Proof. By Proposition 9.2.1, \(\psi _0 = 1\) and \(\psi _j = (\alpha + \beta )\alpha ^{j-1}\) for \(j \ge 1\). Then, as for an MA process, \begin {align*} \gamma (0) &= \sigma _Z^2\sum _{j\ge 0}\psi _j^2 = \sigma _Z^2\Bigl [1 + (\alpha + \beta )^2\sum _{j\ge 1}\alpha ^{2(j-1)}\Bigr ] \\ &= \sigma _Z^2\Bigl [1 + \frac {(\alpha + \beta )^2}{1 - \alpha ^2}\Bigr ] = \sigma _Z^2\frac {1 + 2\alpha \beta + \beta ^2}{1 - \alpha ^2} , \\ \gamma (1) &= \sigma _Z^2\sum _{j\ge 0}\psi _j\psi _{j+1} = \sigma _Z^2\Bigl [(\alpha + \beta ) + (\alpha + \beta )^2\sum _{j\ge 1}\alpha ^{2j-1}\Bigr ] \\ &= \sigma _Z^2(\alpha + \beta )\Bigl [1 + \frac {\alpha (\alpha + \beta )}{1 - \alpha ^2}\Bigr ] , \end {align*}

and \(1 - \alpha ^2 + \alpha ^2 + \alpha \beta = 1 + \alpha \beta \) gives the stated \(\gamma (1)\). For \(k \ge 2\), multiply the model by \(X_{t-k}\) and take expectations: \(X_{t-k}\) involves only shocks up to time \(t - k\), so \(E(X_{t-k}Z_t) = E(X_{t-k}Z_{t-1}) = 0\), leaving \(\gamma (k) = \alpha \gamma (k - 1)\). □

Solution. With \(\alpha = 0.5\) and \(\beta = -0.3\), \(1 + 2\alpha \beta + \beta ^2 = 1 - 0.3 + 0.09 = 0.79\) and \((1 + \alpha \beta )(\alpha + \beta ) = 0.85 \times 0.2 = 0.17\). So \begin {gather*} \gamma (0) = \frac {0.79}{0.75}\sigma _Z^2 = 1.053\sigma _Z^2 , \qquad \rho (1) = \frac {0.17}{0.79} = 0.215 , \\ \rho (k) = 0.215(0.5)^{k-1} . \end {gather*} As a check, \(\sum \psi _j^2 = 1 + 0.04/0.75 = 1.0533\). The MA term with \(\beta = -0.3\) partly cancels the AR term, so \(\rho (1)\) is much smaller than the \(0.5\) of the AR(1) alone.