Product of three or more independent sub-Gaussian varibles

Question

A random variable $X$ is called subgaussian of order $\sigma^2$ if $\log E[exp\{\theta X\}]\leq \frac{1}{2}\theta^2\sigma^2$ for every $\theta\in\mathbb R$.

Given a sequence of independent subgaussian r.w. $X_k, k\in\mathbb N$, of order $\sigma_k^2$. It is well known that we can choose $\theta$ small such that $\theta X_1X_2$(here independence is not necessary) has finite exponential moment: for instance, one can show that $$E[e^{\theta X_1X_2}]\leq 3$$ if $\theta\leq (4\sigma_1\sigma_2)^{-1}$. My question is that is there some analogous conclusion holds for the product of three or more (independent) subgaussian r.w.? Namely, can we find a small $\theta$ such that $\theta X_1X_2\cdots X_k$ has finite exponential moment for $k\geq 3$?

Actually what I need is a particular case. Let $Y_i$ be a sequence of $\textbf{independent} $ Bernoulli r.w. with mean $p_i, 1\leq i\leq n$. Define $X^n_i:=\frac{Y_i-p_i}{n}.$ By Hoeffding lemma we could show that $X_k^n$ is subgaussian of order $O(n^{-2})$. For each $k\leq n$, define $$D^n_k=\sum_{\substack{i_1\neq\cdots\neq i_k\\1\leq i_1,\cdots, i_k\leq n}}X^n_{i_1}X^n_{i_2}\cdots X^n_{i_k}.$$ What is the best(or largest) $\theta_n$ we can set to make $E[e^{\theta_n \sum_{k=3}^nD_k^n}]<C$ where $C>0$ is some constant not depending on $n$. Any hope that $\theta_n=O(n)$? Under the assumption of independence, I can only prove $E[e^{\theta_n D_k^n}]<C$ for $k=2$, if $\theta\leq n\varepsilon$ for some small $\varepsilon>0$. I cannot extend it to any $k\geq 3$, not to mention the desired term with the sum from $3$ to $n$.

Any idea or reference is very appreciated!

Answer 1

This sub-Gaussian technique cannot work in general for any $k\ge3$. In particular, if $k\ge3$ and $X_1,\dots,X_k$ are iid standard normal, then $$E\exp\Big(c\prod_1^k X_i)=\infty$$ for any $c>0$.

Indeed, let $[k]:=\{1,\dots,k\}$, $X:=(X_1,\dots,X_k)$, let $(U_1,\dots,U_k)$ denote a uniformly distributed unit random vector, and let $|\cdot|$ denote the Euclidean norm. Then for some real $C_k>0$ $$\begin{aligned}E\exp\Big(c\prod_1^k X_i\Big) &\ge E\exp\Big(c\prod_1^k X_i\Big)1\big(X_i>\tfrac1{2\sqrt k}\,|X|\ \forall i\in[k]\big) \\ &=C_k\int_0^\infty\exp\big(c\,(\tfrac1{2\sqrt k})^kr^k-r^2/2\big)\,r^{k-1}dr\\ &\times P\big(U_i>\tfrac1{2\sqrt k}\ \forall i\in[k]\big) =\infty\end{aligned}$$ if $c>0$ and $k\ge3$.

Answer 2

As indicated in the comment, the boundedness can be used to prove that the product is sub-Gaussian. Suppose $X_{1}, \ldots, X_{n}$ are bounded such that for $i \in [n]$ there exists a constant such that $|X_{i}| \leq C_{i}$. Then it suffices to prove for some finite $R$, \begin{equation*} (\mathbb{E}|\prod_{i \in [n]}X_{i}|^{p})^{1/p} \leq R\sqrt{p} \end{equation*} We can use Holder's Inequality to prove the claim. \begin{align*} (\mathbb{E}|\prod_{i \in [n]}X_{i}|^{p})^{1/p} &\leq (\prod_{i \in [n]}\mathbb{E}|X_{i}|^{np})^{1/pn} \\ &\leq (\prod_{i\in[n]}C_{i}^{np})^{1/pn} \\ &= \prod_{i \in [n]}C_{i} \end{align*} From Proposition 2.5.2 in High-Dimensional Probability, we see the product is sub-Gaussian.