Unformatted text preview: +1 is chosen from
~ t 6= and Xt 2 R then Xt+1 is chosen from the distribution
~=
the distribution QXt ; . If X
P Xt ; .
In other words, again we have imbedded our given Markov chain in the structure shown
in Figure 1.81, with the transition kernels U and V given by
For x 2 R: U x; f g = ; U x; fxg = 1 ,
For x 2 S , R: U x; fxg = 1
V ; A = A
For x 2 R: V x; A = Qx; A
For x 2 S , R: V x; A = P x; A:
~
~
The sequence X0 ; X0 ; X1 ; X1 ; : : : is a timeinhomogeneous Markov chain; the transition
~
kernel U used in going from Xt to Xt is di erent from the kernel V used in going from
~
~ t to Xt+1 . Note that Xt 2 S and Xt 2 S for all t. The sequence X0 ; X1 ; : : : is a
~
X
timehomogeneous Markov chain on S with transition kernel UV , de ned by
UV x; B = Z U x; dyV y; B : We claim that UV = P . If x 2 S , R then U x; is point mass on x, so that UV x; B =
V x; B = P x; B . If x 2 R then U x; puts probability on the point and probability
1 , on the point x, so that
UV x; B = V ; B + 1 , V x; B
= B + 1 , Qx; B = P x; B :
~~
The sequence X0 ; X1 ; : : : is a timehomogeneous Markov chain, with transition kernel
~.
V U =: P
~
1.84 Exercise. Write down the transition kernel P in terms of the information given in the
problem. Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 146 ~
If Xt has distribution t on S, then Xt has distribution t = t U on e.
~
S
Finally, here is our Basic Limit Theorem for Harris chains. As usual, the statement
involves an aperiodicity condition. Letting G = ft 1 : P fXt,1 2 Rg 0g, we say the
chain is aperiodic if gcdG = 1. For example, as a simple su cient condition, if R 0,
then the set G contains 1, so that the chain is aperiodic.
1.85 theorem. Let fXt g be an aperiodic Harris chain having a stationary distribution . Let t denote the distribution of Xt and let the initial distribution 0 be absolutely
continuous with respect to . Then kt , k ! 0 as t ! 1. Proof: We are given the Harris chain fXt g with transition kernel P . Suppose we are also given a set R, probability measure , and number 2 0; 1 as in the de nition of a Harris
chain. As discussed above, these determine transition kernels U and V with P = UV and
~
~
~
P = V U , and we will study the chain X0 ; X0 ; X1 ; X1 ; : : :. We are assuming that fXt g
~
has a stationary distribution , and we now know that fXt g has corresponding stationary
distribution = U . By the de nition of the Harris chain fXt g, the state is an accessible
~
~
atom for fXt g, and the aperiodicity assumption implies that is aperiodic. ***WHY?
EXPLAIN THIS. De ning 0 = 0 U , we see that 0 is absolutely continuous with respect
~
~
to . Therefore, by Theorem 1.78 we have kt , k ! 0, where t denotes the distribution
~
~~
~
~
of Xt . But
V = U V = UV = P = :
~
Thus, since
kt+1 , k = kt V , V k kt , k;
~
~
~~
we have kt+1 , k ! 0 as t ! 1.
*** NOTE: Argue somewhere that kP , P k k , k. Can use c...
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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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