Example 422 return to o ce hours with the machinery

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Unformatted text preview: 1 2 N ⇡ (N ) ⇡ (n) < 1, so we can pick ⇡ (0) to make the sum = 1. Concrete example. Suppose s = 1 and an = 1/(n + 1). In this case n ··· 0 1 ( /µ)n = n· = µn · · · µ1 µ 1 · 2···n n! n1 To find the stationary distribution we want to take ⇡ (0) = c so that c 1 X ( /µ)n =1 n! n=0 Recalling the formula for the Poisson distribution with mean /µ, we see that c = e /µ and the stationary distribution is Poisson. 4.4 Exit Distributions and Hitting Times In this section we generalize results from Sections 1.8 and 1.9 to continuous time. We will approach this first using the embedded jump chain with transition probability P q (i, j ) r(i, j ) = where i = j 6=i q (i, j ) i Let Vk = min{t 0 : Xt = k } be the time of the first visit to x and let Tk = min{t 0 : Xt = k and Xs 6= k for some s < t } be the time of the first return. The second definition is made complicated by the fact that is X0 = k then the chain stays at k for an an amount of time that is exponential with rate k. 134 CHAPTER 4. CONT...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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