Stochastic

Case 2 now suppose that z x reasoning as above we

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Unformatted text preview: t X0 = x and let Y0 have the stationary distribution ⇡ , then Yn has distribution ⇡ , and Using (1.14) it follows that X |pn (x, y ) ⇡ (y )| 2P (T > n) ! 0 y proving the convergence theorem. 41 1.7. PROOFS OF THE MAIN THEOREMS* Our next topic is the existence of stationary measures Theorem 1.20. Suppose p is irreducible and recurrent. Let x 2 S and let Tx = inf {n 1 : Xn = x}. µx (y ) = 1 X Px (Xn = y, Tx > n) n=0 defines a stationary measure with 0 < µx (y ) < 1 for all y . Why is this true? This is called the “cycle trick.” µx (y ) is the expected number of visits to y in {0, . . . , Tx 1}. Multiplying by p moves us forward one unit in time so µx p(y ) is the expected number of visits to y in {1, . . . , Tx }. Since X (Tx ) = X0 = x it follows that µx = µx p. • @ • A • y A A A• @• A A • A A• @ @ Tx x• 0 Figure 1.2: Picture of the cycle trick. Proof. To formalize this intuition, let pn (x, y ) = Px (Xn = y, Tx > n) and ¯ interchange sums to get X µx (y )p(y,...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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