# Problem Set 1-Solution.pdf - Introductory Econometrics...

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Introductory Econometrics - Problem Set I (Suggested Solutions) 1 . (a) By definition, μ Y = E [ Y ] = Z 1 0 ydy = 1 2 and E [ Y 2 ] = Z 1 0 y 2 dy = 1 3 . So σ 2 Y = E [ Y 2 ] - μ 2 Y = 1 12 . (b) By definition, E [ Y k ] = Z - y k 1 2 π e - y 2 2 dy , which is equal to zero for k = 1, 3, 5, 7, · · · , because the integrand y k 1 2 π e - y 2 2 is an odd function. (c) Let X N ( 0, 1 ) and Y = X 2 . Then Y is a quadratic function of X , but Cov ( X , Y ) = E [ XY ] - E [ X ] E [ Y ] = E [ X 3 ] - E [ X ] E [ X 2 ] = 0 - 0 × 1 = 0 , where E [ X 3 ] = E [ X ] = 0 by (b) and E [ X 2 ] = 1. 2 . Stock and Watson, Exercise 2 . 6 (a) The table shows P ( Y = 0 ) = 0.068 and P ( Y = 1 ) = 0.932. So E [ Y ] = 0 × P ( Y = 0 ) + 1 × P ( Y = 1 ) = P ( Y = 1 ) = 0.932 . (b) Unemployment rate = #unemployed #labor force = P ( Y = 0 ) = 1 - P ( Y = 1 ) = 1 - E [ Y ] . (c) Calculate the conditional probabilities first P ( Y = 0 | X = 0 ) = P ( X = 0, Y = 0 ) P ( X = 0 ) = 0.053 0.639 = 0.083 P ( Y = 1 | X = 0 ) = P ( X = 0, Y = 1 ) P ( X = 0 ) = 0.586 0.639 = 0.917 P ( Y = 0 | X = 1 ) = P ( X = 1, Y = 0 ) P ( X = 1 ) = 0.015 0.361 = 0.042 P ( Y = 1 | X = 1 ) = P ( X = 1, Y = 1 ) P ( X = 1 ) = 0.346 0.361 = 0.958. 1

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The conditional expectations are E ( Y | X = 0 ) = 0 × P ( Y = 0 | X = 0 ) + 1 × P ( Y = 1 | X = 0 ) = P ( Y = 1 | X = 0 ) = 0.917 E ( Y | X = 1 ) = 0 × P ( Y = 0 | X = 1 ) + 1 × P ( Y = 1 | X = 1 ) = P ( Y = 1 | X = 1 ) = 0.958 (d) Similar to (b), Unemployment rate for non-college graduates = 1 - E [ Y | X = 0 ] = P ( Y = 0 | X = 0 ) = 0.083 Unemployment rate for college graduates = 1 - E [ Y | X = 1 ] = P ( Y = 0 | X = 1
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