stat210a_2007_hw8_solutions

stat210a_2007_hw8_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 8- Solutions Fall 2007 Issued: Thursday, October 25 Due: Thursday, December 1 Problem 8.1 (a) For a Bernoulli random variable, we have: E θ X = θ var θ X = θ (1 - θ ) So, for an i.i.d. sequence of Bernoulli random variables: n ( ¯ X - θ ) d → N (0 (1 - θ )) Applying the delta method with a g function yields: n ( g ( ¯ X ) - g ( θ ) ) d → N 0 , £ g 0 ( θ ) / 2 θ (1 - θ ) · So, we want to determine g ( . ) such that, for some constant K > 0: £ g 0 ( θ ) / 2 θ (1 - θ ) = K 2 That is: g 0 ( θ ) = K p θ (1 - θ ) Hence: g ( θ ) = K Z 1 p t (1 - t ) dt Letting w = t , dw = - 1 2 t dt so: g ( θ ) = 2 K Z 1 p (1 - w 2 ) dw = 2 K arcsin( θ ) + K 1 1
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(b) By CLT, n 1 n n i =1 X i Y i - E ( XY ) 1 n n i =1 X 2 i - E ( X 2 ) 1 n n i =1 Y 2 i - E ( Y 2 ) d → N 0 0 0 , Σ Σ = 1 + ρ 2 2 ρ 2 ρ 2 ρ 2 2 ρ 2 2 ρ 2 ρ 2 2 = V ar ( XY ) Cov ( XY,X 2 ) Cov ( XY,Y 2 ) Cov ( X 2 ,XY ) V ar ( X 2 ) Cov ( X 2 ,Y 2 ) Cov ( Y 2 ,XY ) Cov ( Y 2 ,X 2 ) V ar ( Y 2 ) Let h ( x,y,z ) = x yz and use delta method. Then, n n i =1 X i Y i q n i =1 X 2 i n i =1 Y 2 i - ρ d → N ( 0 , (1 - ρ 2 ) 2 ) g 0 ( ρ ) = K 1 - ρ 2 = K 2(1+ ρ ) + K 2(1 - ρ ) Thus, g ( ρ ) = K 2 log 1+ ρ 1 - ρ + K 1 for K > 0. Problem 8.2 (a) Note that n ˆ 1 n n X i =1 X 2 i - 1 - θ 2 ! d → N ( 0 ,V ar ( X 2 1 - 1 - θ 2 ) ) n ( ¯ X n - θ ) d → N (0 , 1)) n ± ( ¯ X n ) 2 - θ 2 - 1 n d → N ( 0 , 4 θ 2 ) ( By Delta and slutsky Method ) Because E [( X i - θ ) 4 ] = μ 4 < , E [( X i - θ ) 3 ] = μ 3 < . V ar
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stat210a_2007_hw8_solutions - UC Berkeley Department of...

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