 Solution

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.