solutions04 - Stat 351 Fall 2007 Solutions to Assignment #4...

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Unformatted text preview: Stat 351 Fall 2007 Solutions to Assignment #4 Problem #3, page 55: Suppose that X + Y = 2. By definition of conditional density, f X | X + Y =2 ( x ) = f X,X + Y ( x, 2) f X + Y (2) . We now find the joint density f X,X + Y ( x, 2). Let U = X and V = X + Y so that X = U and Y = V- U . The Jacobian of this transformation is J = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = 1- 1 1 = 1 . Since X and Y are independent Γ(2 ,a ), the joint density of ( X,Y ) is f X,Y ( x,y ) = f X ( x ) · f Y ( y ) = xy a 4 e- ( x + y ) /a , for x > , y > , , otherwise , The joint density of ( U,V ) is therefore given by f U,V ( u,v ) = f X,Y ( u,v- u ) · | J | = u ( v- u ) a 4 e- v/a provided that u > 0 and v > u . The marginal density for V is therefore f V ( v ) = Z v u ( v- u ) a 4 e- v/a du = a- 4 e- v/a Z v u ( v- u ) du = v 3 6 a 4 e- v/a , u > . Since V = X + Y , we can rewrite these densities as f X,X + Y ( x, 2) = x (2- x ) a 4 e- 2 /a , 0 < x < 2, and f X + Y (2) = 2 3 6 a 4 e- 2 /a . Finally, we conclude f X | X + Y =2 ( x ) = f X,X + Y ( x, 2) f X + Y (2) = x (2- x ) a 4 e- 2 /a 2 3 6 a 4 e- 2 /a = 3 x (2- x ) 4 provided that 0 < x < 2. Problem #9, page 56 : (a) The density function for Y is given by f Y ( y ) = Z ∞ x 2 2 y 3 · e- x y dx provided that 0 < y < 1. Let u =- x y so that du =- 1 y dx , from which it follows that f Y ( y ) = Z ∞ x 2 2 y 3 · e- x y dx = 1 2 Z ∞ u 2 e- u du = 1 2 Γ(3) = 2! 2 = 1 . That is, Y ∈ U (0 , 1). (b) The conditional density of X given Y = y is therefore f X | Y = y ( x ) = f X,Y ( x,y ) f Y ( y ) = x 2 2 y 3 · e- x y 1 = x 2 2 y 3 · e- x y provided that x > 0. That is, X | Y = y ∈ Γ(3 ,y )....
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.

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solutions04 - Stat 351 Fall 2007 Solutions to Assignment #4...

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