Apparently the following statement is true:

Let $X$ be an isotropic random vector, with finite support set $T \subset \mathbf{R}^n$. If $\|X\|_{\mathrm{s.g.}} = O(1)$, then $|T| \geq e^{cn}$.

Note that $\|\cdot\|_{\mathrm{s.g.}}$ is the sub-gaussian norm of a random vector. It can be defined $\|X\|_{\mathrm{s.g.}} = \sup_{u \in S^{n-1}} \|\langle X,u \rangle \|$, with $\|\cdot\|$ denote the sub-gaussian norm of a scalar random variable (i.e., if $Y$ is a scalar random variable, then $\|Y\| = \inf \{ t > 0 : \mathbf{E} \exp(Y^2/t^2) \leq 2\}$).

Questions: How does one start to prove a statement like this? I don't understand really what the $O(1)$ statement really means in this setting. (I do know the definition of Landau's big-O notation, but I just don't get why it is relevant here. Are we interested in the scaling behavior with respect to $n$?)

  • $\begingroup$ What is an isotropic random random? $\endgroup$ – Arin Chaudhuri Feb 19 '18 at 1:18
  • $\begingroup$ @ArinChaudhuri I have fixed the typo! $\endgroup$ – Drew Brady Feb 20 '18 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.