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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 4  Solutions Fall 2008 Issued: Tuesday, September 22, 2009 Due: Tuesday, September 29, 2009 Problem 4.1 (a) The joint distribution is, p ( x ; θ ) = n Y i =1 P θ ( X i = x i ) = n Y i =1 1 √ 2 π exp { ( x i t i θ ) 2 2 } = (2 π ) n 2 exp { ∑ n i =1 ( x i t i θ ) 2 2 } = (2 π ) n 2 exp { ∑ n i =1 x 2 i 2 } exp { n X i =1 t i x i } exp { θ 2 ∑ n i =1 t 2 i 2 } . By factorization theorem, T ( X ) = ∑ t i X i is sufficient. Since this is also a full rank exponential family, T ( X ) is also complete. It is the same with ˆ θ . (b) Let Y i = X i t i θ ∼ N (0 , 1). We can see that, n X i =1 ( X i t i ˆ θ ) 2 = n X i =1 ( Y i + t i θ t i ∑ n j =1 t j ( Y j + t j θ ) ∑ t 2 i ) 2 = n X i =1 ( Y i ∑ n j =1 t i t j Y j ∑ t 2 i ) 2 , which does not involve θ . Hence, this is ancillary. By Basu’s theorem, ˆ θ and ∑ ( X i t i ˆ θ ) 2 are independent. Problem 4.2 (a) To prove that, we massage the density expression into its exponential family form: f ( x ; λ,μ ) = λ 2 1 x μ 2 x + p λμ + 1 2 log λ 2 π x 3 / 2 I ( x > 0) = [ η ( λ,μ ) · T ( x ) + B ( λ,μ )] h ( x ) with: T ( x ) = h 1 x x i η ( λ,μ ) = h λ 2 μ 2 i h ( x ) = x 3 2 I ( x > 0) B ( λ,μ ) = 1 2 log λ 2 π + p λμ 1 (b) n Y i =1 f ( x i ; λ,μ ) = " λ 2 n X i =1 1 x i μ 2 n X i =1 x i + p λμ + 1 2 log λ 2 π # x 3 / 2 I ( x > 0) Since there exist no a and b such that a ∑ n i =1 X i + b ∑ n i =1 X 1 i = 0 almost surely and the interior of the pa rameter space R × R is nonempty, this is a full rank exponential family and hence T ( X ) = ( ∑ i X i , ∑ i X 1 i ) is a complete sufficient statistic. Now notice let ˜ T ( X ) = ¯ X, ∑ i 1 X i 1 ¯ X and notice that ˜ T 1 ( X ) = T 1 ( X ) n and ˜ T 2 ( X ) = T 2 ( X ) n 2 T 1 ( X ) which is a onetoone transform of T ( X ) implying that ˜ T ( X ) is also a complete statistic....
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 Fall '08
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 Statistics, Normal Distribution, Trigraph, exponential family, exp exp

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