chapter14_s

chapter14_s - CHAPTER 14 SOLUTIONS TO PROBLEMS 14.1 First...

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Unformatted text preview: 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( ) – E( u i,t- 1 u i,t+ 1 ) + E( u i,t- 1 u it ) = − E( ) = 2 it u 2 it u 2 u σ − 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 ) = 2 /(2 ) u 2 u σ σ − = − .5. 14.3 (i) E( e it ) = E( v it − i v λ ) = E( v it ) − λ E( i v ) = 0 because E( v it ) = 0 for all t . (ii) Var( v it − i v λ ) = Var( v it ) + λ 2 Var( i v ) − 2 λ ⋅ Cov( v it , i v ) = 2 v σ + λ 2 E( 2 i v ) − 2 λ E( v it ⋅ i v ). Now, 2 2 2 E( ) v i t a v 2 u σ σ σ = = + and E( v it i v ) = = 1 1 ( ) T it is s v v − = ∑ T E 1 T − [ 2 a σ + 2 a σ + … + ( 2 a σ + 2 u σ ) + + … 2 a σ ] = 2 a σ + 2 u σ / T . Therefore, E( 2 i v ) = 1 T t = ∑ 1 it i T E v − ( ) v = 2 a σ + 2 u σ / T . Now, we can collect terms: Var( v it − i v λ ) = . 2 2 2 2 2 2 2 ( ) ( / ) 2 ( / a u a u a u T T σ σ λ σ σ λ σ σ + + + − + ) Now, it is convenient to write λ = 1 − / η γ , where η ≡ 2 u σ / T and γ ≡ 2 a σ + 2 u σ / T . Then Var( v it − i v λ ) = ( 2 a σ + 2 u σ ) − 2 λ ( 2 a σ + 2 u σ / T ) + λ 2 ( 2 a σ + 2 u σ / T ) = ( 2 a σ + 2 u σ ) − 2(1 − / η γ ) γ + (1 − / η γ ) 2 γ = ( 2 a σ + 2 u σ ) − 2 γ + 2 η γ ⋅ + (1 − 2 / η γ + η / γ ) γ = ( 2 a σ + 2 u σ ) − 2 γ + 2 η γ ⋅ + (1 − 2 / η γ + η / γ ) γ = ( 2 a σ + 2 u σ ) − 2 γ + 2 η γ ⋅ + γ − 2 η γ ⋅ + η = ( 2 a σ + 2 u σ ) + η − γ = 2 u σ . This is what we wanted to show. 78 (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 λ )] = E( v it v is ) − λ E( i v v is ) − λ E( v it i v ) + λ 2 E( 2 i v ) = 2 a σ − 2 λ ( 2 a σ + 2 u σ / T ) + λ 2 E( 2 i v ) = 2 a σ − 2 λ ( 2 a σ + 2 u σ / T ) + λ 2 ( 2 a σ + 2 u σ / T ). The rest of the proof is very similar to part (ii): E( e it e is ) = 2 a σ − 2 λ ( 2 a σ + 2 u σ / T ) + λ 2 ( 2 a σ + 2 u σ / T ) = 2 a σ − 2(1 − / η γ ) γ + (1 − / η γ ) 2 γ = 2 a σ − 2 γ + 2 η γ ⋅ + (1 − 2 / η γ + η / γ ) γ = 2 a σ − 2 γ + 2 η γ ⋅ + (1 − 2 / η γ + η / γ ) γ = 2 a σ − 2 γ + 2 η γ ⋅ + γ − 2 η γ ⋅ + η = 2 a σ + η − γ = 0....
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chapter14_s - CHAPTER 14 SOLUTIONS TO PROBLEMS 14.1 First...

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