HW8Soln - IEOR 161 - Introduction to Stochastic Processes...

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Unformatted text preview: IEOR 161 - Introduction to Stochastic Processes Spring 2010 HW8 Solutions ** Note that the numbering is from Ross 9th Edition 4.18 Let the state at time n be the n th coin flipped. Then the transition probabilities are P 1 , 1 = 0 . 6 ,P 1 , 2 = 0 . 4 P 2 , 1 = 0 . 5 ,P 2 , 2 = 0 . 5 (a) The stationary probabilities satisfy 1 = 0 . 6 1 + 0 . 5 2 1 + 2 = 1 which can be solved to get 1 = 5 / 9, 2 = 4 / 9. The proportion of flips that use coin 1 is 5/9. (b) P 4 1 , 2 = 0 . 4444. 4.20 M X i =0 i P ij = M X i =0 1 M + 1 P ij = 1 M + 1 M X i =0 P ij = 1 M + 1 = j M X i =0 i = M X i =0 1 M + 1 = ( M + 1) 1 M + 1 = 1 4.23 We need 4 states(1 for success, 0 for failure), and the transition matrix is as follows: 11 01 10 00 11 0.8 0.2 01 0.5 0.5 10 0.5 0.5 00 0.5 0.5 Solve the system of linear equations, j = 4 X i =1 i p ij , j = 1 , 2 , 3 , 4 1 4 X i =1 i = 1 , we get the stationary distribution = { 5 11 , 2 11 , 2 11 , 2 11 } , and in the long run the propor-...
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HW8Soln - IEOR 161 - Introduction to Stochastic Processes...

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