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The second set of estimates resemble the OLS estimates slightly more than the third. As we discussed above, model (2) is, in fact, the right one. Unfortunately, one would be hard pressed to reach that as a firm conclusion based on just these results. Of course, the results of the tests in the previous exercise are much more convincing. 7. For the model in Exercise 1, suppose , is normally distributed with mean zero and variance σ2(1 + (γx)2). Show that σ2 and γ2 can be consistently estimated by a regression of the least squares residuals on a constant and x2.

For the model in Exercise 1, suppose , is normally distributed with mean zero and variance σ2(1 + (γx)2). Show that σ2 and γ2 can be consistently estimated by a regression of the least squares residuals on a constant and x2. Is this estimator efficient? The residuals from the least squares regression are ei = yi - y = α + εi - (α + ε ) = εi - ε . The expected value of the squared residual is E[ei2] = E[εi2] + E[ ε 2] - 2E[εi ε ] = σi2 + (1/n2)E[(Σiεi)2] - (2/n)E[εi(Σjεj)] = σi2 + (1/n2)ΣiE[εi2] - (2/n)E[εi2] since the disturbances are uncorrelated.

Consider the first difference, vt = εt - εt-1 which has Var[vt] = 2Var[εt] - 2Cov[(εt,εt-1)] = 2σu2[1/(1 - ρ2) - ρ/(1 - ρ2)] = 2σu2/(1 + ρ) and Cov[vt,vt-1] = 2Cov[εt,εt-1] - Var[εt] - Cov[εt,εt-1] = σu2[1/(1 - ρ2)][2ρ - 1 - ρ2] = σu2[(ρ - 1)/(1 + ρ)]. Therefore, the autocorrelation of the differenced process is Cov[vt,vt-1] / Var[vt] = (ρ - 1) / 2.

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