4.19b - [Solution to 4.19(b x1 1 ex1 x2 1 ex2 fX1,X2(x1 x2...

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[Solution to 4.19(b)] f X 1 ,X 2 ( x 1 , x 2 ) = x α 1 - 1 1 e - x 1 Γ( α 1 ) · x α 2 - 1 2 e - x 2 Γ( α 2 ) for 0 < x 1 < , 0 < x 2 < . Let Y 1 = X 1 / ( X 1 + X 2 ) and Y 2 = X 1 + X 2 . The inverse transformation is X 1 = Y 1 Y 2 and X 2 = Y 2 - X 1 = Y 2 - Y 1 Y 2 = Y 2 (1 - Y 1 ). Thus, the transformation is 1-1. The support of ( X 1 , X 2 ) is A = (0 , ) × (0 , ). The support of ( Y 1 , Y 2 ) is B = (0 , 1) × (0 , ). (Clearly the possible values of Y 2 are 0 to . For any fixed value of Y 2 = X 1 + X 2 , the value of X 1 can range from 0 to Y 2 , and thus Y 1 = X 1 /Y 2 can range from 0 to 1.) The Jacobian of the inverse transformation x 1 = y 1 y 2 , x 2 = y 2 (1 - y 1 ) is J = ± ± ± ± ± ∂x 1 ∂y 1 ∂x 1 ∂y 2 ∂x 2 ∂y 1 ∂x 2 ∂y 2 ± ± ± ± ± = ± ± ± ± ± y 2 y 1 - y 2 1 - y 1 ± ± ± ± ± = y 2 (1 - y 1 ) + y 1 y 2 = y 2 . Thus f Y 1 ,Y 2 ( y 1 , y 2 ) = f X 1 ,X 2 ( y 1 y 2 , y 2 (1 - y 1 )) | y 2 | for ( y 1 , y 2 ) (0 , 1) × (0 , ) . = ( y 1 y 2 ) α 1 - 1 e - y 1 y 2 Γ( α 1 ) · ( y 2 (1 - y 1 )) α 2 - 1 e - y 2 (1 - y 1 ) Γ( α 2 ) · y 2 = Γ( α 1 + α 2 ) Γ( α 1 )Γ( α 2 ) y α 1 - 1 1 (1 - y 1 ) α 2 - 1 · y α 1 + α 2 - 1 2 e - y 2 Γ( α 1 + α 2 ) for 0 < y 1 < 1 and 0 < y 2 < . We now see that the joint density factors into a function of y 1 times a function of y 2 valid for all y 1 , y 2 . Thus Y 1 and
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This note was uploaded on 12/15/2011 for the course STAT 5326 taught by Professor Frade during the Fall '10 term at FSU.

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