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In Vershynin's Introduction to the Non-asymptotic Analysis of Random Matrices, the sub-gaussian norm $\|X\|_{\psi_2}$ of a sub-gaussian random variable $X$ is defined by $$ \|X\|_{\psi_2}=\sup_{p\ge1}p^{-1/2}\left(\mathbb{E}|X|^p\right)^{1/p}. $$

According to this post, it is possible that $\|X\|_{\psi_2}^2\neq\text{Var}(X)$. My question is: Do there exist constants $C_2\ge C_1>0$ such that for all sub-gaussian variables $X$, it holds that $$ C_1\text{Var}(X)\le\|X\|_{\psi_2}^2\le C_2\text{Var}(X)? $$

Kittayo
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  • I also had the same question, but I saw in this book in definition 2.5.2 the sub-gaussian norm defined differently. – Dylan Dijk Apr 14 '23 at 09:37
  • I think we can use a slight modification of the proof of proposition 2.1 from here. With the fact that for a subGaussian rv with mean zero we have: $ \mathbb{E}[\exp{(\lambda X})] \leq \exp{(C \lambda^2||X||^2_{\psi_2})}$. To get that $\mathbb{E}[X^2] \leq C ||X||^2_{\psi_2}$, and hence the upper bound you were looking for. – Dylan Dijk Apr 14 '23 at 10:06
  • Burkholder gundy Davis inequality might be helpful – fGDu94 Jul 26 '23 at 18:08

1 Answers1

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One inequality is easy: $\text{Var}(X)\le\mathbb{E}[X^2]\le 2\|X\|_{\psi_2}^2$. The other inequality does not hold. For every $n\in\mathbb{N}$, define a random variable $X_n$ by $$ \mathbb{P}(X_n=-n) = \mathbb{P}(X_n=n) = \frac{1}{2n^2} \quad\text{and}\quad \mathbb{P}(X_n=0) = 1-\frac{1}{n^2}. $$ Then $\mathbb{E}X_n=0$ and $\text{Var}(X_n)=\mathbb{E}[X_n^2]=1$. The sub-gaussian norm of $X_n$ is $$ \|X_n\|_{\psi_2} = \sup_{p\ge 1}p^{-1/2}\left(\mathbb{E}|X_n|^p\right)^{1/p} = \sup_{p\ge 1} p^{-1/2}n^{1-2/p}\in\left[\frac{1}{2}n^{1/2},~n\right]. $$ Since $\|X_n\|_{\psi_2}^2/\text{Var}(X_n)\ge n/4\to\infty$ as $n\to\infty$, then no such $C_2$ exists.

Kittayo
  • 801