{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Wooldridge IE AISE SSM app c

# Wooldridge IE AISE SSM app c - APPENDIX C SOLUTIONS TO...

This preview shows pages 1–2. Sign up to view the full content.

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 120 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

Wooldridge IE AISE SSM app c - APPENDIX C SOLUTIONS TO...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online