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Wooldridge IE AISE SSM ch14

# Wooldridge IE AISE SSM ch14 - CHAPTER 14 SOLUTIONS TO...

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This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 78 CHAPTER 14 SOLUTIONS TO PROBLEMS 14.1 First, for each t > 1, Var( Δ u it ) = Var( u it u i,t- 1 ) = Var( u it ) + Var( u i,t- 1 ) = 2 2 u σ , where we use the assumptions of no serial correlation in { u t } and constant variance. Next, we find the covariance between Δ u it and Δ u i,t+ 1 . Because these each have a zero mean, the covariance is E( Δ u it ⋅Δ u i,t+ 1 ) = E[( u it u i,t -1 )( u i,t+ 1 u it )] = E( u it u i,t+ 1 ) – E( 2 it u ) – E( u i,t- 1 u i,t+ 1 ) + E( u i,t- 1 u it ) = E( 2 it u ) = 2 u because of the no serial correlation assumption. Because the variance is constant across t , by Problem 11.1, Corr( Δ u it , Δ u i,t+ 1 ) = Cov( Δ u it , Δ u i,t+ 1 )/Var( u it ) = 22 /(2 ) uu = .5. 14.3 (i) E( e it ) = E( v it i v λ ) = E( v it ) E( i v ) = 0 because E( v it ) = 0 for all t . (ii) Var( v it i v ) = Var( v it ) + 2 Var( i v ) 2 Cov( v it , i v ) = 2 v + 2 E( 2 i v ) 2 E( v it i v ). Now, 2 2 E( ) vi t a u v σσ == + and E( v it i v ) = 1 1 () T it is s TE v v = = 1 T [ 2 a + 2 a + + ( 2 a + 2 u ) + + 2 a ] = 2 a + 2 u / T . Therefore, E( 2 i v ) = 1 1 T it i t v v = = 2 a + 2 u / T . Now, we can collect terms: Var( v it i v ) = 2 ( / ) 2 ( / ) au TT ++ + + . Now, it is convenient to write = 1 / η γ , where 2 u / T and 2 a + 2 u / T . Then Var( v it i v ) = ( 2 a + 2 u ) 2 ( 2 a + 2 u / T ) + 2 ( 2 a + 2 u / T ) = ( 2 a + 2 u ) 2(1 / ) + (1 / ) 2 = ( 2 a + 2 u ) 2 + 2 + (1 2 / + / ) = ( 2 a + 2 u ) 2 + 2 + (1 2 / + / ) = ( 2 a + 2 u ) 2 + 2 + 2 + = ( 2 a + 2 u ) + = 2 u . This is what we wanted to show.

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This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 79 (iii) We must show that E( e it e is ) = 0 for t s . Now E( e it e is ) = E[( v it i v λ )( v is i v )] = E( v it v is ) E( i vv is ) E( v it i v ) + 2 E( 2 i v ) = 2 a σ 2 ( 2 a + 2 u / T ) + 2 E( 2 i v ) = 2 a 2 ( 2 a + 2 u / T ) + λ 2 ( 2 a + 2 u / T ). The rest of the proof is very similar to part (ii): E( e it e is ) = 2 a 2 λ ( 2 a + 2 u / T ) + λ 2 ( 2 a + 2 u / T ) = 2 a 2(1 / η γ ) + (1 / ) 2 = 2 a 2 + 2 + (1 2 / + / ) = 2 a 2 + 2 + (1 2 / + η / γ ) γ = 2 a 2 + 2 + 2 + = 2 a + = 0. 14.5 (i) For each student we have several measures of performance, typically three or four, the number of classes taken by a student that have final exams. When we specify an equation for each standardized final exam score, the errors in the different equations for the same student are certain to be correlated: students who have more (unobserved) ability tend to do better on all tests. (ii) An unobserved effects model is score sc = θ c + β 1 atndrte sc + 2 major sc + 3 SAT s + 4 cumGPA s + a s + u sc , where a s is the unobserved student effect. Because SAT score and cumulative GPA depend only on the student, and not on the particular class he/she is taking, these do not have a c subscript.
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Wooldridge IE AISE SSM ch14 - CHAPTER 14 SOLUTIONS TO...

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