Recall that we want to show that pft 1g 1 stated in

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Unformatted text preview: er, on the event fT  ng, we have Yn = Xn , so that the two events fXn 2 A; T  ng and fYn 2 A; T  ng are the same, and hence they have the same probability. Therefore, the rst and third terms in the last expression cancel, yielding n A , A = PfXn 2 A; T ng , PfYn 2 A; T Since the last di erence is obviously bounded by PfT ng: ng, we are done. Note the signi cance of the coupling inequality: it reduces the problem of showing that kn , k ! 0 to that of showing that PfT ng ! 0, or equivalently, that PfT 1g = 1. To do this, we consider the bivariate chain" fZn = Xn ; Yn  : n  0g. A bit of thought con rms that Z0 ; Z1 ; : : : is a Markov chain on the state space S  S. Since the X and Y chains are independent, the probability transition matrix PZ of the Z chain can be written as PZ ixiy ; jx jy  = P ix ; jx P iy ; jy : It is easy to check that the Z chain has stationary distribution Z ix iy  = ix iy : Watch closely now; we're about to make an important reduction of the problem. Recall that we want to show that PfT 1g = 1. Stated in terms of the Z chain, we want to show that with probability one, the Z chain hits the diagonal" fj; j  : j 2 Sg in S  S in nite time. To do this, it is su cient to show that the Z chain is irreducible and recurrent why? . However, since we know that the Z chain has a stationary distribution, by Corollary 1.47, to prove the Basic Limit Theorem, it su ces to show that the Z chain is irreducible. This is, strangelyy , the hard part. This is where the aperiodicity assumption comes in. For example, consider a Markov chain fXn g having the type A frog" transition matrix 0 1 P = 1 0 started out in the condition X0 = 0. Then the stationary chain fYng starts out in the uniform distribution: probability 1 2 on each state 0,1. The bivariate chain fXn ; Yng is not irreducible: for example, from the state 0; 0, we clearly cannot reach the state 0; 1. And this ruins everything. For example, if Y0 = 1, which happens with probability 1 2, the X and Y chains can never meet, so that T = 1. Thus, PfT 1g 1. A little number-theoretic result will help us establish irreducibility of the Z chain. 1.55 Lemma. Suppose A is a set of positive integers that is closed under addition and has greatest common divisor gcd one. Then there exists an integer N such that n 2 A for all n  N . Proof: First we claim that A contains at least one pair of consecutive integers. To see this, suppose to the contrary that the minimal spacing" between successive elements of A is s 1. That is, any two distinct elements of A di er by at least s, and there exists an integer n1 such that both n1 2 A and n1 + s 2 A. Let m 2 A be such that s does not y Or maybe not so strangely, in view of Example 1.32. Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-28 divide m; we know that such an m exists because gcdA = 1. Write m = qs + r, where 0 r s. Now observe that, by the closure under addition assumption, the two numbers a1 = q +1n1 + s and a2 = q +1n1 + m are both in A. However, a1 , a2 = s , r 2 0; s, which contradicts th...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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