Dr. Hackney STA Solutions pg 76

Dr. Hackney STA Solutions pg 76 - 5-10 Solutions Manual for...

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5-10 Solutions Manual for Statistical Inference Now let W = Z/Y , Q = Y . Then Y = Q , Z = WQ , and | J | = q . Therefore f W,Q ( w, q ) = n ( n - 1) θ n ( q - wq ) n - 2 q = n ( n - 1) θ n (1 - w ) n - 2 q n - 1 , 0 < w < 1 , 0 < q < θ. The joint pdf factors into functions of w and q , and, hence, W and Q are independent. 5.25 The joint pdf of X (1) , . . . , X ( n ) is f ( u 1 , . . . , u n ) = n ! a n θ an u a - 1 1 · · · u a - 1 n , 0 < u 1 < · · · < u n < θ. Make the one-to-one transformation to Y 1 = X (1) /X (2) , . . . , Y n - 1 = X ( n - 1) /X ( n ) , Y n = X ( n ) . The Jacobian is J = y 2 y 2 3 · · · y n - 1 n . So the joint pdf of Y 1 , . . . , Y n is f ( y 1 , . . . , y n ) = n ! a n θ an ( y 1 · · · y n ) a - 1 ( y 2 · · · y n ) a - 1 · · · ( y n ) a - 1 ( y 2 y 2 3 · · · y n - 1 n ) = n ! a n θ an y a - 1 1 y 2 a - 1 2 · · · y na - 1 n , 0 < y i < 1; i = 1 , . . . , n - 1 , 0 < y n < θ. We see that f ( y 1 , . . . , y n ) factors so Y 1 , . . . , Y n are mutually independent. To get the pdf of Y 1 , integrate out the other variables and obtain that f Y 1 ( y 1 ) = c 1 y a - 1 1 , 0 < y 1 < 1, for some constant c 1 . To have this pdf integrate to 1, it must be that c 1 = a . Thus f Y 1 ( y 1 ) = ay a - 1 1 , 0 < y 1 < 1. Similarly, for i = 2 , . . . , n - 1, we obtain f Y i ( y i ) = iay ia - 1 i , 0 < y i
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