Dr. Hackney STA Solutions pg 179

# Dr. Hackney STA Solutions pg 179 - w n where we set w = 0...

This preview shows page 1. Sign up to view the full content.

Second Edition 11-7 11.19 a. The marginal distributions of the Y i are somewhat straightforward to derive. As X i +1 gamma( λ i +1 , 1) and, independently, i j =1 X j gamma( i j =1 λ j , 1) (Example 4.6.8), we only need to derive the distribution of the ratio of two independent gammas. Let X gamma( λ 1 , 1) and Y gamma( λ 2 , 1). Make the transformation u = x/y, v = y x = uv, y = v, with Jacobian v . The density of ( U,V ) is f ( u,v ) = 1 Γ( λ 1 )Γ( λ 2 ) ( uv ) λ 1 - 1 v λ 2 - 1 ve - uv e - v = u λ 1 - 1 Γ( λ 1 )Γ( λ 2 ) v λ 1 + λ 2 - 1 e - v (1+ u ) . To get the density of U , integrate with respect to v . Note that we have the kernel of a gamma( λ 1 + λ 2 , 1 / (1 + u )), which yields f ( u ) = Γ( λ 1 + λ 2 ) Γ( λ 1 )Γ( λ 2 ) u λ 1 - 1 (1 + u ) λ 1 + λ 2 - 1 . The joint distribution is a nightmare. We have to make a multivariate change of variable. This is made a bit more palatable if we do it in two steps. First transform W 1 = X 1 , W 2 = X 1 + X 2 , W 3 = X 1 + X 2 + X 3 , ..., W n = X 1 + X 2 + ··· + X n , with X 1 = W 1 , X 2 = W 2 - W 1 , X 3 = W 3 - W 2 , ... X n = W n - W n - 1 , and Jacobian 1. The joint density of the W i is f ( w 1 ,w 2 ,...,w n ) = n Y i =1 1 Γ( λ i ) ( w i - w i - 1 ) λ i - 1 e - w n , w 1 w 2 ≤ ··· ≤
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: w n , where we set w = 0 and note that the exponent telescopes. Next note that y 1 = w 2-w 1 w 1 , y 2 = w 3-w 2 w 2 , ... y n-1 = w n-w n-1 w n-1 , y n = w n , with w i = y n Q n-1 j = i (1 + y j ) , i = 1 ,...,n-1 , w n = y n . Since each w i only involves y j with j ≥ i , the Jacobian matrix is triangular and the determinant is the product of the diagonal elements. We have dw i dy i =-y n (1 + y i ) Q n-1 j = i (1 + y j ) , i = 1 ,...,n-1 , dw n dy n = 1 , and f ( y 1 ,y 2 ,...,y n ) = 1 Γ( λ 1 ) ± y n Q n-1 j =1 (1 + y j ) ! λ 1-1 × n-1 Y i =2 1 Γ( λ i ) ± y n Q n-1 j = i (1 + y j )-y n Q n-1 j = i-1 (1 + y j ) ! λ i-1 e-y n × n-1 Y i =1 y n (1 + y i ) Q n-1 j = i (1 + y j ) ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online