Suppose $X$ and $Y$ are independent sub-Gaussian random variables with 0 mean and $\sigma^2$ sub-Gaussian parameter. More specifically, $\mathbb E[\exp(a^T X)]\leq \exp\{\|a\|_2^2\sigma^2/2\}$ for all $a$, and the same holds for $Y$ as well.
I wish to upper bound the tail probability
$$
\Pr\left[|X^T Y|>t\right]
$$
using $\sigma^2$ and dimension $n$ (that is, both $X$ and $Y$ are $d$-dimensional random variables). How can I achieve this? $X^T Y$ does not seem to be either sub-Gaussian or sub-exponential.
Best Answer
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $\mathbb{R}^n$ with parameter $\sigma^2$ and for $r<\sigma^2$, $$ \mathrm{E}\bigg[\exp\Big\{\frac{r||X||^2}2\Big\}\bigg] = \frac{1}{(2\pi r)^{n/2}}\int_{\mathrm{R}^n} \exp\Big\{-\frac{||a||^2}{2r}\Big\}\, \mathbb{E}[e^{a^T X}]da \\ \le \frac{1}{(2\pi r)^{n/2}}\int_{\mathrm{R}^n} \exp\Big\{-\frac{||a||^2}{2r} + \frac{||a^2||\sigma^2}{2}\Big\}da = \frac{1}{\big(2\pi (\sigma^2-r)\big)^{n/2}}. $$ Therefore, thanks to independence, for any $\lambda<1$, $$ \mathrm{E}\big[\exp\{\lambda X^T Y\}\big] \le \mathrm{E}\bigg[\exp\Big\{\frac{\lambda^2\sigma^2||X||^2}2\Big\}\bigg]\le\frac{1}{\big(2\pi \sigma^2(1-\lambda^2)\big)^{n/2}}. $$ Hence you can easily derive an exponential bound for $\mathrm{P}(X^TY>t)$.