Stochastic

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Unformatted text preview: nswer will turn out to be 15/25, Bob’s fraction of the total supply of pennies. To explain this, let Xn be the number of pennies Bob has after n plays. Xn is a fair game, i.e., x = Ex X1 , or in words the expected number of pennies Bob has is constant in time. Let Vy = min{n 0 : Xn = y } be the time of the first visit to y . Taking a leap of faith the expected number he has at the end of the game should be the same as at the beginning so x = N Px (VN < V0 ) + 0Px (V0 < Vn ) and solving gives Px (VN < V0 ) = x/N for 0 x N (1.17) To prove this note that by considering what happens on the first step h(x) = 1 1 h(x + 1) + h(x 2 2 1) Multiplying by 2 and rearranging h(x + 1) h(x) = h(x) h(x 1) or in words, h has constant slope. Since h(0) = 0 and h(N ) = 1 the slope must be 1/N and we must have h(x) = x/N . The reasoning in the last example can be used to study Example 1.9. 47 1.8. EXIT DISTRIBUTIONS Example 1.42. Wright–Fisher model with no mutation. The state space is S = {0, 1, . . . N } and the transition probability is ✓ ◆N y N ⇣ x ⌘y N x p(x, y ) = yN N The right-hand sid...
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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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