Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 208 / 1

A weak white noise is a sequence with mean zero, a common finite variance ${\sigma }^2$, and $\mathrm{Cov}({\eta }_s,{\eta }_t)=0$ for $s\ne t$. A strong white noise is a sequence of independent and identically distributed random variables with mean zero and finite variance. Strong white noise is therefore weak white noise, but the converse need not hold. Neither definition, by itself, requires a Gaussian distribution.

For the spectral questions take the unique weakly stationary process solving the equation, as is customary for a stable autoregressive moving-average model. The equation alone also admits nonstationary solutions differing by $C{4}^{-t}$, for which a spectral density of a stationary process need not exist. Let $B$ be the backshift operator. Since the autoregressive root is $4$, the stationary solution is causal andUsing the convention $\gamma (h)={\int }_{-\pi }^{\pi }{e}^{ih\lambda }f(\lambda )d\lambda$, the spectral density of a stationary process isThe stationary mean is zero, since $\mu -\mu /4=0$.

Here the innovation process consists of linear innovations, the linear one-step prediction errors: ${\epsilon }_t=X_t-{\mathrm{proj}}_{{\mathcal{H}}_{t-1}}{X}_t$, where ${\mathcal{H}}_{t-1}$ is the closed linear span of the past $X$'s in $L^2$. For Gaussian processes this is also the conditional expectation prediction error. For general non-Gaussian strong white noise these two notions can differ.

The given ${\eta }_t$ is not the linear innovation process. The moving-average factor has its zero at $1/3$, inside the unit disk, and is noninvertible as a causal moving-average filter. The identityis the moving-average root reflection that places this zero outside the unit disk. Thus the causal invertible representation has innovation variance $9$, rather than the given variance $1$.

For an explicit verification, defineThe filter has constant squared modulus $9$, so $\epsilon$ is weak white noise with variance $9$. The new moving-average factor has root $3$ and is invertible, while its autoregressive factor is causal. Hence the past spans of $X$ and $\epsilon$ agree, and $X_t-{\epsilon }_t$ belongs to that past span. Orthogonality of ${\epsilon }_t$ to past $\epsilon$ therefore identifies it as the linear innovation process. The variance difference proves that it cannot be ${\eta }_t$. If the original noise is Gaussian, the new linear innovations are independent Gaussian variables; without Gaussianity they need only be uncorrelated random variables.

The root-reflected, invertible autoregressive moving-average model isThis follows by multiplying the filter identity for $\epsilon$ by $1-B/3$. Its moving-average root $3$ and autoregressive root $4$ are both outside the unit disk. Thus it is the representation in terms of the innovation process, rather than merely another noise representation with the same spectrum.

Expanding the stable autoregressive inverse gives the infinite moving-average representation in linear innovations:ThereforeFor comparison, the causal representation in the originally supplied noise isBoth converge in $L^2$, since their coefficients are square summable. The second is causal, but its driving noise is not the linear innovation process.

Use the original-noise coefficients ${\psi }_0=1$ and ${\psi }_j=-11\cdot 4^{-j}$ for $j\ge 1$. The autocovariance function of a causal linear process is ${\gamma }_X(h)={\sum }_{j\ge 0}{\psi }_j{\psi }_{j+|h|}$, since the original noise has variance $1$. ThusFor $h\ge 1$,By covariance symmetry,In particular, all nonzero-lag covariances are negative, despite the positive autoregressive coefficient.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.