# When p q qpn 0 so doing some algebra g x n p x 1 q qpx

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Unformatted text preview: e two ways to prove this. Verify the guess. Let g (x) = x(N x). Clearly, g (0) = g (N ) = 0. If 0 < x < N then by considering what happens on the ﬁrst step we have 1 1 g (x) = 1 + g (x + 1) + g (x 2 2 If g (x) = x(N 1) x) then the right-hand side is 1 1 = 1 + (x + 1)(N x 1) + (x 1)(N x + 1) 2 2 1 1 = 1 + [x(N x) x + N x 1] + [x(N x) + x 2 2 = 1 + x(N x) 1 = x(N x) (N x + 1)] Derive the answer. (1.25) implies that g (x) = 1 + (1/2)g (x + 1) + (1/2)g (x 1) Rearranging gives g (x + 1) Setting g (1) general that g (x) = g (0) = c we have g (2) g (k ) g (k 2 + g (x) g (1) = c 1) = c 2(k g (x 2, g (3) 1) 1) g (2) = c 4 and in 53 1.10. INFINITE STATE SPACES* Using g (0) = 0 and summing we have 0 = g (N ) = N X 2(k c 1) = cN 2· k=1 N (N 2 1) Pm since, as one can easily check by induction, j =1 j = m(m + 1)/2. Solving gives c = (N 1). Summing again, we see that g (x) = x X (N 1) 2(k 1) = x(N 1) x(x + 1) = x(N x) k=1 Example 1.50. Duration of nonfair games. Consider the gambler’s ruin c...
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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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