Grinstead-Snell-Probability · Problem 11.2.24

Markov Chains

Mathematics · absorbing-markov-chains

Problem

Show that wxw_x of Exercise 23 satisfies the following conditions: (a) wx=pwx+1+qwx−1w_x = pw_{x+1} + qw_{x-1} for x=1,2,…,T−1x = 1, 2, \ldots, T-1. (b) w0=0w_0 = 0. (c) wT=1w_T = 1. Show that these conditions determine wxw_x. Show that, if p=q=1/2p = q = 1/2, then wx=xTw_x = \frac{x}{T} satisfies (a), (b), and (c) and hence is the solution. If p≠qp \neq q, show that wx=(q/p)x−1(q/p)T−1w_x = \frac{(q/p)^x - 1}{(q/p)^T - 1} satisfies these conditions and hence gives the probability of the gambler winning.

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