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Unformatted text preview: Hints for Assignment 2 Ling Chen November 5, 2008 2.9. You may assume that you can interchange integration with differentiation. Key steps to prove part (a) are the following: E ( ∇ log f θ ( X )) = E ∇ f θ ( X ) f θ ( X ) = Z ∇ f θ ( X ) f θ ( X ) · f θ ( X ) dX = Z ∇ f θ ( X ) dX = ∇ Z f θ ( X ) dX = ∇ (1) = 0 . Note that I have simplified the notations. For part (b), you need to start with ∇ 2 l ( θ ) = ∇ ∇ f θ ( X ) f θ ( X ) = f θ ( X ) ∇ 2 f θ ( X ) ∇ f θ ( X )( ∇ f θ ( X )) T ( f θ ( X )) 2 . and use the same idea of interchange and the result in part (a). 3.2. Using the notations Y = ( y 1 ,...,y n ) T , X = 1 ··· 1 x 1 ··· x n T , B = ( α,β ) T and E = ( 1 ,..., n ) T , we can write (3.24) as Y = XB + E , and the OLS estimate for B is ˆ B = ( X T X ) 1 X T Y . You can prove (3.27) by following the arguments in Section 2.3.3 and using Definition 2.3 of Wishart distribution. The only thing that might make the multivariate case more complicated is the following: Lemma. Suppose E = ( 1 ,..., n ) T , in which t are i.i.d. N (0 ,V ), W = AE and Z = BE , where A and B are a × n and b × n matrices. Then w i , the ith row of W , and z j , the jth row of Z are jointly normal withe mean 0 and Cov( w i ,z j ) = ( AB T ) ij V , where M ij is the ( i,j )th entry of matrix M . In particular, w i ∼ N (0 , ( AA T ) ii V ). Proof. w i = ∑ n t =1 A it t and z j = ∑ n t =1 B jt t are linear combinations of t . This verifies that w i and z j are jointly normal with mean 0. Furtherare jointly normal with mean 0....
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This document was uploaded on 11/08/2009.
 Fall '09

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