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164CHAPTER 6. FOREIGN EXCHANGE MARKET EFFICIENCYneed not be zero. You can’t say that E(² t ² t−2 )iszeroeither.Youcan,however, say that E(² t ² t−k ) = 0 for k ≥ 3. When the forecast horizonof the forward exchange rate is 3 sampling periods, the error term ispotentially correlated with 2 lags of itself and follows an MA(2) process.If you work with a k − period forward rate, you must be preparedfor the error term to follow an MA(k-1) process.Generalized least squares procedures, such as Cochrane-Orcutt orHildreth-Lu, covered in elementary econometrics texts cannot be usedto handle these serially correlated errors because these estimators areinconsistent if the regressors are not econometrically exogenous. Researchersusually follow Hansen and Hodrick by estimating the coefficientvector by least squares and then calculating the asymptotic covariancematrix assuming that the regression error follows a movingaverage process. Least squares is consistent because the regression error² t , being a rational expectations forecast error under the null, isuncorrelated with the regressors, z t−3 . 1Hansen-Hodrick Regression Tests of UIPHansen and Hodrick ran two sets of regressions. In the Þrst set, theindependent variables were the lagged forward exchange rate forecasterrors (s t−3 −f t−6,3 ) of the own currency plus those of cross rates. In thesecond set, the independent variables were the own forward premiumand those of cross rates (s t−3 −f t−3,3 ). They rejected the null hypothesisat very small signiÞcance levels.Let’s run their second set of regressions using the dollar, pound,1 To compute the asymptotic covariance matrix of the least-squares vector,follow the GMM interpretation of least squares developed in chapter 2.2. Assumethat ² t is conditionally homoskedastic, and let w t = z t−3 ² t . We haveE(w t wt)=E(² 0 2 t z t−3 z 0 t−3 )=E(E[²2 t z t−3 z 0 t−3 |z t−3 ]) = γ 0E(z t−3 z 0 t−3 )=γ 0Q 0 ,whereγ 0 = E(² 2 t ) and Q = E(z t−3 z 0 t−3). Now, E(w t wt−1) 0 = E(² t ² t−1 z t−3 z 0 t−4) =E(E[² t ² t−1 z t−3 z 0 t−4 |z t−3 ,z t−4 ]) = E(z t−3 z0 t−4 E[² t² t−1 |z t−3 ,z t−4 ]) = γ 1 Q 1 , whereγ 1 =E(² t ² t−1 ), and Q 1 =E(z t−3 z t−4 ). By an analogous argument, E(w t wt−2) 0 =γ 2 Q 2 ,andE(w t wt−k 0 )=0, fork ≥ 3. Now, D =E(∂(z t ² t)/∂β 0 )=Q 0 so theasymptotic covariance matrix for the least squares estimator is, (Q 0 0 W−1 Q 0 ) −1where W = γ 0 Q 0 + P 2j=1 γ j(Q j + Q 0 j ). Actually, Hansen and Hodrick used weeklyobservations with the 3-month forward rate which leads the regression error tofollow an MA(11).