Solution. \(\Var (X_t) = 1.28/(1 - 0.36) = 2\); \(\rho (3) = 0.6^3 = 0.216\); \(\Cov (X_t, X_{t+2}) = 0.6^2(2) = 0.72\).

Solution. (a) \(\alpha _1 + \alpha _2 = 0.88 < 1\), \(\alpha _2 - \alpha _1 = -1.52 < 1\), \(\alpha _2 = -0.32 > -1\): stationary. The auxiliary roots are \(0.8\) and \(0.4\). (b) \(\alpha _1 + \alpha _2 = 1.1 > 1\): not stationary; one auxiliary root is \(1.064\). (c) \(\alpha _1 + \alpha _2 = -1.3 < 1\), \(\alpha _2 - \alpha _1 = -0.5 < 1\), \(\alpha _2 = -0.9 > -1\): stationary. Since \(\alpha _1^2 + 4\alpha _2 = 0.16 - 3.6 < 0\) the roots are complex, with modulus \(\sqrt {0.9} = 0.95\), close to \(1\): the ac.f. oscillates and dies away slowly.

Solution. By Theorem 8.2.1, \(\sigma _X^2 = 1/(1 - 0.5(0.6579) - 0.24(0.5689)) = 1/(1 - 0.3289 - 0.1365) = 1/0.5345 = 1.871\).

Solution. The equations for \(k = 1, 2\) are \(\rho (1) = \alpha _1 + \alpha _2\rho (1)\) and \(\rho (2) = \alpha _1\rho (1) + \alpha _2\), that is \(0.6 = \alpha _1 + 0.6\alpha _2\) and \(0.5 = 0.6\alpha _1 + \alpha _2\). Solving, \begin {align*} \alpha _1 &= \frac {\rho (1)(1 - \rho (2))}{1 - \rho (1)^2} = \frac {0.6(0.5)}{0.64} = 0.469 , \\ \alpha _2 &= \frac {\rho (2) - \rho (1)^2}{1 - \rho (1)^2} = \frac {0.5 - 0.36}{0.64} = 0.219 . \end {align*}

These satisfy the triangle conditions. Replacing \(\rho (1)\), \(\rho (2)\) by sample values is how AR models are estimated by the Yule–Walker method, and \(\alpha _2\) here is the partial autocorrelation at lag \(2\) (Section 11).

Solution. \((1 - 0.5B - 0.24B^2)(1 + \psi _1B + \psi _2B^2 + \psi _3B^3 + \dots ) = 1\). The coefficient of \(B\): \(\psi _1 - 0.5 = 0\), so \(\psi _1 = 0.5\). Of \(B^2\): \(\psi _2 - 0.5\psi _1 - 0.24 = 0\), so \(\psi _2 = 0.49\). Of \(B^3\): \(\psi _3 - 0.5\psi _2 - 0.24\psi _1 = 0\), so \(\psi _3 = 0.245 + 0.12 = 0.365\). In general \(\psi _j = 0.5\psi _{j-1} + 0.24\psi _{j-2}\), the same recursion as the ac.f.