Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 218 / 6

An MA($q$) process iswhere $({\epsilon }_t)$ is white noise of variance ${\sigma }^2$. For MA(1), $X_t={\epsilon }_t+\theta {\epsilon }_{t-1}$ andThe transformation $(\theta ,{\sigma }^2)\mapsto (1/\theta ,{\theta }^2{\sigma }^2)$ leaves this autocovariance unchanged. A Gaussian process is determined by its mean and covariance, so the parameters are not identifiable unless one selects, for example, the invertible representative $|\theta |<1$.

An MA($q$) process is invertible when its innovations admit a causal absolutely summable linear representation in present and past observations. With the backshift operator $B$, MA(1) satisfiesIf $|\theta |<1$, the geometric series converges absolutely:Thus the process is invertible.

Since $X_1$ is the sum of two independent centered Gaussian innovations,For any nonzero $a\in {\mathbb{R}}^n$,Expanding each $X_t={\epsilon }_t+\theta {\epsilon }_{t-1}$ expresses this as ${\sigma }^2$ times a sum of squared innovation coefficients. If all coefficients vanished, the coefficient of the latest innovation gives $a_n=0$, and backward induction gives every $a_t=0$, a contradiction. Hence $a^T\Sigma a>0$ and the covariance matrix is positive definite.

A method-of-moments estimator usesUnder $|\theta |<1$, choose the invertible rootwith the continuous value zero when ${\hat{r}}_1=0$. Alternatively maximize the exact Gaussian likelihood using the positive-definite covariance matrix from part (i), producing ${\hat{\theta }}_{\mathrm{ML}}$. Given either estimate,is the moment estimate; likelihood estimation may instead profile ${\sigma }^2$. Under a fixed interior parameter $|\theta |<1$ and standard stationary ergodic finite-moment regularity, both estimators are consistent and asymptotically normal, with Gaussian maximum likelihood asymptotically efficient.

The MA(1) model has much smaller AIC, $2871.79$ versus $3217.1$, so select MA(1). A nominal Wald 95% interval for its non-intercept parameter isThe estimate is close to the noninvertible boundary $\theta =1$, where the regular asymptotic normal approximation becomes poor and likelihood curvature can understate the true one-sided uncertainty. Therefore statement (iii) is the most plausible: the nominal interval is too narrow to attain its stated coverage reliably.