stat210a_2007_hw10_solutions

stat210a_2007_hw10_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 10- Solutions Fall 2007 Issued: Thursday, December 8 Due: Thursday, December 15 Problem 10.1 a) The action space is given by A = { 0 , 1 } . For θ = θ 0 , the loss function is given by: l ( θ 0 ) = ( 0 , if δ = 0 1 , if δ = 1 For θ = θ 1 , the loss function is given by: l ( θ 1 ) = ( 0 , if δ = 1 1 , if δ = 0 It follows that E ( l ( θ,δ ( X )) | θ = θ 0 )) = E ( δ ( X ) | θ = θ 0 )) and E ( l ( θ,δ ( X )) | θ = θ 1 )) = E (1 - δ ( X ) | θ = θ 1 )) As a result: r ( λ,δ ) = E ( l ( θ,δ ( X ))) = E ( l ( θ,δ ( X )) | θ = θ 0 )) P ( θ = θ 0 ) + E ( l ( θ,δ ( X )) | θ = θ 1 )) P ( θ = θ 1 ) = λ 0 E 0 ( δ ( X )) + (1 - λ 0 ) E 1 (1 - δ ( X )) b) To minimize the Bayes risk, it is sufficient to minimize the posterior risk. The posterior risk of taking action δ is given by: E ± L ( θ,δ ) X ) = δ λ 0 P ( X | θ 0 ) λ 0 P ( X | θ 0 ) + (1 - λ 0 ) P ( X | θ 1 ) + (1 - δ ) (1 - λ 0 ) P ( X | θ 1 ) λ 0 P ( X | θ 0 ) + (1 - λ 0 ) P ( X | θ 1 ) = δ λ 0 P ( X | θ 0 ) - (1 - λ 0 ) P ( X | θ 1 ) λ 0 P ( X | θ 0 ) + (1 - λ 0 ) P ( X | θ 1 ) + (1 - λ 0 ) P ( X | θ 1 ) λ 0 P ( X | θ 0 ) + (1 - λ 0 ) P ( X | θ 1 ) Hence, the optimal decision δ * ( X ) is given by: δ * ( X ) = I ((1 - λ 0 ) P ( X | θ 1 ) - λ 0 P ( X | θ 0 ) 0) = I ² P ( X | θ 1 ) P ( X | θ 0 ) λ 0 1 - λ 0 1
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δ * ( X ) = I " 1 n n X i =1 log ± P ( X i | θ 1 ) P ( X i | θ 0 ) 1 n log ± λ 0 1 - λ 0 # Now, under H 0 : 1 n n X i =1 log ± P ( X i | θ 1 ) P ( X i | θ 0 ) p E θ 0 " 1 n n X i =1 log ± P ( X i | θ 1 ) P ( X i | θ 0 ) # = D ( θ 0 || θ 1 ) > 0 while, under H 1 : 1 n n X i =1 log ± P ( X i | θ 1 ) P ( X i | θ 0 ) p E θ 1 " 1 n n X i =1 log ± P ( X i | θ 1 ) P ( X i | θ 0 ) # = - D ( θ 1 || θ 0 ) < 0 Simultaneously, we have that for any λ 0 (0 , 1): 1 n log ± λ 0 1 - λ 0 0 Hence: δ * ( X ) p ( 1 , under H 1 0 , under H 0 Now, from boundedness of δ and the convergence in probability above, we get: E 0 [ δ ( X )] 0 and E 1 [1 - δ ( X )] 0 which in turn yield: r ( λ,δ λ ) = λ E 0 [ δ ( X )] + (1 - λ )(1 - E 1 [ δ ( X )]) 0 as required. Problem 10.2 (a) For θ = ( μ x y 2 x 2 y ), sup θ Ω l ( θ ) = n 2 log ˆ σ 2 x - n 2 log ˆ σ 2 y - 1 σ 2 x n X i =1 ( X i - ˆ μ x ) 2 - 1 σ 2 y n X i =1 ( Y i - ˆ μ y ) 2 - n log(2 π ) sup θ Ω 0 l ( θ ) = - n log ˆ σ 2 - 1 σ 2 ˆ n X i =1 ( X i - ˆ μ ) 2 + n
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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.

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stat210a_2007_hw10_solutions - UC Berkeley Department of...

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