Stochastic

The rst equality is lemma 112 from lemma 111 we see

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

Ask a homework question - tutors are online