# Stochastic

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Unformatted text preview: chain in which the state is the di↵erence of the scores. The state space is 2, 1, 0, 1, 2 with 2 (win for server) and 2 (win for opponent). The transition probability is 2 1 0 -1 -2 2 1 .6 0 0 0 1 0 0 .6 0 0 0 0 .4 0 .6 0 -1 0 0 .4 0 0 -2 0 0 0 .4 1 If we let h(x) be the probability of the server winning when the score is x then X h(x) = p(x, y )h(y ) y with h(2) = 1 and h( 2) = 0. This gives us three equations in three unknowns h(1) = .6 + .4h(0) h(0) = .6h(1) + .4h( 1) h( 1) = .6h(0) 45 1.8. EXIT DISTRIBUTIONS Using the ﬁrst and third equations in the second we have h(0) = .6(.6 + .4h(0)) + .4(.6h(0)) = .36 + .48h(0) so we have h(0) = 0.36/0.52 = 0.6923. The last computation uses special properties of this example. To introduce a general approach, we rearrange the equations to get h(1) .4h(0) + 0h( 1) = .6 .6h(1) + h(0) 0h(1) .4h( 1) = 0 .6h(0) + h( 1) = 0 which can be written in matrix form as 0 10 1 01 1 .4 0 h(1) .6 @ .6 1 .4A @ h(0) A = @ 0 A 0 .6 1 h( 1) 0 Let C = {1, 0, 1} be the nonabsor...
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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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