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Dr. Hackney STA Solutions pg 91

Dr. Hackney STA Solutions pg 91 - Second Edition 6-5 6.19...

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Second Edition 6-5 6.19 To check if the family of distributions of X is complete, we check if E p g ( X ) = 0 for all p , implies that g ( X ) 0. For Distribution 1, E p g ( X ) = 2 x =0 g ( x ) P ( X = x ) = pg (0) + 3 pg (1) + (1 - 4 p ) g (2) . Note that if g (0) = - 3 g (1) and g (2) = 0, then the expectation is zero for all p , but g ( x ) need not be identically zero. Hence the family is not complete. For Distribution 2 calculate E p g ( X ) = g (0) p + g (1) p 2 + g (2)(1 - p - p 2 ) = [ g (1) - g (2)] p 2 + [ g (0) - g (2)] p + g (2) . This is a polynomial of degree 2 in p . To make it zero for all p each coefficient must be zero. Thus, g (0) = g (1) = g (2) = 0, so the family of distributions is complete. 6.20 The pdfs in b), c), and e) are exponential families, so they have complete sufficient statistics from Theorem 6.2.25. For a), Y = max { X i } is sufficient and f ( y ) = 2 n θ 2 n y 2 n - 1 , 0 < y < θ. For a function g ( y ), E g ( Y ) = θ 0 g ( y ) 2 n θ 2 n y 2 n - 1 dy = 0 for all θ implies g ( θ ) 2 2 n - 1 θ 2 n = 0 for all θ by taking derivatives. This can only be zero if g ( θ ) = 0 for all θ , so Y = max { X i } is complete.
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