Discrete-time stochastic processes

The general idea is illustrated in figure 34 but the

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Unformatted text preview: ing that the added generality is not worth much. 92 3.2. STRONG LAW OF LARGE NUMBERS FOR RENEWAL PROCESSES 93 customer is a rv, IID over customers, and independent of arrival times and server. We define a new renewal counting process, {N 0 (t); t ≥ 0}, for which the renewal epochs are those epochs in the original process {N (t); t ≥ 0} at which an arriving customer sees an empty system (i.e., no customer in queue and none in service). To make the time origin probabilistically identical to the renewal epochs in this new renewal process, we consider time 0 as an arrival epoch that starts a busy period. There are two renewal counting processes, {N (t); t ≥ 0} and {N 0 (t); t ≥ 0} in this example, and the renewals in the second process are those particular arrivals in the first process that arrive to an empty system. Throughout our study of renewal processes, we use X and E [X ] interchangeably to denote the mean inter-renewal interval, and use σ 2 to denote the variance of the inter-renewal interval. We will usually assume that X is finite, but, except where explicitly stated, we need not assume that σ 2 is finite. This means, first, that this variance need not be calculated (which is often difficult if renewals are embedded into a more complex process), and second, that the results are relatively robust to modeling errors on the far tails of the inter-renewal distribution. Much of this chapter will be devoted to understanding the behavior of N (t)/t and N (t) as t becomes large. N (t)/t is the time-average renewal rate over the interval (0,t]. It is a random variable (i.e., it is not defective) for each value of t (see Exercise 3.1). One of the ma jor results about renewal theory, which we establish shortly, is that, with probability 1, this family of random variables, {N (t)/t; t > 0}, has a limiting value, limt→1 N (t)/t, equal to 1/ X . This result is called the strong law of large numbers for renewal processes. We shall often refer to it by the less precise statement that the time-average renewal rate is 1/ X . This result is an analog (and direct consequence) of the strong law of large numbers, Theorem 1.5. Another important result is the elementary renewal theorem, which states that E [N (t)/t] also approaches 1/X as t → 1. It seems surprising that this does not follow from the strong law for renewal processes, but in fact it doesn’t, and we shall develop several widely useful results in establishing this theorem. The final ma jor result is Blackwell’s theorem, which shows that, for appropriate values of δ , the expected number of renewals in an interval (t, t + δ ] approaches δ /X as t → 1. We shall thus interpret 1/X as an ensemble-average renewal rate. This rate is the same as the above time-average renewal rate. We shall see the benefits of being able to work with both time-averages and ensemble-averages. 3.2 Strong Law of Large Numbers for renewal processes To get an intuitive idea why N (t)/t should approach 1/X for large t, note that Sn /n is the sample average of n inter-renewal intervals. From the strong law of large numbers, we know that Sn /n approaches X with probability 1 as t → 1. From Figure 3.1, observe that N (Sn ), the number of renewals at the epoch of the nth renewal, is n, and thus N (Sn )/Sn is n/Sn . This is the reciprocal of the sample average of n inter-renewal intervals. Since Sn /n → X as n → 1, we should expect n/Sn to approach2 1/X , and thus we should 2 Note that the fact that E [Sn /n] = X does not imply that E [n/Sn ] = 1/ X . We really need the stronger result that Sn /n → X with probability 1 94 CHAPTER 3. RENEWAL PROCESSES hypothesize that N (Sn )/Sn approaches 1/X as n → 1, and thus that N (t)/t approaches 1/X . To make this precise, we need the following lemma. Essentially, it says that the first renewal occurs eventually, and after that, the second eventually occurs, and so forth. Slope = Slope = N (t) t N (t) SN (t) = n Sn ✘✘✘ ❅ ❄ ✘✘✘ ❅ ✘✘✘ ❘ ❅ ✘✘✘ ✘✘✘ ✘✘ t ✘✘✘ N (t) 0 S3 S1 S2 Figure 3.1: Comparison of n/Sn with N (t)/t for N (t) = n. Lemma 3.1. Let {N (t); t ≥ 0} be a renewal counting process with inter-renewal rv’s {Xn ; n ≥ 1}. Then (whether or not E [Xn ] is finite), limt→1 N (t) = 1 with probability 1 and limt→1 E [N (t)] = 1. Proof: Note that for each sample point ω ∈ ≠, N (t, ω ) is a nondecreasing function of t and thus either has a finite limit or an infinite limit. The probability that this limit is finite with value less than n is lim Pr {N (t) < n} = 1 − lim Pr {N (t) ≥ n} = 1 − lim Pr {Sn ≤ t} t→1 t→1 t→1 Since the Xi are rv’s, the sums Sn are also rv’s (i.e., nondefective) for each n (see Section 1.3.7), and thus limt→1 Pr {Sn ≤ t} = 1 for each n. Thus limt→1 Pr {N (t) < n} = 0 for each n. This shows that the set of sample points ω for which limt→1 N (t) < 1 has probability 0. Thus limt→1 N (t) = 1 with probability 1. Next, E [N (t)] is nondecreasing...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.

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