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

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

This is the end of the preview. Sign up to access the rest of the document.

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 &lt; q then q/p &gt; 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 &gt; 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...
View Full Document

## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online