This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ) =...
View
Full
Document
This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.
 Fall '02
 Staff

Click to edit the document details