# The markov property implies that the two parts of our

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Unformatted text preview: lies that the two parts of our journey are independent. time 0 • m m+n • • aa aa ` ` • `` a • aa i • ` `` ``a `a aa``` `• ` `• aj • aa a• a • • Proof of (1.2). We do this by combining the solutions of Q1 and Q2. Breaking things down according to the state at time m, X P (Xm+n = j, Xm = k |X0 = i) P (Xm+n = j |X0 = i) = k Using the deﬁnition of conditional probability as in the solution of Q1, P (Xm+n = j, Xm = k, X0 = i) P (X0 = i) P (Xm+n = j, Xm = k, X0 = i) P (Xm = k, X0 = i) = · P (Xm = k, X0 = i) P (X0 = i) = P (Xm+n = j |Xm = k, X0 = i) · P (Xm = k |X0 = i) P (Xm+n = j, Xm = k |X0 = i) = 10 CHAPTER 1. MARKOV CHAINS By the Markov property (1.1) the last expression is = P (Xm+n = j |Xm = k ) · P (Xm = k |X0 = i) = pm (i, k )pn (k, j ) and we have proved (1.2). Having established (1.2), we now return to computations. Example 1.11. Gambler’s ruin. Suppose for simplicity that N = 4 in Example 1.1, so that the transition probability is 0 1.0 0.6 0 0 0 0 1...
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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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