Discrete-time stochastic processes

it is convenient in working with stopping rules to

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Unformatted text preview: must approach a limit as t → 1; and third, the limits must be the same. The following example, for a stochastic process very different from a renewal process, shows that equality between time and ensemble averages is not always satisfied for arbitrary processes. Example 3.2.1. Let {Xi ; i ≥ 1} be a sequence of binary IID random variables, each taking the value 0 with probability 1/2 and 2 with probability 1/2. Let {Mn ; n ≥ 1} be the product process in which Mn = X1 X2 · · · Xn . Since Mn = 2n if X1 to Xn each take the value 2 (an event of probability 2−n ) and Mn = 0 otherwise, we see that limn→1 Mn = 0 with probability 1. Also E [Mn ] = 1 for all n ≥ 1. Thus the time-average exists and equals 0 with probability 1 and the ensemble-average exists and equals 1 for all n, but the two are different. The problem is that as n increases, the atypical event in which Mn = 2n has a probability approaching 0, but still has a significant effect on the ensemble-average. Before establishing the results about ensemble-averages, we state and briefly discuss the central limit theorem for renewal processes. Theorem 3.2 (Central Limit Theorem for N (t)). Assume that the inter-renewal intervals for a renewal counting process {N (t); t ≥ 0} have finite standard deviation σ > 0. Then ( ) N (t) − t/X lim Pr < α = Φ(α). (3.3) −3/2 √ t→1 σX t Ry where Φ(y ) = −1 √1 π exp(−x2 /2)dx. 2 3.3. EXPECTED NUMBER OF RENEWALS 97 ✟ ✟✟ ❍ ✛✟ ✻ ❍ n ✟✟ ✲❄ ❍ α√n σ ✟ ■ ❅ X ✟ ❅√ ✟✟ α nσ ✟✟ ✟✟ ✟✟ 1 ✟✟ Slope = X ✟ nX t ✟ E [Sn ] Figure 3.4: Illustration of the central limit theorem for renewal processes. A given integer n is shown on the vertical axis, and the corresponding mean, E [Sn ] = nX is shown on the horizontal axis. The horizontal line with arrows at height n indicates α standard deviations from E [Sn ], and the vertical line with arrows indicates the distance below (t/X ). This says that N (t) tends to Gaussian with mean t/X and standard deviation σ X −3/2 √ t. The theorem can be proved by applying the central limit theorem (CLT) for a sum of IID rv’s, (1.56), to Sn and then using the identity {Sn ≤ t} = {N (t) ≥ n}. The general idea is illustrated in Figure 3.4, but the details are somewhat tedious, and can be found, for example, in [16]. We simply outline the argument here. For any real α, the CLT states that © √™ Pr Sn ≤ nX + α nσ ≈ Φ(α) Rα where Φ(α) = −1 √1 π exp(−x2 /2) dx and where the approximation becomes exact in the 2 limit n → 1. Letting √ t = nX + α nσ, and using {Sn ≤ t} = {N (t) ≥ n}, Pr {N (t) ≥ n} ≈ Φ(α). (3.4) Since t is monotonic in n for fixed α, we can express n in terms of t, getting √ t ασ n t n= − ≈ − ασ t1/2 (X )−3/2 . X X X Substituting this into (3.4) establishes the theorem for −α, which establishes the theorem since α is arbitrary. The omitted details involve handling the approximations carefully. 3.3 Expected number of renewals Let E [N (t)] be denoted by m(t) in what follows. We first find an exact expression for m(t). This is often quite messy for large t, so we then find the asymptotic behavior of m(t). Since N (t)/t approaches 1/X with probability 1, we expect m(t) to grow with a 98 CHAPTER 3. RENEWAL PROCESSES slope that asymptotically approaches 1/X , but we will find that this is not quite true in general. Two somewhat weaker results, however, are true. The first, called the elementary renewal theorem (Theorem 3.4), states that limt→1 m(t)/t = 1/X . The second result, called Blackwell’s theorem (Theorem 3.5), states that, sub ject to some limitations on δ > 0, limt→1 [m(t + δ ) − m(t)] = δ /X . This says essentially that the expected renewal rate approaches steady-state as t → 1. We will find a large number of applications of Blackwell’s theorem throughout the remainder of the text. The exact calculation of m(t) makes use of the fact that the expectation of a non-negative random variable is the integral of its complementary distribution function, m(t) = E [N (t)] = 1 X n=1 Pr {N (t) ≥ n} . Since the event N (t) ≥ n is the same as Sn ≤ t, m(t) is expressed in terms of the distribution functions of Sn , n ≥ 1, as follows. m(t) = 1 X n=1 Pr {Sn ≤ t} . (3.5) Although this expression looks fairly simple, it becomes increasingly complex with increasing t. As t increases, there is an increasing set of values of n for which Pr {Sn ≤ t} is significant, and Pr {Sn ≤ t} itself is not that easy to calculate if the interarrival distribution FX (x) is complicated. The main utility of (3.5) comes from the fact that it leads to an integral equation for m(t). Since Sn = Sn−1 + Xn for each n ≥ 1 (interpreting S0 as 0), and since Xn and Sn−1 are independent, we can use the convolution equation (1.11) to get Zt Pr {Sn ≤ t} = Pr {Sn−1 ≤ t − x} dFX (x) for n ≥ 2. x=0 Substituting this in (3.5) for n ≥ 2 and using the fact th...
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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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