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 $n$?)

Edit (3/19/24): A more precise version of this question. Let us drop the subscript s.g. and without comment use that $\|\cdot\|$ denotes the sub-Gaussian norm of a scalar or vector-valued random variable.

The question is asking the following. Show the following is true: for every $K > 0$, there exists a constant $c = c(K) > 0$ such that: for any integer $n \geq 1$, for every finitely supported random variable $X \in \mathbf{R}^n$ and $\mathbf{E} X \otimes X = I_n$, and $\|X\| = K$, the support of $X$ is at least of cardinality $e^{cn}$.

  • $\begingroup$ 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
    Commented Feb 19, 2018 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
    Commented Feb 20, 2018 at 0:06
  • $\begingroup$ I want to point out that the question above---makes sense. Yes, it perhaps requires more care to state, but I think the intent of the question was quite clear as originally stated. I have updated it to include a more precise statement. $\endgroup$
    – Drew Brady
    Commented Mar 19 at 22:33

1 Answer 1


Note that for any sequence of sub-Gaussian random variables $Y_1, \dots, Y_N$, we have, for an absolute constant $c > 0$, that $$ \mathbf{E}[\max_{n \leq N} |Y_n|^2] \leq c K^2 \log N, $$ where $K = \max_{n \leq N} \|Y_n\|$. (For instance, one may prove this using the fact that the square of a sub-Gaussian random variable is sub-Exponential). Thus, using that $X$ is finitely supported, $$ n = \mathbf{E} \|X\|_2^2 \leq \mathbf{E}\max_{t \in T} |X_{\tilde t}|^2 \leq c \|X\|^2 \log |T|, $$ where for any $t \in T =\mathrm{supp}(X)$ we took $\tilde t = t/\|t\|_2$, so that $\tilde X_t$. The first equality holds by considering $t = X$, so that $X_{\tilde X} = \|X\|_2$. Inverting the bound, $$ |T| \geq e^{C n}, $$ where $C = c'/\|X\|^2$, as required.


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