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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 < 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
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).
- Spring '10
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