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.

  • $\begingroup$ @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. $\endgroup$
    – pluskid
    May 12, 2015 at 19:17

1 Answer 1


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 $\frac{\lambda^2}{2}< \cosh\lambda -1$.


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