Renewal Theory. (iv) For s < t, N(t) N(s) equals the number of events in (s, t].

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1 Renewal Theory Def. A stochastic process {N(t), t 0} is said to be a counting process if N(t) represents the total number of events that have occurred up to time t. X 1, X 2,... times between the events (arrivals). S n = X X n the time of the nth event Example. Poisson process Definition implies: (i) N(t) 0 (ii) N(t) is integer valued (iii) If s < t, then N(s) N(t) (iv) For s < t, N(t) N(s) equals the number of events in (s, t]. 1

2 Poisson process Def. The counting process {N(t), t 0} is called a Poisson process with rate λ, if X 1, X 2,... are independent and have a common exponential distribution P(X n x) = 1 e λx, x 0. Examples: sequence of phone calls, traffic, machine failures,... λt (λt)n For the Poisson process, P(N(t) = n) = e, n = 0, 1,... n! λt (λt)n 1 S n has an Erlang or Gamma distribution: f Sn (t) = λe (n 1)! Expected number of events up to time t is m(t) = E[N(t)] = λt Naturally, λ is called the rate of the process 2

3 Renewal Process Def. A counting process {N(t), t 0} with i.i.d. inter-arrival times is called a renewal process X 1, X 2,... - independent inter-arrival times with a common distribution F µ = E[X n ], n 1 Ex. 1. Lifetimes of bulbs are i.i.d. A failed bulb is immediately replaced by a new one. N(t) # bulbs failed by time t Ex. 2. An inventory system has a stock u. The demands in successive weeks 1, 2,... are i.i.d. The stock will be expired at week N(u) + 1. Let S 0 = 0, S n = n i=1 X i. We have: N(t) = max{n : S n t} and N(t) <, t 0. However, N(t) as t 3

4 Example 7.1 P(X n = i) = p(1 p) i 1, geometric distribution, # trials until the first success. Then S n is # trials until the nth success ( ) k 1 P(S n = k) = p n (1 p) k n, k n n 1 Hence, P {N(t) = n} = [t] k=n ( ) k 1 p n (1 p) k n n 1 [t] k=n+1 ( ) k 1 p n+1 (1 p) k n 1 n 1 Equivalently, since a renewal occurs w.p. p at 1, 2,..., we have ( ) [t] P {N(t) = n} = p n (1 p) [t] n n 1 4

5 One-to-one correspondence b/w renewal process and its mean-value function The renewal function uniquely determines the renewal process Example 7.2. Suppose we have a renewal process whose mean-value function is given by m(t) = 2t What is the distribution of the # renewals occurring by time 10? Solution. Since m(t) = 2t is the mean-value function of a Poisson process with rate 2, it follows from the one-to-one correspondence that N(10) has a Poisson distribution with parameter 20: 20 20n P {N(10) = n} = e n!, n 0 5

6 Example 7.3 Let X 1, X 2,... have uniform distribution on [0, 1], that is, f(x) = 1 if x [0, 1] and f(x) = 0 otherwise. We want to determine m(t) for t 1. The renewal equation becomes: m(t) = F(t) + = t + t 0 t 0 m(t x)f(x) dx = t + m(y) dy Differentiating both sides: m (t) = 1 + m(t) t 0 m(t x) dx Taking h(t) = 1 + m(t), we get: h (t) = h(t) h(t) = Ke t Thus, m(t) = Ke t 1. Since m(0) = 0, it follows that K = 1. The answer: m(t) = e t 1, 0 t 1. 6

7 Limiting Theorems N( ) = lim t N(t) = with probability 1. At which rate does N(t) go to infinity? Proposition 7.1 (Strong Law of Large Numbers for N(t))With probability 1, N(t) 1 t µ as t (renewal process goes to infinity at rate 1/µ) Elementary Renewal Theorem m(t) t 1 µ as t ( where 1 0 ). (renewal process goes to infinity at average rate 1/µ) 7

8 Example 7.7 Single-server bank. Arrivals of customers: Poisson process (λ). A customer enters the bank if the server is available. Otherwise, the customer leaves. The service time has a distribution G. (a) What is the rate at which customers enter the bank? (b) What proportion of potential customers actually enter the bank? Solution. (a) Renewal occurs when a customer enters a bank. Let µ G be the average service time. then µ = µ G + 1/λ (memory-less property). The rate at which customers enter is 1/µ = λ/(1 + λµ G ) (b) Consider a process in discrete time, where 1, 2,... are successive customers. The renewal corresponds to a customer who actually entered the bank. Average number of customers in a cycle is µ = λµ G + 1 (customers rejected during the service plus one accepted customer). The rate is 1/(λµ G + 1). 8

9 Wald s Equation X 1, X 2,... i.i.d. N integer-valued r.v. If N and X 1, X 2,... are independent, then E(X X N ) = E(X)E(N) But independence is a too strong condition. Def. An integer-valued random variable N is said to be a stopping time for the sequence X 1, X 2,... if the event {N = n} is independent of X n+1, X n+2,... for all n = 1, 2,.... Theorem (Wald s equation). If X 1, X 2,... are i.i.d with E[X n ] = E[X] < and N is a stopping time for X 1, X 2,... s.t. E[N] <, then [ N E X n ] = E[N]E[X] n=1 9

10 Stopping times... Def. An integer-valued random variable N is said to be a stopping time for the sequence X 1, X 2,... if the event {N = n} is independent of X n+1, X n+2,... for all n = 1, 2,.... {N(t), t 0} renewal process X 1, X 2,... inter-arrival times Which of the following are stopping times? N=5 N is independent of X 1, X 2,... (e.g. geometric with par-r p) N = N(10) N = N(10) + 1 N = N(t) N = N(t)

11 Renewal application of Wald s equation X 1, X 2,... inter-arrival times N(t) + 1 is a stopping time: the first renewal after t. N(t) + 1 = n X X n 1 t, X X n > t The event {N(t) + 1 = n} does not depend on X n+1, X n+2,.... Hence, E[X X N(t)+1 ] = E[X]E[N(t) + 1] Proposition 7.2. If µ = E(X) <, then E(S N(t)+1 ) = µ(m(t) + 1) or µ(m(t) + 1) = t + E[Y (t)], where Y (t) is the excess time. 11

12 Example 7.9 First, use element of type 1, lifetime is exponential with par-r µ 1. Then use element of type 2, lifetime is exponential with par-r µ 2. Then replace the machine. Find the av. # machines used by time t Solution. Let X(t) = 1, 2 be the type of the element used at time t. Then {X(t), t 0} is a continuous time Markov chain, thus, at time t, the element 1 is in use w.p. (Example 6.11) P 11 (t) = µ 1 µ 1 + µ 2 e (µ 1+µ 2 )t + µ 1 µ 1 + µ 2 Further, for our renewal process, µ = 1/µ 1 + 1/µ 2 and Y (t) = (1/µ 1 + 1/µ 2 )P 11 (t) + (1/µ 2 )(1 P 11 (t)) Combining everything together, we get m(t) = t µ + E[Y (t)] µ 1 = µ 1µ 2 µ 1 + µ 2 t µ 1µ 2 µ 1 + µ 2 [1 e (µ 1+µ 2 )t ] 12

13 Asymptotic Normality of N(t) Central Limit Theorem for Renewal Processes. Let µ and σ 2, assumed finite, represent the mean and variance of an inter-arrival time. Then ( ) N(t) t/µ P σ < y t/µ 3 as t. 1 2π y e x2 /2 dx 13

14 Example Processing times by machines 1 and 2 have a Gamma distribution with par-s n = 4, λ = 2, and uniform [0,4] distribution, respectively. Approximate the prob-ty that the two machines together complete at least 90 jobs by time t = 100 Solution. We have µ 1 = 2, σ1 2 = 1, µ 2 = 2, σ2 2 = 16/12. Thus, N 1 (100) + N 2 (100) N( , 100/ /6) = N(100, 175/6) { } N 1 (100) + N 2 (100) P {N 1 (100)+N 2 (100) > 89.5} = P > 175/6 175/6 1 Φ ( /6 ) = Φ where Φ is a standard normal distr. function ( /6 ) = Φ (1.944) =

Exponential Distribution

Exponential Distribution Definition: Exponential distribution with parameter λ: { λe λx x 0 f(x) = 0 x < 0 The cdf: F(x) = x Mean E(X) = 1/λ. f(x)dx = Moment generating function: φ(t) = E[e tx ] = { 1