In symbols pi i 1 p pi i 1 1 p p0 0 1 when

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Unformatted text preview: hain in which p(i, i + 1)p and p(i, i 1) = q , where p 6= q . Let ⌧ = min{n : Xn 62 (0, N )}. We claim that Ex ⌧ = x q N p q 1 p1 (q/p)x (q/p)N · (1.27) This time the derivation is somewhat tedious, so we will just verify the guess. We want to show that g (x) = 1 + pg (x + 1) + qg (x 1). Plugging the formula into the right-hand side: x+1 x1 N 1 (q/p)x+1 1 (q/p)x 1 =1 + p +q p· +q N qp qp qp 1 (q/p) 1 (q/p)N pq N p + q (q/p)x (q + p) x + =1 + qpqpqp 1 (q/p)N which = g (x) since p + q = 1. To see what this says note that if p < q then q/p > 1 so 1 N !0 (q/p)N and g (x) = x q p (1.28) To see this is reasonable note that our expected value on one play is p q , so we lose an average of q p per play, and it should take an average of x/(q p) to lose x dollars. When p > q , (q/p)N ! 0, so doing some algebra g (x) ⇡ N p x [1 q (q/p)x ] + x p q (q/p)x Using (1.23) we see that the probability of not hitting 0 is 1 (q/p)x . In this case, since our expected winnings per play is p q , it should take about (N x)/(p...
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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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