sfi - Stat 5101 (Geyer) 2002 Final Exam Problem 1...

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Unformatted text preview: Stat 5101 (Geyer) 2002 Final Exam Problem 1 Calculating E ( X | Y ) is like part (a) of additional problem 7 on the homework (Problem 4.7.15 in DeGroot and Schervish). We didnt do a problem like this about var( X | Y ), but it is just another conditional expectation, defined by formula (4.7.5) in DeGroot and Schervish. E ( X k | y ) = Z x k g 1 ( x | y ) dx = Z 1 x k ( x + y ) dx 1 2 + y = 1 1 2 + y Z 1 ( x k +1 + x k y ) dx = 1 1 2 + y x k +2 k + 2 + x k +1 y k + 1 1 = 1 1 2 + y 1 k + 2 + y k + 1 In particular, E ( X | y ) = 1 3 + y 2 1 2 + y E ( X 2 | y ) = 1 4 + y 3 1 2 + y and var( X | y ) = E ( X 2 | y )- E ( X | y ) 2 = 1 4 + y 3 1 2 + y- 1 3 + y 2 1 2 + y 2 = 1 + 6 y + 6 y 2 18(1 + 2 y ) 2 Problem 2 This is like additional problem 6 on the homework. 1 First look up the moments for the brand name distributions E ( X ) = 1- p p var( X ) = 1- p p 2 E ( Y | X ) = Xp var( Y | X ) = Xp (1- p ) Then E ( Y ) = E { E ( Y | X ) } = E { Xp } = pE ( X ) = 1- p and var( Y ) = var { E ( Y | X ) } + E { var( Y | X ) } = var { Xp } + E { Xp (1- p ) } = p 2 var( X ) + p (1- p ) E ( X ) = (1- p ) + (1- p ) 2 Problem 3 This is like Additional Problem 8 on the homework, also 5.4.6 in DeGroot and Schervish. (a) Poisson( t ) with = 1 . 5 and t = 10, that is, Poisson(15). The parameter of the Poisson is the mean, so E ( X ) = 15. For the Poisson var( X ) = , so sd( X ) =...
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This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.

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sfi - Stat 5101 (Geyer) 2002 Final Exam Problem 1...

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