Solution. \(\phi _{11} = 0.5\) and \(\phi _{22} = (0.3 - 0.25)/(1 - 0.25) = 0.067\). For \(\phi _{33}\), solve \[ \begin {pmatrix} 1 & 0.5 & 0.3 \\ 0.5 & 1 & 0.5 \\ 0.3 & 0.5 & 1 \end {pmatrix}\begin {pmatrix} \phi _{31} \\ \phi _{32} \\ \phi _{33} \end {pmatrix} = \begin {pmatrix} 0.5 \\ 0.3 \\ 0.1 \end {pmatrix} , \] which gives \(\phi _{31} = 0.473\), \(\phi _{32} = 0.113\), \(\phi _{33} = -0.098\). (Check the first equation: \(0.473 + 0.5(0.113) + 0.3(-0.098) = 0.500\).)

Solution. \(\rho (1) = -0.6/1.36 = -0.441\), so \(\phi _{11} = -0.441\) and, by Theorem 11.1.4(ii), \(\phi _{22} = -0.1946/(1 - 0.1946) = -0.242\). Both are negative, and the partial ac.f. continues to decay with the same sign, while the ac.f. is zero from lag \(2\).

Solution. By Theorem 11.1.4(i) with \(p = 1\), the system for \(k = 3\) is solved by \((\phi _{31}, \phi _{32}, \phi _{33}) = (\alpha , 0, 0)\). Directly: with \(\rho (j) = \alpha ^j\), \(\sum _i\phi _{3i}\rho (j - i) = \alpha \rho (j - 1) = \alpha ^j = \rho (j)\) for \(j = 1, 2, 3\), so the equations hold and \(\phi _{33} = 0\).

Solution. The band is \(\pm 0.2\). Series A: the ac.f. cuts off after lag \(1\) and the partial ac.f. tails off, alternating: MA(1). Since \(r_1 = 0.62\) exceeds \(\frac 12\), the largest possible for an MA(1) (Theorem 6.2.1), the fit should be checked carefully; sampling error can push \(r_1\) over the bound. Series B: the partial ac.f. cuts off after lag \(1\) and the ac.f. decays roughly geometrically (\(0.75^2 = 0.56\), \(0.75^3 = 0.42\)): AR(1) with \(\hat \alpha \approx 0.75\).