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610-hw5sol

# 610-hw5sol - Homework 5 Solutions(1 By change of variables...

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Homework 5 Solutions (1) By change of variables, E φ ( X 2 ) = Z 1 0 φ ( x 2 ) dx = Z 1 0 φ ( y ) 2 y dy E φ (1 - X 2 ) = Z 1 0 φ (1 - x 2 ) dx = Z 1 0 φ ( y ) 2 1 - y dy. By the generalized Neyman-Pearson Lemma, the maximizer takes the form φ * ( x ) = 1 1 > k 1 2 x + k 2 2 1 - x 0 otherwise , where k 1 and k 2 are determined to satisfy the constraints. Because g ( x ) = k 1 2 x + k 2 2 1 - x is convex, k 1 and k 2 have to be chosen such that { 1 > g ( x ) } = { b 1 < x < b 2 } for some b 1 and b 2 satisfying the constraints. In order to satisfy the constraints, E φ * ( X 2 ) = Z b 2 b 1 1 2 y dy = p b 2 - p b 1 = 1 2 E φ * (1 - X 2 ) = Z b 2 b 1 1 2 1 - y dy = p 1 - b 1 - p 1 - b 2 = 1 2 , from which we obtain b 1 = 1 8 (4 - 7) and b 2 = 1 8 (4 + 7). Hence, we have φ * ( x ) = 1 1 8 (4 - 7) < x < 1 8 (4 + 7) 0 otherwise . (2) (i) The joint density of ( X 1 ,X 2 ,...,X n ) is f ( x 1 ,x 2 ,...,x n ) = λ n e - x . For λ 1 < λ 0 , f 1 (( X 1 ,X 2 ,...,X n )) f 0 (( X 1 ,X 2 ,...,X n )) = ± λ 1 λ 2 ² n e n X ( λ 0 - λ 1 ) , which is increasing in X . 1

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(ii) Reparametrize as θ = - λ . Then, the equivalent form of the test is H 0 : θ = θ 0 against θ = θ 1 > θ 0 . By (i), we know that when θ 1 = - λ 1 > - λ 0 = θ 0 , f 1 /f 0 is increasing in X . Thus, by the Neyman-Pearson lemma, the UMP test rejects when X > c for some c . Note that 2 λn X = n i =1 2 λX i χ 2 2 n . Thus, α = P λ 0 ( X > c ) = P λ 0 (2 λ 0 n X > 2 λ 0 nc ) = 1 - q - 1 2 n (2 λ 0 nc ) . (Note that
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610-hw5sol - Homework 5 Solutions(1 By change of variables...

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