Solution. See Theorem 8.2.1. Substitute \(Y_{(2)} = X_{(2)}\beta + U_{(2)}\) and \[ b_{(1)} = \beta + (X_{(1)}'X_{(1)})^{-1}X_{(1)}'U_{(1)} \] to obtain \[ d = U_{(2)} - X_{(2)}(X_{(1)}'X_{(1)})^{-1}X_{(1)}'U_{(1)} , \] so \(E(d) = 0\), and since \(U_{(1)}\) and \(U_{(2)}\) are uncorrelated, \[ \Cov (d) = \sigma ^2\bigl [I_{n_2} + X_{(2)}(X_{(1)}'X_{(1)})^{-1}X_{(2)}'\bigr ] . \] The first term is the variance of the new disturbances and the second the sampling error in \(b_{(1)}\) propagated into the predictions. Note that \(\Cov (d)\) is not diagonal: all the prediction errors share the same estimation error.

Solution. (a) This is the Chow test with \(k = 2\) and \(n = 74\). \(\text {RSS}_1 + \text {RSS}_2 = 4.7045\), so \[ F = \frac {(8.9160 - 4.7045)/2}{4.7045/70} = \frac {2.1058}{0.06721} = 31.3 . \] The \(5\%\) point of \(F_{2,70}\) is \(3.13\), so the hypothesis of a common relation is rejected overwhelmingly (\(p \approx 2 \times 10^{-10}\)). Separate regressions fit much better.

(b) The test assumes equal disturbance variances in the two groups. \(s_1^2 = 1.2150/38 = 0.0320\) for public schools and \(s_2^2 = 3.4895/32 = 0.1090\) for private schools, a ratio of \(3.41\). The \(5\%\) point of \(F_{32,38}\) is about \(1.75\), so equal variances are clearly rejected: private schools’ spending is much more variable. The Chow \(F\) statistic therefore does not have its nominal distribution. With a ratio of \(31\), the conclusion is unlikely to change, but a version robust to unequal variances should be used: a Wald test using each group’s own variance estimate, or the heteroscedasticity-robust tests of the section on heteroscedasticity.

(c) Fit \(E = \alpha + \gamma D + \beta N + \delta DN + U\) with \(D = 1\) for private schools, and test \(\delta = 0\) with a \(t\) test. If it is not rejected, the types share a slope and differ only in their intercept \(\gamma \). This requires the full data, not just the three residual sums of squares.

Solution. By Theorem 8.2.2, with \(n_2 = 15\) and \(n_1 - k = 12\), \(F = (0.1715/15)/(0.1151/12) = 0.01143/0.00959 = 1.19\). The \(5\%\) point of \(F_{15,12}\) is about \(2.62\), so the forecast test does not reject, although the break test did. The forecast test asks whether the first period’s coefficients predict the second period acceptably; the break test asks whether the coefficients are equal. They are different questions, and the break test is more powerful when both periods can be estimated. The forecast test needs no estimate from the second period, only predictions into it, so it works with any \(n_2 \geq 1\). The break test needs \(\text {RSS}_2\), which requires \(n_2 > k\).

Solution. (a) The auxiliary slope is \(\delta = \sum x_2x_3/\sum x_2^2 = -20/50 = -0.4\), so the bias is \(\delta \beta _3 = -0.4 \times 3 = -1.2\).

(b) \(E(\tilde b_2) = 1 - 1.2 = -0.2\): the short regression gets the sign of the effect wrong. A negatively correlated omitted variable with a positive effect can reverse a true positive relation, as pop15 did for pop75 in Example 8.1.3.

Solution. \(C_t = \alpha + \gamma _2Q_{2t} + \gamma _3Q_{3t} + \gamma _4Q_{4t} + \beta Y_t + u_t\), where \(Q_{jt} = 1\) in quarter \(j\); the first quarter is the base, and \(\gamma _j\) is the difference between quarter \(j\)’s intercept and the first quarter’s. With all four dummies and a constant, \(Q_1 + Q_2 + Q_3 + Q_4 = \iota \), so the columns of \(X\) are linearly dependent, \(X'X\) is singular, and the coefficients are not identified: the dummy variable trap. (Dropping the constant instead and keeping all four dummies is equivalent.) Seasonality is tested by the \(F\) test of \(\gamma _2 = \gamma _3 = \gamma _4 = 0\), comparing this model with the model without dummies, with \(q = 3\).

Solution. Let \(S_t\) be the residual sum of squares from the first \(t\) observations. Adding observation \(t\) to a regression on \(t - 1\) observations increases the residual sum of squares by exactly \(w_t^2\). This follows from the Chow forecast test with \(n_2 = 1\): by Theorem 8.2.2, the increase \(S_t - S_{t-1}\) equals the Wald numerator \(d^2/(1 + x_t'(X_{t-1}'X_{t-1})^{-1}x_t)\) with \(d = Y_t - x_t'b_{t-1}\), which is \(w_t^2\). With \(k\) observations the regression fits exactly, so \(S_k = 0\). Summing, \(e'e = S_n = \sum _{t=k+1}^n (S_t - S_{t-1}) = \sum _{t=k+1}^n w_t^2\). So the recursive residuals are \(n - k\) independent pieces into which the residual sum of squares is decomposed, matching its \(n - k\) degrees of freedom.