Discrete-time stochastic processes

# For residual 2 life rz xn xn z so the integral in

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Unformatted text preview: ≤ t + b. Thus, applying (3.14) to the truncated process, hi hi h i e ee ee E X (m(t) + 1) ≤ E X (m(t) + 1) = E SN (t)+1 ≤ t + b. Combining this equation with (3.15), we have upper and lower bounds on m(t), 1 1 m(t) 1 b 1 −≤ ≤ h i+ h i− . E [X ] t t t e e EX tE X (3.16) hi R √ b e Finally, choose b = t. Then as t → 1, b → 1 also, so that E X = 0 [1−FX (x)] dx→E [X ]. Also, both of the ﬁnal terms on the right of (3.16) approach 0, completing the proof. Note that this theorem (and its proof ) have not assumed ﬁnite variance. The theorem also hi e holds when E [X ] is inﬁnite; E X in (3.16) simply approaches 1 in this case. We have just shown that m(t) has the property that limt→1 [m(t)/t] = 1/E [X ]. Note again that N [t, ω ]/t is the average number of renewals from 0 to t for a sample function ω , and m(t)/t is the average of this over ω . Combining with Theorem 3.1, the limiting time and ensemble-average equals the time-average renewal rate for each sample function except for a set of probability 0. Another interesting question is to determine the expected renewal rate in the limit of large t without averaging from 0 to t. That is, are there some values of t at which renewals are more likely than others for large t? If the inter-renewal intervals have an integer distribution function (i.e., each inter-renewal interval must last for an integer number of time units), then each renewal epoch Sn must also be an integer. This means that N (t) can increase only at integer times and the expected rate of renewals is zero at all non-integer times. An obvious generalization of this behavior for integer valued inter-renewal intervals is that of inter-renewals that occur only at integer multiples of some real number d > 0. Such a distribution is called an arithmetic distribution. The span of an arithmetic distribution is the largest number d such that this property holds. Thus, for example if X takes on only the values 0, 2, and 6, its distribution is arithmetic with span 2. Similarly, X takes on only the values 1/3 and 1/5, then the span is 1/15. The remarkable thing, for our purposes, is that any inter-renewal distribution that is not an arithmetic distribution leads to a uniform expected rate of renewals in the limit of large t. This result is contained in Blackwell’s theorem which we state without proof (see Section 11.1, Theorem 1 of [9]). Recall, however, that for the special case of an inter-renewal density that has a rational Laplace transform, Blackwell’s theorem is a simple consequence of (3.9). 5 This proof can be postponed and read later. 3.4. RENEWAL-REWARD PROCESSES; TIME-AVERAGES 105 Theorem 3.5 (Blackwell). If a renewal process has an inter-renewal distribution that is non-arithmetic, then for any δ > 0, lim [m(t + δ ) − m(t)] = t→1 δ . E [X ] (3.17) If the inter-renewal distribution is arithmetic with span d, then for any integer n ≥ 1 lim [m(t + nd) − m(t)] = t→1 nd . E [X ] (3.18) Eq. (3.17) says that for non-arithmetic distributions, the expected number of arrivals in the interval (t, t + δ ] is equal to δ /E [X ] in the limit t → 1. Since the theorem is true for arbitrarily small δ , the theorem almost, but not quite, says that dm(t)/dt is equal to 1/E [X ] in the limit t → 1. Unfortunately this latter statement is not true, and one can see the reason by looking at an example where X can take on only the values 1 and π . Then no matter how large t is, N (t) can only increase at discrete points of time of the form k + j π where k and j are non-negative integers; thus dm(t)/dt is either 0 or 1 for all t. As t gets larger, however, the jumps in m(t) become both smaller in magnitude and more closely spaced from one to the next. Thus [m(t + δ ) − m(t)]/δ approaches 1/E [X ] as t → 1 for any ﬁxed δ , no matter how small, but as δ gets smaller, the convergence in t gets slower. No matter how large one chooses t, [m(t + δ ) − m(t)]/δ does not approach 1/E [X ] as δ → 0. Since the inter-renewal intervals are positive random variables, multiple renewals cannot occur simultaneously, and thus, as shown in Exercise 3.15, (3.17) implies that the probability of a renewal in a small interval (t, t + δ ] tends to δ /E [X ] + o(δ ) as t → 1. Thus, for a non-arithmetic inter-renewal distribution, the limiting distribution of renewals in a small interval (t, t + δ ] satisﬁes lim Pr {N (t + δ ) − N (t) = 0} = 1 − δ /X + o(δ ) t→1 lim Pr {N (t + δ ) − N (t) = 1} = δ /X + o(δ ) t→1 lim Pr {N (t + δ ) − N (t) ≥ 2} = o(δ ). (3.19) t→1 If we compare this with Eq. (2.17), associating the rate ∏ of a Poisson process with 1/X , we see that, asymptotically, a renewal process with a non-arithmetic inter-renewal distribution satisﬁes two of the three requirements in deﬁnition 3 of a Poisson process. That is, the increments are asymptotically stationary and the renewals do not occur simultaneously. If the renewal process is not Poisson, however, the increments are not independent. For an arithmetic renewal process with span d, (3.18), with n = 1, states that the p...
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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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