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

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Unformatted text preview: IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 10 Solution Chapter 7 Question 69 (a) R e- x e- x dx = 1 2 (b) R e- x x 3 3! e- x dx = 1 96 R e- y y 3 dy = (4) 96 = 1 16 (c) R e- x e- x x 3 3! e- x dx R e- x e- x dx = 2 3 4 = 2 81 Question 75 X is Poisson with mean = 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 = E[ X ]. Then for any a E[( X- a ) 2 ] = E[( X- + - a ) 2 ] = E[( X- ) 2 ] + ( - a ) 2 + 2E[( x- )( - a )] = E[( X- ) 2 ] + ( -...
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This note was uploaded on 03/03/2009 for the course IEOR 172 taught by Professor Righter during the Fall '07 term at University of California, Berkeley.

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

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