Solution. (a) By Theorem 12.1.2, \(\Var (Y_{25}) = 25 \times 4 = 100\), and \(\Corr (Y_{10}, Y_{40}) = \sqrt {10/40} = 0.5\).

(b) \(Y_{30} = Y_{20} + u_{21} + \dots + u_{30}\), so given \(Y_{20} = 7\) it is \(N(7, 40)\), and the interval is \(7 \pm 1.96\sqrt {40} = 7 \pm 12.4\). For \(Y_{21}\) it is \(7 \pm 1.96(2) = 7 \pm 3.9\). The width grows with the square root of the horizon: ten periods ahead the interval is \(\sqrt {10} = 3.2\) times as wide.

Solution. (a) \(\gamma = 1.5 - 0.5 - 1 = 0\) and \(\delta _1 = -\phi _2 = 0.5\), so \(\Delta Y_t = 0.5\Delta Y_{t-1} + u_t\). There is a unit root: \(Y_t\) is \(I(1)\), and its difference is a stationary AR(1). The polynomial \(1 - 1.5z + 0.5z^2 = (1 - z)(1 - 0.5z)\) confirms it.

(b) \(\gamma = 1.2 - 0.3 - 1 = -0.1\) and \(\delta _1 = 0.3\), so \(\Delta Y_t = -0.1Y_{t-1} + 0.3\Delta Y_{t-1} + u_t\). Since \(\gamma \neq 0\) there is no unit root. The roots of \(1 - 1.2z + 0.3z^2\) are \(z = (1.2 \pm \sqrt {0.24})/0.6\), that is \(2.82\) and \(1.18\), both outside the unit circle, so the process is stationary. With \(\gamma = -0.1\), however, it is close enough to a unit root that an ADF test in a moderate sample would probably not reject.

Solution. (a) The levels statistics are above the \(5\%\) trend critical value \(-3.41\), so a unit root is not rejected; the differences are far below \(-2.86\). Both series are \(I(1)\).

(b) With \(N = 2\) the Engle–Granger \(5\%\) critical value is \(-3.34\), and \(-3.82 < -3.34\), so the null of no cointegration is rejected (though not at \(1\%\), \(-3.90\)). Consumption and income share a long-run relation with an elasticity of \(0.97\). The critical value is more negative than the Dickey–Fuller value because \(\hat v_t\) are least squares residuals, chosen to look as stationary as possible.

(c) Not by itself: Theorem 12.2.5 shows that \(R^2\) stays large in spurious regressions. Here the low Durbin–Watson statistic (\(R^2 > d\)) would be a warning, and only the cointegration test of (b) shows that the relation is genuine.

Solution. (a) By Theorem 12.5.1, \(\mu = 2/0.4 = 5\) and \(\theta = (0.3 + 0.1)/0.4 = 1\), so \[ \Delta Y_t = 0.3\Delta X_t - 0.4\bigl (Y_{t-1} - 5 - X_{t-1}\bigr ) + \varepsilon _t , \] and the long run is \(Y^* = 5 + X^*\), with a unit long-run effect.

(b) \(1 - \phi = 0.4\), so \(40\%\) is removed each period; the half-life is \(\log 0.5/\log 0.6 = 1.36\) periods.

(c) The restriction is \(\beta _1 = -\phi \beta _0 = -0.18\), but \(\beta _1 = 0.1\); equivalently the impact effect \(0.3\) differs from the long-run effect \(1\). It is not satisfied.

Solution. (a) \(b = \sum X_tY_t/\sum X_t^2\). By Lemma 12.2.3(ii), \(\sum X_t^2\) is of order \(n^2\). The numerator has mean zero, by independence, and variance \(\sum _t\sum _sE(X_tX_s)E(Y_tY_s)\); with \(E(X_tX_s) = \sigma ^2\min (t, s)\) and \(\sum _s|E(Y_tY_s)|\) bounded, this is of order \(n^2\), so the numerator is of order \(n\). Hence \(b\) is of order \(n/n^2 = 1/n\).

(b) The left side is \(I(0)\) and has bounded variance, while \(X_t\) has variance growing like \(t\). The only way the equation \(Y_t = bX_t + u_t\) can hold with a stationary disturbance is \(b = 0\): the stochastic trend in \(X_t\) has nothing to match on the left side. In a balanced regression both sides have the same order of integration, or the \(I(1)\) regressors are cointegrated so that their combination is \(I(0)\), as in the error correction model.

Solution. Each term \(Y_{t-1}u_t\) has mean zero, since \(u_t\) is independent of \(Y_{t-1}\). For the variance, terms with \(s < t\) are uncorrelated: \(E(Y_{s-1}u_sY_{t-1}u_t) = E(Y_{s-1}u_sY_{t-1})E(u_t) = 0\). So the variance is \(\sum _tE(Y_{t-1}^2)E(u_t^2) = \sigma ^2\sum _t(t - 1)\sigma ^2 = \sigma ^4n(n - 1)/2\). The standard deviation is of order \(n\), in agreement with the scaling by \(n\) in the lemma. In the stationary case \(E(Y_{t-1}^2)\) is constant, the variance is of order \(n\), and the scaling is \(\sqrt n\), which is why the stationary theory gives normal limits at rate \(\sqrt n\).