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appendixc_s - APPENDIX C SOLUTIONS TO PROBLEMS C.1(i This...

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APPENDIX C SOLUTIONS TO PROBLEMS C.1 (i) This is just a special case of what we covered in the text, with n = 4: E( Y ) = μ and Var( Y ) = σ 2 /4. (ii) E( W ) = E( Y 1 )/8 + E( Y 2 )/8 + E( Y 3 )/4 + E( Y 4 )/2 = μ [(1/8) + (1/8) + (1/4) + (1/2)] = μ (1 + 1 + 2 + 4)/8 = μ , which shows that W is unbiased. Because the Y i are independent, Var( W ) = Var( Y 1 )/64 + Var( Y 2 )/64 + Var( Y 3 )/16 + Var( Y 4 )/4 = σ 2 [(1/64) + (1/64) + (4/64) + (16/64)] = σ 2 (22/64) = σ 2 (11/32). (iii) Because 11/32 > 8/32 = 1/4, Var( W ) > Var( Y ) for any σ 2 > 0, so Y is preferred to W because each is unbiased. C.3 (i) E( W 1 ) = [( n – 1)/ n ]E( Y ) = [( n – 1)/ n ] μ , and so Bias( W 1 ) = [( n – 1)/ n ] μ μ = – μ / n . Similarly, E( W 2 ) = E( Y )/2 = μ /2, and so Bias( W 2 ) = μ /2 – μ = – μ /2. The bias in W 1 tends to zero as n , while the bias in W 2 is – μ /2 for all n . This is an important difference. (ii) plim( W 1 ) = plim[( n – 1)/ n ] plim( Y ) = 1 μ = μ . plim( W 2 ) = plim( Y )/2 = μ /2. Because plim( W 1 ) = μ and plim( W 2 ) = μ /2, W 1 is consistent whereas W 2 is inconsistent. (iii) Var( W 1 ) = [( n – 1)/ n ] 2 Var( Y ) = [( n – 1) 2 / n 3 ] σ 2 and Var( W 2 ) = Var( Y )/4 = σ 2 /(4 n ).
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