Solution. \(\gamma (0) = (1 + 0.36 + 0.04)\sigma _Z^2 = 1.40\sigma _Z^2\), \(\gamma (1) = (0.6 + 0.6 \times 0.2)\sigma _Z^2 = 0.72\sigma _Z^2\), \(\gamma (2) = 0.2\sigma _Z^2\). So \(\rho (1) = 0.514\), \(\rho (2) = 0.143\), and \(\rho (k) = 0\) for \(k \ge 3\). The equation \(1 + 0.6B + 0.2B^2 = 0\) has discriminant \(0.36 - 0.8 < 0\), so its roots are complex conjugates; their product is \(1/0.2 = 5\), so each has modulus \(\sqrt 5 > 1\). The process is invertible.

Solution. \(-0.3\beta ^2 - \beta - 0.3 = 0\) gives \(\beta = (1 \pm \sqrt {1 - 0.36})/(-0.6) = (1 \pm 0.8)/(-0.6)\), so \(\beta = -\frac 13\) or \(\beta = -3\). They are reciprocals, as they must be. Use the invertible value \(\hat \beta = -0.333\).

Solution. The original has \(\gamma (0) = 1 + 16 = 17\) and \(\gamma (1) = 4\). The candidate \(X_t = Z_t^* + 0.25Z_{t-1}^*\) with \(\Var (Z_t^*) = 16\) has \(\gamma (0) = 16(1 + 0.0625) = 17\) and \(\gamma (1) = 16(0.25) = 4\). They agree, and \(|0.25| < 1\).

Solution. The band \(\pm 2/\sqrt {200} = \pm 0.14\) contains every \(r_k\) from lag \(3\) on, while \(r_1\) and \(r_2\) lie well outside it: the correlogram cuts off after lag \(2\), suggesting MA(2). An MA(1) would need \(\rho (2) = 0\), but \(r_2 = 0.31\) is more than twice the band. (By itself, \(r_1 = 0.45\) would be possible for an MA(1), since it is below \(\frac 12\).)