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HW3Soln

# HW3Soln - IEOR 161 Introduction to Stochastic Processes...

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IEOR 161 - Introduction to Stochastic Processes Spring 2010 HW3 Solutions ** Note that the numbering is from Ross 9th Edition 3.42 Let X be the number of people who arrives before you. Because you are equally likely to be the first, or second, ..., or eleventh arrival, we have P ( X = i ) = 1 11 , i = 0 , ..., 10 Therefore, E [ X ] = 1 11 10 X i =1 i = 5 and, E [ X 2 ] = 1 11 10 X i =1 i 2 = 35 giving that, V ( X ) = 35 - 25 = 10 3.45 We are given that E [ X n | X n - 1 ] = 0 , V ( X n | X n - 1 ) = βX 2 n - 1 , X 0 = x 0 a) It’s easy to see that, E [ X n ] = E [ E [ X n | X n - 1 ]] = 0 b) From a) we have that V ( X n ) = E [ X 2 n ], where E [ X 2 n ] = E [ E [ X 2 n | X n - 1]] = E [ V ( X n | X n - 1 )] = E [ βX 2 n - 1 ] = β E [ X 2 n - 1 ] = β n X 2 0 = β n x 2 0 3.50 Let’s start with the general case in b), b) P ( N = n ) = 1 3 10 n ( . 3) n ( . 7) 10 - n + 10 n ( . 5) n ( . 5) 10 - n + 10 n ( . 7) n ( . 3) 10 - n 1

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a) P ( N = 0) = 0 . 00974. c) N is clearly not binomial. d) A fair game has an expected winnings of 0. For this game to be a fair game, the expected number of heads needs to be 5 (win 5, lose 5). Now, E [ N ] = 3 1 3 + 5 1 3 + 7 1 3 = 5 therefore this is a fair game.
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HW3Soln - IEOR 161 Introduction to Stochastic Processes...

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