Covariance inequality and strong mixing In this post I conclude that, for a error process $\{\epsilon_t\}$, $\mid Cov(\epsilon_t,\epsilon_l)\mid \leq C \alpha(\mid t-l \mid)^{1/p}$, for some constant $C$ (given conditions on the boundeness of integrals).
Now, my question is why the following statement is true:
Let$\{\epsilon_t\}$ be $\alpha$-mixing (or strongly mixing). If $\mid t-k\mid>M$ for $k\in\{a,b,c\}$, then $\mid Cov(\epsilon_t,\epsilon_a\epsilon_b\epsilon_c)\mid\leq C'\alpha(M)^{1/p}$ for some constant $C'$?
I don't know if there always exists an $k\in\{a,b,c\}:\mid Cov(\epsilon_t,\epsilon_a\epsilon_b\epsilon_c)\mid\leq \mid Cov(\epsilon_t,\epsilon_k)\mid$, because then I could use $\mid Cov(\epsilon_t,\epsilon_k)\mid \leq C \alpha(\mid t-k\mid)^{1/p}\leq C \alpha(M)^{1/p}$, since the dependence coefficient $\alpha$ is nonincreasing. On the other hand  $\alpha (\sigma(\epsilon_t),\sigma(\epsilon_k:k=a,b,c))\geq \alpha(\sigma(\epsilon_t),\sigma(\epsilon_k))$ for each $k\in\{a,b,c\}$, looking at the definition.
I know this question is quite specific (stochastic process), but I would appreciate any help. Thanks!
 A: For simplicity let $k=\min\{a,b,c\}>t$. Then $\epsilon_a\epsilon_b\epsilon_c\in \mathcal{F}_{k}^{\infty}$, and for $p>4$,
\begin{align}
|\operatorname{Cov}(\epsilon_t,\epsilon_a\epsilon_b\epsilon_c)|&\le 8\|\epsilon_t\|_p\|\epsilon_a\epsilon_b\epsilon_c\|_{p/3}\alpha(\mathcal{F}_1^{t},\mathcal{F}_{k}^{\infty})^{1-4/p} \\
&\le 8\sup_{t\ge 1}\|\epsilon_t\|_p^4\times \alpha(k-t)^{1-4/p},
\end{align}
where the first inequality uses Corollary A.2 in Hall & Heyde's Martingale Limit Theory and Applications (1980).
A: Define $f(A,B)=\mid P(A\cap B)-P(A)P(B)\mid$ for measurable sets $A,B$ and let $k=\min(a,b,c)$. Note that 
\begin{align}
\alpha(\sigma(\epsilon_t),\sigma(\epsilon_l:l=a,b,c))&=\sup\{f(A,B):A\in\sigma(\epsilon_t),B\in\sigma(\epsilon_l:l=a,b,c)\}\\
&\leq\sup\{f(A,B):A\in\sigma(\epsilon_t),B\in\sigma(\epsilon_l:l\geq k)\}\\
&\in\{\sup\{f(A,B):A\in\sigma(\epsilon_j),B\in\sigma(\epsilon_l:l\geq j+\mid k-t\mid)\}:j\in\mathbb{Z}\}\\
&\subseteq\{\sup\{f(A,B):A\in\sigma(\epsilon_l:l\leq j),B\in\sigma(\epsilon_l:l\geq j+\mid k-t\mid)\}:j\in\mathbb{Z}\}\\
&:=\{\sup\{f(A,B):A\in\mathcal{F}_1^j,B\in\mathcal{F}_{j+\mid k-t\mid}^\infty\}:j\in\mathbb{Z}\}
\end{align}
By taking the supremum and keeping in mind that $\alpha(n)$ is nonincreasing, it follows that $\alpha(\sigma(\epsilon_t),\sigma(\epsilon_l:l=a,b,c))\leq \alpha(\mid k-t\mid)\leq \alpha(M)$. So, using the Davydov's inequality, it follows that 
$$\mid Cov(\epsilon_t,\epsilon_a\epsilon_b\epsilon_c)\mid\leq C \alpha (M)^{1/p},$$
assuming $\epsilon_t\in L^q, \epsilon_a\epsilon_b\epsilon_c\in L^r$ and $1/q+1/r=1-1/p$.
