# Suppose rst that we start at y the times between

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Unformatted text preview: z ) = y 1 XX pn (x, y )p(y, z ) ¯ n=0 y Case 1. Consider the generic case ﬁrst: z 6= x. X X pn (x, y )p(y, z ) = ¯ Px (Xn = y, Tx &gt; n, Xn+1 = z ) y y = Px (Tx &gt; n + 1, Xn+1 = z ) = pn+1 (x, z ) ¯ Here the second equality holds since the chain must be somewhere at time n, and the third is just the deﬁnition of pn+1 . Summing from n = 0 to 1, we have ¯ 1 XX pn (x, y )p(y, z ) = ¯ n=0 y 1 X pn+1 (x, z ) = µx (z ) ¯ n=0 since p0 (x, z ) = 0. ¯ Case 2. Now suppose that z = x. Reasoning as above we have X X pn (x, y )p(y, x) = ¯ Px (Xn = y, Tx &gt; n, Xn+1 = x) = Px (Tx = n + 1) y y 42 CHAPTER 1. MARKOV CHAINS Summing from n = 0 to 1 we have 1 XX pn (x, y )p(y, x) = ¯ n=0 y 1 X Px (Tx = n + 1) = 1 = µx (x) n=0 since Px (Tx = 0) = 0. To check µx (y ) &lt; 1, we note that µx (x) = 1 and X 1 = µx (x) = µx (z )pn (z, x) µx (y )pn (y, x) z so if we pick n with p (y, x) &gt; 0 then we conclude µx (y ) &lt; 1. To prove that µx (y ) &gt; 0 we note that this is trivial for y = x the point used to deﬁne the measure....
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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