Solution. The differences are \(\Delta x = (1, 2, 0)\) and \(\Delta y = (1, 3, 0)\), so \(b_{\text {FD}} = (1 + 6 + 0)/(1 + 4 + 0) = 7/5 = 1.4\). For the within estimator, the deviations from firm means are \(\pm \frac 12\Delta x\) and \(\pm \frac 12\Delta y\), so \(\sum \tilde x\tilde y = 2 \times \frac 14(1 + 6 + 0) = 3.5\) and \(\sum \tilde x^2 = 2 \times \frac 14(1 + 4 + 0) = 2.5\), and \(b_{\text {FE}} = 3.5/2.5 = 1.4\). Firm 3, whose \(x\) did not change, contributes nothing to either estimate.

Solution. (a) The intraclass correlation is \(0.5/1.5 = 0.33\). For \(T = 4\), \(\theta = 1 - 1/\sqrt {1 + 4(0.5)} = 1 - 1/\sqrt 3 = 0.423\); for \(T = 20\), \(\theta = 1 - 1/\sqrt {11} = 0.698\).

(b) The unit mean \(\bar y_i\) contains \(\alpha _i\) plus an average of \(T\) disturbances, whose variance \(\sigma _\varepsilon ^2/T\) falls with \(T\). As \(T\) grows the mean becomes a precise measure of \(\alpha _i\), and GLS removes more of it, approaching complete demeaning. In the limit the between variation carries no information about \(\beta \) that is not contaminated by \(\alpha _i\), and only the within variation is used.

Solution. Completed schooling is constant over time for each adult, so it is collinear with the individual intercepts and is removed by the within transformation; there is no within variation to identify its coefficient. Alternatives are random effects (if ability is uncorrelated with schooling, which is doubtful), Mundlak’s regression, which estimates the coefficient of time-invariant variables under the weaker assumption that they are uncorrelated with the part of the effect not explained by the means of the time-varying regressors, the Hausman–Taylor estimator, which uses time-varying exogenous regressors as instruments, or a different design, such as twins or siblings, where differences between family members provide the variation.

Solution. \(H = (0.52 - 0.82)^2/(0.028^2 - 0.026^2) = 0.09/0.000108 = 833\), on one degree of freedom, far above \(3.84\). Random effects is rejected; the effects are correlated with the regressor, and the fixed effects estimate \(0.52\) should be reported, with cluster-robust standard errors. (Software 16.7.2, with unrounded values, gives \(673\).)

Solution. \(A = 1 - (1 - 0.8^{10})/(10 \times 0.2) = 1 - 0.8926/2 = 0.5537\). Then \(-(1.8/9)A = -0.1107\), and the correction factor is \(1 - (1.6/(0.2 \times 9))(0.5537) = 1 - 0.4922 = 0.5078\). The bias is \(-0.1107/0.5078 = -0.218\), so \(\plim \hat \rho _{\text {FE}} = 0.582\). The approximation \(-(1 + \rho )/(T - 1) = -0.2\) gives \(0.6\). With persistent series the bias remains large even with ten periods.

Solution. The regressor is constant within states, so \(\rho _x = 1\), and the Moulton factor is \(1 + (30 - 1)(1)(0.05) = 2.45\). The true variance is about \(2.45\) times the conventional one, and the standard error \(\sqrt {2.45} = 1.57\) times as large. A seemingly small intraclass correlation has a large effect when clusters are big and the regressor varies only between clusters. Clustering by state, with \(50\) clusters, is the remedy.