hw7_stat210a_solutions

hw7_stat210a_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 7 Fall 2006 Issued: Thursday, September 14, 2006 Due: Thursday, September 21, 2006 Graded exercises Problem 7.1 Let Z n = Y n X n , we have that Z n p 1 as f ( z ) = z - 1 is a continuous function at 1. Now Y n = Z n X n d X by Slutsky’s theorem. Problem 7.2 (a) First, let Y i = X 2 i and notice that: E X i = μ E X 2 i = μ 2 + σ 2 var X i = σ 2 var X 2 i = E X 4 i - ( μ 2 + σ 2 ) 2 cov( X i , X 2 i ) = E ( X 3 i ) - μ ( μ 2 + σ 2 ) It follows that: n ¯ X n - μ ¯ Y n - ( μ 2 + σ 2 ) d → N 0 0 , σ 2 E X 3 i - μ ( μ 2 + σ 2 ) E X 3 i - μ ( μ 2 + σ 2 ) E X 4 i - ( μ 2 + σ 2 ) 2 We now notice that s 2 n = ¯ Y n - ( ¯ X n ) 2 , so: ¯ X n s 2 n = g ¯ X n ¯ Y n = ¯ X n ¯ Y n - ( ¯ X n ) 2 To apply the delta method, we compute: g ¯ z 1 ¯ z 2 = 1 0 - 2 z 1 1 Denoting z = μ μ 2 + σ 2 , η 3 = E X 3 i - μ ( μ 2 + σ 2 ) and η 4 = E X 4 i - ( μ 2 + σ 2 ) 2 we have, from the delta method: n ¯ X n ¯ s n - g ( z ) d → N 0 0 , g ( z ) σ 2 η 3 η 3 η 4 g ( z ) T d → N 0 0 , σ 2 η 3 - 2 μσ 2 η 3 - 2 μσ 2 4 σ 2 μ 2 - 4 μη 3 + η 4 1
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Finally, notice that: μ 3 = E ( X i - μ ) 3 = E X 3 i - 3 μσ 2 - μ 3 = η 3 - 2 μσ 2 μ 4 = E ( X i - μ
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