Ass4sol - STAT 330 Assignment 4 Solutions 1. a. By...

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STAT 330 Assignment 4 Solutions 1. a. By independence, the joint distribution of Y 1 and Y 2 is the product of the two marginal densities: β α / ) ( 1 2 1 1 ) ( ) ( 1 2 1 2 1 2 1 2 1 1 ) , ( x x e x x x x f a + Γ Γ + = , x 1 0, x 2 0. With U and V as defined, we have that x 1 = u 1 u 2 and x 2 = u 2 (1– u 1 ). Thus, the Jacobian of transformation J = u 2. Thus, the joint density of U 1 and U 2 is 2 / 1 1 2 1 2 1 ) ( ) ( 1 2 1 2 2 1 2 1 1 )] 1 ( [ ) ( ) , ( u e u u u u u u f u a β α α β α Γ α Γ = α + α β α + α α α β α Γ α Γ = α + α / 1 2 1 1 1 1 ) ( ) ( 1 2 2 1 2 1 2 1 1 ) 1 ( u e u u u a , with 0 < u 1 < 1, and u 2 > 0. b. 1 1 1 1 ) ( ) ( ) ( / 1 0 1 1 1 1 1 ) ( ) ( 1 1 2 1 1 1 2 1 2 1 2 1 1 1 ) 1 ( ) 1 ( ) ( α α α Γ α Γ α + α Γ β α + α β α α α Γ α Γ = = α + α u u dv e v u u u f a a a v U , with 0 < u 1 < 1. This is the beta density as defined. c. β α + α α + α Γ β α α α Γ α Γ β α + α β α + α α + α = = / 1 2 ) ( 1 1 1 0 1 1 1 1 ) ( ) ( 1 / 1 2 1 2 2 2 1 2 1 2 1 2 1 1 2 2 1 2 1 2 ) 1 ( ) ( u u
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This note was uploaded on 07/12/2010 for the course STAT 330 taught by Professor Paulasmith during the Winter '08 term at Waterloo.

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Ass4sol - STAT 330 Assignment 4 Solutions 1. a. By...

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