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# chapter10 - 1 Both estimators are unbiased Now V(X 1 = 2/2n...

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1 Chapter 10 10–1. Both estimators are unbiased. Now, V ( X 1 ) = σ 2 / 2 n while V ( X 2 ) = σ 2 /n . Since V ( X 1 ) < V ( X 2 ), X 1 is a more efficient estimator than X 2 . 10–2. E ( ˆ θ 1 ) = μ , E ( ˆ θ 2 ) = (1 / 2) E (2 X 1 - X 6 + X 4 ) = (1 / 2)(2 μ - μ + μ ) = μ . Both estimators are unbiased. V ( ˆ θ 1 ) = σ 2 / 7 , V ( ˆ θ 2 ) = 1 2 2 V (2 X 1 - X 6 + X 4 ) = 1 4 [4 V ( X 1 ) + V ( X 6 ) + V ( X 4 )] = 1 4 6 σ 2 = 3 σ 2 / 2 ˆ θ 1 has a smaller variance than ˆ θ 2 . 10–3. Since ˆ θ 1 is unbiased, MSE ( ˆ θ 1 ) = V ( ˆ θ 1 ) = 10. MSE ( ˆ θ 2 ) = V ( ˆ θ 2 ) + ( Bias ) 2 = 4 + ( θ - θ/ 2) 2 = 4 + θ 2 / 4 . If θ < 24 = 4 . 8990, ˆ θ 2 is a better estimator of θ than ˆ θ 1 , because it would have smaller MSE . 10–4. MSE ( ˆ θ 1 ) = V ( ˆ θ 1 ) = 12, MSE ( ˆ θ 2 ) = V ( ˆ θ 2 ) = 10, MSE ( ˆ θ 3 ) = E ( ˆ θ 3 - θ ) 2 = 6 . ˆ θ 3 is a better estimator because it has smaller MSE . 10–5. E ( S 2 ) = (1 / 24) E (10 S 2 1 + 8 S 2 2 + 6 S 2 3 ) = (1 / 24)(10 σ 2 + 8 σ 2 + 6 σ 2 ) = (1 / 24)24 σ 2 = σ 2 10–6. Any linear estimator of μ is of the form ˆ θ = n i =1 a i X i where a i are constants. ˆ θ is an unbiased estimator of μ only if E ( ˆ θ ) = μ , which implies that n i =1 a i = 1. Now V ( ˆ θ ) = n i =1 a 2 i σ 2 . Thus we must choose the a i to minimize V ( ˆ θ ) subject to the constraint a i = 1. Let λ be a Lagrange multiplier. Then F ( a i , λ ) = n X i =1 a 2 i σ 2 - λ ˆ n X i =1 a i - 1 ! and ∂F/∂a i = ∂F/∂λ = 0 gives 2 a i σ 2 - λ = 0; i = 1 , 2 , . . . , a n X i =1 a i = 1 The solution is a i = 1 /n . Thus ˆ θ = X is the best linear unbiased estimator of μ .

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2 10–7. L ( α ) = n Y i =1 α X i e - α /X i ! = α Σ X i e - , n Y i =1 X i ! n L ( α ) = n X i =1 X i n α - - n ˆ n Y i =1 X i ! ! d ‘ n L ( α ) = n X i =1 X i - n = 0 ˆ α = n X i =1 X i /n = X 10–8. For the Poisson distribution, E ( X ) = α = μ 0 1 . Also, M 0 1 = X . Thus ˆ α = X is the moment estimator of α . 10–9. L ( λ ) = n Y i =1 λe - λt i = λ n e - λ n i =1 t i n L ( λ ) = n ‘ n λ - λ n X i =1 t i d ‘ n L ( λ ) = ( n/λ ) - n X i =1 t i = 0 ˆ λ = n , n X i =1 t i = ( t ) - 1 10–10. E ( t ) = 1 = μ 0 1 , and M 0 1 = t . Thus 1 = t or ˆ λ = ( t ) - 1 . 10–11. If X is a gamma random variable, then E ( X ) = r/λ and V ( X ) = r/λ 2 . Thus E ( X 2 ) = ( r + r 2 ) λ 2 . Now M 0 1 = X and M 0 2 = (1 /n ) n i =1 X 2 i . Equating moments, we obtain r/λ = X, ( r + r 2 ) λ 2 = (1 /n ) n X i =1 X 2 i or, ˆ λ = X ," (1 /n ) n X i =1 X 2 i - X 2 # ˆ r = X 2 ," (1 /n ) n X i =1 X 2 i - X 2 #
3 10–12. E ( X ) = 1 /p , M 0 1 = X . Thus 1 /p = X or ˆ p = 1 / X . 10–13. L ( p ) = n Y i =1 (1 - p ) X i - 1 p = p n (1 - p ) Σ X i - n n L ( p ) = n ‘ n p + ( n i =1 X i - n ) n (1 - p ). From d ‘ n L ( p ) /dρ = 0, we obtain ( n/ ˆ p ) - ˆ n X i =1 X i - n !, (1 - ˆ ρ ) = 0 ˆ p = n , n X i =1 X i = 1 / X 10–14. E ( X ) = p , M 0 1 = X . Thus ˆ p = X . 10–15. E ( X ) = np ( n is known), M 0 1 = X N ( X is based on a sample of N observations.) Thus np = X N or ˆ p = X N /n . 10–16. E ( X ) = np , V ( X ) = np (1 - p ), E ( X 2 ) = np - np 2 + n 2 p 2 M 0 1 = X , M 0 2 = (1 /N ) N i =1 X 2 i . Equating moments, np = X, np - np 2 + n 2 p 2 = (1 /N ) N X i =1 X 2 i ˆ n = X 2 ," X - (1 /N ) N X i =1 ( X i - X ) 2 # , ˆ p = X/ ˆ N 10–17. L ( p ) = N Y i =1 n X i p X i (1 - p ) n - X i = " N Y i =1 n X i # p n i =1 X i (1 - p ) nN - N i =1 X i n L ( p ) = N X i =1 n n X i + ˆ N X i =1 X i !

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