# 4 h0 h k6i k6i 124 chapter 4 continuous time markov

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Unformatted text preview: rocess stays in state Y0 be t1 = ⌧0 / (Y0 ). At time T1 = t1 the process jumps to Y1 , where it should stay for an exponential amount of time with rate (Y1 ), so we let the time the process stays in state Y1 be t2 = ⌧1 / (Y1 ). At time T2 = t1 + t2 the process jumps to Y2 , where it should stay for an exponential amount of time with rate (Y2 ), so we let the time the process stays in state Y2 be t3 = ⌧2 / (Y2 ). Continuing in the obvious way, we can let the amount of time the process stays in Yn be tn+1 = ⌧n / (Yn ), so that the process jumps to Yn+1 at time Tn+1 = t1 + · · · + tn+1 In symbols, if we let T0 = 0, then for n X (t) = Yn 0 we have for Tn t &lt; Tn+1 (4.2) Computer simulation. The construction described above gives a recipe for simulating a Markov chain. Generate independent standard exponentials ⌧i , say, by looking at ⌧i = ln Ui where Ui are uniform on (0, 1). Using another sequence of random numbers, generate the transitions of Yn , then deﬁne ti , Tn , and Xt as above. The good news about the formal construction above...
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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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