F g 0 2 is positive recurrent 3 for all a 2 a 4 for aa

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Unformatted text preview: fXt+1 2 Ag 1. MARKOV CHAINS   I fXt 2 Ag : That is, P x; Adx is the expected number of visits made by the segment X1 ; : : : ; XT  to the set A. Is this the same as A, which is the expected number of visits made by the segment X0 ; : : : ; XT ,1  to the set A? The answer is yes! In fact, since X0 = XT = , the two segments X1 ; : : : ; XT  = X1 ; : : : ; XT ,1 ;  and X0 ; : : : ; XT ,1  =  ; X1 ; : : : ; XT ,1  consist of precisely the same states, just visited in a di erent order. Of course the mere di erence in ordering leaves the number of visits to the set A unchanged between the two segments. 1.89 Proposition. Suppose a Markov chain has an accessible atom distribution . Then f g = 1=E T . and a stationary Proof: By the same proof as the SLLN P before, using the cycle idea, we know that if the chain is started in the state , then 1=n n=1 I fXt = g ! 1=E T  with probability 1. t Combining this with Proposition 1.76, here is what we know. If the chain is started out in the distribution , then with probability 1 it hits at some nite time, after which, with probability 1, the long run fraction of visits to converges to 1=E T . We have used this type of reasoning before: the nite amount of time it takes the chain to hit does not have any e ect on the limiting long-run fraction of time the chain spends in the state . Thus, for a chain started in the distribution ,  P lim 1=n n!1 n X t=1  I fXt = g = 1=E T  = 1: By the Bounded Convergence Theorem,  E as n ! 1. But for each n,  E 1=n n X t=1 1=n n X t=1  I fXt = g ! 1=E T   I fXt = g = 1=n n X t=1 P fXt = g = f g: This, f g = 1=E T . *** ALTERNATIVELY, do it this way..... 1.90 Theorem. distribution . Then Stochastic Processes Suppose the chain fXt g has an accessible atom and a stationary J. Chang, March 30, 1999 1.10. GENERAL STATE SPACE MARKOV CHAINS 1. f...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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