Renewal Processes and Repairable Systems
Renewal Processes and Repairable Systems
Proefschrift
ter verkrijging van de graad van doctor aan de Technische Universiteit Delft, op gezag van de Rector Magnificus prof. dr. ir. J. T. Fokkema, voorzitter van het College voor Promoties, in het openbaar te verdedigen op mandaag 3 februari 2003 om 16.00 uur
door
Suyono
Magister Sains Matematika Universitas Gadjah Mada, Yogyakarta, Indonesi¨e
geboren te Purworejo, Indonesi¨e
Dit proefschrift is goedgekeurd door de promotor: Prof. dr. F. M. Dekking Samenstelling promotiecommissie: Rector Magnificus, Prof. dr. F. M. Dekking, Prof. dr. R. M. Cooke, Prof. dr. M. Iosifescu, Prof. dr. C. L. Scheffer, Prof. dr. R. K. Sembiring, Prof. dr. H. J. van Zuylen, Dr. J. A. M. van der Weide,
voorzitter Technische Universiteit Delft, promotor Technische Universiteit Delft Centre for Mathematical Statistics, Roemeni¨e Technische Universiteit Delft (emeritus) Institut Teknologi Bandung, Indonesi¨e Technische Universiteit Delft Technische Universiteit Delft
The research in this thesis has been carried out under the auspices of the Thomas Stieltjes Institute for Mathematics, at the University of Technology in Delft.
Published and distributed by : DUP Science DUP Science is an imprint of Delft University Press P.O. Box 98 2600 MG Delft The Netherlands Telephone: +31 15 27 85 678 Telefax: +31 15 27 85 706 E-mail:
[email protected] ISBN 90-407-xxx-x Keywords: Poisson point processes, renewal processes, repairable systems c 2002 by Suyono Copyright ° All right reserved. No part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without written permission from the publisher: Delft University Press. Printed in The Netherlands
To my wife Demonti, my son Afif, and my daughters Hana and Nadifa.
Acknowledgments I am very grateful to my supervisor Dr. J. A. M. van der Weide, who introduced me interesting topics to research, for his creative guidance, his idea, his encouragement, and his patience in guiding me doing the research. I also would like to express my gratitude to my promotor Prof. dr. F. M. Dekking not only for giving me the opportunity to carry out my Ph.D research at the Delft University of Technology, but also for his constructive comments. In addition, I wish to thank all members of the Department of Control, Risk, Optimization, Stochastic and Systems (CROSS) for their hospitality and their assistances, especially to Cindy and Diana for their administrative assistance, and to Carl for his computer assistance. Also, I wish to thank Durk Jellema and Rene Tamboer for the arrangement of almost everything I needed during my research and stay in the Netherlands. I am pleased to acknowledge my colleagues at the Department of Mathematics, State University of Jakarta. Their encouragement and support have been invaluable. In this occasion I also would like to thank all of my friends in the Netherlands for their support and their help. This work would not have been possible without the support of a cooperative project between Indonesian and Dutch governments. In this connection I would like to express my appreciation to Prof. dr. R. K. Sembiring and Dr. A. H. P. van der Burgh who organized a research workshop four years ago in Bandung, to Dr. O. Simbolon and Dr. B. Karyadi, former project managers of Proyek Pengembangan Guru Sekolah Menengah (Secondary School Teachers Development Project), and to H. P. S. Althuis, director of Center for International Cooperation in Applied Technology (CICAT). Finally, I would like to thank my dearest wife and our beloved children. Saya menyampaikan banyak terima kasih kepada istriku Dra. Demonti Siswari atas pengorbananya yang begitu besar yang dengan rela dan ikhlas ditinggal menuntut ilmu di Belanda, atas doanya yang senantiasa dipanjatkan kepadaNya sehingga saya alhamdulillah banyak mendapatkan jalan keluar ketika dalam kesulitan, atas motivasinya yang selalu membuat saya kembali bersemangat dikala kehilangan semangat, atas kesabarannya yang dengannya alhamdulillah kita banyak vii
viii
mendapat pertolongan dan kebaikan dari Allah, dan atas jerih payah serta usahanya mengurus dan mendidik anak-anak yang diamanahkan kepada kita sehingga insyaAllah menjadi qurata a’yun bagi kita. Juga untuk anak-anakku, Hana Firdaus, Muhamad Afif Abdurrahim, dan Nadifa Mumtaz, ayah sampaikan banyak terima kasih atas pengertian dan doa kalian. Pengorbanan kalian sungguh sangat besar untuk keberhasilan ayah. Kalian menjalani masa kecil tanpa memperoleh perhatian dan kasih sayang yang sepantasnya dari ayah. Meskipun masih kecil kalian sudah harus ikut membantu perjuangan yang ayah dan ibu lakukan. Semoga ini semua menjadi latihan dan bekal yang berharga bagi kalian untuk menjalani kehidupan di masa yang akan datang. Amiin. Tidak lupa untuk ayah, ibu, adik, mertua dan saudara-saudaraku semua, saya sampaikan terima kasih atas dorongan, doa, dan bantuannya yang telah diberikan padaku dan keluargaku khususnya saat saya tinggalkan anak-anak dan istriku untuk menempuh program Ph.D. selama kurang lebih 4 tahun di Belanda.
Delft, February 3, 2003
Suyono
Contents Acknowledgments
vii
1 Introduction
1
1.1
Related works . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2.1
Point processes . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2.2
Renewal processes . . . . . . . . . . . . . . . . . . . . . .
9
Outline of the thesis . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.3
2 Renewal Reward Processes
13
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
Instantaneous reward processes . . . . . . . . . . . . . . . . . . .
14
2.2.1
A civil engineering example . . . . . . . . . . . . . . . . .
20
2.3
Renewal reward processes . . . . . . . . . . . . . . . . . . . . . .
25
2.4
Asymptotic properties . . . . . . . . . . . . . . . . . . . . . . . .
28
2.5
Covariance structure of renewal processes . . . . . . . . . . . . .
33
2.6
System reliability in a stress-strength model . . . . . . . . . . . .
35
2.6.1
Type I models . . . . . . . . . . . . . . . . . . . . . . . .
36
2.6.2
Type II models . . . . . . . . . . . . . . . . . . . . . . . .
40
3 Integrated Renewal Processes
13
45
3.1
Notations and Definitions . . . . . . . . . . . . . . . . . . . . . .
45
3.2
(N (t)) a Poisson or Cox process . . . . . . . . . . . . . . . . . . .
46
3.3
(N (t)) a renewal process . . . . . . . . . . . . . . . . . . . . . . .
50
3.4
Asymptotic properties . . . . . . . . . . . . . . . . . . . . . . . .
57
3.5
An application . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
ix
x
4 Total Downtime of Repairable Systems 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . 4.2 Distribution of total downtime . . . . . . . . . . 4.3 System availability . . . . . . . . . . . . . . . . . 4.4 Covariance of total downtime . . . . . . . . . . . 4.5 Asymptotic properties . . . . . . . . . . . . . . . 4.6 Examples . . . . . . . . . . . . . . . . . . . . . . 4.7 Systems consisting of n independent components 4.7.1 Exponential failure and repair times . . . 4.7.2 Arbitrary failure or repair times . . . . . A The proof of Theorem 2.5.1
CONTENTS
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
65 65 66 73 76 78 83 88 88 93 95
B Numerical inversions of Laplace transforms 105 B.1 Single Laplace transform . . . . . . . . . . . . . . . . . . . . . . . 105 B.2 Double Laplace transform . . . . . . . . . . . . . . . . . . . . . . 108 Bibliography
113
Samenvatting
117
Summary
119
Ringkasan
121
Curriculum Vitae
123
Chapter 1
Introduction Many situations in our daily life can be modelled as renewal processes. Two examples are arrivals of claims in an insurance company and arrivals of passengers at a train station, if the inter-arrival times between consecutive claims or passengers are assumed to be independent and identically distributed (iid) non-negative random variables. In these examples the number of claims or passengers arrived during the time interval [0, t], t ≥ 0, is usually taken as a formal definition of a renewal process. A mathematical definition of a renewal process is given in Subsection 1.2.2. From renewal processes we may construct other stochastic processes. Firstly, consider the renewal-process example of the insurance claims. Consider the claim sizes as random variables and assume that they are independent and identically distributed. If we interpret the claim size as a reward, then the total amount of all claims during the time interval [0, t] is an example of another process which is known as a renewal reward process. In this thesis we study renewal reward processes in Chapter 2. In renewal reward processes it is usually assumed that the reward is earned all at once at the end of the ’inter-arrival times’ of the corresponding renewal process. It means that only the rewards earned until the last renewal before time t are considered. The reward may also depend on the inter-arrival time. Motivated by an application in the study of traffic, see Subsection 2.2.1, it is interesting to consider a version of renewal reward processes where the reward is a function of an inter-arrival time length, instantaneously earned, and the reward earned in an incomplete time interval is also taken into account. We call this version in this thesis an instantaneous reward process. Now consider the second example of renewal processes about the arrivals of passengers at a train station. Suppose that a train just departed at time 0, and there were no passengers left. We are interested in the waiting time from
1
2
Introduction
when passengers arrive (after time 0) until the departure of the next train at some time t ≥ 0. One of the quantities of interest is the total waiting time of all passengers during the time interval [0, t]. This is an example of a class of stochastic processes which we will call an integrated renewal process. We will study integrated renewal processes in Chapter 3. Integrated renewal processes have a connection with shot noise processes. In the shot noise processes usually it is assumed that shocks occur according to a Poisson process. Then associated with the ith shock, which occurred at time Si > 0, i ≥ 1, is a random variable Xi which represents the ’value’ of that shock. The values of the shocks are assumed to be iid and independent of their arrival process. The value of the ith shock at time t ≥ Si equals Xi ψ(t − Si ), where ψ(x) ≥ 0 for x ≥ 0, and ψ(x) = 0 otherwise. Then the total value of all shocks at time t is a shot noise process. As we will see in Chapter 3, if we take Xi ≡ 1 for all i ≥ 1 and ψ(x) = x1[0,∞) (x), then the corresponding shot noise process is an ’integrated Poisson process’. The next topic that we will study in this thesis is the total downtime of repairable systems. We first consider a system, regarded as a single component, that can be in two states, either up (in operation) or down (under repair). We suppose that the system starts to operate at time 0. After the first failure the system is repaired, and then functions like the new system. Similarly after the second failure the system is repaired, and functions again as good as the new one, and so on. We will assume that the alternating sequence of up and down times form a so called alternating renewal process. One of the interesting quantities to consider is the total downtime of the system during the time interval [0, t]. Note that the total downtime can be considered as a reward process on an alternating renewal process. The total downtime is important because it can be used as a performance measure of the system. We will study the total downtime in Chapter 4. In this chapter we also consider repairable systems comprising n ≥ 2 independent components. Expressions for the marginal distributions of the renewal reward processes (including the instantaneous reward processes), the integrated renewal processes, and the total downtime of repairable systems, in a finite time interval [0, t], are derived in this thesis. Our approach is based on the theory of point processes, especially Poisson point processes. The idea is to represent the processes that we study (including the total downtime of repairable systems) as functionals of Poisson point processes. Important tools we will use are the Palm formula, and the Laplace functional of a Poisson point process. Usually we obtain the marginal distributions of the processes in the form of Laplace transforms. Asymptotic properties of the processes that we study are also investigated. We use Tauberian theorems to derive asymptotic properties of the expected value of the processes from their Laplace-transform expressions. Other asymptotic properties of the processes like asymptotic variances and asymptotic dis-
Related works
3
tributions are studied. The rest of this chapter is organized as follows. In the next section we give an overview of the literature which has a connection with the study of the renewal and instantaneous reward processes, the integrated renewal processes and the total downtime of repairable systems. We also explain the position of our contributions. In Section 1.2 we introduce some basic notions and facts about point processes and summarize some facts about renewal processes which will be used in the subsequent chapters. Finally in Section 1.3 we give the outline of this thesis.
1.1
Related works
Renewal reward processes have been discussed by many authors. Several results are known in the literature. For example, the renewal-reward theorem and its expected-value version, and the asymptotic normality of the processes can be found in Wolff [49] and Tijms [46]. These asymptotic properties are frequently used in applications of these processes, see for example Csenki [6], Popova and Wu [31], Parlar and Perry [29], and Herz et al. [20]. Herz et al. [20] modelled the maximal number of cars that can safely cross the traffic stream on a one-way road during the time interval [0, t] as a renewal reward process. For large t the renewal-reward theorem and its expected-value version have been used. In this application it is interesting to consider the case when t is small. So we need to know the distribution properties of renewal reward processes (and also instantaneous reward processes) in every finite time interval [0, t]. Several authors have discussed this case. An integral equation for the expected value of renewal reward processes was studied by Jack [23]. Mi [25] gave bounds for the average reward over a finite time interval. Erhardsson [12] studied an approximation of a stationary renewal reward process in a finite time interval by a compound Poisson distribution. In this thesis we derive expressions for the marginal distributions of renewal and instantaneous reward processes in a finite time interval. We give an application to the study of traffic. We also reconsider asymptotic properties of renewal reward processes. A proof of the expected-value version of the renewal-reward theorem by means of a Tauberian theorem is given. Asymptotic normality of instantaneous reward processes is proved. Another result that we derive is the covariance structure of a renewal process, which is a special case of renewal reward processes. We also study a topic about system reliability in a stressstrength model which is closely related to renewal reward processes. It seems that the terminology ’integrated renewal process’ has not been known yet in the literature. But its special case where the corresponding renewal process is a homogeneous Poisson process has been considered by several authors. In Example 2.3(A) of Ross [36] the expected value of an ’integrated
4
Introduction
homogeneous Poisson process’ has been calculated. The probability density function (pdf) of an integrated homogeneous Poisson process can be found in Gubner [19]. In this thesis we derive the marginal distributions of integrated renewal processes. We also consider other natural generalizations of the integrated homogeneous Poisson process, where the homogeneous Poisson process is generalized into the non-homogeneous one and a Cox process (doubly stochastic Poisson process). Asymptotic properties of the integrated Poisson and renewal processes are also investigated. The distribution of the total downtime of a repairable system has been widely discussed in a number of papers. An expression for the cumulative distribution function (cdf) of the total downtime up to a given time t has been derived by several authors using different methods. In Tak´acs [44] the total probability theorem has been used. The derivation in Muth [26] is based on consideration of the excess time. Finally in Funaki and Yoshimoto [13] the cdf of the total downtime is derived by a conditioning technique. Srinivasan et al. [37] derived an expression for the pdf of the total uptime, which has an obvious relation to the total downtime, of a repairable system. They also discussed the covariance structure of the total uptime. For longer time intervals, Tak´acs [44] and R´enyi [33] proved that the limiting distribution of the total downtime is a normal distribution. Tak´ acs [44] also discussed asymptotic mean and variance of the total downtime. In all these papers it is assumed that the failure times and the repair times are independent. Usually it is also assumed that the failure times are iid, and the same for the repair times. An exception is in Tak´acs [45], where the iid assumption has been dropped, but still under the assumption of independence between the failure times and repair times. In his paper [45] Tak´acs discussed some possibilities of asymptotic distributions of the total downtime. In this thesis we use a different method for the computation of the distribution of the total downtime. We also consider a more general situation where we allow dependence of the failure time and the repair time. Some asymptotic properties, which are generalizations of the results of Tak´acs [44] and R´enyi [33], of the total downtime are derived. We also discuss the total downtime of repairable systems consisting of n ≥ 2 independent component.
1.2
Basic notions
1.2.1
Point processes
A point process is a random distribution of points in some space E. In this thesis we will assume that E is a locally compact Hausdorff space with a countable base. As an example E is a subset of an Euclidean space of finite dimension. These assumptions ensure the existence of a Poisson point process in an infinite space, among others, which will be used in the next chapters.
Basic notions
5
The concept of a point process is formally described as follows. Let E be the Borel σ-algebra of subsets of E, i.e., the σ-algebra generated by the open sets. For x ∈ E, define the measure δx on (E, E) by ½ 1, x ∈ A δx (A) := 0, x ∈ /A for A ∈ E. The measure δx is called the Dirac measure in x. A point measure on (E, E) is a Radon measure µ, a non-negative measure µ with property µ(K) < ∞ for every compact set K, which has a representation: X δxi , µ= i∈I
where I is a countable index set and xi ∈ E. A point measure µ is called simple if µ({x}) ≤ 1 for all x ∈ E. Designate by Mp (E) be the space of all point measures on (E, E). Let Mp (E) be the smallest σ-algebra making all evaluation maps m ∈ Mp (E) 7→ m(A) ∈ N0 ,
A ∈ E,
measurable, where N0 = N0 ∪ {∞} and N0 denotes the set of all non-negative integers. Definition 1.2.1 A point process on E is a measurable map from a probability space (Ω, F, P) into the measurable space (Mp (E), Mp (E)). So if N is a point process on E and ω ∈ Ω then N (ω) is a point measure on (E, E). The probability distribution PN of the point process N is the measure P ◦ N −1 on (Mp (E), Mp (E)). The following proposition says that a map N from Ω into Mp (E) is a point process on E if and only if N (A) is an extended non-negative-valued random variable for each A ∈ E. Proposition 1.2.1 [34] Let N be a map from a probability space (Ω, F, P) into the space (Mp (E), Mp (E)). The map N is a point process if and only if for every A ∈ E the map ω 7→ N (ω, A) from (Ω, F, P) into (N0 , P(N0 )) is measurable, where P(N0 ) denotes the power set of N0 . The intensity measure of a point process N is the measure ν on (E, E) defined, for A ∈ E, by ν(A)
:= = =
E[N (A)] Z N (ω, A)P(dω) ZΩ µ(A)PN (dµ). Mp (E)
6
Introduction
Example 1.2.1 LetP(Xn , n ≥ 1) be an iid sequence of non-negative random n variables. Let Sn = i=1 Xi . Then N :=
∞ X
δ Sn
n=1
is a point process on [0, ∞). The stochastic process (N (t), t ≥ 0) where N (t) = N ([0, t]), the number of points in the interval [0, t], is called a renewal process. In this context the interval [0, ∞) is usually referred to as the interval of time. The random variable Xn is called the nth inter-arrival time or nth cycle of the renewal process. The intensity measure of the renewal process N (t) is given by ν(dt) = dm(t), where m(t) is the renewal function, see Subsection 1.2.2. Let f be a non-negative measurable function defined on E. Recall that there exist simple functions fn with 0 ≤ fn ↑ f and fn is of the form fn =
kn X
(n)
Ai
i
i=1 (n)
(n)
ci 1A(n) ,
∈ E,
(n)
where ci are constants and Ai , i ≤ kn , are disjoint. Define for the function f and ω ∈ Ω Z N (f )(ω) := f (x)N (ω, dx). E
This is a random variable since by the monotone convergence theorem N (f )(ω) = lim N (fn )(ω) n→∞
and each N (fn )(ω) =
kn X
(n)
(n)
ci N (ω, Ai )
i=1
is a random variable. The Laplace functional of a point process N is defined as the function ψN which takes non-negative measurable functions f on E into [0, ∞) by £ ¤ ψN (f ) := E exp{−N (f )} Z © ª exp − N (f )(ω) P(dω) = ½ Z ¾ ZΩ = exp − f (x)µ(dx) PN (dµ). Mp (E)
E
Proposition 1.2.2 [34] The Laplace functional ψN of a point process N uniquely determines the probability distribution PN of N .
Basic notions
7
Poisson point processes One of the most important examples of point processes is a Poisson point process. Definition 1.2.2 Given a Radon measure µ on (E, E), a point process N on E is called a Poisson point process on E with intensity measure µ if N satisfies (a) For any A ∈ E, and any non-negative integer k ( −µ(A) e µ(A)k if µ(A) < ∞ k! P(N (A) = k) = 0 if µ(A) = ∞, (b) If A1 , ..., Ak are disjoint sets in E then N (A1 ), ..., N (Ak ) are independent random variables. Example 1.2.2 Let E = [0, ∞) and the intensity measure µ satisfies µ([0, t]) = λt for some positive constant λ and for any t ≥ 0. Then the Poisson point process N is called a Rhomogeneous Poisson point process on E with intensity or t rate λ. If µ([0, t]) = 0 λ(x)dx where λ(x) is a non-negative function of x, then the Poisson process N is called a non-homogeneous Poisson process on E with intensity or rate function λ(x). The following theorem concerning the Laplace functional of a Poisson point process will be used in the subsequent chapters. Theorem 1.2.1 [34] The Laplace functional of a Poisson point process N on E with intensity measure µ is given by ½ Z ¾ −f (x) ψN (f ) = exp − (1 − e )µ(dx) . E
The next theorem says that a renewal process with inter-arrival times iid exponential random variables is a homogeneous Poisson process. Theorem 1.2.2 [32] Let (Xn , n ≥ 1) be iid sequence of exponential random Pan P∞ n variables with parameter 1. Let Sn = i=1 Xi and set N = n=1 δSn . Then N is a homogeneous Poisson process on [0, ∞) with rate 1. Starting from a Poisson point process we may construct a new Poisson point process whose points live in a higher dimensional space. Theorem 1.2.3 [35] Let Ei , i = 1, 2 be two locally compact Hausdorff spaces with countable bases. Suppose (Xn , n ≥ 1) are random elements of E1 such that ∞ X n=1
δXn
8
Introduction
is Poisson point process on E1 with intensity measure µ. Suppose (Yn , n ≥ 1) are iid random elements of E2 with common probability distribution Q and suppose the Poisson point process and the sequence (Yn ) are defined on the same probability space and are independent. Then the point process on E1 × E2 ∞ X
δ(Xn ,Yn )
n=1
is Poisson point process with intensity measure µ × Q, where µ × Q(A1 × A2 ) = µ(A1 )Q(A2 ) for Ai measurable subset of Ei , i = 1, 2. Palm distribution Palm distribution plays an important role in the study of point processes. A Palm distribution is defined as a Radon-Nikodym derivative. Let N be a point process on E with distribution PN such that ν = E[N ] a Radon measure. Let B ∈ Mp (E) and let 1B : Mp (E) 7→ {0, 1} be the indicator function, i.e., ½ 1, µ ∈ B 1B (µ) = . 0, µ ∈ /B Consider the measure E[1B (N )N ] which is absolutely continuous with respect to ν. By the Radon-Nikodym theorem there exists a unique almost surely (ν) function Px (B) : E 7→ [0, 1] such that Z Z Px (B)ν(dx) = 1B (µ)µ(A)PN (dµ) A
Mp (E)
for all A ∈ E. The family {Px (B) : x ∈ E, B ∈ Mp (E)} can be chosen so that Px is a probability measure on Mp (E) for all x ∈ E and so that for each B ∈ Mp (E) the function x 7→ Px (B) is measurable, see Grandell [17]. The measure Px is called the Palm distribution (or Palm measure) of the point process N . Let N be a Poisson point process with intensity measure ν and distribution Pν . Theorem 1.2.4 [17] For every non-negative measurable function f on E × Mp (E), Z Z Z Z f (x, µ)µ(dx)Pν (dµ) = f (x, µ + δx )Pν (dµ)ν(dx). (1.1) Mp (E)
E
E
Mp (E)
The equation (1.1) is known as the Palm formula for a Poisson point process. We will frequently use this formula later. The Palm distribution for the Poisson
Basic notions
9
point process N can be obtain by taking f (x, µ) = 1A (x)1B (µ), A ∈ E, B ∈ Mp (E), which gives Px = ∆x ∗ Pν where ∆x ∗ Pν denotes the convolution between ∆x , the Dirac measure in δx , and Pν , i.e., Z Z 1B (µ1 + µ2 )∆x (dµ1 )Pν (dµ2 ) ∆x ∗ Pν (B) := Mp (E)
Z
Mp (E)
1B (µ + δx )Pν (dµ).
= Mp (E)
1.2.2
Renewal processes
In the previous section we saw that a renewal process on the non-negative real line is an example of point processes. In this section we summarize some facts about renewal processes including delayed renewal processes which will be used in the subsequent chapters. Equivalently, the notion of the renewal process in Example 1.2.1 can be stated as follows. Definition 1.2.3 Let (Xi , i ≥ 1) be an iid sequence of non-negative random variables. A renewal process (N (t), t ≥ 0) is a process such that N (t) = sup{n ≥ 0 : Sn ≤ t} where Sn =
n X
Xi ,
n≥1
and
S0 = 0.
i=1
The renewal process (N (t)) represents the number of occurrences of some event in the time interval [0, t]. We commonly interpret Xn as the time between the (n − 1)st and nth event and call it the nth inter-arrival time or nth cycle and interpret Sn as the time of nth event or the time of nth arrival. The link between N (t) and the sum Sn of iid random variables is given by N (t) ≥ n if and only if
Sn ≤ t.
(1.2)
The distribution of N (t) can be represented in terms of the cdfs of the interarrival times. Let F be the cdf of X1 and Fn be the cdf of Sn . Note that Fn is the n-fold convolution of F with itself. From (1.2) we obtain P(N (t) = n) = P(N (t) ≥ n) − P(N (t) ≥ n + 1) = P(Sn ≤ t) − P(Sn+1 ≤ t) = Fn (t) − Fn+1 (t).
(1.3)
10
Introduction
Let m(t) be the expected number of events in the time interval [0, t], i.e., m(t) = E[N (t)]. The function m(t) is called a renewal function. The relationship between m(t) and F is given by the following proposition. Proposition 1.2.3 [18] m(t) =
∞ X
Fn (t).
(1.4)
n=1
The renewal function m(t) also satisfies the following integral equation. Proposition 1.2.4 [18] The renewal function m satisfies the renewal equation Z t m(t) = F (t) + m(t − x)dF (x). (1.5) 0
Let µ = E(X1 ) and σ 2 = Var(X1 ) be the mean and the variance of X1 , and assume that µ and σ are strictly positif. Some asymptotic properties of the renewal process (N (t)) are given in the following. Theorem 1.2.5 [18] Assume that µ < ∞. Then, (a) lim
t→∞
N (t) 1 = t µ
with probability 1,
(b) lim
t→∞
m(t) 1 = , t µ
(1.6)
(c) t σ 2 − µ2 + + o(1) (1.7) µ 2µ2 where o(1) denotes a function of t tending to zero as t → ∞, provided σ 2 is finite. m(t) =
Theorem 1.2.6 [7] Assume that σ 2 < ∞. Then Var[N (t)] σ2 = 3. t→∞ t µ lim
If µ3 = E(X13 ) < ∞, then Var[N (t)] =
σ2 t + µ3
µ
1 5σ 4 2µ3 + 4− 3 12 4µ 3µ
(1.8) ¶ + o(1).
For large t the distribution of N (t) is approximately normal, i.e., N (t) − t/µ d p −→ N (0, 1) as t → ∞. tσ 2 /µ3
(1.9)
Basic notions
11
Delayed renewal processes Let X0 , X1 , X2 , . . . be a sequence of non-negative independent random variables. Let G be the cdf of X0 and F be the common cdfs of Xi , i ≥ 1. Let Sn =
n−1 X
Xi ,
n≥1
and
S0 = 0.
i=0
The stochastic process (ND (t), t ≥ 0) where ND (t) = sup{n ≥ 0 : Sn ≤ t} is called a delayed renewal process. It is easy to see that P(ND (t) = n) = G ∗ Fn−1 (t) − G ∗ Fn (t), where F ∗ G represents the convolution of F and G. The corresponding delayed renewal function satisfies mD (t) := E[ND (t)] =
∞ X
G ∗ Fn−1 (t).
n=1
As for ordinary renewal processes, the delayed renewal processes have the following asymptotic properties: (a) The delayed renewal function mD (t) is asymptotically a linear function of t, i.e., lim
t→∞
mD (t) t
1 , µ
=
(1.10)
where µ = E(X1 ), see Ross [36], (b) If σ 2 = Var(X1 ) < ∞, µ0 = E(X0 ) < ∞ and F is a non-lattice distribution, then mD (t) =
t σ 2 + µ2 µ0 + − + o(1) 2 µ 2µ µ
(1.11)
and if σ 2 < ∞ then Var[ND (t)] t→∞ t lim
see Tak´acs [44].
=
σ2 , µ3
(1.12)
12
1.3
Introduction
Outline of the thesis
The subsequent chapters in this thesis can be read independently, and are mostly based on the results in Suyono and van der Weide [38, 39, 40, 41, 42, 43]. Firstly, in Chapter 2 we discuss renewal reward processes. In Section 2.2 we study a version of renewal reward processes, which we call an instantaneous reward process. In Section 2.3 we consider the marginal distribution of renewal reward processes in a finite time interval. Section 2.4 deals with asymptotic properties of the renewal and instantaneous reward processes. The covariance structure of renewal processes is studied in Section 2.5. Finally Section 2.6 is devoted to a study of system reliability in a stress-strength model, where the amplitude of a stress occurring at a time t can be considered as a reward. Chapter 3 deals with integrated renewal processes. In Section 3.2 we consider the marginal distributions of integrated Poisson and Cox processes. In the next section we consider the marginal distributions of integrated renewal processes. Asymptotic properties of the integrated renewal processes are studied in Section 3.4. Finally in the last section we give an application. In Chapter 4 we discuss the total downtime of repairable systems. We start with investigating the distribution of the total downtime in Section 4.2. Section 4.3 is devoted to the study of system availability, which is closely related to the total downtime. Section 4.4 concerns the covariance of the total downtime. Asymptotic properties of the total downtime for the dependent case is studied in Section 4.5. Two examples are given in the next section. Finally in Section 4.7 we consider the total downtime of repairable systems consisting of n ≥ 2 independent components. Most of the results about the marginal distributions of the processes that we study are in the form of Laplace transforms. In some cases the transforms can be inverted analytically, but mostly the transforms have to be inverted numerically. We give numerical inversions of Laplace transforms in Appendix B.
Chapter 2
Renewal Reward Processes 2.1
Introduction
Consider Pnan iid sequence (Xn , n ≥ 1) of strictly positive random variables. Let Sn = i=1 Xi , n ≥ 1, S0 = 0, and (N (t), t ≥ 0) be the renewal process with renewal cycles Xn . Associated with each cycle length Xn is a reward Yn where we assume that ((Xn , Yn ), n ≥ 1) is an iid sequence of random vectors. The variables Xn and Yn can be dependent. The stochastic process (R(t), t ≥ 0) where N (t)
R(t) =
X
Yn
(2.1)
n=1
(with the usual convention that the empty sum equals 0) is called a renewal reward process. Taking Yn ≡ 1, we see that renewal processes can be considered as renewal reward processes. Motivated by an application in the study of traffic, see Subsection 2.2.1, it is interesting to consider a version of renewal reward processes where the reward is a function of cycle length, i.e, N (t)
Rφ (t) =
X
φ(Xn ) + φ(t − SN (t) )
(2.2)
n=1
where φ is a measurable real-valued function. We will call the process (Rφ (t), t ≥ 0) an instantaneous reward process. We will assume that rewards are nonnegative, i.e., the function φ is a non-negative. Note that in this process we also consider the reward earned in the incomplete renewal cycle (SN (t) , t]. If we only consider the reward until the last renewal before time t, and take Yn = φ(Xn ) then (2.2) reduces to (2.1). 13
14
Renewal Reward Processes
In this chapter we will study these renewal and instantaneous reward processes. Firstly, in Section 2.2, we consider the marginal distribution of the instantaneous reward process defined in (2.2). An application of this process to the study of traffic is given. We give an example where the variables Xn represent the time intervals between consecutive cars on a crossing in a one-way road, and φ(x) represents the number of cars that can cross the traffic stream safely during a time interval of length x between two consecutive cars. In Section 2.3 we consider the marginal distributions of the renewal reward process given by formula (2.1). We will only consider the case where the random variables Yn are non-negative. Asymptotic properties of the renewal and instantaneous reward processes are discussed in Section 2.4. We give an alternative proof for the expected-value version of the renewal-reward theorem using a Tauberian theorem. Section 2.5 deals with the covariance structure of renewal processes. The last section is devoted to the study of system reliability in a stress-strength model, where the amplitude of a stress occurring at a time t can be considered as a reward. Besides considering renewal processes as the occurrence of the stresses, in this last section we also model the occurrence of the stresses as a Cox process (doubly stochastic Poisson process). We will denote the cdfs of the random variables X1 and Y1 by F and G respectively, and denote the joint cdf of X1 and Y1 by H, i.e., H(x, y) = P(X1 ≤ x, Y1 ≤ y). The Laplace-Stieltjes transforms of F and H will be denoted by F ∗ and H ∗ , i.e., Z ∞ F ∗ (β) = e−βx dF (x) 0
and
∞Z
Z H ∗ (α, β) = 0
2.2
∞
e−(αx+βy) dH(x, y).
0
Instantaneous reward processes
In this section we will consider the marginal distributions of the instantaneous reward process as defined in (2.2). We will use the theory of point processes introduced in Chapter 1. Let (Ω, F, P) be the probability space on which the iid sequence (Xn , n ≥ 1) is defined and also an iid sequence (Un , n ≥ 1) of exponentially distributed random variables with parameter 1 such that the sequences (Xn ) and (Un ) are independent. Let (Tn , n ≥ 1) be the sequence of partial sums of the variables Un . By Theorem 1.2.2 ∞ X n=1
δTn
Instantaneous reward processes
15
is a Poisson point process on [0, ∞) with intensity measure ν(dx) = dx, and by Theorem 1.2.3 the map Φ :
ω
7→
P∞
n=1 δ(Tn (ω),Xn (ω))
defines a Poisson point process on E = [0, ∞) × [0, ∞) with intensity measure ν(dtdx) = dtdF (x), where F is the cdf of X1 . Note that for almost all ω ∈ Ω Φ(ω) is a simple point measure on E satisfying Φ(ω)({t} × [0, ∞)) ∈ {0, 1} for every t ≥ 0. Note also that ν([0, t] × [0, ∞)) < ∞ for t ≥ 0. Let Mp (E) be the set of all point measures on E. We will denote the distribution of Φ by Pν , i.e., Pν = P ◦ Φ−1 . Define for t ≥ 0 the functional A(t) on Mp (E) by Z A(t)(µ) = 1[0,t) (s)xµ(dsdx). E
In the sequel we will write A(t, µ) = A(t)(µ). Suppose that the point measure µ ∞ has the support supp(µ) = ((tn , xn ))n=1 with t1 < t2 < . . ., see Figure 2.1 (a). It follows that x6
A(t, µ)6
¾ x3
(t3 , x3 ) (t1 , x1 ) r r (t2 , x2 ) r
¾ x2
¾ x1
t
0
0
t1
t2
(a)
t3 (b)
Figure 2.1: (a). Illustration of supp(µ).
µ=
∞ X
(b). Graph of A(t, µ)
δ(tn ,xn )
n=1
and A(t, µ) can be expressed as A(t, µ) =
∞ X n=1
1[0,t) (tn )xn .
t
16
Renewal Reward Processes
Note that for every t ≥ 0, A(t, µ) is finite almost surely. Figure 2.1 (b) shows the graph of a realization of A(t, µ). For (tn , xn ) ∈ supp(µ) 1[0,xn ) (t − A(tn , µ)) = 1 ⇐⇒ A(tn , µ) ≤ t < A(tn , µ) + xn ⇐⇒ x1 + . . . + xn−1 ≤ t < x1 + . . . + xn . Hence for a measurable, bounded function f on E we have Z 1[0,x) (t − A(s, µ))f (s, x)µ(dsdx) = f (tn , xn ) E
where n is chosen such that x1 + . . . + xn−1 ≤ t < x1 + . . . + xn . Now define for t ≥ 0 the functional Rφ (t) on Mp (E) by Z © ª Rφ (t)(µ) = 1[0,x) (t − A(s, µ)) µ(1[0,s) ⊗ φ) + φ(t − A(s, µ)) µ(dsdx) E
where Z µ(1[0,t) ⊗ φ) =
E
1[0,t) (s)φ(x)µ(dsdx).
Note that if µ=
∞ X
δ(tn ,xn )
n=1
with t1 < t2 < . . ., then Rφ (t)(µ) ) (∞ ∞ ∞ ´ ³ X X X = 1[0,xn ) (t − A(tn , µ)) 1[0,tn ) (ti )φ(xi ) + φ t − 1[0,tn ) (ti )xi n=1
=
n−1 X i=1
i=1 n−1 ³ X ´ φ(xi ) + φ t − xi
i=1
(2.3)
i=1
where n satisfies x1 +. . .+xn−1 ≤ t < x1 +. . .+xn . Then we have a representation for the instantaneous reward process (Rφ (t)) as a functional of the Poisson point process Φ as stated in the following lemma. Lemma 2.2.1 With probability 1, Rφ (t) = Rφ (t)(Φ).
Instantaneous reward processes
17
Proof: Let ω ∈ Ω. Since Φ(ω) =
Rφ (t)(Φ(ω)) =
P∞
n=1 δ(Tn (ω),Xn (ω) ,
i−1 X
then using (2.3) we obtain
φ(Xn (ω)) + φ(t − Si−1 (ω))
n=1
where i satisfies Si−1 (ω) ≤ t < Si (ω). But this condition holds if and only if i = N (t, ω) + 1, where N (t) = sup{n ≥ 0 : Sn ≤ t}, which completes the proof. 2 The following theorem gives the formula for the Laplace transform of the marginal distribution of the instantaneous reward process (Rφ (t), t ≥ 0).
Theorem 2.2.1 Let (Xn , n ≥ 1) be an iid sequence of strictly positive random variables with common cdf F . Let (Sn , n ≥ 0) be the sequence of partial sums of the variables Xn and (N (t), t ≥ 0) be the corresponding renewal process: N (t) = sup{n ≥ 0 : Sn ≤ t}. Let φ : [0, ∞) → [0, ∞) be a measurable function. Let (Rφ (t), t ≥ 0) be the instantaneous reward process defined in (2.2). Then for α, β > 0 Z
R∞
∞
E(e
−αRφ (t)
)e
−βt
dt =
0
[1 − F (t)]e−βt−αφ(t) dt R∞ . 1 − 0 e−βt−αφ(t) dF (t) 0
(2.4)
Proof: By Lemma 2.2.1 Z E(e−αRφ (t) ) =
e−αRφ (t)(µ) Pν (dµ) Mp (E)
½
Z =
exp Mp (E)
=
Z −α E
1[0,x) (t − A(s, µ))
¾ h i µ(1[0,s) ⊗ φ) + φ(t − A(s, µ)) µ(dsdx) Pν (dµ) Z Z 1[0,x) (t − A(s, µ)) Mp (E)
½
exp
E
i¾ − α µ(1[0,s) ⊗ φ) + φ(t − A(s, µ)) µ(dsdx)Pν (dµ). h
Applying the Palm formula for Poisson point processes, see Theorem 1.2.4, we
18
Renewal Reward Processes
obtain E(e−αRφ (t) ) Z ∞Z = 0
½
exp
∞
Z
0
Mp (E)
1[0,x) (t − A(s, µ + δ(s,x) )) i¾
h
− α (µ + δ(s,x) )(1[0,s) ⊗ φ) + φ(t − A(s, µ + δ(s,x) ))
Pν (dµ)dF (x)ds Z ∞Z ∞Z = 0
½
exp
0
Mp (E)
1[0,x) (t − A(s, µ))
h i¾ − α µ(1[0,s) ⊗ φ) + φ(t − A(s, µ)) Pν (dµ)dF (x)ds.
Using Fubini’s theorem and a substitution we obtain Z
∞
0
E(e−αRφ (t) )e−βt dt Z Z ∞Z ∞Z
∞
= 0
=
½
0
1[0,x) (t − A(s, µ))
0
Mp (E)
h i¾ exp − α µ(1[0,s) ⊗ φ) + φ(t − A(s, µ)) e−βt dtPν (dµ)dF (x)ds ½ h Z ∞Z ∞Z i¾ exp − βA(s, µ) + αµ(1[0,s) ⊗ φ) Pν (dµ) 0
·Z
0
Mp (E)
n
x
exp
o ¸ − βt − αφ(t) dt dF (x)ds.
0
The integral with respect to Pν can be calculated as follows. Note that Z βA(s, µ) + αµ(1[0,s) ⊗ φ) =
E
1[0,s) (r)[βu + αφ(u)]µ(drdu).
So we can apply the formula for the Laplace functional of Poisson point processes, see Theorem 1.2.1, to obtain ½
Z exp Mp (E)
= =
h i¾ − βA(s, µ) + αµ(1[0,s) ⊗ φ) Pν (dµ)
½
¾ h i 1 − e−1[0,s) (r)[βu+αφ(u)] dF (u)dr ½ Z0 ∞ 0 ¾ exp −s [1 − e−βu−αφ(u) ]dF (u) . exp
Z
∞
−
0
Z
∞
Instantaneous reward processes
19
It follows that ½ ¾ Z ∞ Z ∞Z ∞ Z ∞ E(e−αRφ (t) )e−βt dt = exp − s [1 − e−βu−αφ(u) ]dF (u) 0 0 ·0Z x 0 n o ¸ exp − βt − αφ(t) dt dF (x)ds 0 ½ ¾ Z ∞ Z ∞ = exp − s [1 − e−βu−αφ(u) ]dF (u) ds 0 Z0 ∞ Z x n o exp − βt − αφ(t) dtdF (x) R0∞ 0 [1 − F (t)]e−βt−αφ(t) dt 0 R = . 2 ∞ 1 − 0 e−βu−αφ(u) dF (u) We can take derivatives with respect to α in (2.4) to find Laplace transforms of the moments of Rφ (t). For example the Laplace transforms of the first and second moments of Rφ (t) are given in the following proposition. Proposition 2.2.1 Suppose that the same assumptions as in the Theorem 2.2.1 hold. Assume also that the function φ(t) is continuous or piecewise continuous in every finite interval (0, T ). Then £ ¤ (a) If E φ(X1 )e−βX1 < ∞ for some β > 0 and φ(t) = o(e−γt ), γ > 0, as t → ∞, then for β > γ Z ∞ E[Rφ (t)]e−βt dt 0 R∞ R∞ φ(t)e−βt dF (t) + β 0 [1 − F (t)]e−βt φ(t)dt 0 = . (2.5) β[1 − F ∗ (β)] ¤ £ (b) If E φ2 (X1 )e−βX1 < ∞ for some β > 0 and φ(t) = o(e−γt/2 ), γ > 0, as t → ∞, then for β > γ R∞ 2 R∞ Z ∞ φ (t)e−βt dF (t) + β 0 [1 − F (t)]φ2 (t)e−βt dt 2 −βt 0 E[Rφ (t)]e dt = β[1 − F ∗ (β)] 0 R∞ R∞ −βt 2 0 φ(t)e dF (t) 0 [1 − F (t)]φ(t)e−βt dt + [1 − F ∗ (β)]2 £R ∞ ¤2 2 0 φ(t)e−βt dF (t) + . (2.6) β[1 − F ∗ (β)]2 Corollary 2.2.1 If we only consider the rewards until the last renewal before time t, then (2.2) simplifies to N (t)
Rφ (t) =
X
n=1
φ(Xn )
(2.7)
20
Renewal Reward Processes
and Z
∞
E(e−αRφ (t) )e−βt dt =
0
1 − F ∗ (β) R∞ . β[1 − 0 e−βt−αφ(t) dF (t)]
As an application of Corollary 2.2.1 consider the function φ(t) = t. In this PN (t) case Rφ (t) = SN (t) = n=1 Xn and the double Laplace transform of SN (t) is given by Z ∞ 1 − F ∗ (β) E(e−αSN (t) )e−βt dt = . β[1 − F ∗ (α + β)] 0 As another application take φ(t) ≡ 1 in (2.7). Then Rφ (t) = N (t) which is a renewal process. In this case we have Z ∞ 1 − F ∗ (β) E(e−αN (t) )e−βt dt = , (2.8) β[1 − e−α F ∗ (β)] 0 from which, using the uniqueness theorem for power series, see Bartle [3], we can derive Z ∞ 1 P[N (t) = n]e−βt dt = (2.9) [1 − F ∗ (β)]F ∗ (β)n . β 0 Also, from (2.8) we can easily deduce that Z ∞ E[N (t)]e−βt dt = 0
F ∗ (β) . β[1 − F ∗ (β)]
(2.10)
Formulae (2.9) and (2.10) are standard, see for example Grimmett and Stirzaker [18], and can be derived directly using (1.3) and (1.4) or (1.5) by taking their Laplace transforms.
2.2.1
A civil engineering example
Consider a traffic stream on a one-way road. The number of cars that can cross the stream on an intersection depends on the time intervals between consecutive cars in the traffic stream. Civil engineers usually model the traffic stream as a homogeneous Poisson process, which means that the distances between consecutive cars are assumed to be independent random variables all with the same exponential distribution, see Herz et al. [20]. The number of cars that can safely cross the traffic stream between the nth and the (n + 1)th cars of the traffic stream equals bxn+1 /ac, where xn+1 is the time distance between the two cars, a > 0 some parameter, and bxc represents the integer part of x. As a more general and more realistic model we consider a renewal process as a model for the traffic stream, i.e. the time intervals between consecutive are iid
Instantaneous reward processes
21
with some arbitrary distribution. The number of cars that can safely cross the traffic stream during the time between two consecutive cars in the traffic stream can be considered as a reward and the total number of cars that can cross the traffic stream up to time t is an instantaneous reward process. We will calculate the distribution of the maximal number of cars that can safely cross the traffic stream during the time interval [0, t]. Suppose that we have 100 synthetic data of the inter-arrival times of cars as in Table 2.1. The average of the data is equal to 5.7422. If we assume that the Table 2.1: The synthetic data of the inter-arrival times of cars. 1.2169 2.5696 2.6913 3.5028 3.9254 4.1170 4.4784 4.7814 5.1833 5.3864 5.5378 5.7809 6.2104 6.6626 6.7373 7.1100 7.9067 8.3575 9.0716 9.6002
1.3508 2.6021 2.7065 3.5474 3.9400 4.1417 4.7046 4.8284 5.2221 5.4620 5.6410 5.8106 6.2269 6.6807 6.7529 7.1933 8.0114 8.3931 9.1862 10.2193
1.5961 2.6447 2.8696 3.5577 4.0549 4.2162 4.7171 4.8364 5.2357 5.4675 5.6628 5.8397 6.3748 6.6835 6.7672 7.3344 8.0606 8.4245 9.4143 10.2391
1.6633 2.6762 3.2053 3.6191 4.0759 4.2280 4.7174 4.8691 5.3068 5.4865 5.6834 5.8755 6.6107 6.7116 6.9731 7.6249 8.2526 8.8314 9.4661 10.7850
2.5308 2.6783 3.4394 3.7724 4.1093 4.2526 4.7585 5.0278 5.3291 5.4907 5.7610 6.1123 6.6587 6.7283 7.0478 7.6311 8.3095 9.0008 9.5850 11.8890
data is exponentially distributed with parameter λ, then the estimate for λ is equal to 0.1741 (=1/5.7422). Suppose that the reward function φ is given by φ(t) = bt/2c. In this case
Z
∞
E[Rφ (t)]e−βt dt =
0
and Z
∞ 0
E(e−αRφ (t) )e−βt dt =
[λ + β][1 −
(2.11)
(β + λ)e−2(β+λ) £ ¤ β 2 1 − e−2(β+λ)
(2.12)
1 − e−2(λ+β) − λ[1 − e−2(λ+β) ]
(2.13)
e−α−2(λ+β) ]
22
Renewal Reward Processes
with λ = 0.1741. Using numerical inversions of Laplace transforms, see Appendix B, we obtain the graph of the mean of Rφ (t), see Figure 2.2 (dashed line), and the distribution of Rφ (t) for t = 10, see the first column of Table 2.2. 8
7
6
E(R(t))
5
4
3
2
1
0
−1
0
2
4
6
8
10 t
12
14
16
18
20
Figure 2.2: Graphs of the mean of Rφ (t): solid line for Gamma(1,6), dotted line for empirical distribution and dashed line for exp(0.1741).
Table 2.2: Distributions of Rφ (10) with Xn ∼ exp(0.1741), Xn ∼ Gamma(1, 6) and with F (x) = Fn (x), the empirical distribution function of the data set in Table 2.1. k 0 1 2 3 4 5
P(Rφ (10) = k); Xn ∼ exp(0.1741) 0.0001 0.0032 0.0452 0.2597 0.6040 0.0877
P(Rφ (10) = k); F (x) = Fn (x) 0.0000 0.0007 0.0026 0.1543 0.8223 0.0200
P(Rφ (10) = k); Xn ∼ Gamma(1, 6) 0.0000 0.0003 0.0013 0.1344 0.8301 0.0338
If we look at the histogram of the data, see Figure 2.3, it does not seem reasonable to assume that the data is exponentially distributed. Without assuming that the data has come from a certain family of parametric distributions we can calculate the distribution of the instantaneous reward process using the
Instantaneous reward processes
23
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
Figure 2.3: Histogram of the data set in Table 2.1. empirical distribution Fn of the data: Fn (x) =
#{Xi ≤ x : i = 1, ..., n} . n
(2.14)
Let X1:n ≤ X2:n ≤ ... ≤ Xn:n be the order statistics corresponding to Xi , i = 1, 2, ..., n, and let X0 = 0. We denote by xj:n the realizations of Xj:n . Using (2.14) we obtain Z ∞ E[Rφ (t)]e−βt dt 0 ¤ −βt Pn Pn R xk:n £ 1 −βxk:n 1 − k−1 + β k=1 xk−1:n e φ(t)dt k=1 φ(xk:n )e n n £ ¤ Pn = (. 2.15) 1 −βx k:n β 1 − n k=1 e Based on the data in Table 2.1, n
1X φ(xk:n )e−βxk:n n k=1 · 22 59 81 X X 1 X −βxk:100 = e +2 e−βxk:100 + 3 e−βxk:100 100 k=5 k=23 k=60 ¸ 96 100 X X +4 e−βxk:100 + 5 e−βxk:100 k=82
=: K1 (β)
k=97
24
Renewal Reward Processes
and β
n Z X k=1
=
·
xk:n
1−
xk−1:n
¸ k − 1 −βt e φ(t)dt n
· ¸ 1 96e−2β + 78e−4β + 41e−6β + 19e−8β + 4e−10β − K1 (β). 100
So the numerator of (2.15) equals · ¸ 1 96e−2β + 78e−4β + 41e−6β + 19e−8β + 4e−10β . 100 It follows that Z ∞ 96e−2β + 78e−4β + 41e−6β + 19e−8β + 4e−10β i h E[Rφ (t)]e−βt dt = . (2.16) P100 0 β 100 − k=1 e−βxk:100 Inverting this transform numerically we obtain the graph of the mean of Rφ (t), see Figure 2.2 (dotted line). Next we calculate the double Laplace transform of Rφ (t) using the empirical distribution of the inter-arrival times. Substituting (2.14) into (2.4) we obtain Z
∞
R∞
E(e−αRφ (t) )e−βt dt =
0
=
[1 − Fn (t)]e−αφ(t)−βt dt R∞ 1 − 0 e−αφ(t)−βt dFn (t) ¤ −αφ(t)−βt Pn R xk:n £ k−1 e dt k=1 xk−1:n 1 − n P . (2.17) n 1 − n1 k=1 e−αφ(xk:n )−βxk:n 0
Based on the data in Table 2.1, the numerator of (2.17) is equal to · ¸ k − 1 −αφ(t)−βt 1− e dt n xk−1:100 · ³ 1 = K2 (β) − 96 + 78e−(α+2β) + 41e−2(α+2β) + 19e−3(α+2β) 100β ¸ ´³ ´ −4(α+2β) −α −2β 1−e e +4e
100 Z X k=1
xk:100
and the denominator of (2.17) is equal to 100
1−
1 X −αφ(xk:100 )−βxk:100 e 100 k=1
=
K2 (β) , 100
Renewal reward processes
25
where K2 (β)
=
4 X
100 −
e−βxk:100 − e−α
k=1
−e−3α
81 X
22 X
e−βxk:100 − e−2α
k=5
e−βxk:100 − e−4α
k=60
59 X
e−βxk:100
k=23
96 X
e−βxk:100 − e−5α
k=82
100 X
e−βxk:100 .
k=97
The distribution of Rφ (t) for t = 10 can be seen in the second column of Table 2.2. The data set in Table 2.1 was generated from a Gamma(1,6) random variable which has a pdf 1 5 −x f (x; 1, 6) = x e , x ≥ 0. 120 Based on this cycle length distribution the graph of the mean of Rφ (t) can be seen in Figure 2.2 (solid line) and the distribution of Rφ (t) for t = 10 can be seen in the last column of Table 2.2. From this table we see that the KolmogorovSmirnov distance (see e.g., Dudewicz [10]) between the cdfs of Rφ (t) based on the exponential and the Gamma cycles equals 0.2183, whereas the KolmogorovSmirnov distance between the cdfs of Rφ (t) based on the empirical distribution function and the Gamma cycle equals 0.0199. So we conclude in this example that approximation for the distribution of Rφ (t) based on the use of the empirical distribution function of the data is better than the use of an exponential distribution with parameter estimated from the data.
2.3
Renewal reward processes
Consider the renewal reward process defined in (2.1), i.e., N (t)
R(t) =
X
Yn .
n=1
We assume that Y1 is a non-negative random variable. In this section we will derive an expression for the distribution of R(t) for finite t. Let (Ω, F, P) be the probability space on which the iid sequence (Xn , Yn ) of random vectors is defined and also an iid sequence (Un , n ≥ 1) of exponentially distributed random variables with parameter 1 such that the sequences (Xn , Yn ) and (Un ) are independent. Let (Tn , n ≥ 1) be the sequence of partial sums of the variables Un . Then the map Φ:
ω ∈ Ω 7→
∞ X n=1
δ(Tn (ω),Xn (ω)),Yn (ω)) ,
26
Renewal Reward Processes
where δ(x,y,z) is the Dirac measure in (x, y, z), defines a Poisson point process on E = [0, ∞) × [0, ∞) × [0, ∞) with intensity measure ν(dtdxdy) = dtdH(x, y), where H is the joint cdf of X1 and Y1 . Let Mp (E) be the set of all point measures on E. We will denote the distribution of Φ over Mp (E) by Pν . Define for t ≥ 0 the functionals AX (t) and AY (t) on Mp (E) by Z 1[0,t) (s)xµ(dsdxdy) AX (t)(µ) = E
and
Z AY (t)(µ) = E
1[0,t) (s)yµ(dsdxdy).
In the sequel we will write AX (t, µ) = AX (t)(µ) and AY (t, µ) = AY (t)(µ). If µ=
∞ X
δ(ti ,xi ,yi )
i=1
with t1 < t2 < . . ., then AX (t, µ) =
∞ X
1[0,t) (ti )xi
and AY (t, µ) =
i=1
∞ X
1[0,t) (ti )yi .
i=1
Note that with probability 1, AX (t, µ) and AY (t, µ) are finite. Define also for t ≥ 0 the functional R(t) on Mp (E) by Z R(t)(µ) = 1[0,x) (t − AX (s, µ))AY (s, µ)µ(dsdxdy). E
Then we can easily prove the following lemma: Lemma 2.3.1 With probability 1, R(t) = R(t)(Φ). The following theorem gives the formula for the distribution of R(t) in the form of double Laplace transform. Theorem 2.3.1 Let ((Xn , Yn ), n ≥ 1) be an iid sequence of random vectors with joint cdf H, where Xn are strictly positive and Yn are non-negative random variables. Let (N (t), t ≥ 0) be the renewal process with renewal cycles Xn . Define for t ≥ 0 N (t)
R(t) =
X
n=1
Yn .
Renewal reward processes
Then for α, β > 0
Z
∞
27
E(e−αR(t) )e−βt dt =
0
1 − F ∗ (β) . β[1 − H ∗ (β, α)]
(2.18)
Proof: By Lemma 2.3.1 E(e−αR(t) ) Z =
e−αR(t)(µ) Pν (dµ)
Mp (E)
½
Z =
exp Mp (E)
Z −α E
Z
Z =
Mp (E)
E
¾ 1[0,x) (t − AX (s, µ))AY (s, µ)µ(dsdxdy) Pν (dµ) ½
1[0,x) (t − AX (s, µ)) exp
¾ − αAY (s, µ) µ(dsdxdy)Pν (dµ).
Applying the Palm formula for Poisson point processes we obtain Z ∞Z ∞Z ∞Z E(e−αR(t) ) = 1[0,x) (t − AX (s, µ)) 0
½
exp
0
0
Mp (E)
¾ − αAY (s, µ) Pν (dµ)dH(x, y)ds.
Using Fubini’s theorem we obtain Z ∞ E(e−αR(t) )e−βt dt 0 ½ h Z ∞Z i¾ 1 = exp − βAX (s, µ) + αAY (s, µ) Pν (dµ)ds. [1 − F ∗ (β)] β 0 Mp (E) Using the Laplace functional of Poisson point processes we obtain ½ h Z i¾ exp − βAX (s, µ) + αAY (s, µ) Pν (dµ) Mp (E)
½
Z
=
exp Mp (E)
½
= =
Z −
¾ 1[0,s) (r)(βu + αv)µ(drdudv) Pν (dµ)
E ∞Z ∞
¾ h i 1 − e−1[0,s) (r)[βu+αv] dH(u, v)dr 0 0 0 ½ ¾ exp − s[1 − H ∗ (β, α)] . exp
Z
∞
Z
−
It follows that Z ∞ E(e−αR(t) )e−βt dt = 0
=
½ ¾ Z ∞ 1 ∗ ∗ exp − s[1 − H (β, α)] ds [1 − F (β)] β 0 1 − F ∗ (β) . 2 β[1 − H ∗ (β, α)]
28
Renewal Reward Processes
The following proposition concerns the Laplace transforms of the first and second moments of R(t), which can be derived by taking derivatives with respect to α in (2.18), and then setting α = 0. Proposition 2.3.1 Under the same assumptions as in the Theorem 2.3.1 we have £ ¤ (a) If E Y1 e−βX1 < ∞ for some β > 0, then R ∞ R ∞ −βx Z ∞ ye dH(x, y) E[R(t)]e−βt dt = 0 0 , (2.19) ∗ (β)] β[1 − F 0 ¤ £ (b) If E Y12 e−βX1 < ∞ for some β > 0, then Z ∞ E[R2 (t)]e−βt dt 0
R∞R∞ =
2 −βx
y e dH(x, y) + β[1 − F ∗ (β)]
0
0
i2 hR R ∞ ∞ 2 0 0 ye−βx dH(x, y) β[1 − F ∗ (β)]2
.(2.20)
Remark 2.3.1 If (Xn , n ≥ 1) and (Yn , n ≥ 1) are independent then (2.18), (2.19), and (2.20) reduce to (a) Z
∞
E(e−αR(t) )e−βt dt =
0
1 − F ∗ (β) , β[1 − F ∗ (β)G∗ (α)]
(b) Z
∞
E[R(t)]e−βt dt =
0
µY F ∗ (β) , β[1 − F ∗ (β)]
(c) Z
∞
2
E[R (t)]e 0
−βt
dt =
· ¸ F ∗ (β) 2µ2Y F ∗ (β) 2 2 σ + µY + , β[1 − F ∗ (β)] Y 1 − F ∗ (β)
where µY = E(Y1 ) and σY2 = Var(Y1 ).
2.4
Asymptotic properties
Asymptotic properties of renewal reward processes like the renewal-reward theorem and its expected-value version, and asymptotic normality of the processes
Asymptotic properties
29
are well known. In this section we will reconsider some of them. We will use Tauberian theorems to derive expected-value version of the renewal reward theorem. We will also investigate other asymptotic properties of the renewal reward processes including asymptotic normality of the instantaneous reward process defined in (2.2). We first consider the renewal reward process (R(t), t ≥ 0) with renewal cycles Xn and rewards non-negative random variables Yn as defined in (2.1), i.e., N (t) X R(t) = Yn . (2.21) n=1
If µX = E(X1 ) and µY = E(Y1 ) are finite, then lim
t→∞
R(t) t
=
µY µX
with probability 1,
(2.22)
and lim
t→∞
E[R(t)] t
=
µY , µX
(2.23)
which are well known as the renewal-reward theorem and its expected-value version respectively, see Tijms [46] for example. The renewal reward theorem can easily be proved using the strong law of large numbers. A proof of (2.23) can be found for example in Ross [36], where he used Wald’s equation. We will give a proof for (2.23) using the following Tauberian theorem which can be found in Widder [48]. Theorem 2.4.1 (Tauberian theorem) If α(t) is non-decreasing and such that the integral Z ∞ e−st dα(t) f (s) = 0
converges for s > 0 and if for some non-negative number γ and some constant C C f (s) ∼ γ as s −→ 0 s then Ctγ α(t) ∼ as t −→ ∞. Γ(γ + 1) The proof of (2.23): In Section 2.3 we have proved that the Laplace transform of E[R(t)] is given by R ∞ R ∞ −βu Z ∞ ve dH(u, v) −βt E[R(t)]e dt = 0 0 , (2.24) ∗ (β)] β[1 − F 0
30
Renewal Reward Processes
see (2.19). Assuming µY is finite, we obtain from this equation R ∞ R ∞ −βu Z ∞ ve dH(u, v) e−βt dE[R(t)] = 0 0 . ∗ (β) 1 − F 0 Using dominated convergence we can prove that Z ∞Z ∞ ve−βu dH(u, v) = µY + o(1) as 0
β −→ 0.
0
Similarly if µX is finite, F ∗ (β) = 1 − βµX + o(β) as β −→ 0. It follows that
Z
∞
µY 1 µX β
e−βt dE[R(t)] ∼
0
as
β −→ 0.
Obviously E[R(t)] is non-decreasing. So we can apply the Tauberian theorem with γ = 1. 2 A stronger version of (2.23) can be derived for the case where X1 has a 2 density in some interval. Assume that σX = Var(X1 ) and σXY = Cov(X1 , Y1 ) are finite. Let fX and fXY denote the density function of X1 and the joint density of X1 and Y1 respectively. Let M (t) = E[R(t)]. Conditioning on X1 , from (2.21) we obtain Z t M (t) = K(t) + M (t − x)fX (x)dx, (2.25) 0
RtR∞
where K(t) = 0 0 yfXY (x, y)dydx. Define for t ≥ 0 the function Z(t) = M (t) −
µY t. µX
From (2.25), we find that Z(t) satisfies the following integral equation Z t Z(t) = a(t) + Z(t − x)fX (x)dx, 0
where
Z ∞ µY (x − t)fX (x)dx. µX t We see that a(t), t ≥ 0, is a finite sum of monotone functions. So we can use the key renewal theorem, see e.g., Tijms [46], to obtain Z ∞ 1 lim Z(t) = a(x)dx t→∞ µX 0 ·Z ∞ ¸ Z ∞Z ∞ 1 µY = [K(x) − µY ]dx + (s − x)fX (s)dsdx . µX 0 µX 0 x
a(t) = K(t) − µY +
Asymptotic properties
31
It can easily be verified that Z ∞ [K(x) − µY ]dx = −E[X1 Y1 ] = −(σXY + µX µY ) 0
and
Z
∞
Z
∞
(s − x)fX (s)dsdx = 0
x
It follows that lim Z(t)
t→∞
= =
1 1 2 E(X12 ) = (σX + µ2X ). 2 2
· ¸ 1 µY 2 2 −(σXY + µX µY ) + (σ + µX ) µX 2µX X 2 µY − 2µX σXY − µ2X µY σX . 2µ2X
From this we conclude that E[R(t)] =
µY σ 2 µY − 2µX σXY − µ2X µY t+ X + o(1) as µX 2µ2X
t → ∞.
(2.26)
Now we will consider asymptotic properties of the instantaneous reward process defined in (2.2), i.e., N (t)
Rφ (t) =
X
φ(Xn ) + φ(t − SN (t) ).
n=1
Putting µY = E[φ(X1 )], we can prove that (2.22) and (2.23) look exactly the same (the contribution of the reward earned in the incomplete renewal cycle (SN (t) , t] disappear in the limit). Next, to obtain a formula close to (2.26) we put σXY = Cov(X1 , φ(X1 )). A similar argument as for proving (2.26) can be used to deduce µY σ 2 µY − 2µX σXY − µ2X µY t+ X + A + o(1) as t → ∞, µX 2µ2X R∞ where A = µ1X 0 [1 − F (t)]φ(t)dt, provided the function φ(t) is Laplace transformable. The extra constant A can be interpreted as a contribution of the reward earned in the incomplete renewal cycle. Under some conditions on the function φ, the limiting distribution of the instantaneous reward process (Rφ (t)) is a normal distribution. To prove this we need the following lemmas. E[Rφ (t)] =
Lemma 2.4.1 [8] If χ(t), ε(t) and δ(t) are random functions (0 < t < ∞), and are such that the asymptotic distribution of χ(t) exists, ε(t) converges in probability to 1 and δ(t) converges in probability to 0 for t → ∞, then the asymptotic distribution of χ(t)ε(t) + δ(t) exists and coincides with that of χ(t).
32
Renewal Reward Processes (1)
(2)
Lemma 2.4.2 [33] If (Xn ), (Xn ), and (Xn ) are sequences of random vari(1) (2) (1) (2) ables such that Xn ≤ Xn ≤ Xn and the sequences (Xn ) and (Xn ) have the same asymptotic distribution for n → ∞, then Xn has also that asymptotic distribution. Lemma 2.4.3 [33] Let ξn denote a sequence of identically distributed random variables having finite second moment. Let ν(t) denote a positive integer-valued random variable for t > 0, for which ν(t) converges in probability to c > 0 for t ξν(t) t → ∞. Then √ converges in probability to 0. ν(t)
Assume that the function φ is non-negative and non-decreasing, or bounded. Let Yn := φ(Xn ). Then N (t)
X
n=1
N (t)
Yn =
X
N (t)+1
φ(Xn ) ≤ Rφ (t) ≤
n=1
X
N (t)+1
φ(Xn ) =
n=1
X
Yn .
(2.27)
n=1
2 Assume that σX and σY2 are finite. Then using the Central Limit Theorem for random sums, see Embrechts et al. [11] we obtain, as t → ∞, · µ ¶ ¸−1/2 N (t) X µY t µY d Var Y1 − X1 Yn − t −→ N (0, 1), (2.28) µX µX µ X n=1
where µ ¶ µY t Var Y1 − X1 µX µX
µ =
2 − 2µX µY σXY µ2X σY2 + µ2Y σX µ3X
Now we will consider the limiting distribution of C= Note that
PN (t)+1 n=1
PN (t)+1 n=1
¶ t.
Yn . Let
2 µ2X σY2 + µ2Y σX − 2µX µY σXY . µ3X
Yn −
√ Ct
µY µX t
PN (t)
Yn − √ Ct
n=1
=
µY µX t
YN (t)+1 + √ Ct
where the first term in the right-hand side converges in distribution to the Y (t)+1 p standard normal random variable. If we can prove that N√ −→ 0 as t → ∞ t then by Lemma 2.4.1 PN (t)+1 n=1
Yn −
√ Ct
µY µ
t
d
−→ N (0, 1) as t → ∞.
(2.29)
Covariance structure of renewal processes
But since σY2 < ∞ it follows, by Lemma 2.4.3 and the fact that 0) as t → ∞, r YN (t)+1 YN (t)+1 N (t) + 1 √ p = t t N (t) + 1
33 N (t) a.s. 1 −→ µX (> t
p
−→ 0 as t → ∞. Finally, combining (2.27), (2.28), (2.29), and Lemma 2.4.2, it follows that Y Rφ (t) − µµX t d √ −→ N (0, 1) as t → ∞. Ct
2.5
Covariance structure of renewal processes
Basically, the covariance structure of renewal reward processes can be derived using point processes, but it will involve more complicated calculations. In this section we will derive the covariance structure of a renewal process, a special case of renewal reward processes. Using the notations in Section 2.2 define for t ≥ the functional N(t) on Mp (E) by Z Z N(t)(µ) = 1[0,x) (t − A(s, µ))1[0,s) (u)µ(dudv)µ(dsdx). E
Let ω ∈ Ω. Then
E
∞ X ∞ ½ X
¾ 1[0,Xn (ω)) (t − A(Tn (ω), Φ(ω)))1[0,Tn (ω)) (Ti (ω))
N(t)(Φ(ω)) =
n=1 i=1 ∞ X
=
1[0,TN (t,ω)+1 (ω)) (Ti (ω))
i=1
= N (t, ω). So we have, with probability 1, N (t) = N(t)(Φ). Using this functional expression of N (t) we derive the double Laplace transform of E[N (t1 )N (t2 )] which is stated in the following theorem. The proof is given in Appendix A. Theorem 2.5.1 Let (Xn , n ≥ 1) be an iid sequence of strictly positive random variables with common cdf F . Let (N (t), t ≥ 0) be the renewal process with renewal cycles Xn . Then for α, β > 0, Z ∞Z ∞ E[N (t1 )N (t2 )]e−αt1 −βt2 dt1 dt2 0
0
=
[1 − F ∗ (α)F ∗ (β)]F ∗ (α + β) . αβ[1 − F ∗ (α)][1 − F ∗ (β)][1 − F ∗ (α + β)]
(2.30)
34
Renewal Reward Processes
Example 2.5.1 Let (N (t), t ≥ 0) be a renewal process with inter-arrival times Xn having Gamma distribution with common pdf f (x; λ, m) =
λe−λx (λx)m−1 , Γ(m)
λ > 0,
x ≥ 0.
Using (2.10) and (2.30) we obtain Z
∞
E[N (t)]e−βt dt =
0
λm β[(β + λ)m − λm ]
and Z
∞
0
Z
∞
E[N (t1 )N (t2 )]e−αt1 e−βt2 dt1 dt2
0
=
[λ(α + λ)m − λ3m ] . αβ[(α + λ)m − λm ][(β + λ)m − λm ][(α + β + λ)m − λm ]
As an example, take m = 2. Transforming back these Laplace transforms we obtain E[N (t)] =
1 1 1 λt − + e−2λt 2 4 4
and for t1 ≤ t2 E[N (t1 )N (t2 )] =
1 [1 − 2λ(t2 − t1 ) + 4λ2 t1 t2 − (1 + 4λt1 − 2λt2 )e−2λt1 16 −(1 + 2λt1 )e−2λt2 + e−2λ(t2 −t1 ) ].
Hence for t1 ≤ t2 Cov[N (t1 ), N (t2 )] =
1 1 λ(t1 − e−2λt1 − e−2λt2 ) + (e−2λ(t2 −t1 ) − e−2λ(t1 +t2 ) ). 4 16
Note that for m = 1 the process (N (t), t ≥ 0) is a homogeneous Poisson process with rate λ, and in this case E[N (t1 )N (t2 )] = λ2 t1 t2 + λ min{t1 , t2 }. This result can also be obtained using (2.30). Moreover the covariance between N (t1 ) and N (t2 ) for t1 < t2 is given by Cov[N (t1 ), N (t2 )]
=
λt1 .
System reliability in a stress-strength model
2.6
35
System reliability in a stress-strength model
In this section we consider a system which is supposed to function during a certain time after which it fails. The failures of the system are of two kinds: proper failures which are due to own wear-out and occur even in an unstressed environment, and failures which are due to random environmental stresses. We will only consider the latter. The system we consider here is generic, and models all kinds of varieties of products, subsystems, or components. The study about system reliability where the failures are only due to random environmental stresses is known as a stress-strength interference reliability or a stress-strength model. Many authors have paid attention to the study of such a model, see for example Xue and Yang [50], Chang [4], and Gaudoin and Soler [14]. In other literature this model is also called a load-capacity interference model, see Lewis and Chen [24]. In Gaudoin and Soler [14] three types of stresses are considered: point, alternating and diffused stresses. The systems they considered may have a memorization: a stress which occurs at a given time can influence the future failures if the system has kept it in memory. Two types of stochastic influence models are proposed: stress-strength duration models (type I) and random environment lifetime models (type II). The type I models are based on the assumption that a system failure occurs at time t if the accumulation of memory of all stresses occurred before time t exceeds some strength threshold of the system whereas the type II models are models for which, conditionally on all stresses that occurred before, the cumulative failure (hazard) rate at time t is proportional to the accumulation of memory of all stresses occurring before time t. In type II models the stresses weaken the system. The accumulation of stresses will cause the system to fail, but this failure is not associated to a given strength threshold. In this section we will consider extensions of some models proposed in Gaudoin and Soler. We will restrict ourselves to point stresses. The point stresses are impulses which occur at random times with random amplitudes. In the type I models Gaudoin and Soler assumed that the occurrence times of the stresses is modelled respectively as a homogeneous Poisson process, a non-homogeneous Poisson process and a renewal process. The two kinds of memory they considered are: (i) the system keeps no memory of the stresses, and (ii) the system keeps a permanent memory of all stresses occurred before. The amplitudes of the stresses and their occurrence times are assumed to be independent. We give two generalizations. Firstly, the occurrence times of the stresses are modelled as a Cox process (doubly stochastic Poisson process) and we keep the independence assumption. Secondly, the occurrence times of the stresses is modelled as a renewal process, but they may depend on the amplitudes of the stresses. We discuss this in Subsection 2.6.1. In the type II models Gaudoin and Soler model the occurrence times of the stresses respectively as a homogeneous Poisson process and a non-homogeneous
36
Renewal Reward Processes
Poisson process. They assumed that the amplitudes of the stresses are independent of their occurrence times, and considered any kind of memory. We give a generalization where the occurrence times of the stresses is modelled as a Cox process. We give also a further generalization where the occurrence times of the stresses are modelled as a renewal process and may depend on their amplitudes, but we only assume that the system keeps a permanent memory of the stresses which occurred. We discuss this in Subsection 2.6.2.
2.6.1
Type I models
Suppose that a system operated at time t0 = 0 is exposed to stresses occurring at random time points S1 , S2 , S3 , ... where S0 := 0 < Si < Si+1 , ∀i ≥ 1. Let N (t) = sup{n ≥ 0 : Sn ≤ t} be the number of stresses that occurred in the time interval [0, t]. Let the amplitude of the stress at time Sn be given by the non-negative random variable Yn . Assume that the sequence (Yn , n ≥ 1) is iid with a common distribution function G and independent of the sequence (Sn , n ≥ 1). After the occurrence of a stress the system may keeps the stress into its memory. In Gaudoin and Soler the memory of the system is represented by a deterministic Stieltjes measure. Here we will represent the memory of the system in terms of a recovery rate. We call a function h the recovery rate of the system if it is non-negative, non-increasing, bounded above from 1, and it vanishes on (−∞, 0). We will assume that at time t the contribution of the stress that has occurred at time Sn ≤ t has an amplitude Yn h(t − Sn ). So the accumulation of the stresses at time t is given by L(t)
=
∞ X
Yn h(t − Sn )
n=1 N (t)
=
X
Yn h(t − Sn ).
(2.31)
n=1
If the strength threshold of the system equals a positive constant u, then the reliability of the system at time t is given by ˜ R(t)
=
P( sup L(s) ≤ u).
(2.32)
0≤s≤t
˜ In general it is difficult to calculate R(t). In the case the system keeps no memory of the stresses, the equation (2.32) simplifies to ˜ R(t)
=
P(max{Y1 , Y2 , ..., YN (t) } ≤ u)
(2.33)
and in the case of the system keeps a permanent memory of the stresses (without
System reliability in a stress-strength model
37
recovery), equation (2.31) reduces to N (t)
L(t) =
X
Yn
(2.34)
P(L(t) ≤ u).
(2.35)
n=1
and (2.32) simplifies to ˜ R(t) =
We see that if (N (t)) is a renewal process, then (L(t)) is a renewal reward process. Gaudoin and Soler [14] consider homogeneous Poisson processes, non-homogeneous Poisson processes and renewal processes as models for (N (t)). A generalization of the non-homogeneous Poisson process is the Cox process, see e.g., Grandell [17]. A Cox process can be considered as a non-homogeneous Poisson process with randomized intensity measure. For a non-homogeneous Poisson process with intensity measure ν we have (ν[0, t])k −ν[0,t] , k = 0, 1, 2, . . . e k! For a Cox process the intensity measure ν is chosen according to some probability measure Π and Z (ν[0, t])k −ν[0,t] P(N (t) = k) = Π(dν), k = 0, 1, 2, . . . . e k! P(N (t) = k)
=
So if (N (t)) is a Cox process then, by conditioning on the number of stresses in the time interval [0, t], the reliability in (2.33) can be expressed as Z ˜ = e−[1−G(u)]ν[0,t] Π(dν). R(t) As an example let ν[0, t] = Λt where Λ is chosen according to uniform distribution in [0, 1]. Then ˜ R(t) =
1 − e−[1−G(u)]t . [1 − G(u)]t
˜ In the case without recovery, note that the reliability R(t) is just the cdf of L(t) at point u. It follows that we only need to calculate the distribution of L(t) in (2.34). By conditioning on the number of stresses in the time interval [0, t] we obtain the Laplace transform of L(t) which is given by ´ ³ ψ(t, α) := E e−αL(t) Z ∗ = e−[1−G (α)]ν[0,t] Π(dν) Z Rt ∗ = e−[1−G (α)] 0 ν(ds) Π(dν).
38
Renewal Reward Processes
As an example let ν(ds) = X(s)ds where (X(t), t ≥ 0) is a continuous time Markov chain on {0, 1}. Suppose that the chain starts at time 0 in state 1 where it stays an exponential time with mean 1/λ1 . Then it jumps to state 0 where it stays an exponential time with mean 1/λ0 , and so on. It follows that ´ ³ Rt ψ(t, α) = E e−c(α) 0 X(s)ds where c(α) = 1 − G∗ (α). Let τi = inf{t ≥ 0|X(t) 6= i}. Starting from 1, the random variable τ1 is the time at which the process leaves the state 1 and P1 (τ1 > t) = e−λ1 t . Similarly for τ0 we have P0 (τ0 > t) = e−λ0 t . Then ψ1 (t, α)
³ ³ ´ ´ Rt Rt = E1 e−c(α) 0 X(s)ds , τ1 > t + E1 e−c(α) 0 X(s)ds , τ1 < t Z t λ1 e−(λ1 +c(α))x ψ0 (t − x, c(α))dx, (2.36) = e−(λ1 +c(α))t + 0
and ψ0 (t, α)
³ ³ ´ ´ Rt Rt = E0 e−c(α) 0 X(s)ds , τ0 > t + E0 e−c(α) 0 X(s)ds , τ0 < t Z t λ0 e−λ0 x ψ1 (t − x, c(α))dx. (2.37) = e−λ0 t + 0
Define for β > 0 and i = 0, 1 Z ψˆi (β, α) =
∞
e−βt ψi (t, α)dt.
0
From (2.36) and (2.37) we get the system of equations ( λ1 1 ψˆ1 (β, α) = λ1 +c(α)+β + λ1 +c(α)+β ψˆ0 (β, α) λ0 ˆ 1 ˆ ψ0 (β, α) = λ0 +β + λ0 +β ψ1 (β, α) It follows that ψˆ1 (β, α) =
λ1 + λ 0 + β . µc(α) + (λ1 + λ0 + c(α))β + β 2
System reliability in a stress-strength model
39
Transforming back this transform we obtain " Ã√ ! à √ !# bt λ − c(α) bt − 21 (λ+c(α))t √ ψ1 (t, α) = e cos + sin 2 2 b
(2.38)
where λ = λ0 + λ1 and b = 4λ0 c(α) − [λ + c(α)]2 . To find the distribution function of L(t) we transform back numerically the Laplace transform in (2.38) with respect to α. For example if λ0 = λ1 = 1 and G(x) = 1 − e−x then we get the distribution function of L(10) as in Figure 2.4. Note that there is a mass 1
0.9
0.8
0.7
F(u)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10 u
12
14
16
18
20
Figure 2.4: The graph of the distribution function of L(10).
1.4
1.2
1
R(t)
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25 t
30
35
40
45
50
˜ Figure 2.5: The graph of R(t) for u = 5. at 0 which corresponds to the event that no stress occurs in the time interval
40
Renewal Reward Processes
˜ [0, t]. The graph of R(t), for u = 5, is given in Figure 2.5. Next we will consider the second generalization of the reliability in (2.35), i.e., ˜ = P(L(t) ≤ u) R(t) PN (t) where L(t) = n=1 Yn . As in Gaudoin and Soler [14] we assume that N (t) is a renewal process, but we allow a dependence between the sequence (Sn ) and (Yn ). Since N (t) is a renewal process then Xn = Sn − Sn−1 , n = 1, 2, . . . , where S0 = 0 are iid random variables. We will assume that ((Xn , Yn ), n ≥ 1) is an iid sequence of random vectors. Note that in this case L(t) is a renewal reward process. So we can use Theorem 2.3.1 to determine the distribution of L(t): Z ∞ 1 − F ∗ (β) E(e−αL(t) )e−βt dt = β[1 − H ∗ (β, α)] 0 where F is the cdf of X1 , and H is the joint cdf of X1 and Y1 .
2.6.2
Type II models
Using the notation in Subsection 2.6.1 we now consider a model for the lifetime of the system where the cumulative failure rate at time t is proportional to the accumulation of all stresses occurred before time t. In this case the system reliability is given by ³ ´ P ˜ = E e−α ∞ n=1 Yn h(t−Sn ) R(t) where α > 0 is a proportionality constant and h is an arbitrary recovery function, see Subsection 2.6.1. PWe see in this case that the reliability is the Laplace ∞ transform of L(t) := n=1 Yn h(t − Sn ). The case where the occurrence times (Sn ) is a non-homogeneous Poisson process has been discussed by Gaudoin and Soler [14]. Here firstly we will consider a generalization where (Sn ) is a Cox process. We assume that the sequences (Sn ) and (Yn ) are independent. We will express L(t) as a functional of a Poisson point process. Let (Ω, F, P) be the probability space on which the random variables Sn and Yn are defined such that the sequences (Sn ) and (Yn ) are independent. Since we assume that (Sn ) is a Cox process, then the map ω ∈ Ω 7→
∞ X
δSn (ω)
n=1
defines a Poisson point process on [0, ∞) with intensity measure ν where the measure ν is chosen randomly according to some probability distribution Π. More formally, the intensity measure ν is chosen from the set M+ ([0, ∞)) of all
System reliability in a stress-strength model
41
Radon measures on [0, ∞). Moreover, since (Sn ) and (Yn ) are independent, the map ∞ X Φ : ω ∈ Ω 7→ δ(Sn (ω),Yn (ω)) n=1
defines a Poisson point process on E = [0, ∞) × [0, ∞) with intensity measure ν × G(dtdy) = ν(dt)dG(y), where G is the cdf of Y1 . Let Mp (E) be the set of simple point measures µ on E. Denote the distribution of Φ over Mp (E) by Pν×G , i.e., Pν×G = P ◦ Φ−1 . As in Lemma 2.2.1 we have, with probability 1, L(t) = L(t)(Φ)
(2.39)
R
where L(t)(µ) = E yh(t − s)µ(dsdy). Using the formula for the Laplace functional of Poisson point processes, see Theorem 1.2.1, we obtain Z Z R ˜ R(t) = e− E αyh(t−s)1[0,t] (s)µ(dsdy) Pν×G (dν)Π(dν) M+ ([0,∞))
Z =
M+ ([0,∞))
Z =
Mp (E)
¾ ½ Z h i −αyh(t−s)1[0,t] (s) 1−e ν × G(dsdy) Π(dν) exp − E
½ Z th ¾ i 1 − G∗ (αh(t − s)) ν(ds) Π(dν). exp − 0
M+ ([0,∞))
The last equality follows from the independence assumption between (Sn ) and (Yn ). As an example let M+ ([0, ∞)) = {µ : µ(dt) = λdt, λ ∈ [0, ∞)} and Π be a probability distribution of an exponential random variable with parameter η on M+ ([0, ∞)). Then ½ Z t h Z ∞ i ¾ ˜ R(t) = exp − λ 1 − G∗ (αh(t − s)) ds ηe−ηλ dλ 0 0 ½ h Z t Z ∞ i ¾ ∗ exp − t + η − G (αh(t − s))ds λ dλ = η 0
=
η+t−
Rt 0
0
η
G∗ (αh(t − s))ds
.
Moreover, if we assume that h(t) = e−t and G(u) = 1 − e−γu , γ > 0, then we get ˜ R(t)
=
η ³ η − ln
αe−t +γ α+γ
´.
Now consider the second generalization where the system keeps a permanent memory of the stresses (h(t) ≡ 1). Suppose that the occurrence times (Sn ) are
42
Renewal Reward Processes
modelled as a renewal process. As in the end of the previous subsection, if Xn = Sn − Sn−1 , n = 1, 2, . . . where S0 = 0 represent the inter-arrival times of the renewal process and (Xn , Yn ) is assumed to be an iid sequence of random vectors then Z ∞ 1 − F ∗ (β) −βt ˜ R(t)e dt = β[1 − H ∗ (β, α)] 0 where F is the cdf of the X1 and H is the joint cdf of X1 and Y1 . As a special case, when the sequences (Xn ) and (Yn ) are independent, we have Z ∞ 1 − F ∗ (β) −βt ˜ R(t)e dt = β[1 − F ∗ (β)G∗ (α)] 0 where G is the cdf of the Y1 . As an example assume that (Xn , Yn ) is an iid sequence of random vectors. Suppose that X1 and Y1 have a joint bivariate exponential distribution with P(X1 > x, Y1 > y) = e−(λ1 x+λ2 y+λ12 max(x,y)) ;
x, y ≥ 0;
λ1 , λ2 , λ12 > 0.
The marginals are given by P(X1 > x)
=
e−(λ1 +λ12 )x
(2.40)
P(Y1 > y)
=
e−(λ2 +λ12 )y .
(2.41)
and
The correlation coefficient ρ between X1 and Y1 is given by ρ= In this case
Z 0
∞
λ12 . λ1 + λ2 + λ12
−βt ˜ R(t)e dt
=
C1 β + C2 , C1 β 2 + C3 β + C 4
where C1 = λ2 + λ12 + α, C2 = C1 (C1 + λ1 ), C3 = λ1 α + C2 , and C4 = α(λ1 + λ12 )(C1 + λ1 ). Inverting this transform we obtain · µ ¶ q C 1 − 2C3 t 2 ˜ 1 R(t) = e cos 4C1 C4 − C3 t 2C1 µ ¶# q 2C2 − C3 1 + p sin 4C1 C4 − C32 t . 2C1 4C1 C4 − C32 In case the sequences (Xn ) and (Yn ) are independent where X1 and Y1 satisfy (2.40) and (2.41), Z ∞ α + λ2 + λ12 −βt ˜ R(t)e dt = . β(α + λ 2 + λ12 ) + α(λ1 + λ12 ) 0
System reliability in a stress-strength model
43
Inverting this transform we obtain ½ ˜ R(t)
=
exp
−
¾ α(λ1 + λ12 ) t . α + λ2 + λ12
Now we will observe the effect of the dependence between X1 and Y1 on the ˜ reliability R(t) in this example. As examples, firstly take α = 1, λ1 = λ2 = 32 , 1 λ12 = 3 . Then for the dependent case, ρ = 0.2 and h i 3 e− 2 t cosh(0.9574t) + 1.2185 sinh(0.9574t) ,
˜ R(t) =
and for the independent case ˜ R(t)
=
1
e− 2 t .
˜ The graphs of R(t) can be seen in Figure 2.6. 1 dependent independent 0.9
0.8
0.7
R(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5 t
6
7
8
9
10
˜ Figure 2.6: The graphs of R(t) for α = 1, λ1 = λ2 = 32 , λ12 = 31 . Secondly, take α = 1, λ1 = λ2 = 31 , λ12 = 32 . Then for the dependent case, ρ = 0.5 and h i ˜ R(t) = e−1.25t cosh(0.6292t) + 1.7219 sinh(0.6292t) , and for the independent case ˜ R(t)
=
1
e− 2 t .
˜ The graphs of R(t) can be seen in Figure 2.7.
44
Renewal Reward Processes
1 dependent independent 0.9
0.8
0.7
R(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5 t
6
7
8
9
10
˜ Figure 2.7: The graphs of R(t) for α = 1, λ1 = λ2 = 31 , λ12 = 32 . 1 dependent independent 0.9
0.8
0.7
R(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5 t
6
7
8
9
10
˜ Figure 2.8: The graphs of R(t) for α = 1, λ1 = λ2 = 91 , λ12 = 98 . Finally, take α = 1, λ1 = λ2 = 91 , λ12 = 98 . Then for the dependent case, ρ = 0.8 and h i ˜ R(t) = e−1.0833t cosh(0.3436t) + 2.9913 sinh(0.3436t) , and for the independent case ˜ R(t) =
1
e− 2 t .
˜ The graphs of R(t) can be seen in Figure 2.8.
Chapter 3
Integrated Renewal Processes 3.1
Notations and Definitions
Consider a locally finite point process on the positive half line [0, ∞). Denote the ordered sequence of points by 0 < S1 < S2 < . . .. We will think of the points Sn as arrival times. We define S0 := 0, but this does not mean that we assume that there is a point in 0. Let N (t) be the number of arrivals in the time interval [0, t], i.e., N (t) = sup{n ≥ 0 : Sn ≤ t}. Define for t ≥ 0 Z t Y (t) = N (s)ds. 0
If (N (t), t ≥ 0) is a renewal process, we call the stochastic process (Y (t), t ≥ 0) an integrated renewal process. Note that we can express Y (t) as N (t)
Y (t) =
X
(t − Si ) = tN (t) − Z(t),
(3.1)
i=1
where
N (t)
Z(t) =
X
Si .
(3.2)
i=1
Figure 3.1 shows the graphs of Y (t) and Z(t). In this chapter we will discuss the distributions of Y (t) and Z(t). In Section 3.2 we discuss the distributions of Y (t) and Z(t) when (N (t)) is a Poisson or a Cox process. In Section 3.3 we discuss the distributions of Z(t) and Y (t) when (N (t)) is a renewal process. Their asymptotic properties are studied in Section 3.4. Finally an application is given in Section 3.5. 45
46
Integrated Renewal Processes
N (t)6 2
¡ ¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ 1 ¡¡ Z(t) ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡ ¡¡ ¡ Y (t) ¡ ¡ ¡ ¡ ¡ 0 S1 S2
t
Figure 3.1: Graphs of Y (t) and Z(t).
3.2
(N (t)) a Poisson or Cox process
Firstly, suppose that the process (N (t), t ≥ 0) is a homogeneous Poisson process with rate λ. It is well known that given N (t) = n, the n arrival times S1 , ..., Sn have the same distribution as the order statistics corresponding to n independent random variables uniformly distributed on the time interval [0, t] (see e.g. Ross [36]). Conditioning on the number of arrivals in the time interval [0, t] we obtain ∞ h i X Pn E(e−αY (t) ) = E e−α i=1 (t−Si ) |N (t) = n P(N (t) = n) =
n=0 ∞ X
i (λt)n h Pn e−αnt E e−α i=1 Vi e−λt n! n=0
where Vi , i = 1, 2, ..., n are independent and identically uniform random variables on [0, t]. Since ¤ £ ¤ 1 £ αt E e−αV1 = e −1 αt it follows that · ¸ ∞ X ¤ n (λt)n −λt 1 £ αt −αY (t) −αnt ) = e E(e e −1 e αt n! n=0 h in λ(1−e−αt ) ∞ X α = e−λt n! n=0 ½ ¾ λ(1 − αt − e−αt ) = exp . (3.3) α From (3.3) we deduce that E[Y (t)] = 21 λt2 and Var[Y (t)] = 31 λt3 . Using a similar argument we can prove that Z(t) has the same Laplace transform as
(N (t)) a Poisson or Cox process
47
Y (t). So by uniqueness theorem for Laplace transforms we conclude that Z(t) has the same distribution as Y (t). The distribution of Y (t) has mass at zero with P(Y (t) = 0) = e−λt . The density function fY (t) of the continuous part of Y (t) can be obtained by inverting the Laplace transform in (3.3). Note that we can express (3.3) as E(e−αY (t) ) = =
∞ X λn (1 − e−αt )n n!αn n=0 ¡n¢ ∞ n X λn X (−1)k k e−ktα −λt e . n! αn n=0
e−λt
k=0
Inverting this transform we obtain, for x > 0, fY (t) (x) =e
−λt
∞ X λn n! n=1
"
# µ ¶ n n−1 X xn−1 k n (x − kt) (−1) 1(0,∞) (x) + 1(kt,∞) (x) k (n − 1)! (n − 1)! k=1
∞ X λn xn−1 = e−λt 1(0,∞) (x) n!(n − 1)! n=1
+λe−λt
∞ ∞ X (−1)k X [λ(x − kt)]n−1 1(kt,∞) (x) k! (n − 1)!(n − k)!
k=1
n=k
√ √ λ = √ e−λt I1 (2 λx)1(0,∞) (x) x ∞ X p 1 (−1)k −λt +λe [λ(x − kt)] 2 (k−1) Ik−1 (2 λ(x − kt))1(kt,∞) (x) k! k=1
where Ik (x) is the Modified Bessel function of the first kind, i.e., Ik (x) = (1/2x)
k
∞ X
(1/4x2 )m , m!Γ(k + m + 1) m=0
see Gradshteyn and Ryzhik [16]. The graphs of the pdf of Y (t) for λ = 2, t = 1, 2, 3 and t = 10 can be seen in Figure 3.2. For large t, the distribution of Y (t) can be approximated with a normal distribution having mean 21 λt2 and variance 31 λt3 . To prove this we will consider the characteristic function of the normalized Y (t). Firstly, note that Y (t)− 1 λt2 µ ¶ √ −iα √ 2 √ iα 3 1 λt3 = e 21 iα 3λt E e− t√λt Y (t) . 3 E e
48
Integrated Renewal Processes
0.7 0.18
0.16
0.6
0.14
0.5 0.12
0.1
f
Y(1)
fY(2)(x)
(x)
0.4
0.08
0.3 0.06
0.04
0.2
0.02
0.1 0
0
0
1
2
3 x
4
5
−0.02
6
0
2
4
6
(a) λ = 2, t = 1
8
10 x
12
14
16
18
20
(b) λ = 2, t = 2
0.1
0.016
0.014
0.08 0.012
0.06
(x)
Y(10)
0.008
f
fY(3)(x)
0.01
0.04
0.006
0.02 0.004
0 0.002
−0.02
0
5
10
15 x
20
25
30
0
0
20
40
(c) λ = 2, t = 3
60
80
100 x
120
140
160
180
200
(d) λ = 2, t = 10
Figure 3.2: The graphs of the pdf of Y (t) for λ = 2, t = 1, 2, 3 and t = 10. √
√ 3 and an expansion we obtain Using (3.3) with α is replaced by iα t λt ( √ " #) √ µ ¶ √ √ −iα λt λt iα 3 √ 3 Y (t) √ 3t − iα t λt λ √ E e = exp 1− √ −e iα 3 λt ( √ " #) √ λt λt 3α2 iα3 3 −3/2 √ √ + o(t = exp − ) iα 3 2λt 2λt λt ( ) √ 1 √ 1 2 λt λt −3/2 √ o(t = exp − iα 3λt − α + ) 2 2 iα 3
as t → ∞. It follows that Y (t)− 1 λt2 −iα √ 2 1 λt3 3 E e
1
−→ e− 2 α
2
as
t→∞
which is the characteristic function of the standard normal distribution. Now consider the case where (N (t)) is a non-homogeneous Poisson process with intensity measure ν. Given N (t) = n, the arrival times Si , i = 1, 2, ..., n
(N (t)) a Poisson or Cox process
49
have the same distribution as the order statistics of n iid random variables having a common cdf ( ν([0,x]) ν([0,t]) , x ≤ t G(x) = 1, x > t. In this case the Laplace transform of Z(t) is given by ´ ³ E e−αZ(t)
= =
∞ X
³ ´ Pn E e−α i=1 Si |N (t) = n P(N (t) = n)
n=0
"R t
∞ X n=0
0
½Z
e−αx dν([0, x]) ν([0, t])
#n
ν([0, t])n −ν([0,t]) e n!
¾ e−αx dν([0, x]) e−ν([0,t]) 0 ½Z t h ¾ i e−αx − 1 dν([0, x]) . = exp t
= exp
(3.4)
0
From this Laplace transform we deduce that Z E[Z(t)]
t
=
xdν([0, x]) 0
and
Z
t
Var[Z(t)] =
x2 dν([0, x]).
0
Similarly, we can prove that ½Z E(e
−αY (t)
) =
¾
t
exp
[e
−α(t−x)
− 1]dν([0, x]) ,
(3.5)
0
Z
t
E[Y (t)] =
(t − x)dν([0, x]), 0
and
Z Var[Y (t)] =
t
(t − x)2 dν([0, x]).
0
Note that in general Y (t) has a different distribution from Z(t) in case (N (t)) is a non-homogeneous Poisson process. Next we consider the distributions of Y (t) and Z(t) when (N (t)) is a Cox process. A Cox process is a generalization of a Poisson process. In a Cox process the intensity measure ν of the Poisson process is chosen randomly according to
50
Integrated Renewal Processes
some probability distribution Π. So if (N (t)) is a Cox process then from (3.4) and (3.5) we obtain ½ Z t ¾ Z ³ ´ E e−αZ(t) = exp − [1 − e−αx ]dν([0, x]) Π(dν) 0
and ³ E e
−αY (t)
Z
´ =
½ Z t ¾ −α(t−x) exp − [1 − e ]dν([0, x]) Π(dν). 0
As an example let the intensity measure ν satisfy ν([0, t]) = Λt for some positive random variable Λ. If Λ is exponentially distributed with parameter η, then ´ ³ ´ ³ αη E e−αY (t) = E e−αZ(t) = . α(η + t) − (1 − e−αt )
3.3
(N (t)) a renewal process
In this section we will consider the distribution of the processes (Y (t)) and (Z(t)) defined in (3.1) and (3.2) for the case that (N (t)) is a renewal process. Let Xn = Sn − Sn−1 , n ≥ 1, be the inter-arrival times of the renewal process. Note that (Xn , n ≥ 1) is an iid sequence of strictly positive random variables. Let F denote the cdf of X1 . As usual we will denote Laplace-Stieltjes transform of F by F ∗ . First we consider the process (Z(t)). Obviously we can express Z(t) as N (t)
Z(t) =
X
[N (t) + 1 − i]Xi .
(3.6)
i=1
We will use point processes to derive the distribution of Z(t). Let (Ω, F, P) be the probability space on which the iid sequence (Xn ) is defined and also an iid sequence (Un , n ≥ 1) of exponentially distributed random variables with parameter 1 such that the sequences (Xn ) and (Un ) are independent. Let (Tn , n ≥ 1) be the sequence of partial sums of the variables Un . Then the map P∞ Φ : ω 7→ (3.7) n=1 δ(Tn (ω),Xn (ω)) , where δ(x,y) is the Dirac measure in (x, y), defines a Poisson point process on E = [0, ∞) × [0, ∞) with intensity measure ν(dtdx) = dtdF (x). Let Mp (E) be the set of all point measures on E. We will denote the distribution of Φ by Pν , i.e., Pν = P ◦ Φ−1 .
(N (t)) a renewal process
51
Define for t ≥ 0 the functional A(t) on Mp (E) by Z A(t)(µ) = 1[0,t) (s)xµ(dsdx). E
In the sequel we write A(t, µ) = A(t)(µ). Define also for t ≥ 0 the functional Z(t) on Mp (E) by Z Z Z(t)(µ) = 1[0,x) (t − A(s, µ))µ([r, s) × [0, ∞))u1[0,s) (r)µ(drdu)µ(dsdx). E
E
Lemma 3.3.1 With probability 1, Z(t)
=
Z(t)(Φ).
Proof: Let ω ∈ Ω. Then Z(t)(Φ(ω)) ∞ X 1[0,Xn (ω)) (t − A(Tn (ω), Φ(ω))) = n=1 ∞ X i=1 ∞ X
=
Φ(ω)([Ti (ω), Tn (ω)) × [0, ∞))Xi (ω)1[0,Tn (ω)) (Ti (ω)) Φ(ω)([Ti (ω), TN (t,ω)+1 (ω)) × [0, ∞))Xi (ω)1[0,TN (t,ω)+1 (ω)) (Ti (ω))
i=1 N (t,ω)
X
=
[N (t, ω) + 1 − i]Xi (ω).
2
i=1
Theorem 3.3.1 Let (Xn , n ≥ 1) be an iid sequence of strictly positive random variables with common distribution function F . Let (Sn , n ≥ 0) be the sequence of partial sums of the variables Xn and (N (t), t ≥ 0) be the corresponding renewal process: N (t) = sup{n ≥ 0 : Sn ≤ t}. Let N (t)
Z(t) =
X
[N (t) + 1 − i]Xi .
i=1
Then for α, β > 0 Z
∞ 0
E(e−αZ(t) )e−βt dt =
∞ Y n X 1 F ∗ (α[n + 1 − i] + β)(3.8) [1 − F ∗ (β)] β n=0 i=1
(with the usual convention that the empty product equals 1).
52
Integrated Renewal Processes
Proof: By Lemma 3.3.1 Z E(e−αZ(t) ) = e−αZ(t)(µ) Pν (dµ) Mp (E)
½
Z =
exp
Z Z −α
Mp (E)
E
E
1[0,x) (t − A(s, µ))
¾ µ([r, s) × [0, ∞))u1[0,s) (r)µ(drdu)µ(dsdx) Pν (dµ) Z Z 1[0,x) (t − A(s, µ))
=
Mp (E)
E
½
¾
Z
exp −α E
µ([r, s) × [0, ∞))u1[0,s) (r)µ(drdu) µ(dsdx)Pν (dµ)
Applying the Palm formula for Poisson point processes, see Theorem 1.2.4, we obtain E(e−αZ(t) ) Z ∞Z = 0
½
exp
∞
0
Z
Z
1[0,x) (t − A(s, µ + δ(s,x) ))
Mp (E)
− E
¾
α(µ + δ(s,x) )([r, s) × [0, ∞))u1[0,s) (r)(µ + δ(s,x) )(drdu)
Pν (dµ)dF (x)ds Z ∞Z ∞Z 1[0,x) (t − A(s, µ))
= 0
½
exp
0
Z
Mp (E)
− E
¾ αµ([r, s) × [0, ∞))u1[0,s) (r)µ(drdu) Pν (dµ)dF (x)ds.
Using Fubini’s theorem and a substitution we obtain Z
∞ 0
E(e−αZ(t) )e−βt dt Z Z ∞Z ∞Z
= 0
½
0
Mp (E)
0
∞
1[0,x) (t − A(s, µ))
¾ αµ([r, s) × [0, ∞))u1[0,s) (r)µ(drdu) e−βt dtPν (dµ)dF (x)ds E Z ∞Z 1 = [1 − F ∗ (β)] β 0 Mp (E) ½ Z h ¾ i αµ([r, s) × [0, ∞)) + β u1[0,s) (r)µ(drdu) Pν (dµ)ds. exp − exp
Z
−
E
(N (t)) a renewal process
53
The integral with respect to Pν can be written as a sum of integrals over the sets Bn := {µ ∈ Mp (E) : µ([0, s) × [0, ∞)) = n}, n = 0, 1, 2, .... Fix a value of n and let µ ∈ Mp (E) be such that µ([0, s) × [0, ∞)) = n and supp(µ) = ((ti , xi ))∞ i=1 . So tn < s ≤ tn+1 . For such a measure µ the integrand with respect to Pν can be written as Z h i αµ([r, s) × [0, ∞)) + β u1[0,s) (r)µ(drdu) E
∞ h i X αµ([ti , s) × [0, ∞)) + β xi 1[0,s) (ti )
=
i=1 n X
=
(α[n + 1 − i] + β)xi .
i=1
Now the measure Pν is the image measure of P under the map Φ, see (3.7). Expressing the integral with respect to Pν over Bn as an integral with respect to P over the subset An := {ω ∈ Ω : Tn (ω) < s ≤ Tn+1 (ω)} of Ω, and using independence of (Tn ) and (Xn ), we obtain Z ¤ R £ e− E αµ([r,s)×[0,∞))+β u1[0,s) (r)µ(drdu) Pν (dµ) Bn
½
Z
=
exp
−
An
=
n X
¾ (α[n + 1 − i] + β)Xi (ω) P(dω)
i=1
· ½ X ¾¸ n E exp − (α[n + 1 − i] + β)Xi P(An ) i=1
= =
¾¸ n · ½ n Y s −s E exp − (α[n + 1 − i] + β)Xi e n! i=1 n Y
F ∗ (α[n + 1 − i] + β)
i=1
Hence
Z e−
R E
sn −s e . n!
[αµ([r,s)×[0,∞))+β]u1[0,s) (r)µ(drdu)
Pν (dµ)
Mp (E)
=
∞ Y n X
F ∗ (α[n + 1 − i] + β)
n=0 i=1
R∞
sn −s e . n!
n
Since for each n, 0 sn! e−s ds = 1, the theorem follows. 2 The Laplace transform of the ¡ ¢ mean of Z(t) can be derived from (3.8) as follows. Let ϕ(α) = E e−αZ(t) and define Wn,i (α) =
F ∗ (α[n + 1 − i] + β).
54
Integrated Renewal Processes
Then Vn (α) =
n Y
F ∗ (α[n + 1 − i] + β) =
i=1
n Y
Wn,i (α).
i=1
Hence Vn0 (α)
n X
=
n Y
0 Wn,j (α)
j=1
Wn,i (α),
i=1,i6=j
where Z 0 Wn,j (α) =
∞
−(n + 1 − j)
xe−(α[n+1−j]+β)x dF (x)
0
¤ £ if E X1 e−βX1 < ∞ for some β > 0. Since Wn,i (0) = F ∗ (β) and Z 0 Wn,j (0) = −(n + 1 − j)
∞
xe−βx dF (x),
0
it follows that Vn0 (0) =
Z
n X
−(n + 1 − j)
Z
∞
= −
xe−βx dF (x)F ∗ (β)n−1
0
1 2
xe−βx dF (x)F ∗ (β)n−1
0
j=1
= −
∞
Z
n X
(n + 1 − j)
j=1 ∞
xe−βx dF (x)F ∗ (β)n−1 n(n + 1).
0
Hence Z
∞
E[Z(t)]e−βt dt 0 Z ∞ = −ϕ0 (0)e−βt dt 0
= = =
Z ∞ ∞ X 1 ∗ −βx xe dF (x) F ∗ (β)n−1 n(n + 1) [1 − F (β)] 2β 0 n=0 Z ∞ 1 1 xe−βx dF (x) [1 − F ∗ (β)] ∗ (β)]3 β [1 − F 0 R ∞ −βx xe dF (x) 0 . β[1 − F ∗ (β)]2
Thus we have the following proposition:
(3.9)
(N (t)) a renewal process
55
Proposition ¤ 3.3.1 Under the same assumptions as Theorem 3.3.1, and if £ E X1 e−βX1 < ∞ for some β > 0, then Z
R∞
∞
E[Z(t)]e
−βt
xe−βx dF (x) . β[1 − F ∗ (β)]2
0
dt =
0
(3.10)
Now we will derive the Laplace transform of the second moment of Z(t). From (3.9) we obtain
Vn00 (α)
n X
=
j=1
+
n Y
00 Wn,j (α)
n X
Wn,i (α)
i=1,i6=j n X
0 Wn,j (α)
j=1
0 Wn,k (α)
k=1,k6=j
n Y
Wn,i (α),
i=1,i6=j6=k
where Z 00 Wn,j (α) =
∞
(n + 1 − j)2
x2 e−(α[n+1−j]+β)x dF (x)
0
¤ £ 0 if E X12 e−βX1 < ∞ for some β > 0,. Since Wn,i (0) = F ∗ (β), Wn,j (0) = R ∞ −βx R ∞ 00 2 2 −βx −(n + 1 − j) 0 xe dF (x), and Wn,j (0) = (n + 1 − j) 0 x e dF (x), then
Vn00 (0)
=
n X
Z 2
x2 e−βx dF (x)F ∗ (β)n−1
0
j=1
+
∞
(n + 1 − j)
n X
Z
∞
−(n + 1 − j) 0
j=1 n X
Z
∞
=
∞
−(n + 1 − k)
xe−βx dF (x)F ∗ (β)n−2
0
k=1,k6=j
Z
xe−βx dF (x)
2 −βx
x e
∗
dF (x)F (β)
0
n−1
n X
(n + 1 − j)2
j=1
hZ
∞
+ 0
i2
xe−βx dF (x) F ∗ (β)n−2
n X (n + 1 − j) j=1
n X k=1,k6=j
(n + 1 − k).
56
So
Integrated Renewal Processes
Z 0
∞
E[Z 2 (t)]e−βt dt Z ∞ = ϕ00 (0)e−βt dt 0
Z ∞ ∞ n X X 1 = [1 − F ∗ (β)] x2 e−βx dF (x)F ∗ (β)n−1 (n + 1 − j)2 β 0 n=0 j=1 hZ +
∞
n i2 X xe−βx dF (x) F ∗ (β)n−2 (n + 1 − j)
0
µZ
j=1 ∞
n X
(n + 1 − k)
k=1,k6=j ∗
1 F (β) + 1 x2 e−βx dF (x) [1 − F ∗ (β)] β [1 − F ∗ (β)]4 0 ¶ Z h ∞ i2 2 + F ∗ (β) +2 xe−βx dF (x) [1 − F ∗ (β)]5 0 µ Z ∞ 1 ∗ = [F (β) + 1] x2 e−βx dF (x) β[1 − F ∗ (β)]3 0 Z i2 ¶ 4 + 2F ∗ (β) h ∞ −βx + . xe dF (x) 1 − F ∗ (β) 0 =
Thus we have the following proposition: Proposition £ ¤ 3.3.2 Under the same assumptions as Theorem 3.3.1, and if E X12 e−βX1 < ∞ for some β > 0, R∞ Z ∞ [1 + F ∗ (β)] 0 x2 e−βx dF (x) 2 −βt E[Z (t)]e dt = β[1 − F ∗ (β)]3 0 R∞ 2[2 + F ∗ (β)][ 0 xe−βx dF (x)]2 + . (3.11) β[1 − F ∗ (β)]4 Remark 3.3.1 If X1 is exponentially distributed with parameter λ, then using (3.10) and (3.11) we obtain Z ∞ λ E[Z(t)]e−βt dt = 3 β 0 and
Z 0
∞
E[Z 2 (t)]e−βt dt =
2λ[3λ + β] . β5
Inverting these transforms we obtain E[Z(t)] = 21 λt2 , E[Z 2 (t)] = 41 λ2 t4 + 31 λt3 , and hence Var[Z(t)] = 31 λt3 . These results are the same as those in the previous section.
Asymptotic properties
57
Now we will consider the marginal distribution of the process (Y (t)) when (N (t)) is a renewal process. It is easy to see that N (t)
Y (t)
=
X
(i − 1)Xi + N (t)[t − SN (t) ].
i=1
Define for t ≥ 0 the functional Y(t) on Mp (E) by Z Z © Y(t)(µ) = 1[0,x) (t − A(s, µ)) µ([0, r) × [0, ∞))u1[0,s) (r) E
E
+µ([0, s) × [0, ∞))(t − A(s, µ))} µ(drdu)µ(dsdx). Then as in Lemma 3.3.1, with probability 1, Y (t) = Y(t)(Φ). The following theorem can be proved using arguments as for Z(t). We omit the proof. Theorem 3.3.2 Let (Xn , n ≥ 1) be an iid sequence of strictly positive random variables with common distribution function F . Let (Sn , n ≥ 0) be the sequence of partial sums of the variables Xn and (N (t), t ≥ 0) be the corresponding renewal process: N (t) = sup{n ≥ 0 : Sn ≤ t}. Let N (t)
Y (t) =
X
(i − 1)Xi + N (t)[t − SN (t) ].
i=1
Then (a) Z
∞
E(e−αY (t) )e−βt dt =
0
∞ n X 1 − F ∗ (αn + β) Y ∗ F (α[i − 1] + β), αn + β n=0 i=1
(b) Z
∞
E[Y (t)]e−βt dt =
0
£
F ∗ (β) , − F ∗ (β)]
β 2 [1
¤
(c) If E X1 e−βX1 < ∞ for some β > 0, then R∞ Z ∞ 2F ∗ (β)[1 − F ∗ (β)2 + β 0 te−βt dF (t)] 2 −βt E[Y (t)]e dt = . β 3 [1 − F ∗ (β)]3 0
3.4
Asymptotic properties
In this section we will discuss asymptotic properties of (Y (t)) and (Z(t)) as defined in Section 3.1 for the case that (N (t)) is a renewal process having interarrival times Xn with common cdf F . We first consider asymptotic properties of the mean of Z(t).
58
Integrated Renewal Processes
Theorem 3.4.1 If µ1 = E[X1 ] < ∞ then as t → ∞ E[Z(t)] ∼
t2 . 2µ1
Proof: In Section 3.3 we have proved that the Laplace transform of E[Z(t)] is given by R ∞ −βx Z ∞ xe dF (x) −βt 0 E[Z(t)]e dt = . ∗ β[1 − F (β)]2 0 Note that Z ∞
Z −βt
e
dE[Z(t)] =
0
−βt
lim E[Z(t)]e
t→∞
∞
+β
E[Z(t)]e−βt dt.
0
Since 0 ≤ Z(t) ≤ tN (t) where N (t) denotes the renewal process corresponding to the sequence (Xn ), it follows that 0 ≤ lim E[Z(t)]e−βt t→∞
lim tE[N (t)]e−βt · ¸ t = lim t + o(1) e−βt t→∞ µ1 = 0. ≤
t→∞
This implies Z
∞
e−βt dE[Z(t)] =
0
=
Z ∞ β E[Z(t)]e−βt dt 0 R ∞ −βx xe dF (x) 0 . ∗ [1 − F (β)]2
By dominated convergence it is easy to see that Z ∞ xe−βx dF (x) = µ1 + o(1) 0
and F ∗ (β) =
1 − µ1 β + o(β)
as β → 0. Hence Z 0
∞
e−βt dE[Z(t)]
∼
1 µ1 β 2
as
β → 0.
Obviously E[Z(t)] is non-decreasing. So we can apply Theorem 2.4.1 (Tauberian theorem) with γ = 2 to get the result. 2
Asymptotic properties
59
Next we will derive a stronger version for the asymptotic form of E[Z(t)]. We will assume that the inter-arrival times Xn are continuous random variables. We also assume that the Laplace transform of E[Z(t)] given in Theorem 3.3.1 is a rational function, i.e., R ∞ −βx xe dF (x) 0 (3.12) β[1 − F ∗ (β)]2 is a rational function of β. This situation holds for example when X1 has a gamma distribution. Since the Laplace transform of E[Z(t)] is a rational function, we can split (3.12) into partial fractions. To do this, firstly observe that F ∗ (0) = 1 and F ∗ 0 (0) = −µ1 < 0. So we conclude that the equation 1 − F ∗ (β) = 0
(3.13)
has a simple root at β = 0. Hence the partial fraction expansion of (3.12) contains terms proportional to 1/β 2 and 1/β. For now, we will consider β as a complex variable and denote its real part by <(β). Assuming µi = E(X1i ), i = 1, 2, 3 are finite we can express the Laplace transform of E[Z(t)] as Z
∞
E[Z(t)]e−βt dt =
0
1 1 + 2 µ1 β 3 µ1
µ
µ3 µ2 + 2 6 4µ1
¶
1 + r(β). β
(3.14)
where r(β) is a rational function of β with non-zero poles at β1 , β2 , .... It follows from (3.13) that for every j, <(βj ) < 0. Also, since F is continuous, then there can be no purely imaginary roots of (3.13). There are some other remarks about the roots βj . If βj is a simple real root then it gives a term proportional to 1/(β −βj ) in the partial fraction of r(β), inverting into eβj t . If βj is a multiple root then it leads to a term proportional to tr eβj t . Since <(βj ) < 0 these terms tend to zero exponentially fast as t → ∞. Note also that the roots βj must occur in conjugate pairs. Otherwise E[Z(t)] contains a term which is a complex number. So from (3.14) we obtain E[Z(t)] =
1 2 1 t + 2 2µ1 µ1
µ
µ3 µ2 + 2 6 4µ1
¶ + o(1) as
t → ∞.
(3.15)
In (3.15) the term o(1) tends to 0 as t → ∞ exponentially fast. Similar arguments as before can be used to obtain the asymptotic variance of Z(t). Assume that µ4 = E[X14 ] < ∞. If the Laplace transform of E[Z 2 (t)] in (3.11) is a rational function of β, then we can prove that Z 0
∞
E[Z 2 (t)]e−βt dt =
6 µ21 β 5
+
2(µ2 − µ21 ) C + + r(β) µ31 β 4 β
60
Integrated Renewal Processes
where C is a constant depending on µi , i = 1, 2, 3, 4 and r(β) is a rational function of β having non-zero poles. Inverting this transform we obtain E[Z 2 (t)] =
1 4 µ2 − µ21 3 t + t + C + o(1) as 4µ21 3µ31
t → ∞.
(3.16)
It follows, from (3.15) and (3.16), that Var[Z(t)] µ2 − µ21 → 3 t 3µ31
as t → ∞.
For the process (Y (t)) we have the following asymptotic properties. The proof is quite similar to that of (Z(t)), and is therefore omitted. Proposition 3.4.1 Let N (t) be a renewal process with inter-arrival times Xn . Rt Let µi = E[X1i ]. Let Y (t) = 0 N (s)ds be the corresponding integrated renewal process. Then (a) If µ1 < ∞ then E[Y (t)]
∼
t2 2µ1
as
t → ∞.
(b) If µi , i = 1, 2, 3 are finite and if the Laplace transform of E[Y (t)] stated in Theorem 3.3.2 is a rational function of β, then µ ¶ µ 2 ¶ 1 2 µ2 µ2 µ3 E[Y (t)] = t + −1 t+ − 2 + o(1) as t → ∞. 2µ1 2µ21 4µ31 6µ1 (c) If µi , i = 1, ..., 5 are finite and if the Laplace transform of E[Y 2 (t)] stated in Theorem 3.3.2 is a rational function of β, then µ ¶ µ ¶ 1 4 5µ2 − 2µ21 3 1 3µ22 21µ2 2µ3 2 E[Y 2 (t)] = t + t + 1 + − − t 4µ21 6µ31 µ21 2µ21 2 3µ1 µ ¶ 1 2µ3 µ2 µ4 3µ2 µ3 3µ32 + 2 − 2+ − + t µ1 3 µ1 4µ1 2µ21 2µ51 +C + o(1) as t → ∞, where C is a constant depending on µi , i = 1, 2, 3, 4, 5. (d) Under the assumptions as in (c), Var[Y (t)] µ2 − µ21 → 3 t 3µ31
as
t → ∞.
An application
3.5
61
An application
Suppose that travellers arrive at a train depot according to a renewal process. Suppose that a train just departed at time 0, and there were no travellers left. If the next train departs at some time t ≥ 0 then the sum of the waiting times PN (t) of all the travellers arriving in the time interval [0, t], i.e., i=1 (t − Si ), is an integrated renewal process. In Ross [36] Example 2.3(A) the special case of this process where the arrival process of the travellers is a homogeneous Poisson process with rate λ has been considered. He showed that the expected sum of the waiting times of the travellers arriving in the time interval [0, t] is equal to 1/2λt2 . The calculation is based on conditioning on the number of arrivals in the interval [0, t]. Using the results in the preceding sections in this example, we give more information about the process. PN (t) Let Y (t) = i=1 (t − Si ). As stated in Section 3.2 the variance of Y (t) is equal to 1/3λt3 , the distribution of Y (t) has mass at 0 with P(Y (t) = 0) = e−λt and the continuous part of Y (t) has a density function, for x > 0, fY (t) (x) √ √ λ = √ e−λt I1 (2 λx)1(0,∞) (x) x ∞ X p 1 (−1)k +λe−λt (λ(x − kt)) 2 (k−1) Ik−1 (2 λ(x − kt))1(kt,∞) (x) k! k=1
where Ik (x) is the Modified Bessel function of the first kind. The graphs of the mean and the variance of Y (t) for λ = 0.5 can be seen in Figure 3.3 and 3.4 (solid line). The graphs of the density function of Y (t) for λ = 0.5, t = 3 and t = 10 can be seen in Figure 3.5 and 3.6 (solid line). Now suppose that the inter-arrival times Xn have a common Gamma(γ,2) distribution having a density function f (x; γ, 2) = γ 2 xe−γx ,
γ > 0,
x ≥ 0.
Note that if γ = 2λ then these inter-arrival times have the same mean as the exponential random variable with parameter λ. Using Theorem 3.3.2 we obtain Z ∞ γ2 E[Y (t)]e−βt dt = , β 3 (β + 2γ) 0 and
Z
∞
E[Y 2 (t)]e−βt dt =
0
2γ 2 (β 3 + 4γβ 2 + 8γ 2 β + 6γ 3 ) . β 5 (β + 2γ)3
Inverting these transforms we obtain E[Y (t)] =
γ 2 1 1 1 −2γt t − t+ − e 4 4 8γ 8γ
62
Integrated Renewal Processes
100
solid line : exp(0.5) dotted line : U(0,4) dashed line : Gamma(1,2)
90
80
70
E(Yt)
60
50
40
30
20
10
0
0
2
4
6
8
10 t
12
14
16
18
20
Figure 3.3: Graphs of the mean of Y (t) when the (Xn ) iid exp(0.5), uniform(0,4) and Gamma(1,2). 1200
solid line : exp(0.5) dotted line : U(0,4) dashed line : Gamma(1,2)
1000
800
t
Var(Y )
600
400
200
0
−200
0
2
4
6
8
10 t
12
14
16
18
20
Figure 3.4: Graphs of the variance of Y (t) when the (Xn ) iid exp(0.5), uniform(0,4) and Gamma(1,2). and E[Y 2 (t)] =
γ2 4 γ 1 1 1 1 1 1 t − t3 + t2 − t+ + ( t2 − t− )e−2γt . 2 16 24 8 8γ 32γ 16 16γ 32γ 2
Hence the variance of Y (t) is given by
An application
63
0.35 exp(0.5) Gamma(1,2) 0.3
0.25
fY(3)(x)
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5 x
6
7
8
9
10
Figure 3.5: Graphs of the probability density function of Y (3) when the (Xn ) are iid exp(0.5) and Gamma(1,2). 0.045 exp(0.5) Gamma(1,2) 0.04
0.035
0.03
fY(10)(x)
0.025
0.02
0.015
0.01
0.005
0
−0.005
0
10
20
30
40 x
50
60
70
80
Figure 3.6: Graphs of the probability density function of Y (10) when the (Xn ) are iid exp(0.5) and Gamma(1,2).
Var[Y (t)] =
γ 3 1 1 1 1 −4γt t − t− + te−2γt − e . 12 6γ 64γ 2 8 64γ 2
The graphs of the mean and the variance of Y (t) for γ = 1 (hence X1 has the same mean as the exponential random variable with parameter 0.5) can be seen in Figure 3.3 and 3.4 (dashed line).
64
Integrated Renewal Processes
The double Laplace transform of Y (t) when the inter-arrival times Xn have a common Gamma(γ, 2) distribution is given by Z ∞ E(e−αY (t) )e−βt dt 0
= γ2
∞ X
n (αn + β + γ)2 − γ 2 Y 1 . 2 (αn + β) (αn + β + γ) [α(i − 1) + β + γ]2 n=0 i=1
The pdf of Y (t) can be approximated by first truncating the infinite sum in this transform and then by inverting the truncated transform. The graphs of the pdfs of Y (3) and Y (10) for γ = 1 can be seen in Figure 3.5 and 3.6 (dashed line). If we assume that the inter-arrival times Xn , are independent and uniformly distributed on [0, 4] (hence X1 has the same mean as an exponential random variable with parameter 0.5), then the Laplace transforms of the first and second moments of Y (t) are given by Z
∞
0
and Z ∞ 0
E[Y 2 (t)]e−βt dt =
E[Y (t)]e−βt dt =
1 − e−4β , β 2 [4β − 1 + e−4β ]
[1 − e−4β ][4β 2 + β − 41 − (4β 2 + β − 21 )e−4β − 41 e−8β ] . β 3 [2β − 21 (1 − e−4β )]3
Using numerical inversions of Laplace transforms in Appendix B we get the graphs of E[Y (t)] and Var[Y (t)], see Figure 3.3 and 3.4 (dotted line). The marginal pdf of Y (t) is more complicated to obtain in this case.
Chapter 4
Total Downtime of Repairable Systems 4.1
Introduction
Consider a repairable system which is at any time either in operation (up) or under repair (down) after failure. The effectiveness of the system can be measured by the total downtime, i.e., the total amount of time the system is down during a given time interval. An expression for the cumulative distribution function (cdf) of the total downtime in the time interval [0, t] has been derived by several authors using different methods. In Tak´ acs [44] the total probability theorem has been used. The derivation in Muth [26] is based on consideration of the excess time. Finally in Funaki and Yoshimoto [13] the cdf of the total downtime is derived by a conditioning technique. Srinivasan et al. [37] derived an expression for the probability density function (pdf) of the total uptime of the system in the time interval [0, t]. They also discussed its covariance structure. For longer time intervals, Tak´ acs [44] and R´enyi [33] proved that the distribution of the total downtime approaches a normal distribution. Tak´ acs [44] also discussed the asymptotic mean and variance of the total downtime. In all these papers it is assumed that the failure time and the repair time are independent. We use a different method for computation of the distribution of the total downtime. We also consider a more general situation where we allow dependence of the failure time and the repair time. Our derivation is based on a representation of the total downtime as a functional of a Poisson point process. This chapter is organized as follows. In Section 4.2 we define the total downtime and derive its distribution in a fixed time interval. In Section 4.3 we discuss the system availability which is closely related to the total downtime. 65
66
Total Downtime of Repairable Systems
In Section 4.4 we study the covariance structure of the total downtime for a dependent case. Asymptotic properties of the total downtime for a dependent case are derived in Section 4.5. We give examples in Section 4.6 and finally in Section 4.7 we consider repairable systems consisting of n ≥ 2 independent components.
4.2
Distribution of total downtime
We consider a repairable system which is at any time either in operation (up) or under repair (down) after failure, denoted as 1 and 0 respectively. Suppose that the system starts to operate at time t = 0. Let (Xi ) and (Yi ), i ≥ 1, denote the time spent in the state 1 and 0 respectively during the ith visit to that state. The random variables Xi and Yi are known as the failure time and the repair time respectively. We assume that the sequence (Xi , Yi ) of random vectors is iid with strictly positive components. However, our set up is more general than that in Tak´acs [44], Muth [26], Funaki and Yoshimoto [13], R´enyi [33], and Srinivasan et al. [37], as we allow that Xi and Yi are dependent. Pn Let Sn = i=1 (Xi + Yi ), n ≥ 1, S0 = 0, and N (t) = sup{n ≥ 0 : Sn ≤ t}. Then the total downtime D(t) can be expressed (with the usual convention that the empty sum equals 0) as ( P N (t) Yi , if SN (t) ≤ t < SN (t) + XN (t)+1 i=1 D(t) = (4.1) PN (t)+1 t − i=1 Xi , if SN (t) + XN (t)+1 ≤ t < SN (t)+1 . Denote the state of the system at time t by Z(t). Then the total downtime D(t) can also be expressed as Z t D(t) = 1{0} (Z(s))ds. (4.2) 0
We will assume that Z(t) is right continuous. Throughout this chapter we will use the following notation for cdfs: F (x) = P(X1 ≤ x), G(y) = P(Y1 ≤ y), H(x, y) = P(X1 ≤ x, Y1 ≤ y), K(w) = P(X1 + Y1 ≤ w). Pn Pn We denote by Fn and Gn the cdfs of i=1 Xi and i=1 Yi , respectively. The Laplace-Stieltjes transforms of a cdf F and a joint cdf H will be denoted by F ∗ and H ∗ , i.e., Z ∞
F ∗ (β) = 0
e−βx F (x)dx
Distribution of total downtime
and
67
Z
∞
∗
Z
H (α, β) = 0
∞
e−(αx+βy) dH(x, y).
0
We will use point processes for the derivation of the distribution of the total downtime D(t). Let (Ω, F, P) be the probability space on which the iid sequence (Xi , Yi ) is defined and also an iid sequence (Ui , i ≥ 1) of exponentially distributed random variables with parameter 1 such that the sequences (Ui ) and (Xi , Yi ) are independent. Let (Tn , n ≥ 1) be the sequence of partial sums of the variables Ui . Then the map P∞ Φ : ω 7→ n=1 δ(Tn (ω),Xn (ω),Yn (ω)) , where δ(x,y,z) is the Dirac measure in (x, y, z), defines a Poisson point process on E = [0, ∞) × [0, ∞) × [0, ∞) with intensity measure ν(dtdxdy) = dtdH(x, y). Note that for almost all ω ∈ Ω, Φ(ω) is a simple point measure on E such that there is at most one point from the support of Φ(ω) on each fibre {t} × [0, ∞) and Φ(ω)([0, t] × [0, ∞)) < ∞ for every t ≥ 0. Let Mp (E) be the set of all point measures on E. We will denote by Pν the distribution of Φ over Mp (E). Define on Mp (E), for t ≥ 0, the functionals Z AX (t)(µ) = x1[0,t) (s)µ(dsdxdy), E
Z AY (t)(µ) = E
y1[0,t) (s)µ(dsdxdy),
and A(t)(µ) = AX (t)(µ) + AY (t)(µ). So for example A(t)(µ) is the sum of the x- and the y-coordinates of the points in the support of µ up to time t. In the sequel it is convenient to write an expression like AX (t)(µ) as AX (t, µ). Define also for t ≥ 0 Z ½ 1[0,x) (t − A(s, µ))AY (s, µ) D(t)(µ) = E ¾ +1[x,x+y) (t − A(s, µ))[t − AX (s+, µ)] µ(dsdxdy). The next lemma motivates the definition of D(t). Lemma 4.2.1 With probability 1, D(t) = D(t)(Φ).
68
Total Downtime of Repairable Systems
Proof: Let ω ∈ Ω. Then
D(t)(Φ(ω)) ∞ · X 1[0,Xi (ω)) (t − A(Ti (ω), Φ(ω)))AY (Ti (ω), Φ(ω)) = i=1
¸ + 1[Xi (ω),Xi (ω)+Yi (ω)) (t − A(Ti (ω), Φ(ω)))[t − AX (Ti (ω)+, Φ(ω))] .
Note that 1[0,Xi (ω)) (t − A(Ti (ω), Φ(ω))) = 1 if and only if SN (t,ω) (ω) ≤ t < SN (t,ω) (ω) + XN (t,ω)+1 (ω), and this last statement implies that i = N (t, ω) + 1. The same conclusion is true when 1[Xi (ω),Xi (ω)+Yi (ω)) (t − A(Ti (ω), Φ(ω))) = 1. Since the intervals {[Si−1 , Si−1 + Xi ), [Si−1 + Xi , Si−1 + Xi + Yi ) : i ≥ 1} partition [0, ∞), for any t > 0 one and only one of the indicators in the sum will be non-zero. So if 1[0,Xi (ω)) (t − A(Ti (ω), Φ(ω))) = 1 then i = N (t, ω) + 1 and D(t)(Φ(ω)) = AY (TN (t,ω)+1 (ω), Φ(ω)) ½ 0, if N (t, ω) = 0 PN (t,ω) = Y (ω), if N (t, ω) ≥ 1, j j=1 and if 1[Xi (ω),Xi (ω)+Yi (ω)) (t − A(Ti (ω), Φ(ω))) = 1 then D(t)(Φ(ω)) = t − AX (TN (t,ω)+1 (ω)+, Φ(ω)) N (t,ω)+1
= t−
X
Xj (ω).
2
j=1
The following theorem gives the distribution of the total downtime D(t) in the form of a double Laplace transform.
Theorem 4.2.1 Let D(t) be as defined in (4.1). Then for α, β > 0 Z 0
∞
E[e−αD(t) ]e−βt dt =
α[1 − F ∗ (β)] + β[1 − H ∗ (β, α + β)] . β(α + β)[1 − H ∗ (β, α + β)]
(4.3)
Distribution of total downtime
69
Proof: By Lemma 4.2.1 E(e−αD(t) ) Z =
e−αD(t)(µ) Pν (dµ)
Mp (E)
½
Z =
exp
Z h
Mp (E)
=
1[0,x) (t − A(s, µ))AY (s, µ)
−α E
¾ i +1[x,x+y) (t − A(s, µ))(t − AX (s+, µ)) µ(dsdxdy) Pν (dµ) Z Z h 1[0,x) (t − A(s, µ))e−αAY (s,µ) Mp (E)
E
i +1[x,x+y) (t − A(s, µ))e−α(t−AX (s+,µ)) µ(dsdxdy)Pν (dµ) =: C1 (α, t) + C2 (α, t). Applying the Palm formula for Poisson point processes, see Theorem 1.2.4, we obtain Z Z C1 (α, t) := 1[0,x) (t − A(s, µ))e−αAY (s,µ) µ(dsdxdy)Pν (dµ) Z
Mp (E) E ∞Z ∞Z ∞
Z
= 0
½
0
0
Mp (E)
0
0
Mp (E)
1[0,x) (t − A(s, µ + δ(s,x,y) ))
¾ exp − αAY (s, µ + δ(s,x,y) ) Pν (dµ)dH(x, y)ds Z ∞Z ∞Z ∞Z 1[0,x) (t − A(s, µ))e−αAY (s,µ)
=
0
Pν (dµ)dH(x, y)ds and Z
Z C2 (α, t) :=
1[x,x+y) (t Mp (E) E Z ∞Z ∞Z ∞Z
− A(s, µ))e−α(t−AX (s+,µ)) µ(dsdxdy)Pν (dµ)
= 0
=
0
0
Mp (E)
0
0
Mp (E)
1[x,x+y) (t − A(s, µ + δ(s,x,y) ))
½ h i¾ exp − α t − AX (s+, µ + δ(s,x,y) ) Pν (dµ)dH(x, y)ds Z ∞Z ∞Z ∞Z 1[x,x+y) (t − A(s, µ)) 0
exp
½
h i¾ − α t − AX (s+, µ) − x Pν (dµ)dH(x, y)ds.
70
Total Downtime of Repairable Systems
Using Fubini’s theorem and a substitution we obtain Z
∞ 0
C1 (α, t)e−βt dt Z ∞Z ∞Z ∞Z
·Z
x
=
−βt
e 0
0
0
¸ dt
0
Mp (E)
i¾ h exp − αAY (s, µ) + βA(s, µ) Pν (dµ)dH(x, y)ds ½ h Z ∞Z i¾ 1 ∗ exp − αAY (s, µ) + βA(s, µ) Pν (dµ)ds. [1 − F (β)] β 0 Mp (E)
=
½
Note that Z αAY (s, µ) + βA(s, µ) = E
1[0,s) (˜ s)(α˜ y + β(˜ x + y˜))µ(d˜ sd˜ xd˜ y ).
So we can use the formula for the Laplace functional of Poisson point processes, see Theorem 1.2.1, to obtain ½
Z exp Mp (E)
= =
h i¾ − αAY (s, µ) + βA(s, µ) Pν (dµ)
½
¾ i 1 − e−1[0,s) (˜s)(α˜y+β(˜x+˜y)) dH(˜ x, y˜)d˜ s 0 0 0 ½ h i¾ ∗ exp − s 1 − H (β, α + β) . exp
Z
∞
Z
∞
Z
∞
h
−
It follows that Z
∞
C1 (α, t)e
−βt
dt =
0
=
½ Z ∞ h i¾ 1 ∗ ∗ exp − s 1 − H (β, α + β) ds [1 − F (β)] β 0 1 − F ∗ (β) . (4.4) β[1 − H ∗ (β, α + β)]
Distribution of total downtime
71
Similarly we calculate the Laplace transform of C2 (α, t) as follows: Z
∞
0
C2 (α, t)e−βt dt Z ∞Z ∞Z ∞Z
Z
∞
= 0
½
0
0
Mp (E)
0
0
Mp (E)
1[x,x+y) (t − A(s, µ))
0
h i¾ − α t − AX (s+, µ) − x e−βt dtPν (dµ)dH(x, y)ds · Z x+y ¸ Z ∞Z ∞Z ∞Z −(α+β)t = e dt exp
0
½
exp
x
¾ − (α + β)A(s, µ) + α[AX (s+, µ) + x] Pν (dµ)dH(x, y)ds · ¸ Z ∞Z ∞Z ∞Z −(α+β)x −(α+β)(x+y) αx e e −e
1 α+β 0 0 0 Mp (E) ½ ¾ exp − (α + β)A(s, µ) + αAX (s+, µ) Pν (dµ)dH(x, y)ds ·Z ∞Z ∞h ¸ i 1 = e−βx − e−(βx+(α+β)y) dH(x, y) α+β 0 0 ½ ¾ Z ∞Z exp − (α + β)A(s, µ) + αAX (s+, µ) Pν (dµ)ds =
0
=
Mp (E)
1 [F ∗ (β) − H ∗ (β, α + β)] α+β ½ ¾ Z ∞Z exp − (α + β)A(s, µ) + αAX (s+, µ) Pν (dµ)ds. 0
Mp (E)
The integral with respect to Pν can be calculated using the Palm formula for Poisson point process as follows: ½
Z exp
¾ − (α + β)A(s, µ) + αAX (s+, µ) Pν (dµ)
Mp (E)
½
Z
=
exp Mp (E)
−
¾ Z h i 1[0,s) (˜ s)(α + β)(˜ x + y˜) − 1[0,s] (˜ s)α˜ x µ(d˜ sd˜ xd˜ y ) Pν (dµ) E
½ Z ∞Z ∞Z ∞h ¾ i −[1[0,s) (˜ s)(α+β)(˜ x+˜ y )−1[0,s] (˜ s)α˜ x] = exp − 1−e dH(˜ x, y˜)d˜ s ½ Z0 ∞ Z0 ∞ h0 ¾ i = exp −s 1 − e−[β x˜+(α+β)˜y] dH(˜ x, y˜) 0 0 ½ h i¾ = exp − s 1 − H ∗ (β, α + β)
72
Total Downtime of Repairable Systems
It follows that Z ∞ C2 (α, t)e−βt dt 0
= =
1 [F ∗ (β) − H ∗ (β, α + β)] α+β F ∗ (β) − H ∗ (β, α + β) . (α + β)[1 − H ∗ (β, α + β)]
Z
½
∞
exp
h i¾ ∗ − s 1 − H (β, α + β) ds
0
(4.5)
Summing (4.4) and (4.5) we get the result. 2 Taking derivatives with respect to α in (4.3) and setting α = 0 we get the Laplace transforms of E[D(t)] and E[D2 (t)] as stated in the following proposition: Proposition 4.2.1 For β > 0, (a) Z
∞
E[D(t)]e−βt dt =
0
F ∗ (β) − H ∗ (β, β) , β 2 [1 − H ∗ (β, β)]
(4.6)
(b) Z
∞
E[D2 (t)]e−βt dt =
0
−
· 2 F ∗ (β) − H ∗ (β, β) β3 1 − H ∗ (β, β) # R∞R∞ β[1 − F ∗ (β)] 0 0 ye−β(x+y) dH(x, y) (4.7) . [1 − H ∗ (β, β)]2
Remark 4.2.1 For the case that (Xi ) and (Yj ) are independent (4.3) simplifies to Z ∞ α[1 − F ∗ (β)] + β[1 − F ∗ (β)G∗ (α + β)] E(e−αD(t) )e−βt dt = . (4.8) β(α + β)[1 − F ∗ (β)G∗ (α + β)] 0 Tak´ acs [44], Muth [26], Funaki and Yoshimoto [13] derived for the independent case the following formula for the distribution function of the total downtime: ½ P∞ P(D(t) ≤ x) =
1,
n=0
Gn (x)[Fn (t − x) − Fn+1 (t − x)], t > x t ≤ x.
Taking double Laplace transforms on both sides of (4.9) we obtain (4.8).
(4.9)
System availability
4.3
73
System availability
This section concerns the system availability of repairable systems, which is closely related to the total downtime. The system availability A11 (t) at time t is defined as the probability that the system is working at time t, i.e., A11 (t) =
P(Z(t) = 1).
The relationship between the system availability A11 (t) and the total downtime D(t) is given by the following equation: Z t E[D(t)] = t − A11 (s)ds, (4.10) 0
which can easily verified using (4.2). In Pham-Gia and Turkkan [30] the system availability of a repairable system where both uptime and downtime are gamma distributed has been considered. They calculate the system availability by computing numerically the renewal density of a renewal process with inter-arrival times the sum of two gamma random variables, and then using the following integral equation: Z t A11 (t) = F¯ (t) + F¯ (t − u)dm(u), (4.11) 0
where m(t) = E[N (t)]. This equation can be found for example in Barlow [2]. In general an expression for the system availability can be derived using our result about the expected value of the total downtime given in (4.6), and using (4.10). Taking Laplace transforms on both sides of (4.10) we obtain Z ∞ Z 1 1 ∞ E[D(t)]e−βt dt = − A11 (t)e−βt dt. 2 β β 0 0 Taking (4.6) into consideration we obtain Z ∞ A11 (t)e−βt dt = 0
1 − F ∗ (β) . β[1 − H ∗ (β, β)]
In particular, if (Xi ) and (Yi ) are independent then Z ∞ 1 − F ∗ (β) A11 (t)e−βt dt = . β[1 − F ∗ (β)G∗ (β)] 0
(4.12)
(4.13)
Remark 4.3.1 The Laplace transform of A11 (t) can also be derived from (4.11). Taking Laplace transform on both sides of this equation we obtain Z ∞ 1 A11 (t)e−βt dt = [1 − F ∗ (β)][1 + m∗ (β)] (4.14) β 0
74
Total Downtime of Repairable Systems
where m∗ is the Laplace-Stieltjes transform of m(t). But it is well known that m∗ (β) =
K ∗ (β) , 1 − K ∗ (β)
where K is the cdf of X1 + Y1 . Substituting this equation into (4.14) and using the fact that K ∗ (β) = H ∗ (β, β) we get (4.12). Example 4.3.1 Let (Xi , i ≥ 1) be an iid sequence of non-negative random variables having a common Gamma(λ, m) distribution with a pdf f (x; λ, m) =
λn xn−1 e−λx , Γ(m)
x ≥ 0.
Let (Yi , i ≥ 1) be an iid sequence of non-negative random variables having a common Gamma(µ, n) distribution. Assume that (Xi ) and (Yi ) are independent. Then using (4.13) we obtain Z ∞ (µ + β)n [(λ + β)m − λm ] A11 (t)e−βt dt = . (4.15) β[(λ + β)m (µ + β)n − λm µn ] 0 The system availability A11 (t) can be obtained by inverting this transform. As an example let m = n = 1. Then X1 and Y1 are exponentially distributed with parameter λ and µ respectively. The system availability is given by A11 (t) =
µ λ −(λ+µ)t + . e λ+µ λ+µ
(4.16)
As another example let m = n = 2, λ = 1 and µ = 2. In this case A11 (t) =
√ √ 2 1 1 5 √ −3t/2 + e−3t + e−3t/2 cos( 7t/2) + 7e sin( 7t/2). 3 12 4 28
For non-integers m and n we can invert numerically the transform in (4.15). As an example let m = 7.6, n = 2.4, λ = 2 and µ = 1.1765. In this case Z ∞ (1.1765 + β)2.4 ((2 + β)7.6 − 27.6 ) A11 (t)e−βt dt = . β[(2 + β)7.6 (1.1765 + β)2.4 − 27.6 1.17652.4 ] 0 The graph of A11 (t) can be seen in Figure 4.1 which is the same as Figure 1 in Pham-Gia and Turkkan [30]. Example 4.3.2 Let ((Xn , Yn ), n ≥ 1) be an iid sequence of non-negative random vectors having a common joint bivariate exponential distribution given by P(X1 > x, Y1 > y) =
e−(λ1 x+λ2 y+λ12 max(x,y)) ;
x, y ≥ 0;
λ1 , λ2 , λ12 > 0.
System availability
75
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0
2
4
6
8
10
12
14
16
18
20
Figure 4.1: Graph of A11 (t). Obviously =
e−(λ1 +λ12 )x
(4.17)
P(Y1 > y) =
e−(λ2 +λ12 )y .
(4.18)
P(X1 > x) and
The correlation coefficient ρXY between X1 and Y1 is given by ρXY =
λ12 . λ1 + λ2 + λ12
The Laplace-Stieltjes transform of the cdf F of X1 and the joint cdf H of X1 and Y1 are given by λ1 + λ12 F ∗ (β) = β + λ1 + λ12 and H ∗ (α, β) =
(λ1 + λ2 + λ12 + α + β)(λ1 + λ12 )(λ2 + λ12 ) + λ12 αβ , (λ1 + λ2 + λ12 + α + β)(λ1 + λ12 + α)(λ2 + λ12 + β)
see Barlow [2]. It follows that Z ∞ A11 (t)e−βt dt = 0
(λ + 2β)(λ2 + λ12 ) β[2β 2 + (3λ + λ12 )β + λ(λ + λ12 )]
76
Total Downtime of Repairable Systems
where λ = λ1 + λ2 + λ12 . Inverting this transform we obtain
A11 (t) =
4.4
λ2 + λ12 λ1 λ12 (λ2 − λ1 ) + e−λt + e−(λ+λ12 )t/2 . λ + λ12 λ1 + λ2 (λ1 + λ2 )(λ + λ12 )
Covariance of total downtime
Let U (t) = t − D(t) be the total uptime of the system up to time t. Obviously Cov(D(t1 ), D(t2 )) = Cov(U (t1 ), U (t2 )). So we might as well study Cov(U (t1 ), U (t2 )). Let 0 ≤ t1 ≤ t2 < ∞. Then ·Z E[U (t1 )U (t2 )] =
t1
Z
t2
E Z
x=0 y=0 t1 Z t1
¸ 1{1} (Z(x))1{1} (Z(y))dydx
P(Z(x) = 1, Z(y) = 1)dydx
= 2 x=0 Z t1
y=x Z t2
P(Z(x) = 1, Z(y) = 1)dydx.
+ x=0
(4.19)
y=t1
Let ϕ(x, y) = P(Z(x) = 1, Z(y) = 1). For 0 ≤ x ≤ y < ∞, ϕ(x, y) =
P(Z(x) = 1, Z(y) = 1, y < X1 ) +P(Z(x) = 1, Z(y) = 1, x < X1 < y) +P(Z(x) = 1, Z(y) = 1, X1 < x).
(4.20)
Obviously P(Z(x) = 1, Z(y) = 1, y < X1 ) = 1 − F (y). For the second term, note that the event ”Z(x) = 1, Z(y) = 1, x < X1 < y” is equivalent toP the event ”x < X1 and forPsome n ≥ 1, Sn < y < Sn + Xn+1 ”, n n where Sn = i=1 (Xi + Yi ). Let Rn = i=2 (Xi + Yi ), n ≥ 2. Then (X1 , Y1 ),
Covariance of total downtime
77
Rn and Xn+1 are independent. Denote by Kn the cdf of Rn . Then P(Z(x) = 1, Z(y) = 1, x < X1 < y) ∞ X = P(x < X1 , Sn ≤ y < Sn + Xn+1 ) n=1
= P(x < X1 , X1 + Y1 ≤ y < X1 + Y1 + X2 ) ∞ X + P(x < X1 , (X1 + Y1 ) + Rn ≤ y < (X1 + Y1 ) + Rn + Xn+1 ) n=2
Z
Z
Z
=
dF (x2 )dH(x1 , y1 ) x1 ∈(x,y] ∞ Z X
y1 ∈[0,y−x1 ]
x2 ∈(y−x1 −y1 ,∞)
Z
Z
Z
+
n=2
x1 ∈(x,y]
y1 ∈[0,y−x1 ]
rn ∈[0,y−x1 −y1 ]
xn+1 ∈(y−x1 −y1 −rn ,∞)
dF (xn+1 )dKn (rn )dH(x1 , y1 ) (Z Z Z =
dF (x2 ) x1 ∈(x,y]
+ Z
w∈[x1 ,y]
∞ Z X
dF (xn+1 )dKn (rn ) dH(x1 , w − x1 ) Z
= Z
)
Z rn ∈[0,y−w]
n=2
x2 ∈(y−w,∞)
x1 ∈(x,y]
xn+1 ∈(y−w−rn ,∞) ∞ X
P(Rn ≤ y − w < Rn + Xn+1 )dH(x1 , w − x1 )
w∈[x1 ,y] n=1
Z
=
P(z(y − w) = 1)dH(x1 , w − x1 ) Z
x1 ∈(x,y]
w∈[x1 ,y]
Z
=
A11 (y − w)dH(x1 , w − x1 ), x1 ∈(x,y]
w∈[x1 ,y]
where A11 (t) denotes the availability of the system at time t starting in state 1 at time 0. Finally, the last term in (4.20) can be obtained by conditioning on X1 + Y1 , i.e., P(Z(x) = 1, Z(y) = 1, X1 ≤ x) = P(Z(x) = 1, Z(y) = 1, X1 + Y1 ≤ x) Z ∞ P(Z(x) = 1, Z(y) = 1, X1 + Y1 ≤ x|X1 + Y1 = w)dK(w) = Z0 x = P(Z(x − w) = 1, Z(y − w) = 1)dK(w) Z0 x = ϕ(x − w, y − w)dK(w). 0
78
Total Downtime of Repairable Systems
So we obtain Z
Z ϕ(x, y)
=
A11 (y − w)dH(x1 , w − x1 )
1 − F (y) + x1 ∈(x,y]
Z
w∈[x1 ,y]
x
+
ϕ(x − w, y − w)dK(w).
(4.21)
0
Taking double Laplace transforms on both sides of (4.21) we obtain Z ∞Z ∞ ϕ(x, y)e−αx−βy dxdy ϕ(α, ˆ β) := 0
=
where
0
α[1 − F ∗ (β)] − β[F ∗ (β) − F ∗ (α + β)] αβ(α + β)[1 − K ∗ (α + β)] Aˆ11 (β)[H ∗ (β, β) − H ∗ (α + β, β)] + . α[1 − K ∗ (α + β)] Z
Aˆ11 (β) :=
∞
A11 (t)e−βt dt =
0
1 − F ∗ (β) , β[1 − H ∗ (β)]
see (4.12). This formula is a generalization of the result in Srinivasan [37], since H ∗ (β) = F ∗ (β)G∗ (β) when Xi and Yi are independent. Now from (4.19) we obtain Z ∞ Z ∞ E[U (t1 )U (t2 )]e−αt1 −βt2 dt2 dt1 t1 =0
= = It follows that Z ∞Z ∞ 0
4.5
t2 =t1
2ϕ(0, ˆ α + β) ϕ(α, ˆ β) − ϕ(0, ˆ α + β) + β(α + β) αβ ϕ(α, ˆ β) [α − β]ϕ(0, ˆ α + β) + . αβ αβ(α + β)
E[U (t1 )U (t2 )]e−αt1 −βt2 dt1 dt2
0
=
1 [ϕ(α, ˆ β) + ϕ(β, ˆ α)]. αβ
Asymptotic properties
In this section we want to address asymptotic properties of the total downtime D(t). To this end we use a method in Tak´acs [44] which is based on a comparison with the asymptotic properties of a delayed renewal process related to the process that we are studying. First we summarize some known results about ˜ (t), t ≥ 0) which will be used in the following. delayed renewal processes (N
Asymptotic properties
79
Let (Vn , n ≥ 1) be an i.i.d. sequence of non-negative random variables. Let V0 be a non-negative random variable which is independent of the sequence Pn−1 ˜ (t) = sup{n ≥ 0 : S˜n ≤ t}. (Vn ). Define S˜0 = 0, S˜n = i=0 Vi , n ≥ 1 and N ˜ (t) The Laplace-Stieltjes transforms of the first and the second moments of N are given by Z
∞
E(e−βV0 ) 1 − E(e−βV1 )
(4.22)
2E(e−βV0 ) E(e−βV0 ) − , [1 − E(e−βV1 )]2 1 − E(e−βV1 )
(4.23)
˜ (t)] e−βt dE[N
=
0
and Z
∞
˜ 2 (t)] = e−βt dE[N
0
respectively, see Tak´ acs [44]. Now we are in a position to derive the asymptotic properties of D(t). The same argument as used in Tak´ acs [44] for the independent case can now be employed to derive asymptotic properties of the total downtime for the dependent case. 2 Let µX = E(X1 ), µY = E(Y1 ), σX = Var(X1 ), σY2 = Var(Y1 ) and σXY = Cov(X1 , Y1 ). Let Vn = Xn + Yn , n = 1, 2, 3.... (4.24) ˜ (t), t ≥ 0) be the delayed renewal process determined by Lemma 4.5.1 Let (N the random variables (Vn ), n = 0, 1, 2, ..., where V0 has the distribution ½ P(V0 ≤ x) =
1 µX
0,
Rx 0
[1 − F (y)]dy,
x≥0 x<0
(4.25)
and Vn , n = 1, 2, ..., are defined as in (4.24). Then ˜ (t)] = t. E[D(t)] + µX E[N Proof: It is easy to verify that E(e−βV0 ) = Substitution in (4.22) yields Z
∞
˜ (t)] = e−βt dE[N
0
1−F ∗ (β) βµX
(4.26) and E(e−βV1 ) = H ∗ (β, β).
1 − F ∗ (β) . βµX [1 − H ∗ (β, β)]
(4.27)
F ∗ (β) − H ∗ (β, β) . β[1 − H ∗ (β, β)]
(4.28)
Now from (4.6) we obtain Z 0
∞
e−βt dE[D(t)] =
80
Total Downtime of Repairable Systems
Taking the Laplace-Stieltjes transform on the left-hand side of (4.26) and taking (4.27) and (4.28) into consideration we obtain Z ∞ Z ∞ −βt ˜ (t)] e dE[D(t)] + µX e−βt dE[N 0
0
F ∗ (β) − H ∗ (β, β) 1 − F ∗ (β) = + ∗ β[1 − H (β, β)] β[1 − H ∗ (β, β)] 1 = β which is equal to the Laplace-Stieltjes transform of t.
2
Remark 4.5.1 The relation (4.26) has been proved by Tak´ acs [44] for the case that (Xi ) and (Yj ) are independent. Theorem 4.5.1 If µX + µY < ∞, then lim
t→∞
E[D(t)] µY = , t µX + µY
(4.29)
2 and if σX and σY2 are finite, and X1 + Y1 is a non-lattice random variable, then µ ¶ 2 µY t µY σX − µX σY2 − 2µX σXY µX µY lim E[D(t)] − = − . t→∞ µX + µY 2(µX + µY )2 2(µX + µY ) (4.30)
Proof: From Lemma 4.5.1 and (1.10) we obtain E[D(t)] t→∞ t lim
˜ (t)] E[N t→∞ t µX = 1− µX + µY µY = . µX + µY =
1 − µX lim
For the second part, from (4.26) we obtain µ ¶ µ ¶ µY t t ˜ (t)] − lim E[D(t)] − = −µX lim E[N (. 4.31) t→∞ t→∞ µX + µY µX + µY Now if X1 + Y1 is a non-lattice random variable, then using (1.11) we obtain µ ¶ t ˜ (t)] − lim E[N t→∞ µX + µY 2 2 σX + σY + 2σXY + (µX + µY )2 µ0 = − (4.32) 2(µX + µY )2 µX + µY
Asymptotic properties
81
Next it is easy to see that µ0 = E(V0 ) =
2 + µ2X σX . 2µX
(4.33)
Substitution (4.32) and(4.33) into (4.31) completes the proof. 2 Remark 4.5.2 The first result (4.29) of Theorem 4.5.1 can be proved using a Tauberian theorem. From (4.28) if µX and µY are finite, we obtain Z ∞ µY e−βt dE[D(t)] ∼ as β → 0. β(µX + µY ) 0 Obviously E[D(t)] is non-decreasing. So we can use Theorem 2.4.1 to conclude (4.29). Now we will derive the asymptotic variance of D(t). ˜ (t), t ≥ 0) be the delayed renewal process determined by Lemma 4.5.2 Let (N the random variables (Vn ), n = 0, 1, 2, ..., where V0 has the Laplace transform R∞R∞ [1 − F ∗ (β)] 0 0 ye−β(x+y) dH(x, y) −βV0 E(e ) = βµX µY and Vn , n = 1, 2..., are defined as in (4.24). Then Z
t
E[D2 (t)] = 2
˜ (t)] + E[N ˜ 2 (t)]). E[D(u)]du − µX µY (E[N
(4.34)
0
Proof: Firstly, it is easy to see from the Laplace transform of V0 that µ0 = E(V0 ) = µX +
2 + µ2X σX σXY + σY2 + µ2Y + . 2µX µY
(4.35)
Then take the Laplace-Stieltjes transforms on both sides of (4.34) and use (4.6), (4.7), (4.22), (4.23), and (4.35). 2 2 Theorem 4.5.2 If σX and σY2 are finite, and X1 + Y1 is a non-lattice random variable, then
Var[D(t)] t→∞ t lim
=
2 − 2µX µY σXY µ2X σY2 + µ2Y σX . (µX + µY )3
Proof: If X1 + Y1 is a non-lattice random variable then, using (1.11) and (1.12) respectively, we can show that the delayed renewal process defined in Lemma
82
Total Downtime of Repairable Systems
4.5.2 has properties as t → ∞ ˜ (t)] = E[N −
t σ 2 + σY2 + 2σXY + (µX + µY )2 + X µX + µY 2(µX + µY )2 2 2 2 µY (σX + µX ) + 2µX µY + 2µX (σXY + σY2 + µ2Y ) + o(1) 2µX µY (µX + µY )
and ˜ 2 (t)] = E[N From (4.30) we deduce Z t 2 E[D(u)]du = 0
+
˜ (t)]2 + E[N
2 + σY2 + 2σXY σX t + o(t) (µX + µY )3
· ¸ 2 µY t2 − µX σY2 − 2µX σXY µY σX µX µY + − t µX + µY (µX + µY )2 µX + µY o(t) as t → ∞.
It follows, using Lemma 4.5.2, that µ2Y t2 (µX + µY )2 · ¸ 2 µX µ3Y + (µ2X − 2σX )µ2Y + (σY2 + 4σXY )µX µY − µ2X σY2 − t (µX + µY )3 + o(t).
E[D2 (t)] =
and hence, by taking (4.30) into consideration, Var[D(t)] = =
E[D2 (t)] − E[D(t)]2 2 µ2X σY2 + µ2Y σX − 2µX µY σXY t + o(t). (µX + µY )3
2
Now we will consider the asymptotic distribution of the total downtime. For the case that (Xi ) and (Yj ) are independent the limiting distribution of the total downtime D(t) is normal as t → ∞, i.e., D(t) − q 2 2
µY t µX +µY
d
2 µX σY +µ2Y σX (µX +µY )3 t
−→ N (0, 1),
(4.36)
2 provided σX and σY2 are finite, see Tak´ acs [44] and R´enyi [33]. In the following theorem we give the limiting distribution of the total downtime D(t) for the dependent case. 2 Theorem 4.5.3 If σX and σY2 are finite then
q
D(t) −
µY t µX +µY
2 +µ2 σ 2 −2µ µ σ µ2X σY X Y XY Y X (µX +µY )3
d
−→ N (0, 1) t
as
t → ∞.
Examples
83
Proof: First note that N (t)
X
N (t)+1
Yi ≤ D(t) ≤
i=1
X
Yi ,
i=1
Pn where N (t) = sup{n ≥ 0 : j=1 (Xj + Yj ) ≤ t}. Using the Central Limit Theorem for random sums, see Embrechts et al. [11], we obtain · µ ¶ ¸−1/2 N (t) X µY (X1 + Y1 ) t t µ d Y −→ Var Y1 − Yi − N (0, 1), µX + µY µX + µY µ + µ X Y i=1 where µ ¶ µY (X1 + Y1 ) t Var Y1 − µX + µY µX + µY µ ¶ 2 µ 2µY t 2 Y = σY + Var(X + Y ) − Cov(Y, X + Y ) 2 (µX + µY ) µX + µY µX + µY 2 σY2 (µX + µY )2 + µ2Y (σX + σY2 + 2σXY ) − 2µY (µX + µY )(σXY + σY2 ) = t (µX + µY )3 2 µ2 σ 2 + µ2Y σX − 2µX µY σXY = X Y t. (µX + µY )3 The proof is complete if we can show that YN (t)+1 P √ −→ 0 as t
t → ∞.
(4.37)
P
1 But by the fact that N t(t) −→ µX +µ (> 0) and the assumption that σY2 < ∞, Y then using Lemma 2.4.3 we obtain
YN (t) P p −→ 0. N (t) Hence (4.37) follows.
4.6
2
Examples
In this section we give two examples. In the first example we will see the effect of dependence of the failure and repair times on the distribution of the total downtime. In the second example we will see that for some cases we have analytic expressions for the first and second moments of the total downtime.
84
Total Downtime of Repairable Systems
Example 4.6.1 Let (Xi , i ≥ 1) and (Yi , i ≥ 1) be the sequences of the failure times and repair times respectively, of a repairable system such that ((Xi , Yi ), i ≥ 1) is an iid sequence of non-negative random vectors. Let X1 and Y1 have a joint bivariate exponential distribution given by P(X1 > x, Y1 > y) =
e−(λ1 x+λ2 y+λ12 max(x,y)) ;
x, y ≥ 0;
λ1 , λ2 , λ12 > 0.
The marginals are given by P(X1 > x) =
e−(λ1 +λ12 )x
(4.38)
P(Y1 > y)
e−(λ2 +λ12 )y .
(4.39)
and =
1 1 1 2 2 λ1 +λ12 , µY = λ2 +λ12 , σX = (λ1 +λ12 )2 , σY = λ12 (λ1 +λ2 +λ12 )(λ1 +λ12 )(λ2 +λ12 ) , and the correlation coefficient
In this case we have µX = 1 (λ2 +λ12 )2 ,
σXY = ρXY between X1 and Y1
ρXY =
λ12 . λ1 + λ2 + λ12
Using (4.6) we obtain Z ∞ E[D(t)]e−βt dt 0
=
(2λ1 + λ12 )β + (λ1 + λ12 )2 + λ2 (λ1 + λ12 ) . β 2 [2β 2 + (3λ1 + 3λ2 + 4λ12 )β + (λ1 + λ2 )2 + 3λ12 (λ1 + λ2 ) + 2λ212 ]
This transform can be inverted analytically. As an example for λ1 = 1, λ2 = 2, and λ12 = 3 we obtain E[D(t)] =
4 13 2 1 t− + e−9t/2 + e−6t . 9 162 81 18
The distribution of D(t) has mass at 0 with P(D(t) = 0) = P(X1 > t) = e−(λ1 +λ12 )t . The pdf of the continuous part of D(t) can be obtained by inverting its double Laplace transform. As an example let λ1 = 1, λ2 = 2, and λ12 = 3. In this case Z
∞
E[e−αD(t) ]e−βt dt =
0
where C(α, β) =
α+β−
4α 4+β
− βC(α, β)
β(α + β) [1 − C(α, β)]
20(6 + α + 2β) + 3β(α + β) . (6 + α + 2β)(4 + β)(5 + β)
(4.40)
Examples
85
0.8
exact density asymptotic density
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
Figure 4.2: The graph of the density of D(10) with λ1 = 1 λ2 = 2 and λ12 = 3. Solid line for exact density; dashed line for normal approximation.
Using the numerical inversion of a double Laplace transform, see Appendix B, we get the graph of the pdf of D(10), see Figure 4.2 (solid line). In this Figure we also compare the pdf of D(10) with its normal approximation (dashed line). We see that the pdf of D(10) is close to its normal approximation. The effect of dependence between the failure and the repair times can be seen in Figure 4.3. In this figure we compare the graphs of the normal approximations of D(10) where (Xi ) and (Yj ) are independent and satisfy (4.38) and (4.39) with the normal approximations of D(10) for various correlation coefficients ρXY . We see that the smaller their correlation coefficient the closer their normal approximations. Example 4.6.2 Let (Xi , i ≥ 1) and (Yi , i ≥ 1) denote the sequences of the failure times and repair times respectively, of a repairable system such that (Xi ) are iid non-negative random variables having a common Gamma(λ, m) distribution with a pdf fX1 (x) =
λm m−1 −λx e , x Γ(m)
x≥0
and (Yi ) are iid non-negative random variables having a common Gamma(µ, n) distribution. Firstly we will consider the case where m = n = 1. In this case X1 and Y1 are exponentially distributed with parameter λ and µ respectively. Using (4.6),
86
Total Downtime of Repairable Systems
1.8
0.7
dependent independent
1.6
dependent independent 0.6
1.4 0.5 1.2 0.4
1
0.8
0.3
0.6 0.2 0.4 0.1 0.2
0
0
1
2
3
4
5
6
7
8
9
0
10
(a) σXY = 0.8 (λ1 = λ2 = 1, λ12 = 8)
0
1
2
3
4
5
6
7
8
9
10
(b) σXY = 0.5 (λ1 = λ2 = 1, λ12 = 2)
0.5
dependent independent
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
(c) σXY = 0.2 (λ1 = λ2 = 2, λ12 = 1)
Figure 4.3: The graphs of the normal approximations of D(10). Solid line for dependent cases and dashed line for the independent cases. (4.7), and (4.3) we obtain Z
∞
E[D(t)]e−βt dt =
0
Z
∞
E[D2 (t)]e−βt dt =
0
β 2 [β
λ , + λ + µ]
2λ(λ + β) , β 3 [β + λ + µ]2
and Z
∞
E[e−αD(t) ]e−βt dt =
0
α+β+λ+µ . (λ + β)(α + β) + µβ
Inverting these transforms we obtain E[D(t)] =
λt λ − (1 − e−(λ+µ)t ), λ + µ (λ + µ)2
Examples
87
λ2 t2 2λ(µ − λ)t + (λ + µ)2 (λ + µ)3
E[D2 (t)] =
+
2λ(λ − 2µ) + 2λ[2µ − λ + µ(λ + µ)t]e−(λ+µ)t , (λ + µ)4
and µ ¶ √ √ α+λ+µ √ e−(α+λ+µ)t/2 cos( ct/2) + sin( ct/2) (4.41) c
E[e−αD(t) ] =
where c = 4λα − (α + λ + µ)2 . The graph of the pdf of D(t) can be obtain by inverting numerically the Laplace transform in (4.41). As an example the graph of the pdf of D(20) for λ = 1 and µ = 5 can be seen in Figure 4.4 (dashed line). 0.7
X1~G(2,2),Y1~G(10,2) X1~exp(1),Y1~exp(5)
0.6
0.5
ft(x)
0.4
0.3
0.2
0.1
0
0
1
2
3
4 x
5
6
7
8
Figure 4.4: The graph of the pdf of D(20) with X1 ∼ exp(1), Y1 ∼ exp(2) (solid line) and X2 ∼ Gamma(2, 2), Y2 ∼ Gamma(10, 2) (dashed line). For integers m and n we have explicit expressions for the first and second moments of D(t). As an example, if m = n = 2, λ = 2 and µ = 10, then using (4.6) and (4.7) we obtain E[D(t)] =
1 1 1 −12t 1 1 t− + + e cos(2t)e−6t + sin(2t)e−6t , 6 18 180 20 10
and E[D2 (t)]
=
1 2 1 1 1 −12t 1 −12t 1 t + t− + + e + t cos(2t)e−6t te 36 216 864 864 108 4 1 −6t − sin(2t)e . 8
88
Total Downtime of Repairable Systems
In this case we also have explicit expressions for E[e−αD(t) ]. The graph of the pdf of D(20) for m = n = 2, λ = 2 and µ = 10 can be seen in Figure 4.4 (solid line). Note that in this case X1 and Y1 have means 1 and 0.2 respectively, which are the same as the means of exponential random variables with parameter 1 and 5 respectively.
4.7
Systems consisting of n independent components
In this section we will consider the distribution of the total uptime (downtime) of a system consisting of n ≥ 2 stochastically independent components. The system we will discuss can be a series, a parallel or a k-out-of-n system, but we will formulate the results only for a series system. The results only concern the total uptime. The corresponding results for the total downtime can be derived similarly, or using the obvious relation between the total uptime and total downtime. Firstly we will discuss the case where both the failure and repair times of the components are exponentially distributed, and later on we will consider the case where the failure or the repair times are arbitrarily distributed.
4.7.1
Exponential failure and repair times
Consider a series system comprising n ≥ 2 stochastically independent two-state components, each of which can be either up or down, denoted as 1 and 0 respectively. Suppose that system starts to operate at time 0. If a component fails, it is repaired and put into operation again. During the repair the unfail component may fail. There are no capacity constrains at the repair shop. Denote by Zi (t), i = 1, 2, ..., n the state of the ith component at time t. Then the total uptime of the system in the time interval [0, t] is given by Z t U (t) = 11n (Z1 (s), ..., Zn (s))ds (4.42) 0
where 1n denotes the vector of ones of length n. Let Xij and Yij , j = 1, 2, ..., denote the consecutive uptimes and downtimes, respectively, of the ith component. Assume that the sequences (Xij ) and (Yij ) are independent. Assume also that for each i, the random variables Xij , j = 1, 2, ..., have a common exponential distribution with parameter λi , and Yij , j = 1, 2, ..., have a common exponential distribution with parameter µi . Then (Zi (t), t ≥ 0), i = 1, 2, . . . , n are independent, continuous time Markov chains on {0, 1} with generators ¶ µ −µi µi , λi , µi > 0. Qi = λi −λi
Systems consisting of n independent components
89
Let Yn (t) = (Z1 (t), Z2 (t), . . . , Zn (t)). Then Yn = (Yn (t), t ≥ 0) is a continuous-time Markov chain on I = {0, 1}n , the set of row vectors of length n with entries zeros and or ones. Let a ∈ I be a state of the Markov chain Yn . Then a has the form a = (²1 (a), ²2 (a), . . . , ²n (a)) where ²j (a) ∈ {0, 1}, j = 1, 2, . . . , n. The generator Q = (qab )a,b∈I of the Markov chain Yn has the following properties. Suppose b = (²1 (b), ²2 (b), . . . , ²n (b)). If a and b have two or more different entries, then qab = qba = 0. Suppose now a and b have only one different entry. Then there exists an index j such that ²i (a) = ²i (b) for all i 6= j and ²j (a) = 1 − ²j (b). Let νi,0 = λi
and νi,1 = µi .
Then qab = νj,²j (b)
and
qba = νj,²j (a) .
Lemma 4.7.1 The vector π = (πa )a∈I where πa = ν1,²1 (a) ν2,²2 (a) . . . νn,²n (a)
n Y
1 λ + µi i=1 i
is the stationary distribution of the Markov chain Yn . Proof: It is clear that πa is non-negative for every a ∈ I. The fact that X πa = 1 a∈I
can be proved by induction. The proof is complete if we can show that πa qab = πb qba
for all a, b ∈ I.
90
Total Downtime of Repairable Systems
If a and b have two or more different entries then qab = qba = 0. Now suppose that a and b have only one different entry, say at jth entry. In this case νi,²i (a) = νj,²j (b) for all i 6= j and if νj,²j (a) = λj then νj,²j (b) = µj and vice versa. The entries πa and πb of π are given by πa = Cνj,²j (a) and πb = Cνj,²j (b) where C = ν1,²1 (a) . . . νj−1,²j−1 (a) νj+1,²j+1 (a) . . . νn,²n (a)
n Y
1 . λ + νi i i=1
It follows that πa qab
=
Cνj,²j (a) νj,²j (b)
=
Cνj,²j (b) νj,²j (a)
=
πb qba .
2
As a consequence of the fact that the Markov chain Yn has the stationary distribution π, we have the following proposition: Proposition 4.7.1 Let U (t) be the total uptime of a series system with n stochastically independent components. Suppose that the ith, i = 1, 2, . . . , n component has up and down times which are independent and exponentially distributed with parameter λi and µi respectively. Then (a) Eπ [U (t)] =
n Y
µi t, λ + µi i=1 i
where Eπ denotes the expectation under initial distribution the stationary distribution π, (b) with probability 1, n
Y µi U (t) −→ , t λ + µi i=1 i
as
t −→ ∞.
Systems consisting of n independent components
91
Proof: Use the fact that if the initial distribution of the Markov chain Yn is its stationary distribution then the distribution of the chain at any time equals the stationary distribution, and the fact that the fraction of time the chain in a state approximately equals the stationary distributions at that state, see e.g. Wolff [49]. 2 Next we will derive the variance of U (t) for the case n = 2. For the bigger n the calculations become more complicated. For n = 2 the Markov chain Y2 has the state space I = {00, 01, 10, 11} and stationary distribution π = (πa )a∈I =
1 (λ1 λ2 (λ1 + µ1 )(λ2 + µ2 )
λ1 µ2
µ1 λ2
µ1 µ2 ).
The transition matrix of the Markov chain Y2 can be calculated explicitly: P (t)
=
P1 + P2 e−(λ1 +µ1 )t + P3 e−(λ2 +µ2 )t + P4 e−(λ1 +µ1 +λ2 +µ2 )t (λ1 + µ1 )(λ2 + µ2 )
where
1 1 P1 = 1 π, 1
µ 1 λ2 µ 1 λ2 P2 = −λ1 λ2 −λ1 λ2
λ1 µ2 −λ1 λ2 P3 = λ1 µ2 −λ1 λ2 and
µ1 µ2 −µ1 λ2 P4 = −λ1 µ2 λ1 λ 2
µ1 µ2 µ1 µ2 −λ1 µ2 −λ1 µ2
−µ1 λ2 −µ1 λ2 λ1 λ2 λ1 λ2
−µ1 µ2 −µ1 µ2 , λ1 µ 2 λ1 µ 2
−λ1 µ2 λ1 λ 2 −λ1 µ2 λ 1 λ2
λ1 µ2 −µ1 λ2 λ1 µ2 −µ1 λ2
−µ1 µ2 µ1 λ2 , −µ1 µ2 µ1 λ2
−µ1 µ2 µ1 λ2 λ1 µ2 −λ1 λ2
−µ1 µ2 µ 1 λ2 λ1 µ2 −λ1 λ2
µ1 µ2 −µ1 λ2 . −λ1 µ2 λ1 λ2
In particular the transition probability from state 11 at time 0 into state 11 at time t is given by P11,11 (t) =
h 1 µ1 µ2 + λ1 µ2 e−(λ1+µ1 )t + µ1 λ2 e−(λ2+µ2 )t (λ1 + µ1 )(λ2 + µ2 ) i (4.43) +λ1 λ2 e−(λ1+λ2 +µ1 +µ2 )t .
92
Total Downtime of Repairable Systems
Under the stationary distribution we have Eπ [U (t)] = π11 t =
µ1 µ2 t . (λ1 + µ1 )(λ2 + µ2 )
For the second moment of U (t), under the stationary distribution, ·Z π
2
E [U (t)] =
t
π
E
0
Z tZ
¸2 1{11} Y2 (s)ds
s
Pπ (Y (r) = 11, Y (s) = 11)drds Z tZ s µ1 µ2 = 2 P11,11 (s − r)drds. (λ1 + µ1 )(λ2 + µ2 ) 0 0
= 2
0
0
Using (4.43) we obtain Eπ [U 2 (t)] = (π11 t)2 + σ 2 t + r(t) where
· ¸ 2µ1 µ2 λ1 λ2 λ1 µ2 µ1 λ2 σ = + + (λ1 + µ1 )(λ2 + µ2 ) (λ1 + λ2 + µ1 + µ2 )2 λ1 + µ1 λ2 + µ2 2
and r(t) =
λ1 λ2 [e−(λ1 +λ2 +µ1 +µ2 )t − 1] λ1 µ2 [e−(λ1 +µ1 )t − 1] µ1 λ2 [e−(λ2 +µ2 )t − 1] + + . (λ1 + λ2 + µ1 + µ2 )2 (λ1 + µ1 )2 (λ2 + µ2 )2
It follows that the variance of U (t), under the stationary distribution, Varπ [U (t)] = σ 2 t + r(t). From this expression we conclude Varπ [U (t)] = σ2 . t→∞ t lim
This limit is also valid under any initial distribution of the Markov chain Y2 , and can be obtained using a formula (8.11) of Iosifescu [21]. Moreover, using another formula on page 256 of Iosifescu, under any initial distribution of the Markov chain Yn , the limiting distribution of U (t) is normal, i.e., U (t) − π11 t d √ −→ N (0, 1) as σ t
t → ∞.
Now we will consider the probability distribution of U (t). Define for α, β > 0 and a ∈ I Z ∞ ˆ Ea [e−αU (t) ]e−βt dt ψa (α, β) := 0
Systems consisting of n independent components
93
and starting from t = 0 in state a τa := inf{t ≥ 0 : Yn (t) 6= a}. Starting from 1n , the random variable τ1n is the time at which the chain leaves the state 1n and Pn P1n (τ1n > t) = e− i=1 λi t . Conditioning on T1n we obtain the system equations (
ψˆ1n (α, β) = ψˆa (α, β) =
P
1 ˆ P b6=1n q1n b ψb (α, β)] α+β+ n λi [1 + i=1P 1 ˆ b6=a qab ψb (α, β)] if a 6= 1n β+qa [1 +
Pn where qa = i=1 νi,1−²i (a) whenever a = (²1 (a), ²2 (a), . . . , ²n (a)). Solving this system equations we get the double Laplace transform of U (t). As an example for n = 2, the solution for ψˆ11 (α, β) is given by ABC + Dβ + Cβ 2 + β 3 µ1 µ2Cα + (ABC + Eα)β + (D + F α)β 2 + (2C + α)β 3 + β 4 (4.44) where A = λ1 + µ1 , B = λ2 + µ2 , C = A + B, D = A2 + B 2 + 3AB, E = AB + µ1 A + µ2 B + 2µ1 µ2 and F = µ1 + µ2 + C. Note that the left-hand side of (4.44) can be written as Z ∞Z t f (x, t)e−(αx+βt) dxdt ψˆ11 (α, β) =
0
0
where f (x, t) is the density function of U (t) at time t. Transforming back the double Laplace transform (4.44) with respect to α, we get Z ∞ 2 +2Cβ 3 +β 4 )x (ABC + Dβ + Cβ 2 + β 3 ) − (ABCβ+Dβ µ1 µ2 C+Eβ+F β 2 +β 3 f (x, t)e−βt dt = e . µ1 µ2 C + Eβ + F β 2 + β 3 0 The probability density function of U (t) can be obtained by inverting numerically this transform.
4.7.2
Arbitrary failure or repair times
In general it is complicated to obtain an explicit expression for the distribution of the total uptime of systems comprising n ≥ 2 components when the failure or repair times of the components are arbitrarily distributed. In some cases it is possible to derive the expression for the mean of the total uptime. Consider the series system in the previous subsection. Assume that for each i, the random variables Xij , j = 1, 2, ..., have a common distribution function
94
Total Downtime of Repairable Systems
Fi , and the random variables Yij , j = 1, 2, ..., have a common distribution function Gi . Denote by Fi∗ and G∗i the Laplace-Stieltjes transforms of Fi and Gi (i) respectively. Let A11 (t) be the availability of the ith component at time t. Then from (4.42), the mean of the total uptime of the series system can be formulated as Z tY n (i) E[U (t)] = A11 (s)ds. (4.45) 0 i=1
In some cases we have analytic expressions for the availability, which can be obtained by inverting its Laplace transform given by Z ∞ 1 − Fi∗ (β) (i) A11 (t)e−βt dt = , (4.46) β[1 − Fi∗ (β)G∗i (β)] 0 see (4.13). As an example let n = 2. Suppose that X1j and Y1j are exponentially distributed with parameter 1 and 2 respectively. Suppose also that X2j ∼ Gamma(1,2) having a pdf f (x) = xe−x ,
x≥0
and Y2j ∼ Gamma(2,2). Then using (4.46), we obtain (1)
A11 (t) =
2 1 −3t + e 3 3
and (2)
A11 (t) =
√ √ √ 2 1 1 + e−3t + e−3t/2 [7 cos( 7t/2) + 5 7 sin( 7t/2)]. 3 12 28
Using (4.45) we obtain E[U (t)] =
4 115 1 −6t 5 t+ − − e−3t e 9 · 396 216 54 ¸ √ 7 −9t/2 1 −3t/2 − + e e cos( 7t/2) 264 6 · ¸ √ √ 19 −9t/2 1 − + e−3t/2 e 7 sin( 7t/2). 1848 42
Appendix A
The proof of Theorem 2.5.1 In Section 2.5 we have proved that, with probability 1, N (t) = N(t)(Φ) where Φ is a Poisson point process having intensity measure ν(dsdx) = dsdF (x) and Z Z 1[0,x) (t − A(s, µ))1[0,s) (u)µ(dudv)µ(dsdx), (A.1) N(t)(µ) = E
E
where
Z A(s, µ) = E
1[0,s) (y)zµ(dydz)
and E = [0, ∞) × [0, ∞). In the sequel we will write I n = [0, ∞)n and M = Mp (E), the set of all point measures on E. The distribution of Φ on M is denoted by Pν . The main tools or arguments that we will use in this proof are: (1) (2) (3) (4)
: : : :
Fubini’s theorem substitution The Palm formula for Poisson point processes (see Theorem 1.2.4) The Laplace functional of Poisson point processes (see Theorem 1.2.1).
We indicate the use of these arguments by writing the corresponding numbers (1,2)
over equality signs. For example the notation = means that we have used Fubini’s theorem and substitution one or several times. We will also use the following notations: For fix α, β ≥ 0 we put P Q R S T
= = = = =
1 − F ∗ (α) 1 − F ∗ (β) 1 − F ∗ (α + β) F ∗ (α) − F ∗ (α + β) F ∗ (β) − F ∗ (α + β) 95
96
The proof of Theorem 2.5.1
where F ∗ denotes the Laplace-Stieltjes transform of F . Note that P + S = Q + T = R. Define for fix α, β ≥ 0 Z e−αA(s,µ)−βA(˜s,µ) Pν (dµ),
L(α, β, s, s˜) :=
s, s˜ ≥ 0.
M
For s > s˜, L(α, β, s, s˜) ½ Z h ¾ Z i = 1[0,s) (y)αz + 1[0,˜s) (y)βz µ(dydz) Pν (dµ) exp − M ¾ ½ Z ∞ ZE ∞ h i (4) 1 − e1[0,s) (y)αz+1[0,˜s) (y)βz dF (z)dy = exp − 0
½ =
exp
Z
0 s˜ Z ∞
[1 − e
− 0
½
Z −(α+β)z
s
Z
]dF (z)dy −
0
[1 − e s˜
−αz
]dF (z)dy
0 ¾
=
− s˜[1 − F ∗ (α + β)] − (s − s˜)[1 − F ∗ (α)] ½ ¾ ∗ ∗ ∗ exp − s[1 − F (α)] − s˜[F (α) − F (α + β)]
=
e−sP −˜sS .
=
¾
∞
exp
In the sequel we will write L(α, β, s, s˜; s > s˜) = L(α, β, s, s˜) when s > s˜. Hence L(α, β, s, s˜; s > s˜) = e−sP −˜sS . Similarly, it can be proved that for s < s˜ L(α, β, s, s˜; s < s˜) = e−˜sQ−sT , and for s = s˜ L(α, β, s, s) := L(α, β, s, s˜; s = s˜) = e−sR . Now we will calculate the double Laplace transform of E[N (t1 )N (t2 )]. Using (A.1) we obtain Z
∞
0
Z
∞
E[N (t1 )N (t2 )]e−αt1 −βt2 dt1 dt2 Z Z N(t1 , µ)N(t2 , µ)Pν (dµ)e−αt1 −βt2 dt1 dt2
0
=
I2
M
97
Z
(2)
=
I2
Z
I4
(1,2,3)
=
Z
·Z
1[0,x) (t1 − A(s, µ))1[0,s) (u)µ(dudv)µ(dsdx) ¸ 1[0,˜x) (t2 − A(˜ s, µ))1[0,˜s) (˜ u)µ(d˜ ud˜ v )µ(d˜ sd˜ x) e−αt1 −βt2 M
I4
Pν (dµ)dt1 dt2 Z Z Z 1 1[0,˜x) (t2 − A(˜ s, µ + δ(s,x) ))1[0,s) (u)1[0,˜s) (˜ u) α I3 M I6 [1 − e−αx ]e−βt2 e−αA(s,µ) (µ + δ(s,x) )(d˜ ud˜ v )(µ + δ(s,x) )(d˜ sd˜ x) µ(dudv)Pν (dµ)dF (x)dsdt2 .
Note that this integral can be split into four terms. For one of these terms, the integration over I 6 is with respect to δ(s,x) (d˜ sd˜ x)δ(s,x) (d˜ ud˜ v ). This integral equals 0, because the integrand contains the factor 1[0,˜s) (˜ u) which with respect to these measures integrates to 1[0,s) (s). So we only need to calculate the three remaining integrals.
Case 1: The integral with µ(d˜ sd˜ x) and δ(s,x) (d˜ ud˜ v ). In this case Z Z Z 1 T1 := 1[0,˜x) (t2 − A(˜ s, µ + δ(s,x) ))1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ] α I3 M I6 (1,2)
=
(3)
=
(3)
=
= (4)
=
= =
e−βt2 e−αA(s,µ) δ(s,x) (d˜ ud˜ v )µ(d˜ sd˜ x)µ(dudv)Pν (dµ)dF (x)dsdt2 Z Z Z 1 1[0,s) (u)1[0,˜s) (s)[1 − e−αx ][1 − e−β x˜ ] αβ I 2 M I 4 e−αA(s,µ)−βA(˜s,µ)−βx µ(dudv)µ(d˜ sd˜ x)Pν (dµ)dF (x)ds Z Z Z T 1[0,s) (u)1[0,˜s) (s)[1 − e−β x˜ ]e−αA(s,µ)−βA(˜s,µ) αβ I 3 M I 2 µ(dudv)Pν (dµ)dF (˜ x)d˜ sds Z Z Z Z QT ∞ ∞ 1[0,s) (u)e−αA(s,µ)−αv−βA(˜s,µ)−βv αβ 0 s I2 M Pν (dµ)dF (v)dud˜ sds Z ∞Z ∞ QT ∗ sL(α, β, s, s˜, s < s˜)d˜ sds F (α + β) αβ 0 s Z ∞Z ∞ QT ∗ se−˜sQ−sT d˜ sds F (α + β) αβ 0 s QT ∗ 1 F (α + β) αβ QR2 F ∗ (α + β)[F ∗ (β) − F ∗ (α + β)] . αβ[1 − F ∗ (α + β)]2
98
The proof of Theorem 2.5.1
Case 2: The integral with δ(s,x) (d˜ sd˜ x) and µ(d˜ ud˜ v ). In this case Z Z Z 1 T2 := 1[0,˜x) (t2 − A(˜ s, µ + δ(s,x) ))1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ] α I3 M I6 =
(1,2,3)
=
=
e−βt2 e−αA(s,µ) µ(d˜ ud˜ v )δ(s,x) (d˜ sd˜ x)µ(dudv)Pν (dµ)dF (x)dsdt2 Z Z Z 1 1[0,x) (t2 − A(s, µ))1[0,s) (u)1[0,s) (˜ u)[1 − e−αx ] α I3 M I4 e−αA(s,µ) e−βt2 µ(d˜ ud˜ v )µ(dudv)Pν (dµ)dF (x)dsdt2 Z Z Z 1 1[0,s) (u)1[0,s) (˜ u)[1 − e−αx ][1 − e−βx ] αβ I 4 M I 2 e−(α+β)A(s,µ)−(α+β)v (µ + δ(u,v) )(d˜ ud˜ v )Pν (dµ)dF (v)dudF (x)ds Z Z Z P −T ∗ 1[0,s) (u)1[0,s) (˜ u)e−(α+β)A(s,µ) F (α + β) αβ I2 M I2 (µ + δ(u,v) )(d˜ ud˜ v )Pν (dµ)duds.
This integral can be split into two terms.
Subcase 21: Using the measure µ(d˜ ud˜ v ) in the inner integral. In this case Z Z Z P −T ∗ T21 := 1[0,s) (u)1[0,s) (˜ u)e−(α+β)A(s,µ) F (α + β) αβ 2 2 I M I µ(d˜ ud˜ v )Pν (dµ)duds Z Z P −T ∗ (3) = 1[0,s) (˜ u)se−(α+β)A(s,µ)−(α+β)˜v F (α + β) αβ I3 M Pν (dµ)dF (˜ v )d˜ uds Z ∞ P −T ∗ 2 = s2 L(α, β, s, s)ds F (α + β) αβ 0 Z ∞ P −T ∗ (4) 2 = s2 e−sR ds F (α + β) αβ 0 P −T ∗ 2 = F (α + β)2 3 αβ R 2F ∗ (α + β)2 [1 − F ∗ (α) − F ∗ (β) + F ∗ (α + β)] = . αβ[1 − F ∗ (α + β)]3
99
Subcase 22: Using the measure δ(u,v) (d˜ ud˜ v ) in the inner integral. In this case
T22
:=
= (4)
=
= =
Z Z Z P −T ∗ 1[0,s) (u)1[0,s) (˜ u)e−(α+β)A(s,µ) F (α + β) αβ 2 2 I M I δ(u,v) (d˜ ud˜ v )Pν (dµ)duds Z P −T ∗ 1[0,s) (u)1[0,s) (u)L(α, β, s, s)duds F (α + β) αβ 2 ZI ∞ P −T ∗ se−sR ds F (α + β) αβ 0 P −T ∗ 1 F (α + β) 2 αβ R F ∗ (α + β)[1 − F ∗ (α) − F ∗ (β) + F ∗ (α + β)] . αβ[1 − F ∗ (α + β)]2
Hence T2
:= T21 + T22 F ∗ (α + β)[1 + F ∗ (α + β)][1 − F ∗ (α) − F ∗ (β) + F ∗ (α + β)] = . αβ[1 − F ∗ (α + β)]3
Case 3: The integral with µ(d˜ sd˜ x) and µ(d˜ ud˜ v ). In this case Z Z Z 1 T3 := 1[0,˜x) (t2 − A(˜ s, µ + δ(s,x) )) α I3 M I6 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ]e−βt2 e−αA(s,µ) (1,2)
=
(3)
=
µ(d˜ ud˜ v )µ(d˜ sd˜ x)µ(dudv)Pν (dµ)dF (x)dsdt2 Z Z Z 1 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ]e−αA(s,µ) αβ I 2 M I 6 e−βA(˜s,µ+δ(s,x) ) µ(d˜ ud˜ v )µ(dudv)µ(d˜ sd˜ x)Pν (dµ)dF (x)ds Z Z Z 1 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 4 M I 4 e−αA(s,µ+δ(˜s,˜x) ) e−βA(˜s,µ+δ(s,x) ) µ(d˜ ud˜ v )(µ + δ(˜s,˜x) )(dudv)Pν (dµ)dF (˜ x)d˜ sdF (x)ds.
We can be split this integral into two terms.
100
The proof of Theorem 2.5.1
Subcase 31: Using the measure δ(˜s,˜x) (dudv). In this case Z Z Z 1 T31 := 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ]e−αA(s,µ+δ(˜s,˜x) ) αβ I 4 M I 4 =
(3)
=
= (4)
=
= =
e−βA(˜s,µ+δ(s,x) ) µ(d˜ ud˜ v )δ(˜s,˜x) (dudv)Pν (dµ)dF (˜ x)d˜ sdF (x)ds Z Z Z 1 1[0,s) (˜ s)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 4 M I 2 e−αA(s,µ)−α˜x e−βA(˜s,µ) µ(d˜ ud˜ v )Pν (dµ)dF (˜ x)d˜ sdF (x)ds Z ∞Z sZ Z PS 1[0,˜s) (˜ u)e−αA(s,µ)−α˜v e−βA(˜s,µ)−β v˜ αβ 0 2 0 I M Pν (dµ)dF (˜ v )d˜ ud˜ sds Z ∞Z s PS ∗ s˜L(α, β, s, s˜; s > s˜)d˜ sds F (α + β) αβ Z0 ∞ Z0 s PS ∗ s˜e−sP −˜sS d˜ sds F (α + β) αβ 0 0 PS ∗ 1 F (α + β) αβ P R2 ∗ ∗ F (α + β)[F (α) − F ∗ (α + β)] . αβ[1 − F ∗ (α + β)]2
Subcase 32: Using the measure µ(dudv). In this case Z Z Z 1 T32 := 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ]e−αA(s,µ+δ(˜s,˜x) ) αβ I 4 M I 2 Z −βA(˜ s,µ+δ(s,x) ) e µ(dudv) µ(d˜ ud˜ v )Pν (dµ)dF (˜ x)d˜ sdF (x)ds E Z Z Z 1 (3) = 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 6 M I 2 e−αA(s,µ+δ(˜s,˜x) )−αv e−βA(˜s,µ+δ(s,x) +δ(u,v) ) (µ + δ(u,v) )(d˜ ud˜ v )Pν (dµ)dF (v)dudF (˜ x)d˜ sdF (x)ds. This integral can be split into two terms. Sub-subcase 321: Using the measure µ(d˜ ud˜ v ) in the inner integral. In this case Z Z Z 1 T321 := 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 6 M I 2 e−αA(s,µ+δ(˜s,˜x) )−αv e−βA(˜s,µ+δ(s,x) +δ(u,v) ) =
µ(d˜ ud˜ v )Pν (dµ)dF (v)dudF (˜ x)d˜ sdF (x)ds A+B
101
where A
1 αβ
=
e (3)
=
Z
Z
s
Z
Z
Z
1[0,s) (u)1[0,˜s) (˜ u)[1 I2 0 I3 M I2 −αA(s,µ+δ(˜ s,µ+δ(s,x) +δ(u,v) ) s,˜ x) )−αv −βA(˜
− e−αx ][1 − e−β x˜ ]
e
µ(d˜ ud˜ v )Pν (dµ)dF (v)dudF (˜ x)d˜ sdF (x)ds Z Z sZ Z 1 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 2 0 I 5 M e−αA(s,µ)−α˜v−α˜x−αv e−βA(˜s,µ+δ(u,v) )−β v˜
=
+
= + (4)
=
+ = =
Pν (dµ)dF (˜ v )d˜ udF (v)dudF (˜ x)d˜ sdF (x)ds Z ∞ Z s Z s˜ Z ∞ Z PS ∗ s˜e−αA(s,µ)−αv e−βA(˜s,µ)−βv F (α + β) αβ 0 0 0 0 M Pν (dµ)dF (v)dud˜ sds Z ∞Z sZ sZ ∞Z PS ∗ s˜e−αA(s,µ)−αv e−βA(˜s,µ) F (α + β) αβ 0 0 s˜ 0 M Pν (dµ)dF (v)dud˜ sds Z ∞Z s PS ∗ 2 s˜2 L(α, β, s, s˜; s > s˜)d˜ sds F (α + β) αβ 0 0 Z ∞Z s PS ∗ s˜(s − s˜)L(α, β, s, s˜; s > s˜)d˜ sds F (α)F ∗ (α + β) αβ Z ∞ Z 0s 0 PS ∗ s˜2 e−sP −˜sS d˜ sds F (α + β)2 αβ 0 0 Z ∞Z s PS ∗ s˜(s − s˜)e−sP −˜sS d˜ sds F (α)F ∗ (α + β) αβ 0 0 PS ∗ 2 PS ∗ 1 F (α + β)2 + F (α)F ∗ (α + β) 2 2 3 αβ PR αβ P R · ¸ F ∗ (α + β)[F ∗ (α) − F ∗ (α + β)] 2F ∗ (α + β) F ∗ (α) + , αβ[1 − F ∗ (α + β)]2 1 − F ∗ (α + β) 1 − F ∗ (α)
and B
=
1 αβ e
(3)
=
Z
Z
∞
Z
Z
Z
1[0,s) (u)1[0,˜s) (˜ u)[1 − I2 s I3 M I2 −αA(s,µ+δ(˜ s,µ+δ(s,x) +δ(u,v) )−βx s,˜ x) )−αv −βA(˜
e−αx ][1 − e−β x˜ ]
e
µ(d˜ ud˜ v )Pν (dµ)dF (v)dudF (˜ x)d˜ sdF (x)ds Z Z ∞Z Z 1 1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 − e−β x˜ ] αβ I 2 s 5 I M s,µ)−βx−βv−β v ˜ ˜ v ) )−αv −βA(˜ e−αA(s,µ+δ(u,˜ e
Pν (dµ)dF (˜ v )d˜ udF (v)dudF (˜ x)d˜ sdF (x)ds
102
The proof of Theorem 2.5.1
=
+
= + (4)
=
+ = =
Z ∞Z ∞Z ∞Z sZ ∞Z QT ∗ se−αA(s,µ)−α˜v e−βA(˜s,µ)−β v˜ F (α + β) αβ 0 s 0 0 0 M Pν (dµ)dF (˜ v )d˜ udF (v)d˜ sds Z ∞ Z ∞ Z ∞ Z s˜ Z ∞ Z QT ∗ se−αA(s,µ) e−βA(˜s,µ)−β v˜ F (α + β) αβ 0 s 0 s 0 M Pν (dµ)dF (˜ v )d˜ udF (v)d˜ sds Z ∞Z ∞ QT ∗ s2 L(α, β, s, s˜; s < s˜)d˜ sds F (α + β)2 αβ 0 s Z ∞Z ∞ QT ∗ s(˜ s − s)L(α, β, s, s˜; s < s˜)d˜ s F (β)F ∗ (α + β) αβ Z ∞ Z 0∞ s QT ∗ s2 e−˜sQ−sT d˜ sds F (α + β)2 αβ 0 s Z ∞Z ∞ QT ∗ s(˜ s − s)e−˜sQ−sT d˜ sds F (β)F ∗ (α + β) αβ 0 s QT ∗ 2 QT ∗ 1 F (α + β)2 + F (β)F ∗ (α + β) 2 2 3 αβ QR αβ Q R · ¸ F ∗ (α + β)[F ∗ (β) − F ∗ (α + β)] 2F ∗ (α + β) F ∗ (β) + . αβ[1 − F ∗ (α + β)]2 1 − F ∗ (α + β) 1 − F ∗ (β)
Sub-subcase 322: Using the measure µ(d˜ ud˜ v ) in the inner integral. In this case
T322
:=
1 αβ e
=
+
= +
Z
Z
Z
1[0,s) (u)1[0,˜s) (˜ u)[1 − e−αx ][1 I6 M I2 −αA(s,µ+δ(˜ s,µ+δ(s,x) +δ(u,v) ) s,˜ x) )−αv −βA(˜
− e−β x˜ ]
e
δ(u,v) (d˜ ud˜ v )Pν (dµ)dF (v)dudF (˜ x)d˜ sdF (x)ds Z Z s Z ∞ Z s˜ Z 1 ∗ du [1 − e−αx ][1 − e−β x˜ ] F (α + β) αβ 2 0 0 0 I M e−αA(s,µ)−α˜x e−βA(˜s,µ) Pν (dµ)dudF (˜ x)d˜ sdF (x)ds Z Z ∞Z ∞Z s Z 1 ∗ du [1 − e−αx ][1 − e−β x˜ ] F (α + β) αβ 0 0 I2 s M e−αA(s,µ) e−βA(˜s,µ)−βx Pν (dµ)dudF (˜ x)d˜ sdF (x)ds Z ∞Z s PS ∗ s˜L(α, β, s, s˜; s > s˜)d˜ sds F (α + β) αβ Z0 ∞ Z0 ∞ QT ∗ sL(α, β, s, s˜; s < s˜)d˜ sds F (α + β) αβ 0 s
103
(4)
=
+ = =
Z ∞Z s PS ∗ s˜e−sP −˜sS d˜ sds F (α + β) αβ 0 0 Z ∞Z ∞ QT ∗ se−˜sQ−sT d˜ sds F (α + β) αβ 0 s PS ∗ 1 QT ∗ 1 F (α + β) + F (α + β) αβ P R2 αβ QR2 · ¸ F ∗ (α + β) ∗ ∗ ∗ ∗ F (α) − F (α + β) + F (β) − F (α + β) . αβ[1 − F ∗ (α + β)]
So we obtain T3
= T31 + A + B + T322 ½ F ∗ (α + β) [1 + F ∗ (α + β)][F ∗ (α) + F ∗ (β) − 2F ∗ (α + β)] = αβ[1 − F ∗ (α + β)] 1 − F ∗ (α + β) ¾ ∗ ∗ ∗ ∗ F (α)[F (α) − F (α + β)] F (β)[F ∗ (β) − F ∗ (α + β)] + . 1 − F ∗ (α) 1 − F ∗ (β)
It follows that Z 0
∞
Z
∞
E[N (t1 )N (t2 )]e−αt1 −βt2 dt1 dt2
0
= T1 + T2 + T3 [1 − F ∗ (α)F ∗ (β)]F ∗ (α + β) = . αβ[1 − F ∗ (α)][1 − F ∗ (β)][1 − F ∗ (α + β)]
2
104
The proof of Theorem 2.5.1
Appendix B
Numerical inversions of Laplace transforms B.1
Single Laplace transform
Let f be a real-valued function defined on the positive half-line. The Laplace transform of f is defined to be Z ∞ f (t)e−βt dt, (B.1) fˆ(β) = 0
where β is a complex variable, whenever this integral exists. Given fˆ, we can retrieve the original function f using the following inversion formula: f (t) =
1 2πi
=
eat 2π
Z Z
a+i∞
etβ fˆ(β)dβ
a−i∞ ∞ itu
e
fˆ(a + iu)du.
(B.2)
−∞
where a is a real number chosen such that fˆ(β) has no singularity on or to the right of the vertical line β = a, see e.g. Abate and Whitt [1]. For some Laplace transforms fˆ we have analytic expressions for f , a table for these is available, see for example Oberhettinger [28]. When the transform cannot be inverted analytically, we can approximate the function f numerically. Several numerical inversion algorithms have been proposed by several authors, see for example Abate and Whitt [1], Weeks [47] and Iseger [22]. Following Abate and Whitt, we will use the trapezoidal rule to approximate the integral in (B.2) and analyze the corresponding discretization error using the Poisson summation formula. 105
106
Numerical inversions of Laplace transforms
The trapezoidal rule approximates the integral of a function g over the bounded interval [c, d] by the integral of the piecewise linear function obtained by connecting the n + 1 evenly spaced points g(c + kh), 0 ≤ k ≤ n where h = (d − c)/n, i.e., " # Z d n−1 g(c) + g(d) X g(x)dx ≈ h g(c + kh) , + 2 c k=1
see Davis and Rabinowitz [9]. In case c = −∞ and d = ∞ we approximate the integral of g over the real line as Z ∞ ∞ X g(x)dx ≈ h1 g(kh1 ) (B.3) −∞
k=−∞
where h1 is a small positive constant. This formula can also be obtained using the trapezoidal rule with obvious modifications. Applying (B.3) to (B.2) with step size h1 = π/t, t > 0, and letting a = A/t at the same time, we get f (t) ≈
∞ eA X (−1)k fˆ([A + iπk]/t). 2t
(B.4)
k=−∞
This approximation can also be obtained by using the Poisson summation formula: For an integrable function g ∞ X
g(t + 2πk/h2 ) =
k=−∞
∞ h2 X ϕ(kh2 )e−ih2 tk 2π
(B.5)
k=−∞
R∞ where h2 is some positive constant and ϕ(u) = −∞ g(x)eiux dx, the Fourier transform of g. Taking g(x) = e−a1 x f (x)1[0,∞) (x) in (B.5) where a1 is chosen such that the function g is integrable, we obtain ∞ X
e−a1 (t+2πk/h2 ) f (t + 2πk/h2 ) =
k=0
∞ h2 X ˆ f (a1 − ikh2 )e−ih2 tk (B.6) 2π k=−∞
where fˆ is the Laplace transform of f , see (B.1). Letting a1 = A/t and h2 = π/t in (B.6) we obtain f (t) =
∞ eA X (−1)k fˆ([A + iπk]/t) − ed 2t k=−∞
where ed =
∞ X k=1
e−2kA f ([2k + 1]t).
(B.7)
Single Laplace transform
107
Comparing (B.4) and (B.7), we conclude that ed is an explicit expression for the discretization error associated with the trapezoidal rule approximation. This discretization error can easily be bounded whenever f is bounded. For example if |f (x)| ≤ C then |ed | ≤ Ce−2A /(1 − e−2A ), and if |f (x)| ≤ Cx then |ed | ≤
(3e−2A − e−4A )Cx . (1 − e−2A )2
We used (B.4) to invert numerically Laplace transforms in this thesis. Note that the formula (B.4) can be expressed as ∞ eA ˆ eA X f (A/t) + (−1)k <(fˆ([A + iπk]/t)) 2t t
f (t) ≈
(B.8)
k=1
¯ = 2<(fˆ(β)), where β¯ and <(β) denote the by using the fact that fˆ(β) + fˆ(β) complex conjugate and real part of β respectively. Below we give a numerical-inversion example of Laplace transform of the expected value of the instantaneous reward process in Chapter 2. We write the programs in Matlab. Example B.1.1 In Subsection 2.2.1 equation (2.12) we have the Laplace transform of the expected value of an instantaneous reward process: Z
∞
E[Rφ (t)]e−βt dt =
0
(β + λ)e−2(β+λ) . β 2 [1 − e−2(β+λ) ]
(B.9)
Denote the right-hand side of (B.9) by fˆ(β). Using (B.8) we get E[Rφ (t)] ≈
M eA ˆ eA X f (A/t) + (−1)k <(fˆ([A + iπk]/t)). 2t t k=1
The following is a Matlab program for approximating E[Rφ (t)] for λ = 0.1741.
function [f]=expmean(t,M) A=5; P=exp(A)/(2*t); B=A/t; C=i*pi/t; lambda=0.1741; D=B+lambda;
108
Numerical inversions of Laplace transforms
T1=P*D*exp(-2*D)/(B^2*(1-exp(-2*D))); m=0; for k=1:M beta=B+C*k; U=(beta+lambda)*exp(-2*(beta+lambda)); V=beta^2*(1-exp(-2*(beta+lambda))); R=real(U/V); m=m+(-1)^k*R; end T2=2*P*m; f=T1+T2; %f=E[R_\phi(t)].
B.2
Double Laplace transform
This section concerns a generalization of the result in the previous section. We refer to Choudhury at al. [5] with some modifications. Let f (t1 , t2 ) be a realvalued function of non-negative real variables t1 and t2 . Denote its double Laplace transform by Z
∞
fˆ(α, β) =
Z
0
∞
f (t1 , t2 )e−(αt1 +βt2 ) dt1 dt2
0
where α and β are complex variables, provided the integral exists. To retrieve numerically the function f from fˆ, we use the two-dimensional Poisson summation formula: For a real-valued function g on R2 , ∞ X
∞ X
j=−∞ k=−∞ ∞ X
=
g(t1 + 2πj/h1 , t2 + 2πk/h2 ) ∞ X h1 h2 ϕ(jh1 , kh2 )e−i(h1 t1 j+h2 t2 k) , 4π 2
j=−∞ k=−∞
where ϕ denotes the bivariate Fourier transform of g, i.e., Z
∞
Z
∞
ϕ(u, v) = −∞
−∞
g(x, y)ei(xu+yv) dxdy,
(B.10)
Double Laplace transform
109
Taking g(x, y) = f (x, y)e−(a1 x+a2 y) when x, y ≥ 0 and g(x, y) = 0 otherwise in (B.10), we obtain ∞ X
∞ X
e−[a1 (t1 +2πj/h1 )+a2 (t2 +2πk/h2 )] f (t1 + 2πj/h1 , t2 + 2πk/h2 )
j=−∞ k=−∞
=
∞ ∞ h1 h 2 X X ˆ f (a1 − ijh1 , a2 − ikh2 )e−i(h1 t1 j+h2 t2 k) . 4π 2 j=−∞ k=−∞
Letting h1 = π/t1 , h2 = π/t2 , a1 = A1 /t1 , a2 = A2 /t2 , we obtain, after some simplifications, f (t1 , t2 ) =
∞ ∞ eA1 +A2 X X (−1)j+k fˆ([A1 − iπj]/t1 , [A2 − iπk]/t2 ) − ed 4t1 t2 j=−∞ k=−∞
(B.11) where ed
=
∞ X ∞ X
e−(A1 j+A2 k) f ([2j + 1]t1 , [2k + 1]t2 )
j=0 k=1
+
∞ X ∞ X
e−(A1 j+A2 k) f ([2j + 1]t1 , [2k + 1]t2 ).
j=1 k=0
If |f (t1 , t2 )| ≤ C for some constant C, then the error ed can be bounded: |ed | ≤
C(e−A1 + e−A2 − e−(A1 +A2 ) ) . (1 − e−A1 )(1 − e−A2 )
¯ = 2<(fˆ(α, β)), from (B.11), for A1 = A2 = A, By noting that fˆ(α, β) + fˆ(¯ α, β) we get an approximation for f (t1 , t2 ): f (t1 , t2 )
M X N X e2A ˆ ≈ −f (A/t1 ) + 2 (−1)j+k <(fˆ([A − iπj]/t1 , [A − iπk]/t2 )) 4t1 t2 j=0 k=0 M X N X +2 (−1)j+k <(fˆ([A − iπj]/t1 , [A + iπk]/t2 )) . (B.12) j=1 k=0
We used this formula to obtain the numerical inversion of the double Laplace transform in (4.40) of the total downtime. So far we have assumed that the variables t1 and t2 are continuous. It is also possible to consider the case where t1 or t2 are discrete variables. Choudhury at al. [5] have discussed the cases where t1 or t2 are non-negative integers. We will
110
Numerical inversions of Laplace transforms
discuss the case where one of the variable is discrete and the other is continuous in the following example. In Chapter 2 we have the following formula: For α, β > 0, Z
∞
R∞ E(e−αRφ (t) )e−βt dt =
0
[1 − F (t)]e−αφ(t)−βt dt R∞ 1 − 0 e−αφ(t)−βt dF (t) 0
(B.13)
where N (t)
Rφ (t) =
X
φ(Xn ) + φ(t − SN (t) )
n=1
where φ is a non-negative measurable function, see (2.4). It can be proved that this formula remains true if we replace α by iα. The random variable Rφ (t) possibly takes values x + lλ, l = 0, ±1, ±2, . . ., depend on the choice of the function φ. In this case, by replacing α with iα, we can write the equation (B.13) as Z 0
∞ ∞X
R∞ −iα(x+lλ)−βt
P(Rφ (t) = x + lλ)e
l=0
dt =
[1 − F (t)]e−iαφ(t)−βt dt R∞ . 1 − 0 e−iαφ(t)−βt dF (t) 0
Denote the right-hand side of this equation by ψ(α, β). Using the inversion formula for Laplace transforms and discrete Fourier transforms, see Abate and Whitt [1], we obtain P(Rφ (t) = x + lλ) =
λeat 4π 2
Z
2π/λ 0
Z
∞
ψ(α, a + iβ)ei[(x+lλ)α+tβ] dβdα (B.14)
−∞
for suitably constant a. Applying the trapezoidal rule twice to (B.14) with step sizes 2π/(λn) and π/t for the integrals with respect to α and β respectively, and letting a = A/t at the same time, we obtain P(Rφ (t) = x + lλ) " ∞ ³ ´ eA 1 X ≈ (−1)k ψ(0, (A + iπk)/t) + ψ(2π/λ, (A + iπk)/t)ei2πx/λ 2nt 2 k=−∞ n−1 ∞ X X (B.15) + (−1)k ψ(2πj/(λn), (A + iπk)/t)ei2πj(l+x/λ)/n . j=1 k=−∞
We used this formula to approximate the probability mass function of Rφ (t) in Subsection 2.2.1 as described in the following example.
Double Laplace transform
111
Example B.2.1 We have the following transform of Rφ (t), after replacing α with iα,: Z ∞ E[e−iαRφ (t) ]e−βt dt 0
=
1 − 2e−2(β+λ) , (β + λ)[1 − e−iα e−2(β+λ) ] − λ[1 − e−2(β+λ) ]
(B.16)
see (2.13). In this example Rφ (t) is a non-negative-integer-valued random variable. Denote the right-hand side of (B.16) by ψ(α, β). Since ψ(0, (A+iπk)/t) = ψ(2π, (A + iπk)/t), using (B.15) we get a formula for approximating the pdf of Rφ (t): P(Rφ (t) = l) ≈
n−1 M eA X X (−1)k ψ(2πj/n, (A + iπk)/t)ei2πjl/n . 2nt j=0 k=−M
The following is a Matlab program for approximating the pdf of Rφ (t) for λ = 0.1741.
function [f]=exppmf(l,t,n,M) A=5; P=exp(A)/(2*n*t); B=A/t; C=i*pi/t; h=2*pi/n; D=i*l*h; lambda=0.1741; m=0; for j=0:(n-1) E=exp(D*j); alpha=j*h; F=exp(-i*alpha); for k=-M:M beta=B+C*k; U1=beta+lambda; U2=exp(-2*U1); U=1-U2; V1=U1*(1-F*U2); V2=lambda*U; V=V1-V2; R=E*U/V;
112
Numerical inversions of Laplace transforms
m=m+(-1)^k*R; end end f=real(P*m); %P(R_\phi(t)=l)
Bibliography [1] J. Abate and W. Whitt, The Fourier-series method for inverting transforms of probability distributions, Queuing Systems, 10, 1992, pp. 5-88. [2] R. E. Barlow and F. Proschan, Statistical theory of reliability and life testing, Holt, Rinehart and Winston, New York, 1975. [3] R. E. Bartle, The Elements of Real Analysis, John Wiley, New York, 1964. [4] D. S. Chang, Reliability bounds for the stress-strength model, Computers ind. Engng, vol. 29 no. 1-4, 1995, pp. 15-19. [5] G. L. Choudhury, D. M. Lucantoni and W. Whitt, Multidimensional transform inversion with applications to the transient M/G/1 queue, The Annals of Applied Probability, vol. 4 no. 3, 1994, pp. 719-740. [6] A. Csenki, Asymptotics for renewal reward processes with retrospective reward structure, Operations Research Letters, 26, 2000, pp. 201-209. [7] D. R. Cox, Renewal Theory, Spottiswoode Ballantyne, London, 1962. [8] H. Cram´er, Mathematical Methods of Statistics, Princeton University Press, Princeton, 1946. [9] P. J. Davis and P. Rabinowitz, Methods of Numerical Integration, 2nd ed., Academic Press, New York, 1984. [10] E. J. Dudewicz, Modern Mathematical Statistics, John Wiley, New York, 1988. [11] P. Embrechts, C. Kl¨ upelberg and T. Mikosch, Modelling extremal events for insurance and finance, Springer-Verlag, Berlin, 1999. [12] T. Erhardsson, On stationary renewal reward processes where most rewards are zero, Probab. Theory Relat. Fields, 117, 2000, pp. 145-161. [13] K. Funaki, K. Yoshimoto, Distribution of total uptime during a given time interval, IEEE Trans. Reliability, vol 43 no. 3, 1994 Sep, pp. 489-492. 113
114
BIBLIOGRAPHY
[14] O. Gaudoin and J. L. Soler, Failure rate behavior of components subjected to random stresses, Reliability Engineering and System Safety, 58, 1997, pp. 19-30. [15] B. V. Gnedenko, Y. K. Belyayev, A. D. Solovyev, Mathematical Methods of Reliability Theory, Academic Press, New York, 1969. [16] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, New York, 1994. [17] J. Grandell, Doubly Stochastic Poisson Processes, Springer-Verlag, Berlin, 1943. [18] G. R. Grimmett and D. R. Stirzaker, Probability and Random Processes, Clarendon Press, Oxford, 1997. [19] J. A. Gubner, Computaion of shot-noise probability distributions and densities, SIAM Journal on Scientific Computing, vol. 17 no. 3, 1996, pp. 750-761. [20] R. Herz, H. G. Schlichter und W. Siegener, Angewandte Statistik f¨ ur Verkehrs- und Regionplaner, 2. auflage, Werner-Ingenieus-Texte 42, Werner-Verlag, D¨ usseldorf, 1992. [21] M. Iosifescu, Finite Markov Processes and Their Applications, John Wiley, New York, 1980. [22] P. W. den Iseger and M. A. J. Smith, A new method for inverting Laplace transforms, Econometric Institute, EUR Rotterdam, 1998, pp. 1-14. [23] N. Jack, Repair replacement modelling over finite time horizons, J. Opl. Res. Soc. vol. 42 no. 9, 1991, pp. 759-766. [24] E. E. Lewis and H. C. Chen, Load-capacity interference and bathtub curve, IEEE Trans. Reliability, vol 43 no. 3, 1994 Sept, pp. 470-475. [25] J. Mi, Average number of events and average reward, Probab. Eng. Inform. Sc. vol. 14 no. 4, 2000, pp. 485-510. [26] E. J. Muth, A method for predicting system downtime, IEEE Trans. Reliability, R-17, 1968 Jun, pp. 97-102. [27] E. J. Muth, Excess time, A measure of system repairability, IEEE Trans. Reliability, R-19, 1970 Feb, pp. 16-19. [28] F. Oberhettinger, Fourier Transforms of Distributions and Their Inverses; a collection of tables, Academic Press, New York, 1973.
BIBLIOGRAPHY
115
[29] M. Parlar and D. Perry, Inventory models of future supply uncertainty with single and multiple suppliers, Naval Research Logistics, 43, 1996, pp. 191-210. [30] T. Pham-Gia and N. Turkkan, System availability in a Gamma alternating renewal process, Naval Research Logistics, vol 46 no. 7, 1999 Oct, pp. 822844. [31] E. Popova and H. C. Wu, Renewal reward processes with fuzzy rewards and their applications to T-age replacement policies, European Journal of Operational Research, 117, 1999, pp. 606-617. [32] R. D. Reiss, A Course on Point Processes, Springer-Verlag, Berlin, 1993. [33] A. Renyi, On the asymptotic distribution of the sum of a random number of independent random variables, Acta Mathematica Academiae Scientiarum Hungaricae, 8, 1957, pp. 193-199. [34] S. I. Resnick, Extreme Values, Reguler Variation, and Point Processes, Springer-Verlag, Berlin, 1987. [35] S. I. Resnick, Adventures in Stochastic Processes, Birkauser, Boston, 1992. [36] S. M. Ross, Stochastic processes, John Wiley, New York, 1996 [37] S. K. Srinivasan, R. Subramanian, and K. S. Ramesh, Mixing of two renewal processes and its applications to reliability theory, IEEE Trans. Reliability, R-20, 1971 May, pp. 51-55. [38] Suyono and J. A. M. van der Weide, Renewal reward processes, Journal of Indonesian Math. Society, vol. 6 no. 5, 2000, pp. 353-357. [39] Suyono and J. A. M. van der Weide, Covariance of a renewal process, Proc. Indonesian Student Scientific Meeting (ISSM), Manchester, UK, 2001, pp. 383-386. [40] Suyono and J. A. M. van der Weide, System reliability in stress-strength models, Prosiding Konferensi Nasional Matematika XI, Malang, Indonesia, 2002, pp. 1135-1139. [41] Suyono and J. A. M. van der Weide, Total uptime of series systems, Proc. ISSM, Berlin, Germany, 2002, pp. 486-489. [42] Suyono and J. A. M. van der Weide, A method for computing total downtime distributions in repairable systems, Submitted to the Journal of Applied Probability, 2002. [43] Suyono and J. A. M. van der Weide, Integrated renewal processes, Submitted to the Stochastic Processes and their Applications, 2002.
116
BIBLIOGRAPHY
[44] L. Tak´ acs, On certain sojourn time problems in the theory of stochastic processes, Acta Mathematica Academiae Scientiarum Hungaricae, 8, 1957, pp. 169-191. [45] L. Tak´ acs, On a sojourn time problem in the theory of stochastic processes, Transactions of American Mathematical Society, 93, 1959, pp. 531-540. [46] H. C. Tijms, Stochastic Models: An Algorithmic Approach, John Wiley, New York, 1994. [47] W. T. Weeks, Numerical inversion of Laplace transforms using Laguerre functions, J.ACM 13, 1966, pp. 419-426. [48] D. V. Widder, The Laplace transform, Princeton University Press, Princeton, 1946. [49] R. W. Wolff, Stochastic Modeling and the Theory of Queues, Prentice-Hall, Englewood Cliffs, NJ, 1989. [50] J. Xue and K. Yang, Upper & lower bounds of stress-strength interference reliability with random strength-degradation, IEEE Trans. reliability vol. 46 no. 1, 1997 March, pp. 142-145.
Samenvatting Vernieuwingsprocessen en repareerbare systemen In dit proefschrift bespreken we de volgende onderwerpen. 1. Renewal reward processen Wij geven een berekening van de marginale verdelingen van renewal reward processen en van een klasse van processen die we in dit proefschrift aanduiden als instantaneous reward processen. Onze benadering is gebaseerd op de theorie van puntprocessen, in het bijzonder Poisson puntprocessen. Hierbij wordt het renewal reward proces (instantaneous reward proces) gerepresenteerd als een functionaal van een Poisson punt proces. Belangrijke hulpmiddelen die gebruikt worden bij de afleiding zijn de Palm formule en de formule voor de Laplace getransformeerde van een Poisson puntproces. Als resultaat vinden we de Laplace getransformeerden van de marginale verdelingen. We geven een toepassing van instantaneous reward processen in een verkeersprobleem. Een aantal asymptotische eigenschappen van renewal reward processen wordt opnieuw onder de loupe genomen. Zo geven we met behulp van een Tauber stelling een bewijs van de versie van de renewal reward stelling met verwachtingswaarden. We leiden vervolgens een tweede orde term af voor de versie van de renewal reward stelling met verwachtingswaarden. Gelijksoortige resultaten onderzoeken we voor instantaneous reward processen. Verder bewijzen we asymptotische normaliteit voor instantaneous reward processen. We berekenen de covariantie structuur van een renewal proces, dat kan worden beschouwd als een bijzonder geval van een renewal reward proces. Verder bestuderen we systeem betrouwbaarheid voor een stress-strength model, waarbij de omvang van de stress ge¨interpreteerd wordt als een ”reward”. De tijdstippen waarop stress optreedt modelleren met behulp van vernieuwingsprocessen en Cox processen. Met de resultaten die we afgeleid hebben voor renewal reward processen onderzoeken we de invloed van de 117
afhankelijkheid tussen stress en strength op de systeem betrouwbaarheid. 2. Ge¨integreerde vernieuwingsprocessen In de literatuur is de kansdichtheid van een ge¨integreerd homogeen Poisson proces bekend. Het is natuurlijk om dit te generaliseren voor niethomogene Poisson processen, Cox processen en vernieuwingsprocessen. In dit proefschrift leiden we formules af voor de marginale verdelingen met behulp van conditionering voor ge¨integreerde Poisson en Cox processen. Voor ge¨integreerde vernieuwingsprocessen gebruiken we puntproces representaties. De resultaten worden gegeven in de vorm van Laplace getransformeerden. De asymptotische eigenschappen van ge¨integreerde vernieuwingsprocessen worden onderzocht. Tenslotte is er een toepassing op een verkeersprobleem. 3. Totale downtime van reprareerbare systemen Verschillende auteurs hebben met een aantal verschillende methodes formules afgeleid voor verdelingsfunctie van de toatle downtime van een repareerbaar systeem, dat bij deze studies als ´e´en component wordt beschouwd. Wij leiden formules af met nog een andere methode (gebaseerd op puntprocessen) en we beschouwen ook het algemenere geval waarbij afhankelijkheid wordt toegelaten tussen faal en reparatie tijden. De covariantie structuur en de asymptotische eigenschappen van de totale downtijd zijn voor het onafhankelijke geval bekend in de literatuur. Wij leiden resultaten af voor het afhankelijke geval. We geven voorbeelden van het effect van afhankelijkheid tussen faaltijd en reparatietijd op de totale downtijd. We bespreken ook de totale downtijd voor repareerbare systemen die uit twee of meer stochastisch onafhankelijke componenten bestaan. We leiden ee uitdrukking af voor de marginale verdeling van de totale uptijd van het systeem als faaltijden en reparatietijden van iedere component exponentieel verdeeld zijn. Voor willekeurige faal- en reparatietijden geven we een uitdrukking voor de verwachte totale uptijd.
118
Summary Renewal processes and repairable systems In this thesis we discuss the following topics. 1. Renewal reward processes The marginal distributions of renewal reward processes and its version, which we call in this thesis instantaneous reward processes, are derived. Our approach is based on the theory of point processes, especially Poisson point processes. The idea is to represent the renewal reward processes and its version as functionals of Poisson point processes. Important tools we use are the Palm formula and the Laplace functional of Poisson point processes. The results are presented in the form of Laplace transforms. An application of the instantaneous reward processes to the study of traffic is given. Some asymptotic properties of the renewal reward processes are reconsidered. A proof of the expected-value version of the renewal reward theorem using the Tauberian theorem is given. A second order term in the expected-value version of the renewal reward theorem is obtained. Similar results for the instantaneous reward processes are investigated. Asymptotic normality of the instantaneous reward processes is proved. The covariance structure of renewal processes, which can be considered as a special case of renewal reward processes, is derived. As an addition, we study system reliability in a stress-strength model, where the amplitudes of stresses can be considered as rewards. We consider renewal and Cox processes as models for the occurrences of the stresses. Using our result on renewal reward processes we investigate the effect of dependence between stress and strengths on system reliability. 2. Integrated renewal processes The marginal probability density function of an integrated homogeneous Poisson Process is known in the literature. It is natural to generalize the 119
integrated homogeneous Poisson process into integrated non homogeneous Poisson, Cox, and renewal processes. In this thesis we derive expressions for the marginal distributions of integrated Poisson and Cox processes using conditioning arguments, and derive the marginal distributions of integrated renewal processes using the theory of point processes. The results are presented in the form of Laplace transforms. Asymptotic properties of the integrated renewal processes are also investigated. An application to the study of traffic is given. 3. Total downtime of repairable systems An expression for the cumulative distribution function of the total downtime of a repairable system, which is regarded as a single component, under an assumption that the failure and the repair times of the system are independent has been derived by several authors using different methods. We use a different method (using point processes) to compute the distribution function of the total downtime. We also consider a more general situation where we allow dependence of the failure and the repair times of the system. The covariance structure and asymptotic properties of the total downtime for the independent case are also known in the literature. We derive the similar results for the dependent case. Examples are given to see the effect of dependence between the failure and the repair times on the total downtime. We also discuss the total downtime of repairable systems consisting of n ≥ 2 stochastically independent components. We derive an expression for the marginal distribution of the total uptime of the system for the case the failure and the repair times of each component are exponentially distributed. For arbitrary failure or repair times of the components we derive an expression for the mean of the total uptime.
120
Ringkasan Proses renewal dan sistem-sistem tereparasi Di tesis ini di bahas topik-topik berikut. 1. Proses renewal reward Distribusi marginal dari proses renewal reward (renewal reward process) dan versinya, yang dinamakan proses instantaneous reward (instantaneous reward process), diturunkan. Dasar teori yang dipakai adalah teori tentang proses titik (point process), khususnya proses titik Poisson. Cara yang digunakan adalah dengan menyatakan proses renewal reward dan versinya sebagai fungsional dari proses-proses titik Poisson. Untuk perhitungan digunakan rumus-rumus Palm dan fungsional Laplace dari proses titik Poisson. Hasil-hasil yang diperoleh disajikan dalam bentuk transformasi Laplace. Penerapan proses instantaneous reward dalam studi tentang lalu lintas diberikan. Beberapa sifat asimtotik dari proses renewal reward dipelajari kembali. Bukti dengan teorema Tauber untuk versi harga harapan dari teorema renewal reward diberikan. Suku konstan dalam versi harga harapan dari teorema renewal reward ditemukan. Sifat-sifat serupa untuk proses instantaneous reward juga di selidiki. Distribusi asimtotik normal untuk proses instantaneous reward dibuktikan. Kovariansi dari proses renewal, yang juga dapat dipandang sebagai proses renewal reward, diturunkan. Sebagai tambahan dipelajari reliabilitas sistem dalam model stress-strength, dimana besarnya stress dapat dipandang sebagai reward. Proses kejadian dari stress dimodelkan dengan proses Cox dan renewal. Efek dari dependen antara stress dan strength terhadap reliabilitas sistem diselidiki menggunakan hasil tentang proses renewal reward. 2. Proses integrated renewal Fungsi kepadatan probabilitas marginal dari sebuah proses integrated renewal (integrated renewal process) dengan proses Poisson homogen se121
bagai proses dasarnya telah dikenal di literatur. Adalah alami untuk menggeneralisasi proses Poisson homogen sebagai proses dasar ke proses Poisson tak homogen, proses Cox, dan ke proses renewal. Di tesis ini distribusi marginal dari proses integrated renewal diturunkan dengan teknik kondisional untuk proses Poisson dan Cox sebagai proses dasarnya, dan dengan memakai teori tentang proses titik untuk proses renewal sebagai proses dasarnya. Hasil-hasilnya disajikan dalam bentuk transformasi Laplace. Sifat-sifat asimtotik dari proses integrated renewal juga diselidiki. Sebuah penerapan dalam studi lalu lintas diberikan. 3. Total downtime sistem-sistem tereparasi Sebuah ekspresi untuk fungsi distribusi kumulatif dari total downtime sebuah sistem yang dapat direparasi, di mana sistem tersebut dipandang sebagai sebuah komponen, telah diturunkan oleh beberapa penulis dengan metode yang berbeda di bawah asumsi bahwa waktu-waktu bekerja dan reparasi dari sistem saling independen. Di tesis ini sebuah metode yang berbeda (yakni dengan menggunakan teori tentang proses titik) digunakan untuk menentukan fungsi distribusi dari total downtime. Situasi yang dipelajari juga lebih umum karena diperbolehkan adanya dependensi antara waktu bekerja dan reparasi dari sistem. Kovariansi dan sifat-sifat asimtotik dari total downtime untuk kasus independen juga telah diketahui di literatur. Hasil-hasil yang serupa untuk kasus dependen diturunkan dalam tesis ini. Contoh-contoh untuk melihat efek dependen antara waktu bekerja dan reparasi terhadap total downtime diberikan. Di tesis ini juga dibahas total downtime dari sistem-sistem dengan dua komponen atau lebih dimana komponen-komponen tersebut saling independen secara stokastik. Sebuah ekspresi untuk distribusi marginal dari total uptime sistem dimana waktu bekerja dan reparasi dari setiap komponennya berdistribusi eksponensial diturunkan. Untuk kasus distribusi waktu bekerja dan reparasi dari komponennya sembarang, harga harapan dari total uptime diturunkan.
122
Curriculum Vitae Suyono was born on December 18, 1967 in Purworejo, Indonesia. After finishing his secondary high school in Purworejo in 1986, he studied Mathematics Education at the Jakarta Institute for Teachers Training and Education and graduated in 1991. In August 1994 he started his Master Program in Mathematics at Gadjah Mada University in Yogyakarta, Indonesia, and obtained his degree in 1998. From October 1998 until February 2003 he carried out a doctoral research at the Department of Control, Risk, Optimization, Stochastic and Systems (CROSS), Faculty of Information Technology and Systems (ITS), Delft University of Technology. Currently he is a lecturer at the Department of Mathematics, Faculty of Mathematics and Natural Sciences, State University of Jakarta, Indonesia.
123
