# Bound variance proxy of a subGaussian random variable by its variance

If I know $X$ is a sub-Gaussian random variable, and I know it has finite variance $\sigma^2$. Can I assert that $\sigma^2$ is a valid variance proxy for $X$?

Definition (sub-Gaussian Random Variable) A random variable $X$ is called sub-Gaussian with variance proxy $\sigma^2$, if

• $E[X] = 0$
• $E[\exp(sX)] \leq \exp(s^2\sigma^2/2), \quad \forall s\in\mathbb{R}$

Note the variance proxy is not unique. Any larger number than a valid variance proxy is still a valid variance proxy.

I can easily show using the moment generating function that the variance proxy of a sub-Gaussian random variable is greater than or equal to its variance. But I'm not sure about the inverse direction.

• @MichaelHardy sorry for making it confusing. I have edited the text. The variance proxy is not unique. I just wonder if the variance could be one valid variance proxy. May 12, 2015 at 19:17

The answer is no. Here is an example. For $$p\in(0,1)$$, define the random variables
$$X_p = \begin{cases}1&\text{with probability p/2;}\\-1&\text{with probability p/2;}\\0&\text{with probability 1-p.}\end{cases}$$
Clearly $$\mathbb E[X_p]=0$$. Since $$X_p$$ is bounded, you have that it is subgaussian by Hoeffding's lemma.
You can evaluate the variance $$\mathbb V[X_p] = p$$. Moreover you have $$\mathbb E[e^{\lambda X_p}] = 1 + (\cosh\lambda-1)p\,.$$
For $$p$$ small enough you can find some $$\lambda$$ such that $$e^{\lambda^2p/2}< 1 + (\cosh\lambda-1)p$$ and hence conclude that $$X_p$$ is not subgaussian with variance proxy $$p$$.
Indeed, Taylor expanding in $$p$$ you get $$e^{\lambda^2p/2} = 1 + \frac{\lambda^2}{2}\,p + o(p)$$ and for $$\lambda$$ large enough we have $$\frac{\lambda^2}{2}< \cosh\lambda -1$$.