Dr. Hackney STA Solutions pg 134

Dr. Hackney STA Solutions pg 134 - Second Edition 8-13 The...

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Second Edition 8-13 The Type II error probability is 1 - P (1 < X < 3 | θ = 1) = 1 - Z 3 1 1 π 1 1+( x - 1) 2 dx = 1 - 1 π arctan( x - 1) ± ± ± ± 3 1 . 6476 . c. We will not have f (1 | θ ) /f (1 | 0) = f (3 | θ ) /f (3 | 0) for any other value of θ 6 = 1. Try θ = 2, for example. So the rejection region 1 < x < 3 will not be most powerful at any other value of θ . The test is not UMP for testing H 0 : θ 0 versus H 1 : θ > 0. 8.30 a. For θ 2 > θ 1 > 0, the likelihood ratio and its derivative are f ( x | θ 2 ) f ( x | θ 1 ) = θ 2 θ 1 θ 2 1 + x 2 θ 2 2 + x 2 and d dx f ( x | θ 2 ) f ( x | θ 1 ) = θ 2 θ 1 θ 2 2 - θ 2 1 ( θ 2 2 + x 2 ) 2 x. The sign of the derivative is the same as the sign of x (recall, θ 2 2 - θ 2 1 > 0), which changes sign. Hence the ratio is not monotone. b. Because f ( x | θ ) = ( θ/π )( θ 2 + | x | 2 ) - 1 , Y = | X | is sufficient. Its pdf is f ( y | θ ) = 2 θ π 1 θ 2 + y 2 , y > 0 . Differentiating as above, the sign of the derivative is the same as the sign of y , which is positive. Hence the family has MLR. 8.31 a. By the Karlin-Rubin Theorem, the UMP test is to reject H 0 if i X i > k , because i X i is sufficient and i X i Poisson( ) which has MLR. Choose the constant k to satisfy
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