A distribution $\nu$ is said to be R-sub-Gaussian, $R>0$, if for all $t\in\mathbb{R}$ we have \begin{align} \mathbb{E}_{X\sim\nu}\left[\exp(tX-t\mathbb{E}(X))\right]\le\exp(R^2t^2/2) \end{align} Is it possible to write a Bernoulli distribution with mean $p$ as an R-sub-Gaussian distribution for some $R$?
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1$\begingroup$ All bounded variables are subgaussian. See Example 2.3 here. $\endgroup$– VHarisopMay 14, 2018 at 16:53
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This is essentially the content of Hoeffding's lemma, which states that a random variable supported on the interval $[a,b]$ is sub-Gaussian with $R = (b-a)/2$. So Bernoulli random variables are sub-Gaussian with $R=1/2$.
answered May 14, 2018 at 16:54
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$\begingroup$ Doesn't Hoeffding's lemma assume that $\mathbb{E}(X)=0$? $\endgroup$– jonemMay 14, 2018 at 17:02
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$\begingroup$ Nevermind, I understand your description now after reading this: blog.wouterkoolen.info/BernoulliSubGaussian/post.html $\endgroup$– jonemMay 14, 2018 at 17:27
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$\begingroup$ @jonem Apply Hoeffding's lemma to the random variable $Y := X - E[X]$. $\endgroup$ May 14, 2018 at 17:34