If not pick a site w 2 s t r1 and repeat the

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Unformatted text preview: nd we have x ! z by definition, so Lemma 1.4 implies y ! z . Cx is closed and irreducible so all states in Cx are recurrent. Let R1 = Cx . If S T R1 = ;, we are done. If not, pick a site w 2 S T R1 and repeat the procedure. ******* The rest of this section is devoted to the proof of Theorem 1.7. To do this, it is enough to prove the following two results. Lemma 1.9. If x is recurrent and x ! y , then y is recurrent. Lemma 1.10. In a finite closed set there has to be at least one recurrent state. To prove these results we need to introduce a little more theory. Recall the time of the k th visit to y defined by k k Ty = min{n > Ty 1 : Xn = y } and ⇢xy = Px (Ty < 1) the probability we ever visit y at some time n 1 when we start from x. Using the strong Markov property as in the proof of (1.4) gives k Px (Ty < 1) = ⇢xy ⇢k 1 . yy (1.5) Let N (y ) be the number of visits to y at times n compute EN (y ). 1. Using (1.5) we can 16 CHAPTER 1. MARKOV CHAINS Lemma 1.11. Ex N (y )...
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