Hw1_sol

# Hw1_sol - h(1 = 0 2 h(1 0 7 h(2 0 1 h(2 = 0 05 h(1 0 1 h(2...

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Homework 1: Solutions Review 1.28 (a) Al’s winning probability p has p = 1 / 5 + (4 / 5)(1 - b ) p or p = 1 / { 5 - 4(1 - b ) } . When b = 1 / 3 this is 1 / (7 / 3) = 3 / 7. (b) if we want the answer to be 1 / 2 we want 4(1 - b ) = 3 or b = 1 / 4. 2.3 In order for the maximum to be m we must choose that number and 2 of the m - 1 smaller numbers, so P ( X = m ) = ( m - 1 2 ) / ( 15 3 ) for 3 m 15. 3.12 Mean 18 · (1 / 3)+12 · (1 / 2) = 12, variance 18 · (1 / 3)(2 / 3)+12 · (1 / 2)(1 / 2) = 7. Chapter 1 9.2 0 1 2 3 4 5 0 0 1 0 0 0 0 1 1 / 25 8 / 25 16 / 25 0 0 0 2 0 4 / 25 12 / 25 9 / 25 0 0 3 0 0 9 / 25 12 / 25 4 / 25 0 4 0 0 0 16 / 25 8 / 25 1 / 25 5 0 0 0 0 1 0 9.5 ( a ) A B C A 0 1 / 2 1 / 2 B 3 / 4 0 1 / 4 C 3 / 4 1 / 4 0 (b) At time 2, A has probability 3/4, while B and C have probability 1/8 each. The probability of B at time 3 is then (3 / 4)(1 / 2) + (1 / 8)(0) + (1 / 8)(1 / 4) = 13 / 32. 9.6 (a) The transition probability is 1 2 3 4 1 0 . 2 0 . 7 0 . 1 0 2 0 . 05 0 . 1 0 . 05 0 . 8 3 0 0 1 0 4 0 0 0 1 1

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(b) Let h ( x ) = P x ( V 3 < V 4 ). Clearly h (3) = 1 and h (4) = 0. To compute the values for the other two states we note that
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Unformatted text preview: h (1) = 0 . 2 h (1) + 0 . 7 h (2) + 0 . 1 h (2) = 0 . 05 h (1) + 0 . 1 h (2) + 0 . 05 Solving gives h (1) = . 125 /. 685 and h (2) = . 045 /. 685. 9.7 (a) 1 → 2 but 2 6→ 1 so 1 is transient. 3 → 2 but 2 6→ 3 so 3 is transient. 5 → 2 but 2 6→ 5 so 5 is transient. { 2 , 4 } is a ﬁnite irreducible closed set so all these states are recurrent. (b) 3 → 6 but 6 6→ 3 so 3 is transient. 2 → 1 but 1 6→ 2 so 1 is transient. { 1 , 4 , 5 , 6 } is a ﬁnite irreducible closed set so all these states are recurrent 2...
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## This note was uploaded on 09/29/2011 for the course MATH 4740 taught by Professor Durrett during the Spring '10 term at Cornell.

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Hw1_sol - h(1 = 0 2 h(1 0 7 h(2 0 1 h(2 = 0 05 h(1 0 1 h(2...

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