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Dr. Hackney STA Solutions pg 88

Dr. Hackney STA Solutions pg 88 - 6-2 Solutions Manual for...

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6-2 Solutions Manual for Statistical Inference 6.6 The joint pdf is given by f ( x 1 , . . . , x n | α, β ) = n i =1 1 Γ( α ) β α x i α - 1 e - x i = 1 Γ( α ) β α n n i =1 x i α - 1 e - Σ i x i . By the Factorization Theorem, ( n i =1 X i , n i =1 X i ) is sufficient for ( α, β ). 6.7 Let x (1) = min i { x 1 , . . . , x n } , x ( n ) = max i { x 1 , . . . , x n } , y (1) = min i { y 1 , . . . , y n } and y ( n ) = max i { y 1 , . . . , y n } . Then the joint pdf is f ( x , y | θ ) = n i =1 1 ( θ 3 - θ 1 )( θ 4 - θ 2 ) I ( θ 1 3 ) ( x i ) I ( θ 2 4 ) ( y i ) = 1 ( θ 3 - θ 1 )( θ 4 - θ 2 ) n I ( θ 1 , ) ( x (1) ) I ( -∞ 3 ) ( x ( n ) ) I ( θ 2 , ) ( y (1) ) I ( -∞ 4 ) ( y ( n ) ) g ( T ( x ) | θ ) · 1 h ( x ) . By the Factorization Theorem, ( X (1) , X ( n ) , Y (1) , Y ( n ) ) is sufficient for ( θ 1 , θ 2 , θ 3 , θ 4 ). 6.9 Use Theorem 6.2.13. a. f ( x | θ ) f ( y | θ ) = (2 π ) - n/ 2 e - Σ i ( x i - θ ) 2 / 2 (2 π ) - n/ 2 e - Σ i ( y i - θ ) 2 / 2 = exp - 1 2 n i =1 x 2 i - n i =1 y 2 i +2 θn y - ¯ x ) . This is constant as a function of θ if and only if ¯ y = ¯ x ; therefore ¯ X is a minimal sufficient statistic for θ . b. Note, for X location exponential( θ ), the range depends on the parameter. Now f ( x | θ ) f ( y | θ ) = n i =1 ( e - ( x i - θ ) I ( θ, ) ( x i ) ) n i =1 ( e - ( y i - θ ) I ( θ, ) ( y i ) ) = e e - Σ i x i n i =1 I ( θ, ) ( x i ) e e - Σ i y i n i =1 I ( θ, ) ( y i ) = e - Σ i x i I ( θ, ) (min x i ) e - Σ i y i I ( θ, ) (min y i ) . To make the ratio independent of
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