hw9sln - and 2 if blue The transition probability matrix is...

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IEOR 161 Operations Research II University of California, Berkeley Spring 2008 Homework 9 Solution Chapter 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 π 1 = 0 . 6 π 1 + 0 . 5 π 2 π 1 + π 2 = 1 which can be solved to get π 1 = 5 / 9, π 2 = 4 / 9. The proportion of flips that use coin 1 is 5/9. (b) P 4 1 , 2 = 0 . 4444. 19. The limiting probabilities are obtained from π 0 = 0 . 7 π 0 + 0 . 5 π 1 π 1 = 0 . 4 π 2 + 0 . 2 π 3 π 2 = 0 . 3 π 0 + 0 . 5 π 1 π 0 + π 1 + π 2 + π 3 = 1 which can be solved to get π 0 = 0 . 25, π 1 = 0 . 15, π 2 = 0 . 15, π 3 = 0 . 45. The proportion of days that it rains is π 0 + π 1 = 0 . 4 20. M i =0 π i P ij = M i =0 1 M + 1 P ij = 1 M + 1 M i =0 P ij = 1 M + 1 = π j M i =0 π i = M i =0 1 M + 1 = ( M + 1) 1 M + 1 = 1 22. Let X n be the remainder when Y n is divided by 13. Therefore X n is a Markov chain with states 0 , 1 , 2 , . . . , 12. It is easy to verify that i P ij = 1 6 + 1 6 + 1 6 + 1 6 + 1 6 + 1 6 = 1 Thus it is doubly stochastic, and by Exercise 20, lim n →∞ P { Y n is a multiple of 13 } = lim n →∞ P { X n = 0 } = π 0 = 1 13 1
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24. Let the state be the color of the last ball selected. Call it 0 if it was red, 1 if white,
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Unformatted text preview: and 2 if blue. The transition probability matrix is P = 1 / 5 4 / 5 2 / 7 3 / 7 2 / 7 3 / 9 4 / 9 2 / 9 The limiting probabilities are obtained from π = 1 / 5 π + 2 / 7 π 1 + 1 / 3 π 2 π 1 = 3 / 7 π 1 + 4 / 9 π 2 π + π 1 + π 2 = 1 which can be solved to get π = 25 / 89, π 1 = 28 / 89, π 2 = 36 / 89. 26. Let the state be the ordering of the deck of n cards, so there are n ! states. The transition probabilities are P ( i 1 , i 2 , · · · , i n ) , ( i j , i 1 , i 2 , · · · , i j-1 i j +1 , · · · , i n ) = 1 n This Markov Chain is doubly stochastic, so in the limit all n ! states are equally likely. 2...
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