# hw10_sol - IEOR 172 Probability and Risk Analysis for...

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IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 10 Solution Chapter 7 Question 69 (a) R 0 e - x e - x dx = 1 2 (b) R 0 e - x x 3 3! e - x dx = 1 96 R 0 e - y y 3 dy = Γ(4) 96 = 1 16 (c) R 0 e - x e - x x 3 3! e - x dx R 0 e - x e - x dx = 2 3 4 = 2 81 Question 75 X is Poisson with mean λ = 2 and Y is Binomial with parameters 10,3/4. Hence (a) P { X + Y = 2 } = P { X = 0 } P { Y = 2 } + P { X = 1 } P { Y = 1 } + P { X = 2 } P { Y = 0 } = e - 2 10 2 ! 3 4 2 1 4 8 + 2 e - 2 10 1 ! 3 4 1 4 9 + 2 e - 2 1 4 10 (b) P { XY = 0 } = P { X = 0 } + P { Y = 0 } - P { X = Y = 0 } = e - 2 + 1 4 10 - e - 2 1 4 10 (c) E[ XY ] = E[ X ]E[ Y ] = 2 · 10 · 3 4 = 15 Question 76 M ( t 1 , t 2 ) = E h e t 1 X + t 2 Y i = X x X y e t 1 x + t 2 y p ( x, y ) = 1 36 6 X x =1 x +6 X y = x +1 e t 1 x + t 2 y . Theoretical Exercise Question 1 Let μ = E[ X ]. Then for any a E[( X - a ) 2 ] = E[( X - μ + μ - a ) 2 ] = E[( X - μ ) 2 ] + ( μ - a ) 2 + 2E[( x - μ )( μ - a )] = E[( X - μ ) 2 ] + ( μ - a ) 2 + 2( μ - a )E[( X - μ )] = E[( X - μ ) 2 ] + ( μ - a ) 2 . 1

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Question 44 Write X n = X n - 1 i =1 Z i where Z i is the number of offspring of the i th individual of the ( n - 1)st generation. Hence (a) E[ X n ] = E[E[ X n | X n - 1 ]] = E[ μX n - 1 ] = μ E[ X n - 1 ] (b) E[ X n ] = μ E[ X n - 1 ] = μ 2 E[ X n - 2 ] = . . . = μ n E[ X 0 ] = μ n .
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• Fall '07
• righter
• Probability, Probability theory, Unif, 76 m, Xn Write Xn

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