Bounding the expectation of random variables' product.

164 Views Asked by At

Let $\{X_i\}$ be a sequence of strongly mixing random variables, not necessarily (strict) stationary. Assume that $E\lvert X_i\rvert^4\leq C<\infty$ and that there is $0<a<1$ such that the mixing coefficient satisfies $\alpha(m)<Ca^{m}$. Here $C>0$ is a generic constant. Is there any chance (maybe, adding some assumptions) that $$\sum_{i,i',j,j'=1}^{T} E(X_i X_{i'} X_{j} X_{j'})\leq CT^2 v_T$$ for some $C>0$ constant and $v_T$ some slowly positive sequence converging to infinity?

This is far from trivial question. If someone know about mixing random processes, could you give me feedbacks?

For a while, I could only obtain a bound of order $O(T^3 v_T)$. This question relates to this.

1

There are 1 best solutions below

7
On BEST ANSWER

Theorem 2.1 in these notes by Emmanuel Rio states the following: for a centered sequence $(X_i)_{i\geqslant 1}$, the following inequality holds. $$ \mathbb E\left[\left(\sum_{i=1}^nX_i\right)^4\right]\leqslant 3\left(\sum_{i,j=1}^n \lvert \mathbb E\left[X_iX_j\right]\rvert\right)^2 +48 \sum_{k=1}^n\int_{0}^1\min\{\alpha^{-1}(u),n\}^3Q_k(u)^4du, $$ where $$ Q_k(u)=\inf\{t>0\mid\mathbb P\{\lvert X_k\rvert>t\}\leqslant u\} $$ and $\alpha^{-1}(u)=\operatorname{Card}\{n,\alpha(n)\geqslant u\}$.

After having expanded $\left(\sum_{t=1}^TX_t\right)^4$ one finds exactly $\sum_{i,i',j,j'=1}^{T} E(X_i X_{i'} X_{j} X_{j'})$. The term $\lvert \mathbb E\left[X_iX_j\right]\rvert$ can be control a the covariance inequality and $\left(\sum_{i,j=1}^T \lvert \mathbb E\left[X_iX_j\right]\rvert\right)^2$ is of order $T^2$.

If we furthermore assume that $\sup_{k\geqslant 1}\int_0^1\alpha^{-1}(u)^2Q_k(u)^4du$ is finite, then $\sum_{i,i',j,j'=1}^{T} E(X_i X_{i'} X_{j} X_{j'})$ is of order $T^2$. When the mixing rates are bounded by $a^n$ for $0<a<1$, $\alpha^{-1}(u)$ behave like $-c\ln u$ but finiteness of $\sup_{k\geqslant 1}\int_0^1\alpha^{-1}(u)^2Q_k(u)^4du$ is more restrictive than uniform boundedness of the moments of order $4$.