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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 ; ) = n Y i =1 P ( X i = x i ) = n Y i =1 1 2 exp { ( x i t i ) 2 2 } = (2 ) n 2 exp { n i =1 ( x i t i ) 2 2 } = (2 ) n 2 exp { n i =1 x 2 i 2 } exp { n X i =1 t i x i } exp { 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 . (b) Let Y i = X i t i N (0 , 1). We can see that, n X i =1 ( X i t i ) 2 = n X i =1 ( Y i + t i  t i n j =1 t j ( Y j + t j ) 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 . Hence, this is ancillary. By Basus theorem, and ( X i t i ) 2 are independent. Problem 4.2 (a) To prove that, we massage the density expression into its exponential family form: f ( x ; , ) = 2 1 x 2 x + p + 1 2 log 2 x 3 / 2 I ( x > 0) = [ ( , ) T ( x ) + B ( , )] h ( x ) with: T ( x ) = h 1 x x i ( , ) = h 2 2 i h ( x ) = x 3 2 I ( x > 0) B ( , ) = 1 2 log 2 + p 1 (b) n Y i =1 f ( x i ; , ) = " 2 n X i =1 1 x i 2 n X i =1 x i + p + 1 2 log 2 # 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

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