Ass4sol

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/12/2010 for the course STAT 330 taught by Professor Paulasmith during the Winter '08 term at Waterloo.

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online