Practice Final Solutions

Practice Final Solutions - Stat 116 Practice Final...

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Unformatted text preview: Stat 116 Practice Final Solutions 1. Let X 1 X 2 X 3 Normal 1 , 4 1 1 1 9 (1) Find f X 1 | X 2 ,X 3 ( X 1 | X 2 = 1 ,X 3 = 5) (2) What is the joint distribution of W 1 = 3 X 1 + X 2 ,W 2 = X 1- X 2 ? (3) Let Y 1 = X 1 + X 2 , Y 2 = X 1 /X 2 . Find the joint p.d.f. of Y 1 , Y 2 : f Y 1 ,Y 2 ( y 1 ,y 2 ). Solution (1) X 1 , X 2 are independent of X 3 , f X 1 | X 2 ,X 3 ( X 1 | X 2 = 1 ,X 3 = 5) = f X 1 | X 2 ( X 1 | X 2 = 1) X 1 X 2 N , 4 1 1 1 f X 1 ,X 2 ( x 1 ,x 2 ) = 1 2 3 exp {- 1 2 x 2 1- 2 x 1 x 2 + 4 x 2 2 3 } f X 2 ( x 2 ) = 1 2 exp {- x 2 2 } f X 1 | X 2 ( x 1 | x 2 ) = 1 2 3 exp {- x 2 1- 2 x 1 x 2 + x 2 2 2 3 } f X 1 | X 2 ( x 1 | x 2 = 1) = 1 2 3 exp {- ( x 1- 1) 2 2 3 } This is a Normal distribution N (1 , 3). (2) W 1 W 2 = 3 1 1- 1 X 1 X 2 W 1 W 2 N 3 1 1- 1 , 3 1 1- 1 4 1 1 1 3 1 1- 1 N , 43 9 9 3 (3) X 1 = Y 1 Y 2 1 + Y 2 ; X 2 = Y 1 1 + Y 2 Jacobian matrix: 1 y 1 x 1 y 1 x 2 y 2 x 1 y 2 x 2- 1 = 1 1 1 x 2- x 1 x 2 2- 1 = x 2 2 x 1 + x 2 Therefore f Y 1 ,Y 2 ( y 1 ,y 2 ) = f X 1 ,X 2 ( x 1 ,x 2 ) x 2 2 x 1 + x 2 = 1 2 3 exp {- 1 2 ( x 1- x 2 ) 2 3 + x 2 2 } x 2 2 x 1 + x 2 = 1 2 3 exp {- 1 2 y 2 1 (1- y 2 ) 2 3(1 + y 2 ) 2 + y 2 1 (1 + y 2 ) 2 } x 2 2 x 1 + x 2 2. Suppose the volume of a raindrop ( V i ) can be modeled as a Gamma distribution with parameter p = 2, = 16 (i.e. V i Gamma(2,16), E ( V i ) = 2 / 16)....
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This note was uploaded on 01/13/2010 for the course STATS 116 taught by Professor Staff during the Spring '07 term at Stanford.

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Practice Final Solutions - Stat 116 Practice Final...

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