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Markov

# Now we complete the proof of the basic limit theorem

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Unformatted text preview: e de nition of s. This proves the claim. Thus, A contains two consecutive integers, say, c and c +1. Now we will nish the proof by showing that n 2 A for all n  c2 . If c = 0 this is trivially true, so assume that c 0. We have, by closure under addition, c2 = cc 2 A c2 + 1 = c , 1c + c + 1 2 A .. . 2 + c , 1 = c + c , 1c + 1 2 A: c Thus, fc2 ; c2 + 1; : : : ; c2 + c , 1g, a set of c consecutive integers, is a subset of A. Now we can add c to all of these numbers to show that the next set fc2 + c; c2 + c +1; : : : ; c2 +2c , 1g of c integers is also a subset of A. Repeating this argument, clearly all integers c2 or above are in A. Let i 2 S, and retain the assumption that the chain is aperiodic. Then since the set fn : P ni; i 0g is clearly closed under addition, and, by the aperiodicity assumption, has greatest common divisor 1, the previous lemma applies to give that P n i; i 0 for all su ciently large n. From this, for any i; j 2 S, since irreducibility implies that P m i; j  0 for some m, it follows that P n i; j  0 for all su ciently large n. Now we complete the proof of the Basic Limit Theorem by showing that the chain fZn g is irreducible. Let ix ; iy ; jx ; jy 2 S. It is su cient to show, in the bivariate chain fZn g, that n jx jy  is accessible from ix iy . To do this, it is su cient to show that PZ ix iy ; jx jy  0 for some n. However, by the assumed independence of fXn g and fYn g, n PZ ix iy ; jx jy  = P nix; jx P n iy ; jy ; which, by the previous paragraph, is positive for all su ciently large n. Of course, this implies the desired result, and we are done. 1.56 Exercise. A little practice with the coupling idea i Consider a Markov chain fXn g having probability transition matrix 0 1=2 1=4 1=4 1 P = @ 1=4 1=2 1=4 A : 1=4 1=4 1=2 Note that fXn g has stationary distribution  = 1=3; 1=3; 1=3. Using the sort of coupling we did in the proof of the Basic Limit Theorem, show that, no matter what the initial distribution 0 of X0 is, we have for all n. Stochastic Processes 2 11 n kn , k  3 16 J. Chang, March 30, 1999 1.9. A SLLN FOR MARKOV CHAINS Page 1-29 ii Do you think the bound you just derived is a good one? In particular, is 11 16 the smallest we can get? What is the best we could do? iii Can you use a more aggressive" coupling to get a better bound? What do I mean? The coupling we used in the proof of the Basic Limit Theorem was not very aggressive, in that it let the two chains evolve independently until they happened to meet, and only then started to use the same uniform random numbers to generate the paths. No attempt was made to get the chains together as fast as possible. A more aggressive coupling would somehow make use of some random numbers in common to both chains in generating their paths right from the beginning. 1.9 A SLLN for Markov chains The usual Strong Law of Large Numbers for independent and identically distributed iid random variables says that if X1 ; X2 ; : : : are iid with mean , the...
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