The rst equality is lemma 112 from lemma 111 we see

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Unformatted text preview: (Xn = y ) n=1 With the two lemmas established we can now state our next main result. Theorem 1.13. y is recurrent if and only if 1 X n=1 pn (y, y ) = Ey N (y ) = 1 Proof. The first equality is Lemma 1.12. From Lemma 1.11 we see that Ey N (y ) = 1 if and only if ⇢yy = 1, which is the definition of recurrence. 17 1.4. STATIONARY DISTRIBUTIONS . With this established we can easily complete the proofs of our two lemmas Proof of Lemma 1.9 . Suppose x is recurrent and ⇢xy > 0. By Lemma 1.6 we must have ⇢yx > 0. Pick j and ` so that pj (y, x) > 0 and p` (x, y ) > 0. pj +k+` (y, y ) is probability of going from y to y in j + k + ` steps while the product pj (y, x)pk (x, x)p` (x, y ) is the probability of doing this and being at x at times j and j + k . Thus we must have ! 1 1 X X j +k+` j k p (y, y ) p (y, x) p (x, x) p` (x, y ) k=0 k=0 P If x is recurrent then k p (x, x) = 1, so implies that y is recurrent. k P m pm (y, y ) = 1 and Theorem 1.13 Proof of Lemma 1.10. If all the states in C are transient then Lemma 1.11 implies that Ex N (y ) < 1 for all x and y in C . Since C...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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