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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.

The $n$th cumulant is the coefficient of $t^n/n!$ in the cumulant-generating function. The first two cumulants are the expected value and variance; the third is the third central moment.

An Edgeworth expansion approximates a probability density function by a normal distribution multiplied by Hermite polynomials encoding higher cumulants. For a centered variable of variance ${\sigma }^2$ and third cumulant ${\kappa }_3$,The correction follows from inverse-transforming the cubic term of the cumulant-generating function. It is an asymptotic approximation near the bulk of the distribution; truncation can give a negative result in far tails.

For centered fast Fourier modes and $V=u\int {\varphi }^4/4!$, the two-fast-field vertex is $(u/4)\int {\phi }^2{\eta }^2$. Its connected second cumulant uses $⟨{\eta }^2(x){\eta }^2(y){⟩}_c=2G_{>}(x-y{)}^2$, producing $-u^2\int {\phi }^2(x){\phi }^2(y)G_{>}(x-y{)}^2/16$. Extracting the local zero-external-momentum quartic coefficient gives $u^{\prime }=b^{4-D}(u-3u^2{I}_2/2)+\cdots$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}K(p{)}^{-2}>0$. At $D=4$ its sign makes weak positive quartic coupling marginally irrelevant for increasing length scale, after tuning the critical mass. This local truncation omits generated derivative and higher-order operators; it is not the exact complete finite-shell effective action.

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.