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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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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