Download Practice Questions for Final - Econometrics II | ECON 581 and more Exams Econometrics and Mathematical Economics in PDF only on Docsity! ECON 581 Startz/ Tsang Econometrics II Winter 2008 1 Answer to ECON581 Final - March 20, 2008 Question 1 – Warming Up [20 points] 1.1 ( ) 2ˆˆ 0xyxu x y x x y x xy xyxβ ⎛ ⎞ = − = − = − =⎜ ⎟⎜ ⎟ ⎝ ⎠ ∑∑ ∑ ∑ ∑ ∑∑ Saying that the RHS and the residual are orthogonal, which is just repeating the question, has 2 point. Otherwise 0 point. 1.2 ( ) ( ) ( ) ( )22 2 2ˆ ˆ ˆ ˆvar 0E E Eβ β β β β= − = − > Using the Jensen’s inequality is fine too. Otherwise 0 point. 1.3 a) Just do the algebra… ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 21 1 2 2 1 1 2 2 2 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 2 1 22 2 2 2 1 2 1 2 2 2 2 2 1 22 2 1 2 2 2 2 1 2 1var E E 2 1 A A A A x x u x x ux y x y x u x u x x x x x x x u x u x u x u x x u u x x x x x x x x x x β β β β β σ σ + + ++ + = = = + + + + ⎛ ⎞ ⎛ ⎞+ = = + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎛ ⎞ = +⎜ ⎟+⎝ ⎠ = + Following the same logic for ( )ˆvar β gives us ( ) 2 1 1 2 2 3 3 4 4 2 2 2 2 1 2 3 4 ˆvar E x u x u x u x u x x x x β ⎛ ⎞+ + + = ⎜ ⎟+ + +⎝ ⎠ Skipping writing out the cross terms and remembering that 1 3x x= and 2 4x x= , we have ECON 581 Startz/ Tsang Econometrics II Winter 2008 2 ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 2 2 1 1 2 2 1 3 2 42 2 1 2 2 2 2 2 2 2 2 1 2 1 22 2 1 2 2 2 2 2 1 2 1ˆvar E 2 1 2 1 4 A B A B x u x u x u x u x x x x x x x x x x β σ σ σ σ ⎛ ⎞ ⎜ ⎟= + + + ⎜ ⎟+⎝ ⎠ ⎛ ⎞ ⎜ ⎟= + + + ⎜ ⎟+⎝ ⎠ + = + b) It turns out that ( ) ( )ˆvar varAβ β< if 2 2 2 4 A B A σ σσ +< c) The GLS estimator is simply to divide each observation by it’s standard deviation. 3 31 1 2 2 4 4 2 2 2 2 22 2 2 31 2 4 2 2 2 2 A A B B GLS A A B B y xy x y x y x xx x x σ σ σ σβ σ σ σ σ + + + = + + + It’s from ECON481, and it’s an all-or-nothing question! Question 2 – MLE for a Nonlinear Regression [10 points] For simplicity, we treat 2σ as known. The log-likelihood function is ( ) ( ) ( )( ) 2 2 2 1 11ln , ln 2 ln 2 2 2 n i i i y xn nL α β α β π σ σ= ⎡ ⎤− + ⎢ ⎥= − − − ⎢ ⎥⎣ ⎦ ∑ Differentiating: ( ) ( )( )( )2 1 ln , 1 1 1 0 n i i i i L A y x xα α β α β β α σ = ∂ = = − + + = ∂ ∑ ( ) ( )( )2 1 ln , 1 1 0 n i i i i L A y x xβ α β α β α β σ = ∂ = = − + = ∂ ∑ With two equations and two unknowns, we can solve out α̂ , β̂ . Next we find the Hessian: ( ) ( ) 2 2 2 1 ln , 1 1 n i i L B xαα α β β α α σ = ∂ = = − + ∂ ∂ ∑ ( )2 2 2 2 1 ln , 1 n i i L B xββ α β α β β σ = ∂ = = − ∂ ∂ ∑ ( ) ( )( )2 1 ln , 1 1 0 n i i i i L A y x xβ α β α β α β σ = ∂ = = − + = ∂ ∑ ( ) ( )( ) 2 2 1 ln , 1 2 1 n i i i i i L B y x x xαβ α β α β β α σ = ∂ = = − + ∂ ∂ ∑
