.00625 = 160 students. If instead ε = 1, we require n > 0.1 = 10 students. 18.

(a) E[Xi ] = E[IB (Zi )] = P(Zi ∈ B). Setting p := P(Zi ∈ B), we see that Xi = IB (Zi ) ∼ Bernoulli(p). Hence, var(Xi ) = p(1 − p).

(b) In fact, the Xi are independent. Hence, they are uncorrelated. 19. Mn = 0 if and only if all the Xi are zero. Hence,   n \ {Xi = 0} = P(Mn = 0) = P i=1

n

∏ P(Xi = 0) i=1

= (1 − p)n .

In particular, if p = 1/1000, then P(M100 = 0) = (1 − p)100 = 0.999100 = 0.905. Hence, the chances are more than 90% that when we run a simulation, M100 = 0 and we learn nothing! 20. If Xi = Z ∼ Bernoulli(1/2) for all i, then Mn =

1 n 1 n Xi = ∑ Z = Z, ∑ n i=1 n i=1

36

Chapter 3 Problem Solutions and m = E[Xi ] = E[Z] = 1/2. So, P(|Mn − m| ≥ 1/4) = P(|Z − 1/2| ≥ 1/4). Now, Z − 1/2 = ±1/2, and |Z − 1/2| = 1/2 with probability one. Thus, P(|Z − 1/2| ≥ 1/4) = P(1/2 ≥ 1/4) = 1 6→ 0.

21. From the discussion of the weak law in the text, we have P(|Mn − m| ≥ εn ) ≤

σ2 . nεn2

If nεn2 → ∞ as n → ∞, then probability on the left will go to zero as n → ∞.
22. We have from the example that with p := λ /(λ + µ ), pX|Z (i| j) = ij pi (1 − p) j−i for i = 0, . . . , j. In other words, as a function of i, pX|Z (i| j) is a binomial( j, p) pmf. Hence, j

E[X|Z = j] =

∑ ipX|Z (i| j)

i=0

is just the mean of a binomial( j, p) pmf. The mean of such a pmf is j p. Hence, E[X|Z = j] = j p = jλ /(λ + µ ).
23. The problem is telling us that P(Y = k|X = i) = nk pki (1 − pi )n−k . Hence P(Y < 2|X = i) = P(Y = 0|X = i) + P(Y = 1|X = i) = (1 − pi )n + npi (1 − pi )n−1 .

24. The problem is telling us that P(X = k|Y = j) = λ jk e−λ j /k!. Hence, P(X > 2|Y = j) = 1 − P(X ≤ 2|Y = j) = 1 − [P(X = 0|Y = j) + P(X = 1|Y = j) + P(X = 2|Y = j)] = 1 − [e−λ j + λ j e−λ j + λ j2 e−λ j /2] = 1 − e−λ j [1 + λ j + λ j2 /2].

25. For the first formula, write pX|Y (xi |y j ) :=

P(X = xi ,Y = y j ) P(X = xi )P(Y = y j ) = = P(X = xi ) = pX (xi ). P(Y = y j ) P(Y = y j )

Similarly, for the other formula, pY |X (y j |xi ) :=

P(Y = y j , X = xi ) P(Y = y j )P(X = xi ) = = P(Y = y j ) = pY (y j ). P(X = xi ) P(X = xi )

Chapter 3 Problem Solutions

37

26. We use the law of total probability to write ∞

∑ P(X −Y = n|Y = k)P(Y = k)

P(T = n) = P(X −Y = n) =

k=0

∞

=

∑ P(X − k = n|Y = k)P(Y = k),

by the substitution law,

k=0 ∞

=

∑ P(X = n + k|Y = k)P(Y = k)

k=0 ∞

=

∑ P(X = n + k)P(Y = k),

by independence,

k=0 ∞

=

∑ (1 − p)pn+k · (1 − q)qk

k=0

∞

= (1 − p)(1 − q)pn ∑ (pq)k = k=0

(1 − p)(1 − q)pn . 1 − pq

27. The problem is telling us that

µ n e−µ n!

P(Y = n|X = 1) =

and

P(Y = n|X = 2) =

ν n e−ν . n!

The problem also tells us that P(X = 1) = P(X = 2) = 1/2. We can now write P(X = 1|Y = 2) =

P(X = 1,Y = 2) P(Y = 2|X = 1)P(X = 1) (µ 2 e−µ /2)(1/2) = = . P(Y = 2) P(Y = 2) P(Y = 2)

It remains to use the law of total probability to compute 2

P(Y = 2) =

∑ P(Y = 2|X = i)P(X = i)

i=1

= [P(Y = 2|X = 1) + P(Y = 2|X = 2)]/2 = [µ 2 e−µ /2 + ν 2 e−ν /2]/2 = [µ 2 e−µ + ν 2 e−ν ]/4. We conclude by writing P(X = 1|Y = 2) =

1 µ 2 e−µ /4 = . [µ 2 e−µ + ν 2 e−ν ]/4 1 + (ν /µ )2 eµ −ν

28. Let X = 0 or X = 1 according to whether message zero or message one is sent. The problem tells us that P(X = 0) = P(X = 1) = 1/2 and that P(Y = k|X = 0) = (1 − p)pk

and

P(Y = k|X = 1) = (1 − q)qk ,

where q 6= p. We need to compute P(X = 1|Y = k) =

(1 − q)qk (1/2) P(Y = k|X = 1)P(X = 1) = . P(Y = k) P(Y = k)

38

Chapter 3 Problem Solutions We next use the law of total probability to compute P(Y = k) = [P(Y = k|X = 0) + P(Y = k|X = 1)]/2 = [(1 − p)pk + (1 − q)qk ]/2. We can now compute P(X = 1|Y = k) =

1 (1 − q)qk (1/2) = . (1−p)pk [(1 − p)pk + (1 − q)qk ]/2 1 + (1−q)qk

29. Let R denote the number of red apples in a crate, and let G denote the number of green apples in a crate. The problem is telling us that R ∼ Poisson(ρ ) and G ∼ Poisson(γ ) are independent. If T = R + G is the total number of apples in the crate, we must compute P(T = k|G = 0)P(G = 0) P(G = 0|T = k) = . P(T = k) We first use the law of total probability, substitution, and independence to write P(T = k|G = 0) = P(R+G = k|G = 0) = P(R = k|G = 0) = P(R = k) = ρ k e−ρ /k!. We also note from the text that the sum of two independent Poisson random variables is a Poisson random variable whose parameter is the sum of the individual parameters. Hence, P(T = k) = (ρ + γ )k e−(ρ +γ ) /k!. We can now write

ρ k e−ρ /k! · e−γ = P(G = 0|T = k) = (ρ + γ )k e−(ρ +γ ) /k!



ρ ρ +γ

k

.

30. We begin with n −λ

e 1 · λ n! P(Y = 1|X = n)P(X = n) = n+1 . P(X = n|Y = 1) = P(Y = 1) P(Y = 1)

Next, we compute ∞

P(Y = 1) = = =

∑ P(Y = 1|X = n)P(X = n)

n=0 e−λ

λ e−λ

λ

∞

λ n+1

∑ (n + 1)!

=

n=0

[eλ − 1] =

e−λ

λ

1 − e−λ

λ

∞

∑

k=1

∞

=

1

∑ n+1 ·

n=0

λk k!

=

e−λ λ



λ n e−λ n!

 λk − 1 ∑ k=0 k! ∞

.

We conclude with n −λ

n −λ

e 1 · λ n! ·λ e λ n+1 P(X = n|Y = 1) = . = n+1 −λn! = λ P(Y = 1) (1 − e )/λ (e − 1)(n + 1)! 1 n+1

Chapter 3 Problem Solutions

39

31. We begin with P(Y = k|X = n)P(X = n) = P(X = n|Y = k) = P(Y = k)

n
k n−k · λ n e−λ /n! k p (1 − p)

P(Y = k)

.

Next,   n k p (1 − p)n−k λ n e−λ /n! P(Y = k) = ∑ P(Y = k|X = n)P(X = n) = ∑ n=0 n=k k ∞

∞

∞

=

pk λ k e−λ k!

pk λ k e−λ [(1 − p)λ ]n−k = (n − k)! k! n=k

=

(pλ )k e−pλ pk λ k e−λ (1−p)λ = e . k! k!

∑

Note that Y ∼ Poisson(pλ ). We continue with P(X = n|Y = k) = =

n
k n−k · λ n e−λ /n! k p (1 − p)

P(Y = k)

=

[(1 − p)λ ]n−k e−(1−p)λ . (n − k)!

∞

[(1 − p)λ ]m m! m=0

∑

n
k n−k · λ n e−λ /n! k p (1 − p) (pλ )k e−pλ /k!

32. First write

P {X > k} ∩ {max(X,Y ) > k} P(X > k)

, P X > k max(X,Y ) > k = = P max(X,Y ) > k P max(X,Y ) > k

since {X > k} ⊂ {max(X,Y ) > k}. We next compute



P max(X,Y ) > k = 1 − P max(X,Y ) ≤ k = 1 − P(X ≤ k)P(Y ≤ k).

If we put θk := P(X ≤ k) and use the fact that X and Y have the same pmf, then
1 − θk 1 1 − θk = = . P X > k max(X,Y ) > k = (1 − θk )(1 + θk ) 1 + θk 1 − θk2

With n = 100 and p = .01, we compute

θ1 = P(X ≤ 1) = P(X = 0) + P(X = 1) = .99100 + .9999 = .366 + .370 = .736. It follows that the desired probability is 1/(1 + θ1 ) = 1/1.736 = 0.576. 33.

(a) Observe that P(XY = 4) = P(X = 1,Y = 4) + P(X = 2,Y = 2) + P(X = 4,Y = 1) = (1 − p)(1 − q)[pq4 + p2 q2 + p4 q].

40

Chapter 3 Problem Solutions (b) Write pZ ( j) =

∑ pY ( j − i)pX (i) i ∞

=

∑ pY ( j − i)pX (i),

since pX (i) = 0 for i < 0,

∑ pY ( j − i)pX (i),

since pY (k) = 0 for k < 0,

i=0 j

=

i=0

j

j

= (1 − p)(1 − q) ∑ pi q j−i = (1 − p)(1 − q)q j ∑ (p/q)i . i=0

i=0

Now, if p = q, pZ ( j) = (1 − p)2 p j ( j + 1). If p 6= q, pZ ( j) = (1 − p)(1 − q)q j

1 − (p/q) j+1 q j+1 − p j+1 = (1 − p)(1 − q) . 1 − p/q q− p

34. For j = 0, 1, 2, 3, j

∑ (1/16)

pZ ( j) =

= ( j + 1)/16.

i=0

For j = 4, 5, 6, 3

∑

pZ ( j) =

(1/16) = (7 − j)/16.

i= j−3

For other values of j, pZ ( j) = 0. 35. We first write

P(Y = j|X = 1) P(X = 0) ≥ P(Y = j|X = 0) P(X = 1)

as

λ1j e−λ1 / j! λ0j e−λ0 / j!

≥

1− p . p

We can further simplify this to λ j 1

λ0

≥

1 − p λ1 −λ0 e . p

Taking logarithms and rearranging, we obtain h 1 − p i. ln(λ1 /λ0 ). j ≥ λ1 − λ0 + ln p

Observe that the right-hand side is just a number (threshold) that is computable from the problem data. If we observe Y = j, we compare j to the threshold. If j is greater than or equal to this number, we decide X = 1; otherwise, we decide X = 0.

Chapter 3 Problem Solutions 36. We first write

as

41

P(Y = j|X = 1) P(X = 0) ≥ P(Y = j|X = 0) P(X = 1) (1 − q1 )q1j

(1 − q0 )q0j

We can further simplify this to q j 1

q0

≥

≥

1− p . p

(1 − p)(1 − q0 ) . p(1 − q1 )

Taking logarithms and rearranging, we obtain h (1 − p)(1 − q ) i. 0 j ≤ ln ln(q1 /q0 ), p(1 − q1 ) since q1 < q0 implies ln(q1 /q0 ) < 0. 37. Starting with P(X = xi |Y = y j ) = h(xi ), we have P(X = xi ,Y = y j ) = P(X = xi |Y = y j )P(Y = y j ) = h(xi )pY (y j ). If we can show that h(xi ) = pX (xi ), then it will follow that X and Y are independent. Now observe that the sum over j of the left-hand side reduces to P(X = xi ) = pX (xi ). The sum over j of the right-hand side reduces to h(xi ). Hence, pX (xi ) = h(xi ) as desired. 38. First write pXY (1, j) = (1/3)3 j e−3 / j!

and

pXY (2, j) = (4/6)6 j e−6 / j!

Notice that 3 j e−3 / j! is a Poisson(3) pmf and 6 j e−6 / j! is a Poisson(6) pmf. Hence, pX (1) = ∑∞j=0 pXY (1, j) = 1/3 and pX (2) = ∑∞j=0 pXY (2, j) = 2/3. It then follows that pY |X ( j|1) is Poisson(3) and pY |X ( j|2) is Poisson(6). With these observations, it is clear that E[Y |X = 1] = 3 and E[Y |X = 2] = 6, and E[Y ] = E[Y |X = 1](1/3) + E[Y |X = 2](2/3) = 3(1/3) + 6(2/3) = 1 + 4 = 5. To obtain E[X|Y = j], we first compute pY ( j) = pXY (1, j) + pXY (2, j) = (1/3)3 j e−3 / j! + (2/3)6 j e−6 / j! and pX|Y (1| j) = and pX|Y (2| j) =

1 (1/3)3 j e−3 / j! = (1/3)3 j e−3 / j! + (2/3)6 j e−6 / j! 1 + 2 j+1 e−3 (2/3)6 j e−6 / j! (1/3)3 j e−3 / j! + (2/3)6 j e−6 / j!

=

1 . 1 + 2−( j+1) e3

42

Chapter 3 Problem Solutions We now have E[X|Y = j] = 1 · =

1 1 + 2 j+1 e−3 1

+2·

2 j+1 e−3 1 + 2 j+2 e−3 = . 1 + 2 j+1 e−3 1 + 2 j+1 e−3

+2·

1 + 2 j+1 e−3

1 1 + 2−( j+1) e3

39. Since Y is conditionally Poisson(k) given X = k, E[Y |X = k] = k. Hence, ∞

∞

E[Y ] =

∑ E[Y |X = k]P(X = k)

∑ kP(X = k)

=

= E[X] =

k=1

k=1

1 , 1− p

since X ∼ geometric1 (p). Next ∞

E[XY ] =

∞

∑ E[XY |X = k]P(X = k)

=

∑ kE[Y |X = k]P(X = k)

=

k=1 ∞

=

k=1 ∞

k=1

= var(X) + (E[X])2 =

∑ E[kY |X = k]P(X = k), by substitution,

∑ k2 P(X = k)

= E[X 2 ]

k=1

p 1 1+ p + = . 2 2 (1 − p) (1 − p) (1 − p)2

Since E[Y 2 |X = k] = k + k2 , E[Y 2 ] =

∞

∑ E[Y 2 |X = k]P(X = k)

k=1

∞

=

∑ (k + k2 )P(X = k)

= E[X] + E[X 2 ]

k=1

1+ p 2 1 + = . = 1 − p (1 − p)2 (1 − p)2 Finally, we can compute var(Y ) = E[Y 2 ] − (E[Y ])2 =

2 1 1 − = . (1 − p)2 (1 − p)2 (1 − p)2

40. From the solution of the example, it is immediate that E[Y |X = 1] = λ and E[Y |X = 0] = λ /2. Next, E[Y ] = E[Y |X = 0](1 − p) + E[Y |X = 1]p = (1 − p)λ /2 + pλ . Similarly, E[Y 2 ] = E[Y 2 |X = 0](1 − p) + E[Y 2 |X = 1]p = (λ /2 + λ 2 /4)(1 − p) + (λ + λ 2 )p. To conclude, we have var(Y ) = E[Y 2 ] − (E[Y ])2 = (λ /2 + λ 2 /4)(1 − p) + (λ + λ 2 )p − [(1 − p)λ /2 + pλ ]2 .

Chapter 3 Problem Solutions

43

41. Write E[(X + 1)Y 2 ] =

1

1

∑ E[(X + 1)Y 2 |X = i]P(X = i)

=

i=0

i=0 1

=

∑ E[(i + 1)Y 2 |X = i]P(X = i)

∑ (i + 1)E[Y 2 |X = i]P(X = i).

i=0

Now, since given X = i, Y is conditionally Poisson(3(i + 1)), = 3(i + 1) + 9(i + 1)2 . E[Y 2 |X = i] = (λ + λ 2 ) λ =3(i+1)

It now follows that

E[(X + 1)Y 2 ] =

1

∑ (i + 1)[3(i + 1) + 9(i + 1)2 ]P(X = i)

i=0 1

=

∑ (i + 1)2 [3 + 9(i + 1)]P(X = i)

i=0

= 12(1/3) + 84(2/3) = 4 + 56 = 60. 42. Write ∞

∞

E[XY ] =

∑ E[XY |X = n]P(X = n)

=

∑ E[nY |X = n]P(X = n)

n=0 ∞

n=0 ∞

λ n e−λ n! n=0 n=0       X +1−1 1 X = E = 1−E . = E X +1 X +1 X +1

=

∑ nE[Y |X = n]P(X = n)

=

1

∑ nn+1

By a problem in the previous chapter, this last expectation is equal to (1 − e−λ )/λ . Hence, 1 − e−λ E[XY ] = 1 − . λ 43. Write ∞

E[XY ] =

∞

∑ E[XY |X = n]P(X = n)

=

∑ nE[Y |X = n]P(X = n)

=

n=1 ∞

=

n=1

∑ E[nY |X = n]P(X = n)

n=1 ∞

n

∑ n 1 − q P(X = n)

n=1

i 1 1 h = var(X) + (E[X])2 E[X 2 ] = 1−q 1−q   p 1 1+ p 1 + = . = 2 2 1 − q (1 − p) (1 − p) (1 − q)(1 − p)2

44

Chapter 3 Problem Solutions

44. Write E[X 2 ] =

∞

∑ E[X 2 |Y = k]P(Y = k)

k=1

∞

=

∑ (k + k2 )P(Y = k)

= E[Y +Y 2 ]

n=1

= E[Y ] + E[Y 2 ] = m + (r + m2 ) = m + m2 + r. 45. Using probability generating functions, we see that GV (z) = E[zX+Y ] = E[zX zY ] = E[zX ]E[zY ] = [(1 − p) + pz]n [(1 − p) + pz]m = [(1 − p) + pz]n+m . Thus, V ∼ binomial(n + m, p). We next compute P(V = 10|X = 4) = P(X +Y = 10|X = 4) = P(4 +Y = 10|X = 4)   m 6 = P(Y = 6|X = 4) = P(Y = 6) = p (1 − p)m−6 . 6 46. Write GY (z) = E[zY ] =

∞

∑ E[zY |X = k]P(X = k)

∞

=

k=1 ∞

=

∑ (ez−1 )k P(X = k)

k=1

= GX (ez−1 ) =

∑ ek(z−1) P(X = k)

k=1

(1 − p)ez−1 . 1 − pez−1

CHAPTER 4

Problem Solutions 1. Let Vi denote the input voltage at the ith sampling time. The problem tells us that the Vi are independent and uniformly distributed on [0, 7]. The alarm sounds if Vi > 5 for i = 1, 2, 3. The probability of this is   3 3 \ {Vi > 5} = ∏ P(Vi > 5). P i=1

i=1

Now, P(Vi > 5) = 8/343 = 0.0233. 2. We must solve

R∞ t

R7 5

(1/7) dt = 2/7. Hence, the desired probability is (2/7)3 =

f (x) dx = 1/2 for t. Now,

1 ∞ 1 2x dx = − 2 = 2 . x t t t √ Solving 1/t 2 = 1/2, we find that t = 2. Z ∞

−3

R

3. To find c, we solve 01 cx−1/2 dx = 1. The left-hand side of this equation is 2cx1/2 |10 = R 2c. Solving 2c = 1 yields c = 1/2. For the median, we must solve t1 (1/2)x−1/2 dx = 1/2 or x1/2 |t1 = 1/2. We find that t = 1/4. 4.

(a) For t ≥ 0, P(X > t) =

R∞ t

λ e−λ x dx = −e−λ x |t∞ = e−λ t .

(b) First, P(X > t + ∆t|X > t) = P(X > t + ∆t, X > t)/P(X > t). Next, observe that {X > t + ∆t} ∩ {X > t} = {X > t + ∆t},

and so P(X > t + ∆t|X > t) = P(X > t + ∆t)/P(X > t) = e−λ (t+∆t) /e−λ t = e−λ ∆t . 5. Let Xi denote the voltage output by regulator i. Then the Xi are i.i.d. exp(λ ) random variables. Now put 10

Y :=

∑ I(v,∞) (Xi )

i=1

so that Y counts the number of regulators that output more than v volts. We must compute P(Y = 3). Now, the I(v,∞) (Xi ) are i.i.d. Bernoulli(p) random variables, where ∞ Z ∞ −λ x −λ x λe p = P(Xi > v) = dx = −e = e−λ v . v

v

Next, we now from the previous chapter that a sum of n i.i.d. Bernoulli(p) random variables is a binomial(n, p). Thus,     n 3 10 −3λ v (1 − e−λ v )7 = 120e−3λ v (1 − e−λ v )7 . P(Y = 3) = p (1 − p)n−3 = e 3 3 45

46

Chapter 4 Problem Solutions R∞

−2λ . λ e−λ x dx = −e−λ x |∞ 2 =e

T (a) P min(X1 , . . . , Xn ) > 2 = P ni=1 {Xi > 2} = ∏ni=1 P(Xi > 2) = e−2nλ .

6. First note that P(Xi > 2) =

2

(b) Write



P max(X1 , . . . , Xn ) > 2 = 1 − P max(X1 , . . . , Xn ) ≤ 2  n  \ = 1−P {Xi ≤ 2} i=1

n

= 1 − ∏ P(Xi ≤ 2) = 1 − [1 − e−2λ ]n . i=1

7.

(a) P(Y ≤ 2) =

R2 0

µ e−µ y dy = 1 − e−2µ .

(b) P(X ≤ 12,Y ≤ 12) = P(X ≤ 12)P(Y ≤ 12) = (1 − e−12λ )(1 − e−12µ ). (c) Write

P({X ≤ 12} ∪ {Y ≤ 12}) = 1 − P(X > 12,Y > 12) = 1 − P(X > 12)P(Y > 12)

= 1 − e−12λ e−12µ = 1 − e−12(λ +µ ) .

8.

(a) Make the change of variable y = λ x p , dy = λ px p−1 dx to get ∞ Z ∞ Z ∞ p λ px p−1 e−λ x dx = e−y dy = −e−y = 1. 0 0 0

(b) The same change of variables also yields P(X > t) =

Z ∞ t

p

λ px p−1 e−λ x dx =

Z ∞

λt p

p

e−y dy = e−λ t .

(c) The probability that none of the Xi exceeds 3 is   n n \ p {Xi ≤ 3} = ∏ P(Xi ≤ 3) = [1 − P(X1 > 3)]n = [1 − e−λ 3 ]n . P i=1

i=1

The probability that at least one of them exceeds 3 is  n   n  [ \ p P {Xi > 3} = 1 − P {Xi ≤ 3} = 1 − [1 − e−λ 3 ]n . i=1

9.

i=1

√ / 2π , f 0 (x) < 0 for x > 0 and f 0 (x) > 0 for x < 0. √ 2 (b) Since f 00 (x) = (x2 −1)e−x /2 / 2π , we see that f 00 (x) > 0 for |x| > 1 and f 00 (x) < 0 for |x| < 1. (a) Since f 0 (x) = −xe−x

(c) Rearrange ex

2 /2

2 /2

≥ x2 /2 to get e−x

2 /2

≤ 2/x2 → 0 as |x| → ∞.

Chapter 4 Problem Solutions

47

10. Following the hint, write f (x) = ϕ ((x − m)/σ )/σ , where ϕ is the standard normal density. Observe that f 0 (x) = ϕ 0 ((x − m)/σ )/σ 2 and f 00 (x) = ϕ 00 ((x − m)/σ )/σ 3 . (a) Since the argument of ϕ 0 is positive for x > m and negative for x < m, f (x) is decreasing for x > m and increasing for x < m. Hence, f has a global maximum at x = m. (b) Since the absolute value of the argument of ϕ 00 is greater than one if and only if |x − m| > σ , f (x) is concave for |x − m| < σ and convex for |x − m| > σ . 11. Since ϕ is bounded, limσ →∞ ϕ ((x − m)/σ )/σ = 0. Hence, limσ →∞ f (x) = 0. For x 6= m, we have 2 . i h  2 exp − x−m σ 2σ 2 √ ≤ √ → 0 f (x) =  2 = √ 2π σ 2π (x − m)2 2π σ x−m σ √ as σ → 0. Otherwise, since f (m) = [ 2π σ ]−1 , limσ →0 f (m) = ∞.

12. (a) f (x) = ∑n pn fn (x) is obviously nonnegative. Also, Z ∞

−∞

f (x) dx =

Z ∞

∑ pn fn (x) dx =

−∞ n

∑ pn n

Z ∞

−∞

fn (x) dx =

∑ pn

= 1.

n

3/4

1/2

(c)

1/4 0

=

n

3/4 (b)

∑ pn · 1

1/2 1/4

x

0

13. Clearly, (g ∗ h)(x) =

1

2

R∞

0

3

x

0

1

2

3

−∞ g(y)h(x − y) dy ≥

0 since g and h are nonnegative. Next,  g(y)h(x − y) dy dx (g ∗ h)(x) dx = −∞ −∞ −∞ Z ∞  Z ∞ g(y) h(x − y) dx dy = −∞ −∞  Z ∞ Z ∞ Z ∞ g(y) dy = 1. = g(y) h(θ ) d θ dy = −∞ −∞ −∞ | {z }

Z ∞

Z ∞ Z ∞

=1

14.

R (a) Let p > 1. On Γ(p) = 0∞ x p−1 e−x dx, use integration by parts with u = x p−1 and

dv = e−x dx. Then du = (p − 1)x p−2 dx, v = −e−x , and Z ∞ ∞ Γ(p) = −x p−1 e−x +(p − 1) x(p−1)−1 e−x dx = (p − 1)Γ(p − 1). 0 0 {z } | =0

48

Chapter 4 Problem Solutions √ R (b) On Γ(1/2) = 0∞ x−1/2 e−x dx,√make the change of variable x = y2 /2 or y = 2x. Then dx = y dy and x−1/2 = 2/y. Hence, Z ∞√ √ Z ∞ −y2 /2 √ √ Z ∞ e−y2 /2 2 −y2 /2 √ y dy = 2 dy = 2 2π e Γ(1/2) = dy e y 0 0 0 2π √ √ √ 1 = 2 2π · = π . 2 (c) By repeatedly using the recursion formula in part (a), we have       2n − 1 2n − 3 2n − 1 2n − 1 2n − 3 2n + 1 = Γ = · Γ Γ 2 2 2 2 2 2 .. . 2n − 1 2n − 3 5 3 1 = · · · · · · Γ(1/2) 2 2 2 2 2 2n − 1 2n − 3 5 3 1 √ = · ··· · · · π 2 2 2 2 2 (2n − 1)!! √ = π. 2n (d) First note that g p (y) = 0 for y ≤ 0, and similarly for gq (y). Hence, in order to have g p (y)gq (x − y) > 0, we need y > 0 and x − y > 0, or equivalently, x > y > 0. Of course, if x ≤ 0 this does not happen. Thus, (g p ∗ gq )(x) = 0 for x ≤ 0. For x > 0, we follow the hint and write (g p ∗ gq )(x) = =

Z ∞

−∞

Z x 0

=

g p (y)gq (x − y) dy

g p (y)gq (x − y) dy

1 Γ(p)Γ(q)

Z x 0

y p−1 e−y · (x − y)q−1 e−(x−y) dy

Z

xq−1 e−x x p−1 y (1 − y/x)q−1 dy = Γ(p)Γ(q) 0 Z 1 xq e−x = (xθ ) p−1 (1 − θ )q−1 d θ , Γ(p)Γ(q) 0 =

x p+q−1 e−x Γ(p)Γ(q)

Z 1 0

ch. of var. θ = y/x,

θ p−1 (1 − θ )q−1 d θ .

(∗)

Now, the left-hand side is a convolution of densities, and is therefore a density by Problem 13. In particular, this means that the left-hand side integrates to R one. On the right-hand side, note that 0∞ x p+q−1 e−x dx = Γ(p + q). Hence, integrating the above equation with respect to x from zero to infinity yields 1 =

Γ(p + q) Γ(p)Γ(q)

Z 1 0

θ p−1 (1 − θ )q−1 d θ .

Solving for the above integral and substituting the result into (∗), we find that (g p ∗ gq )(x) = g p+q (x).

Chapter 4 Problem Solutions 15.

R

∞ (a) In −∞ fλ (x) dx = λ dx to get

R∞

−∞ λ

49

f (λ x) dx, make the change of variable y = λ x, dy =

Z ∞

−∞

fλ (x) dx =

Z ∞

−∞

f (y) dy = 1.

(b) Observe that g1,λ (x) = λ

(λ x)0 e−λ x = λ e−λ x , 0!

which we recognize as the exp(λ ) density. (c) The desired probability is Z ∞

Pm (t) :=

t

R

λ

(λ x)m−1 e−λ x dx. (m − 1)!

Note that P1 (t) = t∞ λ e−λ x dx = e−λ t . For m > 1, apply integration by parts with u = (λ x)m−1 /(m − 1)! and dv = λ e−λ x dx. Then Pm (t) =

(λ t)m−1 e−λ t + Pm−1 (t). (m − 1)!

Applying this result recursively, we find that (λ t)m−1 e−λ t (λ t)m−2 e−λ t + + · · · + e−λ t . (m − 1)! (m − 2)!

Pm (t) = (d) We have g 2m+1 , 1 (x) = 2

2

1 2

( 21 x)m−1/2 e−x/2 (1/2)m (1/2)1/2 xm−1/2 e−x/2 = (2m−1)!! √ Γ((2m + 1)/2) π m 2

xm−1/2 e−x/2 √ . = (2m − 1) · · · 5 · 3 · 1 · 2π 16.

(a) We see that b1,1√ (x) = 1 is the uniform(0, 1) density, b2,2 (x) = 6x(1 − x), and b1/2,1 (x) = 1/(2 x ). 2

b

(x)

1/2,1

b (x) 2,2

1.5 b (x) 1

1,1

0.5

0 0

0.25

0.5 x

0.75

1

50

Chapter 4 Problem Solutions (b) From Problem 14(d) and its hint, we have g p+q (x) = (g p ∗ gq )(x) =

x p+q−1 e−x Γ(p)Γ(q)

Z 1 0

θ p−1 (1 − θ )q−1 d θ .

Integrating the left and right-hand sides with respect to x from zero to infinity yields Z Γ(p + q) 1 p−1 θ 1 = (1 − θ )q−1 d θ , Γ(p)Γ(q) 0 which says that the beta density integrates to one. 17. Starting with Γ(p) Γ(q) = Γ(p + q)

Z 1 0

u p−1 (1 − u)q−1 du,

make the change of variable u = sin2 θ , du = 2 sin θ cos θ d θ . We obtain Γ(p) Γ(q) = Γ(p + q)

Z 1 0

= Γ(p + q)

u p−1 (1 − u)q−1 du

Z π /2 0

= 2Γ(p + q)

(sin2 θ ) p−1 (1 − sin2 θ )q−1 · 2 sin θ cos θ d θ

Z π /2

(sin θ )2p−1 (cos θ )2q−1 d θ .

0

Setting p = q = 1/2 on both sides yields Γ(1/2)2 = 2 √

and it follows that Γ(1/2) =

Z π /2

1 dθ = π ,

0

π.

18. Starting with Γ(p) Γ(q) = Γ(p + q)

Z 1 0

u p−1 (1 − u)q−1 du,

make the change of variable u = sin2 θ , du = 2 sin θ cos θ d θ . We obtain Γ(p) Γ(q) = Γ(p + q) =

Z 1 0

u p−1 (1 − u)q−1 du

Z π /2 0

= 2

(sin2 θ ) p−1 (1 − sin2 θ )q−1 · 2 sin θ cos θ d θ

Z π /2

(sin θ )2p−1 (cos θ )2q−1 d θ .

0

Setting p = (n + 1)/2 and q = 1/2 on both sides yields   n+1 √ π Γ Z π /2 2  = 2  sinn θ d θ , n+2 0 Γ 2 and the desired result follows.

Chapter 4 Problem Solutions

51

19. Starting with the integral definition of B(p, q), make the change of variable u = 1 − e−θ , which implies both du = e−θ d θ and 1 − u = e−θ . Hence, B(p, q) =

Z 1 0

u p−1 (1 − u)q−1 du = =

Z ∞ 0

(1 − e−θ ) p−1 (e−θ )q−1 e−θ d θ

0

(1 − e−θ ) p−1 e−qθ d θ .

Z ∞

20. We first use the fact that the density is even and then makepthe change of variable eθ = 1 + x2 /ν , which implies both eθ d θ = 2x/ν dx and x = ν (eθ − 1). Thus, Z ∞ Z ∞ x2 −(ν +1)/2 x2 −(ν +1)/2 dx = 2 dx 1+ 1+ ν ν −∞ 0 Z ∞ 1 = 2 (eθ )−(ν +1)/2 · ν2 eθ p dθ 0 ν (eθ − 1) √ Z ∞ θ −ν /2 θ 1/2 p (e ) (e ) / eθ − 1 d θ = ν 0

√ Z ∞ θ −ν /2 (e ) (1 − e−θ )−1/2 d θ = ν 0

√ Z∞ = ν (1 − e−θ )1/2−1 e−θ ν /2 d θ . 0

√ By the preceding problem, this is equal to ν B(1/2, ν /2), and we see that Student’s t density integrates to one. 21.

(a) Using Stirling’s formula, 1+ν   1 + ν ν /2 √  1 + ν (1+ν )/2−1/2 −(1+ν )/2 2π Γ e e−1/2 2  ≈ 2 2  = √ √  ν ν /2−1/2 −ν /2 √  ν ν /2−1/2 √ ν e ν 2π ν νΓ 2 2 2  1 + ν ν /2 (ν /2)1/2 1 √ = e−1/2 = [(1 + 1/ν )ν ]1/2 √ ν ν 2 e1/2 1 1 → (e1 )1/2 √ = √ . 1/2 2e 2 (b) First write 

1+

  2 2 x2 1/2 x2 ν 1/2  x2 (ν +1)/2 = 1+ 1+ → [ex ]1/2 11/2 = ex /2 . ν ν ν

It then follows that 1+ν    2 −(ν +1)/2 Γ 1 + xν 1 2 ·  = fν (x) = √  √ √  ν 2 (ν +1)/2 ν B( 12 , ν2 ) π 1 + xν νΓ 2 2

→ √

e−x /2 1 √ . 2 /2 = x 2π e 2π

52

Chapter 4 Problem Solutions

22. Making the change of variable t = 1/(1 + z) as suggested in the hint, note that it is equivalent to 1 + z = t −1 , which implies dz = −t −2 dt. Thus, Z 1 Z 1 Z ∞  p−1 1 1 − t  p−1 p+q−2 z p−1 p+q dt t dt dz = − 1 t = p+q 2 t t t 0 0 0 (1 + z) Z 1

=

0

(1 − t) p−1t q−1 dt = B(q, p) = B(p, q).

Hence, fZ (z) integrates to one. Z ∞

23. E[X] =

1

x·

2 dx = x3

Z ∞ 1

2x−2 dx =

−2 ∞ = 2. x 1

24. If the input-output relation has n levels, then the distance from −Vmax to +Vmax should be n∆; i.e., n∆ = 2Vmax , or ∆ = 2Vmax /n. Next, we have from the example in the text that the performance is ∆2 /12, and we need ∆2 /12 < ε , or 1  2Vmax 2 < ε. 12 n √ Solving this for n yields Vmax / 3ε < n = 2b . Taking natural logarithms, we have √
. ln 2. b > ln Vmax / 3ε 25. We use the change of variable x = z − m as follows: E[Z] = = 26. E[X 2 ] =

Z ∞ 1

Z ∞

−∞

Z ∞

x2 ·

−∞

Z ∞

z fZ (z) dz =

−∞

x f (x) dx + m

2 dx = x3

Z ∞ 2

x

1

Z ∞

−∞

z f (z − m) dz =

Z ∞

−∞

f (x) dx = E[X] + m = 0 + m = m.

∞ dx = 2 ln x = 2(∞ − 0) = ∞. 1

27. First note that since Student’s t density is even, E[|X|k ] = tional to Z ∞ 0

xk dx = (1 + x2 /ν )(ν +1)/2

Z 1 0

(x + m) f (x) dx

xk dx + (1 + x2 /ν )(ν +1)/2

R∞

k −∞ |x| f ν (x) dx

Z ∞ 1

is propor-

xk dx (1 + x2 /ν )(ν +1)/2

With regard to this last integral, observe that Z ∞ 1

xk dx ≤ 2 (1 + x /ν )(ν +1)/2

Z ∞ 1

xk dx = ν (ν +1)/2 2 (x /ν )(ν +1)/2

Z ∞ 1

dx , xν +1−k

the range of which is finite if ν + 1 − k > 1, or k < ν . Next, instead of breaking √ integration at one, we break it at the solution of x2 /ν = 1, or x = ν . Then Z ∞

√ ν

xk dx ≥ 2 (1 + x /ν )(ν +1)/2

Z ∞ √

ν

xk dx = 2 (x /ν + x2 /ν )(ν +1)/2

which is infinite if ν + 1 − k ≤ 1, or k ≥ ν .

Z ∞
ν (ν +1)/2 √ 2 ν

dx , xν +1−k

Chapter 4 Problem Solutions

53

28. Begin with E[Y 4 ] = E[(Z + n)4 ] = E[Z 4 + 4Z 3 n + 6Z 2 n2 + 4Zn3 + n4 ]. The moments of the standard normal were computed in an example in this chapter. Hence E[Y 4 ] = 3 + 4 · 0 · n + 6 · 1 · n2 + 4 · 0 · n3 + n4 = 3 + 6n2 + n4 . 29. E[X n ] =

Z ∞

xn

0

30.

x p−1 e−x 1 dx = Γ(p) Γ(p)

Z ∞ 0

Γ(n + p) . Γ(p)

x(n+p)−1 e−x dx =

(a) First write E[X] =

Z ∞ 0

−x2 /2

x · xe

1 dx = 2

Z ∞

2 −x2 /2

x e

−∞

√ Z 2 2π ∞ 2 e−x /2 dx = x √ dx, 2 −∞ 2π

where the last integral is√the second moment of a standard normal density, which 2π p is one. Hence, E[X] = = π /2. 2 (b) For higher-order moments, first write Z ∞

E[X n ] =

0

xn · xe−x

2 /2

dx =

Z ∞ 0

xn+1 e−x

2 /2

dx.

2 Now √ make the change of variable t = x /2, which implies x = dt/ 2t. Hence,

E[X n ] =

Z ∞ 0

[(2t)1/2 ]n+1 e−t

= 2n/2

Z ∞ 0

√ 2t, or dx =

dt 21/2t 1/2

t [(n/2)+1]−1 e−t dt = 2n/2 Γ(1 + n/2).

31. Let Xi denote the flow on link i, and put Yi := I(β ,∞) (Xi ) so that Yi = 1 if the flow on link i is greater than β . Put Z := ∑ni=1 Yi so that Z counts the number of links with flows greater than β . The buffer overflows if Z > 2. Since the Xi are i.i.d., so are the Yi . Furthermore, the Yi are Bernoulli(p), where p = P(Xi > β ). Hence, Z ∼ binomial(n, p). Thus, P(Z > 2) = 1 − P(Z ≤ 2)        n 2 n n p (1 − p)n−2 p(1 − p)n−1 + (1 − p)n + = 1− 2 1 0 = 1 − (1 − p)n−2 [(1 − p)2 + np(1 − p) + 21 n(n − 1)p2 ].

In remains to compute p = P(Xi > β ) =

Z ∞ β

−x2 /2

xe

∞ −β 2 /2 . = e

x2 /2

dx = −e

β

32. The key is to use the change of variable θ = λ x p , which implies both d θ = λ px p−1 dx and x = (θ /λ )1/p . Hence, E[X n ] =

Z ∞ 0

p

xn · λ px p−1 e−λ x dx = n/p

= (1/λ )

Z ∞ 0

Z ∞ 0

[(θ /λ )1/p ]n e−θ d θ

 θ [(n/p)+1]−1 e−θ d θ = Γ(1 + n/p) λ n/p .

54

Chapter 4 Problem Solutions

33. Write Z ∞

x1/2 e−x dx = 0 √ = Γ(3/2) = (1/2)Γ(1/2) = π /2.

E[Y ] = E[(X 1/4 )2 ] = E[X 1/2 ] =

Z ∞ 0

x3/2−1 e−x dx

34. We have   n   n \ [ {Xi ≥ µ /2} {Xi < µ /2} = 1 − P P i=1

i=1

n

= 1 − ∏ P(Xi ≥ µ /2) i=1

= 1−

Z

∞

µ /2

−λ x

λe

dx

n

,

with λ := 1/µ ,

= 1 − (e−λ µ /2 )n = 1 − e−n/2 . 35. Let Xi ∼ exp(λ ) be i.i.d., where λ = 1/20. We must compute  5   5  [ \ P {Xi > 25} = 1 − P {Xi ≤ 25} i=1

i=1

5

= 1 − ∏ P(Xi ≤ 25) i=1

= 1−

Z

25

−λ x

λe

0

dx

5

= 1 − (1 − e−25λ )5 = 1 − (1 − e−5/4 )5 = 0.815. 36. The first two calculations are h(X) =

Z 2

(1/2) log 2 dx = log 2 and

0

h(X) =

Z 1/2

2 log(1/2) dx = log(1/2).

0

For the third calculation, note that − ln f (x) = 21 [(x − m)/σ ]2 + 12 ln 2πσ 2 . Then Z ∞

  f (x) [(x − m)/σ ]2 + ln 2πσ 2 dx −∞      1 X −m 2 + ln 2πσ 2 = 21 {1 + ln 2πσ 2 } = = E 2 σ

h(X) =

1 2

37. The main difficulty is to compute Z ∞

−∞

x2n (1 + x2 /ν )−(ν +1)/2 dx.

1 2

ln 2πσ 2 e.

Chapter 4 Problem Solutions

55

First use the fact that the integrand is even and then makep the change of variable eθ = 1 + x2 /ν , which implies both eθ d θ = 2x/ν dx and x = ν (eθ − 1). Thus, Z ∞

−∞

x2n (1 + x2 /ν )−(ν +1)/2 dx = ν = ν

Z ∞

x2n−1 (1 + x2 /ν )−(ν +1)/2

0

Z ∞ p 0

2x dx ν

( ν (eθ − 1) )2n−1 (eθ )−(ν +1)/2 eθ d θ

= ν n+1/2 = ν n+1/2

Z ∞ 0

(eθ − 1)n−1/2 e−θ (ν +1)/2 eθ d θ

0

(1 − e−θ )n−1/2 e−θ (ν −2n)/2 d θ

Z ∞ Z ∞

(1 − e−θ )(n+1/2)−1 e−θ (ν −2n)/2 d θ
= ν n+1/2 B n + 1/2, (ν − 2n)/2 , by Problem 19. = ν n+1/2

0

Hence,


E[X 2n ] = ν n+1/2 B n + 1/2, (ν − 2n)/2 · √ = ν n+1/2 38. From MX (s) = eσ

2 s2 /2

ν −2n Γ( 2n+1 2 )Γ( 2 )

Γ( ν +1 2 )

1 ν B( 21 , ν2 )

ν −2n 2n+1 Γ( ν +1 n Γ( 2 )Γ( 2 ) 2 ) ·√ = . ν ν Γ( 12 )Γ( ν2 ) Γ( 12 )Γ( ν2 )

, we have MX0 (s) = MX (s)σ 2 s and then MX00 (s) = MX (s)σ 4 s2 + MX (s)σ 2 .

Since MX (0) = 1, we have MX00 (1) = σ 2 . 2

39. Let M(s) := es /2 denote the moment generating function of the standard normal random variable. For the N(m, σ 2 ) moment generating function, we use the change of variable y = (x − m)/σ , dy = dx/σ to write Z ∞

−∞

esx

2 2 Z ∞ Z ∞ )2 ] exp[− 21 ( x−m e−y /2 e−y /2 √ σ dx = es(σ y+m) √ dy = esm dy esσ y √ −∞ −∞ 2π σ 2π 2π

= esm M(sσ ) = esm+σ −s

40. E[esY ] = E[es ln(1/X) ] = E[eln X ] = E[X −s ] = 41. First note that |x| =



Z ∞

−∞

0

x, x ≥ 0, −x, x < 0.

Then the Laplace(λ ) mgf is E[esX ] =

Z 1

esx · λ2 e−λ |x| dx

2 s2 /2

.

x−s dx =

1 x1−s 1 = . 1 − s 0 1−s

56

Chapter 4 Problem Solutions  Z ∞ Z 0 λ sx λ x sx −λ x dx + = e e dx e e 2 0 −∞  Z ∞ Z 0 λ x(λ +s) −x(λ −s) = dx + dx . e e 2 0 −∞ Of these last two integrals, the one on the left is finite if λ > Re s, while the second is finite if Re s > −λ . For both of them to be finite, we need −λ < Re s < λ . For such s both integrals are easy to evaluate. We get   1 1 2λ λ λ λ2 sX MX (s) := E[e ] = + = · 2 2 = 2 2. 2 λ −s λ +s 2 λ −s λ −s Now, MX0 (s) = 2sλ 2 /(λ 2 − s2 )2 , and so the mean is MX0 (0) = 0. We continue with MX00 (s) = 2λ 2

(λ 2 − s2 )2 + 4s2 (λ 2 − s2 ) (λ 2 − s2 )4 .

Hence, the second moment is MX00 (0) = 2/λ 2 . Since the mean is zero, the second moment is also the variance. 42. Since X is a nonnegative random variable, for s ≤ 0, sX ≤ 0 and esX ≤ 1. Hence, for s ≤ 0, MX (s) = E[esX ] ≤ E[1] = 1 < ∞. For s > 0, we show that MX (s) = ∞. We use the fact that for z > 0, ∞ n z3 z ≥ . ez = ∑ 3! n=0 n! Then for s > 0, sX > 0, and we can write   Z (sX)3 2s3 ∞ x3 s3 MX (s) = E[esX ] ≥ E E[X 3 ] = dx = ∞. = 3! 3! 3! 1 x3 43. We apply integration by parts with u = x p−1 /Γ(p) and dv = e−x(1−s) dx. Then du = x p−2 /Γ(p − 1) dx and v = −e−x(1−s) /(1 − s). Hence, Z ∞

Z

∞ x p−1 x p−1 e−x dx = e−x(1−s) dx Γ(p) 0 Γ(p) 0 Z ∞ 1 x p−1 e−x(1−s) ∞ x p−2 −x(1−s) + = − · e dx. Γ(p) 1 − s 1 − s 0 Γ(p − 1)

Mp (s) =

esx

0

The last term is Mp−1 (s)/(1 − s). The other term is zero if p > 1 and Re s < 1. 44.

(a) In this case, we use the change of variable t = x(1 − s), which implies x = t/(1 − s) and dx = dt/(1 − s). Hence, Z ∞

Z

∞ x p−1 e−x 1 dx = x p−1 e−x(1−s) dx Γ(p) Γ(p) 0 0 Z ∞ 1 t  p−1 −t dt = e Γ(p) 0 1 − s 1−s Z  1 p  1 p 1 ∞ p−1 −t · t e dt = . = 1−s Γ(p) 0 1−s | {z }

Mp (s) =

esx

=1

Chapter 4 Problem Solutions

57

(b) From MX (s) = (1 − s)−p , we find MX0 (s) = p(1 − s)−p−1 , MX00 (s) = p(p + 1)(1 − s)−p−2 , and so on. The general result is Γ(n + p) (1 − s)−p−n . Γ(p)

(n)

MX (s) = p(p + 1) · · · (p + [n − 1])(1 − s)−p−n = Hence, the Taylor series is ∞

sn

(n)

∑ n! MX

MX (s) =

∞

(0) =

n=0

45.

sn Γ(n + p) . Γ(p) n=0

∑ n! ·

(a) Make the change of variable t = λ x or x = t/λ , dx = dt/λ . Thus, E[esX ] =

Z ∞

esx

0

λ (λ x) p−1 e−λ x dx = Γ(p)

Z ∞

e(s/λ )t

0

t p−1 e−t dt, Γ(p)

which is the moment generating function of g p evaluated at s/λ . Hence, 

E[esX ] =

 λ p 1 p , = 1 − s/λ λ −s

and the characteristic function is   λ p p 1 E[e jν X ] = = . 1 − jν /λ λ − jν

 λ m  λ m , and the chf is . λ −s λ − jν  1 k/2 (c) The chi-squared with k degrees of freedom mgf is , and the chf is 1 − 2s  1 k/2 . 1 − 2 jν

(b) The Erlang(m, λ ) mgf is

46. First write

Z ∞

2

−x /2 2e esx √ dx = −∞ 2π √ If we let (1 − 2s) = 1/σ 2 ; i.e., σ = 1/ 1 − 2s, then 2

MY (s) = E[esY ] = E[esX ] =

MY (s) = σ

Z ∞ −x2 (1−2s)/2 e −∞

√ 2π

dx.

Z ∞ −(x/σ )2 /2 e −∞

1 √ . dx = σ = √ 1 − 2s 2π σ

47. First observe that 2

esx e−(x−m)

2 /2

2

2

2

2

2

= e−(x −2xm+m −2sx )/2 = e−[x (1−2s)−2xm]/2 e−m /2 2 2 2 2 = e−(1−2s){x −2xm/(1−2s)+[m/(1−2s)] −[m/(1−2s)] }/2 e−m /2 2 2 2 = e−(1−2s){x−[m/(1−2s)]} /2 em /[2(1−2s)] e−m /2 = e−(1−2s){x−[m/(1−2s)]}

2 /2

2 /(1−2s)

esm

.

58

Chapter 4 Problem Solutions √ If we now let 1 − 2s = 1/σ 2 , or σ = 1/ 1 − 2s, and µ = m/(1 − 2s), then sX 2

sY

E[e ] = E[e

] =

2 /(1−2s)

= esm

Z ∞

−(x−m)2 /2 sx2 e

√

e

−∞

2π

sm2 /(1−2s)

dx = e

σ

2

σ =

esm /(1−2s) √ . 1 − 2s

Z ∞ −[(x−µ )/σ ]2 /2 e −∞

√

2π σ

dx

48. ϕY (ν ) = E[e jνY ] = E[e jν (aX+b) ] = E[e j(ν a)X ]e jν b = ϕX (aν )e jν b . 49. The key observation is that |ν | =



ν , ν ≥ 0, −ν , ν < 0.

It then follows that fX (x) = = = = =

50.

Z

1 ∞ −λ |ν | − jν x e e dν 2π −∞ Z ∞  Z 1 1 0 λ ν − jν x e e e−λ ν e− jν x d ν + dν 2π 0 2π −∞ Z ∞  Z 0 1 e−ν (λ + jx) d ν + eν (λ − jx) d ν 2π 0 −∞ ∞ 0   1 1 −1 −ν (λ + jx) ν (λ − jx) + e e 2π λ + jx λ − jx 0 −∞     1 1 1 2λ 1 λ /π = = 2 + . 2 2 2π λ + jx λ − jx 2π λ + x λ + x2

 2  2 e−x /2 d e−x /2 √ = −x √ (a) = −x f (x). dx 2π 2π Z Z ∞ Z ∞ d ∞ jν x e f (x) dx = j e jν x x f (x) dx = − j e jν x f 0 (x) dx. (b) ϕX0 (ν ) = d ν −∞ −∞ −∞

(c) In this last integral, let u = e jν x and dv = f 0 (x) dx. Then du = jν e jν x dx, v = f (x), and the last integral is equal to ∞ Z ∞ jν x e jν x f (x) dx = − jνϕX (ν ). e f (x) − jν −∞ −∞ | {z } =0

(d) Combining (b) and (c), we have ϕX0 (ν ) = − j[− jνϕX (ν )] = −νϕX (ν ). (e) If K(ν ) := ϕX (ν )eν K 0 (ν ) = ϕX0 (ν )eν

2 /2

2 /2

, then

+ ϕX (ν ) · ν eν

2 /2

= −νϕX (ν )eν

2 /2

+ ϕX (ν ) · ν eν

2 /2

= 0.

Chapter 4 Problem Solutions

59

(f) By the mean-value theorem of calculus, for every ν , there is a ν0 between 0 and ν such that K(ν ) − K(0) = K 0 (ν0 )(ν − 0). Since the derivative is zero, we have 2 K(ν ) = K(0) = ϕX (0) = 1. It then follows that ϕX (ν ) = e−ν /2 . 51. Following the hints, we first write d x p e−x px p−1 e−x − x p e−x d xg p (x) = = = pg p (x) − xg p (x) = (p − x)g p (x). dx dx Γ(p) Γ(p) In

ϕX0 (ν ) =

d dν

Z ∞ 0

e jν x g p (x) dx = j

Z ∞ 0

e jν x xg p (x) dx,

apply integration by parts with u = xg p (x) and dv = e jν x dx. Then du is given above, v = e jν x /( jν ), and   Z xg p (x)e jν x ∞ 1 ∞ jν x 0 ϕX (ν ) = j − jν 0 e (p − x)g p (x) dx jν 0 {z } | =0

  Z∞ Z 1 1 ∞ jν x jν x = − p e g p (x) dx − e ( jx)g p (x) dx ν j 0 0   1 1 = − pϕX (ν ) − ϕX0 (ν ) = −(p/ν )ϕX (ν ) + (1/ jν )ϕX0 (ν ). ν j

Rearrange this to get

ϕX0 (ν )(1 − 1/ jν ) = −(p/ν )ϕX (ν ), and multiply through by − jν to get

ϕX0 (ν )(− jν + 1) = j pϕX (ν ). Armed with this, the derivative of K(ν ) := ϕX (ν )(1 − jν ) p is K 0 (ν ) = ϕX0 (ν )(1 − jν ) p + ϕX (ν )p(1 − jν ) p−1 (− j)

= (1 − jν ) p−1 [ϕX0 (ν )(1 − jν ) − j pϕX (ν )] = 0.

By the mean-value theorem of calculus, for every ν , there is a ν0 between 0 and ν such that K(ν ) − K(0) = K 0 (ν0 )(ν − 0). Since the derivative is zero, we have K(ν ) = K(0) = ϕX (0) = 1. It then follows that ϕX (ν ) = 1/(1 − jν ) p . 52. We use the formula cov(X, Z) = E[XZ] − E[X]E[Z]. The mean of an exp(λ ) random variable is 1/λ . Hence, E[X] = 1. Since Z := X + Y , E[Z] = E[X] + E[Y ]. Since the Laplace random variable has zero mean, E[Y ] = 0. Hence, E[Z] = E[X] = 1. Next, E[XZ] = E[X(X + Y )] = E[X 2 ] + E[XY ] = E[X 2 ] + E[X]E[Y ] by independence. Since E[Y ] = 0, E[XZ] = E[X 2 ] = var(X) + (E[X])2 = 1 + 12 = 2, where we have used the fact that the variance of an exp(λ ) random variable is 1/λ 2 . We can now write cov(X, Z) = 2 − 1 = 1. Since Z is the sum of independent, and therefore uncorrelated, random variables, var(Z) = var(X + Y ) = var(X) + var(Y ) = 1 + 2 = 3, where we have used the fact that the variance of a Laplace(λ ) random variable is 2/λ 2 .

60

Chapter 4 Problem Solutions

53. Since Z = X +Y , where X ∼ N(0, 1) and Y ∼ Laplace(1) are independent, we have var(Z) = var(X +Y ) = var(X) + var(Y ) = 1 + 2 = 3. 54. Write MZ (s) = E[esZ ] = E[es(X−Y ) ] = E[esX e−sY ] = E[esX ]E[e−sY ] = MX (s)MY (−s). If MX (s) = MY (s) = λ /(λ − s), then MZ (s) =

λ λ λ λ λ2 · = · = 2 2, λ − s λ − (−s) λ −s λ +s λ −s

which is the Laplace(λ ) mgf. 55. Because the Xi are independent, we can write  n  n n sXi sYn s(X1 +···+Xn ) MYn (s) := E[e ] = E[e ] = E ∏e = ∏ E[esXi ] = ∏ MXi (s). i=1

i=1

(∗)

i=1

2 2

(a) For Xi ∼ N(mi , σi2 ), MXi (s) = esmi +σi s /2 . Hence,  n  n   n smi +σi2 s2 /2 2 σ 2 s2 /2 2 MYn (s) = ∏ e = exp s ∑ mi + ∑ σi s /2 = e|sm+{z }, i=1

i=1

i=1

N(m,σ 2 ) mgf

provided we put m := m1 + · · · + mn and σ 2 := σ12 + · · · + σn2 .

(b) For Cauchy random variables, we must observe that the moment generating function exists only for s = jν . Equivalently, we must use characteristic functions. In this case, (∗) becomes

ϕYn (ν ) := E[e jνYn ] =

n

∏ ϕXi (ν ). i=1

Now, the Cauchy(λi ) chf is ϕXi (ν ) = e−λi |ν | . Hence,   n   n −λi |ν | ϕYn (ν ) = ∏ e = exp − ∑ λi |ν | = i=1

i=1

provided we put λ := λ1 + · · · + λn .

−λ |ν | |e {z } ,

Cauchy(λ ) chf

(c) For Xi ∼ gamma(pi , λ ), the mgf is MXi (s) = [λ /(λ − s)] pi . Hence, n

MYn (s) =

∏ i=1



λ λ −s

 pi

=



λ λ −s

 p1 +···+pn

=

p λ , λ −s | {z }



gamma(p, λ ) mgf

provided we put p := p1 + · · · + pn .

Chapter 4 Problem Solutions

61

56. From part (c) of the preceding problem, Y ∼ gamma(rp, λ ). The table inside the back cover of the text gives the nth moment of a gamma random variable. Hence, E[Y n ] =

Γ(n + rp) . λ n Γ(rp)

57. Let Ti denote the time to transmit packet i. Then the time to transmit n packets is T := T1 + · · · + Tn . We need to find the density of T . Since the Ti are exponential, we can apply the remark in the statement of Problem 55(c) to conclude that T ∼ Erlang(n, λ ). Hence, λ (λ t)n−1 e−λ t fT (t) = , t ≥ 0. (n − 1)!   n n 1 1 = ∑ ln . By Problem 40, each term is an exp(1) 58. Observe that Y = ln ∏ X X i i i=1 i=1 random variable. Hence, by the remark in the statement of Problem 55(c), Y ∼ Erlang(n, 1); i.e., yn−1 e−y fY (y) = , y ≥ 0. (n − 1)! 59. Consider the characteristic function,

ϕY (ν ) = E[e n

=

jν Y

−λ |νβi |

∏e i=1

] = E[e

jν (β1 X1 +···+βn Xn ) n

=

−λ βi |ν |

∏e i=1

] = E



n

∏e

j(νβi )Xi

i=1



n

=

  n   = exp −λ ∑ βi |ν | .

∏ E[e j(νβi )Xi ] i=1

i=1

This is the chf of a Cauchy random variable with parameter λ ∑ni=1 βi . Hence, fY (y) = 

λ π

∑ni=1 βi . 2 λ ∑ni=1 βi + y2

60. We need to compute P(|X −Y | ≤ 2). If we put Z := X −Y , then we need to compute P(|Z| ≤ 2). We first find the density of Z using characteristic functions. Write

ϕZ (ν ) = E[e jν (X−Y ) ] = E[e jν X e− jνY ] = E[e jν X ]E[e j(−ν )Y ] = e−|ν | e−|−ν | = e−2|ν | , which is the chf of a Cauchy(2) random variable. Since the Cauchy density is even,  Z 2   1  2 2 1 2 π 1 −1 z + = tan−1 (1) = · = . tan P(|Z| ≤ 2) = 2 fZ (z) dz = 2 π 2 2 π π 4 2 0 0

61. Let X := U +V +W be the sum of the three voltages. The alarm sounds if X > x. To find P(X > x), we need the density of X. Since U, V , and W are i.i.d. exp(λ ) random variables, by the remark in the statement of Problem 55(c), X ∼ Erlang(3, λ ). By Problem 15(c), 2 (λ x)k e−λ x P(X > x) = ∑ . k! k=0

62

Chapter 4 Problem Solutions

62. Let Xi ∼ Cauchy(λ ) be the i.i.d. line loads. Let Y := X1 + · · · + Xn be the total load. The substation shuts down if Y > `. To find P(Y > `), we need to find the density of Y . By Problem 55(b), Y ∼ Cauchy(nλ ), and so  Z ∞  y  1  ∞ 1 tan−1 + P(Y > `) = fY (y) dy = π nλ 2 ` `    `    1 1 1 1 ` = − tan−1 + . = 1− tan−1 π nλ 2 2 π nλ 63. Let the Ui ∼ uniform[0, 1] be the i.i.d. efficiencies of the extractors. Let Xi = 1 if extractor i operates with efficiency less than 0.25; in symbols, Xi = I[0,0.25) (Ui ), which is Bernoulli(p) with p = 0.25. Then Y := X1 + · · · + X13 is the number of extractors operating at less than 0.25 efficiency. The outpost operates normally if Y < 3. We must compute P(Y < 3). Since Y is the sum of i.i.d. Bernoulli(p) random variables, Y ∼ binomial(13, p). Thus, P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)       13 2 13 13 0 12 13 p (1 − p)11 p(1 − p) + p (1 − p) + = 2 1 0 = (1 − p)11 [(1 − p)2 + 13p(1 − p) + 78p2 ] = 0.3326.

64. By the remark in the statement of Problem 55(c), R = T + A is chi-squared with k = 2 degrees of freedom. Since the number of degrees of freedom is even, R is Erlang(k/2, 1/2) = Erlang(1, 1/2) = exp(1/2). Hence, P(R > r) =

Z ∞ r

65.

(1/2)e−x/2 dx = e−r/2 .

(a) Since c2n+k is a density, it integrates to one. So, Z ∞ 0

ck,λ 2 (x) dx =

2 Z ∞ ∞ (λ 2 /2)n e−λ /2

∑

n!

0 n=0

∞

=

(λ 2 /2)n e−λ n! n=0

∑ ∞

=

∑

2 /2

2 (λ 2 /2)n e−λ /2

n=0 |

n! {z

Z ∞

|0

c2n+k (x) dx c2n+k (x) dx {z } =1

= 1.

}

Poisson(λ 2 /2) pmf

(b) The mgf is Z ∞

esx ck,λ 2 (x) dx  2 Z ∞  ∞ (λ 2 /2)n e−λ /2 sx c2n+k (x) dx = e ∑ n! 0 n=0

Mk,λ 2 (s) =

0

Chapter 4 Problem Solutions ∞

(λ 2 /2)n e−λ = ∑ n! n=0

2 /2

Z ∞ 0

63

esx c2n+k (x) dx

(2n+k)/2 1 1 − 2s   2 ∞ e−λ /2 1 λ 2 /2 n = ∑ (1 − 2s)k/2 n=0 n! 1 − 2s  h i 1 2 exp (λ 2 /2) 1−2s −1 e−λ /2 (λ 2 /2)/(1−2s) = = e (1 − 2s)k/2 (1 − 2s)k/2  h i 2s exp (λ 2 /2) 1−2s exp[sλ 2 /(1 − 2s)] = . = (1 − 2s)k/2 (1 − 2s)k/2 ∞

(λ 2 /2)n e−λ = ∑ n! n=0

2 /2



(c) If we first note that   d (1 − 2s)λ 2 − sλ 2 (−2) sλ 2 = λ 2, = ds 1 − 2s s=0 1 − 2s s=0

α (s) + β (s) , where (1 − 2s)k 2 2 0 α (0) = λ and β (0) = k. Hence, E[X] = Mk,λ 2 (0) = λ + k.

then it is easy to show that Mk,0 λ 2 (s) has the general form

(d) The usual mgf argument gives MY (s) = E[esY ] = E[es(X1 +···+Xn ) ] =

n

∏ Mki ,λi2 (s) i=1

n

exp[sλi2 /(1 − 2s)] = ∏ (1 − 2s)ki /2 i=1 =

exp[s(λ12 + · · · + λn2 )/(1 − 2s)] . (1 − 2s)(k1 +···+kn )/2

If we put k := k1 + · · · + kn and λ 2 := λ12 + · · · + λn2 , we see that Y is noncentral chi-squared with k degrees of freedom and noncentrality parameter λ 2 . (e) We first consider eλ

√ √ x + e−λ x

2

√ 1 ∞ (λ x )n = ∑ [1 + (−1)n ] = 2 n=0 n! ∞

=

λ 2n xn

2n (λ 2 /2)n (x/2)n

∑ 1 · 3 · 5 · · · (2n − 1) · n!

n=0 ∞

=

λ 2n xn n=0 (2n)!

∑ 1 · 3 · 5 · · · (2n − 1) · 2n · n!

n=0 ∞

=

∞

∑

∑

n=0

√

π (λ 2 /2)n (x/2)n , Γ( 2n+1 2 ) · n!

by Problem 14(c).

64

Chapter 4 Problem Solutions We can now write 2

e−(x+λ )/2 eλ √ · 2π x

√

√ x + e−λ x

2

2

e−(x+λ )/2 ∞ √ = ∑ 2π x n=0 ∞

=

(λ 2 /2)n e−λ n! n=0

∑ ∞

=

(λ 2 /2)n e−λ ∑ n! n=0

√ π (λ 2 /2)n (x/2)n Γ( 2n+1 2 ) · n! 2 /2

·

(1/2)(x/2)n−1/2 e−x/2 Γ( 2n+1 2 )

2 /2

c2n+1 (x) = c1,λ 2 (x).

R

66. First, P(X ≥ a) = a∞ 2x−3 dx = 1/a2 , while, using the result of Problem 23, the Markov bound is E[X]/a = 2/a. Thus, the true probability is 1/a2 , but the bound is 2/a, which decays much more slowly for large a. 67. We begin by noting that P(X ≥ a) = e−a , E[X] = 1, and E[X 2 ] = 2. Hence, the Markov bound is 1/a, and the Chebyshev bound is 2/a2 . To find the Chernoff bound, we must minimize h(s) := e−sa MX (s) = e−sa /(1 − s) over 0 < s < 1. Now, h0 (s) =

(1 − s)(−a)e−sa + e−sa . (1 − s)2

Solving h0 (s) = 0, we find s = (a − 1)/a, which is positive only for a > 1. Hence, the Chernoff bound is valid only for a > 1. For a > 1, the Chernoff bound is h((a − 1)/a) =

e−a·(a−1)/a = ae1−a . 1 − (a − 1)/a

(a) It is easy to see that the Markov bound is smaller than the Chebyshev bound for 0 < a < 2. However, note that the Markov bound is greater than one for 0 < a < 1, and the Chebyshev bound is greater than one for 0 < a < 2. (b) MATLAB. 2

0

10

Chebyshev 2/a2

Markov 1/a −2

1.5

10

2

Chebyshev 2/a −4

10

1 1−a

Chernoff ae

Chernoff ae1−a

−6

10

0.5 Markov 1/a P(X > a) 0 1 2

3

a

4

5

P(X > a)

−8

10

6

6

8

10

12

a

14

16

18

20

The Markov bound is the smallest on [1, 2]. The Chebyshev bound is the smallest from a = 2 to a bit more than a = 5. Beyond that, the Chernoff bound is the smallest.

CHAPTER 5

Problem Solutions 1. For x ≥ 0,

Z x

F(x) =

0

−λ t

λe

x −λ t

dt = −e

For x < 0, F(x) = 0. 2. For x ≥ 0, F(x) =

Z x t 0

λ2

−(t/λ )2 /2

e

F(x) =

Z x 0

For x < 0, F(x) = 0. 4. For x ≥ 0, first write F(x) =

−(x/λ ) = 1−e

dt = −e

2 /2

.

0

x p p p λ pt p−1 e−λ t dt = −e−λ t = 1 − e−λ x .

Z xr 0

0

x −(t/λ )2 /2

For x < 0, F(x) = 0. 3. For x ≥ 0,

−λ x = 1−e .

0

2 t 2 −(t/λ )2 /2 dt = e π λ3

Z x/λ r 2 0

π

θ 2 e−θ

2 /2

dθ ,

where we have used the change of variable θ = t/λ . Next use integration by parts 2 with u = θ and dv = θ e−θ /2 d θ . Then r  x/λ Z x/λ  2 −θ 2 /2 −θ 2 /2 F(x) = dθ −θ e + 0 e π 0 r Z x/λ −θ 2 /2 2 x −(x/λ )2 /2 e √ +2 e dθ = − πλ 0 2π r 2 x −(x/λ )2 /2 + 2[Φ(x/λ ) − 1/2] = − e πλ r 2 x −(x/λ )2 /2 = 2Φ(x/λ ) − 1 − . e πλ For x < 0, F(x) = 0. 5. For y > 0, F(y) = P(Y ≤ y) = P(eZ ≤ y) = P(Z ≤ ln y) = FZ (ln y). Then fY (y) = fZ (ln y)/y for y > 0. Since Y := eZ > 0, fY (y) = 0 for y ≤ 0. 65

66

Chapter 5 Problem Solutions 6. To begin, write FY (y) = P(Y ≤ y) = P(1 − X ≤ y) = P(1 − y ≤ X) = 1 − FX (1 − y). Thus, fY (y) = − fX (1 − y) · (−1) = fX (1 − y). In the case of X ∼ uniform(0, 1), fY (y) = fX (1 − y) = I(0,1) (1 − y) = I(0,1) (y), since 0 < 1 − y < 1 if and only if 0 < y < 1. 7. For y > 0, FY (y) = P(Y ≤ y) = P(ln(1/X) ≤ y) = P(1/X ≤ ey ) = P(X ≥ e−y ) = 1 − FX (e−y ). Thus, fY (y) = − fX (e−y ) · (−e−y ) = e−y , since fX (e−y ) = I(0,1) (e−y ) = 1 for y > 0. Since Y := ln(1/X) > 0, fY (y) = 0 for y ≤ 0. 8. For y ≥ 0, FY (y) = P(λ X P ≤ y) = P(X p ≤ y/λ ) = P(X ≤ (y/λ )1/p ) = FX ((y/λ )1/p ). Thus, fY (y) = fX ((y/λ )1/p ) · 1p (y/λ )(1/p)−1 /λ . Using the formula for the Weibull density, we find that fY (y) = e−y for y ≥ 0. Since Y := λ X p ≥ 0, fY (y) = 0 for y < 0. Thus, Y ∼ exp(1). √ 2 9. For y ≥ 0, write FY (y) = P( X ≤ y) = P(X ≤ y2 ) = FX (y2 ) = 1 − e−y . Thus, 2

fY (y) = −e−y · (−2y) =

√ 2 y √ e−(y/(1/ 2)) /2 , (1/ 2)2

√ which is the Rayleigh(1/ 2 ) density. 10. Recall that the moment generating function of X ∼ N(m, σ 2 ) is MX (s) = E[esX ] = 2 2 esm+s σ /2 . Thus, E[Y n ] = E[(eX )n ] = E[enX ] = MX (n) = enm+n

2 σ 2 /2

.

√ √ √ √ 11. For y > 0, FY (y) = P(X 2 ≤ y) = P(− y ≤ X ≤ y) = FX ( y) − FX (− y). Thus, √ √ fY (y) = fX ( y)( 12 y−1/2 ) − fX (− y)(− 12 y−1/2 ). Since fX is even, e−y/2 √ , fY (y) = y−1/2 fX ( y) = √ 2π y

y > 0.

12. For y > 0, FY (y) = P((X + m)2 ≤ y) √ √ = P(− y ≤ X + m ≤ y) √ √ = P(− y − m ≤ X ≤ y − m) √ √ = FX ( y − m) − FX (− y − m).

Chapter 5 Problem Solutions

67

Thus, √ √ fY (y) = fX ( y − m)( 21 y−1/2 ) − fX (− y − m)(− 12 y−1/2 )   √ √ 2 2 1 e−( y−m) /2 + e−(− 2−m) /2 = √ 2 2π y   √ √ 2 2 1 2 = √ e−(y−2 ym+m )/2 + e−(y+2 ym+m )/2 2π y √   √ 2 e−(y+m )/2 em y + e−m y = √ , y > 0. 2 2π y 13. Using the example mentioned in the hint, we have n

FXmax (z) =

∏ FXk (z) = F(z)n

n

and

k=1

FXmin (z) = 1− ∏ [1−FXk (z)] = 1−[1−F(z)]n . k=1

14. Let Z := max(X,Y ). Since X and Y are i.i.d., we have from the preceding problem that FZ (z) = FX (z)2 . Hence, fZ (z) = 2FX (z) fX (z) = 2(1 − e−λ z ) · λ e−λ z ,

z ≥ 0.

Next, E[Z] =

Z ∞ 0

= 2

z fZ (z) dz = 2

Z ∞ 0

= 2

Z ∞

z · λ e−λ z dz −

1 3 1 − = . λ 2λ 2λ

0

λ ze−λ z (1 − e−λ z ) dz

Z ∞ 0

z · (2λ )e−(2λ )z dz

15. Use the laws of total probability and substitution and the fact that conditioned on X = m, Y ∼ Erlang(m, λ ). In particular, E[Y |X = m] = m/λ . We can now write ∞

E[XY ] =

∞

∑ E[XY |X = m]P(X = m)

=

∑ mE[Y |X = m]P(X = m)

=

∑ E[mY |X = m]P(X = m)

m=0 ∞

=

m=0 ∞

∑ m(m/λ )P(X = m)

m=0

m=0

=

µ + µ2 1 E[X 2 ] = , λ λ

since X ∼ Poisson(µ ).

16. The problem statement tells us that P(Y > y|X = n) = e−ny . Using the law of total probability and the pgf of X ∼ Poisson(λ ), we have ∞

P(Y > y) =

∑ P(Y > y|X = n)P(X = n)

n=0

−yX

= E[e

−y

∞

=

∑ e−ny P(X = n)

n=0

λ (e−y −1)

] = GX (e ) = e

.

68 17.

Chapter 5 Problem Solutions (a) Using the law of substitution, independence, and the fact that Y ∼ N(0, 1), write FZ|X (z|i) = P(Z ≤ z|X = i) = P(X +Y ≤ z|X = i) = P(Y ≤ z − i|X = i) = P(Y ≤ z − i) = Φ(z − i). Next, 1

FZ (z) =

∑ FZ|X (z|i)P(X = i)

= (1 − p)Φ(z) + pΦ(z − 1),

i=0

and so fZ (z) =

(1 − p)e−z

2 /2

+ pe−(z−1) √ 2π

2 /2

.

√ (b) From part (a), it is easy to see that fZ|X (z|i) = exp[−(z − i)2 /2]/ 2π . Hence, fZ|X (z|1) P(X = 0) ≥ fZ|X (z|0) P(X = 1)

becomes

exp[−(z − 1)2 /2] 1− p ≥ , exp[−z2 /2] p

or ez−1/2 ≥ (1 − p)/p. Taking logarithms, we can further simplify this to z ≥

1 1− p + ln . 2 p

18. Use substitution and independence to write FZ|A,X (z|a, i) = P(Z ≤ z|A = a, X = i) = P(X/A +Y ≤ z|A = a, X = i) = P(Y ≤ z − i/a|A = a, X = i) = P(Y ≤ z − i/a) = Φ(z − i/a). 19.

(a) Write √ FZn (z) = P(Zn ≤ z) = P( Yn ≤ z) = P(Yn ≤ z2 ) = FYn (z2 ). For future reference, note that fZn (z) = fYn (z2 ) · (2z). Since Yn is chi-squared with n degrees of freedom, i.e., gamma(n/2, 1/2), fYn (y) =

1 2

(y/2)n/2−1 e−y/2 , Γ(n/2)

and so fZn (z) = z

(z2 /2)n/2−1 e−z Γ(n/2)

y > 0,

2 /2

,

z > 0.

(b) When n = 1, we obtain the folded normal density, fZ1 (z) = z

(z2 /2)−1/2 e−z Γ(1/2)

2 /2

2

e−z /2 = 2√ , 2π

z > 0.

Chapter 5 Problem Solutions

69

(c) When n = 2, we obtain the Rayleigh(1) density, fZ2 (z) = z

(z2 /2)0 e−z Γ(1)

2 /2

= ze−z

2 /2

,

z > 0.

(d) When n = 3, we obtain the Maxwell density, (z2 /2)1/2 e−z fZ3 (z) = z Γ(3/2)

2 /2

2

z2 e−z /2 = √ ·1 1 = 2 2 Γ( 2 )

r

2 2 −z2 /2 z e , π

z > 0.

(e) When n = 2m, we obtain the Nakagami-m density, fZ2m (z) = z

(z2 /2)m−1 e−z Γ(m)

2 /2

=

2 2 z2m−1 e−z /2 , 2m Γ(m)

z > 0.

20. Let Z := X1 + √· · · + Xn . By Problem 55(a) in Chapter 4, Z ∼ N(0, n), and it follows that V := Z/ n ∼ N(0, 1). We can now write Y = Z 2 = nV 2 . By Problem 11, V 2 is chi-squared with one degree of freedom. Hence, FY (y) = P(nV 2 ≤ y) = P(V 2 ≤ y/n), and

21.

e−(y/n)/2 e−(y/n)/2 1 , fY (y) = fV 2 (y/n)/n = p · = √ 2π ny 2π y/n n

y > 0.

(a) Since FY (y) = P(Y ≤ y) = P(X 1/q ≤ y) = P(X ≤ yq ) = FX (yq ), q

q

q−1

fY (y) = fX (y ) · (qy

q−1

) = qy

q

(yq ) p−1 e−y qyqp−1 e−y · = , Γ(p) Γ(p)

y > 0.

q

(b) Since q > 0, as y → 0, yq → 0, and e−y → 1. Hence, the behavior of fY (y) as y → 0 is determined by the behavior of y p−1 . For p > 1, p − 1 > 0, and so y p−1 → 0. For p = 1, y p−1 = y0 = 1. For 0 < p < 1, p − 1 < 0 and so y p−1 = 1/y1−p → ∞. Thus,  0, p > 1,  q/Γ(1/q), p = 1, lim fY (y) =  y→0 ∞, 0 < p < 1. (c) We begin with the given formula

q

fY (y) =

λ q(λ y) p−1 e−(λ y) , Γ(p/q)

y > 0.

(i) Taking q = p and replacing λ with λ 1/p yields fY (y) = λ 1/p p(λ 1/p y) p−1 e−(λ p

= λ py p−1 e−λ y , which is the Weibull(p, λ ) density.

1/p y) p

= λ 1/p pλ 1−1/p y p−1 e−λ y

p

70

Chapter 5 Problem Solutions √ (ii) Taking p = q = 2 and replacing λ with 1/( 2 λ ) yields √ √ √ 2 2 fY (y) = 2/( 2 λ )[y/( 2 λ )]e−[y/( 2 λ )] = (y/λ 2 )e−(y/λ ) /2 , which is the required Rayleigh density. √ (iii) Taking p = 3, q = 2, and replacing λ with 1/( 2 λ ) yields √ √ √ 2 2 2/( 2 λ )[y/( 2 λ )]2 e−[y/( 2 λ )] (y2 /λ 3 )e−(y/λ ) /2 √ 1 1 fY (y) = = Γ(3/2) 2 · 2 Γ( 2 ) r 2 2 y −(y/λ )2 = e , π λ3 which is the required Maxwell density. (d) In

Z ∞

E[Y n ] =

q

yn

0

λ q(λ y) p−1 e−(λ y) dy, Γ(p/q)

make the change of variable t = (λ y)q , dt = q(λ y)q−1 λ dy. Then Z

∞ q 1 (λ y)n (λ y) p−q e−(λ y) λ q(λ y)q−1 dy n Γ(p/q)λ 0 Z ∞ 1 = (t 1/q )n+p−q e−t dt Γ(p/q)λ n 0 Z ∞ Γ((n + p)/q) 1 t (n+p)/q−1 e−t dt = . = Γ(p/q)λ n 0 Γ(p/q)λ n

E[Y n ] =

(e) We use the same change of variable as in part (d) to write FY (y) = =

q Z y λ q(λ θ ) p−1 e−(λ θ )

Γ(p/q)

0

1 Γ(p/q)

Z (λ y)q 0

dθ =

1 Γ(p/q)

Z (λ y)q 0

(t 1/q ) p−q e−t dt

t (p/q)−1 e−t dt = G p/q ((λ y)q ). 2

22. Following the hint, let u = t −1 and dv = te−t /2 dt so that du = −1/t 2 dt and v = 2 −e−t /2 . Then 2 Z ∞ −t 2 /2 Z ∞ e e−t /2 ∞ −t 2 /2 − dt dt = − e t t2 x x x 2

e−x /2 = − x 2

<

e−x /2 . x

Z ∞ −t 2 /2 e x

t2

dt

The next step is to write the integral in (∗) as Z ∞ −t 2 /2 e x

t2

dt =

Z ∞ 1 x

t3

· te−t

2 /2

dt

(∗)

Chapter 5 Problem Solutions

71 2

and apply integration by parts with u = t −3 and dv = te−t /2 dt so that du = −3/t 4 dt 2 and v = −e−t /2 . Then 2 Z ∞ −t 2 /2 Z ∞ −t 2 /2 e e−t /2 ∞ e dt = − 3 − 3 dt 2 t t t4 x x x 2

=

e−x /2 −3 x3

Z ∞ −t 2 /2 e x

t4

dt.

Substituting this into (∗), we find that Z ∞ x

2

−t 2 /2

e

2

≥ 23.

2

e−x /2 e−x /2 dt = − +3 x x3 2

e−x /2 e−x /2 − . x x3

Z ∞ −t 2 /2 e x

t4

dt

(a) Write E[FXc (Z)] = = = = = = =

Z ∞

FXc (z) fZ (z) dz −∞  Z ∞ Z ∞

fX (x) dx fZ (z) dz  I(z,∞) (x) fX (x) dx fZ (z) dz −∞ Z ∞  fX (x) I(z,∞) (x) fZ (z) dz dx −∞ Z ∞  fX (x) I(−∞,x) (z) fZ (z) dz dx −∞  Z x fZ (z) dz dx fX (x)

−∞ z Z ∞ Z ∞ −∞

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

−∞

fX (x)FZ (x) dx = E[FZ (X)].

(b) From Problem 15(c) in Chapter 4, we have m−1

FZ (z) = 1 −

∑

k=0

(λ z)k e−λ z , k!

for z ≥ 0,

and FZ (z) = 0 for z < 0. Hence, 

  m−1 (λ X)k e−λ X E[FZ (X)] = E I[0,∞) (X) 1 − ∑ k! k=0 m−1

= P(X ≥ 0) −

∑

k=0

λk E[X k e−λ X I[0,∞) (X)]. k!

72

Chapter 5 Problem Solutions (c) Let X ∼ N(0, 1) so that E[Q(Z)] = E[FXc (Z)] = E[FZ (X)]. Next, since Z ∼ exp(λ ) = Erlang(1, λ ), we can use the result of part (b) to write E[FZ (X)] = P(X ≥ 0) − E[e−λ X I[0,∞) (X)] 1 = − 2

Z ∞ 0

−x2 /2 −λ x e

2

=

1 eλ /2 −√ 2 2π 2

=

1 eλ /2 −√ 2 2π 2

=

√ dx 2π

e

1 eλ /2 −√ 2 2π

Z ∞ 0

Z ∞ 0

Z ∞ λ

e−(x

2 +2xλ +λ 2 )/2

e−(x+λ ) e−t

2 /2

2 /2

dx

dx

dt =

2 1 − eλ /2 Q(λ ). 2

√ (d) Put Z := σ Y and make the following observations. First, for z ≥ 0, √ FZ (z) = P(σ Y ≤ z) = P(σ 2Y ≤ z2 ) = P(Y ≤ (z/σ )2 ) = FY ((z/σ )2 ), and FZ (z) = 0 for z < 0. Second, since Y is chi-squared with 2m-degrees of freedom, Y ∼ Erlang(m, 1/2). Hence, m−1

FZ (z) = 1 −

∑

k=0

((z/σ )2 /2)k e−(z/σ ) k!

2 /2

,

for z ≥ 0.

Third, with X ∼ N(0, 1), √ E[Q(σ Y )] = E[Q(Z)] = E[FXc (Z)] = E[FZ (X)] is equal to   2 m−1 ((X/σ )2 /2)k e−(X/σ ) /2 , E I[0,∞) (X) 1 − ∑ k! k=0 

which simplifies to m−1

P(X ≥ 0) −

∑

k=0

2 1 E[X 2k e−(X/σ ) /2 I[0,∞) (X)]. σ 2k 2k k!

e 2 := (1 + 1/σ 2 )−1 , we have Now, P(X ≥ 0) = 1/2, and with σ 2k −(X/σ )2 /2

E[X e

I[0,∞) (X)] =

Z ∞ 0

−x2 /2 2k −(x/σ )2 /2 e

x e

Z

√ dx 2π

∞ 2 σe = √ x2k e−(x/σe ) /2 dx 2π σe 0 Z ∞ 2 σe 1 x2k e−(x/σe ) /2 dx = ·√ 2 e −∞ 2π σ σe e 2k . = · 1 · 3 · 5 · · · (2k − 1) · σ 2 Putting this all together, the desired result follows.

Chapter 5 Problem Solutions

73

(e) If m = 1 in part (d), then Z defined in solution of part (d) is Rayleigh(σ ) by the same argument as in the solution of Problem 19(b). Hence, the desired result follows by taking m = 1 in the result of part (d). (f) Since the Vi are independent exp(λi ), the mgf of Y := V1 + · · · +Vm is m

λk

∏ λk − s

MY (s) =

m

=

k=1

λk

∑ ck λk − s ,

k=1

where the ck are the result of expansion by partial fractions. Inverse transforming term by term, we find that m

∑ ck · λk e−λk y ,

fY (y) =

k=1

y ≥ 0.

Using this, we can write m

∑ ck (1 − e−λk y ),

FY (y) =

k=1

Next, put Z :=

y ≥ 0.

√ Y , and note that for z ≥ 0, √ FZ (z) = P( Y ≤ z) = P(Y ≤ z2 ) = FY (z2 ).

Hence,

m

FZ (z) =

2

∑ ck (1 − e−λk z ),

k=1

z ≥ 0.

We can now write that E[Q(Z)] = E[FXc (Z)] = E[FZ (X)] is equal to   m  −λk X 2 E I[0,∞) (X) ∑ ck (1 − e ) . k=1

ek2 := (1 + 2λk )−1 , Now observe that with σ −λk X 2

E[I[0,∞) (X)e

] =

Z ∞ 0

−x2 /2 −λk x2 e

e

1 √ dx = 2 2π

Z

2

Z ∞

−∞

σek 1 ∞ e−(1+2λk )x /2 √ dx = 2 −∞ 2 2π 1 σek . = p = 2 2 1 + 2λk

=

−x2 /2 −λk x2 e

e

√

2π

dx

Z ∞ −(x/σek )2 /2 e −∞

√

ek 2π σ

dx

Putting this all together, the desired result follows.

24. Recall that the noncentral chi-squared density with k degrees of freedom and noncentrality parameter λ 2 is given by ∞

ck,λ 2 (t) :=

(λ 2 /2)n e−λ n! n=0

∑

2 /2

c2n+k (t),

t > 0,

74

Chapter 5 Problem Solutions where c2n+k denotes the central chi-squared density with 2n + k degrees of freedom. Hence, Ck,λ 2 (x) =

Z x 0

ck,λ 2 (t) dt =

∞

= 25.

(λ 2 /2)n e−λ ∑ n! n=0

2 Z x ∞ (λ 2 /2)n e−λ /2

∑

n!

0 n=0

2 /2

Z x 0

c2n+k (t) dt

∞

c2n+k (t) dt =

(λ 2 /2)n e−λ ∑ n! n=0

2 /2

C2n+k (x).

(a) Begin by writing √ FZn (z) = P( Yn ≤ z) = P(Yn ≤ z2 ) = FYn (z2 ). Then fZn (z) = fYn (z2 ) · 2z. Next observe that ∞

(mz/2)2`+n/2−1

∑ `!Γ(` + (n/2) − 1 + 1)

In/2−1 (mz) =

∞

=

`=0

(mz/2)2`+n/2−1 . `=0 `!Γ(` + n/2)

∑

From Problem 65 in Chapter 4, 2 /2

∞

fYn (y) :=

(m2 /2)` e−m `! `=0

∑

2 /2

∞

=

(m2 /2)` e−m ∑ `! `=0

c2`+n (y) · 21

(y/2)(2`+n)/2−1 e−y/2 . Γ((2` + n)/2)

So, 2 +z2 )/2

fZn (z) = fYn (z2 ) · 2z = 2ze−(m 2 +z2 )/2

= ze−(m =

· 21

∞

(m2 /2)` (z2 /2)`+n/2−1 `!Γ(` + n/2) `=0

∑

∞

(mz)2`+n/2−1 m−n/2+1 (1/2)2`+n/2−1 zn/2−1 `!Γ(` + n/2) `=0

∑

zn/2 −(m2 +z2 ) e In/2−1 (mz). mn/2−1

(b) Obvious. (c) Begin by writing FYn (y) = P(Zn2 ≤ y) = P(Zn ≤

√ y ) = FZn (y1/2 ).

Then fYn (y) = fZn (y1/2 ) · y−1/2 /2

(y1/2 )n/2 −(m2 +y)/2 √ e In/2−1 (m y )y−1/2 m√n/2−1  y n/2−1 2 √ = 12 e−(m +y)/2 In/2−1 (m y ). m =

1 2

Chapter 5 Problem Solutions

75

(d) First write Z ∞

t n/2 −(m2 +t 2 )/2 e In/2−1 (mt) dt z mn/2−1 Z ∞ (mt)n/2 −(m2 +t 2 )/2 e = In/2−1 (mt) dt. mn−1 z

FZcn (z) =

Now apply integration by parts with dv = mte−t

u = (mt)n/2−1 In/2−1 (mt) and

2 /2

2 /2

e−m

/mn−1 dt.

Then 2 /2

v = −me−m

e−t

2 /2

/mn−1 , and by the hint, du = (mt)n/2−1 In/2−2 (mt) · m dt.

Thus, FZcn (z) = −(mt)n/2−1 In/2−1 (mt) +

Z ∞ (mt)n/2−1 −(m2 +t 2 )/2 I e

n/2−2 (mt) dt

mn−3

z

=

2 2 e−(m +t )/2 ∞ mn−2 z

 z n/2−1 m

2 +z2 )/2

e−(m

In/2−1 (mz) + FZn−2 (z).

(e) Using induction, this is immediate from part (d). e z) = e−z (f) It is easy to see that Q(0, z) = Q(0,

2 /2

. We then turn to

∂ e ∂ −(m2 +z2 )/2 ∞ Q(m, z) = e ∑ (m/z)k Ik (mz) ∂m ∂m k=0  ∞  −(m2 +z2 )/2 k −2k k = e −m(m/z) I (mz) + z (mz) I (mz)z k k−1 ∑ −(m2 +z2 )/2

= e

k=0 ∞ 

∑

k

−m(m/z) Ik (mz) + (m/z) Ik−1 (mz)z

k=0 ∞ 2 2 −(m +z )/2

k

∑ (m/z)

= ze

2 +z2 )/2

= ze−(m

k

k=0 ∞



 Ik−1 (mz) − (m/z)Ik (mz)



∑ (m/z)k Ik−1 (mz) − (m/z)k+1 Ik (mz)

k=0 −(m2 +z2 )/2

= ze

2 +z2 )/2

I−1 (mz) = ze−(m

I1 (mz).

To conclude, we compute

∂Q ∂ = ∂m ∂m =

Z ∞ z

Z ∞ z

2 +t 2 )/2

te−(m

2 +t 2 )/2

−mte−(m

I0 (mt) dt

I0 (mt) dt +

Z ∞ z

2 +t 2 )/2

te−(m

I−1 (mt) · t dt.

76

Chapter 5 Problem Solutions Write this last integral as Z ∞ z

2 +t 2 )/2

(mt)I1 (mt) · (t/m)e−(m

dt. 2

2

Now apply integration by parts with u = (mt)I1 (mt) and dv = te−(t +m )/2 /m dt. 2 2 Then du = (mt)I0 (mt)m dt and v = −e−(m +t )/2 /m, and the above integral is equal to Z 2 +z2 )/2

ze−(m

∞

I1 (mz) +

z

2 +t 2 )/2

mte−(m

I0 (mt) dt.

2 +z2 )/2

Putting this all together, we find that ∂ Q/∂ m = ze−(m ∞

26. Recall that Iν (x) :=

I1 (mz).

(x/2)2`+ν

∑ `!Γ(` + ν + 1) .

`=0

(a) Write ∞ (x/2)2` 1 1 Iν (x) +∑ → = ν (x/2) Γ(ν + 1) `=1 `!Γ(` + ν + 1) Γ(ν + 1)

as x → 0.

Now write zn/2 −(m2 +z2 )/2 e In/2−1 (mz) mn/2−1 In/2−1 (mz) 2 2 zn/2 (mz/2)n/2−1 = n/2−1 e−(m +z )/2 m (mz/2)n/2−1 In/2−1 (mz) 2 2 zn−1 = n/2−1 e−(m +z )/2 . 2 (mz/2)n/2−1

fZn (z) =

Thus,  2 0, n > 1,  e−m /2 2 p lim fZn (z) = n/2−1 lim zn−1 = e−m /2 2/π , n = 1,  z→0 2 Γ(n/2) z→0 ∞, 0 < n < 1.

(b) First note that

∞

Iν −1 (x) =

(x/2)2`+ν −1 = `=0 `!Γ(` + ν )

∑

∞

(` + ν )(x/2)2`+ν −1 . `!Γ(` + ν + 1) `=0

∑

Second, with the change of index ` = k + 1, ∞

Iν +1 (x) =

(x/2)2k+ν +1 ∑ k!Γ(k + ν + 2) = k=0 ∞

`(x/2)2`+ν −1 = = ∑ `=1 `!Γ(` + ν + 1)

∞

(x/2)2`+ν −1

∑ (` − 1)!Γ(` + ν + 1)

`=1 ∞

`(x/2)2`+ν −1

∑ `!Γ(` + ν + 1) .

`=0

Chapter 5 Problem Solutions

77

It is now easy to see that ∞

(2` + ν )(x/2)2`+ν −1 ∑ `!Γ(` + ν + 1) = 2Iν0 (x). `=0

Iν −1 (x) + Iν +1 (x) = It is similarly easy to see that

∞

ν (x/2)2`+ν −1

∑ `!Γ(` + ν + 1)

Iν −1 (x) − Iν +1 (x) =

= 2(ν /x)Iν (x).

`=0

R

(c) To the integral I˜n (x) := (2π )−1 −ππ ex cos θ cos(nθ ) d θ , apply integration by parts with u = ex cos θ and dv = cos nθ d θ . Then du = ex cos θ (−x sin θ ) d θ and v = sin(nθ )/n. We find that   Z 1 x cos θ sin nθ π x π x cos θ e sin n θ sin θ d θ I˜n (x) = e . + 2π n −π n −π | {z } =0

We next use the identity sin A sin B = 12 [cos(A − B) − cos(A + B)] to get x ˜ I˜n (x) = [In−1 (x) − I˜n+1 (x)]. 2n

(d) Since I˜0 (x) := (1/π ) Then 1 I˜0 (x) = π =

1 π

R π x cos θ d θ , make the change of variable t = θ − π /2. 0 e

Z π /2

−π /2

ex cos(t+π /2) dt =

Z π /2 ∞ (−x sint)k

∑

k!

−π /2 k=0

1 ∞ x2` = ∑ (2`)! π `=0

1 π

Z π /2

dt =

−π /2 ∞

e−x sint dt

(−x)k 1 ∑ π k=0 k!

Z π /2

Z π /2

|

Z

sink t dt {z }

−π /2

= 0 for k odd

π /2 2 ∞ x2` sin2` t dt sin2` t dt = ∑ π `=0 (2`)! 0 −π /2   2` + 1 √ π Γ 2 ∞ x2` 2   = , by Problem 18 in Ch. 4. · ∑ (2`)! 2` + 2 π `=0 2Γ 2

Now, (2`)! = 1 · 3 · 5 · · · (2` − 1) · 2 · 4 · 6 · · · 2` = 1 · 3 · 5 · · · (2` − 1) · 2` `!, and from Problem 14(c) in Chapter 4,   2` + 1 1 · 3 · 5 · · · (2` − 1) √ Γ = π. 2 2` Hence,

∞

I˜0 (x) =

(x/2)2`

∑ `!Γ(` + 1)

`=0

=: I0 (x).

78

Chapter 5 Problem Solutions (e) Begin by writing Z

1 π x cos θ e cos([n ± 1]θ ) d θ 2π −π Z π 1 = ex cos θ [cos nθ cos θ ∓ sin nθ sin θ ] d θ . 2π −π

In±1 (x) =

Then 1 2 [In−1 (x) + In+1 (x)]

Since In0 (x)

=

1 2π

Z π

ex cos θ cos nθ cos θ d θ .

−π

  Z Z 1 π x cos θ ∂ 1 π x cos θ e cos(nθ ) d θ = e cos(nθ ) cos θ d θ , = ∂ x 2π −π 2π −π

we see that 12 [In−1 (x) + In+1 (x)] = In0 (x). 27. MATLAB.

0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

28. See previous problem solution for graph. The probabilities are: k 0 1 2 3 4 29.

P(X = k) 0.0039 0.0469 0.2109 0.4219 0.3164

(a) Sketch of f (t): 1

1/2

1/3 1/6 0

1

Chapter 5 Problem Solutions (b) P(X = 0) = (c) We have

R

{0}

f (t) dt = 1/2, and P(X = 1) = Z 1−

P(0 < X < 1) =

0+ Z 1−

=

0+

Z 1−

f (t) dt = 1 −t 3 e dt

=

0+ Z 1 0

R

{1}

Z ∞

f (t) dt =

1+

Z ∞ 1

1 −t 3 e dt

1 −t 3 e dt

(d) Write

f (t) dt = 1/6.

1 −t 1 1 3 e u(t) + 2 δ (t) + 6 δ (t − 1) dt

and P(X > 1) =

79

1 1 − e−1 = − 31 e−t = 3 0

∞ = − 31 e−t = e−1 /3. 1

P(0 ≤ X ≤ 1) = P(X = 0) + P(0 < X < 1) + P(X = 1) =

1 1 − e−1 1 + + = 1 − e−1 /3 2 3 6

and P(X > 1) = P(X = 1) + P(X > 1) = 1/6 + e−1 /3 =

1 + 2e−1 . 6

(e) Write E[X] = =

Z ∞

−∞

t f (t) dt =

Z ∞ 0

1 3 E[exp

t −t 3 e dt + 0 · P(X

= 0) + 1 · P(X = 1)

RV w/λ = 1] + 1/6 = 1/3 + 1/6 = 1/2. R

∞ x 1 30. For the first part of the problem, we have E[eX ] = −∞ e · 2 [δ (x) + I(0,1] (x)] dx = 1 R 1 1 x 1 0 x 2 e + 2 0 e dx = 1/2 + e /2 0 = 1/2 + (e − 1)/2 = e/2. For the second part, first write

P(X = 0|X ≤ 1/2) =

P({X = 0} ∩ {X ≤ 1/2}) P(X = 0) = . P(X ≤ 1/2) P(X ≤ 1/2)

Since P(X = 0) = 1/2 and P(X ≤ 1/2) =

Z 1/2 −∞

1 2 [δ (x) + I(0,1] (x)] dx

we have P(X = 0|X ≤ 1/2) =

1/2 3/4

= 12 + 21

Z 1/2 0

dx = 21 + 41 = 34 ,

= 2/3. R

∞ 31. The approach is to find the density and then compute E[X] = −∞ x fX (x) dx. The catch is that the cdf has a jump at x = 1/2, and so the density has an impulse there. Put   2x, 0 < x < 1/2, 1, 1/2 < x < 1, f˜X (x) :=  0, otherwise.

80

Chapter 5 Problem Solutions Then the density is fX (x) = f˜X (x) + 41 δ (x − 1/2). Hence, E[X] =

Z ∞

−∞

x fX (x) dx = = =

Z 1/2 0

x · 2x dx +


2 1 3 + 12 [1 − 3 2 1 3 1 12 + 8 + 8 =

Z 1

x · 1 dx + 41 · 12

1/2
1 1 2 2 ]+ 8 11 1 24 + 8 =

7/12.

√ R∞ √ x fX (x) dx. The 32. The approach is to find the density and then compute E[ X ] = −∞ catch is that the cdf has a jump at x = 4, and so the density has an impulse there. Put   x−1/2 /8, 0 < x < 4, f˜X (x) := 1/20, 4 < x < 9,  0, otherwise. Then the density is fX (x) = f˜X (x) + 14 δ (x − 4). To begin, we compute Z ∞

−∞

x1/2 f˜X (x) dx =

Z 4

1/8 dx +

0

Z 9 4

9 x1/2 /20 dx = 1/2 + x3/2 /30 4

= 1/2 + (27 − 8)/30 = 1/2 + 19/30 = 17/15.

The complete answer is √ √ E[ X ] = 17/15 + 4/4 = 17/15 + 1/2 = 34/30 + 15/30 = 49/30. 33. First note that for y < 0, Z y

−∞

and for y ≥ 0, Z y

−∞

e−|t| dt =

Hence,

Z 0

−∞

et dt +

e−|t| dt =

Z y 0

Z y

−∞

et dt = ey ,

y e−t dt = 1 + (−e−t ) = 1 + 1 − e−y = 2 − e−y . 0

 y y < 0,  e /4, (2 − e−y )/4 + 1/3, 0 ≤ y < 7, FY (y) =  (2 − e−y )/4 + 1/3 + 1/6, y ≥ 7,  y y < 0,  e /4, 5/6 − e−y /4, 0 ≤ y < 7, =  1 − e−y /4, y ≥ 7.

1

0.75 0.5 0.25 0

−2 0 2 4 6 8

Chapter 5 Problem Solutions

81

34. Let {Y = 0} = {loose connection}, and {Y = 1} = {Y = 0} c . Then P(Y = 0) = P(Y = 1) = 1/2. Using the law of total probability, 1

FX (x) = P(X ≤ x) =

∑ P(X ≤ x|Y = i)P(Y = i)

i=0

 Z 1 = 2 P(X ≤ x|Y = 0) +

x

−∞

 I(0,1] (t) dt .

Since P(X = 0|Y = 0) = 1, we see that  1, x ≥ 1,  1 FX (x) = (1 + x), 0 ≤ x < 1, 2 0, x < 0.

Since there is a jump at x = 0, we must be careful in computing the density. It is fX (x) =

1

1

1/2

1/2

0

1

cdf

35.

1 2 [I(0,1) (x) + δ (x)].

0

1

density

(a) Recall that as x varies from +1 to −1, cos−1 x varies from 0 to π . Hence,
FX (x) = P(cos Θ ≤ x) = P Θ ∈ [−π , −θx ] ∪ [θx , π ] , where θx := cos−1 x. Since Θ ∼ uniform[−π , π ], FX (x) =

cos−1 x π − θx −θx − (−π ) π − θx + = 2 = 1− . 2π 2π 2π π

(b) Recall that as y varies from +1 to −1, sin−1 y varies from π /2 to −π /2. For y ≥ 0,
FY (y) = P(sin Θ ≤ y) = P Θ ∈ [−π , θy ] ∪ [π − θy , π ] =

θy θy − (−π ) π + 2θy 1 sin−1 y + = = + , 2π 2π 2π 2 π

and for y < 0, FY (y) = P(sin Θ ≤ y) = P(Θ ∈ [−π − θy , θy ]) =

2θy + π 1 sin−1 y = + . 2π 2 π

82

Chapter 5 Problem Solutions (c) Write fX (x) = − and fY (y) = (d) First write

−1 1 1/π ·√ = √ , π 1 − x2 1 − x2

1/π 1 1 = p . ·p 2 π 1−y 1 − y2

Y + 1  FZ (z) = P(Z ≤ z) = P ≤ z = P(Y ≤ 2z − 1) = FY (2z − 1). 2 Then differentiate to get fZ (z) = fY (2z − 1) · 2 = =

Γ( 12 + 21 ) Γ( 12 )Γ( 21 )

1 2 p = √ √ p 2 π π z(1 − z) π 1 − (2z − 1)

z1/2−1 (1 − z)1/2−1 ,

which is the beta density with p = q = 1/2. 36. The cdf is

FY (y) =

and the density is

37. The cdf is

and the density is

   Z

1,

√ −1+ 1+y

y ≥ 3,

1/4 dx, −1 ≤ y < 3, √    −1− 1+y 0, y < −1,  y ≥ 3,  √1, 1 = 1 + y, −1 ≤ y < 3, 2 0, y < −1,

  √1 , −1 < y < 3, fY (y) = 4 1+y  0, otherwise.

 1,   p    2(3 − 2/y ) , FY (y) = 6   1/3,    0, fY (y) =

√

y ≥ 2, 1/2 ≤ y < 2, 0 ≤ y < 1/2, y < 0,

1 1 2 −3/2 y I(1/2,2) (y) + δ (y) + δ (y − 2). 6 3 3

Chapter 5 Problem Solutions

83

38. For 0 ≤ y ≤ 1, we first compute the cdf p FY (y) = P(Y ≤ y) = P( 1 − R2 ≤ y) = P(1 − R2 ≤ y2 ) = P(1 − y2 ≤ R2 ) p √ p 1 2 − 1 − y2 2 √ = P( 1 − y ≤ R) = = 1 − √ (1 − y2 )1/2 . 2 2 We then differentiate to get the density  p y , 0 < y < 1, 2(1 − y2 ) fY (y) =  0, otherwise.

√ 39. The first thing to note is that for 0 ≤ R ≤ 2, 0 ≤ R2 ≤ 2. It is then easy√to see that the minimum value of Z = [R2 (1 − R2 /4)]−1 occurs when R2 = 2 or R = 2. Hence, the random variable Z takes values in the range [1, ∞). So, for z ≥ 1, we write     1 2 2 FZ (z) = P 2 ≤ z = P R (1 − R /4) ≥ 1/z R (1 − R2 /4)   = P (R2 /4)(1 − R2 /4) ≥ 1/(4z) . Put y := 1/(4z) and observe that x(1 − x) ≥ y if and only if 0 ≥ x2 − x + y, with equality if and only if √ 1 ± 1 − 4y . x = 2 Since we will have x = R2 /4 ≤ 1/2, we need the negative root. Thus, √     p 1 − 1 − 4y 2 2 = P R ≥ 2[1 − 1 − 1/z ] FZ (z) = P (R /4) ≥ 2 q p √   q 2 − 2[1 − 1 − 1/z ] p √ = P R ≥ 2[1 − 1 − 1/z ] = 2 q p = 1 − 1 − 1 − 1/z. Differentiating, we obtain

p p
−1/2 d 1 [1 − 1 − 1/z ] 1 − 1 − 1/z 2 dz p
−1/2 1 1 1 = − 1 − 1 − 1/z · − (1 − 1/z)−1/2 · 2 2 2 z p 1 = 2 [(1 − 1 − 1/z )(1 − 1/z)]−1/2 . 4z √ √ √ 40. First note that as R varies from 0 to 2, T varies from π to 2 π . For π ≤ t ≤ 2 π , write    2  π π 2 FT (t) = P(T ≤ t) = P p ≤ t = P 2 ≤ 1 − R /4 t 1 − R2 /4 fZ (z) = −

84

Chapter 5 Problem Solutions = P(R2 /4 ≤ 1 − π 2 /t 2 ) = P(R2 ≤ 4[1 − π 2 /t 2 ]) q   √ = P R ≤ 2 1 − π 2 /t 2 = 2(1 − π 2 /t 2 )1/2 .

Differentiating, we find that

√ √ π2 2 π2 2 2 2 −1/2 √ fT (t) = . = (1 − π /t ) t3 t2 t2 − π2 For the second part of the problem, observe that as R varies between 0 and varies between 1 and e−π . For m in this range, note that ln m < 0, and write √ 2 FM (m) = P(M ≤ m) = P(e−π (R/2)/ 1−R /4 ≤ m)  .q  = P π (R/2) 1 − R2 /4 ≥ − ln m .   = P π 2 (R2 /4) (1 − R2 /4) ≥ (− ln m)2   = P (R2 /4) ≥ 1/[1 + {π /(− ln m)}2 ]   = P R ≥ 2[1 + {π /(− ln m)}2 ]−1/2 √ = 1 − 2[1 + {π /(− ln m)}2 ]−1/2 ,

√ 2, M

Differentiating, we find that √ 2 √ 2 2π 2π 2 −3/2 . [1 + { π /(− ln m)} ] = fM (m) = m(− ln m)3 m[(− ln m)2 + π 2 ]3/2 41.

(a) For X ∼ uniform[−1, 1], FY (y) = and fY (y) =



(b) For X ∼ uniform[−1, 2],

and

 

1, y ≥ 1, √ (y + y )/2, 0 ≤ y < 1,  0, y < 0, √ 1 2 [1 + 1/(2 y )], 0,

   

1, (y + 1)/3, √ FY (y) = (y + y )/3,    0, fY (y) =

  

1/3,

y ≥ 2, 1 ≤ y < 2, 0 ≤ y < 1, y < 0,

√ 1 3 [1 + 1/(2 y )], 0,

0 < y < 1, otherwise.

1
Chapter 5 Problem Solutions (c) For X ∼ uniform[−2, 3],

1, (y + 2)/5, √ FY (y) = (y + y )/5,    0,

and fY (y) =

y ≥ 2, 1 ≤ y < 2, 0 ≤ y < 1, y < 0,

1 √ [I (y) + {1 + 1/(2 y )}I(0,1) (y) + δ (y − 1) + δ (y − 2)]. 5 (1,2)

(d) For X ∼ exp(λ ),

 

1, y ≥ 2, P(X ≤ y) + P(X ≥ 3), 0 ≤ y < 2,  0, y < 0,  1, y ≥ 2,  = (1 − e−λ y ) + e−3λ , 0 ≤ y < 2,  0, y < 0,

FY (y) =

and 42.

   

85

fY (y) = λ e−λ y I(0,2) (y) + e−3λ δ (y) + (e−2λ − e−3λ )δ (y − 2).

(a) If X ∼ uniform[−1, 1], then Y = g(X) = 0, and so FY (y) = u(y) (the unit step function), (b) If X ∼ uniform[−2, 2], FY (y) = and

  

fY (y) = (c) We have FY (y) = and fY (y) = (d) If X ∼ Laplace(λ ),

1, 1 2 (y + 1),

0,

and

fY (y) = δ (y).

y ≥ 1, 0 ≤ y < 1, y < 0,

1 [I (y) + δ (y)]. 2 (0,1)

 

1, y ≥ 1, (y + 1)/3, 0 ≤ y < 1,  0, y < 0,

1 3 [I(0,1) (y) + δ (y) + δ (y − 1)].

 y ≥ 1,   R y+1 1, λ −λ x dx, 0 ≤ y < 1, 2 0 2e FY (y) =   0, y < 0,  1, y ≥ 1,  = 1 − e−λ (y+1) , 0 ≤ y < 1,  0, y < 0,

86

Chapter 5 Problem Solutions and

43.

fY (y) = λ e−λ (y+1) I(0,1) (y) + (1 − e−λ )δ (y) + e−2λ δ (y − 1).

(a) If X ∼ uniform[−3, 2], FY (y) =

and fY (y) =

      

FY (y) =

and fY (y) =

      

FY (y) =

and fY (y) =

fY (y) =

1, 1 1/3 + y + 2), 4 (y 1 [y + 2 − (−y)1/2 ], 4

0,

y ≥ 1, 0 ≤ y < 1, −1 ≤ y < 0, y < −1,

√ 2/3 1 )]I(0,1) (y) + 41 [1 + 1/(2 −y )]I(−1,0) (y). 4 [1 + 1/(3y

(c) If X ∼ uniform[−1, 1],

and

0,

y ≥ 1, 0 ≤ y < 1, −1 ≤ y < 0, y < −1,

√ 2/3 1 )]I(0,1) (y) + 15 [1 + 1/(2 −y )]I(−1,0) (y) + 15 δ (y − 1). 5 [1 + 1/(3y

(b) If X ∼ uniform[−3, 1],

44. We have

1, 1 1/3 + y + 2), 5 (y 1 [y + 2 − (−y)1/2 ], 5

      

1, 1 1/3 + 1), 2 (y 1 1/2 ], [1 − (−y) 2

0,

1 1 I(0,1) (y) + √ I(−1,0) (y). 4 −y 6y2/3

 0,    1 [y + 1 + √y + 1], 6 FY (y) = √ 1    6 [3 + y + 1 ], 1, 1 1 6 δ (y − 1) + [ 6

45. We have

y ≥ 1, 0 ≤ y < 1, −1 ≤ y < 0, y < −1,

y < −1, −1 ≤ y < 1, 1 ≤ y < 8, y ≥ 8,

1 1 + 12 (y + 1)−1/2 ]I(−1,1) (y) + 12 (y + 1)−1/2 I(1,8) (y).

FY (y) =

 

fY (y) =

√ 1 1 4 δ (y) + 4 [2y + 1/(2 y) + 1]I(0,1) (y).

and

0, y < 0, √ (y2 + y + y + 1)/4, 0 ≤ y < 1,  1, y ≥ 1,

Chapter 5 Problem Solutions 46. We have FY (y) =

and

      

1, 1 2 6 [4 + 3y − y ], 1 2 6 [3 + 2y − y ],

0,

87

y ≥ 1, 0 ≤ y < 1, −1 ≤ y < 0, y < −1,

fY (y) = [ 12 − 13 y]I(0,1) (y) + 31 [1 − y]I(−1,0) (y) + 61 δ (y). 47. We have FY (y) =

 

fY (y) =

1 3

and



1 3 [1 + y +

1, y ≥ 1, p y/(2 − y) ], 0 ≤ y < 1, 0, y < 0,

   y −1/2 1 δ (y) + 1 + . 2−y (2 − y)2

48. Observe that g is a periodic sawtooth function with period one. Also note that since 0 ≤ g(x) < 1, Y = g(X) takes values in [0, 1). (a) We begin by writing ∞

∞

k=0

k=0

FY (y) = P(Y ≤ y) = P(g(X) ≤ y) = ∑ P(k ≤ X ≤ k+y) = ∑ FX (k+y)−FX (k). When X ∼ exp(1), we obtain, for 0 ≤ y < 1, ∞

FY (y) =

∑ (1 − e−k e−y ) − (1 − e−k )

k=0

∞

=

∑ e−k (1 − e−y )

k=0

=

1 − e−y . 1 − e−1

Differentiating, we get fY (y) =

e−y , 1 − e−1

0 ≤ y < 1.

We say that Y has a “truncated” exponential density. (b) When X ∼ uniform[0, 1), we obtain Y = X ∼ uniform[0, 1).

(c) Suppose X ∼ uniform[ν , ν + δ ), where ν = m + δ for some integer m ≥ 0 and some 0 < δ < 1. For 0 ≤ y < δ FY (y) = (m + 1 + y) − (m + 1) = y, and for δ ≤ y < 1, FY (y) = [(m + 1 + δ ) − (m + 1)] + [(m + y) − (m + δ )] = y. Since FY (y) = y for 0 ≤ y < 1, Y ∼ uniform[0, 1).

88

Chapter 5 Problem Solutions

49. In the derivation of Property (vii) of cdfs, we started with the formula (−∞, x0 ) =

∞ [

(−∞, x0 − n1 ].

∞ [

(−∞, x0 − n1 ).

n=1

However, we can also write (−∞, x0 ) =

n=1

Hence,   ∞ [ 1 G(x0 ) = P(X < x0 ) = P {X < x0 − n } = lim P(X < x0 − N1 ) = lim G(x0 − N1 ). N→∞

n=1

N→∞

Thus, G is left continuous. In a similar way, we can adapt the derivation of Property (vi) of cdfs to write  ∞ \ {X < x0 + 1n } = lim P(X < x0 + 1n ) = lim G(x0 + 1n ) = G(x0 +). P(X ≤ x0 ) = P N→∞

n=1

N→∞

To conclude, write P(X = x0 ) = P(X ≤ x0 ) − P(X < x0 ) = G(x0 +) − G(x0 ). 50. First note that since FY (t) is right continuous, so is 1−FY (t) = P(Y > t). Next, we use the assumption P(Y > t + ∆t|T > t) = P(Y > ∆t) to show that with h(t) := P(Y > t), h(t + ∆t) = h(t) + h(∆t). To this end, write h(∆t) = ln P(Y > ∆t) = ln P(Y > t + ∆t|Y > t) = ln = ln

P({Y > t + ∆t} ∩ {Y > t}) P(Y > t)

P(Y > t + ∆t) = ln P(Y > t + ∆t) − ln P(Y > t) = h(t + ∆t) − h(t). P(Y > t)

Rewrite this result as h(t + ∆t) = h(t) + h(∆t). Then with ∆t = t, we have h(2t) = 2h(t). With ∆t = 2t, we have h(3t) = h(t) + h(2t) = h(t) + 2h(t) = 3h(t). In general, h(nt) = nh(t). In a similar manner we can show that t t h(t) = h = mh(t/m), +···+ m m

and so h(t/m) = h(t)/m. We now have that for rational a = n/m, h(at) = h(n(t/m)) = nh(t/m) = (n/m)h(t) = ah(t). For general a ≥ 0, let ak ↓ a with ak rational. Then by the right continuity of h, h(at) = lim h(ak t) = lim ak h(t) = ah(t). k→∞

k→∞

We can now write h(t) = h(t · 1) = th(1).

Chapter 5 Problem Solutions

89

Thus, t · h(1) = h(t) = ln P(Y > t) = ln(1 − FY (t)),

and we have 1 − FY (t) = eh(1)t , which implies Y ∼ exp(−h(1)). Of couse, −h(1) = − ln P(Y > 1) = − ln[1 − FY (1)]. 51. We begin with   1 1 n Xi − m = √ E[Yn ] = E √ ∑ n i=1 σ nσ 

n

∑

i=1


E[Xi ] − m = 0.

For the variance, we use the fact that since independent random variables are uncorrelated, the variance of the sum is the sum of the variances. Thus,   n n n Xi − m var(Xi ) σ2 √ var(Yn ) = ∑ var = ∑ = = 1. ∑ 2 2 σ n i=1 i=1 nσ i=1 nσ 52. Let Xi denote the time to transmit the ith packet, where Xi has mean m and variance σ 2 . The total time to transmit n packets is Tn := X1 + · · · + Xn . The expected total time is E[Tn ] = nm. Since we do not know the distribution of the Xi , we cannot know the distribution of Tn . However, we use the central limit theorem to approximate P(Tn > 2nm). Note that the sample mean Mn = Tn /n. Write P(Tn > 2nm) = P( 1n Tn > 2m) = P(Mn > 2m) = P(Mn − m > m)     Mn − m m m √ > √ √ = P = P Yn > σ/ n σ/ n σ/ n √ √ = 1 − FYn (m n/σ ) ≈ 1 − Φ(m n/σ ), by the central limit theorem. 53. Let Xi = 1 if bit i is in error, and Xi = 0 otherwise. Then P(Xi = 1) = p. Although the problem does not say so, let us assume that the Xi are independent. Then Mn = 1 n n ∑i=1 Xi is the fraction of bits in error. We cannot reliably decode of Mn > t. To approximate the probability that we cannot reliably decode, write       t −m t −m Mn − m t −m √ > √ √ √ = P Yn > = 1 − FYn P(Mn > t) = P σ/ n σ/ n σ/ n σ/ n     t −m t−p √ ≈ 1−Φ , = 1−Φ p σ/ n p(1 − p)/n since m = E[Xi ] = p and σ 2 = var(Xi ) = p(1 − p).

54. If the Xi are i.i.d. Poisson(1), then Tn := X1 + · · · + Xn is Poisson(n). Thus,     nk e−n 1 1 1 k−n·1 2 √ √ . ≈ √ exp − k! 2 1· n 1· n 2π √ √ Taking k = n, we obtain nn e−n ≈ n!/ 2π n or n! ≈ 2π nn+1/2 e−n . P(Tn = k) =

90

Chapter 5 Problem Solutions

55. Recall that gn is the density of X1 + · · · + Xn . If the Xi are i.i.d. uniform[−1, 1], then gn is the convolution of (1/2)I[−1,1] (x) with itself n times. From graphical considerations, it is clear that gn (x) = 0 for |x| > n; i.e., xmax = n. 56. To begin, write √   ϕYn (ν ) = E e jν (X1 +···+Xn )/ n = E



n

∏ e j(ν /

√ n )Xi

i=1



n

=

∏ i=1

 √ 2 ν 2 /2 n = cosn (ν / n ) ≈ 1 − → e−ν /2 . n

57.

h

√ √ i 1 jν / n 1 − jν / n e e + 2 2

(a) MTTF = E[T ] = n (from Erlang in table). (b) The desired probability is R(t) := P(T > t) =

Z ∞ n−1 −τ τ e t

(n − 1)!

dτ .

R

Let Pn (t) denote the above integral. Then P1 (t) = t∞ e−τ d τ = e−τ . For n > 1, apply integration by parts with u = τ n−1 /(n − 1)! and dv = e−τ d τ . Then Pn (t) =

t n−1 e−t + Pn−1 (t). (n − 1)!

Applying this result recursively, we find that Pn (t) =

t n−1 e−t t n−2 e−t + + · · · + e−t , (n − 1)! (n − 2)!

which is the desired result. (c) The failure rate is r(t) =

t n−1 t n−1 e−t /(n − 1)! fT (t) . = = n−1 k n−1 R(t) t t k −t (n − 1)! ∑ ∑ e k=0 k! k=0 k!

For n = 2, r(t) = t/(1 + t): 1

0 0

58.

2

4

6

8

10

√ (a) Let λ =√1. For p = 1/2, r(t) = 1/(2 2). For p = 1, r(t) = 1. For p = 3/2, r(t) = 3 t/2. For p = 2, r(t) = 2t. For p = 3, r(t) = 3t 2 .

Chapter 5 Problem Solutions

p=3

2

p=2 p = 3/2 p=1 p = 1/2

1 0 0

91

1

(b) We have from the text that   Zt   Zt p p−1 d τ = e−λ t . R(t) = exp − r(τ ) d τ = exp − λ pτ 0

0

(c) The MTTF is E[T ] =

Z ∞

R(t) dt =

0

Z ∞ 0

p

e−λ t dt.

Now make the change of variable θ = λ t p , or t = (θ /λ )1/p . Then Z ∞ Z ∞  θ 1/p−1 d θ 1 θ 1/p−1 e−θ d θ E[T ] = e−θ = λ λp pλ 1/p 0 0 Γ(1/p + 1) 1 Γ(1/p) = . = 1/p pλ λ 1/p (d) Using the result of part (b), p

fT (t) = −R0 (t) = λ t p−1 e−λ t , 59.

t > 0.

(a) Write  Z R(t) = exp −

∞

r(τ ) d τ

0



 Zt  −1 = exp − pτ d τ t0

= exp[−p(lnt − lnt0 )] = exp[ln(t0 /t) p ] = (t0 /t) p ,

t ≥ t0 .

(b) If t0 = 1 and p = 2, R(t) = 1/t 2 has the form

1

0 0

1

R(t) dt =

Z ∞

2

3

4

5

(c) For p > 1, the MTTF is E[T ] =

Z ∞ 0

t0

(t0 /t) p dt = t0p

 t 1−p  ∞ t0 . = 1 − p t0 p−1

92

Chapter 5 Problem Solutions (d) For the density, write pt0 p  t0  p · = p+1 , t t t p

fT (t) = r(t)R(t) = 60.

t ≥ t0 .

(a) A sketch of r(t) = t 2 − 2t + 2 for t ≥ 0 is:

6 4 2 0 0

1

2

3

(b) We first compute Z t

r(τ ) d τ =

Z t 0

0

τ 2 − 2τ + 2 d τ =

2 1 3 3 t − t + 2t.

Then fT (t) = r(t)e− 61.

Rt

0 r(τ ) d τ

1 3 −t 2 +2t)

= [t 2 − 2t + 2]e−( 3 t

,

t ≥ 0.

(a) If T ∼ uniform[1, 2], then for 0 ≤ t < 1, R(t) = P(T > t) = 1, and for t ≥ 2, R(t) = P(T > t) = 0. For 1 ≤ t < 2, R(t) = P(T > t) =

Z 2 t

1 d τ = 2 − t.

The complete formula and sketch are   1, 0 ≤ t < 1, 2 − t, 1 ≤ t < 2, R(t) =  0, t ≥ 2.

1

0 0

1

2

3

(b) The failure rate is r(t) = −

d 1 d ln R(t) = − ln(2 − t) = , dt dt 2−t

1 < t < 2.

Chapter 5 Problem Solutions

93

(c) Since T ∼ uniform[1, 2], the MTTF is E[T ] = 1.5. 62. Write R(t) := P(T > t) = P(T1 > t, T2 > t) = P(T1 > t)P(T2 > t) = R1 (t)R2 (t). 63. Write R(t) := P(T > t) = P({T1 > t} ∪ {T2 > t}) = 1 − P(T1 ≤ t, T2 ≤ t) = 1 − P(T1 ≤ t)P(T2 ≤ t) = 1 − [1 − R1 (t)][1 − R2 (t)] = 1 − [1 − R1 (t) − R2 (t) + R1 (t)R2 (t)] = R1 (t) + R2 (t) − R1 (t)R2 (t).

64. We follow the hint and write E[Y n ] = E[T ] =

Z ∞

P(T > t) dt =

0

Z ∞

P(Y n > t) dt =

0

Z ∞

P(Y > t 1/n ) dt.

0

We then make the change of variable y = t 1/n , or t = yn , dt = nyn−1 dy, to get E[Y n ] =

Z ∞ 0

P(Y > y) · nyn−1 dy.

CHAPTER 6

Problem Solutions 1. Since the Xi are uncorrelated with common mean m and common variance σ 2 ,     n 1 E[Sn2 ] = E ∑ Xi2 − nE[Mn ] n−1 i=1 n oi 1 h n(σ 2 + m2 ) − n var(Mn ) + (E[Mn ])2 = n−1  1  = n(σ 2 + m2 ) − n{σ 2 /n + m2 } n−1  1  (n − 1)σ 2 + nm2 − nm2 = σ 2 . = n−1 2.

(a) The mean of a Rayleigh(λ ) random variable is λ

Then

p π /2. Consider

p λn := Mn / π /2.

p p p p π /2 = λ . E[λn ] = E[Mn / π /2 ] = E[Mn ]/ π /2 = λ π /2

Thus, λn is unbiased. Next, p p p p λn = Mn / π /2 → E[Mn ] π /2 = λ π /2 π /2 = λ , and we see that λn is strongly consistent.

(b) MATLAB. Add the line of code lambdan=mean(X)/sqrt(pi/2). p (c) MATLAB. Since Mn ≈ λ π /2, we solve for π and put

πn := 2(Mn /λ )2 .

Since λ = 3, add the line of code pin=2*(mean(X)/3)ˆ 2. 3.

(a) The mean of a gamma(p, λ ) random variable is p/λ . We put pn := λ Mn . Then E[pn ] = λ E[Mn ] = λ · p/λ = p. Also pn = λ Mn → λ (p/λ ) = p. Thus, pn is unbiased and strongly consistent. (b) MATLAB. In this problem λ = 1/2 and p = k/2, or k = 2p. We use kn := 2pn = 2(λ Mn ) = 2((1/2)Mn ) = Mn . We therefore add the line of code kn=mean(X). 94

Chapter 6 Problem Solutions 4.

95

(a) Since the mean of a noncentral chi-squared random variable with k degrees of freedom and noncentrality parameter λ 2 is k + λ 2 , we put

λn2 := Mn − k. Then E[λn2 ] = E[Mn − k] = E[Mn ] − k = (k + λ 2 ) − k = λ 2 , and we see that λn2 is an unbiased estimator of λ 2 . Next, since λn2 = Mn − k → E[Mn ] − k = (k + λ 2 ) − k = λ 2 , the estimator is strongly consistent.

(b) MATLAB. Since k = 5, add the line of code lambda2n=mean(X)-5. 5.

(a) Since the mean of a gamma(p, λ ) random variable is p/λ , we put λn := p/Mn . Then λn = p/Mn → p/E[Mn ] = p/(p/λ ) = λ , and we see that λn is a strongly consistent estimator of λ . (b) MATLAB. Since p = 3, add the line of code lambdan=3/mean(X).

6.

(a) Since the variance of a Laplace(λ ) random variable is 2/λ 2 , we put

λn :=

q

2/Sn2 .

Since Sn2 converges to the variance, we have λn → that λn is a strongly consistent estimator of λ .

p

2/(2/λ 2 ) = λ , and we see

(b) MATLAB. Add the line of code lambdan=sqrt(2/var(X)). 7.

(a) The mean of a gamma(p, λ ) random variable is p/λ . The second moment is p(p + 1)/λ 2 . Hence, the variance is p(p + 1) p2 p − 2 = 2. 2 λ λ λ Thus, Mn ≈ p/λ and Sn2 ≈ p/λ 2 . Solving for p and λ suggests that we put

λn :=

Mn Sn2

and

pn := λn Mn .

 Now, Mn → p/λ and Sn2 → p/λ 2 . It follows that λn → (p/λ ) (p/λ 2 ) = λ and then pn → λ · (p/λ ) = p. Hence, λn is a strongly consistent estimator of λ , and pn is a strongly consistent estimator of p. (b) MATLAB. Add the code Mn = mean(X) lambdan = Mn/var(X) pn = lambdan*Mn

8. Using results from the problem referenced in the hint, we have
Γ (q + p)/q Γ(1 + p/q) p/q = = . E[X q ] = Γ(p/q)λ q Γ(p/q)λ q λq

96

Chapter 6 Problem Solutions This suggests that we put 

λn := Then

λn →



p/q 1 n

q

∑ni=1 Xi

p/q (p/q)/λ q

1/q

1/q

.

= λ,

and we see that λn is a strongly consistent estimator of λ . 9. In the preceding problem E[X q ] = (p/q)/λ q . Now consider
Γ (2q + p)/q Γ(2 + p/q) (1 + p/q)(p/q) 2q = = . E[X ] = 2q 2q Γ(p/q)λ Γ(p/q)λ λ 2q In this equation, replace λ q by (p/q)/E[X q ] and solve for (p/q). Thus, p/q =

(E[X q ])2 . var(X q )

This suggests that we first put Xnq := and then pn := q and

λn :=



pn /q Xnq

1/q

1 n−1

h



= 

1 n q ∑ Xi n i=1

(Xnq )2  i q q ∑ni=1 (Xi )2 − n(Xn )2 1/q

Xnq h  i q 2 q 2 1 n (X ) − n(X ) ∑ n i=1 i n−1

.

10. MATLAB. OMITTED. 11. MATLAB. The required script can be created using the code from Problem 2 followed by the lines global lambdan lambdan = mean(X)/sqrt(pi/2)

followed by the script from Problem 10 modifed as follows: The chi-squared statistic Z can be computed by inserting the lines p = CDF(b) - CDF(a); Z = sum((H-n*p).ˆ2./(n*p))

Chapter 6 Problem Solutions

97

after the creation of the right edge sequence in the script given in Problem 10, where CDF is the function function y = CDF(t) % Rayleigh CDF global lambdan y = zeros(size(t)); i = find(t>0); y(i) = 1-exp(-(t(i)/lambdan).ˆ2/2);

In addition, the line defining y in the script from Problem 10 should be changed to y=PDF(t), where PDF is the function function y = PDF(t) % Rayleigh density global lambdan y = zeros(size(t)); i = find(t>0); y(i) = (t(i)/lambdanˆ2).*exp(-(t(i)/lambdan).ˆ2/2);

Finally, the chi-squared statistic Z should be compared with zα = 22.362, since α = 0.05 and since there are m = 15 bins and r = 1 estimated parameter, the degrees of freedom parameter is k = m − 1 − r = 15 − 1 − 1 = 13 in the chi-squared table in the text. 12. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the M ATLAB function chi2cdf or gamcdf to compute the required cdfs for evaluating the chi-squared statistic Z. For the same reasons as in Problem 11, zα = 22.362. 13. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the M ATLAB function ncx2cdf to compute the required cdfs for evaluating the chisquared statistic Z. For the same reasons as in Problem 11, zα = 22.362. 14. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the M ATLAB function gamcdf to compute the required cdfs for evaluating the chi-squared statistic Z. For the same reasons as in Problem 11, zα = 22.362. 15. MATLAB. Similar to the solution of Problem 11. For the same reasons as in Problem 11, zα = 22.362. 16. Since E[H j ] = E



 I (X ) ∑ [e j ,e j+1 ) i = n

i=1

n

n

∑ P(e j ≤ Xi < e j+1 )

i=1



H j − np j E √ np j



=

∑ pj

= np j ,

i=1

= 0.

Since the Xi are i.i.d., the I[e j ,e j+1 ) (Xi ) are i.i.d. Bernoulli(p j ). Hence, n  E[(H j − np j )2 ] = var(H j ) = var ∑ I[e j ,e j+1 ) (Xi ) i=1

n

=

∑ var

i=1

  I[e j ,e j+1 ) (Xi ) = n · p j (1 − p j ),

98

Chapter 6 Problem Solutions and so E



H j − np j √ np j

2 

= 1 − p j.

17. If f is an even density, then F(−x) =

Z −x −∞

f (t) dt = −

Z x ∞

f (−θ ) d θ =

Z ∞ x

f (θ ) d θ = 1 − F(x).

√ 18. The width of any confidence interval is w = 2σ y/ n. If σ = 2 and n = 100, w99% =

2 · 2 · 2.576 = 1.03. 10

To make w99% < 1/4 requires 2σ y √ < 1/4 n

or

n > (8σ y)2 = (16 · 2.576)2 = 1699.

2 19. First observe that with Xi = m+Wi , E[X √i ] = m, and var(Xi ) = var(Wi ) = 4. So, σ = 4. For 95% confidence interval, σ yα /2 / n = 2 · 1.960/10 = 0.392, and so

m = 14.846 ± 0.392

with 95% probability.

The corresponding confidence interval is [14.454, 15.238]. 20. Write   1 n P(|Mn − m| ≤ δ ) = P(−δ ≤ Mn − m ≤ δ ) = P −δ ≤ ∑ (m +Wi ) − m ≤ δ n i=1     n n 1 = P −δ ≤ ∑ Wi ≤ δ = P −nδ ≤ ∑ Wi ≤ nδ n i=1 i=1 | {z } Cauchy(n)

2 = tan−1 (nδ /n), π

which is equal to 2/3 if and only if tan−1 (δ ) = π /3, or δ =

√ 3.

21. MATLAB. OMITTED. √ 22. We use the formula m = Mn ± yα /2 Sn / n = 10.083 ± (1.960)(0.568)/10 to get m = 10.083 ± 0.111

with 95% probability,

and the confidence intervale is [9.972, 10.194]. √ 23. We use the formula m = Mn ± yα /2 Sn / n = 4.422 ± (1.812)(0.957)/10 to get m = 4.422 ± 0.173

with 93% probability,

and the confidence interval is [4.249, 4.595].

Chapter 6 Problem Solutions

99

24. We have√Mn = number defective/n = 10/100 = 0.1. We use the formula m = Mn ± yα /2 Sn / n = 0.1 ± (1.645)(.302)/10 to get m = 0.1 ± 0.0497

with 90% probability.

The number of defectives is 10 000m, or number of defectives = 1000 ± 497

with 90% probability.

Thus, we are 90% sure that the number of defectives is between 503 and 1497 out of a total of 10 000 units. √ 25. We have Mn = 1559/3000. We use the formula m = Mn ± yα /2 Sn / n = 0.520 ± √ (1.645)(.5)/ 3000 to get m = 0.520 ± 0.015

with 90% probability,

and the confidence interval is [0.505, 0.535]. Hence, the probability is at least 90% that more than 50.5% of the voters will vote for candiate A. So we are 90% sure that candidate A will win. The 99% confidence interval is given by m = 0.520 ± 0.024

with 99% probability,

and the confidence interval is [0.496, 0.544]. Hence, we are not 99% sure that candidate A will win. 26. We have Mn = √ = 0.0960. We use the formula m = √ number defective/n = 48/500 Mn ± yα /2 Sn / n = 0.0960 ± (1.881)(.295)/ 500 to get m = 0.0960 ± 0.02482

with 94% probability.

The number of defectives is 100 000m, or number of defectives = 9600 ± 2482

with 94% probability.

Thus, we are 94% sure that the number of defectives is between 7118 and 12 082 out of a total of 100 000 units. √ 27. (a) We have Mn = 6/100. We use the formula m = Mn ± yα /2 Sn / n = 0.06 ± (2.170)(.239)/10 to get m = 0.06 ± 0.0519

with 97% probability.

We are thus 97% sure that p = m lies in the interval [0.0081, 0.1119]. Thus, we are not 97% sure that p < 0.1. √ (b) We have Mn = 71/1000. We use the formula m = Mn ± yα /2 Sn / n = 0.071 ± √ (2.170)(.257)/ 1000 to get m = 0.071 ± 0.018

with 97% probability.

We are thus 97% sure that p = m lies in the interval [0.053, 0.089]. Thus, we are 97% sure that p < 0.089 < 0.1.

100 28.

Chapter 6 Problem Solutions (a) Let Ti denote the time to transmit the ith packet. Then we need to compute    n  n \ [ {Ti ≤ t} = 1 − FT1 (t)n = 1 − (1 − e−t/µ )n . {Ti > t} = 1 − P P i=1

i=1

(b) Using the notation from part (a), T = T1 + · · · + Tn . Since the Ti are i.i.d. exp(1/µ ), T is Erlang(n, 1/µ ) by Problem 55(c) in Chapter 4 and the remark following it. Hence, fT (t) = (1/µ )

(t/µ )n−1 e−t/µ , (n − 1)!

t ≥ 0.

(c) We have yα /2 Sn 1.960(1.798) µ = Mn ± √ = 1.994 ± = 1.994 ± 0.352 n 10 and confidence interval [1.642, 2.346] with 95% probability. 29. MATLAB. OMITTED. 30. By the hint, ∑ni=1 Xi is Gaussian with mean nm and variance nσ 2 . Since Mn =
n ∑i=1 Xi /n, it is easy to see that Mn is still Gaussian, and its mean is (nm)/n = m. Gaussian but with mean zero Its variance is (nσ 2 )/n2 = σ 2 /n. Next, Mn − m remains p and the same variance σ 2 /n. Finally, (M σ /n remains Gaussian and with − m)/ n p mean zero, but its variance is (σ 2 /n)/( σ 2 /n )2 = 1. √ 31. We use the fomula m = Mn ± yα /2 Sn / n, where in this Gaussian case, yα /2 is taken from the tables using Student’s t distribution with n = 10. Thus, yα /2 Sn 2.262 · 1.904 √ m = Mn ± √ = 14.832 ± = 14.832 ± 1.362, n 10 and the confidence interval is [13.470, 16.194] with 95% probability. 32. We use [nVn2 /u, nVn2 /`], where u and ` are chosen from the appropriate table. For a 95% confidence interval, ` = 74.222 and u = 129.561. Thus,  2    nVn nVn2 100(4.413) 100(4.413) = = [3.406, 5.946]. , , u ` 129.561 74.222 33. We use [(n − 1)Sn2 /u, (n − 1)Sn2 /`], where u and ` are chosen from the appropriate table. For a 95% confidence interval, ` = 73.361 and u = 128.422. Thus,     (n − 1)Sn2 (n − 1)Sn2 99(4.736) 99(4.736) , = , = [3.651, 6.391]. u ` 128.422 73.361 34. For the two-sided test at the 0.05 significance level, we compare |Zn | with yα /2 = 1.960. Since |Zn | = 1.8 ≤ 1.960 = yα /2 , we accept the null hypothesis. For the one-sided test of m > m0 at the 0.05 significance level, we compare Zn with −yα = −1.645. Since it is not the case that Zn = −1.80 > −1.645 = −yα , we do not accept the null hypothesis.

Chapter 6 Problem Solutions

101

35. Suppose Φ(−y) = α . Then by Problem 17, Φ(−y) = 1 − Φ(y), and so 1 − Φ(y) = α , or Φ(y) = 1 − α . 36.

(a) Since Zn = 1.50 ≤ yα = 1.555, the Internet service provider accepts the null hypothesis. (b) Since Zn = 1.50 > −1.555 = −yα , we accept the null hypothesis; i.e., we reject the claim of the Internet service provider.

37. The computer vendor would take the null hypothesis to be m ≤ m0 . To give the vendor the benefit of the doubt, the consumer group uses m ≤ m0 as the null hypothesis. To accept the null hypothesis would require Zn ≤ yα . Only by using the sigificance level of 0.10, which has yα = 1.282, can the consumer group give the benefit of the doubt to the vendor and still reject the vendor’s claim. 38. Giving itself the benfit of the doubt, the company uses the null hypothesis m > m0 and uses a 0.05 significance level. The null hypothesis will be accepted if Zn > −yα = −1.645. Since Zn = −1.6 > −1.645, the company believes it has justified its claim. 39. Write n

e(b g) =

∑ |Yk − (baxk + bb)|2

k=1 n

= =

∑ |Yk − (baxk + [Y − abx])|2

k=1

∑ |(Yk −Y ) + ab(xk − x)|2

k=1 n h

∑

k=1

40. Write

n

=

a(xk − x)(Yk −Y ) + ab2 (xk − x)2 (Yk −Y )2 − 2b

i

= SYY − 2b aSxY + ab2 Sxx S  xY 2 = SYY − 2b aSxY + ab Sxx = SYY − abSxY = SYY − SxY /Sxx . Sxx E[Y |X = x] = E[g(X) +W |X = x] = E[g(x) +W |X = x]

= g(x) + E[W |X = x] = g(x) + E[W ] = g(x).

41. MATLAB. OMITTED. 42. MATLAB. OMITTED. 43. MATLAB. If z = c/t q , then ln z = ln c − q lnt. If y = ln z and x = lnt, then y = (−q)x + ln c. If y ≈ a(1)x + a(2), then q = −a(1) and c = exp(a(2)). Hence, the two lines of code that we need are qhat = -a(1) chat = exp(a(2))

44. Obvious.

102

Chapter 6 Problem Solutions

45. Write

e−(z−s) /2 e−z /2 . s2 /2 √ = fZe (z) = esz fZ (z)/MZ (s) = esz √ . e 2π 2π 2

2

e = t, put s = t. Then Ze ∼ (t, 1). To make E[Z]

46. Write

Then

fZe (z) = esz fZ (z)/MZ (s) = esz

e = E[Z]

λ (λ z) p−1 e−λ z . λ  p , Γ(p) λ −s

z > 0.

Z ∞  λ −p Z ∞ λ (λ z) p−1 e−z(λ −s) (λ − s) p z p−1 e−z(λ −s) z dz = z dz, λ −s Γ(p) Γ(p) 0 0

e = p/(λ − s). To make which is the mean of a gamma(p, λ − s) density. Hence, E[Z] e E[Z] = t, we need p/(λ − s) = t or s = λ − p/t.

47. MATLAB. OMITTED. 48. First,

Then

. pZe (zi ) = eszi pZ (zi ) [(1 − p) + pes ]. pZe (1) = es p/[(1 − p) + pes ] and

pZe (0) = (1 − p)/[(1 − p) + pes ].

CHAPTER 7

Problem Solutions 1. We have FZ (z) = P(Z ≤ z) = P(Y − X ≤ z) = P((X,Y ) ∈ Az ), where Az := {(x, y) : y − x ≤ z} = {(x, y) : y ≤ x + z}. y

z −z

x

2. We have  h i FZ (z) = P(Z ≤ z) = P(Y /X ≤ z) = P {Y /X ≤ z} ∩ {X < 0} ∪ {X > 0} + = P((X,Y ) ∈ D− z ∪ Dz ) = P((X,Y ) ∈ Az ),

where + Az := D− z ∪ Dz ,

and D− z := {(x, y) : y/x ≤ z and x < 0} = {(x, y) : y ≥ zx and x < 0}, and

z

D+ z := {(x, y) : y/x ≤ z and x > 0} = {(x, y) : y ≤ zx and x > 0}.

lin

eo

fs

lo

pe

y

+

−

Dz

Dz

3.

(a) R := (a, b] × (c, d] is 103

x

104

Chapter 7 Problem Solutions

d

c a

b

(b) A := (−∞, a] × (−∞, d] is

d

c a

b

(c) B := (−∞, b] × (−∞, c] is

d c a

b

(d) C := (a, b] × (−∞, c] is

d

c a

b

(e) D := (−∞, a] × (c, d] is

d

c a

b

Chapter 7 Problem Solutions

105

(f) A ∩ B is

d

c a

b

4. Following the hint and then observing that R and A ∪ B are disjoint, we have


P (X,Y ) ∈ (−∞, b] × (−∞, d] = P (X,Y ) ∈ R + P (X,Y ) ∈ A ∪ B .

(∗)

Next, by the inclusion–exclusion formula,



P (X,Y ) ∈ A ∪ B = P (X,Y ) ∈ A + P (X,Y ) ∈ B − P (X,Y ) ∈ A ∩ B = FXY (a, d) + FXY (b, c) − FXY (a, c).

Hence, (∗) becomes
FXY (b, d) = P (X,Y ) ∈ R + FXY (a, d) + FXY (b, c) − FXY (a, c),

5.

which is easily rearranged to get the rectangle formula,
P (X,Y ) ∈ R = FXY (b, d) − FXY (a, d) − FXY (b, c) + FXY (a, c). (a) {(x, y) : |x| ≤ y ≤ 1} is NOT a product set.

(b) {(x, y) : 2 < x ≤ 4, 1 ≤ y < 2} = (2, 4] × [1, 2). (c) {(x, y) : 2 < x ≤ 4, y = 1} = (2, 4] × {1}.

(d) {(x, y) : 2 < x ≤ 4} = (2, 4] × IR. (e) {(x, y) : y = 1} = IR × {1}.

(f) {(1, 1), (2, 1), (3, 1)} = {1, 2, 3} × {1}.

(g) The union of {(1, 3), (2, 3), (3, 3)} and the set in (f) is equal to {1, 2, 3} × {1, 3}.

(h) {(1, 0), (2, 0), (3, 0), (0, 1), (1, 1), (2, 1), (3, 1)} is NOT a product set. 6. We have   1, x ≥ 2, x − 1, 1 ≤ x < 2, FX (x) =  0, x < 1,

and

FY (y) =

(

1− e

−y −e−2y

y

0,

, y ≥ 0, y < 0,

where the quotient involving division by y is understood as taking its limiting value of one when y = 0. Since FX (x)FY (y) 6= FXY (x, y) when 1 ≤ x ≤ 2 and y > 0, X and Y are NOT independent.

106

Chapter 7 Problem Solutions

7. We have   1, x ≥ 3, 2/7, 2 ≤ x < 3, FX (x) =  0, x < 2,

FY (y) =

and



1 −2y − 5e−3y ], 7 [7 − 2e

0,

y ≥ 0, y < 0.

8. Let y > 0. First compute   (y + 1){1 + e−x(y+1) (−x)} − {y + e−x(y+1) }(1) ∂ y + e−x(y+1) = ∂y y+1 (y + 1)2 =

(y + 1) − x(y + 1)e−x(y+1) − y − e−x(y+1) (y + 1)2

=

1 − e−x(y+1) {1 + x(y + 1)} . (y + 1)2

Then compute    {1 + x(y + 1)}e−x(y+1) (y + 1) − e−x(y+1) (y + 1) ∂ ∂ y + e−x(y+1) = ∂x ∂y y+1 (y + 1)2 =

xe−x(y+1) (y + 1)2 = xe−x(y+1) , (y + 1)2

x, y > 0.

9. The first step is to recognize that fXY (x, y) factors into 2

exp[−|y − x| − x2 /2] e−x /2 e−|y−x| √ fXY (x, y) = = √ · . 2 2 2π 2π When integrating this last factor with respect to y, make the change of variable θ = y − x to get fX (x) =

Z ∞

2

−∞

e−x /2 fXY (x, y) dy = √ 2π

2

e−x /2 = √ 2π

Z ∞ 0

2

Z ∞

−∞

2

1 −|y−x| dy 2e

e−x /2 = √ 2π

e−x /2 e−θ d θ = √ . 2π

Z ∞

−∞

1 −|θ | dθ 2e

Thus, X ∼ N(0, 1). 10. The first step is to factor fXY (x, y) as 2

fXY (x, y) =

2

4 e−(x−y) /2 4e−(x−y) /2 √ = 5· √ . 5 y y 2π 2π

Regarding this last factor a function of x, it is an N(y, 1) density. In other words, when integrated with respect to x, the result is one. In symbols, fY (y) =

Z ∞

−∞

fXY (x, y) dx =

4 y5

Z ∞ −(x−y)2 /2 e −∞

√ 2π

dx =

4 , y5

y ≥ 1.

Chapter 7 Problem Solutions

107

11. We first analyze U := max(X,Y ). Then FU (u) = P(max(X,Y ) ≤ u) = P(X ≤ u and Y ≤ u) = FXY (u, u), and the density is fU (u) =

∂ FXY (x, y) ∂ FXY (x, y) + . ∂x ∂y x=u, y=u x=u, y=u

If X and Y are independent, then FU (u) = FX (u)FY (u),

and

fU (u) = fX (u)FY (u) + FX (u) fY (u).

If in addition X and Y have the same density, say f , (and therefore the same cdf, say F), then FU (u) = F(u)2 , and fU (u) = 2F(u) f (u). We next analyze V := min(X,Y ). Using the inclusion–exclusion formula, FV (v) = P(min(X,Y ) ≤ v) = P(X ≤ v or Y ≤ v)

= P(X ≤ v) + P(Y ≤ v) − P(X ≤ v and Y ≤ v) = FX (v) + FY (v) − FXY (v, v).

The density is ∂ FXY (x, y) ∂ FXY (x, y) − . fV (v) = fX (v) + fY (v) − ∂x ∂y x=v, y=v x=v, y=v

If X and Y are independent, then

FV (v) = FX (v) + FY (v) − FX (v)FY (v), and fV (v) = fX (v) + fY (v) − fX (v)FY (v) − FX (v) fY (v). If in addition X and Y have the same density f and cdf F, then FV (v) = 2F(v) − F(v)2 ,

and

fV (v) = 2[ f (v) − F(v) f (v)].

12. Since X ∼ gamma(p, 1) and Y ∼ gamma(q, 1) are independent, we have from Problem 55(c) in Chapter 4 that Z ∼ gamma(p + q, 1). Since p = q = 1/2, we further have Z ∼ gamma(1, 1) = exp(1). Hence, P(Z > 1) = e−1 . 13. We have from Problem 55(b) in Chapter 4 that Z ∼ Cauchy(λ + µ ). Since λ = µ = 1/2, Z ∼ Cauchy(1). Thus, P(Z ≤ 1) =

1π 1 3 1 1 + = . tan−1 (1) + = π 2 π4 2 4

108

Chapter 7 Problem Solutions

14. First write FZ (z) =

Z ∞  Z z−y −∞

−∞

 fXY (x, y) dx dy.

It then follows that fZ (z) =

∂ FZ (z) = ∂z

Z ∞ Z ∂ z−y −∞

∂z

fXY (x, y) dx dy =

−∞

Z ∞

−∞

fXY (z − y, y) dy.

15. First write − FZ (z) = P((X,Y ) ∈ Az ) = P((X,Y ) ∈ B+ z ∪ Bz )

ZZ

=

fXY (x, y) dx dy +

B+ z

Z ∞  Z z/y

=

−∞

0

ZZ

fXY (x, y) dx dy

B− z

 Z fXY (x, y) dx dy +

0

−∞

Z

∞

z/y

 fXY (x, y) dx dy.

Then

∂ FZ (z) = ∂z

fZ (z) = =

Z ∞ 0

=

Z ∞ 0

fXY (z/y, y)/y dy − Z 0

fXY (z/y, y)/|y| dy +

Z ∞

−∞

−∞

Z 0

−∞

fXY (z/y, y)/y dy

fXY (z/y, y)/|y| dy

fXY (z/y, y)/|y| dy.

16. For the cdf, write FZ (z) =

Z ∞  Z z+x −∞

−∞

 fXY (x, y) dy dx.

Then fZ (z) =

∂ FZ (z) = ∂z

Z Z ∞ ∂ z+x −∞

∂z

−∞

fXY (x, y) dy dx =

Z ∞

−∞

fXY (x, z + x) dx.

17. For the cdf, write FZ (z) = P(Z ≤ z) = P(Y /X ≤ z) = P(Y ≤ Xz, X > 0) + P(Y ≥ Xz, X < 0)   Z 0 Z ∞ Z ∞  Z xz fXY (x, y) dy dx. fXY (x, y) dy dx + = −∞

−∞

0

xz

Then fZ (z) = 18.

∂ FZ (z) = ∂z

(a) The region D:

Z ∞ 0

fXY (x, xz)x dx −

Z 0

−∞

fXY (x, xz)x dx =

Z ∞

−∞

fXY (x, xz)|x| dx.

Chapter 7 Problem Solutions

1

109

D

0 −1 −1

0

1

(b) Since fXY (x, y) = Kxn ym for (x, y) ∈ D and since D contains negative values of x, we must have n even in order that the density be nonnegative. In this case, the integral of the density over the region D must be one. Hence, K must be such that  m+1 1  ZZ Z 1 Z 1 Z 1 n m n y dx 1 = fXY (x, y) dy dx = Kx y dy dx = Kx m + 1 |x| −1 |x| −1 D Z 1  Z 1 Kxn K = xn − |x|n+m+1 dx [1 − |x|m+1 ] dx = m + 1 −1 −1 m + 1   Z 1 2K 1 1 2K = − xn − xn+m+1 dx = m+1 0 m+1 n+1 n+m+2 2K = . (n + 1)(n + m + 2) Hence, K = (n + 1)(n + m + 2)/2. (c) A sketch of Az with z = 0.3:

A

1

z

0 −1 Az −1

0

1

0

1

(d) A sketch of Az ∩ D with z = 0.3:

1 0 −1 −1 (e) P((X,Y ) ∈ Az ) =

Z √z  Z 1 z

z/x

  Z 1 Z 1 Kxn ym dy dx + √ Kxn ym dy dx z

x

110

Chapter 7 Problem Solutions Z √z

=

Kxn

z

K m+1

=

K = m+1 K = m+1

Z

 Z 1  Z 1 ym dy dx + √ Kxn ym dy dx

z/x √ z n

Z

z

Z

√ z

1

z

x [1 − (z/x)m+1 ] dx + n

m+1 n−m−1

n

x dx −

 Z K 1 − zn − m+1 n+1 z

=

x

z

x −z

z

Z

1

dx +

x

Z

√ z

z √ z

z

Z 1

xn [1 − xm+1 ] dx

√ z

Z 1 √

m+1 n−m−1

x

n

z

x −x

dx −

zm+1 xn−m−1 dx −

n+m+1

Z 1

√ x z



dx

n+m+1



dx



 1 − z(n+m+2)/2 . n+m+2

If n 6= m, the remaining integral is equal to √ ( z )n+m+2 − zn+1 . n−m Otherwise, the integral is equal to −

zm+1 ln z. 2

19. Let X ∼ uniform[0, w] and Y ∼ uniform[0, h]. We need to compute P(XY ≥ λ wh). Before proceeding, we make a few observations. First, since X ≥ 0, we can write for z > 0, Z ∞Z ∞

P(XY ≥ z) =

0

z/x

fXY (x, y) dy dx =

Z ∞ 0

fX (x)

Z ∞

z/x

fY (y) dy dx.

Since Y ∼ uniform[0, h], the inner integral will be zero if z/x > h. Since z/x ≤ h if and only if x ≥ z/h, P(XY ≥ z) =

Z ∞

z/h

fX (x)

Z h 1 z/x

h

dy dx =

Z ∞

z/h

 z fX (x) 1 − dx. xh

We can now write Z ∞

Z ∞   λ wh  λw dx = dx fX (x) 1 − fX (x) 1 − xh x λ wh/h λw Z w  w 1 λw 1− dx = (1 − λ ) − λ ln = (1 − λ ) + λ ln λ . = x λw λw w

P(XY ≥ λ wh) =

20. We first compute E[XY ] = E[cos Θ sin Θ] = =

1 2 E[sin 2Θ]

cos(−2π ) − cos(2π ) = 0. 8π

=

1 4π

Z π

−π

sin 2θ d θ

Chapter 7 Problem Solutions

111

Similarly, E[X] = E[cos Θ] =

1 2π

1 2π

Z π

Z π

−π

cos θ d θ =

sin(π ) − sin(−π ) 0−0 = = 0, 2π 2π

and E[Y ] = E[sin Θ] =

−π

sin θ d θ =

cos(−π ) − cos(π ) (−1) − (−1) = = 0. 2π 2π

To prove that X and Y are not independent, we argue by contradiction. However, before we begin, observe that since (X,Y ) satisfies X 2 +Y 2 = cos2 Θ + sin2 Θ = 1, (X,Y ) always lies on the unit circle. Now consider the square of side one centered at the origin, S := {(x, y) : |x| ≤ 1/2, and |y| ≤ 1/2}. Since this region lies strictly inside the unit circle, P((X,Y ) ∈ S) = 0. Now, to obtain a contradiction suppose that X and Y are independent. Then fXY (x, y) = fX (x) fY (y), where fX and fY are both arcsine densities by Problem 35 in Chapter 5. Hence, for |x| < 1 and |y| < 1, fXY (x, y) ≥ fXY (0, 0) = 1/π 2 . We can now write P((X,Y ) ∈ S) =

ZZ S

fXY (x, y) dx dy ≥

ZZ

1/π 2 dx dy = 1/π 2 > 0,

S

which is a contradiction. 21. If E[h(X)k(Y )] = E[h(X)]E[k(Y )] for all bounded continuous functions h and k, then we may specialize this equation to the functions h(x) = e jν1 x and k(y) = e jν2 y to show that the joint characteristic function satisfies

ϕXY (ν1 , ν2 ) := E[e j(ν1 X+ν2Y ) ] = E[e jν1 X e jν2Y ] = E[e jν1 X ]E[e jν2Y ] = ϕX (ν1 )ϕY (ν2 ). Since the joint characteristic function is the product of the marginal characteristic functions, X and Y are independent. 22.

(a) Following the hint, let D denote the half-plane D := {(x, y) : x > x0 },

distance of point from origin is x0 / cos θ y

D

θ x0

x

112

Chapter 7 Problem Solutions and write P(X > x0 ) = P(X > x0 ,Y ∈ IR) = P((X,Y ) ∈ D), where X and Y are both N(0, 1). Then ZZ

P((X,Y ) ∈ D) =

D

2

e−(x +y 2π

2 )/2

dx dy.

Now convert to polar coordinates and write Z π /2 Z ∞

2

e−r /2 r dr d θ −π /2 x0 / cos θ 2π ∞ Z 2 1 π /2 = dθ −e−r /2 2π −π /2

P((X,Y ) ∈ D) =

1 = 2π =

1 π

Z π /2

−π /2

Z π /2

x0 / cos θ

exp[−(x0 / cos θ )2 /2] d θ

exp

0



 −x02 dθ . 2 cos2 θ

(b) In the preceding integral, make the change of variable θ = π /2 − t, d θ = −dt. Then cos θ becomes cos(π /2 − t) = sint, and the preceding integral becomes   Z −x02 1 π /2 dt. exp π 0 2 sin2 t 23. Write Z ∞

−∞

fY |X (y|x) dy =

Z ∞ fXY (x, y) −∞

fX (x)

dy =

1 fX (x)

Z ∞

−∞

fXY (x, y) dy =

fX (x) = 1. fX (x)

24. First write P((X,Y ) ∈ A|x < X ≤ x + ∆x) = =

=

=

   P (X,Y ) ∈ A ∩ x < X ≤ x + ∆x

P(x < X ≤ x + ∆x)    P (X,Y ) ∈ A ∩ (X,Y ) ∈ (x, x + ∆x] × IR

P(x < X ≤ x + ∆x) n h io P (X,Y ) ∈ A ∩ (x, x + ∆x] × IR 

ZZ

P(x < X ≤ x + ∆x) 

fXY (t, y) dy dt

A∩ (x,x+∆x]×IR

Z x+∆x x

fX (τ ) d τ

Chapter 7 Problem Solutions

=

ZZ

=

Z x+∆x Z ∞

113

IA (t, y)I(x,x+∆x]×IR (t, y) fXY (t, y) dy dt Z x+∆x x

fX (τ ) d τ

IA (t, y) fXY (t, y) dy dt −∞ Z x+∆x

x

fX (τ ) d τ

x

Z Z 1 x+∆x ∞

∆x

=

x

1 ∆x

IA (t, y) fXY (t, y) dy dt −∞ Z x+∆x fX (τ ) d τ .

x

It now follows that lim P((X,Y ) ∈ A|x < X ≤ x + ∆x) =

∆t→∞

=

Z ∞

IA (x, y) fXY (x, y) dy

Z ∞

IA (x, y) fY |X (y|x) dy.

−∞

−∞

fX (x)

25. We first compute, for x > 0, xe−x(y+1) = xe−xy , y > 0. e−x As a function of y, this is an exponential density with parameter x. This is very different from fY (y) = 1/(y + 1)2 . We next compute, for y > 0, fY |X (y|x) =

fX|Y (x|y) =

xe−x(y+1) = (y + 1)2 xe−(y+1)x , 1/(y + 1)2

x > 0.

As a function of x this is an Erlang(2, y + 1) density, which is not the same as fX ∼ exp(1). 26. We first compute, for x > 0, xe−x(y+1) = xe−xy , y > 0. e−x As a function of y, this is an exponential density with parameter x. Hence, fY |X (y|x) =

E[Y |X = x] =

Z ∞ 0

y fY |X (y|x) dy = 1/x.

We next compute, for y > 0, fX|Y (x|y) =

xe−x(y+1) = (y + 1)2 xe−(y+1)x , 1/(y + 1)2

x > 0.

As a function of x this is an Erlang(2, y + 1) density. Hence, E[X|Y = y] =

Z ∞ 0

x fX|Y (x|y) dx = 2/(y + 1).

114

Chapter 7 Problem Solutions

27. Write Z ∞

−∞

 fX|Y (x|y) dx fY (y) dy −∞ B  Z ∞ Z ∞ = IB (x) fX|Y (x|y) dx fY (y) dy −∞ −∞ Z ∞  Z ∞ = IB (x) fXY (x, y) dy dx

P(X ∈ B|Y = y) fY (y) dy =

Z ∞ Z

−∞

=

Z

B

−∞

fX (x)dx = P(X ∈ B).

28. For z ≥ 0, fZ (z) =

2 )/(2σ 2 ) Z √z e−(z−y √

√ − z

e−z/(2σ = πσ 2 =

p 2)

z − y2

Z √z 0

−z/(2σ 2 )

e

πσ 2

2

e−(y/σ ) /2 e−z/(2σ · √ dy = 2πσ 2 2πσ

2πσ

p

1 z − y2

dy =

e−z/(2σ πσ 2

2)

Z 1 0

2

2)

Z √z

√ − z

p

1 z − y2

dy

1 √ dt 1 − t2

e−z/(2σ ) , 2σ 2

[sin−1 (1) − sin−1 (0)] =

which is an exponential density with parameter 1/(2σ 2 ). 29. For z ≥ 0,

Z ∞

fZ (z) =

0

= 2

x · λ e−λ xz · λ e−λ x dx +

Z ∞ 0

0

y · λ e−λ yz · λ e−λ y dy

x · λ e−λ xz · λ e−λ x dx

2λ 2 λz+λ

=

Z ∞

Z ∞ 0

x · [λ z + λ ]e−x[λ z+λ ] dx.

Now, this last integral is the expectation of an exponential density with parameter λ z + λ . Hence, fZ (z) =

2λ 2 1 2 , · = λz+λ λz+λ (z + 1)2

z ≥ 0.

30. Using the law of total probability, substitution, and independence, we have P(X ≤ Y ) = =

Z ∞

Z0 ∞ 0

= λ

P(X ≤ Y |X = x) fX (x) dx = P(Y ≥ x) fX (x) dx =

Z ∞ 0

e−x(λ +µ ) dx =

Z ∞ 0

Z ∞ 0

P(x ≤ Y |X = x) fX (x) dx

e−µ x · λ e−λ x dx

λ  −(λ +µ )  ∞ λ . −e = λ +µ λ +µ 0

Chapter 7 Problem Solutions

115

31. Using the law of total probability, substitution, and independence, we have P(Y / ln(1 + X 2 ) > 1) = =

Z ∞

−∞ Z 2 1

=

Z 2

P(Y / ln(1 + X 2 ) > 1|X = x) fX (x) dx

P(Y > ln(1 + x2 )|X = x) · 1 dx 2

P(Y > ln(1 + x )) dx =

1

=

Z 2

eln(1+x

2 )−1

dx =

1

Z 2 1

= tan−1 (2) − tan−1 (1).

Z 2 1

2

e− ln(1+x ) dx

1 dx 1 + x2

32. First find the cdf using the law of total probability and substitution. Then differentiate to obtain the density. (a) For Z = eX Y , FZ (z) = P(Z ≤ z) = P(eX Y ≤ z) = =

Z ∞

−∞

Z ∞

−∞

P(eX Y ≤ z|X = x) fX (x) dx

P(Y ≤ ze−x |X = x) fX (x) dx =

Then fZ (z) = and so

Z ∞

−∞

fZ (z) =

Z ∞

−∞

FY |X (ze−x |x) fX (x) dx.

fY |X (ze−x |x)e−x fX (x) dx,

Z ∞

−∞

fXY (x, ze−x )e−x dx.

(b) Since Z = |X + Y | ≥ 0, we know that FZ (z) and fZ (z) are zero for z < 0. For z ≥ 0, write FZ (z) = P(Z ≤ z) = P(|X +Y | ≤ z) = = = = = Then

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

P(|X +Y | ≤ z|X = x) fX (x) dx

P(|x +Y | ≤ z|X = x) fX (x) dx P(−z ≤ x +Y ≤ z|X = x) fX (x) dx P(−z − x ≤ Y ≤ z − x|X = x) fX (x) dx

Z ∞ −∞

 FY |X (z − x|x) − FY |X (−z − x|x) fX (x) dx.

fZ (z) = = =

Z ∞ −∞

Z ∞ −∞

Z ∞

−∞

 fY |X (z − x|x) − fY |X (−z − x|x)(−1) fX (x) dx

 fY |X (z − x|x) + fY |X (−z − x|x) fX (x) dx

fXY (x, z − x) + fXY (x, −z − x) dx.

116 33.

Chapter 7 Problem Solutions (a) First find the cdf of Z using the law of total probability, substitution, and independence. Then differentiate to obtain the density. Write FZ (z) = P(Z ≤ z) = P(Y /X ≤ z) = = = = =

Z ∞

−∞

Z 0

−∞ Z 0

−∞ Z 0 −∞

Z ∞

−∞

P(Y /X ≤ z|X = x) fX (x) dx

P(Y /x ≤ z|X = x) fX (x) dx = P(Y /x ≤ z) fX (x) dx + P(Y ≥ zx) fX (x) dx +

Z ∞

Z

[1 − FY (zx)] fX (x) dx +

0 ∞

Z ∞

−∞

P(Y /x ≤ z) fX (x) dx

P(Y /x ≤ z) fX (x) dx

P(Y ≤ zx) fX (x) dx

0

Z ∞ 0

FY (zx) fX (x) dx.

Then fZ (z) = = =

Z 0

−∞ Z 0 −∞

Z ∞

−∞

− fY (zx)x fX (x) dx + fY (zx)|x| fX (x) dx +

Z ∞

0 Z ∞ 0

fY (zx)x fX (x) dx fY (zx)|x| fX (x) dx

fY (zx) fX (x)|x| dx.

(b) Using the result of part (a) and the fact that the integrand is even, write Z

Z ∞ −|zx|2 /(2σ 2 ) −(x/σ )2 /2 e e

2 ∞ x −(x/σ )2 [1+z2 ]/2 √ dx. |x| dx = e 2π 0 σ 2 −∞ 2πσ 2πσ √ Now make the change of variable θ = (x/σ ) 1 + z2 to get fZ (z) =

fZ (z) =

√

1/π 1 + z2

Z ∞ 0

θ e−θ

2 /2

which is the Cauchy(1) density.

dθ =

1/π 1/π  −θ 2 /2  ∞ −e , = 2 1+z 1 + z2 0

(c) Using the result of part (a) and the fact that the integrand is even, write fZ (z) = =

Z ∞

−∞

λ −λ |zx| λ −λ |x| |x| dx 2e 2e

λ 2 /2 λ (|z| + 1)

Z ∞ 0

=

λ2 2

Z ∞ 0

xe−xλ (|z|+1) dx

x · λ (|z| + 1)e−λ (|z|+1)x dx,

where this last integral is the mean of an exponential density with parameter λ (|z| + 1). Hence, fZ (z) =

1 1 λ 2 /2 · = . λ (|z| + 1) λ (|z| + 1) 2(|z| + 1)2

Chapter 7 Problem Solutions

117

(d) For z > 0, use the result of part (a) and the fact that the integrand is even, write fZ (z) =

Z ∞

−∞

2 e−x /2 1 √ |x| dx 2 I[−1,1] (zx)

2π

Z 1/z −x2 /2 e

=

x√

0

dx

2π

2 1/z 2 1 1 − e−1/(2z ) √ . = √ (−e−x /2 ) 0 = 2π 2π

√ The same formula holds for z < 0, and it is easy to check that fZ (0) = 1/ 2π . (e) Since Z ≥ 0, for z ≥ 0, we use the result from part (a) to write Z ∞

fZ (z) =

0

= z

zx −(zx/λ )2 /2 x −(x/λ )2 /2 e e x dx λ2 λ2

Z ∞ 0

θ 3 e−θ

z 2 (z + 1)2

=

2 (z2 +1)/2

Z ∞ 0

dθ = z

Z ∞ 0

t 3 e−t

2 /2

dt =

= z

Z ∞ 0

√

t z2 + 1

2z 2 (z + 1)2

Z ∞ 0


x 3 −(x/λ )2 (z2 +1)/2 λ e

3

e−t

2 /2

√

dx λ

dt

z2 + 1

se−s ds.

This last integral is the mean of an exp(1) density, which is one. Hence fZ (z) = 2z/(z2 + 1)2 . 34. For the cdf, use the law of total probability, substitution, and independence to write Z ∞

FZ (z) = P(Z ≤ z) = P(Y / ln X ≤ z) = = =

Z ∞

0

0

P(Y / ln x ≤ z|X = x) fX (x) dx =

0

P(Y ≥ z ln x) fX (x) dx +

Z 1

Z ∞ 1

P(Y / ln X ≤ z|X = x) fX (x) dx Z ∞ 0

P(Y / ln x ≤ z) fX (x) dx

P(Y ≤ z ln x) fX (x) dx.

Then fZ (z) = =

Z 1

Z0 ∞

− fY (z ln x)(ln x) fX (x) dx +

0

Z ∞ 1

fY (z ln x)(ln x) fX (x) dx

fX (x) fY (z ln x)| ln x| dx.

35. Use the law of total probability, substitution, and independence to write E[e(X+Z)U ] = =

Z 1/2

−1/2

Z 1/2

−1/2

E[e(X+Z)U |U = u] du = E[e(X+Z)u ] du =

1 1/2 = 1−u

−1/2

=

Z 1/2

−1/2

Z 1/2

−1/2

E[e(X+Z)u |U = u] du

E[eXu ]E[eZu ] du =

Z 1/2

1 du 2 −1/2 (1 − u)

1 2 4 1 − = 2− = . 1 − 1/2 1 + 1/2 3 3

118

Chapter 7 Problem Solutions

36. Use the law of total probability and substitution to write E[X 2Y ] = =

Z 2 1

E[X 2Y |Y = y] dy =

Z 2

1/y dy = 2 ln 2.

1

y · (2/y2 ) dy = 2

Z 2

E[X 2 y|Y = y] dy =

1

Z 2

Z 2 1

yE[X 2 |Y = y] dy

1

37. Use the law of total probability and substitution to write E[X nY r ] =

Z ∞ 0

Z ∞

E[X nY r |Y = y] fY (y) dy =

Z ∞

0 Z ∞

E[X n yr |Y = y] fY (y) dy

Γ(n + p) fY (y) dy yn Γ(p) Z Γ(n + p) Γ(n + p) (r − n)! Γ(n + p) ∞ r−n y fY (y) dy = E[Y r−n ] = · r−n . = Γ(p) Γ(p) Γ(p) λ 0

=

0

38.

yr E[X n |Y = y] fY (y) dy =

yr

0

(a) Use the law of total probability, substitution, and independence to find the cdf. Write FY (y) = P(Y ≤ y) = P(eVU ≤ y) = =

Z ∞ 0

=

Z ∞ 0

vU

P(e

Z ∞

P(eVU ≤ y|V = v) fV (v) dv

0

≤ y|V = v) fV (v) dv =

Z ∞ 0

P(evU ≤ y) fV (v) dv

P(U ≤ 1v ln y) fV (v) dv.

Then fY (y) =

Z ∞ 0

1 1 vy fU ( v

ln y) fV (v) dv.

To determine when fU ( 1v ln y) is nonzero, we consider the cases y > 1 and y < 1 separately. For y > 1, 1v ln y ≥ 0 for all v ≥ 0, and 1v ln y ≤ 1/2 for v ≥ 2 ln y. Thus, fU ( 1v ln y) = 1 for v ≥ 2 ln y, and we can write fY (y) =

Z ∞

2 ln y

1 yv

· ve−v dv =

e−2 ln y 1 = 3. y y

For y < 1, fU ( 1v ln y) = 1 for −1/2 ≤ 1v ln y, or v ≥ −2 ln y. Thus, fY (y) =

Z ∞

−2 ln y

1 yv

· ve−v dv =

1 2 ln y e = y. y

Putting this all together, we have   1/y3 , y ≥ 1, y, 0 ≤ y < 1, fY (y) =  0, y < 0.

Chapter 7 Problem Solutions

119

(b) Using the density of part (a), Z 1

E[Y ] =

y2 dy +

Z ∞ 1 1

0

y2

1 4 +1 = . 3 3

dy =

(c) Using the law of total probability, substitution, and independence, we have VU

E[e

] = =

39.

Z 1/2

VU

E[e

−1/2

Z 1/2

−1/2

|U = u] du =

Z 1/2

−1/2

E[eVu |U = u] du

1 4 1 1/2 = . du = 2 1 − u −1/2 3 −1/2 (1 − u)

Z 1/2

E[eVu ] du =

(a) This problem is interesting because the answer does not depend on the random variable X. Assuming X has a density fX (x), first write E[cos(X +Y )] = =

Z ∞

−∞

Z ∞

−∞

E[cos(X +Y )|X = x] fX (x) dx E[cos(x +Y )|X = x] fX (x) dx.

Now use the conditional density of Y given X = x to write E[cos(x +Y )|X = x] =

Z x+π x−π

cos(x + y)

dy = 2π

Z 2x+π 2x−π

cos θ

dθ = 0, 2π

since we are integrating cos θ over an interval of length 2π . Thus, E[cos(X + Y )] = 0 as well. (b) Write P(Y > y) = =

Z ∞

−∞ Z 2 1

Z 2

P(Y > y|X = x) fX (x) dx =

e−xy dx =

P(Y > y|X = x) dx

1

e−y − e−2y . y

(c) Begin in the usual way by writing E[XeY ] = =

Z ∞

−∞ Z ∞ −∞

E[XeY |X = x] fX (x) dx =

Z ∞

−∞

E[xeY |X = x] fX (x) dx

xE[eY |X = x] fX (x) dx.

Now observe that

E[eY |X = x] = E[esY |X = x]

s=1

= es

2 x2 /2

Then continue with E[XeY ] =

Z ∞

−∞

xex

2 /2

fX (x) dx =

e49/2 − e9/2 = . 4

1 4

Z 7 3

xex

2 /2



s=1

dx =

= ex

2 /2

.

1 x2 /2
7 e 3 4

120

Chapter 7 Problem Solutions (d) Write E[cos(XY )] = =

Z ∞

−∞ Z 2

E[cos(XY )|X = x] fX (x) dx =

E[Re(e jxY )|X = x] dx = Re

1

= Re

Z 2 1

e−x

2 (1/x)/2

Z 2

dx =

1

Z 2 1

Z 2

E[cos(xY )|X = x] dx

1

E[e jxY |X = x] dx

e−x/2 dx = 2(e−1/2 − e−1 ).

40. Using the law of total probability, substitution, and independence, MY (s) = E[esY ] = E[esZX ] = =

Z ∞ 0

=

Z ∞

=

0

0

E[esZX |Z = z] fZ (z) dz

E[eszX |Z = z] fZ (z) dz = e(sz)

2 σ 2 /2

0

Z ∞

Z ∞

ze−(1−s

Z ∞

fZ (z) dz =

2 σ 2 )z2 /2

Z ∞

E[eszX ] fZ (z) dz

0

e(sz)

2 σ 2 /2

0

ze−z

2 /2

dz

dz.

√ Now make the change of variable t = z 1 − s2 σ 2 to get Z ∞

MY (s) =

0

2 dt t 1/σ 2 1 √ = . e−t /2 √ = 2 2 1−s σ 1/σ 2 − s2 1 − s2 σ 2 1 − s2 σ 2

Hence, Y ∼ Laplace(1/σ ). 41. Using the law of total probability and substitution, Z ∞

n m

E[X Y ] =

0

E[X Y |Y = y] fY (y) dy =

0

ym E[X n |Y = y] fY (y) dy =

Z ∞

=

Z ∞

n m

Z 0∞ 0

E[X n ym |Y = y] fY (y) dy

ym 2n/2 yn Γ(1 + n/2) fY (y) dy

= 2n/2 Γ(1 + n/2)E[Y n+m ] = 2n/2 Γ(1 + n/2) 42.

(n + m)! . β n+m

(a) We use the law of total probability, substitution, and independence to write Z ∞

FZ (z) = P(Z ≤ z) = P(X/Y ≤ z) = = =

Z ∞

Z0 ∞ 0

0

P(X/y ≤ z|Y = y) fY (y) dy = P(X ≤ zy) fY (y) dy =

Z ∞ 0

P(X/Y ≤ z|Y = y) fY (y) dy Z ∞ 0

P(X ≤ zy|Y = y) fY (y) dy

FX (zy) fY (y) dy.

Differentiating, we have fZ (z) =

Z ∞ 0

fX (zy)y · fY (y) dy =

Z ∞ 0

λ

(λ zy) p−1 e−λ zy (λ y)q−1 e−λ y ·λ · y dy. Γ(p) Γ(q)

Chapter 7 Problem Solutions

121

Making the change of variable w = λ y, we obtain fZ (z) = =

Z ∞ (zw) p−1 e−zw wq−1 e−w

·

Γ(p)

0

z p−1 Γ(p)Γ(q)

Z ∞ 0

Γ(q)

· w dw

w p+q−1 e−w(1+z) dw.

Now make the change of variable θ = w(1 + z) so that d θ = (1 + z) dw and w = θ /(1 + z). Then  p+q−1 Z ∞ z p−1 θ dθ fZ (z) = e−θ Γ(p)Γ(q) 0 1 + z (1 + z) =

=

z p−1 Γ(p)Γ(q)(1 + z) p+q z p−1 B(p, q)(1 + z) p+q

.

Z ∞

|

0

θ p+q−1 e−θ d θ {z } = Γ(p+q)

(b) Starting with V := Z/(1 + Z), we first write   Z FV (v) = P ≤ v = P(Z ≤ v + vZ) = P(Z(1 − v) ≤ v) 1+Z = P(Z ≤ v/(1 − v)). Differentiating, we have  v  1  v  (1 − v) + v = fZ . fV (v) = fZ 2 1 − v (1 − v) 1 − v (1 − v)2 Now apply the formula derived in part (a) and use the fact that 1+ to get fV (v) =

1 v = 1−v 1−v

[v/(1 − v)] p−1 v p−1 (1 − v)q−1 1 = , · 1 p+q (1 − v)2 B(p, q) B(p, q)( 1−v )

which is the beta density with parameters p and q. 43. Put q := (n − 1)p and Zi := ∑ j6=i X j , which is gamma(q, λ ) by Problem 55(c) in Chapter 4. Now observe that Yi = Xi /(Xi + Zi ), which has a beta density with parameters p and q := (n − 1)p by Problem 42(b). 44. Using the law of total probability, substitution, and independence, p p FZ (z) = P(Z ≤ z) = P(X/ Y /k ≤ z) = P(X ≤ z Y /k ) Z ∞ Z ∞ p p P(X ≤ z y/k |Y = y) fY (y) dy = P(X ≤ z Y /k |Y = y) fY (y) dy = =

Z0 ∞ 0

P(X ≤ z

p

0

y/k ) fY (y) dy.

122

Chapter 7 Problem Solutions Then fZ (z) =

Z ∞ 0

p p fX (z y/k ) y/k fY (y) dy

Z ∞ −(z2 y/k)/2 p e

1 (y/2)k/2−1 e−y/2 √ dy y/k 2 Γ(k/2) 0 2π Z ∞ 2 1 = √ √ (y/2)k/2−1/2 e−y(1+z /k)/2 dy. 2 π k Γ(k/2) 0

=

Now make the change of variable θ = y(1 + z2 /k)/2, d θ = (1 + z2 /k)/2 dy to get 1 fZ (z) = √ √ π k Γ(k/2) = =

Z ∞ 0

θ 1 + z2 /k

k/2−1/2

1 √ Γ(1/2) k Γ(k/2)(1 + z2 /k)k/2+1/2

e−θ

Z ∞ 0

dθ 1 + z2 /k

θ k/2+1/2−1 e−θ d θ

(1 + z2 /k)−(k+1)/2 1 √ Γ(k/2 + 1/2) = √ , Γ(1/2) k Γ(k/2)(1 + z2 /k)k/2+1/2 k B(1/2, k/2)

which is the required Student’s t density with k degrees of freedom. 45. We use the law of total probability, substitution, and independence to write FZ (z) = P(Z ≤ z) = P(X/Y ≤ z) = = =

Z ∞

Z0 ∞ 0

Z ∞ 0

P(X/Y ≤ z|Y = y) fY (y) dy

P(X/y ≤ z|Y = y) fY (y) dy = P(X ≤ zy) fY (y) dy =

Z ∞ 0

Z ∞ 0

P(X ≤ zy|Y = y) fY (y) dy

FX (zy) fY (y) dy.

Differentiating, we have fZ (z) =

Z ∞ 0

fX (zy)y · fY (y) dy =

Z ∞

r

λr

0

r

(λ zy) p−1 e−(λ zy) (λ y)q−1 e−(λ y) ·λr · y dy. Γ(p/r) Γ(q/r)

Making the change of variable w = λ y, we obtain fZ (z) = =

r Z ∞ (zw) p−1 e−(zw)

r

0

Γ(p/r)

r2 z p−1

Γ(p/r)Γ(q/r)

Z ∞ 0

r

·r

wq−1 e−w · w dw Γ(q/r) r

r

w p+q−1 e−w (1+z ) dw.

Now make the change of variable θ = wr (1 + zr ) so that d θ = rwr−1 (1 + zr ) dw and w = (θ /[1 + zr ])1/r . Then fZ (z) =

r2 z p−1 Γ(p/r)Γ(q/r)

Z ∞ 0

θ 1 + zr

(p+q−1)/r

e−θ

dθ r(1 + zr )


(r−1)/r θ 1+zr

Chapter 7 Problem Solutions

=

123

Z

∞ rz p−1 θ (p+q)/r−1 e−θ d θ (p+q)/r r Γ(p/r)Γ(q/r)(1 + z ) 0 | {z } = Γ((p+q)/r)

=

rz p−1

B(p/r, q/r)(1 + zr )(p+q)/r

46. For 0 < z ≤ 1, fZ (z) =

1 4

Z z 0

.

y−1/2 (z − y)−1/2 dy =

1 4z

Z z 0

p

1 (y/z)(1 − (y/z))

dy.

Now make the change of variable t 2 = y/z, 2t dt = dy/z to get fZ (z) = Next, for 1 < z ≤ 2,

Z 1

1 2

Z 1 0

√

1 1 − t2

dt =

1 −1 1 sin t = π /4. 2 0 Z

√

1 1 1 1/ z √ dt dy = √ 2 1−1/z z−1 (y/z)(1 − (y/z)) 1 − t2 √ p  1  −1 √ 1 −1 1/ z = = sin t √ sin (1/ z ) − sin−1 ( 1 − 1/z ) . 2 2 1−1/z

fZ (z) =

1 4z

p

Putting this all together yields  π /4, p    0 < z ≤ 1, √ −1 −1 1 fZ (z) = sin (1/ z ) − sin ( 1 − 1/z ) , 1 < z ≤ 2, 2 0, otherwise.

47. Let ψ denote the N(0, 1) density. Using

  v − ρu 1 fUV (u, v) = ψρ (u, v) = ψ (u) · p ψ p , 1 − ρ2 1 − ρ2 {z } | N(ρ u, 1−ρ 2 ) density in v

we see that

Z ∞

Z ∞



v − ρu fU (u) = fUV (u, v) dv = ψ (u) · p ψ p 2 −∞ −∞ 1−ρ 1 − ρ2   Z ∞ 1 v − ρu p ψ p dv = ψ (u). = ψ (u) −∞ 1 − ρ2 1 − ρ2 | {z } 1

density in v integrates to one

Similarly writing



u − ρv fUV (u, v) = ψρ (u, v) = p ψ p 2 1−ρ 1 − ρ2 1



· ψ (v),



dv

124

Chapter 7 Problem Solutions we have Z ∞

48. Using

Z ∞



u − ρv p fV (v) = fUV (u, v) du = ψ p 2 −∞ −∞ 1−ρ 1 − ρ2   Z ∞ 1 v − ρu p ψ p = ψ (v) du = ψ (v). −∞ 1 − ρ2 1 − ρ2

ψρ (u, v) = ψ (u) · p

we can write

1



· ψ (v) du

  v − ρu ψ p , 1 − ρ2 1 − ρ2 1

  1 x − mX y − mY ψρ , σX σY σX σY ! ! y−m   x−mX Y 1 1 x − mX σY − ρ σX p ψ ψ = ·p . σX σY σX 1 − ρ2 1 − ρ2

fXY (x, y) =

Then in to get

R∞

−∞ f XY (x, y) dy,

fX (x) =

Z ∞

−∞

(∗)

make the change of variable v = (y − mY )/σY , dv = dy/σY

! Z ∞  X v − ρ x−m 1 1 x − mX σX p fXY (x, y) dy = ψ ψ p dv σX σX −∞ 1 − ρ2 1 − ρ2 | {z }   1 x − mX = ψ . σX σX

density in v integrates to one

Thus, fX ∼ N(mX , σX2 ). Using this along with (∗), we obtain 1 fXY (x, y) p ψ = fY |X (y|x) = fX (x) σY 1 − ρ 2

y−mY σY

p

X − ρ x−m σX

!

1 − ρ2 ! y − [mY + σσYX ρ (x − mX )] 1 p p = ψ . σY 1 − ρ 2 σY 1 − ρ 2   Thus, fY |X (· |x) ∼ N mY + σσYX ρ (x − mX ), σY2 (1 − ρ 2 ) . Proceeding in an analogous way, using   u − ρv 1 ψρ (u, v) = p ψ p · ψ (v), 1 − ρ2 1 − ρ2 we can write

  1 x − mX y − mY fXY (x, y) = ψρ , σX σY σX σY =

σX σY

1 p

1 − ρ2

ψ

x−mX σX

p

Y − ρ y−m σY

1 − ρ2

!



 y − mY ·ψ . σY

(∗∗)

Chapter 7 Problem Solutions Then in to get

R∞

−∞ f XY (x, y) dx,

fY (y) =

Z ∞

−∞

125

make the change of variable u = (x − mX )/σX , du = dx/σX

! Z ∞  Y u − ρ y−m 1 1 y − mY σY p ψ ψ p du fXY (x, y) dx = σY σY −∞ 1 − ρ2 1 − ρ2 | {z }   1 y − mY = . ψ σY σY

density in u integrates to one

Thus, fY ∼ N(mY , σY2 ). Using this along with (∗∗), we obtain 1 fXY (x, y) p = ψ fX|Y (x|y) = fY (y) σX 1 − ρ 2

x−mX σX

p

Y − ρ y−m σY

1 − ρ2 ! x − [mX + σσYX ρ (y − mY )] 1 p p ψ . = σX 1 − ρ 2 σX 1 − ρ 2   Thus, fX|Y (· |y) ∼ N mX + σσYX ρ (y − mY ), σX2 (1 − ρ 2 ) .

!

49. From the solution of Problem 48, we have that   σY fY |X (· |x) ∼ N mY + ρ (x − mX ), σY2 (1 − ρ 2 ) σX and

  σX fX|Y (· |y) ∼ N mX + ρ (y − mY ), σX2 (1 − ρ 2 ) . σY

Hence,

E[Y |X = x] =

Z ∞

y fY |X (y|x) dx = mY +

σY ρ (x − mX ), σX

E[X|Y = y] =

Z ∞

x fX|Y (x|y) dx = mX +

σX ρ (y − mY ). σY

and

−∞

−∞

50. From Problem 48, we know that fX ∼ N(mX , σX2 ). Hence, E[X] = mX and E[X 2 ] = var(X) + m2X = σX2 + m2X . To compute cov(X,Y ), we use the law of total probability and substitution to write cov(X,Y ) = E[(X − mX )(Y − mY )] = = =

Z ∞

−∞

Z ∞

−∞

Z ∞

Z ∞

−∞

E[(X − mX )(Y − mY )|Y = y] fY (y) dy

E[(X − mX )(y − mY )|Y = y] fY (y) dy (y − mY )E[(X − mX )|Y = y] fY (y) dy

 (y − mY ) E[X|Y = y] − mX fY (y) dy −∞ Z ∞ nσ o X = (y − mY ) ρ (y − mY ) fY (y) dy σY −∞ =

126

Chapter 7 Problem Solutions Z

∞ σX σX ρ ρ · E[(Y − mY )2 ] (y − mY )2 fY (y) dy = σY σY −∞ σX ρ · σY2 = σX σY ρ . = σY

=

It then follows that

51.

cov(X,Y ) = ρ. σX σY

(a) Using the results of Problem 47, we have fU (u) = =

Z ∞

fUV (u, v) dv =

Z ∞

fUV (u, v) du =

−∞

1 2

Z ∞

ψρ1 (u, v) + ψρ2 (u, v) dv

Z ∞

ψρ1 (u, v) + ψρ2 (u, v) du

−∞

1 [ψ (u) + ψ (u)] = ψ (u), 2

and fV (v) = =

−∞

1 2

−∞

1 [ψ (v) + ψ (v)] = ψ (v). 2

Thus, fU and fV are N(0, 1) densities. (b) Write

ρ := E[UV ] = =

Z ∞Z ∞ −∞

Z ∞Z ∞

−∞ −∞

uv fUV (u, v) du dv

1 uv · [ψρ1 (u, v) + ψρ2 (u, v)] du dv 2 −∞

Z ∞ Z ∞  Z ∞Z ∞ 1 = uvψρ1 (u, v) du dv + uvψρ2 (u, v) du dv 2 −∞ −∞ −∞ −∞ ρ1 + ρ 2 . = 2 (c) If indeed fUV (u, v) =

1 [ψρ (u, v) + ψρ2 (u, v)] 2 1

is a bivariate normal density, then exp fUV (u, v) = In particular then,





−1 [u2 − 2ρ uv + v2 ] 2(1−ρ 2 )

2π

p

1−ρ

2

.

1 [ψρ (u, u) + ψρ2 (u, u)], 2 1 # " 2 2 1 e−u /(1+ρ1 ) e−u /(1+ρ2 ) q = + q . 2 2π 1 − ρ 2 2π 1 − ρ 2

fUV (u, u) = or

2

e−u /(1+ρ ) p 2π 1 − ρ 2

1

2

Chapter 7 Problem Solutions

127

Since t := u2 ≥ 0 is arbitrary, part (iii) of the hint tells us that 2π which is false.

p

1 1−ρ

=

2

−1 −1 q q = = 0, 2 4π 1 − ρ 1 4π 1 − ρ 22

(d) First observe that fV |U (v|u) =

fUV (u, v) 1 = fU (u) 2



 ψρ1 (u, v) ψρ2 (u, v) + . ψ (u) ψ (u) | {z } | {z }

N(ρ1 u,1−ρ12 )

Hence, Z ∞

−∞

N(ρ2 u,1−ρ22 )

1 [(1 − ρ12 ) + (ρ1 u)2 + (1 − ρ22 ) + (ρ2 u)2 ] 2 1 = [2 − ρ12 − ρ22 + (ρ12 + ρ22 )u2 ], 2

v2 fV |U (v|u) dv =

which depends on u unless ρ1 = ρ2 = 0. 52. For u0 , v0 ≥ 0, let D := {(u, v) : u ≥ u0 , v ≥ v0 }. Then P(U > u0 ,V > v0 ) =

ZZ

ψρ (u, v) du dv

D

=

Z tan−1 (v0 /u0 ) Z ∞

v0 / sin θ

0

+

Z π /2

Z ∞

ψρ (r cos θ , r sin θ )r dr d θ

tan−1 (v0 /u0 ) u0 / cos θ

Now, since exp

ψρ (r cos θ , r sin θ ) =



ψρ (r cos θ , r sin θ )r dr d θ . (##) 

−r2 [1 − ρ sin 2θ ] 2(1−ρ 2 )

2π

(#)

p 1 − ρ2

,

we can express the anti-derivative of rψρ (r cos θ , r sin θ ) with respect to r in closed form as p   − 1 − ρ2 −r2 [1 − ρ sin 2θ ] . exp 2π (1 − ρ sin 2θ ) 2(1 − ρ 2 ) Hence, the double integral in (#) reduces to Z tan−1 (v0 /u0 ) 0

h(v20 , θ ) d θ .

The double integral in (##) reduces to p   Z π /2 −u20 1 − ρ2 exp [1 − ρ sin 2 θ ] dθ . 2(1 − ρ 2 ) cos2 θ tan−1 (v0 /u0 ) 2π (1 − ρ sin 2θ )

128

Chapter 7 Problem Solutions Applying the change of variable t = π /2 − θ , dt = −d θ , we obtain Z π /2−tan−1 (v0 /u0 ) 0

hρ (u20 ,t) dt.

53. If ρ = 0 in Problem 52, then U and V are independent. Taking u0 = v0 = x0 , Q(x0 )2 =

Z π /4 0

= 2

h0 (x02 , θ ) d θ +

Z π /4 0 2

h0 (x02 , θ ) d θ

Z π /4 exp[−x02 /(2 sin θ )]

2π

0

54. Factor fXY Z (x, y, z) =

1 dθ = π

Z π /4 0



 −x02 exp dθ . 2 sin2 θ

2 exp[−|x − y| − (y − z)2 /2] √ , z5 2π

as

z ≥ 1,

2

4 e−(y−z) /2 1 −|x−y| . · √ · 2e z5 2π Now, the second factor on the right is an N(z, 1) density in the variable y, and the third factor is a Laplace(1) density that has been shifted to have mean y. Hence, the integral of the third factor with respect to x yields one, and we have by inspection that fXY Z (x, y, z) =

2

fY Z (y, z) =

4 e−(y−z) /2 · √ , z5 2π

z ≥ 1.

We then easily see that fX|Y Z (x|y, z) :=

fXY Z (x, y, z) = fY Z (y, z)

1 −|x−y| , 2e

z ≥ 1.

Next, since the right-hand factor in the formula for fY Z (y, z) is an N(z, 1) density in y, if we integrate this factor with respect to y, we get one. Thus, 4 fZ (z) = 5 , z ≥ 1. z We can now see that 2

fY Z (y, z) e−(y−z) /2 √ = , fZ (z) 2π

fY |Z (y|z) :=

z ≥ 1.

55. To find fXY (x, y), first write fXY Z (x, y, z) := = = =

e−(x−y)

2 /2

e−(x−y)

2 /2

√ 2 2 ) /2

e−(x−y)

2 /2

√ 2 2 ) /2

e−(x−y)

2 /2

√ 2 2 ) /2

e−(y−z) (2π )3/2 e−(y/ 2π

e−(y/ √ 2π 2 e−(y/ √ 2π 2

2 /2

e−z

2 /2

=

e−(x−y)

e−(z−y/2) · √ 2π

2 /2

2

e−(z−y/2) e−y (2π )3/2

2

2

e−(z−y/2) √ ·√ 2π / 2

√

e−[(z−y/2)/(1/ 2 )] √ √ · 2π / 2

2 /2

.

2 /4

Chapter 7 Problem Solutions

129

Now the right-hand factor is an N(y/2, 1/2) density in the variable z. Hence, its integral with respect to z is one. We thus have e−(x−y)

fXY (x, y) =

2 /2

e−(y/ √ 2π 2

√ 2 2 ) /2

√

2

2

e−(y/ 2 ) /2 e−(x−y) /2 = √ √ · √ , 2π 2π 2

which shows that Y ∼ N(0, 2), and given Y = y, X is conditionally N(y, 1). Thus, E[Y ] = 0 and

var(Y ) = 2.

Next, E[X] =

Z ∞

−∞

Z ∞

E[X|Y = y] fY (y) dy =

−∞

y fY (y) dy = E[Y ] = 0,

and var(X) = E[X 2 ] =

Z ∞

−∞

E[X 2 |Y = y] fY (y) dy =

Z ∞

= 1 + E[Y 2 ] = 1 + var(Y ) = 1 + 2 = 3.

−∞

(1 + y2 ) · fY (y) dy

Finally, E[XY ] = =

Z ∞

−∞

Z ∞

−∞

E[XY |Y = y] fY (y) dy = yE[X|Y = y] fY (y) dy =

Z ∞

−∞

Z ∞

−∞

E[Xy|Y = y] fY (y) dy

y2 fY (y) dy = E[Y 2 ] = var(Y ) = 2.

56. First write E[XY ] = = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞

−∞ −∞

E[XY |Y = y, Z = z] fY Z (y, z) dy dz E[Xy|Y = y, Z = z] fY Z (y, z) dy dz yE[X|Y = y, Z = z] fY Z (y, z) dy dz.

Since fX|Y Z (· |y, z) ∼ N(y, z2 ), the preceding conditional expectation is just y. Hence, E[XY ] =

Z ∞Z ∞

−∞ −∞

y2 fY Z (y, z) dy dz = E[Y 2 ] =

Z ∞

−∞

E[Y 2 |Z = z] fZ (z) dz.

Since fY |Z (· |z) ∼ exp(z), the preceding conditional expectation is just 2/z2 . Thus, E[XY ] =

Z ∞ 2 −∞

z2

fZ (z) dz =

Z 2 2 1

3 6 · z2 dz = . z2 7 7

A similar analysis yields E[Y Z] = =

Z ∞

−∞

Z ∞

−∞

E[Y Z|Z = z] fZ (z) dz = zE[Y |Z = z] fZ (z) dz =

Z ∞

−∞

Z ∞

−∞

E[Y z|Z = z] fZ (z) dz

z(1/z) fZ (z) dz =

Z ∞

−∞

fZ (z) dz = 1.

130

Chapter 7 Problem Solutions

57. Write E[XY Z] = = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

E[XY Z|Y = y, Z = z] fY Z (y, z) dy dz E[Xyz|Y = y, Z = z] fY Z (y, z) dy dz yzE[X|Y = y, Z = z] fY Z (y, z) dy dz.

Since fX|Y Z (· |y, z) is a shifted Laplace density with mean y, the preceding conditional expectation is just y. Hence, Z ∞Z ∞

E[XY Z] =

2

2

y z fY Z (y, z) dy dz = E[Y Z] =

−∞ −∞ Z ∞ 2

=

−∞

E[Y z|Z = z] fZ (z) dy =

Z ∞

−∞

Z ∞

−∞

E[Y 2 Z|Z = z] fZ (z) dy

zE[Y 2 |Z = z] fZ (z) dy.

Since fY |Z (· |z) ∼ N(z, 1), the preceding conditional expectation is just 1 + z2 . Thus, E[XY Z] =

Z ∞

−∞

z(1 + z2 ) fZ (z) dz =

= 4(1/3 + 1) = 16/3.

Z ∞ 1

[z + z3 ] · 4/z5 dz = 4

Z ∞ 1

z−4 + z−2 dz

58. Write E[XY Z] = = = = = =

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞ Z ∞ 2 −∞ Z ∞

−∞

E[XY Z|X = x,Y = y] fXY (x, y) dx dy E[xyZ|X = x,Y = y] fXY (x, y) dx dy xyE[Z|X = x,Y = y] fXY (x, y) dx dy x2 y fXY (x, y) dx dy = E[X 2Y ] =

E[x Y |X = x] fX (x) dx = x3 fX (x) dx =

Z 2

Z ∞

−∞

Z ∞

−∞

E[X 2Y |X = x] fX (x) dx

x2 E[Y |X = x] fX (x) dx

x3 dx = 15/4.

1

59. We use the law of total probability, substitution, and independence to write N

ϕY (ν ) = E[e jνY ] = E[e jν ∑i=1 Xi ] =

∞

n=1 ∞

=

∑ E[e jν ∑i=1 Xi |N = n]P(N = n)

n=1 ∞

=

n

n

∑ E[e jν ∑i=1 Xi ]P(N = n)

n=1

N

∑ E[e jν ∑i=1 Xi |N = n]P(N = n)

Chapter 7 Problem Solutions ∞

= =



n

∑ E ∏e

i=1 n=1 ∞  n

∑ ∏ E[e

n=1 i=1 ∞

=

jν Xi

131

 P(N = n)

jν Xi

 ] P(N = n)

∑ ϕX (ν )n P(N = n)

= GN (ϕX (ν )).

n=1

Now, if N ∼ geometric1 (p), GN (z) = [(1 − p)z]/[1 − pz], and if X ∼ exp(λ ), ϕX (ν ) = λ /(λ − jν ). Then

ϕY (ν ) =

(1 − p)λ (1 − p)ϕX (ν ) (1 − p)λ /(λ − jν ) (1 − p)λ = = = , 1 − pϕX (ν ) 1 − pλ /(λ − jν ) (λ − jν ) − pλ (1 − p)λ − jν

which is the exp((1 − p)λ ) characteristic function. Thus, Y ∼ exp((1 − p)λ ).

CHAPTER 8

Problem Solutions 1. We have

    230 280 330 10 40   20 50  7 8 9 =  340 410 480  4 5 6 450 540 630 30 60 

and

 230 280 330 tr  340 410 480  = 1270. 450 540 630 

2. MATLAB. See the answer to the previous problem. 3. MATLAB. We have



 7 4 A0 =  8 5  . 9 6

4. Write r

r

tr(AB) = ∑ (AB)ii = ∑

i=1

i=1

5.



n

∑ Aik Bki

k=1



n

=



r

∑ ∑ Bki Aik

k=1

i=1



n

=

∑ (BA)kk = tr(BA).

k=1

(a) Write 0

tr(AB ) =

r

0

∑ (AB )ii

i=1

r

=



n

0

∑ ∑ Aik (B )ki

i=1

k=1



r

=

n

∑ ∑ Aik Bik .

i=1 k=1

(b) If tr(AB0 ) = 0 for all B, then in particular, it is true for B = A; i.e., 0 = tr(AA0 ) =

r

n

∑ ∑ A2ik ,

i=1 k=1

which implies Aik = 0 for all i and k. In other words, A is the zero matrix of size r × n. 6. Following the hint, we first write 0 ≤ kx − λ yk2 = hx − λ y, x − λ yi = kxk2 − 2λ hx, yi + λ 2 kyk2 . Taking λ = hx, yi/kyk2 yields 0 ≤ kxk2 − 2

|hx, yi|2 |hx, yi|2 |hx, yi|2 + kyk2 = kxk2 − , 2 4 kyk kyk kyk2 132

Chapter 8 Problem Solutions

133

which can be rearranged to get |hx, yi|2 ≤ kxk2 kyk2 . Conversely, suppose |hx, yi|2 = kxk2 kyk2 . There are two cases to consider. If y 6= 0, then reversing the above sequence of observations implies 0 = kx − λ yk2 , which implies x = λ y. On the other hand, if y = 0 and if |hx, yi| = kxk kyk,

then we must have kxk = 0; i.e., x = 0 and y = 0, and in this case x = λ y for all λ . 7. Consider the i j component of E[XB]. Since   (E[XB])i j = E[(XB)i j ] = E ∑ Xik Bk j =

∑ E[Xik ]Bk j

=

k

k

∑(E[X])ik Bk j k

= (E[X]B)i j holds for all i j, E[XB] = E[X]B. 8. tr(E[X]) =

∑(E[X])ii = i

∑ E[Xii ] = E i



∑ Xii i



= E[tr(X)].

9. Write E[kX − E[X]k2 ] = E[(X − E[X])0 (X − E[X])], which is a scalar, n o = tr E[(X − E[X])0 (X − E[X])] h n oi = E tr (X − E[X])0 (X − E[X]) , by Problem 8, oi h n = E tr (X − E[X])(X − E[X])0 , by Problem 4, n o = tr E[(X − E[X])(X − E[X])0 ] , by Problem 8, n

= tr(C) =

∑ Cii

i=1

n

=

∑ var(Xi ).

i=1

   0 10. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and     E[X]2 E[X]E[Y ] E[X]E[Z] E[X 2 ] E[XY ] E[XZ] cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  , E[Z]E[X] E[Z]E[Y ] E[Z]2 E[ZX] E[ZY ] E[Z 2 ] we begin by computing all the entries of these matrices. For the mean vector, Z 2 Z 2 2 3 2 3 E[Z] = z · 7 z dz = 7 z3 dz = 37 · 41 z4 = 3 · 15/28 = 45/28. 1

1

1

Next,

E[Y ] =

Z 2 1

E[Y |Z = z] fZ (z) dz =

Z 2 1

1 z

·

3 2 7 z dz

=

Since E[U] = 0 and since U and Z are independent,

3 7

Z 2 1

z dz =

3 7

2 · 12 z2 = 9/14.

E[X] = E[ZU +Y ] = E[Z]E[U] + E[Y ] = E[Y ] = 9/14.

1

134

Chapter 8 Problem Solutions Thus, the desired mean vector is   E [X,Y, Z]0 = [9/14, 9/14, 45/28]0 .

We next compute the correlations. First, E[Y Z] =

Z 2 1

=

Z 2 1

E[Y Z|Z = z] fZ (z) dz = zE[Y |Z = z] fZ (z) dz =

Next,

Z 2

E[Z 2 ] =

1

z2 · 37 z2 dz =

3 7

Z 2 1

E[Y z|Z = z] fZ (z) dz

Z 2

z(1/z) fZ (z) dz =

Z 2

3 7

1

z4 dz =

1

Again using the fact that E[U] = 0 and independence,

Z 2 1

fZ (z) dz = 1.

2 · 51 z5 = 93/35. 1

E[XZ] = E[(ZU +Y )Z] = E[Z 2 ]E[U] + E[Y Z] = E[Y Z] = 1. Now, E[Y 2 ] =

Z 2 1

E[Y 2 |Z = z] fZ (z) dz =

We can now compute

Z 2 1

(2/z2 ) · 73 z2 dz = 6/7.

E[XY ] = E[(ZU +Y )Y ] = E[ZY ]E[U] + E[Y 2 ] = 6/7, and E[X 2 ] = E[(ZU +Y )2 ] = E[Z 2 ]E[U 2 ] + 2E[U]E[ZY ] + E[Y 2 ] = E[Z 2 ] + E[Y 2 ] = 93/35 + 6/7 = 123/35. We now have that    81/196 81/196 405/392 123/35 6/7 1 cov([X,Y, Z]0 ) =  6/7 6/7 1  −  81/196 81/196 405/392  405/392 405/392 2025/784 1 1 93/35   3.1010 0.4439 −0.0332 =  0.4439 0.4439 −0.0332  . −0.0332 −0.0332 0.0742 

   0 11. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and     E[X 2 ] E[XY ] E[XZ] E[X]2 E[X]E[Y ] E[X]E[Z] cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  , E[ZX] E[ZY ] E[Z 2 ] E[Z]E[X] E[Z]E[Y ] E[Z]2

we compute all the entries of these matrices. To make this job easier, we first factor fXY Z (x, y, z) =

2 exp[−|x − y| − (y − z)2 /2] √ , z5 2π

z ≥ 1,

Chapter 8 Problem Solutions

135

as fX|Y Z (x|y, z) fY |Z (y|z) fZ (z) by writing 2

e−(y−z) /2 4 √ fXY Z (x, y, z) = · 5 , z ≥ 1. z 2π We then see that as a function of x, fX|Y Z (x|y, z) is a shifted Laplace(1) density. Similarly, as a function of y, fY |Z (y|z) is an N(z, 1) density. Thus, 1 −|x−y| · 2e

E[X] =

Z ∞Z ∞

−∞ −∞

= E[Y ] = Z ∞

E[X|Y = y, Z = z] fY Z (y, z) dy dz = Z ∞

Z ∞

E[Y |Z = z] fZ (z) dz =

−∞

−∞

Z

Z ∞Z ∞

−∞ −∞

y fY Z (y, z) dy dz

z fZ (z) dz = E[Z]

∞ 4 4 z · 5 dz = dz = 4/3. 4 z 1 1 z   Thus, E [X,Y, Z]0 = [4/3, 4/3, 4/3]0 . We next compute

=

E[X 2 ] = =

where E[Y 2 ] = Now,

Z ∞

−∞

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞

−∞ −∞

E[X 2 |Y = y, Z = z] fY Z (y, z) dy dz

(2 + y2 ) fY Z (y, z) dy dz = 2 + E[Y 2 ],

E[Y 2 |Z = z] fZ (z) dz = Z ∞

Z ∞

−∞

Z ∞ 4

4 dz = z5 1 Thus, E[Y 2 ] = 3 and E[X 2 ] = 5. We next turn to E[Z 2 ] =

E[XY ] = = = = We also have E[XZ] = = = = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞

−∞ Z ∞ −∞

z2

(1 + z2 ) fZ (z) dz = 1 + E[Z 2 ].

z3

1

dz = 2.

E[XY |Y = y, Z = z] fY Z (y, z) dy dz E[Xy|Y = y, Z = z] fY Z (y, z) dy dz yE[X|Y = y, Z = z] fY Z (y, z) dy dz y2 fY Z (y, z) dy dz = E[Y 2 ] = 3.

E[XZ|Y = y, Z = z] fY Z (y, z) dy dz E[Xz|Y = y, Z = z] fY Z (y, z) dy dz yz fY Z (y, z) dy dz = E[Y Z]

E[Y Z|Z = z] fZ (z) dz =

Z ∞

−∞

z2 fZ (z) dz = E[Z 2 ] = 2.

zE[Y |Z = z] fZ (z) dz

136

Chapter 8 Problem Solutions We now have that

   5 3 2 1 1 1 16 cov([X,Y, Z]0 ) =  3 3 2  −  1 1 1  9 2 2 2 1 1 1     3.2222 1.2222 0.2222 29 11 2 1 11 11 2  =  1.2222 1.2222 0.2222  . = 9 0.2222 0.2222 0.2222 2 2 2 

   0 12. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and     E[X]2 E[X]E[Y ] E[X]E[Z] E[X 2 ] E[XY ] E[XZ] cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  , E[Z]E[X] E[Z]E[Y ] E[Z]2 E[ZX] E[ZY ] E[Z 2 ] we compute all the entries of these matrices. We begin with E[Z] = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

E[Z|Y = y, X = x] fXY (x, y) dy dx x fXY (x, y) dy dx = E[X] = 3/2.

Next, E[Y ] =

Z ∞

−∞

E[Y |X = x] fX (x) dx =

Z ∞

−∞

x fX (x) dx = E[X] = 3/2.

We now compute E[XY ] = =

Z ∞

−∞

Z ∞

−∞

E[XY |X = x] fX (x) dx = xE[Y |X = x] fX (x) dx =

Z ∞

−∞

Z ∞

E[xY |X = x] fX (x) dx

x2 fX (x) dx = E[X 2 ]

−∞ 2

= var(X) + E[X]2 = 1/12 + (3/2) = 7/3. Then E[XZ] = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

E[XZ|Y = y, X = x] fXY (x, y) dy dx xE[Z|Y = y, X = x] fXY (x, y) dy dx

=

Z ∞Z ∞

x2 fXY (x, y) dy dx = E[X 2 ] = 7/3,

E[Y Z] =

Z ∞Z ∞

E[Y Z|Y = y, X = x] fXY (x, y) dy dx

and

= =

−∞ −∞

−∞ −∞

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞ −∞ −∞

yE[Z|Y = y, X = x] fXY (x, y) dy dx xy fXY (x, y) dy dx = E[XY ] = 7/3.

Chapter 8 Problem Solutions

137

Next, E[Y 2 ] =

Z ∞

E[Z 2 ] =

Z ∞Z ∞

and

=

−∞

E[Y 2 |X = x] fX (x) dx =

−∞ −∞ Z ∞Z ∞

−∞ −∞

Z ∞

−∞

2x2 fX (x) dx = 2E[X 2 ] = 14/3,

E[Z 2 |Y = y, X = x] fXY (x, y) dy dx (1 + x2 ) fXY (x, y) dy dx = 1 + E[X 2 ] = 1 + 7/3 = 10/3.

We now have that

   7 7 7  3 2 1 1 1 1 1 cov([X,Y, Z]0 ) =  7 14 7  − 3 2 1 1 7 7 10    0.0833 1 1 1 1  1 29 1  =  0.0833 = 12 0.0833 1 1 13

 1 1 1

 0.0833 0.0833 2.4167 0.0833  . 0.0833 1.0833

   0 13. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and     E[X]2 E[X]E[Y ] E[X]E[Z] E[X 2 ] E[XY ] E[XZ] cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  , E[Z]E[X] E[Z]E[Y ] E[Z]2 E[ZX] E[ZY ] E[Z 2 ]

we compute all the entries of these matrices. In order to do this, we first note that Z ∼ N(0, 1). Next, as a function of y, fY |Z (y|z) is an N(z, 1) density. Similarly, as a function of x, fX|Y Z (x|y, z) is an N(y, 1) density. Hence, E[Z] = 0, E[Y ] =

Z ∞

−∞

E[Y |Z = z] fZ (z) dz =

and E[X] = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞

−∞

z fZ (z) dz = E[Z] = 0,

E[X|Y = y, Z = z] fY Z (y, z) dy dz y fY Z (y, z) dy dz = E[Y ] = 0.

We next compute E[XY ] = = =

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞ −∞ −∞

Z ∞Z ∞

−∞ −∞

= E[Y 2 ] =

E[XY |Y = y, Z = z] fY Z (y, z) dy dz E[Xy|Y = y, Z = z] fY Z (y, z) dy dz yE[X|Y = y, Z = z] fY Z (y, z) dy dz = Z ∞

−∞

E[Y 2 |Z = z] fZ (z) dz =

= 1 + E[Z 2 ] = 2.

Z ∞

−∞

Z ∞Z ∞

−∞ −∞

y2 fY Z (y, z) dy dz

(1 + z2 ) fZ (z) dz

138

Chapter 8 Problem Solutions Then E[Y Z] = =

Z ∞

−∞

Z ∞

−∞

E[Y Z|Z = z] fZ (z) dz =

Z ∞

−∞

zE[Y |Z = z] fZ (z) dz

z2 fZ (z) dz = E[Z 2 ] = 1,

and E[XZ] = = =

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

E[XZ|Y = y, Z = z] fY Z (y, z) dy dz zE[X|Y = y, Z = z] fY Z (y, z) dy dz yz fY Z (y, z) dy dz = E[Y Z] = 1.

Next, E[Y 2 ] =

Z ∞

−∞

E[Y 2 |Z = z] fZ (z) dz =

Z ∞

−∞

(1 + z2 ) fZ (z) dz = 1 + E[Z 2 ] = 2,

and E[X 2 ] = =

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞

−∞ −∞

E[X 2 |Y = y, Z = z] fY Z (y, z) dy dz (1 + y2 ) fY Z (y, z) dy dz = 1 + E[Y 2 ] = 3.

Since E[Z 2 ] = 1, we now have that 

 3 2 1 cov([X,Y, Z]0 ) =  2 2 1  . 1 1 1 14. We first note that Z ∼ N(0, 1). Next, as a function of y, fY |Z (y|z) is an N(z, 1) density. Similarly, as a function of x, fX|Y Z (x|y, z) is an N(y, 1) density. Hence, E[e j(ν1 X+ν2Y +ν3 Z) ] = = = =

Z ∞Z ∞

−∞ −∞ Z ∞Z ∞ −∞ −∞ Z ∞Z ∞

−∞ −∞ Z ∞Z ∞ −∞ −∞ 2

= e−ν1 /2 2

E[e j(ν1 X+ν2Y +ν3 Z) |Y = y, Z = z] fY Z (y, z) dy dz E[e j(ν1 X+ν2 y+ν3 z) |Y = y, Z = z] fY Z (y, z) dy dz E[e jν1 X |Y = y, Z = z]e jν2 y e jν3 z fY Z (y, z) dy dz 2

e jν1 y−ν1 /2 e jν2 y e jν3 z fY Z (y, z) dy dz

Z ∞Z ∞

−∞ −∞

e j(ν1 +ν2 )y e jν3 z fY Z (y, z) dy dz

= e−ν1 /2 E[e j(ν1 +ν2 )Y e jν3 Z ]

Chapter 8 Problem Solutions 2

= e−ν1 /2 2

= e−ν1 /2 2

= e−ν1 /2 2

Z ∞

−∞ Z ∞ −∞ Z ∞

139

E[e j(ν1 +ν2 )Y e jν3 Z |Z = z] fZ (z) dz e jν3 z E[e j(ν1 +ν2 )Y |Z = z] fZ (z) dz e jν3 z e j(ν1 +ν2 )z−(ν1 +ν2 )

−∞

= e−ν1 /2 e−(ν1 +ν2 )

2 /2

2

2 /2

2

2

= e−ν1 /2 e−(ν1 +ν2 )

Z ∞

−∞

2 /2

fZ (z) dz

e j(ν1 +ν2 +ν3 )z fZ (z) dz

E[e j(ν1 +ν2 +ν3 )z ]

= e−ν1 /2 e−(ν1 +ν2 ) /2 e−(ν1 +ν2 +ν3 ) 2 2 2 = e−[ν1 +(ν1 +ν2 ) +(ν1 +ν2 +ν3 ) ]/2 .

2 /2

15. RY := E[YY 0 ] = E[(AX)(AX)0 ] = E[AXX 0 A0 ] = AE[XX 0 ]A0 = ARX A0 . 16. First, since R = E[XX 0 ], we see that R0 = E[XX 0 ]0 = E[(XX 0 )0 ] = E[XX 0 ] = R. Thus, R is symmetric. Next, define the scalar Y := c0 X. Then 0 ≤ E[Y 2 ] = E[YY 0 ] = E[(c0 X)(c0 X)0 ] = E[c0 XX 0 c] = c0 E[XX 0 ]c = c0 Rc, and we see that R is positive semidefinite. 17. Use the Cauchy–Schwarz inequality to write (CXY )i j = E[(Xi − mX,i )(Y j − mY, j )] q q ≤ E[(Xi − mX,i )2 ]E[(Y j − mY, j )2 ] = (CX )ii (CY ) j j . 18. If P = then 

U V



:= P0



X Y



=





 cos θ − sin θ , sin θ cos θ

cos θ sin θ − sin θ cos θ



X Y



=



 X cos θ +Y sin θ . −X sin θ +Y cos θ

We now have to find θ such that E[UV ] = 0. Write E[UV ] = E[(X cos θ +Y sin θ )(−X sin θ +Y cos θ )] = E[(Y 2 − X 2 ) sin θ cos θ + XY (cos2 θ − sin2 θ )] =

σY2 − σX2 sin 2θ + E[XY ] cos 2θ , 2

which is zero if and only if tan 2θ = Hence,

2E[XY ] . σX2 − σY2

  1 −1 2E[XY ] . θ = tan 2 σX2 − σY2

140 19.

Chapter 8 Problem Solutions (a) Write (ei e0i )mn = (ei )m (e0i )n , which equals one if and only if m = i and n = i. Hence, ei e0i must be all zeros except at position i i where it is one. (b) Write  0   e1 e1 e4 e5  e04  E 0E = e5 = e1 e01 + e4 e04 + e5 e05

= diag(1, 0, 0, 0, 0) + diag(0, 0, 0, 1, 0) + diag(0, 0, 0, 0, 1) = diag(1, 0, 0, 1, 1). 20.

(a) We must solve P0CX P = diagonal. With X := U +V , we have CX = E[XX 0 ] = E[(U +V )(U +V )0 ] = CU + E[UV 0 ] + E[VU 0 ] +CV = QMQ0 + I = Q(M + I)Q0 . Hence, Q0CX Q = M + I, which is diagonal. The point here is that we may take P = Q. (b) We now put Y := P0 X = QX = Q(U +V ). Then CY = E[YY 0 ] = E[Q(U +V )(U 0 +V 0 )Q0 ] = Q{CU + E[UV 0 ] + E[VU 0 ] +CV }Q0 = QCU Q0 + QIQ0 = M + I.

21. Starting with u = x + y and v = x − y, we have x =

u+v 2

#

"

u−v . 2

y =

and

We can now write dH =

"

∂x ∂u ∂y ∂u

∂x ∂v ∂y ∂v

=

1/2

1/2

1/2 −1/2

#

,

det dH = −1/2,

and | det dH | = 1/2, and so fUV (u, v) = fXY

u+v u−v 1 , · . 2 2 2

22. Starting with u = xy and v = y/x, write y = xv. Then u = x2 v, and x = (u/v)1/2 . We also have y = (u/v)1/2 v = (uv)1/2 . Then ∂x ∂u ∂y ∂u

√ = (1/2)/ uv, p = (1/2) v/u,

∂x ∂v ∂y ∂v

√ = (−1/2) u/v3/2 , p = (1/2) u/v.

Chapter 8 Problem Solutions In other words, dH = and so

"

141

# √ √ (1/2)/ uv (−1/2) u/v3/2 , p p (1/2) v/u (1/2) u/v

1 1 1 | det dH | = + = . 4v 4v 2|v|

Thus,

fUV (u, v) = fXY For fU , write fU (u) =

Z ∞ 0

p

√ 1 , u/v, uv 2v

fUV (u, v) dv =

Z ∞ 0

fXY

u, v > 0.

p

√ 1 u/v, uv dv 2v

√ Now make the change of variable y = uv, or y2 = uv. Then 2y dy = u dv, and fU (u) =

Z ∞ 0

fXY

u 1 , y dy. y y

For fV , write fV (v) =

Z ∞ 0

fUV (u, v) du =

Z ∞ 0

fXY

p

p

√ 1 u/v, uv du. 2v

This time make the change of variable x = u/v or x2 = u/v. Then 2x dx = du/v, and Z ∞ fV (v) = fXY (x, vx)x dx. 0

23. Starting with u = x and v = y/x, we find that y = xv = uv. Hence, " ∂x ∂x # # " 1 0 ∂u ∂v dH = ∂ y ∂ y = , det dH = u, and | det dH | = |u|. v u ∂u ∂v Then fUV (u, v) = fXY (u, uv)|u| =

λ −λ |u| λ −λ |uv| · 2e · |u|, 2e

and fV (v) = =

Z ∞

−∞

−λ (1+|v|)|u| λ2 du 4 |u|e

λ 2(1 + |v|)

Z ∞ 0

=

λ2 2

Z ∞ 0

ue−λ (1+|v|)u du

u · λ (1 + |v|)e−λ (1+|v|)u du.

Now, this last integral is the mean of an exponential density with parameter λ (1+|v|). Hence, 1 1 λ · = . fV (v) = 2(1 + |v|) λ (1 + |v|) 2(1 + |v|)2

142

Chapter 8 Problem Solutions

24. Starting with u =

√ √ −2 ln x cos(2π y) and v = −2 ln x sin(2π y), we have

u2 + v2 = (−2 ln x)[cos2 (2π y) + sin2 (2π y)] = −2 ln x. Hence, x = e−(u

2 +v2 )/2

. We also have

v = tan(2π y) or u

1 tan−1 (v/u). 2π

y =

We can now write ∂x ∂u ∂y ∂u

In other words,

and so

= −ue−(u =

2 +v2 )/2

1 1 2π 1+(u/v)2



dH = 

·

∂x ∂v ∂y ∂v

,

−v , u2

−ue−(u

2 +v2 )/2

1 1 2π 1+(u/v)2

· −v u2

= −ve−(u =

2 +v2 )/2

1 1 2π 1+(u/v)2

−ve−(u

2 +v2 )/2

1 1 2π 1+(u/v)2

· 1u

·

,

1 u.



,

2 2 −1 e−(u2 +v2 )/2 1 e−(u +v )/2 v2 | det dH| = − · 2π 1 + (v/u)2 2π 1 + (v/u)2 u2 2 2 1 (v/u)2 e−(u +v )/2 + = 1 + (v/u)2 1 + (v/u)2 2π 2

2

e−u /2 e−v /2 √ . = √ 2π 2π

2

2

We next use the formula fUV (u, v) = fXY (x, y) · | det dH|, where x = e−(u +v )/2 and y = tan−1 (v/u)/(2π ). Fortunately, since these formulas for x and y lie in (0, 1], fXY (x, y) = I(0,1] (x)I(0,1] (y) = 1, and we see that 2

2

e−u /2 e−v /2 √ fUV (u, v) = 1 · | det dH| = √ . 2π 2π √ 2 Integrating out v shows that fU (u) = e−u /2 / 2π , and integrating out v shows that √ 2 fV (v) = e−v /2 / 2π . It now follows that fUV (u, v) = fU (u) fV (v), and we see that U and V are independent. 25. Starting with u = x + y and v = x/(x + y) = x/u, we see that v = x/u, or x = uv. Next, from y = u − x = u − uv, we get y = u(1 − v). We can now write ∂x ∂u ∂y ∂u

= v, = 1 − v,

In other words, dH =



∂x ∂v ∂y ∂v

= u, = −u.

 v u , 1 − v −u

Chapter 8 Problem Solutions

143

and so | det dH| = | − uv − u(1 − v)| = |u|. We next write fUV (u, v) = fXY (uv, u(1 − v))|u|. If X and Y are independent gamma RVs, then for u > 0 and 0 < v < 1,

λ (λ uv) p−1 e−λ uv λ (λ u(1 − v))q−1 e−λ u(1−v) · ·u Γ(p) Γ(q)

fUV (u, v) =

λ (λ u) p+q−1 e−λ u Γ(p + q) p−1 · v (1 − v)q−1 , Γ(p + q) Γ(p)Γ(q)

=

which we recognize as the product of a gamma(p + q, λ ) and a beta(p, q) density. Hence, it is easy to integrate out either u or v and show that fUV (u, v) = fU (u) fV (v), where fU ∼ gamma(p + q, λ ) and fV ∼ beta(p, q). Thus, U and V are independent. 26. Starting with u = x + y and v = x/y, write x = yv and then u = yv + y = y(v + 1). Solve for y = u/(v + 1) and x = u − y = u − u/(v + 1) = uv/(v + 1). Next, ∂x ∂u ∂y ∂u

∂x ∂v ∂y ∂v

= v/(v + 1), = 1/(v + 1),

In other words, dH =



= u/(v + 1)2 , = −u/(v + 1)2 .

 v/(v + 1) u/(v + 1)2 , 1/(v + 1) −u/(v + 1)2

and so −u(v + 1) −uv |u| u = = | det dH | = − . (v + 1)3 (v + 1)3 (v + 1)3 (v + 1)2

We next write

fUV (u, v) = fXY



 uv u |u| , . v + 1 v + 1 (v + 1)2

When X ∼ gamma(p, λ ) and Y ∼ gamma(q, λ ) are independent, then U and V are nonnegative, and fUV (u, v) = λ

u [λ uv/(v + 1)] p−1 e−λ uv/(v+1) [λ u/(v + 1)]q−1 e−λ u/(v+1) ·λ · Γ(p) Γ(q) (v + 1)2

= λ

v p−1 (λ u) p+q−1 e−λ u Γ(p + q) · · Γ(p + q) Γ(p)Γ(q) (v + 1) p+q

= λ

v p−1 (λ u) p+q−1 e−λ u · , Γ(p + q) B(p, q)(v + 1) p+q

which shows that U and V are independent with the required marginal densities.

144

Chapter 8 Problem Solutions

27. From solution of the Example at the end of the section, fR,Θ (r, θ ) = fXY (r cos θ , r sin θ )r. What is different in this problem is that X and Y are correlated Gaussian random variables; i.e., 2 2 2 e−(x −2ρ xy+y )/[2(1−ρ )] p fXY (x, y) = . 2π 1 − ρ 2

Hence,

fR,Θ (r, θ ) = =

re−(r

2 cos2 θ −2ρ (r cos θ )(r sin θ )+r2 sin2 θ )/[2(1−ρ 2 )]

2π re−r

p

1 − ρ2

2 (1−ρ sin 2θ )/[2(1−ρ 2 )]

2π

p 1 − ρ2

.

To find the density of Θ, we must integrate this with respect to r. Notice that the 2 2 integrand is proportional to re−λ r /2 , whose anti-derivative is −e−λ r /2 /λ . Here we 2 have λ = (1 − ρ sin 2θ )/(1 − ρ ). We can now write fΘ (θ ) = =

Z ∞ 0

2π

fR,Θ (r, θ ) dr =

p

1 − ρ2

1 p 2π 1 − ρ 2

1 − ρ 2 (1 − ρ sin 2θ )

=

Z ∞ 0

re−λ r

2 /2

p

dr =

1 − ρ2 . 2π (1 − ρ sin 2θ )

1 1 p · 2 λ 2π 1 − ρ

28. Since X ∼ N(0, 1), we have E[X] = E[X 3 ] = 0, E[X 4 ] = 3, and E[X 6 ] = 15. Since W ∼ N(0, 1), E[W ] = 0 too. Hence, E[Y ] = E[X 3 + W ] = 0. It then follows that mY , mX and b = mX − Amy are zero. We next compute CXY = E[XY ] = E[X(X 3 +W )] = E[X 4 ] + E[X]E[W ] = 3 + 0 = 3, and CY = E[Y 2 ] = E[X 6 + 2X 3W +W 2 ] = 15 + 0 + 1 = 16. Hence, ACY = CXY implies A = 3/16, and then Xb = A(Y − mY ) + mX = (3/16)Y .

29. First note that since X and W are zero mean, so is Y . Next,

CXY = E[XY ] = E[X(X +W )] = E[X 2 ] + E[XW ] = E[X 2 ] + E[X]E[W ] = E[X 2 ] = 1, and CY = E[Y 2 ] = E[(X +W )2 ] = E[X 2 + 2XW +W 2 ] = E[X 2 ] + 2E[X]E[W ] + E[W 2 ] = 1 + 0 + 2/λ 2 = 1 + 2/λ 2 . Then A = CXY /CY = 1/[1 + 2/λ 2 ], and Xb =

λ2 Y. 2+λ2

Chapter 8 Problem Solutions

145

30. We first have E[Y ] = E[GX +W ] = GmX + 0 = GmX . Next, CXY = E[(X − mX )(Y − mY )0 ] = E[(X − mX )(GX +W − GmX )0 ] = E[(X − mX )(G{X − mX } +W )0 ] = CX G0 +CXW = CX G0 , and CY = E[(Y − mY )(Y − mY )0 ] = E[(G{X − mX } +W )(G{X − mX } +W )0 ] = GCX G0 + GCXW +CW X G0 +CW = GCX G0 +CW , since X and W are uncorrelated. Solving ACY = CXY implies A = CX G0 (GCX G0 +CW )−1

and

Xb = CX G0 (GCX G0 +CW )−1 (Y − GmX ) + mX .

31. We begin with the result of Problem 30 that

A = CX G0 (GCX G0 +CW )−1 . Following the hint, we make the identifications α = CW , γ = CX , β = G, and δ = G0 . Then A = CX G0 (α + β γδ )−1 −1 −1 −1 −1 = CX G0 [CW −CW G(CX−1 + G0CW G)−1 G0CW ]

−1 −1 −1 −1 = CX G0CW −CX G0CW G(CX−1 + G0CW G)−1 G0CW

−1 −1 −1 = [CX −CX G0CW G(CX−1 + G0CW G)−1 ]G0CW

−1 −1 −1 −1 = [CX (CX−1 + G0CW G) −CX G0CW G](CX−1 + G0CW G)−1 G0CW

−1 −1 −1 −1 = [I +CX G0CW G −CX G0CW G](CX−1 + G0CW G)−1 G0CW −1 −1 = (CX−1 + G0CW G)−1 G0CW .

32. We begin with Xb = A(Y − mY ) + mX ,

where

ACY = CXY .

Next, with Z := BX, we have mZ = BmX and CZY = E[(Z − mZ )(Y − mY )0 ] = E[B(X − mX )(Y − mY )0 ] = BCXY . e Y = CZY . Starting with ACY = CXY , multiply this equation by B to We must solve AC e := BA solves the required equation. Hence, get (BA)CY = BCXY = CZY . We see that A the linear MMSE estimate of X based on Z is b (BA)(Y − mY ) + mZ = (BA)(Y − mY ) + BmX = B{A(Y − mY ) + mX } = BX.

33. We first show that the orthogonality condition

E[(CY )0 (X − AY )] = 0,

for all C,

146

Chapter 8 Problem Solutions implies A is optimal. Write E[kX − BY k2 ] = E[k(X − AY ) + (AY − BY )k2 ]

= E[k(X − AY ) + (A − B)Y )k2 ] = E[kX − AY k2 ] + 2E[{(A − B)Y }0 (X − AY )] + E[k(A − B)Y k2 ]

= E[kX − AY k2 ] + E[k(A − B)Y k2 ] ≥ E[kX − AY k2 ],

where the cross terms vanish by taking C = A − B in the orthogonality condition. Next, rewrite the orthogonality condition as E[(CY )0 (X − AY )] = tr{E[(CY )0 (X − AY )]} = E[tr{(CY )0 (X − AY )}] = E[tr{(X − AY )(CY )0 }] = tr{E[(X − AY )(CY )0 ]} = tr{(RXY − ARY )C0 }.

Now, this expression must be zero for all C, including C = RXY − ARY . However, since tr(DD0 ) = 0 implies D = 0, we conclude that the optimal A solves ARY = RXY . Next, the best constant estimator is easily found by writing E[kX − bk2 ] = E[k(X − mX ) + (mX − b)k2 ] = E[kX − mX k2 ] + kmX − bk2 . Hence, the optimal value of b is b = mX . 34. To begin, write b b 0 ] = E[{(X − mX ) − A(Y − mY )}{(X − mX ) − A(Y − mY )}0 ] E[(X − X)(X − X) = CX − ACY X −CXY A0 + ACY A0 . (∗) We now use the fact that ACY = CXY . If we multiply ACY = CXY on the right by A0 , we obtain ACY A0 = CXY A0 . Furthermore, since ACY A0 is symmetric, we can take the transpose of the above expression and obtain ACY A0 = ACY X . By making appropriate substitutions in (∗), we find that the error covariance is also given by CX − ACY X , CX −CXY A0 , and CX − ACY A0 . 35. Write n o b 2 ] = E[(X − X) b 0 (X − X)] b = tr E[(X − X) b 0 (X − X)] b E[kX − Xk

b 0 (X − X)}] b b b 0 }] = E[tr{(X − X) = E[tr{(X − X)(X − X) b b 0 ]} = tr{CX − ACY X }. = tr{E[(X − X)(X − X)

Chapter 8 Problem Solutions

147

36. MATLAB. We found 

 −0.0622 0.0467 −0.0136 −0.1007  0.0489 −0.0908 −0.0359 −0.1812   A =   −0.0269 0.0070 −0.0166 0.0921  0.0619 0.0205 −0.0067 0.0403

and MSE = 0.0806.

37. Write X in the form X = [Y 0 , Z 0 ]0 , where Y := [X1 , . . . , Xm ]0 . Then    0 0       Y CY CY Z C1 C2 Y Z CX = E = = , Z CZY CZ C20 C3 and CXY = E Solving ACY = CXY becomes AC1 =

 

      Y CY C1 0 Y = = . Z CZY C20

 C1 , C20

or

A =



 I . C20 C1−1

The linear MMSE estimate of X = [Y 0 , Z 0 ]0 is     Y I . Y = AY = C20 C1−1Y C20 C1−1 In other words, Yb = Y and Zb = C20 C1−1Y . Note that the matrix required for the linear MMSE estimate of Z based on Y is the solution of BCY = CZY or BC1 = C20 ; i.e., B = C20 C1−1 . Next, the error covariance for estimating X based on Y is      I C1 C2 b b 0 ] = CX − ACY X = C10 C2 − E[(X − X)(X − X) C2 C3 C20 C1−1     C1 C2 C1 C2 − = C20 C3 C20 C20 C1−1C2   0 0 , = 0 C3 −C20 C1−1C2 and the MSE is b 2 ] = tr(CX − ACY X ) = tr(C3 −C20 C−1C2 ). E[kX − Xk 1

38. Since P0 decorrelates Y , the covariance matrix CZ of Z := P0Y is diagonal. Writing ˜ Z = CXZ in component form, and using the fact that CZ is diagonal, AC

∑ A˜ ik (CZ )k j

= (CXZ )i j

k

becomes A˜ i j (CZ ) j j = (CXZ )i j

148

Chapter 8 Problem Solutions If (CZ ) j j 6= 0, then A˜ i j = (CXZ )i j /(CZ ) j j . If (CZ ) j j = 0, then A˜ i j (CZ ) j j = (CXZ )i j can be solved only if (CXZ )i j = 0, which we now show to be the case by using the Cauchy–Schwarz inequality. Write |(CXZ )i j | = E[(Xi − (mX )i )(Z j − (mZ ) j )] q ≤ E[(Xi − (mX )i )2 ]E[(Z j − (mZ ) j )2 ] q = (CX )ii (CZ ) j j .

Hence, if (CZ ) j j = 0 then (CXZ )i j = 0, and any value of A˜ i j solves A˜ i j (CZ ) j j = ˜ Z = CXZ , observe (CXZ )i j . Now that we have shown that we can always solve AC that this equation is equivalent to ˜ 0CY P) = CXY P or A(P

˜ 0 )CY = CXY . (AP

˜ 0 solves the original problem. Thus, A = AP 39. Since X has the form X = [Y 0 , Z 0 ]0 , if we take G = [I, 0] and W ≡ 0, then    I 0 Y GX +W = = Y. Z 40. Write



1 n 2 E ∑ Xk n k=1

41. Write E





1 n ∑ Xk Xk0 n k=1

=



=

1 n 1 n E[Xk2 ] = ∑ σ 2 = σ 2 . ∑ n k=1 n k=1 1 n 1 n E[Xk Xk0 ] = ∑ C = C. ∑ n k=1 n k=1

42. MATLAB. Additional code: Mn = mean(X,2) MnMAT = kron(ones(1,n),Mn); Chat = (X-MnMAT)*(X-MnMAT)’/(n-1)

43. We must first find fY |X (y|x). To this end, use substitution and independence to write P(Y ≤ y|X = x) = P(X +W ≤ y|X = x) = P(x +W ≤ y|X = x) = P(W ≤ y − x). Then fY |X (y|x) = fW (y − x) = (λ /2)e−λ |y−x| . For fixed y, the maximizing value of x is x = y. Hence, gML (y) = y. 44. By the same argument as in the solution of Problem 43, fY |X (y|x) = (λ /2)e−λ |y−x| . When X ∼ exp(µ ) and we maximize over x, we must impose the constraint x ≥ 0. Hence,  y, y ≥ 0, gML (y) = argmax λ2 e−λ |y−x| = 0, y < 0. x≥0

Chapter 8 Problem Solutions When X ∼ uniform[0, 1], gML (y) = argmax λ2 e−λ |y−x| 0≤x≤1

149

  y, 0 ≤ y ≤ 1, 1, y > 1, =  0, y < 0.

45. By the same argument as in the solution of Problem 43, fY |X (y|x) = (λ /2)e−λ |y−x| . For the MAP estimator, we must maximize fY |X (y|x) fX (x) =

λ −λ |y−x| · µ e−µ x , 2e

x ≥ 0.

By considering separately the cases x ≤ y and x > y, ( µλ −λ y e(λ −µ )x , 0 ≤ x ≤ y, 2 e fY |X (y|x) fX (x) = µλ λ y −(λ +µ )x , x > y, x ≥ 0. 2 e e When y ≥ 0, observe that the two formulas agree at x = y and have the common value (µλ /2)e−µ y ; in fact, if λ > µ , the first formula is maximized at x = y, while the second formula is always maximized at x = y. If y < 0, then only the second formula is valid, and its region of validity is x ≥ 0. This formula is maximized at x = 0. Hence, for λ > µ ,  y, y ≥ 0, gMAP (y) = 0, y < 0.

We now consider the case λ ≤ µ . As before, if y < 0, the maximizing value of x is zero. If y ≥ 0, then the maximum value of fY |X (y|x) fX (x) for 0 ≤ x ≤ y occurs at x = 0 with a maximum value of (µλ /2)e−λ y . The maximum value of fY |X (y|x) fX (x) for x ≥ y occurs at x = y with a maximum value of (µλ /2)e−µ y . For λ < µ , max{(µλ /2)e−λ y , (µλ /2)e−µ y } = (µλ /2)e−λ y ,

which corresponds to x = 0. Hence, for λ < µ , gMAP (y) = 0,

−∞ < y < ∞.

46. From the formula fXY (x, y) = (x/y2 )e−(x/y)

2 /2

· λ e−λ y ,

x, y > 0,

we see that Y ∼ exp(λ ), and p that given Y = y, X ∼ Rayleigh(y). Hence, the MMSE estimator is E[X|Y = y] = π /2 y. To compute the MAP estimator, we must solve argmax(x/y2 )e−(x/y)

2 /2

.

x≥0

We do this by differentiating with respect to x and setting the derivative equal to zero. Write   2 ∂ e−(x/y) /2 x2 2 −(x/y)2 /2 = (x/y )e 1− 2 . ∂x y2 y Hence, gMAP (y) = y.

150

Chapter 8 Problem Solutions

47. Suppose that E[(X − g1 (Y ))h(Y )] = 0

and

E[(X − g2 (Y ))h(Y )] = 0.

Subtracting the second equation from the first yields E[{g2 (Y ) − g1 (Y )}h(Y )] = 0. Since h is an arbitrary bounded function, put h(y) := sgn[g2 (y) − g1 (y)], where   1, x > 0, 0, x = 0, sgn(x) :=  −1, x < 0.

Note also that x · sgn(x) = |x|. Then (∗) becomes E[|g2 (Y ) − g1 (Y )|] = 0.

(∗)

CHAPTER 9

Problem Solutions 1. We first compute 

σ12 σ1 σ2 ρ detC = det σ1 σ2 ρ σ22 p √ and detC = σ1 σ2 1 − ρ 2 . Next, C−1 =

1 detC





= σ12 σ22 − (σ1 σ2 ρ )2 = σ12 σ22 (1 − ρ 2 ),

−σ1 σ2 ρ σ22 −σ1 σ2 ρ σ12



and

 1 −ρ  σ 2 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 )  , 1 =    −ρ 1 2 2 2 σ1 σ2 (1 − ρ ) σ2 (1 − ρ ) 

  x ρy   − 2 2      σ (1 − ρ 2 ) σ1 σ2 (1 − ρ )  x x y C−1 = x y  1 −ρ x  y y + 2 2 2 σ1 σ2 (1 − ρ ) σ1 (1 − ρ )

x2 y2 ρ xy ρ xy − − + σ12 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 ) σ12 (1 − ρ 2 )  y 2 i 1 h x 2 xy = ρ + − 2 , 1 − ρ 2 σ1 σ1 σ2 σ2

=

and the result follows. 2. Here is the plot:

n=4 1

n=1

0 −4 −2 3.

0

2

4

(a) First write c1 X + c2Y = c1 X + c2 (3X) = (c1 + 3c2 )X, which is easily seen to be N(0, (c1 + 3c2 )2 ). Thus, X and Y are jointly Gaussian. (b) Observe that E[XY ] = E[X(3X)] = 3E[X 2 ] = 3 and E[Y 2 ] = E[(3X)2 ] = 9E[X 2 ] = 9. Since X and Y have zero means,     E[X 2 ] E[XY ] 1 3 cov([X,Y ]0 ) = = . 3 9 E[Y X] E[Y 2 ] 151

152

Chapter 9 Problem Solutions (c) The conditional cdf of Y given X = x is FY |X (y|x) = P(Y ≤ y|X = x) = P(3X ≤ y|X = x). By substitution, this last conditional probability is P(3x ≤ y|X = x). The event {3x ≤ y} is deterministic and therefore independent of X. Hence, we can drop the conditioning and get FY |X (y|x) = P(3x ≤ y). If 3x ≤ y, then {3x ≤ y} = Ω, and {3x ≤ y} = ∅ otherwise. Hence, the above probability is just u(y − 3x).

4. If Y = ∑ni=1 ci Xi , its characteristic function is  n  jν ci Xi jν (∑ni=1 ci Xi ) ] = E ∏e = ϕY (ν ) = E[e i=1

n

=

2 2

∏ e j(ν ci )mi −(ν ci ) σi /2

n

n

∏ E[e jν ci Xi ]

n

=

i=1

= e jν (∑i=1 ci mi )−ν

2 ( n c2 σ 2 )/2 ∑i=1 i i

∏ ϕXi (ν ci ) i=1

,

i=1

which is the characteristic function of an N(∑ni=1 ci mi , ∑ni=1 c2i σi2 ) random variable. 5. First, E[Y ] = E[AX + b] = AE[X] + b = Am + b. Second, E[{Y − (Am + b)}{Y − (Am + b)}0 ] = E[A(X − m)(X − m)0 A0 ] = ACA0 . 6. Write out Y1 = X1 Y2 = X1 + X2 Y3 = X1 + X2 + X3 .. . In general, Yn = Yn−1 + Xn , or Xn = Yn − Yn−1 , which we can write in matrix-vector notation as      X1 Y1 1 0 ··· 0  X2     −1 1 0 0       Y2   X3    0 −1 1 0  0   Y3       ..  =  .   .. ..  .     . .       Xn−1    0  Yn−1  0 −1 1 0 Xn Yn 0 · · · 0 −1 1 {z } | =: A

Since X = AY and Y is Gaussian, so is X.

7. Since X ∼ N(0,C), the scalar Y := ν 0 X is also Gaussian and has zero mean. Hence, E[(ν 0 XX 0 ν )k ] = E[(Y 2 )k ] = E[Y 2k ] = (2k − 1) · · · 5 · 3 · 1 · (E[Y 2 ])k . Now observe that E[Y 2 ] = E[ν 0 XX 0 ν ] = ν 0Cν and the result follows.

Chapter 9 Problem Solutions

153

8. We first have E[Y j ] = E[X j − X] = m − E



1 n ∑ Xi n i=1



= m−

1 n ∑ m = 0. n i=1

Next, since E[XY j ] = E[X(X j − X)], we first compute   n 1 1 n n 2 2 2 E[ X X ] = 2 ∑ ∑ E[Xi X j ] = 2 ∑ (σ + m ) + ∑ m n i=1 j=1 n i=1 i6= j o o n n 1 1 σ2 + m2 , = 2 n(σ 2 + m2 ) + n(n − 1)m2 = 2 nσ 2 + n2 m2 = n n n

and

E[ X X j ] =

o 1n 2 1 n σ2 2 2 + m ) + (n − 1)m = E[X X ] = ( + m2 . σ i j ∑ n i=1 n n

It now follows that E[ X Y j ] = 0. 9. Following the hint, in the expansion of E[(ν 0 X)2k ], the sum of all the coefficients of νi1 · · · νi2k is (2k)!E[Xi1 · · · Xi2k ]. The corresponding sum of coeficients in the expansion of (2k − 1)(2k − 3) · · · 5 · 3 · 1 · (ν 0Cν )k is (2k − 1)(2k − 3) · · · 5 · 3 · 1 · 2k k!

∑

j1 ,..., j2k

C j1 j2 · · ·C j2k−1 j2k ,

where the sum is over all j1 , . . . , j2k that are permutations of i1 , . . . , i2k and such that the product C j1 j2 · · ·C j2k−1 j2k is distinct. Since (2k − 1)(2k − 3) · · · 5 · 3 · 1 · 2k k! = (2k − 1)(2k − 3) · · · 5 · 3 · 1 · (2k)(2k − 2) · · · 4 · 2 = (2k)!, Wick’s Theorem follows. 10. Write E[X1 X2 X3 X4 ] =

∑

C j1 j2 C j3 j4 = C12C34 +C13C24 +C14C23 .

j 1 , j2 , j3 , j4

E[X1 X32 X4 ] = C13C34 +C14C33 . 2 E[X12 X22 ] = C11C22 +C12 . 11. Put a := [a1 , . . . , an ]0 . Then Y = a0 X, and 0

0

ϕY (η ) = E[e jηY ] = E[e jη (a X) ] = E[e j(η a) X ] = ϕX (η a) 0

0

0

0

= e j(η a) m−(η a) C(η a)/2 = e jη (a m)−(a Ca)η

2 /2

,

which is the characteristic function of a scalar N(a0 m, a0Ca) random variable.

154

Chapter 9 Problem Solutions 0

0

12. With X = [U 0 ,W 0 ]0 and ν = [α 0 , β 0 ]0 , we have ϕX (ν ) = e jν m−ν Cν /2 , where     mU ν 0m = α 0 β 0 = α 0 mU + β 0 mW , mW and

0

ν Cν = Thus,



α0

β0

      0 0  Sα  S 0 α = α β = α 0 Sα + β 0 T β . Tβ β 0 T

0

ϕX (ν ) = e j(α mU +β

0 m )−(α 0 Sα +β 0 T β )/2 W

0

0

= e jα mU −α Sα /2 e jβ

0 m −β 0 T β /2 W

,

which has the required form ϕU (α )ϕW (β ) of a product of Gaussian characteristic functions. 13.

(a) Since X is N(0,C) and Y := C−1/2 X, Y is also normal. It remains to find the mean and covariance of Y . We have E[Y ] = E[C−1/2 X] = C−1/2 E[X] = 0 and E[YY 0 ] = E[C−1/2 XX 0C−1/2 ] = C−1/2 E[XX 0 ]C−1/2 = C−1/2CC−1/2 = I. Hence, Y ∼ N(0, I). (b) Since the covariance matrix of Y is diagonal, the components of Y are uncorrelated. Since Y is also Gaussian, the components of Y are independent. Since the covariance matrix of Y is the indentity, each Yk ∼ N(0, 1). Hence, each Yk2 is chisquared with one degree of freedom by Problem 46 in Chapter 4 or Problem 11 in Chapter 5. (c) By the Remark in Problem 55(c) in Chapter 4, V is chi-squared with n degrees of freedom.

14. Since Z := det



X −Y Y X



= X 2 +Y 2 ,

where X and Y are independent N(0, 1), observe that X 2 and Y 2 are chi-squared with one degree of freedom. Hence, Z is chi-squared with two degrees of freedom, which is the same as exp(1/2). 15. Begin with fX (x) =

1 (2π )n

Z

0

IRn

0

0

e− jν x e jν m−ν Cν /2 d ν =

1 (2π )n

Z

IRn

0

0

e− j(x−m) ν e−ν Cν /2 d ν .

Now make the multivariate change of variable ζ = C1/2 ν , d ζ = detC1/2 d ν . Then Z

0 −1/2 0 1 dζ e− j(x−m) C ζ e−ζ ζ /2 (2π )n IRn detC1/2 Z −1/2 0 0 1 dζ e− j{C (x−m)} ζ e−ζ ζ /2 √ = . (2π )n IRn detC

fX (x) =

Chapter 9 Problem Solutions Put t = C−1/2 (x − m) so that fX (x) =

1 (2π )n

Z

IRn

0

e− jt ζ e−ζ

0 ζ /2

155

 Z n  1 ∞ − jti ζi −ζ 2 /2 1 dζ i √ e =√ d e ζ i . ∏ detC detC i=1 2π −∞

2

Observe that e−ζi /2 is the characteristic function of a scalar N(0, 1) random variable. Hence, n −ti2 /2 1 e 1 √ √ fX (x) = √ exp[−t 0t/2]. = ∏ n/2 detC i=1 2π (2π ) detC Recalling that t = C−1/2 (x − m) yields fX (x) = 16. First observe that

exp[− 12 (x − m)0C−1 (x − m)] √ . (2π )n/2 detC 

X Y Z := det U V



= XV −YU.

Then consider the conditional cumulative distribution function, FZ|UV (z|u, v) = P(Z ≤ z|U = u,V = v) = P(XV −YU ≤ z|U = u,V = v) = P(Xv −Yu ≤ z|U = u,V = v). Since [X,Y ]0 and [U,V ]0 are jointly Gaussian and uncorrelated, they are independent. Hence, we can drop the conditioning and get FZ|UV (z|u, v) = P(Xv −Yu ≤ z). Next, since X and Y are independent and N(0, 1), Xv −Yu ∼ N(0, u2 + v2 ). Hence, fZ|UV (· |u, v) ∼ N(0, u2 + v2 ). 17. We first use the fact that A solves ACY = CXY to show that (X − mX ) − A(Y − mY ) and Y are uncorrelated. Write E[{(X − mX ) − A(Y − mY )}Y 0 ] = CXY − ACY = 0. We next show that (X − mX ) − A(Y − mY ) and Y are jointly Gaussian by writing them as an affine transformation of the Gaussian vector [X 0 ,Y 0 ]0 ; i.e.,        (X − mX ) − A(Y − mY ) I −A X AmY − mX = + . Y 0 I Y 0 It now follows that (X − mX ) − A(Y − mY ) and Y are independent. Using the hints on substitution and independence, we compute the conditional characteristic function,  0  0 0 E[eν X |Y = y] = E e jν [(X−mX )−A(Y −mY )] · e jν [mX +A(Y −mY )] Y = y  0  0 = e jν [mX +A(y−mY )] E e jν [(X−mX )−A(Y −mY )] Y = y  0  0 = e jν [mX +A(y−mY )] E e jν [(X−mX )−A(Y −mY )] .

156

Chapter 9 Problem Solutions This last expectation is the characteristic function of the zero-mean Gaussian random vector (X − mX ) − A(Y − mY ). To compute its covariance matrix first observe that since ACY = CXY , we have ACY A0 = CXY A0 . Then E[{(X − mX ) − A(Y − mY )}{(X − mX ) − A(Y − mY )}0 ]

= CX −CXY A0 − ACY X + ACY A0 = CX − ACY X .

We now have 0

0

0

E[eν X |Y = y] = e jν [mX +A(y−mY )] e−ν [CX −ACY X ]ν /2 . $
Thus, given Y = y, X is conditionally N mX + A(y − mY ),CX − ACY X .

18. First observe that



X Y Z := det U V



= XV −YU.

Then consider the conditional cumulative distribution function, FZ|UV (z|u, v) = P(Z ≤ z|U = u,V = v) = P(XV −YU ≤ z|U = u,V = v) = P(Xv −Yu ≤ z|U = u,V = v). Since [X,Y,U,V ]0 is Gaussian, given U = u and V = v, [X,Y ]0 is conditionally     u ,C[X,Y ]0 − AC[U,V ]0 ,[X,Y ]0 , N A v where A solves AC[U,V ]0 = C[X,Y ]0 ,[U,V ]0 . We now turn to the conditional distribution of Xv −Yu. Since the conditional distribution of [X,Y ]0 is Gaussian, so is the conditional distribution of the linear combination Xv − Yu. Hence, all we need to find are the conditional mean and the conditional variance of Xv −Yu; i.e.,       X U = u,V = v E[Xv −Yu|U = u,V = v] = E v u Y      X U = u,V = v = v u E Y     u = v u A , v

and

E

 

   2 X u v u −A U = u,V = v Y v   2      U X U = u,V = v v u −A = E V Y         0   X U X U U = u,V = v v = v u E −A −A u Y V Y V      v = v u C[X,Y ]0 − AC[U,V ]0 ,[X,Y ]0 , u 



Chapter 9 Problem Solutions

157

where the last step uses the fact that [X,Y ]0 − A[U,V ]0 is independent of [U,V ]0 . If [X,Y ]0 and [U,V ]0 are uncorrelated, i.e., C[X,Y ]0 ,[U,V ]0 = 0, then A = 0 solves AC[U,V ]0 = 0; in this case, the conditional mean is zero, and the conditional variance simplifies to         v v v u C[X,Y ]0 = v2 + u2 . = v u I u u 19. First write Z − E[Z] = (X + jY ) − (mX + jmY ) = (X − mX ) + j(Y − mY ). Then cov(Z) := E[(Z − E[Z])(Z − E[Z])∗ ] = E[{(X − mX ) + j(Y − mY )}{(X − mX ) + j(Y − mY )}∗ ]

= E[{(X − mX ) + j(Y − mY )}{(X − mX ) − j(Y − mY )}] = var(X) − j cov(X,Y ) + j cov(Y, X) + var(Y ) = var(X) + var(Y ).

20.

(a) Write K := E[(Z − E[Z])(Z − E[Z])H ]

= E[{(X − mX ) + j(Y − mY )}{(X − mX ) + j(Y − mY )}H ] = E[{(X − mX ) + j(Y − mY )}{(X − mX )H − j(Y − mY )H }]

= CX − jCXY + jCY X +CY = (CX +CY ) + j(CY X −CXY ). (b) If CXY = −CY X , the (CXY )ii = −(CY X )ii implies E[(Xi − (mX )i )(Yi − (mY )i )] = −E[(Yi − (mY )i )(Xi − (mX )i )] = −E[(Xi − (mX )i )(Yi − (mY )i )].

Hence, E[(Xi − (mX )i )(Yi − (mY )i )] = 0, and we see that Xi and Yi are uncorrelated. (c) By part (a), if K is real, then CY X = CXY . Circular symmetry implies CY X = −CXY . It follows that CXY = −CXY , and then CXY = 0; i.e., X and Y are uncorrelated. 21. First, fX (x) =

0

e−x (2I)x/2 (2π )n/2 (1/2)n/2

Then fXY (x, y) = fX (x) fY (y) = 22.

fY (y) =

and 0

e−(x x+y πn

0 y)

0

e−y (2I)y/2 . (2π )n/2 (1/2)n/2 H

=

e−(x+ jy) (x+ jy) . πn

(a) Immediate from Problem 20(a). (b) Since ν 0 Qν is a scalar, (ν 0 Qν )0 = ν 0 Qν . Since Q0 = −Q, (ν 0 Qν )0 = ν 0 Q0 ν = −ν 0 Qν . Thus, ν 0 Qν = −ν 0 Qν , and it follows that ν 0 Qν = 0.

158

Chapter 9 Problem Solutions (c) Begin by observing that if K = R + jQ and w = ν + jθ , then Kw = (R + jQ)(ν + jθ ) = Rν + jRθ + jQν − Qθ . Next, wH (Kw) = (ν 0 − jθ 0 )(Rν + jRθ + jQν − Qθ ) = ν 0 Rν + jν 0 Rθ + jν 0 Qν − ν 0 Qθ − jθ 0 Rν + θ 0 Rθ + θ 0 Qν − jθ 0 Qθ = ν 0 Rν + jν 0 Rθ + jν 0 Qν + θ 0 Qν − jν 0 Rθ + θ 0 Rθ + θ 0 Qν − jθ 0 Qθ = ν 0 Rν + θ 0 Rθ + 2θ 0 Qν ,

where we have used the result of parts (a) and (b). 23. First, AZ = (α + jβ )(X + jY ) = (α X − β Y ) + j(β X + α Y ). Second,



α −β β α





X Y

=



 αX − βY . β X + αY

Now assume that circular symmetry holds; i.e., CX = CY and CXY = −CY X . Put U := α X − β Y and V := β X + α Y . Assuming zero means to simplify the notation, CU = E[(α X − β Y )(α X − β Y )0 ] = α CX α 0 − β CY X α 0 − α CXY β 0 + β 0CY β = α CX α 0 − β CY X α 0 + α CY X β 0 + β 0CX β . Similarly, CV = E[(β X + α Y )(β X + α Y )0 ] = β CX β 0 + α CY X β 0 + β CXY α 0 + α CY α 0 = β CX β 0 + α CY X β 0 − β CY X α 0 + α CX α 0 . Hence, CU = CV . It remains to compute CUV = E[(α X − β Y )(β X + α Y )0 ] = α CX β 0 + α CXY α 0 − β CY X β 0 − β CY α 0

= α CX β 0 − α CY X α 0 − β CY X β 0 − β CX α 0

and CVU = E[(β X + α Y )(α X − β Y )0 ] = β CX α 0 − β CXY β 0 + α CY X α 0 − α CY β 0

= β CX α 0 + β CY X β 0 + α CY X α 0 − α CX β 0 ,

which shows that CU = −CV . Thus, if Z is circularly symmetric, so is AZ.

24. To begin, note that with R = [X 0 ,U 0 ]0 and I = [Y 0 ,V 0 ]0 ,     CX CXU CY CYV CR = , CI = , CUX CU CVY CV and CRI =



CXY CXV CUY CUV



,

CIR =



 CY X CYU . CV X CVU

Also, Θ is circularly symmetric means CR = CI and CRI = −CIR .

Chapter 9 Problem Solutions

159

(a) We assume zero means to simplify the notation. First, KZW = E[ZW H ] = E[(X + jY )(U + jV )H ] = E[(X + jY )(U H − jV H )] = CXU − jCXV + jCYU +CYV = 2(CXU − jCXV ),

since Θ is circularly symmetric.

Second, CZeWe =



CXU CXV CYU CYV



=



 CXU CXV , −CXV CXU

since Θ is circularly symmetric.

It is now clear that KZW = 0 if and only if CZeWe = 0.

(b) Assuming zero means again, we compute

KW = E[WW H ] = E[(U + jV )(U + jV )H ] = E[(U + jV )(U H − jV H )] = CU − jCUV + jCVU +CV = 2(CU − jCUV ).

We now see that AKW = KZW becomes 2(α + jβ )(CU − jCUV ) = 2(CXU − jCXV ) or (α CU + β CUV ) + j(β CU − α CUV ) = CXU − jCXV . We also have CWe =



CU CUV CVU CV

(∗)



e e = C e e becomes so that AC W ZW    CU CUV α −β = CZeWe CVU CV β α

or

or





α CU − β CVU α CUV − β CV β CU + α CVU β CUV + α CV

α CU + β CUV α CUV − β CU β CU − α CUV β CUV + α CU



=





= CZeWe

 CXU CXV , −CXV CXU

which is equivalent to (∗). e solves AC e e = C e e . Hence, by (c) If A solves AKW = KZW , then by part (b), A W ZW e = w, e Problem 17, given W   ew e ee . e − mWe ),CZe − AC Ze ∼ N mZe + A( WZ e e e is equivalent to Next, CZe − AC WZ      CX CXY α −β CUX CUY − , CY X CY β α CV X CVY

160

Chapter 9 Problem Solutions which, by the circular symmetry of Θ, becomes      CX CXY CUX CUY α −β − , −CXY CX −CUY CUX β α

or



CX CXY −CXY CX

which is equivalent to



−



 α CUX + β CUY α CUY − β CUX , β CUX − α CUY β CUY + α CUX

2(CX − jCXY ) − (α + jβ ) · 2(CUX − jCUY ), which is exactly KZ − AKW Z . Thus, given W = w, $
Z ∼ N mZ + A(w − mW ), KZ − AKW Z .

25. Let Z = X + jY with X and Y independent N(0, 1/2) as in the text. (a) Since X and Y are zero mean, cov(Z) = E[ZZ ∗ ] = E[X 2 +Y 2 ] = 12 + 12 = 1. √ √ √ √ (b) First write 2|Z|2 = 2(X 2 +Y 2 ) = ( 2 X)2 + ( 2Y )2 . Now, 2 X and 2Y are both N(0, 1). Hence, their squares are chi-squared with one degree of freedom by Problem 46 in Chapter 4 or Problem 11 in Chapter 5. Hence, by Problem 55(c) in Chapter 4 and the remark following it, 2|Z|2 is chi-squared with two degrees of freedom. 26. With X ∼ N(mr , 1) and Y ∼ N(mi , 1), it follows either from Problem 47 in Chapter 4 or from Problem 12 in Chapter 5 that X 2 and Y 2 are noncentral chi-squared with one degree of freedom and respective noncentrality parameters m2r and m2i . Since X and Y are independent, it follows from Problem 65 in Chapter 4 that X 2 +Y 2 is noncentral chi-squared with two degrees of freedom and noncentrality√parameter m2r + m2i . It is now immediate from Problem 26 in Chapter 5 that |Z| = X 2 +Y 2 has the orginal Rice density. 27.

(a) The covariance matrix of W is E[WW H ] = E[K −1/2 ZZ H K −1/2 ] = K −1/2 E[ZZ H ]K −1/2 = K −1/2 KK −1/2 = I. Hence, H

fW (w) =

e−w w = πn

2

e−|wk | ∏ π . k=1 n

(b) By part (a), the Wk = Uk + jVk are i.i.d. N(0, 1) with 2

2

√

2

2

e−|wk | e−(uk +vk ) e−[(uk /(1/ 2 )) +(vk /(1/ fWk (w) = = $p $p
2 =
2 π 2π /2 2π /2 √

e−[uk /(1/ 2 )] p = 2π /2

2 /2

√

2 /2

e−[vk /(1/ 2 )] p · 2π /2

Hence, Uk and Vk are independent N(0, 1/2).

√ 2 2 )) ]/2

= fUkVk (uk , vk ).

Chapter 9 Problem Solutions (c) Write

n

2kW k2 =

161

√ √ 2Uk )2 + ( 2Vk )2 .

∑(

k=1

√ √ Since 2Uk and 2Vk are independent N(0, 1), their squares are chi-squared with one degree √ of freedom √ by Problem 46 in Chapter 4 or Problem 11 in Chapter 5. Next, ( 2Uk )2 + ( 2Vk )2 is chi-squared with two degrees of freedom by Problem 55(c) in Chapter 4 and the remark following it. Similarly, since the Wk are indepdendent, 2kW k2 is chi-squared with 2n degrees of freedom. 28.

(a) Write 0 = (u + v)0 M(u + v) = u0 Mu + v0 Mu + u0 Mv + v0 Mv = 2v0 Mu, since M 0 = M. Hence, v0 Mu = 0. (b) By part (a) with v = Mu we have 0 = v0 Mu = (Mu)0 Mu = kMuk2 . Hence, Mu = 0 for all u, and it follows that M must be the zero matrix.

29. We have from the text that

is equal to



ν0 θ 0





CX CXY CY X CY



ν θ



ν 0CX ν + ν 0CXY θ + θ 0CY X ν + θ 0CY θ , which, upon noting that ν 0CXY θ is a scalar and therefore equal to its transpose, simplifies to ν 0CX ν + 2θ 0CY X ν + θ 0CY θ . (∗) We also have from the text (via Problem 22) that wH Kw = ν 0 (CX +CY )ν + θ 0 (CX +CY )θ + 2θ 0 (CY X −CXY )ν . If (∗) is equal to wH Kw/2 for all ν and all θ , then in particular, this must hold for all ν when θ = 0. This implies

ν 0CX ν = ν 0

CX +CY ν 2

or

ν0

CX −CY ν = 0. 2

Since ν is arbitrary, (CX −CY )/2 = 0, or CX = CY . This means that we can now write wH Kw/2 = ν 0CX ν + θ 0CY θ + θ 0 (CY X −CXY )ν . Comparing this with (∗) shows that 2θ 0CY X ν = θ 0 (CY X −CXY )ν

or

θ 0 (CY X +CXY )ν = 0.

Taking θ = ν arbitrary and noting that CY X +CXY is symmetric, it follows that CY X + CXY = 0, and so CXY = −CY X .

162 30.

Chapter 9 Problem Solutions (a) Since Γ is 2n × 2n, det(2Γ) = 22n det Γ. From the hint it follows that det Γ = (det K)2 /22n . (b) Write VV −1 = (A + BCD)[A−1 − A−1 B(C−1 + DA−1 B)−1 DA−1 ] = (A + BCD)A−1 [I − B(C−1 + DA−1 B)−1 DA−1 ] = (I + BCDA−1 )[I − B(C−1 + DA−1 B)−1 DA−1 ] = I + BCDA−1 − B(C−1 + DA−1 B)−1 DA−1

− BCDA−1 B(C−1 + DA−1 B)−1 DA−1 = I + BCDA−1

− B[I +CDA−1 B](C−1 + DA−1 B)−1 DA−1 = I + BCDA−1 − BC[C−1 + DA−1 B](C−1 + DA−1 B)−1 DA−1 = I + BCDA−1 − BCDA−1 = I. (c) To begin, write ΓΓ−1 =





CX −CY X CY X CX



∆−1 CX−1CY X ∆−1 −1 −1 ∆−1 −∆ CY X CX



CX ∆−1 +CY X ∆−1CY X CX−1 CY X ∆−1 −CY X ∆−1 −1 −1 −1 CY X ∆ −CX ∆ CY X CX CY X CX−1CY X ∆−1 +CX ∆−1   CX ∆−1 +CY X ∆−1CY X CX−1 0 = CY X ∆−1 −CX ∆−1CY X CX−1 (CY X CX−1CY X +CX )∆−1   CX ∆−1 +CY X ∆−1CY X CX−1 0 = . CY X ∆−1 −CX ∆−1CY X CX−1 I

=



Using the hint that ∆−1 = CX−1 −CX−1CY X ∆−1CY X CX−1 , we easily obtain CX ∆−1 = I −CY X ∆−1CY X CX−1 , from which it follows that −1

ΓΓ

=



I CY X ∆−1 −CX ∆−1CY X CX−1

 0 . I

To show that the lower-left block is also zero, use the hint to write CY X ∆−1 −CX ∆−1CY X CX−1 = CY X [CX−1 −CX−1CY X ∆−1CY X CX−1 ] −CX ∆−1CY X CX−1 = CY X CX−1 −CY X CX−1CY X ∆−1CY X CX−1 −CX ∆−1CY X CX−1 = CY X CX−1 − [CY X CX−1CY X +CX ]∆−1CY X CX−1 = CY X CX−1 − ∆∆−1CY X CX−1 = 0.

Chapter 9 Problem Solutions

163

(d) Write KK −1 = 2(CX + jCY X )(∆−1 − jCX−1CY X ∆−1 )/2 = CX ∆−1 +CY X CX−1CY X ∆−1 + j(CY X ∆−1 −CY X ∆−1 ) = ∆∆−1 = I.

(e) We begin with 

x0

y0



−1

Γ

      0 0 x ∆−1 CX−1CY X ∆−1 x = x y y y −∆−1CY X CX−1 ∆−1  −1  −1   ∆ x +CX CY X ∆−1 y = x0 y0 −∆−1CY X CX−1 x + ∆−1 y

= x 0 ∆−1 x + x 0CX−1CY X ∆−1 y − y 0 ∆−1CY X CX−1 x + y 0 ∆−1 y.

Now, since each of the above terms on the third line is a scalar, each term is equal to its transpose. In particular, y 0 ∆−1CY X CX−1 x = x 0CX−1CXY ∆−1 y = −x 0CX−1CY X ∆−1 y. Hence, 1 2



   x x 0 y 0 Γ−1 = y

0 −1 −1 0 −1 1 0 −1 2 (x ∆ x + 2x CX CY X ∆ y + y ∆ y).

(∗)

We next compute zH K −1 z = = =

1 0 2 (x − 1 0 2 (x − 1 0 2 (x −

jy 0 )(∆−1 − jCX−1CY X ∆−1 )(x + jy)

jy 0 )[(∆−1 x +CX−1CY X ∆−1 y) + j(∆−1 y −CX−1CY X ∆−1 x)] jy 0 )[(∆−1 x +CX−1CY X ∆−1 y) + j(∆−1 y − ∆−1CY X CX−1 x)]

by the hint that CX−1CY X ∆−1 = ∆−1CY X CX−1 . We continue with zH K −1 z = =

0 −1 −1 −1 0 −1 −1 −1 1 2 [{x (∆ x +CX CY X ∆ y) + y (∆ y − ∆ CY X CX x)} + j{x 0 (∆−1 y − ∆−1CY X CX−1 x) − y 0 (∆−1 x +CX−1CY X ∆−1 y)}] 0 −1 0 −1 −1 0 −1 0 −1 −1 1 2 [{x ∆ x + x CX CY X ∆ y + y ∆ y − y ∆ CY X CX x} + j{x 0 ∆−1 y − x 0 ∆−1CY X CX−1 x − y 0 ∆−1 x − y 0CX−1CY X ∆−1 y}].

We now use the fact that since each of the terms in the last line is a scalar, it is equal to its transpose. Also CXY = −CY X . Hence, zH K −1 z =

0 −1 0 −1 −1 0 −1 1 2 [{x ∆ x + 2x CX CY X ∆ y + y ∆ y} − j{x 0 ∆−1CY X CX−1 x + y 0CX−1CY X ∆−1 y}].

Since x 0 ∆−1CY X CX−1 x = (x 0 ∆−1CY X CX−1 x)0 = −x 0CX−1CY X ∆−1 x = −x 0 ∆−1CY X CX−1 x, and similarly for y 0CX−1CY X ∆−1 y, the two imaginary terms above are zero.

CHAPTER 10

Problem Solutions 1. Write mX (t) := E[Xt ] = E[g(t, Z)] = g(t, 1)P(Z = 1) + g(t, 2)P(Z = 2) + g(t, 3)P(Z = 3) = p1 a(t) + p2 b(t) + p3 c(t), and RX (t, s) := E[Xt Xs ] = E[g(t, Z)g(s, Z)] = g(t, 1)g(s, 1)p1 + g(t, 2)g(s, 2)p2 + g(t, 3)g(s, 3)p3 = a(t)a(s)p1 + b(t)b(s)p2 + c(t)c(s)p3 . 2. Imitating the derivation of the Cauchy–Schwarz inequality for random variables in Chapter 2 of the text, write 0 ≤ =

Z ∞

−∞ Z ∞

−∞

|g(θ ) − λ h(θ )|2 d θ 2

|g(θ )| d θ − λ

−λ∗

Z ∞

−∞

λ =

−∞

|g(θ )|2 d θ −

h(θ )g(θ )∗ d θ Z ∞

−∞

|h(θ )|2 d θ .

R∞

∗ −∞ g(θ )h(θ ) d θ R∞ 2 −∞ |h(θ )| d θ

to get

0 ≤

−∞

g(θ )h(θ )∗ d θ + |λ |2

Then put

Z ∞

Z ∞

R 2 ∞ −∞ g(θ )h(θ )∗ d θ R∞

2 −∞ |h(θ )| d θ

2 R ∞ −∞ g(θ )h(θ )∗ d θ

2 R ∞ −∞ g(θ )h(θ )∗ d θ Z ∞ R |h(θ )|2 d θ − + R 2 ∞ 2 ∞ −∞ 2 −∞ |h(θ )| d θ −∞ |h(θ )| d θ R 2 ∞ Z ∞ −∞ g(θ )h(θ )∗ d θ 2 = |g(θ )| d θ − R ∞ . 2 −∞ −∞ |h(θ )| d θ

Rearranging yields the desired result.

164

Chapter 10 Problem Solutions

165

3. Write CX (t1 ,t2 ) = E[(Xt1 − mX (t1 ))(Xt2 − mX (t2 ))]

= E[Xt1 Xt2 ] − mX (t1 )E[Xt2 ] − E[Xt1 ]mX (t2 ) + mX (t1 )mX (t2 ) = E[Xt1 Xt2 ] − mX (t1 )mX (t2 ) − mX (t1 )mX (t2 ) + mX (t1 )mX (t2 ) = RX (t1 ,t2 ) − mX (t1 )mX (t2 ).

Similarly, CXY (t1 ,t2 ) = E[(Xt1 − mX (t1 ))(Yt2 − mY (t2 ))] = E[Xt1 Yt2 ] − mX (t1 )E[Yt2 ] − E[Xt1 ]mY (t2 ) + mX (t1 )mY (t2 )

= E[Xt1 Xt2 ] − mX (t1 )mY (t2 ) − mX (t1 )mY (t2 ) + mX (t1 )mY (t2 ) = RXY (t1 ,t2 ) − mX (t1 )mY (t2 ).

4. Write 2  ∗   n  n  n 0 ≤ E ∑ ci Xti = E ∑ ci Xti c X = ∑ k tk n

=

i=1 n

i=1

k=1

n

n

∑ ∑ ci E[Xti Xtk ]c∗k

i=1 k=1

∑ ∑ ci RX (ti ,tk )c∗k .

i=1 k=1

5. Since Xt has zero mean, var(Xt ) = RX (t,t) = t. Thus, Xt ∼ N(0,t), and 2

e−x /(2t) √ . 2π t

fXt (x) =

6. First note that by making the change of variable k = n − i, we have ∞

Yn =

∑

h(k)Xn−k .

k=−∞

(a) Write mY (n) = E[Yn ] = E



∞

∑

h(k)Xn−k

k=−∞



∞

=

∑

h(k)E[Xn−k ]

∑

h(k)mX (n − k).

k=−∞ ∞

=

k=−∞

(b) Write   E[XnYm ] = E Xn

∞

∑

k=−∞

h(k)Xm−k



∞

=

∑

h(k)E[Xn Xm−k ]

∑

h(k)RX (n, m − k).

k=−∞ ∞

=

k=−∞

166

Chapter 10 Problem Solutions (c) Write E[YnYm ] = E



∞

∑

l=−∞

∞

∑

=

l=−∞

  h(l)Xn−l Ym =

h(l)RXY (n − l, m) =

∞

∑

h(l)E[Xn−l Ym ]

l=−∞ ∞

∑

l=−∞

 h(l)

∞

∑

k=−∞

 h(k)RX (n − l, m − k) .

7. Let Xt = cos(2π f t + Θ), where Θ ∼ uniform[−π , π ]. (a) Consider choices t1 = 0 and t2 = −(π /2)/(2π f ). Then Xt1 = cos(Θ) and Xt2 = sin(Θ), which are not jointly continuous. (b) Write E[g(Xt )] = E[g(cos(2π f t + Θ))] = =

Z π +2π f t

−π +2π f t

g(cos(τ ))

Z π

dτ = 2π

−π

g(cos(2π f t + θ ))

Z π

−π

g(cos(τ ))

dθ 2π

dτ , 2π

since the integrand has period 2π and the range of integration has length 2π . Thus, E[g(Xt )] does not depend on t. 8.

(a) Using independence and a trigonometric identity, write E[Yt1 Yt2 ] = E[Xt1 Xt2 cos(2π f t1 + Θ) cos(2π f t2 + Θ)] = E[Xt1 Xt2 ]E[cos(2π f t1 + Θ) cos(2π f t2 + Θ)]   = 21 RX (t1 − t2 )E cos(2π f [t1 − t2 ]) + cos(2π f [t1 + t2 ] + 2Θ)    = 21 RX (t1 − t2 ) cos(2π f [t1 − t2 ]) + E cos(2π f [t1 + t2 ] + 2Θ) {z } | =0

=

1 2 RX (t1 − t2 ) cos(2π

f [t1 − t2 ]).

(b) A similar argument yields E[Xt1 Yt2 ] = E[Xt1 Xt2 cos(2π f t2 + Θ)] = E[Xt1 Xt2 ] E[cos(2π f t2 + Θ)] = 0. | {z } =0

(c) It is clear that Yt is zero mean. Together with part (a) it follows that Yt is WSS. 9. By Problem 7(b), FXt (x) = P(Xt ≤ x) = E[I(−∞,x] (Xt )] does not depend on t, and so we can restrict attention to the case t = 0. Since √ X0 = cos(Θ) has the arcsine density of Problem 35 in Chapter 5, f (x) = (1/π )/ 1 − x2 for |x| < 1. 10. Second-order strict stationarity means that for every two-dimensional set B, for every t1 ,t2 , and ∆t, P((Xt1 +∆t , Xt2 +∆t ) ∈ B) does not depend on ∆t. In particular, this is true whenever B has the form B = A × IR for any one-dimensional set A; i.e., P((Xt1 +∆t , Xt2 +∆t ) ∈ B) = P((Xt1 +∆t , Xt2 +∆t ) ∈ A × IR) = P(Xt1 ∈ A, Xt2 ∈ IR) = P(Xt1 ∈ A). does not depend on ∆t. Hence Xt is first-order strictly stationary.

Chapter 10 Problem Solutions 11.

167

(a) If p1 = p2 = 0 and p3 = 1, then Xt = c(t) = −1 with probability one. Then E[Xt ] = −1 does not depend on t, and E[Xt1 Xt2 ] = (−1)2 = 1 depends on t1 and t2 only through their difference t1 − t2 ; in fact the correlation function is a constant function of t1 − t2 . Thus, Xt is WSS.

(b) If p1 = 1 and p2 = p3 = 0, then Xt = e−|t| with probability one. Then E[Xt ] = e−|t| depends on t. Hence, Xt is not WSS. (c) First, the only way to have X0 = 1 is to have Xt = a(t) = e−|t| , which requires Z = 1. Hence, P(X0 = 1) = P(Z = 1) = p1 . Second, the only way to have Xt ≤ 0 for 0 ≤ t ≤ 0.5 is to have Xt = c(t) = −1, which requires Z = 3. Hence, P(Xt ≤ 0, 0 ≤ t ≤ 0.5) = P(Z = 3) = p3 . Third, the only way to have Xt ≤ 0 for 0.5 ≤ t ≤ 1 is to have Xt = b(t) = sin(2π t) or Xt = c(t) = −1. Hence, P(Xt ≤ 0, 0.5 ≤ t ≤ 1) = P(Z = 2 or Z = 3) = p2 + p3 . 12. With Yk := q(Xk , Xk+1 , . . . , Xk+L−1 ), write E[e j(ν1Y1+m +···+νnYn+m ) ] = E[e j{ν1 q(X1+m ,...,Xm+L )+···+νn q(Xn+m ,...,Xn+m+L−1 )} ]. The exponential on the right is just a function of X1+m , . . . , Xn+L−1+m . Since the Xk process is strictly stationary, the expectation on the right is unchanged if we replace X1+m , . . . , Xn+L−1+m by X1 , . . . , Xn+L−1 ; i.e., the above right-hand side is equal to E[e j{ν1 q(X1 ,...,XL )+···+νn q(Xn ,...,Xn+L−1 )} ] = E[e j(ν1Y1 +···+νnYn ) ]. Hence, the Yk process is strictly stationary. 13. We begin with E[g(X0 )] = E[X0 I[0,∞) (X0 )] =

Z ∞

−∞

xI[0,∞) (x) ·

λ −λ |x| dx 2e

=

1 2

Z ∞ 0

x · λ e−λ x dx,

which is just 1/2 times the expectation of an exp(λ ) random variable. We thus see that E[g(X0 )] = 1/(2λ ). Next, for n 6= 0, we compute Z ∞

2

Z

∞ 2 1 e−x /2 dx = √ xe−x /2 dx E[g(Xn )] = xI[0,∞) (x) · √ −∞ 2π 2π 0 2
∞ 1 1 = √ −e−x /2 = √ . 0 2π 2π

Hence, E[g(X0 )] 6= E[g(Xn )] for n 6= 0, and it follows that Xk is not strictly stationary.

168

Chapter 10 Problem Solutions

14. First consider the mean function, E[q(t + T )] =

1 T0

Z T0

q(t + θ ) d θ =

0

1 T0

Z t+T0

q(τ ) d τ =

t

1 T0

Z T0

q(τ ) d τ ,

0

where we have used the fact that since q has period T0 , the integral of q over any interval of length T0 yields the same result. The second thing to consider is the correlation function. Write Z

1 T0 q(t1 + θ )q(t2 + θ ) d θ T0 0 Z 1 t2 +T0 = q(t1 + τ − t2 )q(τ ) d τ T0 t2 Z 1 T0 = q([t1 − t2 ] + τ )q(τ ) d τ , T0 0

E[q(t1 + T )q(t2 + T )] =

where we have used the fact that as a function of τ , the product q([t1 − t2 ] + τ )q(τ ) has period T0 . Since the mean function does not depend on t, and since the correlation function depends on t1 and t2 only through their difference, Xt is WSS. 15. For arbitrary functions h, write 
 E[h(Xt1 +∆t , . . . , Xtn +∆t )] = E h q(t1 + ∆t + T ), . . . , q(tn + ∆t + T )

Z
1 T0 h q(t1 + ∆t + θ ), . . . , q(tn + ∆t + θ ) d θ T0 0 Z
1 ∆t+T0 = h q(t1 + τ ), . . . , q(tn + τ ) d τ T0 ∆t Z
1 T0 = h q(t1 + τ ), . . . , q(tn + τ ) d τ , T0 0

=

where the last step follows because we are integrating a function of period T0 over an interval of length T0 . Hence, E[h(Xt1 +∆t , . . . , Xtn +∆t )] = E[h(Xt1 , . . . , Xtn )], and we see that Xt is strictly stationary. 16. First write E[Yn ] = E[Xn − Xn−1 ] = E[Xn ] − E[Xn−1 ] = 0 since E[Xn ] does not depend on n. Next, write E[YnYm ] = E[(Xn − Xn−1 )(Xm − Xm−1 )]

= RX (n − m) − RX ([n − 1] − m) − RX (n − [m − 1]) + RX (n − m) = 2RX (n − m) − RX (n − m − 1) − RX (n − m + 1),

which depends on n and m only through their difference. Hence, Yn is WSS. √ 2 17. From the Fourier transform table, SX ( f ) = 2π e−(2π f ) /2 . 18. From the Fourier transform table, SX ( f ) = π e−2π | f | .

Chapter 10 Problem Solutions 19.

169

(a) Since correlation functions are real and even, we can write SX ( f ) =

Z ∞

RX (τ )e− j2π f τ d τ =

−∞ Z ∞

= 2

0

Z ∞

−∞

RX (τ ) cos(2π f τ ) d τ

RX (τ ) cos(2π f τ ) d τ = 2 Re

Z ∞ 0

RX (τ )e− j2π f τ d τ .

(b) OMITTED. (c) The requested plot is at the left; at the right the plot is focused closer to f = 0.

2

2

1

1

0 −6 −3

0

3

0 −1 −0.5

6

0

0.5

1

(d) The requested plot is at the left; at the right the plot is focused closer to f = 0.

4

4

2

2

0 −6 −3 20.

0

3

0 −1 −0.5

6

0

0.5

1

(a) RX (n) = E[Xk+n Xk ] = E[Xk Xk+n ] = RX (−n). (b) Since RX (n) is real and even, we can write ∞

SX ( f ) =

∑

∞

RX (n)e− j2π f n =

∑

n=−∞

n=−∞

∞

∞

=

∑

n=−∞

RX (n) cos(2π f n) − j

∞

=

∑

RX (n)[cos(2π f n) − j sin(2π f n)]

RX (n) cos(2π f n),

∑

n=−∞

|

odd function of n

}| { z RX (n) sin(2π f n) {z

=0

n=−∞

which is a real and even function of f . 21.

(a) Since correlation functions are real and even, we can write ∞

SX ( f ) =

∑

n=−∞

RX (n)e− j2π f n

}

170

Chapter 10 Problem Solutions ∞

= RX (0) + ∑ RX (n)e− j2π f n +

−1

∑

n=1 ∞

n=−∞ ∞

n=1 ∞

k=1

RX (n)e− j2π f n

= RX (0) + ∑ RX (n)e− j2π f n + ∑ RX (−k)e j2π f k = RX (0) + 2 ∑ RX (n) cos(2π f n) n=1

∞

= RX (0) + 2 Re ∑ RX (n)e− j2π f n . n=1

(b) OMITTED. (c) Here is the plot:

2 1 0 −0.5

0

0.5

0

0.5

(d) Here is the plot:

4 2 0 −0.5 22. Write

Z ∞

−∞

h(−t)e− j2π f t dt = =

Z ∞

−∞ Z ∞ −∞

h(θ )e− j2π f (−θ ) d θ h(θ )e j2π f θ d θ

∗ h(θ )∗ e− j2π f θ d θ −∞ Z ∞ ∗ − j2π f θ dθ , h(θ )e =

=

Z

∞

since h is real,

−∞ ∗

= H( f ) .

23. We begin with SXY ( f ) = H( f )∗ SX ( f ) = ( j2π f )∗ SX ( f ) = − j2π f SX ( f ). It then follows that 2 2 d d RXY (τ ) = − RX (τ ) = − e−τ /2 = τ e−τ /2 . dτ dτ

Chapter 10 Problem Solutions

171

Similarly, since SY ( f ) = |H( f )|2 SX ( f ) = −( j2π f )( j2π f )SX ( f ), RY (τ ) = −

2 d2 d d −τ 2 /2 = e−τ /2 (1 − τ 2 ). τe RX (τ ) = RXY (τ ) = 2 dτ dτ dτ

24. Since RX (τ ) = 1/(1 + τ 2 ), we have from the transform table that SX ( f ) = π e−2π | f | . Similarly, since h(t) = 3 sin(π t)/(π t), we have from the transform table that H( f ) = 3I[−1/2,1/2] ( f ). We can now write SY ( f ) = |H( f )|2 SX ( f ) = 9I[−1/2,1/2] ( f ) · π e−2π | f | = 9π e−2π | f | I[−1/2,1/2] ( f ). √ 2 2π e−(2π f ) /2 . √ √  2 2 2 (a) SXY ( f ) = H( f )∗ SX ( f ) = e−(2π f ) /2 ]∗ 2π e−(2π f ) /2 = 2π e−(2π f ) .

25. First note that since RX (τ ) = e−τ

2 /2

, SX ( f ) =

(b) Writing

√ 2 2 1 √ √ SXY ( f ) = √ 2π 2e−( 2) (2π f ) /2 , 2 we have from the transform table that

√ 2 2 1 1 RXY (τ ) = √ e−(τ / 2) /2 = √ e−τ /4 . 2 2

(c) Write 2 1 E[Xt1 Yt2 ] = RXY (t1 − t2 ) = √ e−(t1 −t2 ) /4 . 2 √ √ 2 2 2 (d) SY ( f ) = |H( f )|2 SX ( f ) = e−(2π f ) · 2π e−(2π f ) /2 = 2π e−3(2π f ) /2 .

(e) Writing

√ 2 2 1 √ √ SY ( f ) = √ 2π 3e−( 3) (2π f ) /2 , 3

we have from the transform table that √ 2 2 1 1 RY (τ ) = √ e−(τ / 3) /2 = √ e−τ /6 . 3 3

26. We have from the transform table that SX ( f ) = [sin(π f )/(π f )]2 . The goal is to choose a filter H( f ) so that RY (τ ) = sin(πτ )/(πτ ); i.e., so that SY ( f ) = I[−1/2,1/2] ( f ). Thus, the formula SY ( f ) = |H( f )|2 SX ( f ) becomes I[−1/2,1/2] ( f ) = |H( f )|2 We therefore take

 



sin(π f ) πf

2

πf , | f | ≤ 1/2, H( f ) = sin(π f )  0, | f | > 1/2.

.

172

Chapter 10 Problem Solutions

27. Since Yt and Zt are responses of LTI systems to a WSS input, Yt and Zt are individually WSS. If we can show that E[Yt1 Zt2 ] depends on t1 and t2 only throught their difference, then Yt and Zt will be J-WSS. We show this to be the case. Write  Z ∞ Z ∞ E[Yt1 Zt2 ] = E h(θ )Xt1 −θ d θ g(τ )Xt2 −τ d τ −∞

= =

−∞

Z ∞Z ∞

−∞ −∞

Z ∞Z ∞

−∞ −∞

h(θ )g(τ )E[Xt1 −θ Xt2 −τ ] d τ d θ h(θ )g(τ )RX ([t1 − θ ] − [t2 − τ ]) d τ d θ ,

which depends on t1 − t2 as claimed. 28. Observe that if h(t) := δ (t) − δ (t − 1), then

R∞

−∞ h(τ )Xt−τ d τ

= Xt − Xt−1 .

(a) Since Yt := Xt − Xt−1 is the response of an LTI system to a WSS input, Xt and Yt is J-WSS. (b) Since H( f ) = 1 − e− j2π f , |H( f )|2 = (1 − e− j2π f )(1 − e j2π f ) = 2 − e j2π f − e− j2π f = 2−2

e j2π f + e− j2π f = 2[1 − cos(2π f )], 2

we have SY ( f ) = |H( f )|2 SX ( f ) = 2[1 − cos(2π f )] · 29. In Yt =

Rt

t−3 Xτ

4[1 − cos(2π f )] 2 = . 1 + (2π f )2 1 + (2π f )2

d τ , make the change of variable θ = t − τ , d θ = −d τ to get Yt =

Z 3 0

Xt−θ d θ =

Z ∞

−∞

I[0,3] (θ )Xt−θ d θ .

This shows that Yt is the response to Xt of the LTI system with impulse h(t) = I[0,3] (t). Hence, Yt is WSS. 30. Apply the formula H(0) = (1/π )[(sint)/t]2 .

with h(t) = and we find that 1 1 = π 31.

Z ∞ sint 2 −∞

t

Z

 h(t)e− j2π f t dt −∞ ∞

f =0

Then from the table, H( f ) = (1 − π | f |)I[−1/π ,1/π ] ( f ),

dt,

which is equivalent to π =

Z ∞ sint 2 −∞

t

dt.

(a) Following the hint, we have ∞

E[XnYm ] =

∑

k=−∞

∞

h(k)RX (n, m − k) =

∑

k=−∞

h(k)RX (n − m + k).

Chapter 10 Problem Solutions

173

(b) Similarly, ∞

∑

E[YnYm ] =

l=−∞ ∞

=

∑

 h(l)

∑

h(l)

l=−∞ ∞

=

 h(l)

 h(k)RX (n − l, m − k)

∞

∑

k=−∞ ∞

 h(k)RX (n − l − [m − k])

∑

k=−∞  ∞

 h(k)RX ([n − m] − [l − k]) .

∑

k=−∞

l=−∞

(c) By part (a),

∞

∑

RX (n) =

h(k)RX (n + k),

k=−∞

and so ∞

SXY ( f ) =

∑

RX (n)e

∑

 h(k)

∑

h(k)

n=−∞ ∞

=

k=−∞ ∞

= = =

∞

∑

− j2π f n

RX (n + k)e

∑

RX (m)e− j2π f (m−k)

m=−∞ j2π f k



h(k)e

∑

h(k)e− j2π f k

k=−∞ ∗

∞

 h(k)RX (n + k) e− j2π f n



∑

∑

k=−∞  ∞

∑

=

∞

n=−∞ k=−∞

n=−∞  ∞

k=−∞ ∞





∞

− j2π f n



− j2π f m

∑

RX (m)e

m=−∞ ∗

SX ( f ),



since h(k) is real,

= H( f ) SX ( f ). (d) By part (b), ∞

∑

RY (n) =

l=−∞

and so

∑



∑

h(l)

∑

h(l)

∑

h(l)e− j2π f l

∞

SY ( f ) =

n=−∞ ∞

=

∞

∑

l=−∞ ∞

=

l=−∞

 h(l)

l=−∞ ∞

l=−∞ ∞

=

 h(l)

∞

∑

k=−∞

∞

∑

k=−∞ ∞

∑

h(k)

∑

h(k)

 h(k)RX (n − [l − k]) ,

 h(k)RX (n − [l − k]) e− j2π f n

∑

RX (n − [l − k])e− j2π f n

∑

RX (m)e− j2π f (m+[l−k])

n=−∞ ∞

k=−∞ ∞ k=−∞

m=−∞ ∞

∑

k=−∞

h(k)e j2π f k

∞

∑

m=−∞

= H( f )H( f )∗ SX ( f ) = |H( f )|2 SX ( f ).

RX (m)e− j2π f m

174

Chapter 10 Problem Solutions

32. By the hint,

Z

1 T |x(t)|2 dt = 2T 0 We first observe that since τ T0 ≤ T0 + = n

nE0 1 + 2T 2T

Z τ 0

|x(t)|2 dt.

T T0 ≤ T0 + → T0 , n n

T /n → T0 . It follows that nE0 E0 E0 . = → 2T 2(T /n) 2T0 We next show that the integral on the right goes to zero. Write 1 2T

Z τ 0

|x(t)|2 dt ≤

1 2T

Z T0 0

|x(t)|2 dt =

E0 → 0. 2T

A similar argument shows that 1 2T

Z 0

−T

|x(t)|2 dt → E0 /(2T0 ).

Putting this all together shows that (1/2T ) 33. Write E

Z

∞

−∞

Xt2 dt R



=

Z ∞

−∞

RT

−T

|x(t)|2 dt → E0 /T0 .

E[Xt2 ] dt =

Z ∞

−∞

RX (0) d τ = ∞.

34. If the function q(W ) := 0W S1 ( f ) − S2 ( f ) d f is identically zero, then so is its derivative, q0 (W ) = S1 ( f ) − S2 ( f ). But then S1 ( f ) = S2 ( f ) for all f ≥ 0. 35. If h(t) = I[−T,T ] (t), and white noise is applied to the corresponding system, the cross power spectral density of the input and output is SXY ( f ) = H( f )∗ N0 /2 = 2T

sin(2π T f ) N0 · , 2π T f 2

which is real and even, but not nonnegative. Similarly, if h(t) = e−t I[0,∞) (t), SXY ( f ) =

N0 /2 , 1 + j2π f

which is complex valued. 36. First write 2

SY ( f ) = |H( f )| N0 /2 =



(1 − f 2 )2 N0 /2, | f | ≤ 1, 0, | f | > 1.

Then Z ∞

Z

1 N0 2 (1 − 2 f 2 + f 4 ) d f 2 −∞ 0 i 1 h = N0 f − 32 f 3 + 15 f 5 = N0 [1 − 23 + 51 ] = 8N0 /15.

PY =

SY ( f ) d f =

0

Chapter 10 Problem Solutions

175

2

/2 has power spectral density S ( f ) = e− f 37. First note that RX (τ ) = e−(2πτ ) p X Using the definition of H( f ) = | f | I[−1,1] ( f ),

E[Yt2 ] = RY (0) =

Z ∞

Z ∞

2 /2

Z 1

√ / 2π .

SY ( f ) d f = |H( f )|2 SX ( f ) d f = | f |SX ( f ) d f −∞ −∞ −1 r r Z 1 Z 2 1 − f 2 /2 2  − f 2 /2  1 = 2 f SX ( f ) d f = −e fe df = π 0 π 0 0 r 2 . = (1 − e−1/2 ) π

38. First observe that 2

SY ( f ) = |H( f )| SX ( f ) =



sin π f πf

2

N0 /2.

This is the Fourier transform of RY (τ ) = (1 − |τ |)I[−1,1] (τ ) N20 . Then PY = RY (0) = N0 /2. 39.

(a) First note that H( f ) =

1 1/(RC) = . 1/(RC) + j2π f 1 + j(2π f )RC

Then SXY ( f ) = H( f )∗ SX ( f ) =

N0 /2 . 1 − j(2π f )RC

(b) The inverse Fourier transform of SXY ( f ) = H( f )∗ N0 /2 is

RXY (τ ) = h(−τ )∗ N0 /2 = h(−τ )N0 /2, where the last step follows because h is real. Hence, RXY (τ ) =

N0 τ /(RC) u(−τ ). e 2RC

(c) E[Xt1 Yt2 ] = RXY (t1 − t2 ) = (N0 /(2RC))e(t1 −t2 )/(RC) u(t2 − t1 ). (d) SY ( f ) = |H( f )|2 SX ( f ) =

N0 /2 . 1 + (2π f RC)2

(e) Since N0 /2 N0 SY ( f ) = = · 1 + (2π f RC)2 2(RC)2 we have that RY (τ ) =

2(1/(RC))
1 2 + (2π f )2 RC

N0 −|τ |/(RC) e . 4RC



 RC , 2

176

Chapter 10 Problem Solutions (f) PY = RY (0) = N0 /(4RC).

40. To begin, write E[Yt+1/2Yt ] = RY (1/2). Next, since the input has power spectral density N0 /2 and since h(t) = 1/(1 + t 2 ) has transform H( f ) = π e−2π | f | , we can write SY ( f ) = |H( f )|2 SX ( f ) = |π e−2π | f | |2 N20 = π 2 e−4π | f | N20 =

π N0 2

· π e−2π (2)| f | .

From the transform table, we conclude that RY (τ ) =

π N0 π N0 2 · = , 2 4 + τ2 4 + τ2

and so E[Yt+1/2Yt ] = RY (1/2) = 4π N0 /17. 41. Since H( f ) = sin(π T f )/(π T f ), we can write SY ( f ) = |H( f )|2 Then RY (τ ) =

  sin π T f 2 N0 N0 = T . 2 πT f 2T

N0 (1 − |τ |/T )I[−T,T ] (τ ). 2T

42. To begin, write −t

Yt = e

Z t

θ

−∞

e Xθ d θ =

Z t

−∞

−(t−θ )

e

Xθ d θ =

Z ∞

−∞

e−(t−θ ) u(t − θ )Xθ d θ ,

where u is the unit-step function. We then see that Yt is the response to Xt of the LTI system with impulse response h(t) := e−t u(t). Hence, we know from the text that Xt and Yt are jointly wide-sense stationary. Next, since SX ( f ) = N0 /2, RX (τ ) = (N0 /2)δ (τ ). We then compute in the time domain, RXY (τ ) =

Z ∞

−∞

h(−α )RX (τ − α ) d α =

N0 2

N0 τ e u(−τ ). = 2

Z ∞

−∞

h(−α )δ (τ − α ) d α =

Next, SXY ( f ) = H( f )∗ SX ( f ) = and SY ( f ) = |H( f )|2 SX ( f ) =

N0 /2 , 1 − j2π f N0 /2 . 1 + (2π f )2

It then follows that RY (τ ) = (N0 /4)e−|τ | . 43. Consider the impulse response ∞

h(τ ) :=

∑

n=−∞

hn δ (τ − n).

N0 h(−τ ) 2

Chapter 10 Problem Solutions

177

Then Z ∞

−∞

h(τ )Xt−τ d τ = =

Z ∞ ∞

∑

 hn δ (τ − n) Xt−τ d τ

−∞ n=−∞ Z ∞ ∞

∑

hn

n=−∞

−∞

Xt−τ δ (τ − n) d τ =

∞

∑

hn Xt−n =: Yt .

n=−∞

(a) Since Yt is the response of the LTI system with impulse response h(t) to the WSS input Xt , Xt and Yt are J-WSS. (b) Since Yt is the response of the LTI system with impulse response h(t) to the WSS input Xt , SY ( f ) = |H( f )|2 SX ( f ), where  Z ∞ ∞ Z ∞ δ ( τ − n) e− j2π f τ d τ h H( f ) = h(τ )e− j2π f τ d τ = ∑ n −∞ ∞

=

∑

n=−∞

hn

Z ∞

−∞

−∞ n=−∞

δ (τ − n)e− j2π f τ d τ =

∞

∑

hn e− j2π f n

n=−∞

has period one. Hence, P( f ) = |H( f )|2 is real, nonnegative, and has period one. 44. When the input power spectral density is SW ( f ) = 3, the output power spectral density 2 is |H( f )|2 · 3. We are also told that this output power spectral density is equal to e− f . 2 2 2 Hence, |H( f )|2 · 3 = e− f , or |H( f )|2 = e− f /3. Next, if SX ( f ) = e f I[−1,1] ( f ), then 2

2

SY ( f ) = |H( f )|2 SX ( f ) = (e− f /3) · e f I[−1,1] ( f ) = (1/3)I[−1,1] ( f ). It then follows that 1 sin(2πτ ) 2 sin(2πτ ) RY (τ ) = · 2 = · . 3 2πτ 3 2πτ

45. Since H( f ) = GI[−B,B] ( f ) and Yt is the response to white noise, the output power spectral density is SY ( f ) = G2 I[−B,B] ( f ) · N0 /2, and so RY (τ ) =

sin(2π Bτ ) G2 N0 sin(2π Bτ ) · 2B = G2 BN0 · . 2 2π Bτ 2π Bτ

Note that RY (k∆t) = RY (k/(2B)) = G2 BN0 ·

sin(2π Bk/(2B)) sin(π k) = G2 BN0 · , 2π Bk/(2B) πk

which is G2 BN0 for k = 0 and zero otherwise. It is obvious that the Xi are zero mean. Since E[Xi X j ] = RY (i − j), and the Xi are uncorrelated with variance E[Xi2 ] = RY (0) = G2 BN0 . 46.

∗ ])∗ = (a) First write RX (τ ) = E[Xt+τ Xt∗ ]. Then RX (−τ ) = E[Xt−τ Xt∗ ] = (E[Xt Xt− τ ∗ RX (τ ) .

(b) Since SX ( f ) =

Z ∞

−∞

RX (τ )e− j2π f τ d τ ,

178

Chapter 10 Problem Solutions we can write SX ( f )∗ =

Z ∞

−∞

RX (τ )∗ e j2π f τ d τ = =

Z ∞

−∞ Z ∞ −∞

RX (−t)∗ e− j2π f t dt RX (t)e− j2π f t dt,

by part (a),

= SX ( f ). Since SX ( f ) is equal to its complex conjugate, SX ( f ) is real. (c) Write  Z E[Xt1 Yt∗2 ] = E Xt1 =

Z ∞

∞

∗ 

h(θ )Xt2 −θ d θ

−∞

=

h(θ )∗ RX ([t1 − t2 ] + θ ) d θ =

−∞

(d) By part (c), RXY (τ ) =

Z ∞

Z ∞

h(θ )∗ E[Xt1 Xt∗2 −θ ] d θ

−∞ Z ∞

h(−β )∗ RX ([t1 − t2 ] − β ) d β .

−∞

h(−β )∗ RX (τ − β ) d β ,

−∞ h(−·)∗ and

which is the convolution of RX . Hence, the transform of this equation is the product of the transform of h(−·)∗ and SX . We just have to observe that Z ∞ ∗ Z ∞ Z ∞ h(−β )∗ e− j2π f β d β = h(t)e− j2π f t dt = H( f )∗ . h(t)∗ e j2π f t dt = −∞

−∞

−∞

Hence, SXY ( f ) = H( f )∗ SX ( f ). Next, since    Z ∞ ∗ RY (τ ) = E[Yt+τ Yt ] = E h(θ )Xt+τ −θ d θ Yt∗ =

Z ∞

−∞

−∞

h(θ )E[Xt+τ −θ Yt∗ ] d θ

=

Z ∞

−∞

h(θ )RXY (τ − θ ) d θ ,

is a convolution, its transform is SY ( f ) = H( f )SXY ( f ) = H( f )H( f )∗ SX ( f ) = |H( f )|2 SX ( f ). 47.

(a) RX (τ ) = (b) Write

Z ∞

−∞

RX (τ ) =

SX ( f )e j2π f τ d f = Z ∞

−∞

Z ∞

−∞

δ ( f )e j2π f τ d f = e j2π 0τ = 1.

[δ ( f − f0 ) + δ ( f + f0 )]e j2π f τ d f = e j2π f0 τ + e− j2π f0 τ = 2 cos(2π f0 τ ).

(c) First write SX ( f ) = e− f

2 /2

h 1 2 2 1 √ i 2π = e−( 2π ) (2π f ) /2 · · 2π √ 2π 2π i√ h √ 1 2 2 1 = e−( 2π ) (2π f ) /2 · · 2π 2π . 2π

Chapter 10 Problem Solutions

179

From the Fourier transform table with σ = 1/(2π ), RX (τ ) =

√

2π e−(2πτ )

2 /2

.

(d) From the transform table with λ = 1/(2π ), RX (τ ) =

2 1 1/(2π ) = . · π (1/(2π ))2 + τ 2 1 + (2πτ )2

48. Write E[Xt2 ] = RX (0) = 49.

Z

 SX ( f )e j2π f τ d f −∞ ∞

τ =0

=

Z ∞

SX ( f ) d f =

−∞

Z W

−W

1 d f = 2W.

(a) e− f u( f ) is not even. 2

(b) e− f cos( f ) is not nonnegative. (c) (1 − f 2 )/(1 + f 4 ) is not nonnegative.

(d) 1/(1 + j f 2 ) is not real valued. 50.

(a) Since sin τ is odd, it is NOT a valid correlation function. (b) Since the Fourier transform of cos τ is [δ ( f − 1) + δ ( f + 1)]/2, which is real, even, and nonnegative, cos τ IS a valid correlation function. √ 2 2 (c) Since the Fourier transform of e−τ /2 is 2π e−(2π f ) /2 , which is real, even, and 2 nonnegative, e−τ /2 IS a valid correlation function. (d) Since the Fourier transform of e−|τ | is 2/[1 + (2π f )2 ], which is real, even, and nonnegative, e−|τ | IS a valid correlation function. (e) Since the value of τ 2 e−|τ | at τ = 0 is less than the value for other values of τ , τ 2 e−|τ | is NOT a valid correlation function. (f) Since the Fourier transform of I[−T,T ] (τ ) is (2T ) sin(2π T f )/(2π T f ) is not nonnegative, I[−T,T ] (τ ) is NOT a valid correlation function.

51. Since R0 (τ ) is a correlation function, S0 ( f ) is real, even, and nonnegative. Since R(τ ) = R0 (τ ) cos(2π f0 τ ), S( f ) =

1 2 [S0 ( f

− f0 ) + S0 ( f + f0 )],

which is obviously real and nonnegative. It is also even since S(− f ) =

1 2 [S0 (− f − f 0 ) + S0 (− f + f 0 )] 1 2 [S0 ( f + f 0 ) + S0 ( f − f 0 )], since

= = S( f ).

S0 is even,

Since S( f ) is real, even, and nonnegative, R(τ ) is a valid correlation function.

180

Chapter 10 Problem Solutions

52. First observe that the Fourier transform of R(τ ) = R(τ − τ0 ) + R(τ + τ0 ) is S( f ) = 2S( f ) cos(2π f τ0 ). Hence, the answer cannot be (a) because it is possible to have S( f ) > 0 and cos(2π f τ0 ) < 0 for some values of f . Let S( f ) = I[−1/(4τ0 ),1/(4τ0 )] ( f ), which is real, even, and nonnegative. Hence, its inverse transform, which we denote by R(τ ), is a correlation function. In this case, S( f ) = 2S( f ) cos(2π f τ0 ) ≥ 0 for all f , and is real and even too. Hence, for this choice of R(τ ), R(τ ) is a correlation function. Therefore, the answer is (b). 53. To begin, write R(τ ) =

Z ∞

−∞

S( f )e

j2π f τ

df =

Z ∞

−∞

S( f )[cos(2π f τ ) − j sin(2π f τ )] d f .

Since S is real and even, the integral of S( f ) sin(2π f τ ) is zero, and we have R(τ ) =

Z ∞

−∞

S( f ) cos(2π f τ ) d f ,

which is a real and even function of τ . Finally, Z ∞ Z |R(τ )| = S( f )e j2π f τ d f ≤ −∞

=

Z ∞

−∞

|S( f )| d f =

Z ∞

−∞

∞

−∞

S( f )e j2π f τ d f

S( f )d f = R(0).

54. Let S0 ( f ) denote the Fourier transform of R0 (τ ), and let S( f ) denote the Fourier transform of R(τ ). (a) The derivation in the text showing that the transform of a correlation function is real and even uses only the fact that correlation functions are real and even. Hence, S0 ( f ) is real and even. Furthermore, since R is the convolution of R0 with itself, S( f ) = S0 ( f )2 , which is real, even, and nonnegative. Hence, R(τ ) is a correlation function. (b) If R0 (τ ) = I[−T,T ] (τ ), then sin(2π T f ) S0 ( f ) = 2T 2π T f

and



sin(2π T f ) S( f ) = 2T · 2T 2π T f


Hence, R(τ ) = 2T · 1 − |τ |/(2T ) I[−2T,2T ] (τ ).

2

55. Taking α = N0 /2 as in the text,

h(t) = v(t0 − t) = sin(t0 − t)I[0,π ] (t0 − t). Then h is causal for t0 ≥ π .

56. Since v(t) = e(t/

√ 2 √ √ 2 ) /2 , V ( f ) = 2π 2e−2(2π f )2 /2

√ 2 = 2 π e−(2π f ) . Then

√ 2 V ( f )∗ e− j2π f t0 2 π e−(2π f ) e− j2π f t0 H( f ) = α = α 2 SX ( f ) e−(2π f ) /2 √ √ √ 2 2 = 2α π e−(2π f ) /2 e− j2π f t0 = 2 α · 2π e−(2π f ) /2 e− j2π f t0 , √ 2 and it follows that h(t) = 2 α e−(t−t0 ) /2 .

.

Chapter 10 Problem Solutions

181

57. Let v0 (n) := ∑k h(n − k)v(k) and Yn := ∑k h(n − k)Xk . The SNR is v0 (n0 )2 /E[Yn20 ]. We have Z Z 1/2

E[Yn20 ] =

−1/2

1/2

SY ( f ) d f =

−1/2

|H( f )|2 SX ( f ) d f .

Let V ( f ) := ∑k v(k)e− j2π f k . Then Z 1/2 2 j2π f n0 H( f )V ( f )e |v0 (n0 )| = d f −1/2 Z 1/2 p V ( f )e j2π f n0 2 H( f ) SX ( f ) · p = d f −1/2 SX ( f ) 2

≤

Z 1/2

−1/2

|H( f )|2 SX ( f ) d f

Z 1/2 |V ( f )∗ e− j2π f n0 |2

SX ( f )

−1/2

d f,

with equality if and only if p V ( f )∗ e− j2π f n0 H( f ) SX ( f ) = α p SX ( f )

(#)

for some constant α . It is now clear that the SNR is upper bounded by Z 1/2 |V ( f )∗ e− j2π f n0 |2 −1/2

SX ( f )

df

and that the SNR equals the bound if and only if (#) holds with equality for some constant α . Hence, the matched filter transfer function is H( f ) = α

V ( f )∗ e− j2π f n0 . SX ( f )

58. We begin by writing E[Ut ] = E[Vt + Xt ] = E[Vt ] + E[Xt ], which does not depend on t since Vt and Xt are each individually WSS. Next write E[Ut1 Vt2 ] = E[(Vt1 + Xt1 )Vt2 ] = RV (t1 − t2 ) + RXV (t1 − t2 ).

(∗)

Now write E[Ut1 Ut2 ] = E[Ut1 (Vt2 + Xt2 ] = E[Ut1 Vt2 ] + E[Ut1 Xt2 ]. By (∗), the term E[Ut1 Vt2 ] depends on t1 and t2 only through their difference. Since E[Ut1 Xt2 ] = E[(Vt1 + Xt1 )Xt2 ] = RV X (t1 − t2 ) + RX (t1 − t2 ), it follows that E[Ut1 Ut2 ] depends on t1 and t2 only through their difference. Hence, Ut and Vt are J-WSS.

182

Chapter 10 Problem Solutions

59. The assumptions in the problem imply that Vt and Xt are J-WSS, and by the preceding problem, it follows that Ut and Vt are J-WSS. We can therefore apply the formulas for the Wiener filter derived in the text. It just remains to compute the quantities used in the formulas. First, RVU (τ ) = E[Vt+τ Ut ] = E[Vt+τ (Vt + Xt )] = RV (τ ) + RXV (τ ) = RV (τ ), which implies SVU ( f ) = SV ( f ). Similarly, RU (τ ) = E[Ut+τ Ut ] = E[(Vt+τ + Xt+τ )Ut ] = RVU (τ ) + E[Xt+τ (Vt + Xt )] = RV (τ ) + RXV (τ ) + RX (τ ) = RV (τ ) + RX (τ ), and so SU ( f ) = SV ( f ) + SX ( f ). We then have H( f ) =

SV ( f ) SVU ( f ) = . SU ( f ) SV ( f ) + SX ( f )

60. The formula for RV (τ ) implies SV ( f ) = (1 − | f |)I[−1,1] ( f ). We then have (1 − | f |)I[−1,1] ( f ) SV ( f ) = SV ( f ) + SX ( f ) (1 − | f |)I[−1,1] ( f ) + 1 − I[−1,1] ( f )

H( f ) =

(1 − | f |)I[−1,1] ( f ) = I[−1,1] ( f ), 1 − | f |I[−1,1] ( f )

= and so

h(t) = 2

sin(2π t) . 2π t

61. To begin, write E[|Vt − Vbt |2 ] = E[(Vt − Vbt )(Vt − Vbt )] = E[(Vt − Vbt )Vt ] − E[(Vt − Vbt )Vbt ] = E[(Vt − Vbt )Vt ], by the orthogonality principle, = E[Vt2 ] − E[Vbt Vt ] = RV (0) − E[Vbt Vt ] =

Next observe that  Z E[Vbt Vt ] = E =

= =

∞

−∞

Z ∞

−∞ Z ∞ −∞ Z ∞

  Z h(θ )Ut−θ d θ Vt =

∞

−∞

Z ∞

−∞

h(θ )E[Vt Ut−θ ] d θ

h(θ )RVU (θ ) d θ =

Z ∞

H( f )SVU ( f )∗ d f ,

by Parseval’s formula,

−∞

SVU ( f ) SVU ( f )∗ d f = −∞ SU ( f )

h(θ )RVU (θ )∗ d θ ,

Z ∞ |SVU ( f )|2 −∞

SV ( f ) d f − E[Vbt Vt ].

SU ( f )

since RVU (θ ) is real,

d f.

Putting these two observations together yields E[|Vt − Vbt |2 ] =

Z ∞

−∞

SV ( f ) d f −

Z ∞ |SVU ( f )|2 −∞

SU ( f )

df =

Z ∞ −∞

SV ( f ) −

 |SVU ( f )|2 d f. SU ( f )

Chapter 10 Problem Solutions

183

62. Denote the optimal estimator by Vbn = ∑∞ k=−∞ h(k)Un−k , and denote any other estima˜ h(k)U . The discrete-time orthogonality principle says that if tor by Ven = ∑∞ n−k k=−∞  E (Vn − Vbn )

∞

˜ h(k)U n−k

∑

k=−∞



= 0

(∗)

˜ then h is optimal in that E[|Vn − Vbn |2 ] ≤ E[|Vn − Ven |2 ] for every h. ˜ To esfor every h, tablish the orthogonality principle, assume the above equation holds for every choice ˜ Then we can write of h. E[|Vn − Ven |2 ] = E[|(Vn − Vbn ) + (Vbn − Ven )|2 ] = E[|Vn − Vbn |2 + 2(Vn − Vbn )(Vbn − Ven ) + |Vbn − Ven |2 ] = E[|Vn − Vbn |2 ] + 2E[(Vn − Vbn )(Vbn − Ven )] + E[|Vbn − Ven |2 ].

(∗∗)

Now observe that

  E[(Vn − Vbn )(Vbn − Ven )] = E (Vn − Vbn )  = E (Vn − Vbn )

∞

∑

∞

k=−∞ ∞ 

∑

k=−∞

h(k)Un−k −

∑

˜ h(k)U n−k

k=−∞

 ˜ h(k) − h(k) Un−k



= 0,

 by (∗).

We can now continue (∗∗) writing

E[|Vn − Ven |2 ] = E[|Vn − Vbn |2 ] + E[|Vbn − Ven |2 ] ≥ E[|Vn − Vbn |2 ],

and thus h is the filter that minimizes the mean-squared error.

The next task is to characterize the filter h that satisfies the orthogonality condition ˜ Write the orthogonality condition as for every choice of h.  0 = E (Vn − Vbn ) ∞

=

∑

k=−∞ ∞

=

∑

k=−∞

∞

˜ h(k)U t−k

∑

k=−∞



= E

˜ bn )Ut−k ] = E[h(k)(V n −V

˜ h(k)[R VU (k) − RVU b (k)].



∞

∞

∑

∑

k=−∞

k=−∞

˜ bn )Ut−k h(k)(V n −V



˜ bn )Ut−k ] h(k)E[(V n −V

˜ take h(k) ˜ Since this must hold for all h, = RVU (k) − RVU b (k) to get ∞

∑

RVU (k) − R b (k) 2 = 0. VU

k=−∞

Thus, the orthogonality condition holds for all h˜ if and only if RVU (k) = RVU b (k) for all k.

184

Chapter 10 Problem Solutions b The next task is to analyze RVU b . Recall that Vn is the response of an LTI system to input Un . Applying the result of Problem 31(a) with X replaced by U and Y replaced by Vb , we have, also using the fact that RU is even, ∞

RVU b (m) = RU Vb (−m) =

∑

k=−∞

h(k)RU (m − k).

Taking discrete-time Fourier transforms of

∞

yields

RVU (m) = RVU b (m) = SVU ( f ) = H( f )SU ( f ),

∑

k=−∞

h(k)RU (m − k)

and so H( f ) =

SVU ( f ) . SU ( f )

63. We have 2λ /[λ 2 + (2π f )2 ] 2λ SV ( f ) = = SV ( f ) + SX ( f ) 2λ /[λ 2 + (2π f )2 ] + 1 2λ + λ 2 + (2π f )2 λ 2A = · , A A2 + (2π f )2 √ where A := 2λ + λ 2 . Hence, h(t) = (λ /A)e−A|t| . H( f ) =

64. To begin, write K( f ) =

λ + j2π f λ 1 = + j2π f . A + j2π f A + j2π f 1 + j2π f

Then d −At e u(t) dt = λ e−At u(t) − Ae−At u(t) + e−At δ (t) = (λ − A)e−At u(t) + δ (t),

k(t) = λ e−At u(t) +

since e−At δ (t) = δ (t) for both t = 0 and for t 6= 0. This is a causal impulse response. 65. Let Zt := Vt+∆t . Then the causal Wiener filter for Zt yields the prediction or smoothing filter for Vt+∆t . The Wiener–Hopf equation for Zt is RZU (τ ) =

Z ∞ 0

h∆t (θ )RU (τ − θ ) d θ ,

τ ≥ 0.

Now, RZU (τ ) = E[Zt+τ Ut ] = E[Vt+τ +∆t Ut ] = RVU (τ + ∆t), and so we must solve RVU (τ + ∆t) =

Z ∞ 0

h∆t (θ )RU (τ − θ ) d θ ,

τ ≥ 0.

Chapter 10 Problem Solutions

185

For white noise with RU (τ ) = δ (τ ), this reduces to RVU (τ + ∆t) =

Z ∞ 0

h∆t (θ )δ (τ − θ ) d θ = h∆t (τ ),

τ ≥ 0.

If h(t) = RVU (t)u(t) denotes the causal Wiener filter, then for ∆t ≥ 0 (prediction), we can write h∆t (τ ) = RVU (τ + ∆t) = h(τ + ∆t), τ ≥ 0.

If ∆t < 0 (smoothing), we can write h∆t (τ ) = h(τ + ∆t) only for τ ≥ −∆t. For 0 ≤ τ < −∆t, h(τ + ∆t) = 0 while h∆t (τ ) = RVU (τ + ∆t). 66. By the hint, the limit of the double sums is the desired double integral. If we can show that each of these double sums is nonnegative, then the limit will also be nonnegative. To this end put Zi := Xti e− j2π f ti ∆t i . Then  n  n ∗   n 2  n n = ∑ ∑ E[Zi Zk∗ ] 0 ≤ E ∑ Zi = E ∑ Zi ∑ Zk n

=

i=1 n

i=1

k=1

i=1 k=1

∑ ∑ E[Xti Xtk ]e− j2π f ti e j2π f tk ∆t i ∆t k

i=1 k=1 n n

=

∑ ∑ RX (ti − tk )e− j2π f ti e j2π f tk ∆t i ∆t k .

i=1 k=1

67.

(a) The Fourier transform of CY (τ ) = e−|τ | is 2/[1 + (2π f )2 ], which is continuous 1 RT at f = 0. Hence, we have convergence in mean square of 2T Y −T t dt to E[Yt ].

(b) The Fourier transform of CY (τ ) = sin(πτ )/(πτ ) is I[−1/2,1/2] ( f ), which is con1 RT tinuous at f = 0. Hence, we have convergence in mean square of 2T −T Yt dt to E[Yt ]. 68. We first point out that this is not a question about mean-square convergence. Write 1 2T

Z T

−T

cos(2π t + Θ) dt =

sin(2π T + Θ) − sin(2π (−T ) + Θ) . 2T · 2π

Since | sin x | ≤ 1, we can write 1 ZT 2 2T −T cos(2π t + Θ) dt ≤ 4π T → 0, and so the limit in question exists and is equal to zero.

69. As suggested by the hint, put Yt := Xt+τ Xt . It will be sufficient if Yt is WSS and if the Fourier transform of the covariance function of Yt is continuous at the origin. First, since Xt is WSS, the mean of Yt is E[Yt ] = E[Xt+τ Xt ] = RX (τ ), which does not depend on t. Before examining the correlation function of Yt , we assume that Xt is fourth-order strictly stationary so that E[Yt1 Yt2 ] = E[Xt1 +τ Xt1 Xt2 +τ Xt2 ]

186

Chapter 10 Problem Solutions must be unchanged if on the right-hand side we subtract t2 from every subscript expression to get E[Xt1 +τ −t2 Xt1 −t2 Xτ X0 ]. Since this depends on t1 and t2 only through their difference, we see that Yt is WSS if Xt is fourth-order strictly stationary. Now, the covariance function of Yt is C(θ ) = E[Xθ +τ Xθ Xτ X0 ] − RX (τ )2 . If the Fourier transform of this function of θ is continuous at the origin, then 1 2T

Z T

−T

Xt+τ Xt dt → RX (τ ).

70. As suggested by the hint, put Yt := IB (Xt ). It will be sufficient if Yt is WSS and if the Fourier transform of the covariance function of Yt is continuous at the origin. We assume at the outset that Xt is second-order strictly stationary. Then the mean of Yt is E[Yt ] = E[IB (Xt )] = P(Xt ∈ B), which does not depend on t. Similarly, E[Yt1 Yt2 ] = E[IB (Xt1 )IB (Xt2 )] = P(Xt1 ∈ B, Xt2 ∈ B) must be unchanged if on the right-hand side we subtract t2 from every subscript to get P(Xt1 −t2 ∈ B, X0 ∈ B). Since this depends on t1 and t2 only through their difference, we see that Yt is WSS if Xt is second-order strictly stationary. Now, the covariance function of Yt is C(θ ) = P(Xθ ∈ B, X0 ∈ B) − P(Xt ∈ B)2 . If the Fourier transform of this function of θ is continuous at the origin, then 1 2T

Z T

−T

IB (Xt ) dt → P(Xt ∈ B).

71. We make the following definition and apply the hints: Z

1 T RXY (τ + θ , θ ) d θ T →∞ 2T −T   Z Z ∞ 1 T = lim h(α )RX (τ + θ , θ − α ) d α d θ T →∞ 2T −T −∞   Z Z ∞ 1 T RX (τ + θ , θ − α ) d θ d α = h(α ) lim T →∞ 2T −T −∞   Z ∞ Z 1 T −α = RX (τ + α + β , β ) d β d α h(α ) lim T →∞ 2T −T −α −∞   Z Z ∞ 1 T RX (τ + α + β , β ) d β d α . = h(α ) lim T →∞ 2T −T −∞ | {z }

RXY (τ ) := lim

= RX (τ +α )

Chapter 10 Problem Solutions

187

72. Write Z

1 T RY (τ + θ , θ ) d θ T →∞ 2T −T   Z ∞ Z 1 T h(β )RXY (τ + θ − β , θ ) d β d θ = lim T →∞ 2T −T −∞   Z ∞ Z 1 T = h(β ) lim RXY ([τ − β ] + θ , θ ) d θ d β . T →∞ 2T −T −∞ | {z }

RY (τ ) := lim

= RXY (τ −β )

73. Let SXY ( f ) denote the Fourier transform of RXY (τ ), and let SY ( f ) denote the Fourier transform of RY (τ ). Then by the preceding two problems, SY ( f ) = H( f )SXY ( f ) = H( f )H(− f )SX ( f ) = |H( f )|2 SX ( f ), where, since h is real, H(− f ) = H( f )∗ . 74. This is an instance of Problem 32.

CHAPTER 11

Problem Solutions 1. With λ = 3 and t = 10, P(Nt = 0) = e−λ t = e−3·10 = e−30 = 9.358 × 10−14 . 2. With λ = 12 per minute and t = 20 seconds, λ t = 4. Thus,   (λ t)2 (λ t)3 −λ t P(Nt > 3) = 1 − P(Nt ≤ 3) = 1 − e 1 + λt + + 2! 3!   3 2 4 4 = 1 − e−4 (5 + 8 + 32/3) = 1 − e−4 1 + 4 + + 2 6 = 1 − e−4 (39/3 + 32/3) = 1 − e−4 (71/3) = 0.5665.

3.

(a) P(N5 = 10) =

(2 · 5)10 e−2·5 = 0.125. 10!

(b) We have   5 \ {Ni − Ni−1 = 2} = P i=1

5

5

∏ P(Ni − Ni−1 = 2) i=1

−2 5

−10

=

∏ i=1



22 e−2 2!



= 1.453 × 10−3 .

= (2e ) = 32e

4. Let Nt denote the number of crates sold through time t (in days). Then Ni − Ni−1 is the number of crates sold on day i, and so  5  \ P {Ni − Ni−1 ≥ 3} = i=1

5

∏ P(Ni − Ni−1 ≥ 3)

5

=

i=1

∏ i=1

h

i 1 − P(Ni − Ni−1 ≤ 2)

5 h
i = ∏ 1 − e−λ 1 + λ + λ 2 /2! i=1


i5 = 1 − e−3 1 + 3 + 9/2 = 0.06385. h

5. Let Nt denote the number of fishing rods sold through time t (in days). Then Ni − Ni−1 is the number of crates sold on day i, and so    5  5 \ [ {Ni − Ni−1 ≤ 2} {Ni − Ni−1 ≥ 3} = 1 − P P i=1

i=1

5

= 1 − ∏ P(Ni − Ni−1 ≤ 2) i=1

188

Chapter 11 Problem Solutions

189

5   = 1 − ∏ e−λ (1 + λ + λ 2 /2!) i=1

 5 = 1 − e−2 (1 + 2 + 4/2!) = 1 − e−10 · 55 = 0.858.

6. Since the average time between hit songs is 7 months, the rate is λ = 1/7 per month. (a) Since a year is 12 months, we write P(N12 > 2) = 1 − P(N12 ≤ 2) = 1 − e−12λ [1 + 12λ + (12λ )2 /2!] = 1 − e−12/7 [1 + 12/7 + (12/7)2 /2] = 0.247.

(b) Let Tn denote the time of the nth hit song. Since Tn = X1 + · · · + Xn , E[Tn ] = nE[X1 ] = 7n. For n = 10, E[T10 ] = 70 months. 7.

(a) Since N0 ≡ 0, Nt = Nt − N0 . Since (0,t] ∩ (t,t + ∆t] = ∅, Nt − N0 and Nt+∆t − Nt are independent. (b) Write P(Nt+∆t = k + `|Nt = k) = P(Nt+∆t − Nt = `|Nt = k),

by substitution,

= P(Nt+∆t − Nt = `|Nt − N0 = k), since N0 ≡ 0, = P(Nt+∆t − Nt = `), by independent increments.

(c) Write P(Nt+∆t = k + `|Nt = k)P(Nt = k) P(Nt+∆t = k + `) P(Nt+∆t − Nt = `)P(Nt = k) , by part (b), = P(Nt+∆t = k + `)

P(Nt = k|Nt+∆t = k + `) =

=

=

(λ ∆t)` e−λ ∆t (λ t)k e−λ t · `! k! [λ (t+∆t)]k+` e−λ (t+∆t) (k+`)!  k  

k+` k

t t + ∆t

∆t t + ∆t

(d) In part (c), put ` = n − k and put p = t/(t + ∆t). Then   n k P(Nt = k|Nt+∆t = n) = p (1 − p)n−k , k

`

.

k = 0, . . . , n.

8. The nth customer arrives at time Tn ∼ Erlang(n, λ ). Hence, E[Tn ] =

Γ(1 + n) nΓ(n) n = = . λ Γ(n) λ Γ(n) λ

Alternatively, since Tn = X1 +· · ·+Xn , where the Xi are i.i.d. exp(λ ), E[Tn ] = nE[Xi ] = n/λ .

190 9.

Chapter 11 Problem Solutions (a) E[Xi ] = 1/λ = 0.5 weeks. (b) P(N2 = 0) = e−λ ·2 = e−4 = 0.0183. (c) E[N12 ] = λ · 12 = 24 snowstorms. (d) Write  12   12  [ \ P {Ni − Ni−1 ≥ 5} = 1 − P {Ni − Ni−1 ≤ 4} i=1

i=1

= 1 − [e−λ (1 + λ + λ 2 /2 + λ 3 /6 + λ 4 /24)]12 = 1 − [e−2 (1 + 2 + 2 + 4/3 + 2/3)]12 = 1 − [7e−2 ]12 = 1 − 0.523 = 0.477.

10. First observe that since 1/λ = 2 months, λ = 1/2 per month. (a) P(N4 = 0) = e−λ ·4 = e−4/2 = e−2 = 0.135. (b) Write  4   4  [ \ P {Ni − Ni−1 ≥ 2} = 1 − P {Ni − Ni−1 ≤ 1} i=1

i=1

4

= 1 − ∏ P(Ni − Ni−1 ≤ 1) i=1 4

= 1 − ∏[e−λ ·1 + e−λ ·1 (λ · 1)] i=1

= 1 − [e−λ (1 + λ )]4 = 1 − [ 32 e−1/2 ]4

81 −2 = 1 − 16 e = 0.315.

11. We need to find var(Tn ) = var(X1 + · · · + Xn ). Since the Xi are independent, they are uncorrelated, and so the variance of the sum is the sum of the variances. Since Xi ∼ exp(λ ), var(Tn ) = n var(Xi ) = n/λ 2 . An alternative approach is to use the fact that Tn ∼ Erlang(n, λ ). Since the moments of Tn are available, var(Tn ) = E[Tn2 ] − (E[Tn ])2 =

(1 + n)n  n 2 n = 2. − 2 λ λ λ

12. To begin, use the law of total probability, substitution, and independence to write Yt

E[z ] = E[z =

Nln(1+tU)

Z 1

] =

Z 1

E[zNln(1+tu) ] du =

0

=

Z 1 0

E[z

0

Nln(1+tU)

Z 1 0

(1 + tu)z−1 du.

|U = u] du =

Z 1 0

E[zNln(1+tu) |U = u] du

exp[(z − 1) ln(1 + tu)] du =

Z 1 0

eln(1+tu)

z−1

du

Chapter 11 Problem Solutions

191

To compute this integral, we need to treat the cases z = 0 and z 6= 0 separately. We find that  ln(1 + t)  , z = 0,  t Yt E[z ] = z   (1 + t) − 1 , z 6= 0. tz

13. Denote the arrival times of Nt by T1 , T2 , . . . , and let Xk := Tk − Tk−1 denote the interarrival times. Similarly, denote the arrival times of Mt by S1 , S2 , . . . . (As it turns out, we do not need the interarrival times of Mt .) Then for arbitrary k > 1, we use the law of total probability, substitution, and independence to compute P(MTk − MTk−1 = m) = = =

Z ∞Z ∞ Z0 ∞ Zθ∞ Z0 ∞ Zθ∞ 0

θ

P(MTk − MTk−1 = m|Tk = t, Tk−1 = θ ) fTk Tk−1 (t, θ ) dt d θ P(Mt − Mθ = m|Tk = t, Tk−1 = θ ) fTk Tk−1 (t, θ ) dt d θ P(Mt − Mθ = m) fTk Tk−1 (t, θ ) dt d θ

Z ∞Z ∞ [µ (t − θ )]m e−µ (t−θ )

fTk Tk−1 (t, θ ) dt d θ m!   [µ (Tk − Tk−1 )]m e−µ (Tk −Tk−1 ) = E m!   Z ∞ (µ Xk )m e−µ Xk (µ x)m e−µ x = E = · λ e−λ x dx m! m! 0 Z ∞ µ mλ xm · (λ + µ )e−(λ +µ )x dx = (λ + µ )m! 0 | {z } =

0

θ

mth moment of exp(λ + µ ) density

m! µ mλ λ = · = m (λ + µ )m! (λ + µ ) λ +µ



µ λ +µ

m

,

which is a geometric0 (µ /(λ + µ )) pmf in m. 14. It suffices to show that the probability generating function GMt (z) has the form eµ (z−1) for some µ > 0. We use the law of total probability, substitution, and independence to write ∞

Nt

Nt

∑ E[z∑i=1 Yi |Nt = n]P(Nt = n)

GMt (z) = E[zMt ] = E[z∑i=1 Yi ] =

n=0 ∞

= =

n

∑ E[z∑i=1 Yi |Nt = n]P(Nt = n)

n=0 ∞  n

∑

n=0

 Yi ∏ E[z ] P(Nt = n) = i=1

n

∑ E[z∑i=1 Yi ]P(Nt = n)

n=0 ∞

∑ [pz + (1 − p)]n P(Nt = n)

n=0 λ t([pz+(1−p)]−1)


= GNt pz + (1 − p) = e

Thus, Mt ∼ Poisson(pλ t).

∞

=

= e pλ t(z−1) .

192

Chapter 11 Problem Solutions

t Vi denote the total energy throught time t, with Mt = 0 for Nt = 0. We 15. Let Mt = ∑Ni=1 use the law of total probability, substitution, and independence to write  ∞ ∞  Nt E[Mt ] = ∑ E[Mt |Nt = n]P(Nt = n) = ∑ E ∑ Vi Nt = n P(Nt = n)

n=0 ∞

n=1

 = ∑ E ∑ Vi Nt = n P(Nt = n) = n=1 ∞



n

i=1

∑ nE[V1 ]P(Nt = n)

=

n=1

i=1

∞

∑



n=1

∞

 ∑ E[Vi ] P(Nt = n) n

i=1

= E[V1 ] ∑ nP(Nt = n) = E[V1 ]E[Nt ] = E[V1 ](λ t). n=1

The average time between lightning strikes is 1/λ minutes. 16. We have 2.132(1.96) = 5.170 ± 0.418 with 95% probability. 10

λ = 5.170 ±

In other words, λ ∈ [4.752, 5.588] with 95% probability. 17. To begin, observe that ∞

∞

Y = ∑ g(Ti ) = ∑

i=1 k=1

i=1

n

∞

n

k=1

i=1

k=1

n

∑ gk I(tk−1 ,tk ] (Ti ) = ∑ gk ∑ I(tk−1 ,tk ] (Ti ) = ∑ gk (Ntk − Ntk−1 )

is a sum of independent random variables. Then n

n

E[Y ] =

∑ gk E[Ntk − Ntk−1 ]

k=1

=

∑ gk λ (tk − tk−1 )

k=1

=

Z ∞

g(τ )λ d τ ,

0

since g is piecewise constant. Next, n

ϕY (ν ) = E[e jνY ] = E[e jν ∑k=1 gk (Ntk −Ntk−1 ) ] =

n

∏ E[e jν gk (Ntk −Ntk−1 ) ]

k=1

 n  i = ∏ exp λ (tk − tk−1 )(e jν gk − 1) = exp ∑ λ (tk − tk−1 )(e jν gk − 1) n

k=1

= exp

Z

h

∞

0

k=1


jν g(τ ) e − 1 λ dτ ,

since e jν g(τ ) − 1 is piecewise constant with values e jν gk − 1 (or the value zero if τ does not lie in any (tk−1 ,tk ]). We now compute the correlation,  n  n  E[Y Z] = E ∑ gk (Ntk − Ntk−1 ) ∑ hl (Ntl − Ntl−1 ) k=1

=

l=1

∑ gk hl E[(Ntk − Ntk−1 )(Ntl − Ntl−1 )] + ∑ gk hk E[(Ntk − Ntk−1 )2 ]

k6=l

=

∑ gk hl λ

k6=l

k

2

(tk − tk−1 )(tl − tl−1 ) + ∑ gk hk [λ (tk − tk−1 ) + λ 2 (tk − tk−1 )2 ] k

Chapter 11 Problem Solutions =

∑ gk hl λ 2 (tk − tk−1 )(tl − tl−1 ) + ∑ gk hk λ (tk − tk−1 )

=



k,l

=

193

k

  ∑ gk λ (tk − tk−1 ) ∑ hl λ (tl − tl−1 ) + ∑ gk hk λ (tk − tk−1 ) k

Z ∞

g(τ )λ d τ

Z ∞

l

h(τ )λ d τ +

0

0

Z ∞

k

g(τ )h(τ )λ d τ .

0

Hence, cov(Y, Z) = E[Y Z] − E[Y ]E[Z] =

Z ∞

g(τ )h(τ )λ d τ .

0

18. The key is to use g(τ ) = h(t − τ ) in the preceding problem. It then immediately follows that  Z ∞ Z ∞
jν h(t−τ ) E[Yt ] = h(t − τ )λ d τ , ϕYt (ν ) = exp e − 1 λ dτ , 0

0

and

cov(Yt ,Ys ) =

Z ∞ 0

h(t − τ )h(s − τ )λ d τ .

19. MATLAB. Replace the line X = -log(rand(1))/lambda; % Generate exp(lambda) RV with X = randn(1)ˆ2; % Generate chi-squared RV 20. If Nt is a Poisson process of rate λ , then FX is the exp(λ ) cdf. Hence, P(X1 < Y1 ) =

Z ∞ 0

FX (y) fY (y) dy =

Z ∞ 0

[1 − e−λ y ] fY (y) dy = 1 − MY (−λ ).

If Mt is a Poisson process of rate µ , then MY is the exp(µ ) mgf, and P(X1 < Y1 ) = 1 −

µ µ λ = . = 1− µ − (−λ ) µ +λ λ +µ

21. Since Tn := X1 + · · · + Xn is the sum of i.i.d. random variables, var(Tn ) = n var(X1 ). Since X1 ∼ uniform[0, 1], var(X1 ) = 1/12, and var(Tn ) = n/12. 22. In the case of a Poisson process, Tk is Erlang(k, λ ). Hence,    ∞  ∞ k−1 (λ t)l e−λ t (λ t)l e−λ t = ∑ ∑ ∑ Fk (t) = ∑ 1 − ∑ l! l! k=1 k=1 l=k k=1 l=0    ∞  ∞ ∞ l − l λ t (λ t) e (λ t) e−λ t ∞ = ∑ ∑ I[k,∞) (l) = ∑ ∑ I[k,∞) (l) l! l! k=1 l=0 l=0 k=1 ∞

∞

∞

=

∑l

l=0

(λ t)l e−λ t = E[Nt ] = λ t. l!

194

Chapter 11 Problem Solutions

23. To begin, write  E[Nt |X1 = x] = E ∑ I[0,t] (Tn ) X1 = x = 

∞

n=1

∞

∞





∑ E I[0,t] (X1 + · · · + Xn ) X1 = x

n=1

  = ∑ E I[0,t] (x + X2 + · · · + Xn ) X1 = x .



n=1

(a) If x > t, then x + X2 + · · · + Xn > t, and so I[0,t] (x + X2 + · · · + Xn ) = 0. Thus, E[Nt |X1 = x] = 0 for x > t.

(b) First, for n = 1 and x ≤ t, I[0,t] (x) = 1. Next, for n ≥ 2 and x ≤ t, I[0,t] (x + X2 + · · · + Xn ) = I[0,t−x] (X2 + · · · + Xn ). So, ∞   E[Nt |X1 = x] = 1 + ∑ E I[0,t−x] (X2 + · · · + Xn ) n=2 ∞

  = 1 + ∑ E I[0,t−x] (X1 + · · · + Xn ) n=1

= 1 + E[Nt−x ], where we have used fact that the Xi are i.i.d. (c) By the law of total probability, E[Nt ] = =

=

Z ∞ 0

E[Nt |X1 = x] f (x) dx

0

E[Nt |X1 = x] f (x) dx +

Z t

Z t 0

Z ∞ t

E[Nt |X1 = x] f (x) dx {z } | = 0 for x>t

Z t  1 + E[Nt−x ] f (x) dx = F(t) + E[Nt−x ] f (x) dx. 0

24. With the understanding that m(t) = 0 for t < 0, we can write the renewal equation as m(t) = F(t) +

Z ∞ 0

m(t − x) f (x) dx,

where the last term is a convolution. Hence, taking the Laplace transform of the renewal equation yields M(s) :=

Z ∞

m(t)est dt =

0

Z ∞ 0

F(t)est dt + M(s)MX (s).

Using integration by parts, we have Z ∞ 0

Thus,

F(t)est dt =

Z F(t)est ∞ 1 ∞ − f (t)est dt. s 0 s 0

1 M(s) = − MX (s) + M(s)MX (s), s

Chapter 11 Problem Solutions

195

which we can rearrange as 1 M(s)[1 − MX (s)] = − MX (s), s or

1 MX (s) 1 1 λ λ /(λ − s) λ M(s) = − · = − · = − · = 2. s 1 − MX (s) s 1 − λ /(λ − s) s −s s

It follows that m(t) = λ tu(t).

25. For 0 ≤ s < t < ∞, write   Z t Z s Z t Z s E[Xτ Xθ ] d τ d θ Xθ d θ = E[Vt Vs ] = E Xτ d τ 0 0 0 0    Z s Z t Z s Z t 2 = δ (τ − θ ) d τ d θ RX (τ − θ ) d τ d θ = σ 0

= σ

2

Z s

0

0

0

2

d θ = σ s.

0

26. For 0 ≤ s < t < ∞, write E[Wt Ws ] = E[(Wt −Ws )Ws ] + E[Ws2 ]

= E[(Wt −Ws )(Ws −W0 )] + σ 2 s = E[Wt −Ws ]E[Ws −W0 ] + σ 2 s = 0 · 0 + σ 2 s = σ 2 s.

27. For t2 > t1 , write E[Yt1 Yt2 ] = E[eWt1 eWt2 ] = E[eWt2 −Wt1 e2Wt1 ] = E[eWt2 −Wt1 e2(Wt1 −W0 ) ] = E[eWt2 −Wt1 ]E[e2(Wt1 −W0 ) ] = eσ

2 (t

2 −t1 )/2

e4σ

2t

1 /2

= eσ

2 (t

2 +3t1 )/2

.

28. Since cov(Wti ,Wt j ) = σ 2 min(ti ,t j ), 

t1  t1   cov(X) = σ 2  t1  ..  .

t1 t2 t2 .. .

t1 t2 t3 .. .

··· ··· ··· .. .

t1 t2 t3

 t1 t2   t3  . ..  .  tn

29. Let 0 ≤ t1 < · · · < tn+1 < ∞ and 0 ≤ s1 < · · · < sm+1 < ∞, and suppose that n

g(τ ) =

∑ gi I(ti ,ti+1 ] (τ )

m

and

h(τ ) =

i=1

∑ h j I(s j ,s j+1 ] (τ ).

j=1

Denote the distinct points of {ti } ∪ {s j } in increasing order by θ1 < · · · < θ p . Then p

g(τ ) =

∑ g˜k I(θk ,θk+1 ] (τ )

k=1

p

and

h(τ ) =

∑ h˜ k I(θk ,θk+1 ] (τ ),

k=1

196

Chapter 11 Problem Solutions where the g˜k are taken from the gi , and the h˜ k are taken from the h j . We can now write Z ∞ 0

g(τ ) dWτ +

Z ∞ 0

p

h(τ ) dWτ =

p

∑ g˜k (Wθk+1 −Wθk ) + ∑ h˜ k (Wθk+1 −Wθk )

k=1 p

= =

k=1

∑ [g˜k + h˜ k ](Wθk+1 −Wθk )

k=1 Z ∞ 0

[g(τ ) + h(τ )] dWτ .

30. Following the hint, we first write 2   Z ∞ 2   Z ∞ Z ∞ = E g(τ ) dWτ E g(τ ) dWτ − h(τ ) dWτ 0 0 0 Z ∞  Z ∞ − 2E g(τ ) dWτ h(τ ) dWτ 0

+E = σ2

− 2E

∞

0

0

h(τ ) dWτ

2 

= E

Z

∞

0

∞

0

Z ∞ 0

= σ2

Z ∞ 0

− 2σ

2

31.

g(τ ) dWτ

Z

Z ∞ 0 ∞

0

h(τ )2 d τ  h(τ ) dWτ .

[g(τ ) − h(τ )] dWτ

g(τ )2 d τ + σ 2 Z ∞

Z ∞

h(τ )2 d τ

0

g(τ )h(τ ) d τ .

0

∞

g(τ )h(τ ) d τ .

0

(a) Following the hint, Yt :=

Z t 0

g(τ ) dWτ =

Z ∞

g(τ )I[0,t] (τ ) dWτ ,

2

2

0

it follows that E[Yt2 ]

= σ

2

Z ∞ 0

2 

[g(τ ) − h(τ )]2 d τ

Comparing these two formulas, we see that  Z ∞ Z ∞ Z h(τ ) dWτ = σ 2 E g(τ ) dWτ 0

h(τ ) dWτ

2 

g(τ )2 d τ + σ 2

 Z

= σ2

0

∞

0

Z ∞ 0

Second, we write  Z ∞ Z g(τ ) dWτ − E

0

 Z

[g(τ )I[0,t] (τ )] d τ = σ

Z t 0

g(τ )2 d τ .

Chapter 11 Problem Solutions (b) For t1 ,t2 ≥ 0, write Z E[Yt1 Yt2 ] = E

∞

0

= σ2

g(τ )I[0,t1 ] (τ ) dWτ

Z ∞ 0

= σ2

Z ∞ 0

Z ∞ 0

197

g(τ )I[0,t2 ] (τ ) dWτ

 g(τ )I[0,t1 ] (τ ) g(τ )I[0,t2 ] (τ ) d τ

g(τ )2 I[0,t1 ] (τ )I[0,t2 ] (τ ) d τ = σ 2



Z min(t1 ,t2 )

g(τ )2 d τ .

0

32. By independence of V and the Wiener process along with the result of the previous problem, RY (t1 ,t2 ) = e−λ (t1 +t2 ) E[V 2 ] + σ 2 e−λ (t1 +t2 )

Z min(t1 ,t2 )

e2λ τ d τ .

0

If t1 ≤ t2 , this last integral is equal to Z t1

e2λ τ d τ =

0

1 2 λ t1 (e − 1), 2λ

and so

 σ 2  σ 2 −λ (t2 −t1 ) . E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 − e + 2λ 2λ Similarly, if t2 < t1 ,  σ 2  σ 2 −λ (t1 −t2 ) E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 − . e + 2λ 2λ

In either case, we can write

 σ 2  σ 2 −λ |t1 −t2 | E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 − . e + 2λ 2λ 33. Write E[Yt1 Yt2 ] =

e−λ (t1 +t2 ) e−λ (t1 +t2 ) 2 E[We2λ t1 We2λ t2 ] = · σ min(e2λ t1 , e2λ t2 ). 2λ 2λ

For t1 ≤ t2 , this reduces to

σ 2 −λ (t2 −t1 ) e−λ (t1 +t2 ) 2 2λ t1 . ·σ e = e 2λ 2λ If t2 < t1 , we have

σ 2 −λ (t1 −t2 ) e−λ (t1 +t2 ) 2 2λ t2 ·σ e = e . 2λ 2λ

We conclude that E[Yt1 Yt2 ] =

σ 2 −λ |t1 −t2 | e . 2λ

198

34.

Chapter 11 Problem Solutions

(a) P(t) := E[Yt2 ] = E

Z

t

0

g(τ ) dWτ

2 

Z t

=

g(τ )2 d τ .

0

(b) If g(t) is never zero, then for 0 ≤ t1 < t2 < ∞, P(t2 ) − P(t1 ) =

Z t2

g(τ )2 d τ > 0.

t1

Thus, P(t1 ) < P(t2 ). (c) First, E[Xt ] = E[YP−1 (t) ] = E

Z

P−1 (t)

0

g(τ ) dWτ



= 0

since Wiener integrals have zero mean. Second, Z P−1 (t) 2  Z P−1 (t) 2 E[Xt ] = E g(τ ) dWτ g(τ )2 d τ = P(P−1 (t)) = t. = 0

35.

0

(a) For t > 0, E[Wt2 ] = E[(Wt −W0 )2 ] = t.

(b) For s < 0, E[Ws2 ] = E[(W0 −Ws )2 ] = −s.

(c) From parts (a) and (b) we see that no matter what the sign of t, E[Wt2 ] = |t|. Whether t > s or s < t, we can write |t − s| = E[(Wt −Ws )2 ] = E[Wt2 ] − 2E[Wt Ws ] + E[Ws2 ], and so E[Wt2 ] + E[Ws2 ] − |t − s| |t| + |s| − |t − s| = . 2 2

E[Wt Ws ] = 36.

(a) Write P(X = xk ) = P((X,Y ) ∈ {xk } × IR) = =

Z ∞ −∞

∑ i

Z ∞

−∞

=1

z }| { I{xk } (xi ) IIR (y) fXY (xi , y) dy

 Z ∑ I{xk } (xi ) fXY (xi , y) dy =

∞

−∞

i

fXY (xk , y) dy.

(b) Write
P(Y ∈ C) = P (X,Y ) ∈ IR ×C = =

Z ∞

−∞

∑ i

Z ∞

−∞

IIR (xi )IC (y) fXY (xi , y) dy

   Z  IC (y) ∑ fXY (xi , y) dy = ∑ fXY (xi , y) dy. i

C

i

(c) First write P(Y ∈ C|X = xk ) =

P((X,Y ) ∈ {xk } ×C) P(X = xk ,Y ∈ C) = . P(X = xk ) P(X = xk )

Chapter 11 Problem Solutions Then since P((X,Y ) ∈ {xk } ×C) =

∑ i

Z ∞

−∞

we have

Z

I{xk } (xi )IC (y) fXY (xi , y) dy =

C

P(Y ∈ C|X = xk ) =

199

fXY (xk , y) dy =

pX (xk )

Z  C

Z

C

fXY (xk , y) dy,

 fXY (xk , y) dy. pX (xk )

(d) Write Z ∞

−∞

 I (x )p (x |y) ∑ B i X|Y i fY (y) dy

Z ∞

P(X ∈ B|Y = y) fY (y) dy =

−∞

i

∑ IB (xi )

=

i

∑ IB (xi )

=

i

∑ IB (xi )

=

i

∑

=

i

Z ∞

−∞

Z ∞ fXY (xi , y)

fY (y)

−∞

Z ∞

−∞

Z ∞

−∞

fY (y) dy

fXY (xi , y) dy IIR (y) fXY (xi , y) dy

IB×IR (xi , y) fXY (xi , y) dy

= P((X,Y ) ∈ B × IR) = P(X ∈ B,Y ∈ IR) = P(X ∈ B). 37. The key observations are that since F is nondecreasing, F −1 (U) ≤ x ⇒ U ≤ F(x), and since F −1 is nondecreasing, U ≤ F(x) ⇒ F −1 (U) ≤ x. Hence, {F −1 (U) ≤ x} = {U ≤ F(x)}, and with X := F −1 (U) we can write P(X ≤ x) = P(F −1 (U) ≤ x) = P(U ≤ F(x)) = 38.

Z F(x) 0

(a) The cdf is

F(x) 1 3/4

x/2

1/2 1/4 0

1 2

1

2

x

1 du = F(x).

200

Chapter 11 Problem Solutions (b) For 1/2 ≤ u < 1, Bu = [2u, ∞), and G(u) = 2u. For 1/4 < u < 1/2, Bu = [1, ∞), and G(u) √ = 1. For u = 1/4, √Bu = [1/2, ∞), and G(u) = 1/2. For 0 ≤ u < 1/4, Bu = [ u, ∞), and G(u) = u. Hence,

G(u) 2 2u 1 1/2 0

1 4

1 2

3 4

u

1

39. Suppose G(u) ≤ x. Since F is nondecreasing, F(G(u)) ≤ F(x). By the definition of G(u), F(G(u)) ≥ u. Thus, F(x) ≥ F(G(u)) ≥ u. Now suppose u ≤ F(x). Then by the definition of G(u), G(u) ≤ x. 40. With X := G(U), write P(X ≤ x) = P(G(U) ≤ x) = P(U ≤ F(x)) =

Z F(x)

1 du = F(x).

0

41. To compute G(u), we used the M ATLAB function function x = G(u) i1 = find(u <= .25); i2 = find(.25 < u & u <= .5); i3 = find(.5 < u & u <= 1); x(i1) = sqrt(u(i1)); x(i2) = 1; x(i3) = 2*u(i3);

We obtained the histogram

3 2 1 0 0

0.5

1

2

This is explained by noting that the density corresponding to F(x) is impulsive:

Chapter 11 Problem Solutions

201

f (x) 1 3/4 1/2 1/4 2

x

0

1 2

∑

IB ( jm , . . . , jn )pm,n ( jm , . . . , jn ).

1

42. For an integer-valued process,

µm,n (B) =

jm ,..., jn

If B = {im } × · · · × {in }, then

µm,n+1 (B × IR) = =

∑

IB×IR ( jm , . . . , jn , j)pm,n+1 ( jm , . . . , jn , j)

jm ,..., jn , j

∑ pm,n+1 (im , . . . , in , j), j

µm−1,n (IR × B) = =

∑

IIR×B ( j, jm , . . . , jn )pm−1,n ( j, jm , . . . , jn )

j, jm ,..., jn

∑ pm−1,n ( j, im , . . . , in ), j

and µm,n (B) = pm,n (im , . . . , in ). Hence, if the conditions

µm,n+1 (B × IR) = µm,n (B) and

µm−1,n (IR × B) = µm,n (B)

hold for all B and we take B = {im } × · · · × {in }, then we obtain

∑ pm,n+1 (im , . . . , in , j)

= pm,n (im , . . . , in )

∑ pm−1,n ( j, im , . . . , in )

= pm,n (im , . . . , in ).

j

and j

Conversely, suppose these two formulas hold. Fix an arbitrary B ⊂ IRn−m+1 . If we multiply both formulass by IB (im , . . . , in ) and sum over all im , . . . , in , we obtain the original consistency conditions. 43. For the first condition, write

∑ pm,n+1 (im , . . . , in , j) j

=

∑ q(im )r(im+1 |im ) · · · r(in |in−1 )r( j|in ) j

= q(im )r(im+1 |im ) · · · r(in |in−1 ) | {z } = pm,n (im ,...,in )

∑ r( j|in ) . j

| {z } =1

202

Chapter 11 Problem Solutions For the second condition, write

∑ pm−1,n ( j, im , . . . , in )

=

∑ q( j)r(im | j)r(im+1 |im ) · · · r(in |in−1 )

=



j

j

|

|

44. If

 ∑ q( j)r(im | j) r(im+1 |im ) · · · r(in |in−1 ) . j

{z

= q(im )

}

{z

= pm,n (im ,...,in )

}

µn (Bn ) = µn+1 (Bn × IR) = =

Z ∞

−∞

Z

···

··· Bn

Z

Z ∞

IBn (x1 , . . . , xn )IIR (y) fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1 −∞ Z ∞  −∞

fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1 ,

then necessarily the quantity in square brackets is the joint density fn (x1 , . . . , xn ). Conversely, if the quantity in square brackets is equal to fn (x1 , . . . , xn ), we can write

µn+1 (Bn × IR) = = =

Z ∞

−∞

Z

Z

···

··· Bn

··· Bn

Z ∞

−∞

IBn (x1 , . . . , xn )IIR (y) fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1  fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1

Z Z ∞

Z

−∞

fn (x1 , . . . , xn )dxn · · · dx1 = µn (Bn ).

45. The key is to write

µt1 ,...,tn+1 (Bn,k ) = =

Z ∞

···

−∞

Z

Z ∞

··· Bn

IBn (x1 , . . . , xk−1 , xk+1 , . . . , xn )IIR (xk ) ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxn+1 · · · dx1  ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxk dxn+1 · · · dxk+1 dxk−1 · · · dx1 .

−∞

Z Z ∞

−∞

Then if µt1 ,...,tn+1 (Bn,k ) = µn (Bn ), we must have ft1 ,...,tk−1 ,tk+1 ,...,tn (x1 , . . . , xk−1 , xk+1 , . . . , xn ) =

Z ∞

−∞

ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxk

and conversely. 46. By the hint, [Wt1 , . . . ,Wtn ]0 is a linear transformation of a vector of independent Gaussian increments, which is a Gaussian vector. Hence, [Wt1 , . . . ,Wtn ]0 is a Gaussian vector. Since n and the times t1 < · · · < tn are arbitrary, Wt is a Gaussian process.

Chapter 11 Problem Solutions

203

47. Let X = [Wt1 −W0 , . . . ,Wtn −Wtn−1 ]0 , and let Y = [Wt1 , . . . ,Wtn ]0 . Then Y = AX, where A denotes the matrix given in the statement of Problem 46. Since the components of X are independent, 2 n e−xi /[2(ti −ti−1 )] , fX (x) = ∏ p 2π (ti − ti−1 ) i=1 where it is understood that t0 := 0. Using the example suggested in the hint, we have fX (x) . fY (y) = | det A| x=A−1 y Fortunately, det A = 1, and by definition, Xi = Wti −Wti−1 . Hence, 2

e−(wi −wi−1 ) /[2(ti −ti−1 )] ∏ p2π (t − t ) , i i−1 i=1 n

ft1 ,...,tn (w1 , . . . , wn ) = where it is understood that w0 := 0.

48. By the hint, it suffices to show that C is positive semidefinite. Write a0Ca =

n

n

∑ ∑ ai akCik

n

=

i=1 k=1

n

∑ ∑ ai ak R(ti − tk )

i=1 k=1 n n

=

∑ ai ak

∑

=

Z ∞

i=1 k=1

−∞

Z ∞

−∞

S( f )e j2π f (ti −tk ) d f

n 2 S( f ) ∑ ai e j2π f ti d f ≥ 0. i=1

CHAPTER 12

Problem Solutions 1. If P(A|X = i,Y = j, Z = k) = h(i), then P(A, X = i,Y = j, Z = k) = P(A|X = i,Y = j, Z = k)P(X = i,Y = j, Z = k) = h(i)P(X = i,Y = j, Z = k). Summing over j yields P(A, X = i, Z = k) = h(i)P(X = i, Z = k),

(∗)

and thus h(i) = P(A|X = i, Z = k). Next, summing (∗) over k yields P(A, X = i) = h(i)P(X = i), and then h(i) = P(A|X = i) as well. 2. Write X1 = g(X0 , Z1 ) X2 = g(X1 , Z2 ) = g(g(X0 , Z1 ), Z2 ) X3 = g(X2 , Z3 ) = g(g(g(X0 , Z1 ), Z2 ), Z3 ) .. . In general, Xn is a function of X0 , Z1 , . . . , Zn , and thus (X0 , . . . , Xn ) is a function of (X0 , Z1 , . . . , Zn ), which is independent of Zn+1 . Now observe that P(Xn+1 = in+1 |Xn = in , . . . , X0 = i0 ) = P(g(Xn , Zn+1 ) = in+1 |Xn = in , . . . , X0 = i0 ) = P(g(in , Zn+1 ) = in+1 |Xn = in , . . . , X0 = i0 ) = P(g(in , Zn+1 ) = in+1 ),

where we have used substitution and independence. Since P(Xn+1 = in+1 |Xn = in , . . . , X0 = i0 ) does not depend on in−1 , . . . , i0 , Xn is a Markov chain. 3. Write P(A ∩ B|C) =

P(A ∩ B ∩C) P(A ∩ B ∩C) P(B ∩C) = · = P(A|B ∩C)P(B|C). P(C) P(B ∩C) P(C) 204

Chapter 12 Problem Solutions

205

4. Write P(X0 = i, X1 = j, X2 = k, X3 = l) = P(X3 = l|X2 = k, X1 = j, X0 = i) · P(X2 = k|X1 = j, X0 = i)P(X1 = j|X0 = i)P(X0 = i) = pkl p jk pi j νi .

5. The first equation we write is

π0 =

∑ πk pk0

π0 = π0 p00 + π1 p10 = π0 (1 − a) + π1 b.

as

k

This tells us that π1 = (a/b)π0 . Then we use the fact that π0 + π1 = 1 to write π0 + (a/b)π0 = 1, or π0 = 1/(1 + a/b) = b/(a + b). We also have π1 = (a/b)π0 = a/(a + b). 6. The state transition diagram is

1/2

0 1/2

1/2 1/4 1/2

1

2 3/4

The stationary distribution is π0 = 5/12, π1 = 1/3, π2 = 1/4. 7. The state transition diagram is

0 1/2

1/4

1/4 1/2 1 3/4

2 3/4

The stationary distribution is π0 = 1/5, π1 = 7/10, π2 = 1/10.

206

Chapter 12 Problem Solutions

8. The state transition diagram is

1/2 1/2 0

1

9/10

1/10

1/10 1/2 3

2 9/10

1/2 The stationary distribution is π0 = 9/28, π1 = 5/28, π2 = 5/28, π3 = 9/28. 9. The first equation is

π0 = π0 p00 + π1 p10 = π0 (1 − a) + π1 b, which tells us that π1 = (a/b)π0 . The next equation is

π1 = π0 p01 + π1 p11 + π2 p21 = π0 a + π1 (1 − [a + b]) + π2 b, which tells us that (a + b)π1 = π0 a + π2 b or

(a + b)(a/b)π0 = π0 a + π2 b,

from which it follows that π2 = (a/b)2 π0 . Now suppose that πi = (a/b)i π0 holds for i = 0, . . . , j < N. Then from

π j = π j−1 p j−1, j + π j p j j + π j+1 p j+1, j = π j−1 a + π j (1 − [a + b]) + π j+1 b, we obtain (a + b)π j = π j−1 a + π j+1 b

or

(a + b)(a/b) j π0 = (a/b) j−1 a + π j+1 b,

from which it follows that π j+1 = (a/b) j+1 π0 . To find π0 , we use the fact that π0 + · · · + πN = 1. Using the finite geometric series formula, we have N

1 =

∑ (a/b) j π0

= π0

j=0

Hence,

πj =

1 − a/b  a  j , 1 − (a/b)N+1 b

1 − (a/b)N+1 . 1 − a/b

j = 0, . . . , N,

a 6= b.

If a = b, then π j = π0 for j = 0, . . . , N implies π j = 1/(N + 1).

Chapter 12 Problem Solutions

207

10. Let π and πb denote the two solutions in the example. For 0 ≤ λ ≤ 1, put

πej := λ π j + (1 − λ )πbj ≥ 0.

Then 

∑ πej = ∑ λ π j + (1 − λ )πbj j

j



= λ ∑ π j + (1 − λ ) ∑ πbj = λ · 1 + (1 − λ ) · 1 = 1.

Hence, πe is a valid pmf. Finally, note that

∑ πei pi j i

=

j

j

∑[λ πi + (1 − λ )πbi ]pi j i

= λ π j + (1 − λ )πbj = πej .

= λ ∑ πi pi j + (1 − λ ) ∑ πbi pi j i

i

11. MATLAB. OMITTED. 12. Write

∞

E[T1 ( j)|X0 = i] =

∑ kP(T1 ( j) = k|X0 = i) + ∞ · P(T1 ( j) = ∞|X0 = i)

k=1 ∞

=

k=1

13.

(k)

∑ k fi j

+ ∞(1 − fi j ).

(a) First note that {T1 = 2} = {X2 = j, X1 6= j} and that h {T2 = 5} = {X5 = j} ∩ {X4 = j, X3 6= j, X2 6= j, X1 6= j} ∪{X4 6= j, X3 = j, X2 6= j, X1 6= j} ∪{X4 6= j, X3 6= j, X2 = j, X1 = 6 j} i ∪{X4 6= j, X3 6= j, X2 6= j, X1 = j} . Then

{T2 = 5} ∩ {T1 = 2} = {X5 = j, X4 6= j, X3 6= j, X2 = j, X1 6= j}. It now follows that P(T2 = 5|T1 = 2, X0 = i) = P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 6= j, X0 = i). (b) Write P(X5 = j, X4 6= j, X3 6= j, X2 = j, X1 6= j, X0 = i) as

  [ {X5 = j, X4 6= j, X3 6= j, X2 = j, X1 = l, X0 = i} , P l6= j

which is equal to

∑ P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 = l, X0 = i)P(X2 = j, X1 = l, X0 = i).

l6= j

208

Chapter 12 Problem Solutions By the Markov property, this simplifies to

∑ P(X5 = j, X4 6= j, X3 6= j|X2 = j)P(X2 = j, X1 = l, X0 = i),

l6= j

which is just P(X5 = j, X4 6= j, X3 6= j|X2 = j)P(X2 = j, X1 6= j, X0 = i). It now follows that P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 6= j, X0 = i) is equal to P(X5 = j, X4 6= j, X3 6= j|X2 = j). (c) Write P(X5 = j, X4 6= j, X3 6= j|X2 = j) =

∑

P(X5 = j, X4 = k, X3 = l|X2 = j)

∑

P(X3 = j, X2 = k, X1 = l|X0 = j)

k6= j,l6= j

=

k6= j,l6= j

= P(X3 = j, X2 6= j, X1 6= j|X0 = j). 14. Write

 ∞  \ P(V ( j) = ∞|X0 = i) = P {V ( j) ≥ L} X0 = i L=1

 M  \ {V ( j) ≥ L} X0 = i , = lim P M→∞

= lim P(V ( j) ≥ M|X0 = i),

decreasing events.

M→∞

15. First write

 E[V ( j)|X0 = i] = E ∑ I{ j} (Xn ) X0 = i = 

∞

n=1

∑ P(Xn = j|X0 = i)

n=1

∞

∑ E[I{ j} (Xn )|X0 = i]

n=1

∞

∞

=

limit property of P,

L=1

=

(n)

∑ pi j

.

(∗)

n=1

Next, P(V ( j) = ∞|X0 = i) = lim P(V ( j) ≥ L|X0 = i), L→∞

= lim fi j ( f j j )L−1 , L→∞

from preceding problem,

from the text.

Now, if j is recurrent ( f j j = 1), the above limit is fi j . If fi j > 0, then V ( j) = ∞ with positive probability, and so the expectation on the left in (∗) must be infinite, and hence, so is the sum on the far right in (∗). If fi j = 0, then for any n = 1, 2, . . . , we can write  ∞  [ 0 = fi j := P(T1 ( j) < ∞|Xi = 0) ≥ P {Xm = j} Xi = 0 m=1

(n)

≥ P(Xn = j|X0 = i) = pi j ,

Chapter 12 Problem Solutions

209

(n)

and it follows that ∑∞ n=1 pi j = 0. In the transient case ( f j j < 1), we have from the text that P(V ( j) ≥ L|X0 = i) = fi j ( f j j )L−1 . Using the preceding problem along with f j j < 1, we have P(V ( j) = ∞|X0 = i) = lim fi j ( f j j )L−1 = 0. L→∞

Now, we also have from the text that P(V ( j) = L|X0 = i) =



fi j ( f j j )L−1 (1 − f j j ), L ≥ 1, 1 − fi j , L = 0,

where we use the fact that the number of visits to state j is zero if and only if we never visit state j, and this happens if and only if T1 ( j) = ∞. We can therefore write ∞

∑ L · P(V ( j) = L|X0 = i)

E[V ( j)|X0 = i] =

=

L=0

fi j < ∞, 1− fjj

and it follows that the last sum in (∗) is finite too. 16. From mλ − 1 ≤ bmλ c ≤ mλ , 1 1 1 ≤ . ≤ mλ bmλ c mλ − 1 Then

17.

1 m 1 1 ≤ ≤ → . λ bmλ c λ − 1/m λ

(a) Write h( j) = IS ( j) = ∑nl=1 I{sl } ( j). Then   1 m n 1 m lim lim ∑ h(Xk ) = m→∞ ∑ ∑ I{sl } (Xk ) = m→∞ m m k=1 l=1 k=1 n

=

lim Vm (sl )/m ∑ m→∞

∑ IS ( j)π j j

=

1

m

l=1

k=1

n

=

l=1

=

n

lim I (Xk ) ∑ m→∞ m ∑ {sl }

∑ πs l ,

by Theorems 3 and 4,

l=1

∑ h( j)π j . j

(b) With h( j) = ∑nl=1 cl ISl ( j), where each Sl is a finite subset of states, we can write   n 1 m n 1 m 1 m cl ISl (Xk ) = ∑ cl lim h(Xk ) = lim lim ∑ ∑ ∑ ∑ ISl (Xk ) m→∞ m m→∞ m m→∞ m k=1 k=1 l=1 k=1 l=1   n = ∑ cl ∑ ISl ( j)π j , by part (a) l=1

=



∑ j

j

 c I ( j) ∑ l Sl π j = n

l=1

∑ h( j)π j . j

210

Chapter 12 Problem Solutions (c) If h( j) = 0 for all but finitely many states, then there are at most finitely many distinct nonzero values that h( j) can take, say c1 , . . . , cn . Put Sl := { j : h( j) = cl }. By the assumption about h, each Sl is a finite set. Furthermore, we can write n

∑ cl ISl ( j).

h( j) =

l=1

Hence, the desired result is immediate by part (b). 18. MATLAB. OMITTED. 19. MATLAB. OMITTED. 20. Suppose k ∈ Ai ∩ A j . Since k ∈ Ai , i ↔ k. Since k ∈ A j , j ↔ k. Hence, i ↔ j. Now, if l ∈ Ai , l ↔ i ↔ j and we see that l ∈ A j . Similarly, if l ∈ A j , l ↔ j ↔ i and we see that l ∈ Ai . Thus, Ai = A j . 21. Yes. As pointed out in the discussion at the end of the section, by combining Theorem 8 and Theorem 6, the chain has a unique stationary distribution. Now, by Theorem 4, all states are positive recurrent, which by definition means that the expected time to return to a state is finite. 22. Write (λ ∆t)n e−λ ∆t λ = lim (λ ∆t)n−1 e−λ ∆t = 0, ∆t↓0 ∆t · n! n! ∆t↓0

gi,i+n = lim

for n ≥ 2.

23. The state transition diagram is

0 1

5 2 4

1

2 3

The stationary distribution is π0 = 11/15, π1 = 1/5, π2 = 1/15. 24. MATLAB. Change the line

A=P-In;

to

A=P;.

25. The forward equation is j+1

p0i j (t) =

∑ pik (t)gk j k

=

∑

pik gk j

k= j−1

= pi, j−1 (t)λ j−1 − pi j (t)[λ j + µ j ] + pi, j+1 (t)µ j+1 .

Chapter 12 Problem Solutions

211

The backward equation is p0i j (t) =

∑ gik pk j (t)

i+1

=

k

∑

gik pi j (t)

k=i−1

= µi pi−1, j (t) − (λi + µi )pii (t) + λi pi+1, j (t). The chain is conservative because gi,i−1 +gi,i+1 = µi + λi = −[−(λi + µi )] = −gii < ∞. 26. We use the formula 0 = ∑k πk gk j . For j = 0, we have 0 =

∑ πk gk0

= π0 (−λ0 ) + π1 µ1 ,

k

which implies π1 = (λ0 /µ1 )π0 . For j = 1, we have 0 =

∑ πk gk1

= π0 λ0 + π1 [−(λ1 + µ1 )] + π2 µ2

k

= π0 λ0 + (λ0 /µ1 )π0 [−(λ1 + µ1 )] + π2 µ2 = π0 (λ0 λ1 /µ1 ) + π2 µ2 , which implies π2 = π0 λ0 λ1 /(µ1 µ2 ). Now suppose that πi = π0 λ0 · · · λi−1 /(µ1 · · · µi ) for i = 1, . . . , j. Then from 0 =

∑ πk gk j k

=

= π j−1 λ j−1 − π j (λ j + µ j ) + π j+1 µ j+1

λ0 · · · λ j−2 λ0 · · · λ j−1 π0 λ j−1 − π0 (λ j + µ j ) + π j+1 µ j+1 , µ1 · · · µ j−1 µ1 · · · µ j

and it follows that

π j+1 = Now, let

λ0 · · · λ j π0 . µ1 · · · µ j+1

∞

B :=

λ0 · · · λ j−1 < ∞. j=1 µ1 · · · µ j

∑

From the condition ∑∞j=0 π j = 1, we have

∞

λ0 · · · λ j−1 = π0 (1 + B). j=1 µ1 · · · µ j

1 = π 0 + π 1 + · · · = π0 + π 0 ∑

It then follows that π0 = 1/(1 + B), and   λ0 · · · λ j−1 1 πj = , 1 + B µ1 · · · µ j

j ≥ 1.

If λi = λ and µi = µ , then B = ∑∞j=1 (λ /µ ) j , which is finite if and only if λ < µ . In in this case, ∞  k 1 λ 1+B = ∑ = , µ 1 + λ /µ k=0 and

π j = (1 − λ /µ )(λ /µ ) j ∼ geometric0 (λ /µ ).

212

Chapter 12 Problem Solutions

27. The solution is very similar the that of the preceding problem. In this case, put N

λ0 · · · λ j−1 < ∞, j=1 µ1 · · · µ j

∑

BN := and then π0 = 1/(1 + BN ), and 1 πj = 1 + BN



 λ0 · · · λ j−1 , µ1 · · · µ j

j = 1, . . . , N.

If λi = λ and µi = µ , then N

∑ (λ /µ )k

1 + BN =

=

k=0

and

1 − (λ /µ )N+1 , 1 − λ /µ

1 − λ /µ (λ /µ ) j , 1 − (λ /µ )N+1

πj =

j = 0, . . . , N.

If λ = µ , then π j = 1/(N + 1). 28. To begin, write

mi (t) := E[Xt |X0 = i] =

∑ jP(Xt = j|X0 = i)

=

j

∑ j pi j (t). j

Then m0i (t) =

∑ j p0i j (t) = j

∑j j



∑ pik (t)gk j k



=

  p (t) jg ik k j ∑ ∑ k

h i = ∑ pik (t) (k − 1)gk,k−1 + kgkk + (k + 1)gk,k+1 k

=

h i (k − 1)(k µ ) − k(kλ + α + k µ ) + (k + 1)(kλ + α )

∑ pik (t) k

=

j

∑ pik (t)[−kµ + kλ + α ] k

= −µ mi (t) + λ mi (t) + α .

We must solve m0i (t) + (µ − λ )mi (t) = α , If µ 6= λ , it is readily verified that  mi (t) = i −

mi (0) = i.

 α α e−(µ −λ )t + µ −λ µ −λ

solves the equation. If µ = λ , then mi (t) = α t + i solves m0i (t) = α with mi (0) = i. 29. To begin, write p0i j (t) =

∑ pik (t)gk j k

= pi, j−1 (t)g j−1, j + pi j (t)g j j = λ pi, j−1 (t) − λ pi j (t). (∗)

Chapter 12 Problem Solutions

213

Observe that for j = i, pi,i−1 (t) = 0 since the chain can not go from state i to a lower state. Thus, p0ii (t) = −λ pii (t). Also, pii (0) = P(Xt = i|X0 = i) = 1. Thus, pii (t) = e−λ t , and it is easily verified that (λ t)n e−λ t , n!

pi,i+n (t) =

n = 0, 1, 2, . . . ,

solves (∗). Thus, Xt is a Poisson process of rate λ with X0 = i instead of X0 = 0. 30.

(a) The assumption that Dˆ > −∞ implies that the πk are not all zero, in which case, the πk would not sum to one and would not be a pmf. Aside from this, it is clear ˆ Dˆ = 1. Next, since p j j = 0, write that πk ≥ 0 for all k and that ∑k πk = D/

∑ πk pk j k

−1 ∑ πˆk gk j Dˆ k6= j k6= j    −1 ˆ = π j. ˆ ˆ = π π − g g j j j = πˆ j g j j /D ∑ k kj Dˆ k | {z }

=

gk j

ˆ ∑ [πˆk gkk /D] −gkk

=

=0

(b) The assumption that Dˇ > −∞ implies that the πk are not all zero. Aside from ˇ Dˇ = 1. Next, write this, it is clear that πk ≥ 0 for all k and that ∑k πk = D/

∑ πk gk j k

=

gk j −1 πˇk ∑ ˇ −g D k kk   − πˇ j

ˇ kj ∑[(πˇk /gkk )/D]g k

−1 = Dˇ −1 = Dˇ = 0.



gk j

∑ πˇk −gkk

k6= j



∑ πˇk pk j

|

k

{z

= πˇ j



}

=

 −πˇ j ,

since p j j = 0,

(c) Since πˆk and πˇk are pmfs, they sum to one. Hence, if gii = g for all i, Dˆ = g and Dˇ = 1/g. In (a), πk = πˆk gkk /Dˆ = πˆk g/g = πˆk . In (b), πk = (πˇk /gkk )/Dˇ = (πˇk /g)/(1/g) = πˇk . 31. Following the hints, write P(T > t + ∆t|T > t, X0 = i) = P(Xs = i, 0 ≤ s ≤ t + ∆t|Xs = i, 0 ≤ s ≤ t) = P(Xs = i,t ≤ s ≤ t + ∆t|Xs = i, 0 ≤ s ≤ t) = P(Xs = i,t ≤ s ≤ t + ∆t|Xt = i) = P(Xs = i, 0 ≤ s ≤ ∆t|X0 = i) = P(T > ∆t|X0 = i).

214

Chapter 12 Problem Solutions The exponential parameter is lim

∆t↓0

1 − P(T > ∆t|X0 = i) 1 − P(Xs = i, 0 ≤ s ≤ ∆t|X0 = i) = lim ∆t↓0 ∆t ∆t 1 − P(X∆t = i|X0 = i) = lim =: −gii . ∆t↓0 ∆t

32. Using substitution in reverse, write P(Wt ≤ y|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ) = P(Wt − x ≤ y − x|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ) = P(Wt −Ws ≤ y − x|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ). Now, using the fact that W0 ≡ 0, this last conditional probability is equal to P(Wt −Ws ≤ y − x|Ws −Wsn−1 = x − xn−1 , . . . ,Ws1 −Ws0 = x1 − x0 ,Ws0 −W0 = x0 ). Since the Wiener process has independent increments that are Gaussian, this last expression reduces to √ Z y−x exp[−{θ /[σ t − s ]}2 /2] p dθ . P(Wt −Ws ≤ y − x) = −∞ 2πσ 2 (t − s)

Since this depends on x but not on xn−1 , . . . , x0 ,

P(Wt ≤ y|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ) = P(Wt ≤ y|Ws = x). Hence, Wt is a Markov process. 33. Write

∑ P(X = x|Y = y, Z = z)P(Y = y|Z = z) y

=

∑ y

P(X = x,Y = y, Z = z) P(Y = y, Z = z) · P(Y = y, Z = z) P(Z = z)

=

1 P(X = x,Y = y, Z = z) P(Z = z) ∑ y

=

P(X = x, Z = z) = P(X = x|Z = z). P(Z = z)

34. Using the law of total conditional probability, write Pt+s (B) = P(Xt+s ∈ B|X0 = x) = = = =

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

Z ∞

−∞

P(Xt+s ∈ B|Xs = z, X0 = x) fXs |X0 (z|x) dz P(Xt+s ∈ B|Xs = z) fs (x, z) dz P(Xt ∈ B|X0 = z) fs (x, z) dz Pt (z, B) fs (x, z) dz.

CHAPTER 13

Problem Solutions 1. First observe that E[|Xn | p ] = n p/2 P(U ≤ 1/n) = n p/2 · (1/n) = n p/2−1 , which goes to zero if and only if p < 2. Thus, 1 ≤ p < 2. 2   Nt var(Nt ) λt λ 1 = 2 = → 0. 2. E − λ = 2 E[|Nt − λ t|2 ] = t t t2 t t

3. Given ε > 0, let n ≥ N imply |C(k)| ≤ ε /2. Then for such n,

1 n−1 1 N−1 1 n C(k) = C(k) + ∑ ∑ ∑ C(k) n k=0 n k=0 n k=N and

n−1 N−1 n N−1 1 ∑ C(k) ≤ 1 ∑ |C(k)| + 1 ∑ ε /2 ≤ 1 ∑ |C(k)| + ε /2. n n k=0 n k=N n k=0 k=0

For large enough n, the right-hand side will be less that ε .

4. Applying the hint followed by the Cauchy–Schwarz inequality shows that n−1 q 1 ∑ C(k) ≤ E[(X1 − m)2 ]E[(Mn − m)2 ]. n k=0

Hence, if Mn converges in mean square to m, the left-hand side goes to zero as n → ∞. 5. Starting with the hint, we write E[ZIA ] = E[(Z − Zn )IA ] + E[Zn IA ] ≤ E[Z − Zn ] + nE[IA ], since Zn ≤ Z and Zn ≤ n, = E[|Z − Zn |] + nP(A).

Since Zn converges in mean to Z, given ε > 0, we can let n ≥ N imply E[|Z − Zn |] ≤ ε /2. Then in particular we have E[ZIA ] < ε /2 + NP(A). Hence, if 0 < δ < ε /(2N), P(A) < δ

implies

E[ZIA ] < ε /2 + N ε /(2N) = ε .

6. Following the hint, we find that given ε > 0, there exists a δ > 0 such that P(U ≤ ∆x) = ∆x < δ

implies

E[ f (x +U)I{U≤∆x} ] = 215

Z ∆x 0

f (x + t) dt < ε .

216

Chapter 13 Problem Solutions Now make the change of variable θ = x + t to get Z ∆x

f (x + t) dt =

Z x+∆x x

0

f (θ ) d θ = F(x + ∆x) − F(x).

For 0 < −∆x < δ , take A = {U > 1 + ∆x} and Z = f (x − 1 + U). Then P(U > 1 + ∆x) = −∆x implies E[ f (x + 1 +U)I{U>1+Dx} ] =

Z 1

1+∆x

f (x − 1 + t) dt < ε .

In this last integral make the change of variable θ = x − 1 + t to get Z x

x+∆x

f (θ ) d θ = F(x) − F(x + ∆x).

Thus, F(x + ∆x) − F(x) > −ε . We can now write that given ε > 0, there exists a δ such that |∆x| < δ implies |F(x + ∆x) − F(x)| < ε . 7. Following the hint, we can write h1 i p q 1 1 1 exp ln(|X|/α ) p + ln(|Y |/β )q ≤ eln(|X|/α ) + eln(|Y |/β ) p q p q or

    i h 1 |X| p 1 |Y | q exp ln(|X|/α ) + ln(|Y |/β ) ≤ + p α q β

or

|XY | 1 |X| p 1 |Y |q ≤ + . αβ p αp q βq

Hence,

1 αp 1 βq 1 E[|X| p ] 1 E[|Y |q ] E[|XY |] + = + = 1. ≤ αβ p αp q βq p αp q βq

The hint assumes neither α nor β is zero or infinity. However, if either α or β is zero, then both sides of the inequality are zero. If neither is zero and one of them is infinity, then the right-hand side is infinity and the inequality is trivial. 8. Following the hint, with X = |Z|α , Y = 1, and p = β /α , we have E[|Z|α ] = E[|XY |] ≤ E[|X| p ]1/p E[|Y |q ]1/q = E[(|Z|α )β /α ]α /β · 1 = E[|Z|β ]α /β . Raising both sides to the 1/α yields E[|Z|α ]1/α ≤ E[|Z|β ]1/β .

9. By Lyapunov’s inequality, E[|Xn − X|α ]1/α ≤ E[|Xn − X|β ]1/β . Raising both sides to the α power yields E[|Xn − X|α ] ≤ E[|Xn − X|β ]α /β . Hence, if E[|Xn − X|β ] → 0, then E[|Xn − X|α ] → 0 too. 10. Following the hint, write E[|X +Y | p ] = E[|X +Y | |X +Y | p−1 ] ≤ E[|X| |X +Y | p−1 ] + E[|Y | |X +Y | p−1 ]

≤ E[|X| p ]1/p E[(|X +Y | p−1 )q ]1/q + E[|Y | p ]1/p E[(|X +Y | p−1 )q ]1/q ,

Chapter 13 Problem Solutions

217

where 1/q := 1 − 1/p. Hence, 1/q = (p − 1)/p and q = p/(p − 1). Now divide the above inequality by E[|X +Y | p ](p−1)/p to get E[|X +Y | p ]1/p ≤ E[|X| p ]1/p + E[|Y | p ]1/p . 11. For a Wiener process, Wt −Wt0 ∼ N(0, σ 2 |t −t0 |). To simplify the notation, put σ˜ 2 := σ 2 |t − t0 |. Then Z ∞

2

Z

2

∞ e−(x/σ˜ ) /2 e−(x/σ˜ ) /2 |x| √ x √ dx = 2 dx −∞ 0 2π σ˜ 2π σ˜ r Z ∞ 2σ˜  −t 2 /2  ∞ 2σ˜ |t − t0 | −t 2 /2 dt = √ = √ −e , te = 2σ 2π 0 0 2π 2π

E[|Wt −Wt0 |] =

which goes to zero as |t − t0 | → 0.

12. Let t0 > 0 be arbitrary. Since E[|Nt − Nt0 |2 ] = λ |t − t0 | + λ 2 |t − t0 |2 , it is clear that as t → t0 , E[|Nt − Nt0 |2 ] → 0. Hence, Nt is continuous in mean square. 13. First write R(t, s) − R(τ , θ ) = R(t, s) − R(t, θ ) + R(t, θ ) − R(τ , θ ) = E[Xt (Xs − Xθ )] + E[(Xt − Xτ )Xθ ].

Then by the Cauchy–Schwarz inequality, q q |R(t, s) − R(τ , θ )| ≤ E[Xt2 ]E[(Xs − Xθ )2 ] + E[(Xt − Xτ )2 ]E[Xθ2 ],

which goes to zero as (t, s) → (τ , θ ). Note that we need the boundedness of E[Xt2 ] for t near τ . 14.

(a) E[|Xt+T − Xt |2 ] = R(0) − 2R(T ) + R(0) = 0.

(b) Write

q |R(t + T ) − R(t)| = E[(Xt+T − Xt )X0 ] ≤ E[|Xt+T − Xt |2 ]E[X02 ],

which is zero by part (a). Hence, R(t + T ) = R(t) for all t. 15. Write

k(Xn +Yn ) − (X +Y )k p = k(Xn − X) + (Yn −Y )k p ≤ kXn − Xk p + kYn −Y k p → 0. 16. Let t0 be arbitrary Since kZt − Zt0 k p = k(Xt +Yt ) − (Xt0 +Yt0 )k p = k(Xt − Xt0 ) + (Yt −Yt0 )k p

≤ kXt − Xt0 k p + kYt −Yt0 k p ,

it is clear that if Xt and Yt are continuous in mean of order p, then so is Zt := Xt +Yt .

218

Chapter 13 Problem Solutions

17. Let ε > 0 be given. Since kXn − Xk p → 0, there is an N such that for n ≥ N, kXn − Xk p < ε /2. Thus, for n, m ≥ N, kXn − Xm k p = k(Xn − X) + (X − Xm )k p ≤ kXn − Xk p + kX − Xm k p < ε /2 + ε /2 = ε . 18. Suppose Xn is Cauchy in L p . With ε = 1, there is an N such that for all n, m ≥ N, kXn − Xm k p < 1. In particular, with m = N we have from kXn k p − kXN k p ≤ kXn − XN k p that

kXn k p ≤ kXn − XN k p + kXN k p < 1 + kXN k p ,

To get a bound that also holds for n = 1, . . . , N − 1, write

for n ≥ N.

kXn k p ≤ max(kX1 k p , . . . , kXN−1 k p , 1 + kXN k p ). 19. Since Xn converges, it is Cauchy and therefore bounded by the preceding two problems. Hence, we can write kXn k p ≤ B < ∞ for some constant B. Given ε > 0, let n ≥ N imply kXn − Xk p < ε /(2kY kq ) and kYn −Y kq < ε /(2B). Then kXnYn − XY k1 = E[|XnYn − XnY + XnY − XY |] ≤ E[|Xn (Yn −Y )|] + E[|(Xn − X)Y |] ≤ kXn k p kYn −Y kq + kXn − Xk p kY kq , ε ε + kY kq = ε . < B· 2B 2kY kq

by H¨older’s inequality,

20. If Xn converges in mean of order p to both X and Y , write kX −Y k p = k(X − Xn ) + (Xn −Y )k p ≤ kX − Xn k p + kXn −Y k p → 0. Since kX −Y k p = 0, E[|X −Y | p ] = 0. 21. For the rightmost inequality, observe that kX −Y k p = kX + (−Y )k p ≤ kXk p + k −Y k p = kXk p + kY k p . For the remaining inequality, first write kXk p = k(X −Y ) +Y k p ≤ kX −Y k p + kY k p , from which it follows that kXk p − kY k p ≤ kX −Y k p .

(∗)

Similarly, from kY k p = k(Y − X) + Xk p ≤ kY − Xk p + kXk p , it follows that kY k p − kXk p ≤ kY − Xk p = kX −Y k p .

From (∗) and (∗∗) it follows that kXk p − kY k p ≤ kX −Y k p .

(∗∗)

Chapter 13 Problem Solutions

219

22. By taking pth roots, we see that lim E[|Xn | p ] = E[|X| p ]

(#)

n→∞

is equivalent to kXn k p → kXk p . Since kXk p − kY k p ≤ kX −Y k p ,

we see that convergence in mean of order p implies (#). 23. Write

kX +Y k2p + kX −Y k2p = hX +Y, X +Y i + hX −Y, X −Y i = hX, Xi + 2hX,Y i + hY,Y i + hX, Xi − 2hX,Y i + hY,Y i = 2(kXk2p + kY k2p ).

24. Write |hXn ,Yn i − hX,Y i| = |hXn ,Yn i − hX,Yn i + hX,Yn i − hX,Y i| ≤ khXn − X,Y i| + |hX,Yn −Y i|

≤ kXn − Xk2 kYn k2 + kXk2 kYn −Y k2 → 0.

Here we have used the Cauchy–Schwarz inequality and the fact that since Yn converges, it is bounded. 25. As in the example, for n > m, we can write kYn −Ym k22 ≤

n

n

∑ ∑

k=m+1 l=m+1

|hk | |hl | |hXk , Xl i|.

In this problem, hXk , Xl i = 0 for k 6= l. Hence, kYn −Ym k22 ≤

n

∑

k=m+1

B|hk |2 ≤ B

n

∑

k=m+1

|hk |2 → 0

2 as n > m → ∞ on account of the assumption that ∑∞ k=1 |hk | < ∞.

26. Put Yn := ∑nk=1 hk Xk . It suffices to show that Yn is Cauchy in L p . Write, for n > m,

n n n

1/p

|h | kX k = B ≤ |hk | → 0 kYn −Ym k p = ∑ hk Xk p k k ∑ ∑

p

k=m+1

k=m+1

k=m+1

as n > m → ∞ on account of the assumption that ∑∞ k=1 |hk | < ∞.

27. Write E[Y Z] = E



n

=

 ν  X (t − t ) (s − s ) X ∑ τi i i−1 ∑ θ j j j−1 n

i=1

ν

j=1

∑ ∑ R(τi , θ j )(ti − ti−1 )(s j − s j−1 ).

i=1 j=1

220

Chapter 13 Problem Solutions

28. First note that if ν

n

Y :=

∑ g(τi )Xτi (ti − ti−1 )

and

Z :=

i=1

then

∑ g(θ j )Xθ j (s j − s j−1 ),

j=1

ν

n

∑ ∑ g(τi )R(τi , θ j )g(θ j )(ti − ti−1 )(s j − s j−1 ).

E[Y Z] =

i=1 j=1

Hence, given finer and finer partitions, with Ym defined analogously to Y above, we see that E[|Ym −Yk |2 ] = E[Ym2 ] − 2E[YmYk ] + E[Yk2 ] →

Z bZ b a

+

Z

a

g(t)R(t, s)g(s) dt ds − 2

a bZ b

Z bZ b a

g(t)R(t, s)g(s) dt ds

a

g(t)R(t, s)g(s) dt ds = 0.

a

Thus, Ym is Cauchy in L2 , and there exists a Y ∈ L2 with kYm −Y k2 → 0. Furthermore, since Ym converges in mean square to Y , E[Ym2 ] → E[Y 2 ], and it is clear that E[Ym2 ] →

Z bZ b a

g(t)R(t, s)g(s) dt ds.

a

29. Consider the formula Z T 0

R(t − s)ϕ (s) ds = λ ϕ (t),

0 ≤ t ≤ T.

(∗)

Since R is defined for all t, we can extend the definition of ϕ on the right-hand side in the obvious way. Furthermore, since R has period T , so will the extended definition of ϕ . Hence, both R and ϕ have Fourier series representations, say R(t) =

∑ rn e j2π nt/T

and

ϕ (s) =

n

∑ ϕn e j2π ns/T . n

Substituting these into (∗) yields

∑ rn e j2π nt/T n

Z T

|

0

e− j2π ns/T ϕ (s) ds = {z }

∑ λ ϕn e j2π nt/T . n

= T ϕn

It follows that Trn ϕn = λ ϕn . Now, if ϕ is an eigenfunction, ϕ cannot be the zero function. Hence, there is at least one value of n with ϕn 6= 0. For all n with ϕn 6= 0, λ = Trn . Thus, ϕ (t) = ∑ ϕn e j2π nt/T . n:rn =λ /T

Chapter 13 Problem Solutions 30. If

Rb a

221

R(t, s)ϕ (s) ds = λ ϕ (t) then Z bZ b

0 ≤

a

a

Z b

=

R(t, s)ϕ (t)ϕ (s) dt ds = Z b

ϕ (t)[λ ϕ (t)] dt = λ

a

Z b a

Z b  ϕ (t) R(t, s)ϕ (s) ds dt a

ϕ (t)2 dt.

a

Hence, λ ≥ 0. 31. To begin, write

λk

Z b a

 Z b Z b λk ϕk (t)ϕm (t) dt = R(t, s)ϕk (s) ds ϕm (t) dt a a a Z b  Z b R(s,t)ϕm (t) dt ds, since R(t, s) = R(s,t), ϕk (s) =

ϕk (t)ϕm (t) dt =

Z b

=

Z b

a

a

a

ϕk (s) · λm ϕm (s) ds = λm

We can now write (λk − λm ) If λk 6= λm , we must have 32.

(a) Write Z T

Rb a

Z b a

Z b a

ϕk (s)ϕm (s) ds.

ϕk (t)ϕm (t) dt = 0.

ϕk (t)ϕm (t) dt = 0.

R(t, s)g(s) ds =

0

Z T ∞

 ∑ λk ϕk (t)ϕk (s) g(s) ds

0

k=1

∞

=

∑ λk ϕk (t)

k=1

Z T 0

g(s)ϕk (s) ds =

∞

∑ λk gk ϕk (t).

k=1

(b) Write Z T

R(t, s)ϕ (s) ds =

Z T 0

0

=

Z T 0

=

  ∞ R(t, s) g(s) − ∑ gk ϕk (s) ds k=1 ∞

R(t, s)g(s) ds − ∑ gk k=1

∞

∞

k=1

k=1

Z T

|0

∑ λk gk ϕk (t) − ∑ λk gk ϕk (t)

R(t, s)ϕk (s) ds {z } = λk ϕk (t)

= 0.

(c) Write E[Z 2 ] =

Z TZ T 0

0

ϕ (t)R(t, s)ϕ (s) dt ds =

Z T 0

Z ϕ (t)

0

T

 R(t, s)ϕ (s) ds dt = 0.

222

Chapter 13 Problem Solutions

33. Consider the equation

λ ϕ (t) =

Z T 0

=

Z t 0

−t

= e

e−|t−s| ϕ (s) ds

(#) Z T

e−(t−s) ϕ (s) ds + Z t

t

t

s

e ϕ (s) ds + e

e−(s−t) ϕ (s) ds

Z T t

0

e−s ϕ (s) ds.

Differentiating yields

λ ϕ 0 (t) = −e−t = −e−t

Z t 0

Z t

es ϕ (s) ds + e−t et ϕ (t) + et es ϕ (s) ds + et

Z T t

0

Z T t

e−s ϕ (s) ds + et (−e−t ϕ (t))

e−s ϕ (s) ds.

(##)

Differentiating again yields

λ ϕ 00 (t) = e−t =

Z T 0

Z t 0

es ϕ (s) ds − e−t et ϕ (t) + et

Z T t

e−s ϕ (s) ds + et (−e−t ϕ (t))

e−|t−s| ϕ (s) ds − 2ϕ (t)

= λ ϕ (t) − 2ϕ (t),

by (#),

= (λ − 2)ϕ (t). We can now write

ϕ 00 (t) = (1 − 2/λ )ϕ (t) = −(2/λ − 1)ϕ (t). p For 0 < λ < 2, put µ := 2/λ − 1. Then ϕ 00 (t) = −µ 2 ϕ (t). Hence, we must have ϕ (t) = A cos µ t + B sin µ t

for some constants A and B. 34. The required orthogonality principle is that   L E (Xt − Xbt ) ∑ ci Ai = 0 i=1

for all constants ci , where

Xbt =

L

∑ cbj A j .

j=1

In particular, we must have E[(Xt − Xbt )Ai ] = 0. Now, we know from the text that E[Xt Ai ] = λi ϕi (t). We also have E[Xbt Ai ] =

L

∑ cbj E[A j Ai ]

j=1

= cbi λi .

Chapter 13 Problem Solutions Hence, cbi = ϕi (t), and we have

Xbt =

223

L

∑ Ai ϕi (t).

i=1

35. Since kYn − Y k2 → 0, by the hint, E[Yn ] → E[Y ] and E[Yn2 ] → E[Y 2 ]. Since gn is piecewise constant, we know that E[Yn ] = 0, and so E[Y ] = 0 too. Next, an argument analogous to the one in Problem 21 tells us that if kgn − gk → 0, then kgn k → kgk. Hence, 2

E[Y ] =

lim E[Yn2 ] n→∞

= lim σ n→∞

2

Z ∞ 0

2

gn (t) dt = σ

2

Z ∞

g(t)2 dt.

0

36. First write kY − Ye k2 = kY −Yn k2 + kYn − Yen k2 + kYen − Ye k2 ,

where the first and last terms on the right go to zero. As for the middle term, write 2   Z ∞ Z ∞ gen (t) dWt gn (t) dWt − kYn − Yen k22 = E = E

Z

0

We can now write

0

0

∞

[gn (t) − gen (t)] dWt

2 

= σ2

Z ∞ 0

[gn (t) − gen (t)]2 dt.

kYn − Yen k2 = σ kgn − gen k ≤ σ kgn − gk + σ kg − gen k → 0.

37. We know from our earlier work that the Wiener integral is linear on piecewise-conR R stant functions. To analyze theRgeneral case, let Y = 0∞ g(t) dWt and Z = 0∞ h(t) dWt . We must show that aY + bZ = 0∞ ag(t) + bh(t) dWt . Let gn (t) and hn (t) be piecewiseconstant functions suchR that kgn − gk → 0 and khn − hk → 0 and such that Yn := R∞ ∞ g (t) dW n t and Zn := 0 hn (t) dWt converge in mean square to Y and Z, respectively. 0 Now observe that aYn + bZn = a

Z ∞ 0

gn (t) dWt + b

Z ∞ 0

hn (t) dWt =

Z ∞ 0

agn (t) + bhn (t) dWt ,

(∗)

since gn and hn are piecewise constant. Next, since k(agn + bhn ) − (ag + bh)k ≤ |a| kgn − gk + |b| khn − hk → 0,

R

it follows that the right-hand side of (∗) converges in mean square to 0∞ ag(t) + bh(t) dWt . Since the left-hand side of (∗) converges in mean square to aY + bZ, the desired result follows. 38. We must find all values of β for which E[|Xt /t|2 ] → 0. First compute Z t 2  Z t t 2β +1 2 β E[Xt ] = E = . τ dWτ τ 2β d τ = 2β + 1 0 0 Then t 2β +1 /t 2 = t 2β −1 → 0 if and only if 2β − 1 < 0; i.e., 0 < β < 1/2.

224

Chapter 13 Problem Solutions

39. Using the law of total probability, substitution, and independence, write 2  2   Z T Z ∞  Z T 2 n n = E τ dWτ τ dWτ T = t fT (t) dt E[YT ] = E 0

=

Z ∞  Z t

E

0

0

0

0

2 2   Z ∞  Z t n E fT (t) dt. τ dWτ T = t fT (t) dt = τ n dWτ 0 0

Now use properties of the Wiener process to write  Z ∞ Z t Z ∞ 2n+1 t E[T 2n+1 ] 2 2n E[YT ] = τ d τ fT (t) dt = fT (t) dt = 2n + 1 0 0 0 2n + 1 = 40.

(2n)! (2n + 1)!/λ 2n+1 = 2n+1 . 2n + 1 λ

(a) Write g(t + ∆t) − g(t) = E[ f (Wt+∆t )] − E[ f (Wt )]

= E[ f (Wt+∆t ) − f (Wt )] ≈ E[ f 0 (Wt )(Wt+∆t −Wt )] + 21 E[ f 00 (Wt )(Wt+∆t −Wt )2 ]

= E[ f 0 (Wt −W0 )(Wt+∆t −Wt )] + 12 E[ f 00 (Wt −W0 )(Wt+∆t −Wt )2 ]

= E[ f 0 (Wt −W0 )] · 0 + 21 E[ f 00 (Wt −W0 )] · ∆t.

It then follows that g0 (t) = 12 E[ f 00 (Wt )]. (b) If f (x) = ex , then g0 (t) = 21 E[eWt ] = 12 g(t). In this case, g(t) = et/2 , since g(0) = E[eW0 ] = 1. 2 (c) We have by direct calculation that g(t) = E[eWt ] = es t/2 s=1 = et/2 .

41. Let C be the ball of radius r, C := {Y ∈ L p : kY k p ≤ r}. For X ∈ / C, i.e., kXk p > r, we show that r Xb = X. kXk p

b p = r so that Xb ∈ C as To begin, note that the proposed formula for Xb satisfies kXk required. Now observe that





b p = X − r X = 1 − r kXk p = kXk p − r. kX − Xk



kXk p p kXk p

Next, for any Y ∈ C,

kX −Y k p ≥ kXk p − kY k p

b Thus, no Y ∈ C is closer to X than X.

= kXk p − kY k p ≥ kXk p − r b p. = kX − Xk

Chapter 13 Problem Solutions

225

b i = 0 for 42. Suppose that Xb and Xe are both elements of a subspace M and that hX − X,Y e all Y ∈ M and hX − X,Y i = 0 for all Y ∈ M. Then write e 22 = hXb − X, e Xb − Xi e = h(Xb − X) + (X − X), e Xb − Xei = 0 + 0 = 0. kXb − Xk | {z } ∈M

43. For XbN is to be the projection of XbM onto N, it is sufficient that the orthogonality principle be satisfied. In other words, it suffices to show that hXbM − XbN ,Y i = 0,

Observe that

for all Y ∈ N.

hXbM − XbN ,Y i = h(XbM − X) + (X − XbN ),Y i = −hX − XbM ,Y i + hX − XbN ,Y i.

Now, the last term on the right is zero by the orthogonality principle for the projection of X onto N, since Y ∈ N. To show that hX − XbM ,Y i = 0, observe that since N ⊂ M, Y ⊂ N implies Y ⊂ M. By the orthogonality principle for the projection of X onto M, hX − XbM ,Y i = 0 for Y ∈ M.

44. In the diagram, M is the disk and N is the horizontal line segment. The ◦ is XbM , the projection of X onto the disk M. The  is XbN , the projection of X onto line segment N. The × is ([ XbM )N , the projection of the circle XbM onto the line segment N. We see [ that (Xb ) 6= Xb . M N

N

X

XbM

XbN

d (XbM )N

45. Suppose that gn (Y ) ∈ M and gn (Y ) converges in mean square to some X. We must show that X ∈ M. Since gn (Y ) converges, it is Cauchy. Writing kgn (Y ) − gm (Y )k22 = E[|gn (Y ) − gm (Y )|2 ] =

Z ∞

−∞

|gn (y) − gm (y)|2 fY (y) dy = kgn − gm kY ,

we see that gn is Cauchy in G, which is complete. Hence, there exists a g ∈ G with kgn − gkY → 0. We claim X = g(Y ). Write kg(Y ) − Xk2 = kg(Y ) − gn (Y ) + gn (Y ) − Xk2 ≤ kg(Y ) − gn (Y )k2 + kgn (Y ) − Xk2 , where the last term goes to zero by assumption. Now observe that kg(Y ) − gn (Y )k22 = E[|g(Y ) − gn (Y )|2 ] =

Z ∞

−∞

|g(y) − gn (y)|2 fY (y) dy

= kg − gn kY2 → 0. Thus, X = g(Y ) ∈ M as required.

226

Chapter 13 Problem Solutions R

46. We claim that the required projection is 01 f (t) dWt . Note that this is an element of M since Z 1 Z ∞ f (t)2 dt ≤ f (t)2 dt < ∞. 0

0

Consider the orthogonality condition Z ∞ Z 1  Z Z 1 E f (t) dWt − f (t) dWt g(t) dWt = E 0

0

0

∞

f (t) dWt

1

Z

1

0

 g(t) dWt .

Now put f˜(t) :=



f (t), t > 1, 0, 0 ≤ t ≤ 1,

g(t) ˜ :=

and



so that the orthgonality condition becomes Z ∞ Z ∞  Z E f˜(t) dWt g(t) ˜ dWt = 0

0

∞

0, t > 1, g(t), 0 ≤ t ≤ 1,

f˜(t)g(t) ˜ dt,

0

which is zero since f˜(t)g(t) ˜ = 0 for all t ≥ 0. 47. The function g(t) will be optimal if the orthogonality condition Z ∞   Z ∞ g( ˜ τ ) dWτ = 0 g(τ ) dWτ E X− 0

0

R∞

holds for all g˜ with 0 g( ˜ τ )2 d τ . In particular, this must be true for g( ˜ τ ) = I[0,t] (τ ). In this case, the above expectation reduces to  Z∞  Z ∞  Z ∞ E X I[0,t] (τ ) dWτ − E I[0,t] (τ ) dWτ . g(τ ) dWτ 0

Now, since

0

0

Z ∞ 0

I[0,t] (τ ) dWτ =

Z t 0

dWτ = Wt ,

we have the further simplification E[XWt ] − σ 2

Z ∞ 0

g(τ )I[0,t] (τ ) d τ = E[XWt ] − σ 2

Z t

g(τ ) d τ .

0

Since this must be equal to zero for all t ≥ 0, we can differentiate and obtain g(t) = 48.

1 d · E[XWt ]. σ 2 dt

(a) Using the methods of Chapter 5, it is not too hard to show that  y ≥ 1/4,   √ 1,  [2 − 1 − 4y ]/2, 0 ≤ y < 1/4, √ FY (y) = 1 − 4y ]/2, −2 ≤ y < 0, [3/2 − (1/2)    0, y < −2.

Chapter 13 Problem Solutions

227

It then follows that

 √  1/ √1 − 4y, 0 < y < 1/4, 1/(2 1 − 4y ), −2 < y < 0, fY (y) =  0, otherwise.

(b) We must find a gb(y) such that

E[v(X)g(Y )] = E[b g(Y )g(Y )],

for all bounded g.

For future reference, note that E[v(X)g(Y )] = E[v(X)g(X(1 − X))] =

1 2

Z 1

−1

v(x)g(x(1 − x)) dx.

Now, by considering the problem of solving g(x) = y for x in the two cases 0 ≤ y ≤ 1/4 and −2 ≤ y < 0 suggests that we try  √     √  1 v 1+ 1−4y + 1 v 1− 1−4y , 0 ≤ y ≤ 1/4, 2 2 2 2  √  gb(y) = 1− 1−4y  v , −2 ≤ y < 0. 2

To check, we compute

E[b g(Y )g(Y )] = E[b g(X(1 − X))g(X(1 − X))]  Z 0 Z 1/2 Z 1  1 = + + gb(x(1 − x))g(x(1 − x)) dx 2 −1 0 1/2 Z 0 1 v(x)g(x(1 − x)) dx = 2 −1 +

Z 1/2 v(1 − x) + v(x)

2

0

+

Z 1 v(x) + v(1 − x) 1/2

1 = 2 = 49.

Z

Z 1

−1

2

g(x(1 − x)) dx  g(x(1 − x)) dx

0

−1

v(x)g(x(1 − x)) dx +

Z 1 0

v(x)g(x(1 − x)) dx

v(x)g(x(1 − x)) dx.

(a) Using the methods of Chapter 5, it is not too hard to show that ( FΘ (sin−1 (y)) + 1 − FΘ (π − sin−1 (y)), 0 ≤ y ≤ 1, FY (y) = FΘ (sin−1 (y)) − FΘ (−π − sin−1 (y)), −1 ≤ y < 0. Hence,

 fΘ (sin−1 (y)) + fΘ (π − sin−1 (y))    p , 0 ≤ y < 1,  1 − y2 fY (y) = f (sin−1 (y)) + fΘ (−π − sin−1 (y))    Θ p , −1 < y < 0.  1 − y2



228

Chapter 13 Problem Solutions (b) Consider E[v(X)g(Y )] = E[v(cos Θ)g(sin Θ)] =

Z π

−π

v(cos θ )g(sin θ ) fΘ (θ ) d θ .

Next, write Z π /2 0

v(cos θ )g(sin θ ) fΘ (θ ) d θ =

Z 1

v(cos θ )g(sin θ ) fΘ (θ ) d θ =

Z 0

0

and Z 0

−π /2

p fΘ (sin−1 (y)) dy v( 1 − y2 )g(y) p 1 − y2

p fΘ (sin−1 (y)) v( 1 − y2 )g(y) p dy. −1 1 − y2

Similarly, we can write with a bit more work Z π

π /2

v(cos θ )g(sin θ ) fΘ (θ ) d θ = =

Z π /2 0

Z π /2 0

=

Z 1 0

and Z −π /2 −π

v(cos(π − t))g(sin(π − t)) fΘ (π − t) dt v(− cost)g(sint) fΘ (π − t) dt

p fΘ (π − sin−1 (y)) p v(− 1 − y2 )g(y) dy, 1 − y2

v(cos θ )g(sin θ ) fΘ (θ ) d θ = = =

Z 0

−π /2 Z 0 −π /2

v(cos(−π − t))g(sin(−π − t)) fΘ (−π − t) dt v(− cost)g(sint) fΘ (−π − t) dt

Z 0

p fΘ (−π − sin−1 (y)) p v(− 1 − y2 )g(y) dy. −1 1 − y2

Putting all this together, we see that E[v(X)g(Y )] =

p Z 1 p v( 1 − y2 ) fΘ (sin−1 (y)) + v(− 1 − y2 ) fΘ (π − sin−1 (y)) 0

fΘ (sin−1 (y)) + fΘ (π − sin−1 (y))

· g(y) fY (y) dy p Z 0 p v( 1 − y2 ) fΘ (sin−1 (y)) + v(− 1 − y2 ) fΘ (−π − sin−1 (y)) + fΘ (sin−1 (y)) + fΘ (−π − sin−1 (y)) −1 · g(y) fY (y) dy.

Chapter 13 Problem Solutions

229

We conclude that E[v(X)|Y = y]  √ √ −1 −1 2 2   v( 1−y ) fΘ (sin −1(y))+v(− 1−y−1) fΘ (π −sin (y)) , 0 < y < 1, fΘ (sin (y))+ fΘ (π −sin (y)) √ √ = −1 −1 2 2   v( 1−y ) fΘ (sin −1(y))+v(− 1−y )−1fΘ (−π −sin (y)) , −1 < y < 0. f (sin (y))+ f (−π −sin (y)) Θ

Θ

50. Since X ≥ 0 and g(y) = IB (E[X|Y ]) ≥ 0, E[Xg(Y )] ≥ 0. On the other hand, E[X|Y ]I(−∞,−1/n) (E[X|Y ]) ≤

−1 I (E[X|Y ]), n (−∞,−1/n)

and so 0 ≤ E[Xg(Y )] = E[E[X|Y ]g(Y )] ≤

−1 P(E[X|Y ] < −1/n) < 0, n

which is a contradiction. Hence, P(E[X|Y ] < 0) = 0. 51. Write E[Xg(Y )] = E[(X + − X − )g(Y )] = E[X + g(Y )] − E[X − g(Y )]     = E E[X + |Y ]g(Y ) − E E[X − |Y ]g(Y ) 
 = E E[X + |Y ] − E[X − |Y ] g(Y ) .

By uniqueness, we conclude that E[X|Y ] = E[X + |Y ] − E[X − |Y ].

52. Following the hint, we begin with E[X|Y ] = E[X + |Y ] − E[X − |Y ] ≤ E[X + |Y ] + E[X − |Y ] = E[X + |Y ] + E[X − |Y ]. Then

      E E[X|Y ] ≤ E E[X + |Y ] + E E[X − |Y ] = E[X + ] + E[X − ] < ∞.

53. To show that E[h(Y )X|Y ] = h(Y )E[X|Y ], we have to show that the right-hand side satisfies the characterizing equation of the left-hand side. Since the characterizing equation for the left-hand side is   E[{h(Y )X}g(Y )] = E E[h(Y )X|Y ]g(Y ) , for all bounded g, we must show that

  E[{h(Y )X}g(Y )] = E {h(Y )E[X|Y ]}g(Y ) ,

for all bounded g.

The only other thing we know is the characterizing equation for E[X|Y ], which is   E[Xg(Y )] = E E[X|Y ]g(Y ) , for all bounded g.

(∗)

230

Chapter 13 Problem Solutions Since g in the above formula is an arbitrary and bounded function, and since h is also bounded, we can rewrite the above formula by replacing g(Y ) with g(Y )h(Y ) for arbitrary bounded g. We thus have   E[X{g(Y )h(Y )}] = E E[X|Y ]{g(Y )h(Y )} ,

for all bounded g,

which is equivalent to (∗).

  54. We must show that E[X|q(Y )] satisfies the characterizing equation   of E E[X|Y ] q(Y ) . To write down the characterizing equation for E E[X|Y ] q(Y ) , it is convenient to use the notation Z := E[X|Y ]. Then the characterizing equation for E[Z|q(Y )] is   E[Zg(q(Y ))] = E E[Z|q(Y )]g(q(Y )) ,

for all bounded g.

We must show that this equation holds when E[Z|q(Y )] is replaced by E[X|q(Y )]; i.e., we must show that   E[Zg(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,

for all bounded g.

Replacing Z by its definition, we must show that

  E[E[X|Y ]g(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,

for all bounded g.

(∗∗)

We begin with the characterizing equation for E[X|Y ], which is   E[Xh(Y )] = E E[X|Y ]h(Y ) ,

for all bounded h.

Since h is bounded and arbitrary, we can replace h(Y ) by g(q(Y )) for arbitrary bounded g. Thus,   E[Xg(q(Y ))] = E E[X|Y ]g(q(Y )) ,

for all bounded g.

We next consider the characterizing equation of E[X|q(Y )], which is   E[Xg(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,

for all bounded g.

Combining these last two equations yields (∗∗). 55. The desired result,

E[{h(Y )X}g(Y )] = E[{h(Y )E[X|Y ]}g(Y )], can be rewritten as E[(X − E[X|Y ])g(Y )h(Y )] = 0, where g is a bounded function and h(Y ) ∈ L2 . But then g(Y )h(Y ) ∈ L2 , and therefore this last equation must hold by the orthogonality principle since E[X|Y ] is the projection of X onto M = {v(Y ) : E[v(Y )2 ] < ∞}.

Chapter 13 Problem Solutions

231

56. Let hn (Y ) be bounded and converge to h(Y ). Then for bounded g, h i E[{h(Y )X}g(Y )] = E lim hn (Y )Xg(Y ) n→∞

= lim E[X{hn (Y )g(Y )}] n→∞   = lim E E[X|Y ]{hn (Y )g(Y )} n→∞ h i = E lim hn (Y )E[X|Y ]g(Y ) n→∞   = E {h(Y )E[X|Y ]}g(Y ) .

By uniqueness, E[h(Y )X|Y ] = h(Y )E[X|Y ]. 57. First write

n

Y = E[Y |Y ] = E[X1 + · · · + Xn |Y ] =

∑ E[Xi |Y ].

i=1

By symmetry, we must have E[Xi |Y ] = E[X1 |Y ] for all i. Then Y = nE[X1 |Y ], or E[X1 |Y ] = Y /n. 58. Write E[Xn+1 |Y1 , . . . ,Yn ] = E[Yn+1 +Yn + · · · +Y1 |Y1 , . . . ,Yn ]

= E[Yn+1 |Y1 , . . . ,Yn ] +Yn + · · · +Y1 = E[Yn+1 ] + Xn , by indep. & def. of Xn , = Xn ,

59. For n ≥ 1,

since E[Yn+1 ] = 0.

  E[Xn+1 ] = E E[Xn+1 |Yn , . . . ,Y1 ] = E[Xn ],

where the second equality uses the definition of a martingale. Hence, E[Xn ] = E[X1 ] for n ≥ 1. 60. For n ≥ 1,

  E[Xn+1 ] = E E[Xn+1 |Yn , . . . ,Y1 ] ≤ E[Xn ],

where the inequality uses the definition of a supermartingale. Since Xn ≥ 0, E[Xn ] ≥ 0. Hence, 0 ≤ E[Xn ] ≤ E[X1 ] for n ≥ 1. 61. Since Xn+1 := E[Z|Yn+1 , . . . ,Y1 ],   E[Xn+1 |Yn , . . . ,Y1 ] = E E[Z|Yn+1 , . . . ,Y1 ] Yn , . . . ,Y1 = E[Z|Yn , . . . ,Y1 ], =: Xn .

by the smoothing property,

62. Since Xn := w(Yn ) · · · w(Y1 ), observe that Xn is a function of Y1 , . . . ,Yn and that Xn+1 = w(Yn+1 )Xn . Then E[Xn+1 |Yn , . . . ,Y1 ] = E[w(Yn+1 )Xn |Yn , . . . ,Y1 ] = Xn E[w(Yn+1 )|Yn , . . . ,Y1 ] = Xn E[w(Yn+1 )],

by independence.

232

Chapter 13 Problem Solutions It remains to compute E[w(Yn+1 )] =

Z ∞

−∞

w(y) f (y) dy =

Z ∞ ˜ f (y) −∞

f (y)

f (y) dy =

Z ∞

−∞

f˜(y) dy = 1.

Hence, E[Xn+1 |Yn , . . . ,Y1 ] = Xn ; i.e., Xn is a martingale with respect to Yn . 63. To begin, write f˜Yn+1 ···Y1 (yn+1 , . . . , y1 ) fYn+1 ···Y1 (yn+1 , . . . , y1 ) f˜Y |Y ···Y (yn+1 |yn , . . . , y1 ) f˜Yn ···Y1 (yn , . . . , y1 ) · . = n+1 n 1 fYn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 ) fYn ···Y1 (yn , . . . , y1 )

wn+1 (y1 , . . . , yn+1 ) =

If we put wˆ n+1 (yn+1 , . . . , y1 ) :=

f˜Yn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 ) , fYn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 )

then Xn+1 := wn+1 (Y1 , . . . ,Yn+1 ) = wˆ n+1 (Yn+1 , . . . ,Y1 )Xn , where Xn is a function of Y1 , . . . ,Yn . We can now write E[Xn+1 |Yn , . . . ,Y1 ] = E[wˆ n+1 (Yn+1 , . . . ,Y1 )Xn |Yn , . . . ,Y1 ] = Xn E[wˆ n+1 (Yn+1 , . . . ,Y1 )|Yn , . . . ,Y1 ]. We now show that this last factor is equal to one. Write E[wˆ n+1 (Yn+1 ,Yn , . . . ,Y1 )|Yn = yn , . . . ,Y1 = y1 ] = E[wˆ n+1 (Yn+1 , yn . . . , y1 )|Yn = yn , . . . ,Y1 = y1 ] = =

Z ∞

−∞ Z ∞

−∞

wˆ n+1 (y, yn , . . . , y1 ) fYn+1 |Yn ···Y1 (y|yn , . . . , y1 ) dy f˜Yn+1 |Yn ···Y1 (y|yn , . . . , y1 ) dy = 1.

64. Since Wk depends on Yk and Yk−1 , and since Xn = W1 + · · · + Wn , Xn depends on Y0 , . . . ,Yn . Note also that Xn+1 = Xn +Wn+1 . Now write E[Xn+1 |Yn , . . . ,Y0 ] = Xn + E[Wn+1 |Yn , . . . ,Y0 ]. Next,   Z ∞ E[Wn+1 |Yn = yn , . . . ,Y0 = y0 ] = E Yn+1 − zp(z|Yn ) dz Yn = yn , . . . ,Y0 = y0 −∞   Z ∞ = E Yn+1 − zp(z|yn ) dz Yn = yn , . . . ,Y0 = y0 −∞

= E[Yn+1 |Yn = yn , . . . ,Y0 = y0 ] − =

Z ∞

−∞

Z ∞

−∞

z fYn+1 |Yn ···Y0 (z|yn , . . . , y0 ) dz −

zp(z|yn ) dz

Z ∞

−∞

zp(z|yn ) dz

Chapter 13 Problem Solutions = =

Z ∞

−∞ Z ∞ −∞

233

z fYn+1 |Yn (z|yn ) dz − zp(z|yn ) dz −

Z ∞

−∞

Z ∞

−∞

zp(z|yn ) dz

zp(z|yn ) dz = 0.

Hence, E[Xn+1 |Yn , . . . ,Y0 ] = Xn ; i.e., Xn is a martingale with respect to Yn . 65. First write   E[Yn+1 |Xn = in , . . . , X0 = i0 ] = E ρ Xn+1 Xn = in , . . . , X0 = i0 =

∑ ρ j P(Xn+1 = j|Xn = in , . . . , X0 = i0 ) j

=

∑ ρ j P(Xn+1 = j|Xn = in ). j

Next, for 0 < i < N,

∑ ρ j P(Xn+1 = j|Xn = i) j

= ρ i−1 (1 − a) + ρ i+1 a  1 − a i−1

 1 − a i+1 (1 − a) + a a a  1 − a 2 i  1 − a i−1 h 1−a+ a = a a  1 − a i−1 h (1 − a)2 i = 1−a+ a a 1−ai (1 − a)i h 1+ = ρ i. = ai−1 a =

If Xn = 0 or Xn = 1, then Xn+1 = Xn , and so

E[ρ Xn+1 |Xn , . . . , X0 ] = E[ρ Xn |Xn , . . . , X0 ] = ρ Xn = Yn . Hence, in all cases, E[Yn+1 |Xn , . . . , X0 ] = Yn , and so Yn is a martingale with respect to Xn . 66. First, since Xn is a submartingale, it is clear that An+1 := An + (E[Xn+1 |Yn , . . . ,Y1 ] − Xn ) ≥ 0 and is a function of Y1 , . . . ,Yn . To show that Mn is a martingale, write E[Mn+1 |Yn , . . . ,Y1 ] = E[Xn+1 − An+1 |Yn , . . . ,Y1 ] = E[Xn+1 |Yn , . . . ,Y1 ] − An+1 h i = E[Xn+1 |Yn , . . . ,Y1 ] − An + (E[Xn+1 |Yn , . . . ,Y1 ] − Xn ) = Xn − An = Mn .

67. Without loss of generality, assume f1 ≤ f2 . Then E[Z f1 Z ∗f2 ] = E[Z f1 (Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ]

234

Chapter 13 Problem Solutions = E[(Z f1 − Z−1/2 )(Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ]

= E[Z f1 − Z−1/2 ]E[(Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ] Z f1

= 0 · 0 + E[|Z f1 |2 ] =

−1/2

S(ν ) d ν .

68. Using the geometric series formula,
n+1

1 n j2π f n 1 e j2π f − e j2π f e = · ∑ n k=1 n 1 − e j2π f =

e j2π f 1 − e j2π f n · n 1 − e j2π f

=

e j2π f e jπ f n e− jπ f n − e jπ f n e j2π f e jπ f n sin(π f n) · jπ f · − jπ f = · jπ f · . n e e − e jπ f n e sin(π f )

69. Following the hint, write E[|Y |2 ] = E[YY ∗ ] = E N

N

N

∑

dn Xn

n=−N



n=−N k=−N N

N

n=−N k=−N

=

Z 1/2

−1/2

∑

dk Xk

k=−N N

∑ ∑

n=−N k=−N

dn dk∗

∑ ∑

=

N

N

dn dk∗ E[Xn Xk ] =

∑ ∑

=



S( f )

Z 1/2

−1/2

N

∑

dn e

n=−N

∗ 

dn dk∗ R(n − k)

S( f )e j2π f (n−k) d f

2 j2π f n

d f.

Hence, if ∑Nn=−N dn e j2π f n = 0, then E[|Y |2 ] = 0. 70. Write E[T (G0 )T (H0 )] = E



N

gn Xn

n=−N

N

=

∑ N

∑ ∑

n=−N k=−N

= = 71.

Z 1/2

−1/2

Z 1/2

−1/2



gn h∗n

 S( f )

N

∑

k=−N

Z 1/2

−1/2

N

∑

n=−N

hk Xk

gn e

∗ 

N

=

N

∑ ∑

n=−N k=−N

gn h∗n R(n − k)

S( f )e j2π f (n−k) d f j2π f n



N

∑

k=−N

hk e

j2π f k

∗

df

S( f )G0 ( f )H0 ( f )∗ d f .

(a) Write en ) + T (G en ) −Y k2 kT (G) −Y k2 = kT (G) − T (Gn ) + T (Gn ) − T (G en )k2 + kT (G en ) −Y k2 . ≤ kT (G) − T (Gn )k2 + kT (Gn ) − T (G

Chapter 13 Problem Solutions

235

On the right-hand side, the first and third terms go to zero. To analyze the middle term, write en )k2 = kT (Gn − G en )k2 = kGn − G en k kT (Gn ) − T (G en k → 0. ≤ kGn − Gk + kG − G

(b) To show T is norm preserving on L2 (S), let Gn → G with T (Gn ) → T (G). Then by Problem 21, kT (Gn )k2 → kT (G)k2 , and similarly, kGn k → kGk. Now write kT (G)k2 = lim kT (Gn )k2 = lim kGn k2 , n→∞

n→∞

since Gn is a trig. polynomial,

= kGk. (c) To show T is linear on L2 (S), fix G, H ∈ L2 (S), and let Gn and Hn be trigonometric polynomials converging to G and H, respectively. Then

α Gn + β Hn → α G + β H,

(∗)

and we can write kT (α G + β H) − {α T (G) + β T (H)}k2

= kT (α G + β H) − T (α Gn + β Hn )k2 + kT (α Gn + β Hn ) − {α T (Gn ) + β T (Hn )}k2

+ k{α T (Gn ) + β T (Hn )} − {α T (G) + β T (H)}k2 .

Now, the first term on the right goes to zero on account of (∗) and the defintion of T . The second term on the right is equal to zero because T is linear on trigonometric polynomials. The third term goes to zero upon observing that k{α T (Gn ) + β T (Hn )} − {α T (G) + β T (H)}k2 ≤ |α | kT (Gn ) − T (G)k2 + |β | kT (Hn ) − T (H)k2 . (d) Using parts (b) and (c), kT (G) − T (H)k2 = kT (G − H)k2 = kG − Hk implies that T is actually uniformly continuous. 72. It suffices to show that I[−1/2, f ] ∈ L2 (S). Write Z 1/2 −1/2

73.

Z I[−1/2, f ] (ν ) 2 S(ν ) d ν =

f

S(ν ) d ν ≤

−1/2

Z 1/2

S(ν ) d ν = R(0) = E[Xn2 ] < ∞.

−1/2

(a) We know that for trigonometric polynomials, E[T (G)] = 0. Hence, if Gn is a sequence of trigonometric polynomials converging to G in L2 (S), then T (Gn ) → T (G) in L2 , and then E[T (Gn )] → E[T (G)].

(b) For trigonometric polynomials G and H, we have hT (G), T (H)i := E[T (G)T (H)∗ ] =

Z 1/2

−1/2

G( f )H( f )∗ S( f ) d f .

236

Chapter 13 Problem Solutions If Gn and Hn are sequences of trigonometric polynomials converging to G and H in L2 (S), then we can use the result of Problem 24 to write Z 1/2

hT (G), T (H)i = lim hT (Gn ), T (Hn )i = lim n→∞

n→∞ −1/2

Z 1/2

=

−1/2

Gn ( f )Hn ( f )∗ S( f ) d f

G( f )H( f )∗ S( f ) d f .

(c) For −1/2 ≤ f1 < f2 ≤ f3 < f4 ≤ 1/2, write E[(Z f2 − Z f1 )(Z f4 − Z f3 )∗ ] = E[T (I( f1 , f2 ] )T (I( f3 , f4 ] )∗ ] Z 1/2

= 74.

(a) We take L2 (SY ) :=



−1/2

Z 1/2

G:

−1/2

I( f1 , f2 ] ( f )I( f3 , f4 ] ( f )S( f ) d f = 0.

 |G( f )|2 SY ( f ) d f < ∞ .

(b) Write

∑

∞

∑

gn

n=−N

N

gn

n=−N

=

∑

gnYn =

n=−N

=

∞

N

N

TY (G0 ) =

Z 1/2

−1/2

∑

hk

k=−∞

Z 1/2

−1/2

∑

hk Xn−k

k=−∞

e j2π f (n−k) dZ f

G0 ( f )H( f ) dZ f .

(c) For G ∈ L2 (SY ), let Gn be a sequence of trigonometric polynomials converging to G in L2 (SY ). Since TY is continuous, TY (G) = lim TY (Gn ) = lim n→∞

Z 1/2

n→∞ −1/2

Gn ( f )H( f ) dZ f

= lim T (Gn H) = T (GH), =

n→∞ Z 1/2

−1/2

since T is continuous,

G( f )H( f ) dZ f .

(d) Using part (c), V f := TY (I[−1/2, f ] ) = =

Z 1/2

−1/2 Z f −1/2

I[−1/2, f ] (ν )H(ν ) dZν H(ν ) dZν .

(∗)

Chapter 13 Problem Solutions

237

(e) A slight generalization of (∗) establishes the result for piecewise-constant functions. For general G ∈ L2 (SY ), approximate G by a sequence of piecewiseconstant functions Gn and write Z 1/2

−1/2

G( f ) dV f = lim

Z 1/2

n→∞ −1/2

Gn ( f ) dV f = lim

n→∞ −1/2

= lim T (Gn H) = T (GH) = n→∞

Z 1/2

Z 1/2

−1/2

Gn ( f )H( f ) dZ f G( f )H( f ) dZ f .

CHAPTER 14

Problem Solutions 1. We must show that P(|Xn | ≥ ε ) → 0. Since Xn ∼ Cauchy(1/n) has an even density, we can write P(|Xn | ≥ ε ) = 2

Z ∞ ε

1/(nπ ) dx = 2 (1/n)2 + x2

Z ∞ 1/π nε

1 + y2

dy → 0.

2. Since |cn − c| → 0, given ε > 0, there is an N such that for n ≥ N, |cn − c| < ε . For such n, {ω ∈ Ω : |cn − c| ≥ ε } = ∅, and so P(|cn − c| ≥ ε ) = 0. 3. We show that Xn converges in probability to zero. Observe that Xn takes only the values n and zero. Hence, for √ 0 < ε < 1, |Xn | ≥ ε if and only if Xn = n, which happens if and only if U ≤ 1/ n. We can now write √ √ P(|Xn | ≥ ε ) = P(U ≤ 1/ n ) = 1/ n → 0. 4. Write P(|Xn | ≥ ε ) = P(|V | ≥ ε cn ) = 2

Z ∞

ε cn

fV (v) dv → 0,

since cn → ∞. 5. Using the hint,  ∞  [ P(X 6= Y ) = P {|X −Y | ≥ 1/k} = lim P(|X −Y | ≥ 1/K). K→∞

k=1

We now show that the above limit is zero. To begin, observe that |Xn − X| < 1/(2K)

and

|Xn −Y | < 1/(2K)

imply |X −Y | = |X − Xn + Xn −Y | ≤ |X − Xn | + |Xn −Y | < 1/(2K) + 1/(2K) = 1/K. Hence, |X −Y | ≥ 1/K implies |Xn − X| ≥ 1/(2K) or |Xn −Y | ≥ 1/(2K), and we can write
P(|X −Y | ≥ 1/K) ≤ P {|Xn − X| ≥ 1/(2K)} ∪ {|Xn −Y | ≥ 1/(2K)}

≤ P {|Xn − X| ≥ 1/(2K)} + P {|Xn −Y | ≥ 1/(2K)} ,

which goes to zero as n → ∞

238

Chapter 14 Problem Solutions

239

6. Let ε > 0 be given. We must show that for every η > 0, for all sufficiently large n, P(|Xn − X| ≥ ε ) < η . Without loss of generality, assume η < ε . Let n be so large that P(|Xn − X| ≥ η ) < η . Then since η < ε , P(|Xn − X| ≥ ε ) ≤ P(|Xn − X| ≥ η ) < η . 7.

(a) Observe that P(|X| > α ) = 1 − P(|X| ≤ α ) = 1 − P(−α ≤ X ≤ α ) = 1 − [FX (α ) − FX ((−α )−)] = 1 − FX (α ) + FX ((−α )−) ≤ 1 − FX (α ) + FX (−α ).

Now, since FX (x) → 1 as x → ∞ and FX (x) → 0 as x → −∞, for large α , 1 − FX (α ) < ε /8 and FX (−α ) < ε /8, and then P(|X| > α ) < ε /4. Similarly, for large β , P(|Y | > β ) < ε /4.

(b) Observe that if the four conditions hold, then

|Xn | = |Xn − X + X| ≤ |Xn − X| + |X| < δ + α < 2α , and similarly, |Yn | ≤ 2β . Now that both (Xn ,Yn ) and (X,Y ) lie in the rectangle, |g(Xn ,Yn ) − g(X,Y )| < ε . (c) By part (b), observe that {|g(Xn ,Yn ) − g(X,Y )| ≥ ε } ⊂ {|Xn − X| ≥ δ } ∪ {|Yn −Y | ≥ δ } ∪ {|X| > α } ∪ {|Y | > β }.

Hence, P(|g(Xn ,Yn ) − g(X,Y )| ≥ ε ) ≤ P(|Xn − X| ≥ δ ) + P(|Yn −Y | ≥ δ ) + P(|X| > α ) + P(|Y | > β ) < ε /4 + ε /4 + ε /4 + ε /4 = ε .

8.

(a) Since the Xi are i.i.d., the Xi2 are i.i.d. and therefore uncorrelated and have common mean E[Xi2 ] = σ 2 + m2 and common variance 
2 
2 E Xi2 − E[Xi2 ] = E[Xi4 ] − E[Xi2 ] = γ 4 − (σ 2 + m2 )2 .

By the weak law of large numbers, Vn converges in probability to E[Xi2 ] = σ 2 + m2 . (b) Observe that Sn2 = g(Mn ,Vn )n/(n − 1), where g(m, v) := v − m2 is continuous. Hence, by the preceding problem, g(Mn ,Vn ) converges in probability to g(m, v) = (σ 2 + m2 ) − m2 = σ 2 . Next, by Problem 2, n/(n − 1) converges in probability to 1, and then the product [n/(n − 1)]g(Mn ,Vn ) converges in probability to 1 · g(m, v) = σ 2 .

240 9.

Chapter 14 Problem Solutions (a) Since Xn converges in probability to X, with ε = 1 we have P(|Xn − X| ≥ 1) → 0 as n → ∞. Now, if |X − Xn | < 1, then |X| − |Xn | < 1, and it follows that |X| < 1 + |Xn | ≤ 1 + |Y |.

Equivalently, Hence,

|X| ≥ |Y | + 1 implies

|Xn − X| ≥ 1.

P(|X| ≥ |Y | + 1) ≤ P(|Xn − X| ≥ 1) → 0. (b) Following the hint, write E[|Xn − X|] = E[|Xn − X|IAn ] + E[|Xn − X|IAnc ] 
 ≤ E |Xn | + |X| IAn + ε P(Anc ) 
 ≤ E Y + |X| IAn + ε ≤ ε + ε, 
 since for large n, P(An ) < δ implies E Y + |X| IAn < ε . Hence, E[|Xn − X|] < 2ε . 10.

(a) Suppose g(x) is bounded, nonnegative, and g(x) → 0 as x → 0. Then given ε > 0, there exists a δ > 0 such that g(x) < ε /2 for all |x| < δ . For |x| ≥ δ , we use the fact that g is bounded to write g(x) ≤ G for some positive, finite G. Since Xn converges to zero in probability, for large n, P(|Xn | ≥ δ ) < ε /(2G). Now write E[g(Xn )] = E[g(Xn )I[0,δ ) (|Xn |)] + E[g(Xn )I[δ ,∞) (|Xn |)] ≤ E[(ε /2)I[0,δ ) (|Xn |)] + E[GI[δ ,∞) (|Xn |)] ε = P(|Xn | < δ ) + G P(|Xn | ≥ δ ) 2 ε ε < +G = ε. 2 2G

(b) By applying part (a) to the function g(x) = x/(1 + x), it follows that if Xn converges in probability to zero, then   |Xn | = 0. lim E n→∞ 1 + |Xn | Now we show that if the above limit holds, then Xn must converge in probability to zero. Following the hint, we use the fact that g(x) = 1/(1 + x) is an increasing function for x ≥ 0. Write E[g(Xn )] = E[g(Xn )I[ε ,∞) (|Xn |)] + E[g(Xn )I[0,ε ) (|Xn |)] ≥ E[g(Xn )I[ε ,∞) (|Xn |)] ≥ E[g(ε )I[ε ,∞) (|Xn |)] = g(ε )P(|Xn | ≥ ε ).

Thus, if g(x) is nonnegative and nondecreasing, if E[g(Xn )] → 0, then Xn converges in distribution to zero.

Chapter 14 Problem Solutions

241

11. First note that for the constant random variable Y ≡ c, FY (y) = I[c,∞) (y). Similarly, for Yn ≡ cn , FYn (y) = I[cn ,∞) (y). Since the only point at which FY is not continuous is y = c, we must show that I[cn ,∞) (y) → I[c,∞) (y) for all y 6= c. Consider a y with c < y. For all sufficiently large n, cn will be very close to c — so close that cn < y, which implies FYn (y) = 1 = FY (y). Now consider y < c. For all sufficiently large n, cn will be very close to c — so close that y < cn , which implies FYn (y) = 0 = FY (y). 12. For 0 < c < ∞, FY (y) = P(cX ≤ y) = FX (y/c), and FYn (y) = FX (y/cn ). Now, y is a continuity point of FY if and only if y/c is a continuity point of FX . For such y, since y/cn → y/c, FX (y/cn ) → FX (y/c). For c = 0, Y ≡ 0, and FY (y) = I[0,∞) (y). For y 6= 0,  +∞, y > 0, y/cn → −∞, y < 0, and so FYn (y) = FX (y/cn ) → which is exactly FY (y) for y 6= 0. 13. Since Xt ≤ Yt ≤ Zt ,



1, y > 0, 0, y < 0,

{Zt ≤ y} ⊂ {Yt ≤ y} ⊂ {Xt ≤ y},

we can write P(Zt ≤ y) ≤ P(Yt ≤ y) ≤ P(Xt ≤ y),

or FZt (y) ≤ FYt (y) ≤ FXt (y), and it follows that

lim FZt (y) ≤ lim FYt (y) ≤ lim FXt (y) . t→∞ t→∞ | {z } | {z }

t→∞

= F(y)

Thus, FYt (y) → F(y).

= F(y)

14. Since Xn converges in mean to X, the inequality E[Xn ] − E[X] = E[Xn − X] ≤ E[|Xn − X|]

shows that E[Xn ] → E[X]. We now need the following implications: conv. in mean ⇒ conv. in probability ⇒ conv. in distribution ⇒ ϕXn (ν ) → ϕX (ν ).

Since Xn is exponential, we also have

ϕXn (ν ) =

1/E[Xn ] 1/E[X] → . 1/E[Xn ] − jν 1/E[X] − jν

Since limits are unique, the above right-hand side must be ϕX (ν ), which implies X is an exponential random variable. 15. Since Xn and Yn each converge in distribution to constants x and y, respectively, they also converge in probability. Hence, as noted in the text, Xn +Yn converges in probability to x + y. Since convergence in probability implies convergence in distribution, Xn +Yn converges in distribution to x + y.

242

Chapter 14 Problem Solutions

16. Following the hint, we note that each Yn is a finite linear combination of independent Gaussian increments. Hence, each Yn is Gaussian. Since Y is the mean-square limit of the Gaussian Yn , the distribution of RY is also Gaussian by the example cited in theRhint. Furthermore, since each Yn = 0∞ gn (τ ) dWτ , Yn has zero mean and variance σ 2 0∞ gn (τ )2 d τ = σ 2 kgn k2 . By the cited example, Y has zero mean. Also, since kgn k − kgk ≤ kgn − gk → 0, we have

var(Y ) = E[Y 2 ] = lim E[Yn2 ] = lim var(Yn ) n→∞

n→∞

= lim σ 2 kgn k2 = σ 2 kgk2 = σ 2 n→∞

Z ∞

g(τ )2 d τ .

0

17. For any constants c1 , . . . , cn , write n

∑ ci Xti

i=1

n

=

∑ ci

i=1

Z ∞ 0

g(ti , τ ) dWτ =

 ∑ ci g(ti , τ ) dWτ ,

Z ∞ n 0

which is normal by the previous problem.

|

i=1

{z

=: g(τ )

}

18. The plan is to show that the increments are Gaussian and uncorrelated. It will then follow that the increments are independent. For 0 ≤ u < v ≤ s < t < ∞, write       Xu  Xv − Xu −1 1 0 0   Xv  . =  Xt − Xs 0 0 −1 1 Xs  Xt By writing

Xt :=

Z t 0

g(τ ) dWτ =

Z ∞ 0

g(τ )I[0,t] (τ ) dWτ =

Z ∞ 0

h(t, τ ) dWτ ,

where h(t, τ ) := g(τ )I[0,t] (τ ), we see that by the preceding problem, the vector on the right above is Gaussian, and hence, so are the increments in the vector on the left. Next,  Z t Z v g(τ ) dWτ E[(Xt − Xs )(Xv − Xu )] = E g(τ ) dWτ s u  Z ∞ Z ∞ = E g(τ )I(s,t] (τ ) dWτ g(τ )I(u,v] (τ ) dWτ 0

= σ

2

Z ∞ 0

0

2

g(τ ) I(s,t] (τ )I(u,v] (τ ) d τ = 0.

19. Given a linear combination ∑nk=1 ck Ak , put g(τ ) := ∑nk=1 ck ϕk (τ ). Then  Zb  n  Z b Z b n n ∑ ck Ak = ∑ ck Xτ ϕk (τ ) d τ = Xτ ∑ ck ϕk (τ ) d τ = g(τ )Xτ d τ , k=1

k=1

a

a

k=1

a

Chapter 14 Problem Solutions

243

which is a mean-square limit of sums of the form

∑ g(τi )Xτi (ti − ti−1 ). i

Since Xτ is a Gaussian process, these sums are Gaussian, and hence, so is their meansquare limit. 20. If MXn (s) → MX (s), then this holds when s = jν ; i.e., ϕXn (ν ) → ϕX (ν ). But this implies that Xn converges in distribution to X. Similarly, if GXn (z) → GX (z), then this holds when z = e jν ; i.e., ϕXn (ν ) → ϕX (ν ). But this implies Xn converges in distribution to X. 21.

(a) Write pn (k) = FXn (k + 1/2) − FXn (k − 1/2) → FX (k + 1/2) − FX (k − 1/2) = p(k), where we have used the fact that since X is integer valued, k ±1/2 is a continuity point of FX . (b) The continuity points of FX are the noninteger values of x. For such x > 0, suppose k < x < k + 1. Then k

k

i=0

i=0

FXn (x) = P(Xn ≤ x) = P(Xn ≤ k) = ∑ pn (i) → ∑ p(i) = P(X ≤ k) = FX (x). 22. Let   n k n! pk (1 − pn )n−k p (1 − pn )n−k = pn (k) := P(Xn = k) = k!(n − k)! n k n and p(k) := P(X = k) = λ k e−λ /k!. Next, by Stirling’s formula, as n → ∞, √ 2π nn+1/2 e−n (n − k)! qn := → 1, → 1 and rn (k) := √ n! 2π (n − k)n−k+1/2 e−(n−k) and so qn rn (k) → 1 as well. If we can show that pn (k)qn rn (k) → p(k), then lim pn (k) = lim

n→∞

n→∞

lim pn (k)qn rn (k) p(k) pn (k)qn rn (k) = n→∞ = = p(k). qn rn (k) lim qn rn (k) 1 n→∞

Now write lim pn (k)qn rn (k) = lim pn (k)

n→∞

n→∞

=

√

(n − k)! 2π nn+1/2 e−n ·√ n! 2π (n − k)n−k+1/2 e−(n−k)

e−k nn+1/2 lim pkn (1 − pn )n−k · k! n→∞ (n − k)n−k+1/2

244

Chapter 14 Problem Solutions

= = = =

nn+1/2 e−k lim pkn (1 − pn )n−k · n−k+1/2 k! n→∞ n (1 − k/n)n−k+1/2

(npn )k e−k (1 − pn )n−k lim · k! n→∞ (1 − k/n)−k+1/2 (1 − k/n)n

[1 − (npn )/n]n (1 − pn )−k e−k (npn )k · lim k! n→∞ (1 − k/n)−k+1/2 (1 − k/n)n

λ k e−λ e−k λ k e−λ · 1 · · −k = = p(k). k! 1 k! e

Note that since npn → λ , pn → 0, and so (1 − pn )−k → 1. 23. First write   npn (z − 1) n . GXn (z) = [(1 − pn ) + pn z]n = [1 + pn (z − 1)]n = 1 + n Since npn (z − 1) → λ (z − 1), GXn (z) → eλ (z−1) = GX (z). 24.

(a) Here are the sketches:

I(−∞,a] (t) 1 a

b

I(−∞,b] (t) 1

ga,b (t) 1 t

a

b

t

a

b

t

(b) From part (a) we can write I(−∞,a] (Y ) ≤ ga,b (Y ) ≤ I(−∞,b] (Y ), from which it follows that E[I(−∞,a] (Y )] ≤ E[ga,b (Y )] ≤ E[I(−∞,b] (Y )], | {z } {z } | = P(Y ≤a)

or

= P(Y ≤b)

FY (a) ≤ E[ga,b (Y )] ≤ FY (b). (c) Since FXn (x) ≤ E[gx,x+∆x (Xn )], lim FXn (x) ≤ lim E[gx,x+∆x (Xn )] = E[gx,x+∆x (X)] ≤ FX (x + ∆x).

n→∞

n→∞

(d) Similarly, FX (x − ∆x) ≤ E[gx−∆x,x (X)] = lim E[gx−∆x,x (Xn )] ≤ lim FXn (x). n→∞

n→∞

(e) If x is a continuity point of FX , then given any ε > 0, for sufficiently small ∆x, FX (x) − ε < FX (x − ∆x) and

FX (x + ∆x) < FX (x) + ε .

Chapter 14 Problem Solutions

245

Combining this with parts (c) and (d), we obtain FX (x) − ε ≤ lim FXn (x) ≤ lim FXn (x) < FX (x) + ε . n→∞

n→∞

Since ε > 0 is arbitrary, the liminf and the limsup are equal, limn→∞ FXn (x) exists and is equal to FX (x). 25. If Xn converges in distribution to zero, then Xn converges in probability to zero, and by Problem 10,   |Xn | E = 0. 1 + |Xn | Conversely, if the above limit holds, then by Problem 10, Xn converges in probability to zero, which implies convergence in distribution to zero. 26. Observe that fn (x) = n f (nx) implies   1, x > 0, F(0), x = 0, Fn (x) = F(nx) →  0, x < 0.

So, for x 6= 0, Fn (x) → I[0,∞) (x), which is the cdf of X ≡ 0. In other words, Xn converges in distribution to zero, which implies convergence in probability to zero. 27. Since Xn converges in mean square to X, Xn converges in distribution to X. Since 2 g(x) := x2 e−x is a bounded continuous function, E[g(Xn )] → E[g(X)]. 28. Let 0 < δ < 1 be given. For large t, we have |u(t) − 1| < δ , or −δ < u(t) − 1 < δ

or

1 − δ < u(t) < 1 + δ .

Hence, for z > 0, P(Zt ≤ z(1 − δ )) ≤ P(Zt ≤ zu(t)) ≤ P(Zt ≤ z(1 + δ )). Rewrite this as FZt (z(1 − δ )) ≤ P(Zt ≤ zu(t)) ≤ FZt (z(1 + δ )). Then F(z(1 − δ )) = lim FZt (z(1 − δ )) ≤ lim P(Zt ≤ zu(t)) t→∞

t→∞

and lim P(Zt ≤ zu(t)) ≤ lim FZt (z(1 + δ )) = FZ (z(1 + δ )).

t→∞

t→∞

Since 0 < δ < 1 is arbitrary, and since F is continuous, we must have lim P(Zt ≤ zu(t)) = F(z).

t→∞

The case for z < 0 is similar.

246

Chapter 14 Problem Solutions

29. Rewrite P(c(t)Zt ≤ z) as P(Zt ≤ z/c(t)) = P(Zt ≤ (z/c)(c/c(t)). Then if we put u(t) := c/c(t), we have by the preceding problem that P(c(t)Zt ≤ z) → F(z/c). 30. First write FXt (x) = P(Zt + s(t) ≤ x) = P(Zt ≤ x − s(t)). Let ε > 0 be given. Then s(t) → 0 implies that for large t, |s(t)| < ε , or −ε < s(t) < ε

− ε < −s(t) < ε .

or

Then FZt (x − ε ) = P(Zt ≤ x − ε ) ≤ FXt (x) ≤ P(Zt ≤ x + ε ) = FZt (x + ε ). It follows that F(x − ε ) ≤ lim FXt (x) ≤ lim FXt (x) ≤ F(x + ε ). t→∞

t→∞

Since F is continuous and ε > 0 is arbitrary, lim FXt (x) = F(x).

t→∞

31. Starting with Nbtc ≤ Nt ≤ Ndte , it is easy to see that Nbtc

t Xt := p

−λ

λ /t

Ndte

t ≤ Yt ≤ p

−λ

λ /t

=: Zt .

According to Problem 13, it suffices to show that Xt and Zt converge in distribution to N(0, 1) random variables. By the preceding two problems, the distribution limit of Zt is the same as that of c(t)Zt + s(t) if c(t) → 1 and s(t) → 0. We take  q t dte c(t) := q  = t dte → 1 t dte and

 
λ t dte t dte − 1 q λ √ dte → 0. = λ s(t) := q  − q  λ λ dte λ dte

Finally, observe that

Ndte

−λ dte c(t)Zt + s(t) = q  , λ dte

goes through the values of Yn and therefore converges in distribution to an N(0, 1) random variable. It is similar to show that the distribution limit of Xt is also N(0, 1). 32.

(a) Let G := {Xn → X}. For ω ∈ G, 1 1 → . 1 + Xn (ω )2 1 + X(ω )2 Since P(G c ) = 0, 1/(1 + Xn2 ) converges almost surely to 1/(1 + X 2 ).

Chapter 14 Problem Solutions

247

(b) Since almost sure convergence implies convergence in probability, which implies convergence in distribution, we have 1/(1 + Xn2 ) converging in distribution to 1/(1 + X 2 ). Since g(x) = 1/(1 + x2 ) is bounded and continuous, E[g(Xn )] → E[g(X)]; i.e.,     1 1 lim E = E . n→∞ 1 + Xn2 1 + X2 33. Let GX := {Xn → X} and GY := {Yn → Y }, where P(GXc ) = P(GYc ) = 0. Let G := {g(Xn ,Yn ) → g(X,Y )}. We must show that P(G c ) = 0. Our plan is to show that GX ∩ GY ⊂ G. It follows that G c ⊂ GXc ∪ GYc , and we can then write P(G c ) ≤ P(GXc ) + P(GYc ) = 0. For ω ∈ GX ∩ GY , (Xn (ω ),Yn (ω )) → (X(ω ),Y (ω )). Since g(x, y) is continuous, for such ω , g(Xn (ω ),Yn (ω )) → g(X(ω ),Y (ω )). Thus, GX ∩ GY ⊂ G.

c ) = 0. Put G := 34. Let GX := {Xn → X} and GXY := {X = Y }, where P(GXc ) = P(GXY Y c {Xn → Y }. We must show that P(GY ) = 0. Our plan is to show that GX ∩ GXY ⊂ GY . c , and we can then write It follows that GYc ⊂ GXc ∪ GXY c P(GYc ) ≤ P(GXc ) + P(GXY ) = 0.

For ω ∈ GX ∩ GXY , Xn (ω ) → X(ω ) and X(ω ) = Y (ω ). Hence, for such ω , Xn (ω ) → Y (ω ); i.e., ω ∈ GY . Thus, GX ∩ GXY ⊂ GY .

35. Let GX := {Xn → X} and GY := {Xn → Y }, where P(GXc ) = P(GYc ) = 0. Put GXY := c ) = 0. Our plan is to show that G ∩ G ⊂ G . {X = Y }. We must show that P(GXY X Y XY c c c It follows that GXY ⊂ GX ∪ GY , and we can then write c P(GXY ) ≤ P(GXc ) + P(GYc ) = 0.

For ω ∈ GX ∩ GY , Xn (ω ) → X(ω ) and Xn (ω ) → Y (ω ). Since limits of sequences of numbers are unique, for such ω , X(ω ) = Y (ω ); i.e., ω ∈ GXY . Thus, GX ∩ GY ⊂ GXY . T

c 36. Let GX := {Xn → X}, GY := {Yn → Y }, and GI := ∞ n=1 {Xn ≤ Yn }, where P(GX ) = c c P(GY ) = P(GI ) = 0. This last equality follows because

P(GIc ) ≤

∞

∑ P(Xn > Yn )

= 0.

n=1

Let G := {X ≤ Y }. We must show that P(G c ) = 0. Our plan is to show that GX ∩ GY ∩ GI ⊂ G. It follows that G c ⊂ GXc ∪ GYc ∪ GIc , and we can then write P(G c ) ≤ P(GXc ) + P(GYc ) + P(GIc ) = 0. For ω ∈ GX ∩ GY ∩ GI , Xn (ω ) → X(ω ), Yn (ω ) → Y (ω ), and for all n, Xn (ω ) ≤ Yn (ω ). By properties of sequences of real numbers, for such ω , we must have X(ω ) ≤ Y (ω ). Thus, GX ∩ GY ∩ GI ⊂ G. 37. Suppose Xn converges almost surely to X, and Xn converges in mean to Y . Then Xn converges in probability to X and to Y . By Problem 5, X = Y a.s.

248

Chapter 14 Problem Solutions

38. If X(ω ) > 0, then cn X(ω ) → ∞. If X(ω ) < 0, then cn X(ω ) → −∞. If X(ω ) = 0, then cn X(ω ) = 0 → 0. Hence,   +∞, if X(ω ) > 0, 0, if X(ω ) = 0, Y (ω ) =  −∞, if X(ω ) < 0. 39. By a limit property of probability, we can write  ∞ ∞   ∞  \ [ [ P {Xn = j} X0 = i = lim P {Xn = j} X0 = i M→∞ N=1 n=N

n=M

∞

∑ P(Xn = j|X0 = i) M→∞

≤ lim

∞

∑ M→∞

= lim

n=M

(n)

pi j ,

n=M

(n)

which must be zero since the pi j are summable. 40. Following the hint, write  ∞  \ E[S] P(S = ∞) = P {S > n} = lim P(S > N) ≤ lim = 0. N→∞ N→∞ N n=1 41. Since Sn converges in mean to S, by the problems referred to in the hint, n

n

∞

k=1

k=1

k=1

lim ∑ |hk |E[|Xk |] ≤ B1/p ∑ |hk | < ∞. ∑ E[|hk ||Xk |] = n→∞ n→∞

E[S] = lim E[Sn ] = lim n→∞

42. From the inequality Z n+1 1

t

n

dt ≥ p

Z n+1 n

1 1 dt = , p (n + 1) (n + 1) p

we obtain ∞ >

Z ∞ 1 1

which implies

t

∞

dt = p

p ∑∞ n=1 1/n

∑

Z n+1 1

n=1 n

t

dt ≥ p

∞

1 ∑ (n + 1) p = n=1

∞

1

∑ np ,

n=2

< ∞.

43. Write  n  n  n  n  1 E X X X X i j m l ∑ ∑ ∑ ∑ n4 i=1 j=1 m=1 l=1 n n n n n 1 = 4 ∑ E[Xi4 ] + ∑ ∑ E[Xi Xi Xl Xl ] + ∑ ∑ E[Xi X j Xi X j ] n i=1 i=1 l=1,l6=i i=1 j=1, j6=i  n n n n 3 + ∑ ∑ E[Xi X j X j Xi ] + 4 ∑ ∑ E[Xi X j ] i=1 j=1, j6=i i=1 j=1, j6=i | {z }

E[Mn4 ] =

=0

1 = 4 [nγ + 3n(n − 1)σ 4 ]. n

Chapter 14 Problem Solutions

249

44. It suffices to show that ∑∞ n=1 P(|Xn | ≥ ε ) < ∞. To this end, write P(|Xn | ≥ ε ) = 2 Then

Z ∞ ε

∞

∑ P(|Xn | ≥ ε ) =

n=1

n −nx dx 2e

∞

∞ = −e−nx = e−nε . ε

e−ε < ∞. 1 − e−ε

∑ (e−ε )n =

n=1

45. First use the Rayleigh cdf to write P(|Xn | ≥ ε ) = P(Xn ≥ ε ) = e−(nε )

2 /2

.

Then ∞

∑ P(|Xn | ≥ ε )

∞

=

n=1

2 2

∑ e−ε n /2

n=1

∞

≤

2

∑ (e−ε /2 )n

=

n=1

1 < ∞. 2 1 − e−ε /2

Thus, Xn converges almost surely to zero. 46. For n > 1/ε , write Z ∞ p − 1 −p P(|Xn | ≥ ε ) = x dx = p−1 ε

Then

∞

∑ P(|Xn | ≥ ε )

n>1

=

∑

n<1/ε

n

P(|Xn | ≥ ε ) +

1 . n p−1 ε p−1

1

ε p−1

∞

1 < ∞ p−1 n n>1/ε

∑

since p − 1 > 1. Thus, Xn converges almost surely to zero. 47. To begin, first observe that E[|Xn − X|] = E[|n2 (−1)nYn |] = n2 E[Yn ] = n2 pn . Also observe that P(|Xn − X| ≥ ε ) = P(n2Yn ≥ ε ) = P(Yn ≥ ε /n2 ) = P(Yn = 1) = pn . In order to have Xn converge almost surely to X, it is sufficient to consider pn such that ∑∞ n=1 pn < ∞. 2 1/2 6→ 0. For this choice of p , (a) If pn = 1/n3/2 , then ∑∞ n n=1 pn < ∞, but n pn = n Xn converges almost surely but not in mean to X. 2 (b) If pn = 1/n3 , then ∑∞ n=1 pn < ∞ and n pn = 1/n → 0. For this choice of pn , Xn converges almost surely and in mean to X.

48. To apply the weak law of large numbers of this section to (1/n) ∑ni=1 Xi2 requires only that the Xi2 be i.i.d. and have finite mean; there is no second-moment requirement on Xi2 (which would be a requirement on Xi4 ).

250

Chapter 14 Problem Solutions

49. By writing 1 n Nn = ∑ Nk − Nk−1 , n n k=1 which is a sum of i.i.d. Poisson(λ ) random variables, we see that Nn /n converges almost surely to λ by the strong law of large numbers. We next observe that Nbtc ≤ Nt ≤ Ndte . Then Nbtc Ndte Nt ≤ ≤ , t t t and it follows that Nbtc Ndte Nt ≤ ≤ , dte t btc and then

Ndte dte Nt btc Nbtc · ≤ . ≤ · dte btc t dte btc |{z} |{z} |{z} |{z} →1

→λ

→λ

→1

Hence Nt /t converges almost surely to λ . 50.

(a) By the strong law of large numbers, for ω not in a set of probability zero, 1 n ∑ Xk (ω ) → µ . n k=1 Hence, for ε > 0, for all sufficiently large n, n 1 ∑ Xk (ω ) − µ < ε , n k=1 which implies

1 n ∑ Xk (ω ) − µ < ε , n k=1

from which it follows that n

∑ Xk (ω )

< n(µ + ε ).

k=1

(b) Given M > 0, let ε > 0 and choose n in part (a) so that both n(µ + ε ) > M and n ≥ M hold. Then n

Tn (ω ) =

∑ Xk (ω )

< n(µ + ε ).

k=1

Now, for t > n(µ + ε ) ≥ Tn (ω ), ∞

Nt (ω ) =

∑ I[0,t] (Tk (ω ))

k=1

≥ n ≥ M.

Chapter 14 Problem Solutions

251

(c) As noted in the solution of part (a), the strong law of large numbers implies Tn 1 n = ∑ Xk → µ n n k=1

a.s.

Hence, n/Tn → 1/µ a.s.

(d) First observe that

Nt Nt ≥ , TNt t

YNt = and so

Nt 1 = lim YNt ≥ lim . t→∞ t→∞ t µ

Next, YNt +1 = and so

Nt 1 Nt + 1 Nt + 1 = + , ≤ TNt +1 t t t

1 Nt + 0. = lim YNt +1 ≤ lim µ t→∞ t→∞ t

Hence, lim

t→∞

Nt 1 = . t µ

51. Let Xn := nI(0,1/n] (U), where U ∼ uniform(0, 1]. Then for every ω , U(ω ) ∈ (0, 1]. For n > 1/U(ω ), or U(ω ) > 1/n, Xn (ω ) = 0. Thus, Xn (ω ) → 0 for every ω . However, E[Xn ] = nP(U ≤ 1/n) = 1. Hence, Xn does not converge in mean to zero. 52. Suppose Dε contains n points, say a ≤ x1 < · · · < xn ≤ b. Then n

nε <

∑ G(xk +) − G(xk −)

k=1

n−1

n

= G(xn +) + ∑ G(xk +) − ∑ G(xk −) − G(x1 −) k=1 n−1

k=2

n

≤ G(b) + ∑ G(xk+1 ) − ∑ G(xk−1 ) − G(a) k=1 n−1

k=2

= G(b) + ∑ [G(xk+1 ) − G(xk )] − G(a) k=1

≤ G(b) + G(xn ) − G(x1 ) − G(a) ≤ 2[G(b) − G(a)]. 53. Write

   ∞  ∞ [ [ P U∈ Kn = P {U ∈ Kn } ≤ n=1

n=1

∞

∞

∑ P(U ∈ Kn )

n=1

≤

Since ε > 0 is arbitrary, the probability in question must be zero.

2ε

∑ 2n

n=1

= 2ε .

CHAPTER 15

Problem Solutions 1. Using the formula FWt (x) = FW1 (t −H x), we see that  FW1 (∞) = 1, x > 0, lim FWt (x) = FW1 (−∞) = 0, x < 0, t↓0 which is the cdf of the zero random variable for x 6= 0. Hence, Wt converges in distribution to the zero random variable. Next, X(ω ) := lim t H W1 (ω ) = t→∞

Thus, P(X = ∞) = P(W1 > 0) = 1 − FW1 (0),

 

∞, if W1 (ω ) > 0, 0, if W1 (ω ) = 0,  −∞, if W1 (ω ) < 0. P(X = −∞) = P(W1 < 0) = FW1 (0−),

and P(X = 0) = P(W1 = 0) = FW1 (0) − FW1 (0−). 2. Since the joint characteristic function of a zero-mean Gaussian process is completely determined by the covariance matrix, we simply observe that E[Wλ t1 Wλ t2 ] = σ 2 min(λ t1 , λ t2 ) = λ σ 2 min(t1 ,t2 ), and E[(λ 1/2Wt1 )(λ 1/2Wt2 )] = λ σ 2 min(t1 ,t2 ). 3. Fix τ > 0, and consider the process Zt := Wt − Wt−τ . Since the Wiener process is Gaussian with zero mean, so is the process Zt . Hence, it suffices to consider the covariance E[Zt1 Zt2 ] = E[(Wt1 −Wt1 −τ )(Wt2 −Wt2 −τ )]. The time intervals involved do not overlap if t1 < t2 − τ or if t2 < t1 − τ . Hence,  0, |t2 − t1 | > τ , E[Zt1 Zt2 ] = σ 2 (|t2 − t1 | + τ ), |t2 − t1 | ≤ τ , which depends on t1 and t2 only through their difference. 4. For H = 1/2, qH (θ ) = I(0,∞) (θ ), and CH = 1. So, BH (t) − BH (s) =

Z ∞

−∞

[I(−∞,t) (τ ) − I(−∞,s) (τ )] dWτ = Wt −Ws . 252

Chapter 15 Problem Solutions

253

5. It suffices to show that Z ∞ 1

[(1 + θ )H−1/2 − θ H−1/2 ]2 d θ < ∞.

Consider the function f (t) := t H−1/2 . By the mean-value theorem of calculus, (1 + θ )H−1/2 − θ H−1/2 = f 0 (tˆ),

for some tˆ ∈ (θ , θ + 1).

Since f 0 (t) = (H − 1/2)t H−3/2 ,

(1 + θ )H−1/2 − θ H−1/2 ≤ |H − 1/2|/θ 3/2−H .

Then Z ∞ 1

[(1 + θ )H−1/2 − θ H−1/2 ]2 d θ ≤ (H − 1/2)2

Z ∞ 1

1

θ 3−2H

d θ < ∞,

since 3 − 2H > 1. 6. The expression

  M − µ n P ≤ y = 1−α 1−H σ /n

says that

yσ with probability 1 − α . n1−H Hence, the width of the confidence interval is

µ = Mn ±

2yσ 2yσ = √ · nH−1/2 . 1−H n n 7. Suppose E[Yk2 ] = σ 2 k2H for k = 1, . . . , n. Substituting this into the required formula yields

2 E[Yn+1 ] − σ 2 n2H − σ 2 n2H − σ 2 (n − 1)2H = 2C(n), which we are assuming is equal to   σ 2 (n + 1)2H − 2n2H + (n − 1)2H .

2 ] = σ 2 (n + 1)2H . It follows that E[Yn+1

8.

(a) Clearly,

is zero mean. Also,

(m) Xeν :=

 (m) (m)  E Xeν Xen =

νm

∑

νm

∑

(Xk − µ )

k=(ν −1)m+1

nm

∑

k=(ν −1)m+1 l=(n−1)m+1

C(k − l)

254

Chapter 15 Problem Solutions νm

∑

=

m

∑ C(k − [i + (n − 1)m])

k=(ν −1)m+1 i=1 m m

∑ ∑ C([ j + (ν − 1)m] − [i + (n − 1)m])

=

i=1 j=1 m m

∑ ∑ C( j − i + m(ν − n)).

=

i=1 j=1 (m)

Thus, Xeν

is WSS.

(b) From the solution of part (a), we see that Ce(m) (0) =

m

m

∑ ∑ C( j − i)

i=1 j=1

= mC(0) + 2

m−1

∑ C(k)(m − k)

k=1 2 2H

= E[Ym2 ] = σ m

,

by Problem 7.

Thus, C(m) (0) = σ 2 m2H /m2 = σ 2 m2H−2 . 9. Starting with  σ∞2  C(m) (n) |n + 1|2H − 2|n|2H + |n − 1|2H , = m→∞ m2H−2 2 lim

put n = 0 to get

C(m) (0) = σ∞2 . m→∞ m2H−2 lim

Then observe that (m)

E[(X1 − µ )2 ] C(m) (0) = = m2H−2 m2H−2

2   m 1 E ∑ (Xk − µ ) m k=1 m2H−2

.

10. Following the hint, observe that

2 2 2 2 Ce(m) (n) 1 E[Y(n+1)m ] − E[Ynm ] − E[Ynm ] − E[Y(n−1)m ] = m2H 2 m2H  2 2 2 E[Y(n+1)m ] E[Y(n−1)m ]  1 2H E[Ynm ] 2H (n + 1)2H − 2n + (n − 1) = 2 [(n + 1)m]2H (nm)2H [(n − 1)m]2H →

σ∞2 [(n + 1)2H − 2n2H + (n − 1)2H ]. 2

11. If the equation cited in the text holds, then in particular, Ce(m) (0) = σ∞2 , m→∞ m2H−2 lim

Chapter 15 Problem Solutions

255

and Ce(m) (n)/m2H−2 Ce(m) (n) = lim m→∞ m→∞ C m→∞ C e(m) (0) e(m) (0)/m2H−2 1 = [|n + 1|2H − 2|n|2H + |n − 1|2H ]. 2 Conversely, if both conditions hold, lim ρ (m) (n) = lim

Ce(m) (0)ρ (m) (n) Ce(m) (n) = lim 2H−2 m→∞ m→∞ m m2H−2 (m) Ce (0) = lim 2H−2 lim ρ (m) (n) m→∞ m m→∞ 2 σ = ∞ [|n + 1|2H − 2|n|2H + |n − 1|2H ]. 2 lim

12. To begin, write e jπ f − e− jπ f −2d − jπ f jπ f − jπ f −2d − j2π f −2d = 2 j = e [e − e ] S( f ) = 1 − e 2j −2d   −d = 2 sin(π f ) = 4 sin2 (π f ) .

Then

C(n) = =

Z 1/2  −1/2

−d 4 sin2 (π f ) e j2π f n d f

Z 1/2  −1/2

= 2

−d 4 sin2 (π f ) cos(2π f n) d f

Z 1/2  0

Z π

−d cos(2π f n) d f 4 sin2 (π f )

−d dν cos(nν ) 4 sin2 (ν /2) 2π 0 Z −d 1 π 2 = cos(nν ) d ν . 4 sin (ν /2) π 0

= 2

Next, as suggested in the hint, apply the change of variable θ = 2π − ν to 1 π to get

Z 2π  π

−d cos(nν ) d ν 4 sin2 (ν /2)

Z −d 1 π 4 sin2 ([2π − θ ]/2) cos(n[2π − θ ]) d θ , π 0 which, using a trigonometric identity, is equal to (∗). Thus, Z −d 1 2π  cos(nν ) d ν 4 sin2 (ν /2) 2π 0 Z −d 1 π 4 sin2 (t) cos(2nt) dt. = π 0

C(n) =

(∗)

256

Chapter 15 Problem Solutions By the formula provided in the hint, cos(nπ )Γ(2 − 2d)2 p−1 21−p (1 − 2d)Γ((2 − 2d + 2n)/2)Γ((2 − 2d − 2n)/2) (−1)n Γ(1 − 2d) . = Γ(1 − d + n)Γ(1 − d − n)

C(n) =

13. Following hint (i), let u = sin2 θ and dv = θ α −3 d θ . Then du = 2 sin θ cos θ d θ and v = θ α −2 /(α − 2). Hence, Z r

θ α −3 sin2 θ d θ =

ε

Next, 1 2−α

Z r ε

θ

Now write Z 2r 2ε

α −2

Z r 1 θ α −2 sin2 θ r − θ α −2 sin 2θ d θ . α − 2 ε α − 2 ε Z

2r 1 dt (t/2)α −2 sint sin 2θ d θ = 2 − α 2ε 2 Z 2r α −1 (1/2) = t α −2 sint dt 2−α 2ε 2r   Z 2r 21−α t α −1 1 sint + = t α −1 cost dt . 2−α α −1 1 − α 2ε 2ε

t α −1 cost dt = Re

Z 2r 2ε

t α −1 e− jt dt → Re e− jαπ /2 Γ(α ) = cos(απ /2)Γ(α )

as ε → 0 and r → ∞ by hint (iii). To obtain the complete result, observe that

θ α −2 sin2 θ = θ α

 sin θ 2

θ

and t α −1 sint = t α

sint t

both tend to zero as their arguments tend to zero or to infinity. 14.

(a) First observe that ∞

Q(− f ) = Hence,

and then

1 ∑ |i − f |2H+1 = i=1

−1

1 ∑ | − l − f |2H+1 = l=−∞

h S( f ) = Q(− f ) +

1 | f |2H+1

−1

1 . 2H+1 i=−∞ |i + f |

∑

i + Q( f ) sin2 (π f ),

S( f ) sin2 (π f ) 2H−1 2 = | f | [Q(− f ) + Q( f )] sin ( → π 2. π f ) + | f |1−2H f2

Chapter 15 Problem Solutions

257

(b) We have Z 1/2

−1/2

S( f ) d f = σ 2 = π 2 · =

4 cos([1 − H]π )Γ(2 − 2H) (2π )2−2H (2H − 1)2H

(2π )2H cos(π H)Γ(2 − 2H) . 2H(1 − 2H)

15. Write (−1)n Γ(1 − 2d) Γ(n + 1 − d)Γ(1 − d − n) (−1)n Γ(1 − 2d) = Γ(n + 1 − d)(−1)n Γ(d)Γ(1 − d)/Γ(n + d) Γ(n + d) Γ(1 − 2d) · . = Γ(1 − d)Γ(d) Γ(n + 1 − d)

C(n) =

Now, with ε = 1 − d, observe that Γ(n + 1 − ε ) (n + 1 − ε )n+1−ε −1/2 e−(n+1−ε ) ∼ Γ(n + ε ) (n + ε )n+ε −1/2 e−(n+ε ) = e1−2d

nn+1/2−ε [1 + (1 − ε )/n)]n+1/2−ε nn+ε −1/2 (1 + ε /n)n+ε −1/2

= e1−2d n2d−1

[1 + (1 − ε )/n)]n+1/2−ε . (1 + ε /n)n+ε −1/2

Thus, 1−2d Γ(n + 1 − ε )

n

Γ(n + ε )

1−2d

→ e


1/2−ε e1−ε 1−2d 1/2−ε = e1/2−d . e
ε −1/2 = e eε

Thus, α = 1 − 2d and c = e1/2−d Γ(1 − 2d)/[Γ(1 − d)Γ(d)]. 16. Evaluating the integral, we have In n1−α

=

n1−α − k1−α 1 − (k/n)1−α 1 = → . (1 − α )n1−α 1−α 1−α

Now, given a small ε > 0, for large n, In /n1−α > 1/(1 − α ) − ε , which implies In /n−α > n(1/(1 − α ) − ε ) → ∞. With Bn as in the hint, we have from Bn + n−α − k−α ≤ In ≤ Bn that 1 In Bn Bn k−α + ≤ 1−α ≤ 1−α + α α 1− 1− n n n n n

or

In 1− n α

Thus, Bn /n1−α → 1/(1 − α ).

≤

Bn In 1 k−α ≤ 1−α − − 1−α . α 1− n n n n

258

Chapter 15 Problem Solutions

17. We begin with the inequality n−1

∑ ν 1−α ≤

ν =k

| {z }

Z n

n−1

t 1−α dt ≤ ∑ (ν + 1)1−α . k ν =k | {z } | {z } =: In

=: Bn

= Bn −k1−α +n1−α

Next, since In = (n2−α − k2−α )/(2 − α ), In n2−α

1 − (k/n)2−α 1 → . 2−α 2−α

=

Also, In n2−α

≤

and

Bn k1−α 1 − 2−α + 2− α n n n

In Bn ≤ 2−α n2−α n

imply Bn /n2−α → 1/(2 − α ) as required. 18. Observe that ∑ |hn | = ∑ ∞

∞

n=q

n

∑

n=q k=n−q ∞ ∞

≤

=

αk bn−k

∑ ∑ |αk ||bn−k |I[n−q,n] (k)

n=q k=0 ∞

∞

∑ |αk | ∑ |bn−k |I[0,q] (n − k)

k=0

≤ M





n=q

 ∞ |b | ∑ i ∑ (1 + δ /2)−k q

i=0 q

 = M ∑ |bi | i=0

k=0

1 < ∞. 1 − 1/(1 + δ /2)

19. Following the hint, we first compute   ∑ αm−nYn = ∑ αm−n ∑ ak Xn−k = n

n

=

k

∑ Xl ∑ αm−n an−l l

n

=

∑ αm−n ∑ an−l Xl n

l

∑ Xl ∑ αm−(ν +l) aν , l

ν

|

{z

= δ (m−l)

}

where the reduction to the impulse follows because the convolution of αn and an corresponds in the z-transform domain to the product [1/A(z)] · A(z) = 1, and 1 is the transform of the unit impulse. Thus, ∑n αm−nYn = Xm .

Chapter 15 Problem Solutions

259

Next,

∑ αm−n

∑ αm−nYn =

n

n



∑ bk Zn−k k

∑ Zl ∑ αm−n bn−l

=

n

l



=

∑ αm−n ∑ bn−l Zl

=

n

l

∑ Zl ∑ αm−(l+k) bk l

k

|

{z

= hm−l

}

since this last convolution corresponds in the z-transform domain to the product [1/A(z)] · B(z) =: H(z).

20. Since Xn is WSS, E[|Xn |2 ] is a finite constant. Since ∑∞ k=0 |hk | < ∞, we have by an example in Chapter 13 or by Problem 26 in Chapter 13 that ∑m k=0 hk Xn−k converges in mean square as m → ∞. By another example in Chapter 13, ∞  ∞ E[Yn ] = E ∑ hk Xn−k = ∑ hk E[Xn−k ]. k=0

k=0

If E[Xn−k ] = µ , then

∞

E[Yn ] =

∑ hk µ

k=0

is finite and does not depend on n. Next, by the continuity of the inner product (Problem 24 in Chapter 13),   ∞    m  m E Xl ∑ hk Xn−k = lim E Xl ∑ hk Xn−k = lim ∑ hk E[Xl Xn−k ] m→∞

k=0

m→∞

k=0

= lim

m→∞

∑ hk RX (l − n + k)

k=0

∞

m

=

k=0

∑ hk RX ([l − n] + k).

k=0

Similarly, E[YnYl ] = E



= lim

m→∞ ∞

=

   m   ∑ hk Xn−k Yl = lim E ∑ hk Xn−k Yl ∞

m→∞

k=0 m

∑ hk E[Xn−kYl ]

= lim

k=0

∑ hk RXY ([n − l] − k).

k=0

Thus, Xn and Yn are J-WSS.

k=0

m

m→∞

∑ hk RXY (n − k − l)

k=0