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stat210a_2007_hw9_solutions

stat210a_2007_hw9_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 9- Solutions Fall 2007 Issued: Thursday, December 1 Due: Thursday, December 8 Problem 9.1 We want to prove that, for all θ Θ, P θ ( θ S ( X )) 1 - α . To prove that, we first notice that, according to the definition of S ( X ), θ S ( X ) X A ( θ ). It follows that: P θ ( θ S ( X )) = P θ ( X A ( θ )) 1 - α, for all θ Θ where the inequality follows from the definition of a acceptance region for a level α test. Problem 9.2 a) Let T ( X ) = max 1 i n X i and φ ( X ) ∈ { 0 , 1 } be any function such that: H 0 is rejected if φ ( X ) = 1 and accepted otherwise; E θ ( φ ( X )) α for all θ θ 0 ; We want to prove: E θ 1 ( φ ( X )) E θ 1 ( δ ( X )) , θ 1 > θ 0 To prove that, first notice that for any θ : E θ ( φ ( X )) = E θ ( φ ( X ) I ( T ( X ) θ 0 )) + E θ ( φ ( X ) I ( T ( X ) > θ 0 )) E θ ( δ ( X )) = E θ ( δ ( X ) I ( T ( X ) θ 0 )) + E θ ( δ ( X ) I ( T ( X ) > θ 0 )) Since φ ( X ) 1 and T ( X ) > θ 0 implies δ ( X ) = 1, we have: E θ 1 ( δ ( X ) I ( T ( X ) > θ 0 )) E θ 1 ( φ ( X ) I ( T ( X ) > θ 0 )) It is then enough to prove that: E θ 1 [( δ ( X ) - φ ( X )) I ( T ( X ) θ 0 )] 0 , for all θ 1 > θ 0 1
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We have: E θ 1 [( δ ( X ) - φ ( X )) I ( T ( X ) θ 0 )] = R ( δ ( X ) - φ ( X )) θ n 1 I ( x i θ 0 ) I ( x i θ 1 ) dx = 1 θ n 1 •Z ( δ ( X ) - φ ( X )) I ( x i θ 0 ) dx | {z } h ( δ,φ,θ 0 ) = h ( δ, φ, θ 0 ) θ n 1 and so, it is enough to prove that h ( δ, φ, θ 0 ) 0. To prove that, notice: h ( δ, φ, θ 0 ) θ n 0 = E θ 0 [( δ ( X ) - φ ( X )) I ( T ( X ) θ 0 )] = E θ 0 [( δ ( X ) - φ ( X ))] 0 where: the second equality is due to the fact that P θ 0 ( X i θ 0 ) = 0; the inequality follows from the fact that E θ 0 φ ( X ) E θ 0 δ ( X ) = α . b) First, we show that E θ 0 ( δ ( X )) = α . To do that notice that: P θ ( X ( n ) x ) = 0 , for x ( -∞ , 0); ( x θ ) n , for x [0 , 1]; 1 , for x (1 , ); so: E θ 0 ( δ ( X )) = P θ 0 T ( X ) θ 0 α 1 n · + P θ 0 ( T ( X ) > θ 0 ) | {z } =0 = ˆ θ 0 α 1 n θ 0 !
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