ISM_chapter8 - Chapter 8: Estimation 8.1 Let B = B() ....

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158 Chapter 8: Estimation 8.1 Let ) ˆ ( θ = B B . Then, [] [ ] ( ) ) 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 = . 8.2 a. The estimator θ ˆ is unbiased if E ( θ ˆ ) = θ . Thus, B ( θ ˆ ) = 0. b. E ( θ ˆ ) = θ + 5. 8.3 a. Using Definition 8.3, B ( θ ˆ ) = a θ + b θ = ( a – 1) θ + b . b. Let a b / ) ˆ ( ˆ * θ = θ . 8.4 a. They are equal. b. ) ˆ ( ) ˆ ( θ > θ V MSE . 8.5 a. Note that θ = θ ) ˆ ( * E and 2 * / ) ˆ ( ] / ) ˆ [( ) ˆ ( a V a b V V θ = θ = θ . Then, 2 * * / ) ˆ ( ) ˆ ( ) ˆ ( MSE a V V θ = θ = θ . b. Note that 2 ] ) 1 [( ) ˆ ( ) ˆ ( ) ˆ ( ) ˆ ( MSE b a V B V + θ + θ = θ + θ = θ . A sufficiently large value of a will force ) ˆ ( MSE ) ˆ ( MSE * θ < θ . Example: a = 10. c. A amply small value of a will make ) ˆ ( MSE ) ˆ ( MSE * θ > θ . Example: a = .5, b = 0. 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 c c a 2 2 2 2 1 2 2 σ + σ σ = .
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Chapter 8: Estimation 159 Instructor’s Solutions Manual 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 . From Ex. 6.81 it was shown that 4 ˆ θ has an exponential distribution with mean θ /3, so this estimator is biased. b. It is easily shown that V ( 1 ˆ θ ) = θ 2 , V ( 2 ˆ θ ) = θ 2 /2, V ( 3 ˆ θ ) = 5 θ 2 /9, and V ( 5 ˆ θ ) = θ 2 /9, so the estimator 5 ˆ θ is unbiased and has the smallest variance. 8.9 The density is in the form of the exponential with mean θ + 1. We know that Y is unbiased for the mean θ + 1, so an unbiased estimator for θ is simply Y – 1. 8.10 a. For the Poisson distribution, E ( Y ) = λ and so for the random sample, E ( Y ) = λ . Thus, the estimator Y = λ ˆ is unbiased. b. The result follows from E ( Y ) = λ and E ( Y 2 ) = V ( Y ) + λ 2 = 2 λ 2 , so E ( C ) = 4 λ + λ 2 . c. Since E ( Y ) = λ and E ( 2 Y ) = V ( Y ) + [ E ( Y )] 2 = λ 2 / n + λ 2 = ) / 1 1 ( 2 n + λ . Then, we can construct an unbiased estimator ) / 1 4 ( ˆ 2 n Y Y + = θ . 8.11 The third central moment is defined as 54 ) ( 9 ) ( ] ) 3 [( ] ) [( 2 3 3 3 + = = μ Y E Y E Y E Y E . Using the unbiased estimates 2 ˆ θ and 3 ˆ θ , it can easily be shown that 3 ˆ θ – 9 2 ˆ θ + 54 is an unbiased estimator. 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. c. Note that ) 12 /( 1 ) ( n Y V = so 25 . ) 12 /( 1 ) ( MSE + = n Y .
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ISM_chapter8 - Chapter 8: Estimation 8.1 Let B = B() ....

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