Cornell University Department of Economics Econ 60 - Spring 004 Instructor: Prof. Kiefer Solution to Problem set # 5. Autocorrelation function is defined as ρ h = γ h γ 0 where γ h =Cov X t,x t h =E[X t E X t X t h E X t h for h =0,,,. ahis is an AR process, and it was shown in lecture 3 that α h γ h =σε for h =,,. α γ 0 = σε α ρ h = [ whenh =0 α h when h =,,... b his is an MA process. From lecture 3 we have, γ h = σ ε θ + θ θ whenh = σεθ when h = 0whenh 3 γ 0 = σε [ +θ + θ θ +θ θ when h = +θ +θ θ ρ h = when h = +θ +θ 0whenh 3 c Let s calculate the variance of this ARMA, process. ote that the mean of the process is 0. hen, γ 0 = VarZ t =E Zt = E [ρz t + ε t + θε t = E [ ρ + ε t + θ ε t +ρz t ε t +ρθz t ε t +θε t ε t = ρ E Zt + σ ε + θ σε +ρθe Z t ε t = ρ γ 0 + σε + θ σε +ρθe [ε t ρz t + ε t + θε t = ρ γ 0 + σε + θ σε +ρθσ ε
herefore, [ γ 0 = σ ε +ρθ + θ ρ ote that the process is stationary. We will use the fact that Cov X t,x t h = Cov X s,x s h for all s, t, and h since the process is stationary. since E Z t Z t =ρe Z t Z t +Eε t Z t +Eθε t Z t γ = ργ 0 + θσε E ε t Z t =0 E θε t Z t =θe [ε t ρz t + ε t + θε t = θσε [ [ σ γ = ρ ε +ρθ + θ ρ + θσε For h, we have since E Z t Z t h =ρe Z t Z t h +Eε t Z t h +Eθε t Z t h γ h =ργ h E ε t Z t h =0 E θε t Z t h =0whenh Solving the first order homogenous difference equation, we have γ h =ρ h γ for h herefore, we have θ ρ + [+ρθ+θ when h = ρ ρ h = ρ h {ρ +. a he OLS estimator is given by θ [+ρθ+θ ρ } when h β = β +Z Z Z u where Z t = [ X t Y t
hen, β = β + Z Z Z u Z = β + Q u where Q = Z Z. ote that Z u = t= u t t= X tu t t= Y t u t We now apply a proper version of WLL to conclude that n u t = E u t =0 t= n X t u t = E X t u t =X t E u t =0 t= n Y t u t = E Y t u t =E[β + β X t + β 3 Y t + u t u t t= = β E u t +β X t E u t +β 3 E Y t u t +Eu t u t = β 3 E β + β X t + β 3 Y t 3 + u t u t +Eu t u t = β3e Y t 3 u t +β 3 E u t u t +Eu t u t = β3 3 E Y t 4u t +β3 E u t 3u t +β 3 E u t u t +Eu t u t = β j 3 E u t j u t ow, recall that t= n Y t u t = ρ j E u t j u t =σε ρ β j 3 σε ρ j = ρσ ε ρ ρβ 3 as long as ρβ 3 <. herefore, β = β β + Q β 3 3 ρ = ρσ ε ρ 0 0 σ ε ρ ρ ρβ 3 ρβ 3 j β β β 3
he presence of lagged dependent variable on the right hand side makes all coefficients inconsistent when the error terms show some serial correlation. b How do we cure the problem? he rescue comes from the instrumental variables estimatoriv. If we can find a variable w t which is closely correlated with Y t but not with u t, we can obtain a consistent estimator of β through β IV =W Z W y where W t = [ X t w t IV estimator is consistent since β IV = β +W Z W u hen, β IV = β + W Z W u = β +Π 0 = β where W Z 3. a = Π and W u = 0 by a proper choice of instrument. s = e e k = ε Mε k where M = [I X X X X. hen, E s = E ow, consider ε Mε k E ε Mε=E [tr ε Mε since ε Mε is a scalar = E [tr Mεε since tr AB =tr BA = tr [MEεε = tr [ σ MΩ since E εε =σ Ω [ = σ tr I X X X X Ω [ = σ tr Ω tr X X X X Ω = σ tr Ω σ tr X X X ΩX = σ σ tr X X X ΩX = k E ε Mε. 4
E s = σ σ tr X X X ΩX k k = σ k σ tr [ X X k k X ΩX k b It is straight forward to get lim E s σ = lim k lim = σ σ tr [ Q L lim k = σ c First of all, we will expand s as; s = hen, e e k = ε Mε k = [ σ X tr X k k ε ε k ε X X X X ε = k k X ΩX k s = [ ε ε k ε X X X X ε = ε ε ε X X X X ε = ε ε ε X X X = ε ε 0 Q 0 = ε ε On the other hand, ε i = k k i= herefore, consistency requires that ε ε = σ ε i = i= [ ε ε ε X X X X ε X ε i= ε i = ε ε 5
4. GLS for SURE system is β GLS = σ X X σ X X σ X X σ X X σ m X X m σ m X X m σ m X mx σ m X mx σ mm X mx m m σ j X y j m σ j X y j m σ mj X m y j If σij =0fori j, then β GLS = = σ X X 0 0 0 σ X X 0 0 0 σ mm X m X m X X X y X X X y X mx m X my m = β OLS i.e., GLS is equivalent to OLS for individual equations. If X = X = = X m = X m, β GLS = X V X X V y [ I = X V I I X I X V I y = V X X I X V I y [ = V X X I X V I y [ = I X X X y = β OLS recall the product and inverse rules of Kronecker product. a LetBL= 0.6+L 0.6L+0.5L. hen, we can write BL as : BL =α 0 + α L + α L + α 3 L 3 +... Multiply both sides by 0.6L +0.5L,so σ X y σ X y σ mm X m y m α 0 + α L + α L + α 3 L 3 +... 0.6L +0.5L =0.6+L 6
Since both sides of the equation are equal, it follows that, α 0 =0.6 α 0.6α 0 = α j 0.6α j +0.5α j =0 forj =, 3,... For the specified coefficients, we can find recursively that α 0 =0.6 α =.36 α =.66 α 3 = 0.404 α 4 = 0.885 b Just do the following: multiply both sides by the denominator δl ηl y t = α δ η+β + γlx t + δl ηl ε t y t = δy t + ηy t + α δ η+βx t + γx t + ε t δε t ηε t his model cannot be estimated consistently using OLS because lags of the dependent variable appear on the LHS, i.e. as regressors, and these will be correlated with the errors. o obtain consistent estimators we need to use the instrumental variable method, and here we could use lags of x to estimate the lagged dependent variable and use the predictors in the original regression. 7