Discrete-time stochastic processes

In the other states pii1 and pii1 0 1

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Unformatted text preview: rrent chain; this ensures that the steady-state probabilities for the embedded chain exist and are positive. It is possible 5.7. SEMI-MARKOV PROCESSES 225 for the denominator in (5.62) to be infinite. This can happen either because U (i) is infinite for some i, or because the sum does not converge (for example, there could be a countably infinite number of states and U (i) could be proportional to 1/πi ). When the denominator is infinite, we say that the probabilities {pi } do not exist; this is a bizarre special case, and it is discussed further in Section 6.1, but it doesn’t have to be specifically excluded from the following analysis. Lemma 5.6. Consider a semi-Markov process with an irreducible recurrent embedded chain {Xn ; n ≥ 0}. Given X0 = i, let {Mij (t); t ≥ 0} be the number of transitions into a given state j in the interval (0, t]. Then {Mij (t); t ≥ 0} is a delayed renewal process (or, if j = i, is an ordinary renewal process). Proof: Let M (t) be the total number of state transitions over all states that occur in the interval (0, t]. From Lemma 5.5, limt→1 M (t) = 1 with probability 1. Let Nij (n) be the number of transitions into state j that occur in the embedded Markov chain by the nth transition of the embedded chain. From Lemma 5.4, {Nij (n); n ≥ 0} is a delayed renewal process. It follows from Lemma 3.2 of Chapter 3 that limn→1 Nij (n) = 1 with probability 1. Since Mij (t) is the number of transitions into j during the first M (t) transitions of the embedded chain, we have Mij (t) = Nij (M (t)). Thus, lim Mij (t) = lim Nij (M (t)) = lim Nij (n) = 1. t→1 t→1 t→1 It follows that the time W1 at which the first transition into state j occurs, and the subsequent interval W2 until the next transition, are both finite with probability 1. Subsequent intervals have the same distribution as W2 , and all intervals are independent, so {Mij (t); t ≥ 0} is a delayed renewal process with inter-renewal intervals {Wk ; k ≥ 1}. If i = j , then all Wk are identically distributed and we have an ordinary renewal process, completing the proof. Let W (j ) be the mean inter-renewal interval between successive transitions into state j (i.e., the mean of the inter-renewal intervals W2 , W3 , . . . in {Mij (t); t ≥ 0}). Consider a delayed renewal-reward process defined on {Mij (t); t ≥ 0} for which R(t) = 1 whenever X (t) = j (see Figure 9). Define pj as the time-average fraction of time spent in state j . Then, if U (i) < 1, Theorems 3.8 and 3.12 of Chapter 3 state that R1 R(τ )dτ U (j ) pj = lim 0 = with probability 1. (5.63) t→1 t W (j ) We can also investigate the limit, as t → 1, of the probability that X (t) = j . This is equal to E [R(t)] for the renewal-reward process above. From Equation (3.72) of Chapter 3, if the distribution of the inter-renewal time is non-arithmetic, then pj = lim E [R(t)] = t→1 U (j ) . W (j ) (5.64) Next we must express the mean inter-renewal time, W (j ), in terms of more accessible quantities. From Theorem 3.9 of Chapter 3, lim Mij (t)/t = 1/W (j ) t→1 with probability 1. (5.65) 226 CHAPTER 5. COUNTABLE-STATE MARKOV CHAINS t t ✛ Un Xn =j ✛ ✲ t t Xn+1 6=j Xn+2 6=j Wk t Xn+3 =j ✲ t Xn+4 6=j Figure 5.9: The delayed renewal-reward process for time in state j . The reward is one from an entry into state j , say at the nth transition of the embedded chain, until the next transition out of state j . The expected duration of such an interval Un is U (j ). The inter-renewal interval Wk , assuming the kth occurrence of state j happens on the the nth transition of the embedded chain, lasts until the next entry into state j , with expected duration W (j ). As before, Mij (t) = Nij (M (t)) where, given X0 = i, M (t) is the total number of transitions in (0, t] and Nij (n) is the number of transitions into state j in the embedded Markov chain by the nth transition. Lemma 5.5 shows that limt→1 M (t) = 1 with probability 1, so lim t→1 Mij (t) Nij (M (t)) Nij (n) = lim = lim = πj . n→1 t→1 M (t) M (t) n (5.66) Combining (5.65) and (5.66), we have 1 W (j ) Mij (t) t→1 t Mij (t) M (t) = lim t→1 M (t) t Mij (t) M (t) = lim lim t→1 M (t) t→1 t M (t) = πj lim . t→1 t = lim (5.67) (5.68) Substituting this in (5.63), we see that pj = πj U (j ) limt→1 {M (t)/t}. Since limt→1 {M (t)/t} P is independent of j , and since j pj = 1, we see that limt→1 {M (t)/t} must be equal to P { j πj U (j )}−1 , thus yielding (5.62). Summarizing, we have the following theorem: Theorem 5.9. Assume that the embedded Markov chain of a semi-Markov process is irP reducible and positive-recurrent. If i πi U (i) <P , then, with probability 1, the limiting 1 fraction of time spent in state j is pj = πj U (j )/ i πi U (i). From (5.64), pj is also equal to limt→1 Pr {X (t) = j } if the distribution of the inter-renewal interval between transistions into j is non-arithmetic. A sufficient condition for this (assumP ing that i πi U i < 1) is that Gkj (u) be a non-arithmetic distribution for at least one pair of states k,...
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