Markov

# let t denote the distribution of xt and let the

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Unformatted text preview: +1 is chosen from ~ t 6= and Xt 2 R then Xt+1 is chosen from the distribution ~= the distribution QXt ; . If X P Xt ; . In other words, again we have imbedded our given Markov chain in the structure shown in Figure 1.81, with the transition kernels U and V given by For x 2 R: U x; f g = ; U x; fxg = 1 , For x 2 S , R: U x; fxg = 1 V  ; A = A For x 2 R: V x; A = Qx; A For x 2 S , R: V x; A = P x; A: ~ ~ The sequence X0 ; X0 ; X1 ; X1 ; : : : is a time-inhomogeneous Markov chain; the transition ~ kernel U used in going from Xt to Xt is di erent from the kernel V used in going from ~ ~ t to Xt+1 . Note that Xt 2 S and Xt 2 S for all t. The sequence X0 ; X1 ; : : : is a ~ X time-homogeneous Markov chain on S with transition kernel UV , de ned by UV x; B  = Z U x; dyV y; B : We claim that UV = P . If x 2 S , R then U x;  is point mass on x, so that UV x; B  = V x; B  = P x; B . If x 2 R then U x;  puts probability on the point and probability 1 , on the point x, so that UV x; B  = V  ; B  + 1 , V x; B  = B  + 1 , Qx; B  = P x; B : ~~ The sequence X0 ; X1 ; : : : is a time-homogeneous Markov chain, with transition kernel ~. V U =: P ~ 1.84 Exercise. Write down the transition kernel P in terms of the information given in the problem. Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-46 ~ If Xt has distribution t on S, then Xt has distribution t = t U on e. ~ S Finally, here is our Basic Limit Theorem for Harris chains. As usual, the statement involves an aperiodicity condition. Letting G = ft  1 : P fXt,1 2 Rg 0g, we say the chain is aperiodic if gcdG = 1. For example, as a simple su cient condition, if R 0, then the set G contains 1, so that the chain is aperiodic. 1.85 theorem. Let fXt g be an aperiodic Harris chain having a stationary distribution . Let t denote the distribution of Xt and let the initial distribution 0 be absolutely continuous with respect to . Then kt , k ! 0 as t ! 1. Proof: We are given the Harris chain fXt g with transition kernel P . Suppose we are also given a set R, probability measure , and number 2 0; 1 as in the de nition of a Harris chain. As discussed above, these determine transition kernels U and V with P = UV and ~ ~ ~ P = V U , and we will study the chain X0 ; X0 ; X1 ; X1 ; : : :. We are assuming that fXt g ~ has a stationary distribution , and we now know that fXt g has corresponding stationary distribution  = U . By the de nition of the Harris chain fXt g, the state is an accessible ~ ~ atom for fXt g, and the aperiodicity assumption implies that is aperiodic. ***WHY? EXPLAIN THIS. De ning 0 = 0 U , we see that 0 is absolutely continuous with respect ~ ~ to . Therefore, by Theorem 1.78 we have kt , k ! 0, where t denotes the distribution ~ ~~ ~ ~ of Xt . But V = U V = UV  = P = : ~ Thus, since kt+1 , k = kt V , V k  kt , k; ~ ~ ~~ we have kt+1 , k ! 0 as t ! 1. *** NOTE: Argue somewhere that kP , P k  k , k. Can use c...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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