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For $X_t={\sum }_{j\ge 0}{\psi }_j{\eta }_{t-j}$ driven by centered unit-variance iid noise with a finite fourth moment, put ${\kappa }_4=E{\eta }^4-3$ and $m_3=E{\eta }^3$. Independence givesThe sum of the two linear-quadratic cross covariances is $m_3{\sum }_{j\ge 0}({\psi }_j^2{\psi }_{j+|h|}+{\psi }_j{\psi }_{j+|h|}^2)$. The cumulant term disappears for Gaussian noise. Merely assuming strong white noise does not justify the Gaussian formula; a fourth moment is needed for the variance of the quadratic transform.

With absolutely summable autocovariance, the asymptotic variance of the sample mean satisfiesThe sum is signed, rather than a sum of absolute values. It can be zero: the first difference of strong white noise has telescoping partial sums. A central limit theorem needs further dependence assumptions.

Reflecting an inside-unit-circle zero across the unit circle produces an invertible moving-average model with the same spectral density of a stationary process, after rescaling the driving white noise variance. For real $|\theta |>1$, $|1-\theta e^{-i\lambda }{|}^2={\theta }^2|1-{\theta }^{-1}{e}^{-i\lambda }{|}^2$. The transformed driving sequence is a linear innovation process; without Gaussianity it need not be strong white noise.

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 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.

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 an integer period $S$, ${\Delta }_S{X}_t=X_t-X_{t-S}$ uses the backshift operator as $1-B^S$. It annihilates a deterministic period-$S$ mean. If $X_t=m_t+{\epsilon }_t$ with $m_{t+S}=m_t$ and strong white noise, the result is the stationary moving average ${\epsilon }_t-{\epsilon }_{t-S}$.

If $X_t=c_t{\epsilon }_t$ with a deterministic periodic scale $c_t=c_{t-S}$ and strong white noise of variance ${\sigma }^2$, then ${\Delta }_S{X}_t=c_t({\epsilon }_t-{\epsilon }_{t-S})$. Its variance is $2{\sigma }^2{c}_t^2$, still seasonal when $c_t^2$ varies. A periodic scale model or variance standardization is more appropriate than blindly applying differencing.