Solution. This is Theorem 7.2.1. (i) \(E(a'Y) = a'X\beta = c'\beta \) for all \(\beta \) requires \(X'a = c\). (ii) \(\sigma ^2a'a\). (iii) The Lagrangian \(a'a - 2\lambda '(X'a - c)\) gives \(a = X\lambda \); then \(X'X\lambda = c\), so \(a = X(X'X)^{-1}c\). The resulting estimator is \(a'Y = c'(X'X)^{-1}X'Y = c'b\), so the minimum variance linear unbiased estimator of \(c'\beta \) is \(c'b\), with variance \(\sigma ^2c'(X'X)^{-1}c\).

Solution. (i) \(E(b) = \beta \) and \(\Cov (b) = \sigma ^2(X'X)^{-1}\) (Theorem 7.1.1). (ii) \(Y \sim N(X\beta , \sigma ^2I)\) and \(b \sim N(\beta , \sigma ^2(X'X)^{-1})\), since both are linear in \(U\). (iii) \(\Var (Rb - r) = R\Var (b)R' = \sigma ^2R(X'X)^{-1}R'\), and \(Rb - r \sim N(R\beta - r, \sigma ^2R(X'X)^{-1}R')\). (iv) Under \(H_0\) the mean is zero, and a \(N(0, \Sigma )\) vector \(z\) with \(\Sigma \) pd has \(z'\Sigma ^{-1}z \sim \chi ^2_q\) (Theorem 5.4.2); here \(\Sigma = \sigma ^2R(X'X)^{-1}R'\) is pd because \(R\) has full row rank.

Solution. See Theorem 7.4.2. The unrestricted regression has \(\text {RSS}_U = e'e\) on \(n - k\) degrees of freedom. The restricted model is \(Y_i = \beta _1 + u_i\), fitted by \(\bar Y\), with \(\text {RSS}_R = \sum (Y_i - \bar Y)^2\). There are \(q = k - 1\) restrictions, so \[ F = \frac {(\text {RSS}_R - \text {RSS}_U)/(k - 1)}{\text {RSS}_U/(n - k)} = \frac {R^2/(k - 1)}{(1 - R^2)/(n - k)}, \] which under \(H_0\) has the \(F\) distribution with \(k - 1\) and \(n - k\) degrees of freedom. The independence of numerator and denominator comes from the independence of \(b\) and \(e'e\) (Theorem 5.5.1). Reject for large \(F\).

Solution. (a) \(H_0\colon \beta _L + \beta _K = 1\). \(b_L + b_K = 0.9\), and \[ \widehat \Var (b_L + b_K) = 0.102^2 + 0.102^2 + 2(-0.0061) = 0.0208 - 0.0122 = 0.0086 , \] so the standard error is \(0.0928\) and \(t = (0.9 - 1)/0.0928 = -1.08\). Since \(|t| < 2.026\), constant returns are not rejected.

(b) Ignoring the covariance gives a standard error of \(\sqrt {0.0208} = 0.144\) and \(t = -0.69\). The conclusion is the same here, but the standard error is \(55\%\) too large. The covariance is negative because labour and capital move together in the data. An overestimate of one elasticity tends to come with an underestimate of the other, so their sum is estimated more precisely than either alone. Ignoring a negative covariance makes a test of the sum too conservative; ignoring a positive covariance would make it too liberal.

(c) Substitute \(\beta _K = 1 - \beta _L\): \(\log Q - \log K = \alpha + \beta _L(\log L - \log K) + u\), so regress \(\log (Q/K)\) on \(\log (L/K)\). Its residual sum of squares is \(\text {RSS}_R\), and \(F = (\text {RSS}_R - \text {RSS}_U)/(\text {RSS}_U/37)\) with \((1, 37)\) degrees of freedom; this \(F\) equals \(t^2\) from (a).

Solution. By Corollary 6.4.2, the slope estimates come from the deviation form, so their covariance matrix is \(\sigma ^2\) times the inverse of \(\begin {pmatrix} S_{22} & S_{23} \\ S_{23} & S_{33} \end {pmatrix}\), which is \(\frac {1}{D}\begin {pmatrix} S_{33} & -S_{23} \\ -S_{23} & S_{22} \end {pmatrix}\) with \(D = S_{22}S_{33} - S_{23}^2\). Hence \[ \Corr (b_2, b_3) = \frac {-S_{23}}{\sqrt {S_{33}S_{22}}} = -r_{23} . \] Positively correlated regressors have negatively correlated estimates, and vice versa: if the two regressors move together, the data can pin down their combined effect but not how to divide it, so an overestimate of one coefficient goes with an underestimate of the other. pop15 and pop75 are correlated \(-0.91\), so their estimates are positively correlated. In Software 7.7.2 the correlation is \(0.76\) rather than \(0.91\) because the model has two further regressors. The relevant correlation is then the partial correlation of pop15 and pop75 given dpi and ddpi, with the sign reversed.

Solution. By Theorem 6.4.1, \(b_j = \tilde x'Y/\tilde x'\tilde x\), where \(\tilde x = M_1X_j\) is the residual vector from regressing \(X_j\) on the other regressors (including the intercept). Then \(b_j = \beta _j + \tilde x'U/\tilde x'\tilde x\) and \(\Var (b_j) = \sigma ^2/\tilde x'\tilde x\). But \(\tilde x'\tilde x\) is the residual sum of squares of that auxiliary regression, which is \((1 - R_j^2)\) times its total sum of squares \(\sum (X_{ji} - \bar X_j)^2\). The factor \(1/(1 - R_j^2)\) is the variance inflation factor: it is \(1\) when \(X_j\) is uncorrelated with the other regressors, and it grows without bound as \(X_j\) becomes nearly a linear combination of them. That is the multicollinearity problem, quantified.

Solution. \(\hat Y_0 = 28.566 - 0.4612(35) - 1.6915(2.5) - 0.000337(1000) + 0.4097(4) = 28.566 - 16.142 - 4.229 - 0.337 + 1.639 = 9.497\). The standard error of prediction is \(s\sqrt {1 + 0.0249} = 3.803(1.0124) = 3.850\), and with \(t_{0.025,45} = 2.014\) the interval is \(9.50 \pm 7.75 = (1.74,\ 17.25)\), which is what predict(fit, new, interval = "prediction") returns. Almost all of the width comes from the disturbance variance, the \(1\) in \(1 + x_0'(X'X)^{-1}x_0\): the regression line is estimated far more precisely than an individual country’s savings rate can be predicted.