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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:

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

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?

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  • $\begingroup$ What is $p$? I think there is something wrong with the definition of $\|X\|_{\infty}$ $\endgroup$ Commented Sep 16, 2018 at 0:32
  • $\begingroup$ @KaviRamaMurthy sorry I fix that $\endgroup$
    – ShaoyuPei
    Commented Sep 16, 2018 at 0:35

1 Answer 1

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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 $X$.

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