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HW2Soln

# HW2Soln - IEOR 161 Introduction to Stochastic Processes...

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IEOR 161 - Introduction to Stochastic Processes Spring 2010 HW2 Solutions ** Note that the numbering is from Ross 9th Edition 3.29 Let q i = 1 - p i , i = 1 , 2. Also, let h stand for hit and m for miss. a) μ 1 = E [ N | h ] p 1 + E [ N | m ] q 1 = p 1 ( E [ N | h, h ] p 2 + E [ N | h, m ] q 2 ) + (1 + μ 2 ) q 1 = 2 p 1 p 2 + (2 + μ 1 ) p 1 q 2 + (1 + μ 2 ) q 1 The preceding equation simplifies to μ 1 (1 - p 1 q 2 ) = 1 + p 1 + μ 2 q 1 Similarly, we have that μ 2 (1 - p 2 q 1 ) = 1 + p 2 + μ 1 q 2 Solving these equations gives the solution: μ 1 = (1 + p 1 )(1 - p 2 q 1 ) + q 1 (1 + p 2 ) (1 - p 1 q 2 )(1 - p 2 q 1 ) - q 1 q 2 , μ 2 = (1 + p 2 )(1 - p 1 q 2 ) + q 2 (1 + p 1 ) (1 - p 2 q 1 )(1 - p 1 q 2 ) - q 2 q 1 b) h 1 = E [ H | h ] p 1 + E [ H | m ] q 1 = p 1 ( E [ H | h, h ] p 2 + E [ H | h, m ] q 2 ) + h 2 q 1 = 2 p 1 p 2 + (1 + h 1 ) p 1 q 2 + h 2 q 1 Similarly, we have that h 2 = 2 p 1 p 2 + (1 + h 2 ) p 2 q 1 + h 1 q 2 and we solve these equations to find h 1 and h 2 . 3.31 Let L i denote the length of run i . Conditioning on X , the initial value gives, a) E [ L 1 ] = E [ L 1 | X = 1] p + E [ L 1 | X = 0](1 - p ) = (1 + 1 1 - p - 1) p + (1 + 1 p - 1)(1 - p ) = p 1 - p + 1 - p p 1

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b) E [ L 2 ] = E [ L 2 | X = 1] p + E [ L 2 | X = 0](1 - p ) = (1 + 1 p - 1) p + (1 + 1 1 - p - 1)(1 - p ) = 2 3.37 a) E [ X ] = (2 . 6 + 3 + 3 . 4) / 3 = 3 b) E
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HW2Soln - IEOR 161 Introduction to Stochastic Processes...

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