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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online