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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 125 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 ni ! 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 ni 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 jA , Aj = 2 ji , ij = 1 , minfi; ig:
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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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 Multiplication, Markov Chains

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