At first, here is a brief introduction to sub-Gaussian RVs. A random variable $X\in \mathbb{R}$ is said to be sub-Gaussian with variance proxy $\sigma^2$ if $\mathbb{E}(X) = 0$ and its moment generating function satisfies, for $\forall s\in\mathbb{R}$, $\mathbb{E}(e^{sX})\leq\exp(\frac{\sigma^2s^2}{2})$.

Next this is my question, given a $p\in(0,1)$, we have a general Bernoulli RV: \begin{equation*} T= \left\{ \begin{aligned} &1 &\text{ with probability of }1-p,\\ &1-\frac{1}{p} &\text{ with probability of }p. \end{aligned} \right. \end{equation*} I guess it is a sub-Gaussian RV. I just wonder what does the parameter of variance proxy $\sigma^2$ look like? Is it a function of $p$, $\sigma^2(p)$ or just only can find a universal bound $\sigma_0^2$, which is $\sigma_0^2>=\sigma^2$ for any $p\in(0,1)$??

Thank you guys!


In general, if $X$ is a mean zero random variable that takes values in $[a,b]$, then one can prove that $X$ is sub-Gaussian with variance proxy $(b-a)^2/4$. (See Lemma 1.8 in these notes for the standard proof.) So your random variable $T$ [after centering by $E[T]$] is sub-Gaussian with variance proxy $\frac{1}{4p^2}$.

  • $\begingroup$ Yes. But this bound is too loose. Anyway, thank you for your reply. $\endgroup$ – peter Jan 22 '18 at 3:23

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