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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 Z-score X we get: P - z 1- / 2 ( X- P ( Y a ))...
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