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.


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 $\mathrm{P}(X^TY>t)$.

  • $\begingroup$ You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $\mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty. $\endgroup$ Aug 25 '17 at 17:06
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    $\begingroup$ Thank you for this answer! For anyone looking for a reference in the literature for this result, see Lemma 2.7.7 in the book "High-Dimensional Probability" by Roman Vershynin (1st ed, 2018). $\endgroup$
    – OtZman
    Oct 17 '19 at 2:16
  • $\begingroup$ @OtZman, thanks for the reference! In fact, Vershynin proves this without the independence assumption (and easier imo). $\endgroup$
    – zhoraster
    Oct 17 '19 at 11:21

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 $A$.

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    $\begingroup$ Thanks for your answer. Unfortunately, I think this upper bound is very loose because $\|X\|$ is far from 0 because $\mathbb E\|X\|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $\mathbb EX^T Y = 0$. In particular, in the $P(\|X\|>\sqrt{t})$ bound the $t$ can never be arbitrarily small. $\endgroup$ Jul 8 '17 at 1:36
  • $\begingroup$ And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0. $\endgroup$ Jul 8 '17 at 1:37

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.


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