2
$\begingroup$

I will use the same notation and definitions in High Dimensional Probability, by Roman Vershynin.

I have a sub-Gaussian vector $y$, in $\mathbb{R}^n$ and sub-Gaussian norm $C$ non dependent on $n$. I also assume that $y$ respects the concentration of the norm property, which I define as

$\frac{1}{\sqrt n} (|y|^2 - \mathbb{E}|y|^2)$ is a sub-exponential random variable with sub-exponential norm $C'$ non dependent on $n$, where $|\cdot|$ is the Euclidean norm;

(note that this property holds, for example, for standard Gaussian vectors).

The question is the following: does such a vector respect also the following property?

$\sum_i s_i \,(y_i^2 - \mathbb{E}y_i^2)$ is a sub-exponential random variable with sub-exponential norm $C''$ non dependent on $n$, where $|s| = 1$, and $s_i \geq 0 \; \forall i$.

Note that we are not assuming the independence of the entries of $y$, and that both sub-Gaussianity and concentration of the norm are necessary properties for the thesis to hold. In particular, here is an example of a sub-Gaussian vector that does not respect the concentration of the norm property:

$v = Z g$, where $g$ is a standard Gaussian vector, and $Z$ is a scalar random variable uniform in $[0, 1]$.

However, I'm not sure the result is true in general, since I wasn't able neither to prove it nor to find a counterexample.

$\endgroup$

1 Answer 1

0
$\begingroup$

Your desired conclusion holds, even without the additional assumption that $\frac1{\sqrt n}(|y|^2-E|y|^2)$ is a sub-exponential random variable with a sub-exponential norm not dependent on $n$.

We only need to assume that $y$ is a sub-Gaussian vector $y$ in $\mathbb{R}^n$ with a sub-Gaussian norm $C$ not dependent on $n$: $$E\exp\Big(\sum_1^n y_i^2/C^2\Big)\le2.$$ Then, by Jensen's inequality, $$\exp\Big(\sum_1^n Ey_i^2/C^2\Big)\le E\exp\Big(\sum_1^n y_i^2/C^2\Big)\le2.$$ So, for any $s\in\mathbb{R}^n$ with $|s|=1$, $$ \begin{aligned} E\,&\exp\Big(\Big|\sum_1^n s_i(y_i^2-Ey_i^2)\Big|/(2C^2)\Big) \\ &\le E\exp\Big(\sum_1^n (y_i^2+Ey_i^2)/(2C^2)\Big) \\ &\le\sqrt{E\exp\Big(\sum_1^n (y_i^2+Ey_i^2)/C^2\Big)} \\ &=\sqrt{E\exp\Big(\sum_1^n y_i^2/C^2\Big) \;\exp\Big(\sum_1^n Ey_i^2/C^2\Big)} \\ &\le\sqrt{2\times2}=2. \end{aligned}$$ So, the sub-exponential norm of $\sum_1^n s_i(y_i^2-Ey_i^2)$ is $\le2C^2$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for your reply. However, I believe there is a misunderstanding on the definition of the sub-Gaussian norm in your solution. In your first equation, you assumed the sub-Gaussianity of the squared norm of $y$, which is not our hypothesis. For a vector to be sub-Gaussian with norm $C$, the following needs to hold: $C = \sup_{|u| = 1} | \langle u, y \rangle |_{\psi_2}$, where $|\cdot |_{\psi_2}$ is the usual scalar sub-Gaussian norm. $\endgroup$
    – SiMohani
    Commented Jul 6, 2021 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.