Are Bernoulli distributions R-sub-Gaussian?

Question

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 $R$?

Answer 1

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 $R=1/2$.