hw10_stat210a_solutions

hw10_stat210a_solutions - UC Berkeley Department of...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 10 Fall 2006 Issued: Thursday, November 9, 2006 Due: Thursday, November 16, 2006 Graded exercises Problem 10.1 a) For each i , Z i = I ( Y i > ) follows a Bernoulli distribution with parameter 1- F ( ). Given that Y i are i.i.d., so are Z i and hence S = n i =1 follows a Binomial( n, 1- F ( )) distribution. This is useful in performing a statistical hypothesis test because the distribution of S is known under H . Furthermore, as it does not involve any nuisance parameter, the threshold for a -level test can be easily computed. b) For large n and under the null hypothesis, the distribution of S can be approximated by a N ( n 2 , n 4 ). As a result S- n 2 n 2 has an approximately standard normal distribution for large enough n . E ( s ( Y )) = P ( S > s ) = P S- n 2 n 2 > s- n 2 n 2 1- s- n 2 n 2 = n- 2 s n Problem 10.2 a) The action space is given by A = { , 1 } . For = , the loss function is given by: l ( , ) = , if = 0 1 , if = 1 For = 1 , the loss function is given by: l ( 1 , ) = , if = 1 1 , if = 0 It follows that E ( l ( , ( X )) | = )) = E ( ( X ) | = )) and E ( l ( , ( X )) | = 1 )) = E (1- ( X ) | = 1 )) As a result: r ( , ) = E ( l ( , ( X ))) = E ( l ( , ( X )) | = )) P ( = ) + E ( l ( , ( X )) | = 1 )) P ( = 1 ) = E ( ( X )) + (1- ) E 1 (1- ( X )) b) To minimize the Bayes risk, it is sufficient to minimize the posterior risk. The posterior risk of taking action is given by: E ( L ( , ) X ) = P ( X | ) P ( X | ) + (1- ) P ( X | 1 ) + (1- ) (1- ) P ( X | 1 ) P ( X | ) + (1- ) P ( X | 1 ) = P ( X | )- (1- ) P ( X | 1 ) P ( X | ) + (1- ) P ( X | 1 ) + (1- ) P ( X | 1 ) P ( X | ) + (1- ) P ( X | 1 ) 1 Hence, the optimal decision * ( X ) is given by: * ( X ) = I ((1- ) P ( X | 1 )- P ( X | ) 0) = I P ( X | 1 ) P ( X | ) 1- c) From item b: * ( X ) = I " 1 n n X i =1 log P ( X i | 1 ) P ( X i | ) 1 n log 1- # Now, under H : 1 n n X i =1 log P ( X i | 1 ) P ( X i | ) p E " 1 n n X i =1 log P ( X i | 1 ) P ( X i | ) # = D ( || 1 ) > while, under H 1 : 1 n n X i =1 log P ( X i | 1 ) P ( X i | ) p E 1 " 1 n n X i =1 log P ( X i |...
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hw10_stat210a_solutions - UC Berkeley Department of...

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