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Unformatted text preview: 1.76 Proposition. Suppose a Markov chain has an accessible atom distribution . Then P fT 1g = 1. and a stationary Proof: Let B = fx : Px fT = 1g 0g; these are the states from which it is possible to
go forever without hitting . We want to show that B = 0. Since is recurrent, if the
chain starts from state , then with probability 1 it will return to in nitely many times.
Therefore, the set B cannot be accessible from , for if it were, there would be positive
probability, starting from , of eventually entering the set B and then never returning to
. Thus, by the previous proposition, B = 0.
1.77 Definition. Let and be two probability measures on a set S. We say that is absolutely continuous with respect to if A = 0 for all A S such that A = 0,
that is, each set having probability 0 under also has probability 0 under . Suppose a chain fXt g with transition kernel P and an aperiodic,
accessible atom has a stationary distribution . Let t denote the distribution of Xt and
start the chain in any distribution 0 that is absolutely continuous with respect to . Then
kt , k ! 0 as t ! 1.
1.78 Theorem. Proof: We use the coupling technique from before; much of the reasoning remains the
same, so I'll just give a sketch. Again, we run two independent copies of the chain, fXt g
and fXt g, starting in the initial distributions 0 and , respectively. We want to show that
Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 140 with probability 1 the two chains eventually couple; in fact we claim that they eventually
visit the state at the same time. By using the aperiodicity assumption together with
the numbertheoretic lemma from before, we see that the bivariate chain fXt ; Xt : t =
0; 1; : : :g has an accessible atom ; . The bivariate chain has a stationary distribution:
the obvious product distribution . So by Proposition 1.76, if the bivariate chain were
started out in its stationary distribution , it would eventually hit its atom ;
with probability 1. That is, letting A denote the set of pairs of states x; y such that
Px;y fT ;
1g = 1, we have A = 1. From this, the absolute continuity of
0 with respect to implies that 0 A = 1 observe that Ac = 0 implies
0 A = 0 . Thus, P0 fT ; 1g = 1, as claimed.
1.79 Exercise. Do we really need the hypothesis about the absolute continuity of 0 ? Here is an example although somewhat technical and arti cial that shows how things can go wrong
without it. Let the state space S be the unit interval 0; 1 . Let B = f2,n : n = 1; 2; : : :g.
De ne the distribution to have probability mass 1 2 on the point 1 and density 1 2 on the
rest of the interval, 0; 1. For each state x 2 B , take the nextstate distribution P x; to
=
,n 2 B , de ne P 2,n ; to have mass 2n+1 ,2 on the point 2,n+1 and the
be . For x = 2
2n+1 ,1
remaining mass 1=2n+1 , 1 on the point 1. Show that the state 1 is an accessible atom, and
that is a stationary distribution for the chain. But what happens if we start...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 Multiplication, Markov Chains

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