HW4_2006 - 4-2 There is a typo in this problem = 1[3 X X 2...

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4-2. There is a typo in this problem, ( ) [ ] 4 6 1 2 2 3 2 1 ˆ X X X + = θ E ( ) [ ] μ μ θ = = = + + + = ) 9 ( 9 1 )) ( 9 ( 9 1 ) ( ) ( ) ( 9 1 ˆ 9 2 1 1 X E X E X E X E L E ( ) [ ] μ μ μ μ θ = + = + = ] 2 3 [ 2 1 ) 2 ( ) ( ) 3 ( 2 1 ˆ 4 6 1 2 X E X E X E a) Both $ θ 1 and $ θ 2 are unbiased estimates of μ since the expected values of these statistics are equivalent to the true mean, μ . b) V ( ) ( ) 2 2 9 2 1 2 9 2 1 1 9 1 ) 9 ( 81 1 ) X ( V ) X ( V ) X ( V 9 1 9 X ... X X V ˆ σ = σ = + + + = + + + = θ L 9 ) ˆ ( V 2 1 σ = θ V ( ) ( ) )) ( 4 ) ( ) ( 9 ( 4 1 ) 2 ( ) ( ) 3 ( 2 1 2 2 3 ˆ 4 6 1 4 6 1 2 4 6 1 2 X V X V X V X V X V X V X X X V + + = + + = + = θ = ( ) 2 2 2 4 9 4 1 σ + σ + σ = ) 14 ( 4 1 2 σ 2 7 ) ˆ ( V 2 2 σ = θ Since both estimators are unbiased, the variances can be compared to decide which is the better estimator. The variance of $ θ 1 is smaller than that of $ θ 2 , $ θ 1 is the better estimator. 4-10. Show that X 2 is a biased estimator of μ . i.e, show bias = 0 bias = ( ) 2 2 μ X E Let Y be random varisble, then Using ( ) [ ] 2 2 ) ( ) ( Y E Y V Y E + = (see pg. 58) here we have Y= X so ( ) [ ] 2 2 ) ( ) ( X E X V X E + = X is a linear combination of i X i.e.,
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