stat210a_2007_hw2_solutions

# stat210a_2007_hw2_solutions - UC Berkeley Department of...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 2- Solutions Fall 2007 Issued: Thursday, September 6 Due: Thursday, September 13 Problem 2.1 (a) Let f i ( x ) be the density of X i , X = ( X 1 ,...,X n ) and x = ( x 1 ,...,x n ). Then, n Y i =1 f i ( x i ) = ( 2 πσ 2 )- n/ 2 exp (- 1 2 σ 2 n X i =1 x 2 i ) · exp ( μ σ 2 n X i =1 x i- nμ 2 2 σ 2 ) = h ( x ,σ ) · g ˆ n X i =1 x i ,σ,μ ! where h ( x ,σ ) = ( 2 πσ 2 )- n/ 2 exp (- 1 2 σ 2 n X i =1 x 2 i ) and g ˆ n X i =1 x i ,σ,μ ! = exp ( μ σ 2 n X i =1 x i- nμ 2 2 σ 2 ) . Thus, by factorization theorem, T = n X i =1 X i is a sufficient statistic. (b) It is well known that ∑ n i =1 X i ∼ N ( nμ,nσ 2 ). Thus, P μ ( X = x | T ( X ) = t ) = P μ ( X = x ,T ( X ) = t ) P ( T ( X ) = t ) = h ( x ,σ )exp n μ σ 2 t- nμ 2 2 σ 2 o 1 ( t = ∑ n i =1 x i ) (2 πnσ 2 )- 1 / 2 exp n- t 2 2 nσ 2 o exp n μ σ 2 t- nμ 2 2 σ 2 o = h ( x ,σ ) 1 ( t = ∑ n i =1 x i ) (2 πnσ 2 )- 1 / 2 exp n- t 2 2 nσ 2 o : NOT depend on μ Thus, T = n X i =1 X i is a sufficient statistic. 1 Alternative argument: Suppose without loss of generality that σ 2 = 1 (to simplify notation). Consider the transformation ( X 1 ,...,X n ) ←→ ( T,X 2- X 1 ,...,X n- X 1 ), and note that both vectors are jointly Gaussian. Defining the ( n- 1)-vector V = ( X 2- X 1 ,...,X n- X 1 ), we then have E [ V i | T ] = 0 , for all i , and cov( V i ,V j | T ) = 2 if i = j 1 otherwise . Thus, the mean and covariance matrix of ( V | T ) are independent of μ . Since ( V | T ) is Gaussian, this shows that its distribution is independent of μ ....
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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 Berkeley.

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stat210a_2007_hw2_solutions - UC Berkeley Department of...

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