Homework 6 Solution - Economics 506-Sections M & M1...

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Solution to Homework 6 (The problem numbers refer to numbers in 7 th edition) 8.1 : Let ) ˆ ( θ = B B . Then, [ ] [ ] ( 29 [ ] ) ˆ ( ˆ 2 ) ( ) ˆ ( ˆ ) ) ˆ ( ˆ ( ) ˆ ( ) ˆ ( 2 2 2 2 θ - θ × + + θ - θ = + θ - θ = θ - θ = θ E E B B E E E B E E E MSE 2 ) ˆ ( B V + θ = . 8.6 : a. θ = θ - + θ = θ - + θ = θ ) 1 ( ) ˆ ( ) 1 ( ) ˆ ( ) ˆ ( 2 1 3 a a E a aE E . b. 2 2 2 1 2 2 2 1 2 3 ) 1 ( ) ˆ ( ) 1 ( ) ˆ ( ) ˆ ( σ - + σ = θ - + θ = θ a a V a V a V , since it was assumed that 1 ˆ θ and 2 ˆ θ are independent. To minimize ) ˆ ( 3 θ V , we can take the first derivative (with respect to a ), set it equal to zero, to find 2 2 2 1 2 2 σ + σ σ = a . (One should verify that the second derivative test shows that this is indeed a minimum.) 8.7 : Following Ex. 8.6 but with the condition that 1 ˆ θ and 2 ˆ θ are not independent, we find c a a a a V ) 1 ( 2 ) 1 ( ) ˆ ( 2 2 2 1 2 3 - + σ - + σ = θ . Using the same method w/ derivatives, the minimum is found to be 8.8 : a. Note that 1 ˆ θ , 2 ˆ θ , 3 ˆ θ and 5 ˆ θ are simple linear combinations of Y 1 , Y 2 , and Y 3 . So, it is easily shown that all four of these estimators are unbiased . b. It is easily shown that V ( 1 ˆ θ ) = θ 2 , V ( 2 ˆ θ ) = θ 2 /2, V ( 3 ˆ θ ) = 5θ 2 /9, and V ( 5 ˆ θ ) = θ 2 /9, so the estimator 5 ˆ θ is unbiased and has the smallest variance. 8.12 : a. For the uniform distribution given here, E ( Y i ) = θ + .5. Hence, E ( Y ) = θ + .5 so that B ( Y ) = .5. b. Based on , Y the unbiased estimator is Y – .5. 1
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Homework 6 Solution - Economics 506-Sections M & M1...

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