Solution. (a) \(d = 42.3/48.6 = 0.870\). From the expansion in the proof of Theorem 11.3.2, \begin {align*} 2\sum e_te_{t-1} &= 2\sum e_t^2 - e_1^2 - e_n^2 - \sum (e_t - e_{t-1})^2\\ &= 97.2 - 1.44 - 4.41 - 42.3 = 49.05 , \end {align*}

so \(\sum e_te_{t-1} = 24.525\) and \(\hat r = 24.525/48.6 = 0.505\). Then \[ 2(1 - \hat r) - \frac {e_1^2 + e_n^2}{\sum e_t^2} = 0.991 - \frac {5.85}{48.6} = 0.991 - 0.120 = 0.870 = d . \] The end correction is not negligible with \(n = 20\).

(b) \(d = 0.870 < d_L = 1.10\), so the hypothesis of no autocorrelation is rejected in favour of positive autocorrelation.

Solution. \[ \Omega = \frac {1}{1 - \rho ^2}\begin {bmatrix} 1 & \rho & \rho ^2\\ \rho & 1 & \rho \\ \rho ^2 & \rho & 1\end {bmatrix}, \] \[ P'P = \begin {bmatrix} 1 & -\rho & 0\\ -\rho & 1 + \rho ^2 & -\rho \\ 0 & -\rho & 1\end {bmatrix} . \] The first row of \(P'P\) times the columns of \((1 - \rho ^2)\Omega \) gives \[ (1 - \rho ^2,\ \rho - \rho ,\ \rho ^2 - \rho ^2) = (1 - \rho ^2, 0, 0) . \] The second row gives \begin {align*} &(-\rho + \rho + \rho ^3 - \rho ^3,\ -\rho ^2 + 1 + \rho ^2 - \rho ^2,\ -\rho ^3 + \rho + \rho ^3 - \rho )\\ &\qquad = (0, 1 - \rho ^2, 0), \end {align*}

and the third row, by symmetry, \((0, 0, 1 - \rho ^2)\). Dividing by \(1 - \rho ^2\) gives \(I\).

Solution. (a) The regressors include the lagged dependent variable, so \(d\) is biased towards \(2\) (Theorem 11.2.2) and the bounds do not apply.

(b) \(\hat r \approx 1 - d/2 = 0.19\) and \(n\widehat \Var (\hat \gamma ) = 50(0.0081) = 0.405\), so \(h = 0.19\sqrt {50/0.595} = 1.74\). Against the two-sided \(5\%\) point \(1.96\) this is not significant, but against the one-sided alternative \(\rho > 0\), with critical value \(1.645\), it is. The evidence is borderline.

(c) Then \(n\widehat \Var (\hat \gamma ) = 50(0.0225) = 1.125 > 1\), and \(h\) cannot be computed. Use the Breusch–Godfrey test: regress the residuals on a constant, \(Y_t\), \(C_{t-1}\) and \(e_{t-1}\), and refer \(nR^2\) to \(\chi ^2_1\).

Solution. (a) \(\Var (u_t) = \sigma _\varepsilon ^2(1 + \theta ^2)\). \(\Cov (u_t, u_{t-1}) = E\bigl ((\varepsilon _t + \theta \varepsilon _{t-1})(\varepsilon _{t-1} + \theta \varepsilon _{t-2})\bigr ) = \theta \sigma _\varepsilon ^2\), and the covariance is zero for \(s \geq 2\), because \(u_t\) and \(u_{t-s}\) then share no \(\varepsilon \). So \(\Corr (u_t, u_{t-1}) = \theta /(1 + \theta ^2)\), and \(\Omega \) is tridiagonal: \(1 + \theta ^2\) on the diagonal and \(\theta \) next to it. Its inverse, unlike that of the AR(1) matrix, is full.

(b) \(1 + \theta ^2 - 2|\theta | = (1 - |\theta |)^2 \geq 0\), so \(|\theta |/(1 + \theta ^2) \leq \frac 12\), with equality at \(\theta = \pm 1\). With exogenous regressors \(\hat r\) converges to \(\theta /(1 + \theta ^2)\), so \(\plim d = 2\bigl (1 - \theta /(1 + \theta ^2)\bigr )\) lies between \(1\) and \(3\). A value of \(d\) well below \(1\) in a large sample therefore points to an autoregressive rather than a first-order moving average scheme.

Solution. (a) \(\sigma ^2 = 0.5/(1 - 0.4) = 0.833\); kurtosis \(3(1 - 0.16)/(1 - 0.48) = 2.52/0.52 = 4.85\).

(b) \(h_t = 0.5 + 0.4(9) = 4.1\), a conditional standard deviation of \(2.025\). Given yesterday, \(u_t \sim N(0, 4.1)\), so \[ P(|u_t| > 3) = 2\bigl (1 - \Phi (3/2.025)\bigr ) = 2\bigl (1 - \Phi (1.48)\bigr ) = 0.139 . \] A normal distribution with the unconditional variance would give \[ 2\bigl (1 - \Phi (3/0.913)\bigr ) = 2\bigl (1 - \Phi (3.29)\bigr ) = 0.001 . \] After a large shock, another large move is a hundred times more likely than the unconditional model says.

(c) \(3\alpha _1^2 < 1\), that is \(0 \leq \alpha _1 < 1/\sqrt 3 = 0.577\). For \(0.577 \leq \alpha _1 < 1\) the variance is finite but the kurtosis is infinite.

Solution. (a) By Theorem 11.4.2, \(\delta _0 = \beta _1(1 - \rho )\), \(\delta _1 = \rho \), \(\delta _2 = \beta _2\) and \(\delta _3 = -\rho \beta _2\). The four \(\delta \)s are functions of three parameters, and the restriction is the common factor restriction \(\delta _3 = -\delta _1\delta _2\), that is \(\delta _3 + \delta _1\delta _2 = 0\).

(b) As in Theorem 9.3.1, \[ \text {LR} = n\log \frac {\text {RSS}_R}{\text {RSS}_U} = 80\log \frac {13.9}{12.4} = 80(0.1142) = 9.14 , \] compared with \(\chi ^2_1\), whose \(5\%\) point is \(3.84\) (and \(1\%\) point \(6.63\)). The common factor restriction is rejected: the dynamics are not those of a static relation with AR(1) disturbances, and the dynamic model should be kept.

Solution. (a) \(\hat \gamma _0 = 1 + 1 + 1 = 3\) and \(\hat \gamma _1 = g_2g_1 + g_3g_2 = -1 - 1 = -2\), so \(\hat \gamma _0 + 2\hat \gamma _1 = -1 < 0\): a negative “variance”.

(b) Summing over \(t = 0, \dots , n\), \[ \sum _t(g_t + g_{t+1})^2 = \sum _tg_t^2 + \sum _tg_{t+1}^2 + 2\sum _tg_tg_{t+1} = 2\hat \gamma _0 + 2\hat \gamma _1 , \] since each \(g_t^2\) appears once in each of the first two sums. Hence \(\hat \gamma _0 + \hat \gamma _1 = \frac 12\sum _t(g_t + g_{t+1})^2 \geq 0\). This is the case \(L = 1\) of the proof of Theorem 11.4.3, where the weight on \(\hat \gamma _1\) is \(1 - \frac 12 = \frac 12\), applied to both \(\hat \gamma _1\) and its transpose.