Dr. Hackney STA Solutions pg 119

# Dr. Hackney STA Solutions pg 119 - 7-22 Solutions Manual...

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Unformatted text preview: 7-22 Solutions Manual for Statistical Inference 7.56 Because T is sufficient, (T ) = E[h(X1 , . . . , Xn )|T ] is a function only of T . That is, (T ) is an estimator. If E h(X1 , . . . , Xn ) = (), then E h(X1 , , Xn ) = E [E ( h(X 1 , . . . , X n )| T )] = (), so (T ) is an unbiased estimator of (). By Theorem 7.3.23, (T ) is the best unbiased estimator of (). 7.57 a. T is a Bernoulli random variable. Hence, n Ep T = Pp (T = 1) = Pp i=1 n+1 i=1 Xi > Xn+1 = h(p). b. Xi is a complete sufficient statistic for , so E T estimator of h(p). We have n+1 n n+1 n+1 i=1 Xi is the best unbiased E T i=1 Xi = y = P i=1 n Xi > Xn+1 i=1 n+1 Xi = y n+1 = P i=1 Xi > Xn+1 , i=1 Xi = y P i=1 Xi = y . The denominator equals n+1 y py (1 - p)n+1-y . If y = 0 the numerator is n n+1 P i=1 Xi > Xn+1 , i=1 Xi = 0 = 0. If y > 0 the numerator is n n+1 n n+1 P i=1 Xi > Xn+1 , i=1 Xi = y, X n+1 = 0 +P i=1 Xi > Xn+1 , i=1 Xi = y, X n+1 = 1 which equals n n n n P i=1 Xi > 0, i=1 Xi = y P (Xn+1 = 0) + P i=1 Xi > 1, i=1 Xi = y - 1 P (Xn+1 = 1). For all y > 0, n n n P i=1 Xi > 0, i=1 Xi = y =P i=1 Xi = y = n y p (1 - p)n-y . y If y = 1 or 2, then n n P i=1 Xi > 1, i=1 Xi = y - 1 = 0. And if y > 2, then n n n P i=1 Xi > 1, i=1 Xi = y - 1 =P i=1 Xi = y - 1 = n py-1 (1 - p)n-y+1 . y-1 ...
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