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