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First justify the dependence structure rather than assume that adjacent squares behave like an ordinary lag-one ARCH model. Write $Z_t=X_t^2$. Iterating its nonnegative recurrence givesThe product coefficient has expected value ${\alpha }_2^m$, so it tends to zero in probability. The stationary $Z_{t-2m}$ are a tight family; consequently the remainder tends to zero in probability, without needing independence between that remainder's factors. The increasing partial sums therefore give the representationThis proves the parity decomposition of a lag-two ARCH process: even and odd observations are functions of disjoint noise families. It also shows that the positive scale ${\sigma }_t$ is a function of past noise, independent of ${\epsilon }_t$. The conditional expectation of $X_t$ given past noise is zero, and the second-moment recursion is $m_2={\alpha }_0+{\alpha }_2{m}_2$. independence of the parity families gives the adjacent-square covariance. ThusTo establish finiteness of the fourth moment before using its recursion, note that the $L^2$ norm of the $k$th term in the positive series for $Z_t$ is ${\alpha }_0\sqrt{3}({\alpha }_2\sqrt{3}{)}^k$. The triangle inequality in L2 space makes the series square-integrable when $3{\alpha }_2^2<1$. independence of the current noise and past scale then yieldsand henceIf $3{\alpha }_2^2\ge 1$, a finite $m_4$ would make $(1-3{\alpha }_2^2)m_4=3{\alpha }_0^2+6{\alpha }_0{\alpha }_2{m}_2>0$, which is impossible. Its fourth moment is then infinite. The adjacent-square product remains integrable by parity independence, even in that case.

This covariance is also zero, but symmetry is essential to the proof. Write ${\epsilon }_t={\eta }_t|{\epsilon }_t|$, where ${\eta }_t$ is an independent fair sign. The normal distribution makes that sign independent of its magnitude and of every other driving variable. Conditional on the magnitudes and all noise except this sign, changing ${\eta }_t$ flips $X_t$ and leaves $X_t^2$ unchanged. Every future conditional scale uses only squared past observations, so $X_{t+h}$ is unchanged for $h>0$.

Thus $f(X_{t+h})$ is independent of the remaining fair sign in $X_t$. Averaging that sign gives $\mathbb{E}[X_{t}f(X_{t+h})]=0$, and henceThe Cauchy-Schwarz inequality again guarantees integrability. For odd $h$, the parity decomposition of a lag-two ARCH process also gives independence directly. For even $h$, the sign symmetry of an ARCH process supplies the argument; the martingale difference sequence property alone would not justify this reversed covariance.