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Dr. Hackney STA Solutions pg 93

# Dr. Hackney STA Solutions pg 93 - Second Edition 6-7 6.29...

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Second Edition 6-7 6.29 Let f j = logistic( α j , β j ), j = 0 , 1 , . . . , k . From Theorem 6 . 6 . 5, the statistic T ( x ) = n i =1 f 1 ( x i ) n i =1 f 0 ( x i ) , . . . , n i =1 f k ( x i ) n i =1 f 0 ( x i ) = n i =1 f 1 ( x ( i ) ) n i =1 f 0 ( x ( i ) ) , . . . , n i =1 f k ( x ( i ) ) n i =1 f 0 ( x ( i ) ) is minimal sufficient for the family { f 0 , f 1 , . . . , f k } . As T is a 1 - 1 function of the order statistics, the order statistics are also minimal sufficient for the family { f 0 , f 1 , . . . , f k } . If F is a nonparametric family, f j ∈ F , so part ( b ) of Theorem 6 . 6 . 5 can now be directly applied to show that the order statistics are minimal sufficient for F . 6.30 a. From Exercise 6.9b, we have that X (1) is a minimal sufficient statistic. To check completeness compute f Y 1 ( y ), where Y 1 = X (1) . From Theorem 5.4.4 we have f Y 1 ( y ) = f X ( y ) (1 - F X ( y )) n - 1 n = e - ( y - μ ) e - ( y - μ ) n - 1 n = ne - n ( y - μ ) , y > μ. Now, write E μ g ( Y 1 ) = μ g ( y ) ne - n ( y - μ ) dy . If this is zero for all μ , then μ g ( y ) e - ny dy = 0 for all μ (because ne > 0 for all μ and does not depend on y ). Moreover, 0 = d μ g ( y ) e - ny dy = - g ( μ ) e - for all μ . This implies g ( μ ) = 0 for all μ , so X (1) is complete.
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