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Markov

# De ning p i i 1 pi and p i i 1 qi assume that pi

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Unformatted text preview: oupling argument? In our example, we wanted to establish a statement about two distributions: the distributions of random graphs with edge probabilities p1 and p2 . To do this, we showed how to construct" i.e., simulate using uniform random numbers! random objects having the desired distributions in such a way that the desired conclusion became obvious. The trick was to make appropriate use of the same uniform random variables in constructing the two objects. I think this is a general feature of coupling arguments: somewhere in there you will nd the same set of random variables used to construct two di erent objects about which one wishes to make some probabilistic statement. The term coupling" re ects the fact that the two objects are related in this way. 1.50 Exercise. Consider a Markov Chain on the nonnegative integers S = f0; 1; 2; : : : g. De ning P i; i + 1 = pi and P i; i , 1 = qi , assume that pi + qi = 1 for all i 2 S, and also p0 = 1, q0 = 0, and both pi and qi are positive for all i  1. Use what you know about the simple, symmetric random walk to show that the given Markov chain is recurrent. Stochastic Processes J. Chang, March 30, 1999 1.8. PROOF OF THE BASIC LIMIT THEOREM Page 1-25 1.8 Proof of the Basic Limit Theorem The Basic Limit Theorem says that if an irreducible, aperiodic Markov chain has a stationary distribution , then for each initial distribution 0 , as n ! 1 we have ni ! i for all states i. Let me start by pointing something out, just in case the wording of the statement strikes you as a bit strange. Why does the statement read . . . a stationary distribution"? For example, what if the chain has two stationary distributions? The answer is that this is impossible: the assumed conditions imply that a stationary distribution is in fact unique. In fact, once we prove the Basic Limit Theorem, we will know this to be the case. Clearly if the Basic Limit Theorem is true, an irreducible and aperodic Markov chain cannot have two di erent stationary distributions  and , since obviously ni cannot ~ approach both i and  i for all i. ~ An equivalent but conceptually useful reformulation is to de ne a distance between probability distributions, and then to show that as n ! 1, the distance between the distribution n and the distribution  converges to 0. The notion of distance that we will use is called total variation distance." 1.51 Definition. Let  and  be two probability distributions on the set S. Then the total variation distance k , k between  and  is de ned by k , k = sup A , A : A S 1.52 Exercise. Show that k , k may also be expressed in the alternative forms X 1X k , k = sup jA , Aj = 2 ji , ij = 1 , minfi; ig: AS i2S i2S Two probability distributions  and  assign probabilites to all possible events. The total variation distance between  and  is the largest possible discrepancy between the probabilities assigned by  and  to any event. For example, let 7 denote the distribution of the ordering of a deck of cards after 7 shu es, and let  denote the uniform distribution on all 52! permutations of the dec...
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