Proof. (i) \(1 + \beta ^2 - 2|\beta | = (1 - |\beta |)^2 \ge 0\), so \(|\beta | \le \frac 12(1 + \beta ^2)\) and \(|\rho (1)| = |\beta |/(1 + \beta ^2) \le \frac 12\), with equality exactly when \(|\beta | = 1\).

(ii) If \(|r| \le \frac 12\) and \(r \ne 0\), the equation \(\beta /(1 + \beta ^2) = r\), that is \(r\beta ^2 - \beta + r = 0\), has the real root \(\beta = (1 - \sqrt {1 - 4r^2})/(2r)\), since \(1 - 4r^2 \ge 0\); the MA(1) process with this \(\beta \) has the required ac.f. (For \(r = 0\), take white noise.) Conversely, suppose the function is an ac.f. By Theorem 3.3.1(iv), for any \(n\) and weights \(a_1, \dots , a_n\), \[ 0 \le \sum _i\sum _ja_ia_j\rho (i - j) = \sum _ia_i^2 + 2r\sum _{i=1}^{n-1}a_ia_{i+1} . \] Taking \(a_i = (-1)^i\) gives \(0 \le n - 2r(n - 1)\), so \(r \le \frac {n}{2(n - 1)}\) for every \(n\), and letting \(n \to \infty \), \(r \le \frac 12\). Taking all \(a_i = 1\) gives \(0 \le n + 2r(n - 1)\), and in the same way \(r \ge -\frac 12\). □

Solution. Solve \(\beta /(1 + \beta ^2) = 0.4\), that is \(0.4\beta ^2 - \beta + 0.4 = 0\): \[ \beta = \frac {1 \pm \sqrt {1 - 4(0.16)}}{0.8} = \frac {1 \pm 0.6}{0.8} , \qquad \beta = 0.5 \ \text { or } \ \beta = 2 . \] Both values give \(\rho (1) = 0.4\): \(0.5/1.25 = 0.4\) and \(2/5 = 0.4\). The data cannot distinguish them, because the two models have exactly the same ac.f. The next section explains which one to choose: \(\hat \beta = 0.5\).