Sub-gaussian norm vs. variance proxy

Question

When studying sub-gaussian variables, I have come across several definitions. One of the most common uses the concept of a "variance proxy," i.e. a (mean-zero) random variable $X$ is sub-gaussian with variance proxy $\sigma^2$ if $$ \mathbb{E}\exp(sX) \leq \exp \left(\frac{\sigma^2s^2}{2}\right)$$ for all $s \in \mathbb{R}$. Now, in other texts, e.g. Roman Vershynin's "High-dimensional Probability," sub-gaussian random variables are introduced alongside a kind of Orlicz norm, namely $$ \|X\|_{\psi_2} = \inf\left\{k>0 \vert \mathbb{E}\exp\left(\frac{X^2}{k^2} \right) \leq 2 \right\}$$ which then satisfies a set of equivalent properties characterizing sub-gaussianity, including $$ \mathbb{E}\exp(\lambda X) \leq \exp(C \lambda^2 \|X\|_{\psi_2}^2)$$ for all $\lambda \in \mathbb{R}$. Is there some explicit relation between the sub-gaussian norm and the variance proxy? Or are they perhaps even the same thing? If not, in which scenarios may they be equal?

Answer 1

One has to start with the integral representation $$e^{\frac{X^2}{2r^2}}=\int_{-\infty}^{\infty}e^{sX-\frac{r^2s^2}{2}}\frac{r}{\sqrt{2\pi}}ds.$$ Therefore if $E(e^{sX})\leq e^{\frac{\sigma^2 s^2}{2}}$ we get $E(e^{\frac{X^2}{2r^2}})\leq \frac{r}{\sqrt{r^2-\sigma^2}}.$