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Unformatted text preview: f (Xm ) ! f (x)⇡ (x) n m=1 x The key idea here is that by breaking the path at the return times to x we get a seqeunce of random variables to which we can apply the law of large numbers. Sketch of proof. Suppose that the chain starts at x. Let T0 = 0 and Tk = min{n > Tk 1 : Xn = x} be the time of the k th return to x. By the strong Markov property, the random variables Yk = Tk X m=Tk f (Xm ) 1 +1 are independent and identically distributed. By the cycle trick in the proof of Theorem 1.20 X EYk = µx (y )f (y ) x Using the law of large numbers for i.i.d. variables TL L X 1X 1X f (Xm ) = Yk ! µx (y )f (y ) L m=1 L x k=1 Taking L = Nn (x) = max{k : Tk n} and ignoring the contribution from the last incomplete cycle (Nn (x), n] Nn (x) n X 1X Nn (x) 1 f (Xm ) ⇡ · Yk n m=1 n Nn (x) k=1 Using Theorem 1.21 and the law of large numbers the above ! X 1X µx (y )f (y ) = ⇡ (y )f (y ) Ex T x y y 44 1.8 CHAPTER 1. MARKOV CHAINS Exit Distributions To motivate developments, we begin with an example. Example 1.39. Two year college. At a local two year...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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