= Solution
The selected zero-mean <autoregressive moving-average process> is
$$
X_t=\phi X_{t-1}+\varepsilon_t+\theta\varepsilon_{t-1},
\qquad
\varepsilon_t\overset{\mathrm{iid}}\sim N(0,\sigma^2).
$$
The reported <maximum-likelihood estimates> are
$$
\widehat\phi=0.6997,
\qquad \widehat\theta=0.9510,
\qquad \widehat\sigma^2=0.4944.
$$
Using the displayed asymptotic standard error gives the <Wald confidence interval>
$$
0.9510\pm1.96(0.3287)=[0.307,1.595].
$$
This normal interval is unreliable and likely too narrow because the series has only about twenty observations, the moving-average estimate is near the noninvertibility boundary $\theta=1$, and the same data were used to select the model. The finite-sample likelihood is consequently skewed and model-selection uncertainty is omitted.
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