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Apparently the following statement is true:

If $X$ is an isotropic random vector, with finite support set $T \subset \mathbf{R}^n$, then 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 $n$?)

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  • $\begingroup$ I'm having trouble even making sense of the statement to prove. In what sense is "$O(1)$" meant, given that $X$ has finite support? In what sense is $X$ "isotropic" (which ordinarily means invariant under all rotations, which implies non-finite support unless $X$ is identically $0$)? What does "$c$" refer to in the conclusion? Given that $T$ is finite, then evidently "$|T|$" is the cardinality of $T$, but then isn't it trivial that there exists some $c$ for which $|T|\gt e^{cn}$? (Any $c\lt 0$ obviously works.) $\endgroup$ – whuber Feb 19 '18 at 15:39
  • $\begingroup$ @whuber Thank you for taking a look at this. I think (but am not entirely sure) what this means is that $\|X\|_{s.g.}$ should be bounded above by a constant which is independent of $n$. $X$ being isotropic means that $\mathbf{E}XX^T = I_n$. I don't think the statement is trivial as you say, but if you still think it is, perhaps you can explain why. $\endgroup$ – Drew Brady Feb 20 '18 at 0:06

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