We want to show that stochastic processes j chang

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Unformatted text preview: 1.76 Proposition. Suppose a Markov chain has an accessible atom distribution . Then P fT 1g = 1. and a stationary Proof: Let B = fx : Px fT = 1g 0g; these are the states from which it is possible to go forever without hitting . We want to show that B  = 0. Since is recurrent, if the chain starts from state , then with probability 1 it will return to in nitely many times. Therefore, the set B cannot be accessible from , for if it were, there would be positive probability, starting from , of eventually entering the set B and then never returning to . Thus, by the previous proposition, B  = 0. 1.77 Definition. Let  and  be two probability measures on a set S. We say that  is absolutely continuous with respect to  if A = 0 for all A S such that  A = 0, that is, each set having probability 0 under  also has probability 0 under . Suppose a chain fXt g with transition kernel P and an aperiodic, accessible atom has a stationary distribution . Let t denote the distribution of Xt and start the chain in any distribution 0 that is absolutely continuous with respect to . Then kt , k ! 0 as t ! 1. 1.78 Theorem. Proof: We use the coupling technique from before; much of the reasoning remains the same, so I'll just give a sketch. Again, we run two independent copies of the chain, fXt g and fXt g, starting in the initial distributions 0 and , respectively. We want to show that Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-40 with probability 1 the two chains eventually couple; in fact we claim that they eventually visit the state at the same time. By using the aperiodicity assumption together with the number-theoretic lemma from before, we see that the bivariate chain fXt ; Xt  : t = 0; 1; : : :g has an accessible atom  ; . The bivariate chain has a stationary distribution: the obvious product distribution   . So by Proposition 1.76, if the bivariate chain were started out in its stationary distribution   , it would eventually hit its atom  ;  with probability 1. That is, letting A denote the set of pairs of states x; y such that Px;y fT ;  1g = 1, we have   A = 1. From this, the absolute continuity of 0 with respect to  implies that 0  A = 1 observe that   Ac  = 0 implies 0  A = 0 . Thus, P0  fT ;  1g = 1, as claimed. 1.79 Exercise. Do we really need the hypothesis about the absolute continuity of 0 ? Here is an example although somewhat technical and arti cial that shows how things can go wrong without it. Let the state space S be the unit interval 0; 1 . Let B = f2,n : n = 1; 2; : : :g. De ne the distribution  to have probability mass 1 2 on the point 1 and density 1 2 on the rest of the interval, 0; 1. For each state x 2 B , take the next-state distribution P x;  to = ,n 2 B , de ne P 2,n ;  to have mass 2n+1 ,2 on the point 2,n+1 and the be . For x = 2 2n+1 ,1 remaining mass 1=2n+1 , 1 on the point 1. Show that the state 1 is an accessible atom, and that  is a stationary distribution for the chain. But what happens if we start...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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