Stochastic

# Since p is irreducible and recurrent vxx 1 with

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = ⇢xy /(1 1= X x ⇡ (x) ⇢yy ), so ⇢xy 1 1 ⇢yy 1 ⇢yy the second inequality following from the facts that ⇢xy 1 and ⇡ is a probability measure. This shows that ⇢yy = 1, i.e., y is recurrent. With Lemma 1.26 in hand we are ready to tackle the proof of: Theorem 1.19. Convergence theorem. Suppose p is irreducible, aperiodic, and has stationary distribution ⇡ . Then as n ! 1, pn (x, y ) ! ⇡ (y ). 40 CHAPTER 1. MARKOV CHAINS Proof. Let S be the state space for p. Deﬁne a transition probability p on S ⇥ S ¯ by p((x1 , y1 ), (x2 , y2 )) = p(x1 , x2 )p(y1 , y2 ) ¯ In words, each coordinate moves independently. Step 1. We will ﬁrst show that if p is aperiodic and irreducible then p is ¯ irreducible. Since p is irreducible, there are K, L, so that pK (x1 , x2 ) > 0 and pL (y1 , y2 ) > 0. Since x2 and y2 have period 1, it follows from Lemma 1.16 that if M is large, then pL+M (x2 , x2 ) > 0 and pK +M (y2 , y2 ) > 0, so pK +L+M ((x1 , y1 ), (x2 , y2 )) > 0 ¯ Step 2. Since the two coordinate...
View Full Document

## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

Ask a homework question - tutors are online