6.262.Lec10

# 6.262.Lec10 - DISCRETE STOCHASTIC PROCESSES Lecture 10...

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Lecture 10 - 3/5/2010 Discrete Stochastic Processes 1 DISCRETE STOCHASTIC PROCESSES Lecture 10 Renewal Processes - Chapter 3 Review: Strong Law for Renewal Processes Time Averages vs. Ensemble Averages Central Limit Theorem for Renewal Processes Renewal - Reward Processes Example: Long-Term Time Averages of Residual Life of Interarrival Interval Age of Interarrival Interval Duration of Interarrival Interval General Renewal-Reward Processes Time-Averaged Renewal-Reward Theorem Stopping Times Wald's Theorem

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Lecture 10 - 3/5/2010 Discrete Stochastic Processes 2 R ENEWAL P ROCESSES N(t) t SS S X X 1 1 2 2 3 3 X R ENEWAL P ROCESS : A counting process Nt t a f lq ; 0 with the property that the positive ( PX i == 00 af ) interrenewal times X 1 , X 2 ,... are IID . (Ross allows P (X i = 0) < 1.) The arrival epochs S 1 , S 2 are sums of IID rv's SX in n i = = 1 . Crucial relationship: for every integer 1 n , all t > 0, Nt () n { } = S n t { } . The homogeneous Poisson process is a (very special) renewal process.
Lecture 10 - 3/5/2010 Discrete Stochastic Processes 3 ( ) , 0 N(t) Poisson distributed with 1 [ ] E[N(t)] = var [N(t)] = t. E[N(t)/t] = , x X fx e x EX λ =≥ = var (N(t)/t) = 0. t S TRONG L AW FOR R ENEWAL P ROCESSES Theorem 1 : For a renewal process with mean interarrival time X (possibly not finite) , lim t Nt t X →∞ = a f 1 w. p. 1. ( This result describes a long-term time average, not an expectation. Poisson Case: In Poisson case, theorem is true for each t > 0, not just in limit as t . Outline of Proof: Let 1 + + X . nn SX = L By the strong law of large numbers, () (|l i m )1 S n PX n n ω == →∞ and, since the function: 1 y y is continuous at each y > 0, you showed on the homework that 1 i m n n S X P n = = →∞ .

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Lecture 10 - 3/5/2010 Discrete Stochastic Processes 4 () () () 1 1 ()1 () 1 1 Nt St S SS t SN t S t S + + ++ < ≤< + <≤ + We now need to show that lim( ) = lim( ). tn n n tS →∞ We use the SLLN applied to the arrival epochs S n . First define S as the epoch of the most recent arrival before (or at) time t . From the picture we have Since for every outcome ω , ()/ Nt S and () 1/ S + + run through the same set of values as t as / n nS does as n , the proof is complete provided , so that ( ) 1. 1 t +
Lecture 10 - 3/5/2010 Discrete Stochastic Processes 5 ( ) ( ) lim / t Nt t →∞ Ensemble Averages vs. Time Average s We now understand (?) the time average , which is 1/ X WP1. What's Next? We will also want to understand the ensemble average, E N t () / t [ ] . Is this 1/ X for all t ? (Answer is no except for Poisson) Is this 1/ X in limit t → ∞ ? (Answer is yes, but it doesn't follow from time average. E Nt t t lim →∞ F H G I K J a f doesn't mean the same as lim t E t F H G I K J a f . ) t

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Lecture 10 - 3/5/2010 Discrete Stochastic Processes 6 Central Limit Theorem for N(t) If the iid interrenewal intervals Xn have a finite mean and variance , then where is the standard (i.e., mean = 0, variance = 1) normal cdf.
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## This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.

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6.262.Lec10 - DISCRETE STOCHASTIC PROCESSES Lecture 10...

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