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Unformatted text preview: 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 , P ( S ( X )) 1 . To prove that, we first notice that, according to the definition of S ( X ), S ( X ) X A ( ). It follows that: P ( S ( X )) = P ( X A ( )) 1 , for all where the inequality follows from the definition of a acceptance region for a level test. Problem 9.2 a) Let T ( X ) = max 1 i n X i and ( X ) { , 1 } be any function such that: H is rejected if ( X ) = 1 and accepted otherwise; E ( ( X )) for all ; We want to prove: E 1 ( ( X )) E 1 ( ( X )) , 1 > To prove that, first notice that for any : E ( ( X )) = E ( ( X ) I ( T ( X ) )) + E ( ( X ) I ( T ( X ) > )) E ( ( X )) = E ( ( X ) I ( T ( X ) )) + E ( ( X ) I ( T ( X ) > )) Since ( X ) 1 and T ( X ) > implies ( X ) = 1, we have: E 1 ( ( X ) I ( T ( X ) > )) E 1 ( ( X ) I ( T ( X ) > )) It is then enough to prove that: E 1 [( ( X ) ( X )) I ( T ( X ) )] , for all 1 > 1 We have: E 1 [( ( X ) ( X )) I ( T ( X ) )] = R ( ( X ) ( X )) n 1 I ( x i ) I ( x i 1 ) dx = 1 n 1 Z ( ( X ) ( X )) I ( x i ) dx  {z } h ( ,, ) = h ( ,, ) n 1 and so, it is enough to prove that h ( ,, ) 0. To prove that, notice: h ( ,, ) n = E [( ( X ) ( X )) I ( T ( X ) )] = E [( ( X ) ( X ))] where: the second equality is due to the fact that P ( X i ) = 0; the inequality follows from the fact that E ( X ) E ( X ) = ....
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This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Staff
 Statistics

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