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
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.
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