Solution. (a) \(g = y - \theta /2\), so \(\hat \theta = 2\bar y\). Here \(G = -\frac 12\) and \(S = \Var (y) = \theta ^2/12\), so the asymptotic variance of \(\sqrt n(\hat \theta - \theta )\) is \(G^{-2}S = 4\theta ^2/12 = \theta ^2/3\).

(b) \(\bar y = 10.5/5 = 2.1\), so \(\hat \theta = 4.2\). This exceeds the largest observation \(3.9\), as it must for the data to be possible, but a method of moments estimate can also fall below the maximum, giving an impossible value. The maximum likelihood estimator, the sample maximum, never does, and it converges at the faster rate \(1/n\): the method of moments is not always efficient.

Solution. \(\bar g_n(\beta ) = \frac 1nX'(Y - X\beta ) = 0\) are the normal equations, with solution \(b\). The derivative is \(\partial g/\partial \beta ' = -xx'\), so \(G = -E(xx') = -Q\), and \(S = E(u^2xx')\). The asymptotic variance of \(\sqrt n(b - \beta )\) is \(Q^{-1}E(u^2xx')Q^{-1}\), estimated by \(n(X'X)^{-1}\bigl (\sum e_i^2x_ix_i'\bigr )(X'X)^{-1}\), which is \(n\) times White’s estimator. Under homoscedasticity \(S = \sigma ^2Q\) and the variance becomes \(\sigma ^2Q^{-1}\), the classical result.

Solution. (a) \(Q(\mu ) = (\bar y_1 - \mu )^2 + (\bar y_2 - \mu )^2\) for \(W = I\), minimised by \(\hat \mu = \frac 12(\bar y_1 + \bar y_2)\). Here \(S = \operatorname {diag}(\sigma _1^2, \sigma _2^2)\), and with \(W = S^{-1}\) the objective is \((\bar y_1 - \mu )^2/\sigma _1^2 + (\bar y_2 - \mu )^2/\sigma _2^2\), minimised by the inverse-variance weighted mean \(\hat \mu = (\bar y_1/\sigma _1^2 + \bar y_2/\sigma _2^2)/(1/\sigma _1^2 + 1/\sigma _2^2)\).

(b) The simple mean has variance \((\sigma _1^2 + \sigma _2^2)/(4n) = 5/(4n) = 1.25/n\). The efficient estimator has variance \(1/\bigl (n(1/\sigma _1^2 + 1/\sigma _2^2)\bigr ) = 1/(1.25n) = 0.8/n\), which is \((G'S^{-1}G)^{-1}/n\) with \(G = -(1, 1)'\). It is better even than using \(y_1\) alone (variance \(1/n\)): the noisier measurement still adds information, provided it is weighted correctly.

Solution. (a) There are \(\ell = 1 + 2 + 2 = 5\) instruments and one Euler equation, so five moment conditions; with \(k = 2\) parameters, \(J\) has \(\ell - k = 3\) degrees of freedom.

(b) The \(5\%\) and \(1\%\) points of \(\chi ^2_3\) are \(7.81\) and \(11.34\), so \(J = 13.8\) rejects at \(1\%\). The lagged variables help to predict the pricing error, which the theory says is impossible. Either the model of preferences is wrong (for instance, habits or non-separable utility), or the representative-agent assumption fails, or the consumption data, which are time-aggregated and measured with error, do not match the model’s timing. The test does not say which.

Solution. With \(\ell = k\), \(Z'X\) is square and non-singular, so \[ b(W) = (X'ZWZ'X)^{-1}X'ZWZ'Y = (Z'X)^{-1}W^{-1}(X'Z)^{-1}X'ZWZ'Y = (Z'X)^{-1}Z'Y . \] Equivalently, the \(k\) equations \(Z'(Y - Xb) = 0\) can be solved exactly, so \(Q_n(b) = 0\) at their solution whatever the weight, and that solution is the minimiser.

Solution. \(E(y^r) = \int _0^\infty y^r\frac {\lambda ^\kappa }{\Gamma (\kappa )}y^{\kappa - 1}e^{-\lambda y}\,dy = \frac {\lambda ^\kappa }{\Gamma (\kappa )}\cdot \frac {\Gamma (\kappa + r)}{\lambda ^{\kappa + r}} = \frac {\Gamma (\kappa + r)}{\Gamma (\kappa )\lambda ^r}\). With \(r = 2\), \(\Gamma (\kappa + 2) = (\kappa + 1)\kappa \Gamma (\kappa )\), giving \(\kappa (\kappa + 1)/\lambda ^2\). Since \(E(y^r) = E(e^{r\log y})\), its derivative at \(r = 0\) is \(E(\log y)\); differentiating \(\log E(y^r) = \log \Gamma (\kappa + r) - \log \Gamma (\kappa ) - r\log \lambda \) at \(r = 0\), where \(E(y^0) = 1\), gives \(E(\log y) = \psi (\kappa ) - \log \lambda \).