Let $\{X_i\}$ be an $\alpha$-mixing random process with coefficients $\alpha(k)$ satisfying $\alpha(k)\leq Ca^k$ for some positive constants $a<1$ and $C$. Given that for any $i\in \mathbb{N}$, $E\lvert X_i\rvert^q\leq C$ and $E\lvert X_i\rvert^r\leq C$ by some constant $C>0$, Davydov's inequality states that $$\lvert \operatorname{Cov}( X_j,X_i)\rvert\leq C\alpha(\lvert i-j\rvert)^{1-1/q-1/r}.$$
Now, let the index $i$ be such that $\exists M>0:\lvert i-l\rvert>M, \forall l\in\{i',j,j'\}$. Why the following inequality is valid, using Davydov's result: $$\lvert \operatorname{Cov}( X_i,X_{i'}X_{j}X_{j'})\rvert\leq C\alpha(M)^{1-1/q-1/r} ?$$ I do not know how to deal with the random variable $X_{i'}X_{j}X_{j'}$ and its sigma-algebra generated. If anyone knows about mixing properties, then please, give me advices!
*obs. If $\sigma(X_{i'}X_{j}X_{j'})=\sigma(X_l:l\in\{i',j,j'\})$, then I can obtain the result. However, I don't think that a sequence of random variables generates the same sigma-algebra as its product.
Thanks in advance.
Indeed, in general, the equality $\sigma(X_{i'}X_{j}X_{j'})=\sigma(X_l:l\in\{i',j,j'\})$ may not hold. However, the inclusion $$ \sigma(X_{i'}X_{j}X_{j'})\subset \sigma(X_l:l\in\{i',j,j'\}) $$ holds, because letting $f(x,y,z)=xyz$, the equality hold $\left\{X_{i'}X_{j}X_{j'}\in B\right\}= \left\{ \left(X_{i'},X_{j},X_{j'}\right)\in f^{-1}(B)\right\}$ and the latter is $\sigma(X_l:l\in\{i',j,j'\})$-measurable.