Concentration of two independent sub-Gaussian random variables 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.
 A: 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 $\mathrm{P}(X^TY>t)$.
A: By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[\| X\|>\sqrt t] \cup[\|Y\|>\sqrt t]$, so its probability is bounded above by $P(\| X\|>\sqrt t)+P(\| Y\|>\sqrt t)$, for which $O(\exp{-At})$ bounds can probably be derived, for some $\sigma$-dependent $A$.
A: Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $\sigma=1$ without loss of generality (otherwise, consider $X/\sigma$ and $Y/\sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
\[
E[e^{uX^TY} | X ] = E[e^{u^2\|X\|^2/2} ] = E[e^{u X^\top g}].
\]
Now by the law of total expectation, and a similar argument for $Y$,
\[
E[e^{uX^TY}] \le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2\|g\|^2/2} | g]] = E[e^{u g^Th}].
\]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
