# Upper bound of sub-gaussian norm of bounded random variable?

I am reading the High-Dimensional Probability by Dr.Roman Vershynin , where I stuck on some statement at page 28. where state as below:

Any bounded random variable $X$ is sub-gaussian with: $$\newcommand\norm[1]{\left\lVert#1\right\rVert} \norm{X}_{\psi_2}\leq \frac{\norm{X}_{\infty}}{\sqrt{log2}}$$

where $\newcommand\norm[1]{\left\lVert#1\right\rVert} \norm{X}_{\psi_2}$ is the sub-gaussian norm define as:

$$\newcommand\norm[1]{\left\lVert#1\right\rVert} \norm{X}_{\psi_2} =inf \left\{ t>0 : \mathbb{E} \exp{(\frac{X^2}{t^2}) \leq 2} \right\}$$

where $\newcommand\norm[1]{\left\lVert#1\right\rVert} \norm{X}_{\infty} :=( \mathbb{E} |X|^p)^{1/p}$ as $p \to \infty$

I can see how why the bounded random variable is sub-gaussian (hoeffing lemma ),but How could I see this upper bound of sub-gaussian norm?

• What is $p$? I think there is something wrong with the definition of $\|X\|_{\infty}$ – Kavi Rama Murthy Sep 16 '18 at 0:32
• @KaviRamaMurthy sorry I fix that – ShaoyuPei Sep 16 '18 at 0:35

If $t=(\sqrt {\log 2})^{-1} \|X\|_{\infty}$ then $Ee^{\frac {X^{2}} {t^{2}}} \leq e^{\log 2}= 2$ and hence $\|X\|_{\psi_2} \leq t$. I have used the fact that $\|X\|_{\infty}$ is nothing but the essential supremum of $X$.