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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 α/ 2 solve P ( Z ≥ z 1 α/ 2 ) = α/ 2 and we Zscore ¯ X we get: P ˆ z 1 α/ 2 ≤ ( ¯ X P ( Y ≤ a ))...
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This note was uploaded on 12/05/2010 for the course ECON 120A taught by Professor M.abajian during the Spring '10 term at San Diego.
 Spring '10
 M.Abajian
 Econometrics

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