Since p is irreducible and recurrent vxx 1 with

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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...
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