This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem Set 4 Answer Key  Econometrics 120A 1. (a) Since X = 1 if Y a and X = 0 otherwise, we have: E [ X ] = P ( X = 1) 1 + P ( X = 0) 0 = P ( X = 1) = P ( Y a ) Similarly, using that E [ X ] = P ( Y a ), and V ar ( X ) = E [ X 2 ] ( E [ X ]) 2 , the variance of X can be found by: E [ X 2 ] ( E [ X ]) 2 = P ( X = 1) 1 2 + P ( X = 0) 2 P 2 ( Y a ) = P ( Y a )(1 P ( Y a )) (b) Since E [ X ] = P ( Y a ), a natural estimator for P ( Y a ) is X . (c) Because X is a sample mean, we know its variance is given by: V ar ( X ) = V ar ( X ) n = P ( Y a )(1 P ( Y a )) n where the last equality follows from our answer to part b). Since X is a natural estimator for P ( Y a ), a good estimator for the variance of X is given by replacing X for P ( Y a ), so we get X (1 X ) /n . Notice that this is problem is actually identical to our discussion for estimating proportions. (d) Our estimator X is approximately normally distributed since it is a sample mean of identically distributed random variables. (e) Combining b), c) and d) we have: X approx N P ( Y a ) , P ( Y a )(1 P ( Y a )) n Therefore, if we let z 1 / 2 solve P ( Z z 1 / 2 ) = / 2 and we Zscore X we get: P  z 1 / 2 ( X P ( Y a ))...
View Full
Document
 Spring '10
 M.Abajian
 Econometrics

Click to edit the document details