Solution. (a) \(\bar u_j\) is the average of \(n_j\) independent disturbances, so \(\Var (\bar u_j) = \sigma ^2/n_j\).

(b) GLS multiplies each group’s equation by \(\sqrt {n_j}\), which is weighted least squares with weights \(n_j\): \[ b_{\text {GLS}} = \frac {\sum _jn_j(\bar X_j - \bar X_w)(\bar Y_j - \bar Y_w)}{\sum _jn_j(\bar X_j - \bar X_w)^2}, \] where \(\bar X_w = \sum n_j\bar X_j/\sum n_j\) and \(\bar Y_w\) are the weighted means, the overall household means. This is exactly the least squares estimate on the individual data with each household’s values replaced by its group means.

(c) \(E(\bar u_j) = 0\), so least squares is unbiased; but it gives every group the same weight, whatever its size, so it is not BLUE (Aitken). Its conventional variance assumes a common variance across groups, while the true variances differ by the factor \(1/n_j\).

Solution. (a) Divide by \(X_i\): \(Y_i/X_i = \beta + \alpha (1/X_i) + u_i/X_i\), with \(\Var (u_i/X_i) = \sigma ^2\).

(b) \(\beta \) becomes the intercept and \(\alpha \) the coefficient on \(1/X_i\), as in Example 10.4.2. A common error is to forget this exchange of roles and report the transformed intercept as if it were \(\alpha \).

Solution. Each subset has \(24 - 3 = 21\) degrees of freedom, so \(\text {GQ} = (61.3/21)/(24.8/21) = 2.47\). The \(5\%\) point of \(F_{21,21}\) is about \(2.08\), so homoscedasticity is rejected: the larger firms have a larger disturbance variance. The test is one-sided, because the alternative specified that the variance increases with size.

Solution. The fitted values are the group means, so the residuals are deviations from the group means. With the regressors \((1, D)\) the slope is \(\bar Y_2 - \bar Y_1\), a linear function \(\sum c_iY_i\) with \(c_i = -1/n_1\) in group 1 and \(1/n_2\) in group 2. The sandwich variance of such a function is \(\sum c_i^2e_i^2\), which is \(\sum _{i\in 1}e_i^2/n_1^2 + \sum _{i\in 2}e_i^2/n_2^2 = \hat \sigma _1^2/n_1 + \hat \sigma _2^2/n_2\). The conventional formula is \(s^2(1/n_1 + 1/n_2)\), with \(s^2\) the pooled variance. If the groups’ variances differ, the robust version is the right one: it is the Welch form of the two-sample \(t\) test, while the conventional form is the pooled \(t\) test, which assumes equal variances. This applies directly to the schools data of Section 8, whose variances differed by a factor of more than three.

Solution. White’s test regresses the squared residuals on the regressors, their squares and cross-products. Anything that makes the squared residuals vary systematically with these will be detected, including misspecification of the mean. Suppose the true model is \(Y = \alpha + \beta X + \gamma X^2 + u\) with homoscedastic \(u\), but \(Y\) is regressed on \(X\) alone. The residuals then contain the neglected term \(\gamma (X^2 - \text {fitted part})\), whose square is a polynomial in \(X\), so \(e_i^2\) is strongly related to \(X\) and \(X^2\) and the test rejects, although \(\Var (u)\) is constant. The test should be read as “something is wrong with the conditional mean or variance”, and the functional form checked (for example with RESET) before heteroscedasticity is concluded.

Solution. \(e = MU\), so \(e_i = m_i'U\), with \(m_i'\) the \(i\)th row of \(M\), and \(E(e_i^2) = \sigma ^2m_i'm_i = \sigma ^2(M^2)_{ii} = \sigma ^2m_{ii} = \sigma ^2(1 - p_{ii})\). So \(e_i^2/(1 - p_{ii})\) is unbiased for \(\sigma ^2\) under homoscedasticity, and HC2 uses it to remove the downward bias of the raw squared residuals. HC3 approximates a jackknife. The residual of observation \(i\) when it is left out of the fit is \(e_i/(1 - p_{ii})\), so \(e_i^2/(1 - p_{ii})^2\) is the squared leave-one-out prediction error. It overcorrects slightly under homoscedasticity, but it guards against high-leverage points, whose residuals understate their disturbances most, and it performs best in small samples in simulation studies.