p2_ak - Econ 410 - Fall 2006 Practice Questions for the...

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

View Full Document Right Arrow Icon
Econ 410 - Fall 2006 Practice Questions for the Second Midterm - Answer Key 1. (a) e β 1 is a linear estimator, because we can write it as a linear function of the Y 0 i s : e β 1 = n P i =1 ¡ d i d ¢¡ Y i Y ¢ = n P i =1 ¡ d i d ¢ Y i Y n P i =1 ¡ d i d ¢ = but n P i =1 ¡ d i d ¢ =0 ,so : e β 1 = n P i =1 ¡ d i d ¢ Y i so a i = d i d (b) Since Y i = β 0 + β 1 X i + u i ,wehave : e β 1 = n P i =1 ¡ d i d ¢ Y i = n P i =1 ¡ d i d ¢ ( β 0 + β 1 X i + u i )= = β 0 n P i =1 ¡ d i d ¢ + β 1 n P i =1 ¡ d i d ¢ X i + n P i =1 ¡ d i d ¢ u i But n P i =1 ¡ d i d ¢ =0 So: e β 1 = β 1 n P i =1 ¡ d i d ¢ X i + n P i =1 ¡ d i d ¢ u i (c) e β 1 is conditionally unbiased if: E ³ e β 1 | X 1 , ..., X n ´ = β 1 . Because the least squares assumptions hold, we know: E ( u i | X 1 , ..., X n )= E ( u i | X i )=0 , so we have: E ³ e β 1 | X 1 , ..., X n ´ = β 1 n P i =1 £ d i d ¤ X i which is equal to β 1 if n P i =1 ¡ d i d ¢ X i =1 (d) When the three least squares assumptions hold and the error term is homoskedastic, the Gauss-Markov Theorem says that the OLS es- timator b β
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

p2_ak - Econ 410 - Fall 2006 Practice Questions for the...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online