$\newcommand{\V}{\text{Var}}$
$\newcommand{\E}{\mathbb E}$
Definition:
A mean zero random variable $X$ is $\sigma$ sub-Gaussian if for all
$\lambda \in \mathbb R$,\begin{align} \E\left[\exp\left(\lambda X\right)\right] \leq
\exp\left(\frac{\lambda^2 \sigma^2}{2}\right). \end{align}
Problem:
If $X$ has mean zero and is $\sigma$ sub-Gaussian, show that $\V(X)
\leq \sigma^2$.
My attempt:
I am looking at Proposition 2.1 in these notes and 3.2 in these notes, which both supposedly solve this problem, but my analysis chops aren't quite good enough to justify what I'm doing.
We have that for all $\lambda \in \mathbb R$,
\begin{align}
\E\left[\exp\left(\lambda X\right)\right] &\leq \exp\left( \frac{\lambda^2 \sigma^2}{2}\right)\\
\implies 1 + \lambda \E[X] + \frac{\lambda^2 \E[X^2]}{2!} &\leq 1 + \frac{\lambda^2 \sigma^2}{2} + o(\lambda^2)
\end{align}
as $\lambda \to 0$. After subtracting and canceling certain terms from both sides of the inequality, letting $\lambda \to 0$, and using that $\E[X] = 0$ by assumption, we get that
\begin{align}
\E[X^2] &\leq \sigma^2\\
\implies \V(X) &\leq \sigma^2
\end{align}
since in this case $\E[X^2] = \V(X)$.
Question:
Is this a legitimate solution? I am not totally comfortable with the manipulations or little-$o$ notation. I did a bunch of Googling but didn't find anything that seemed to help.
Thanks.
Best Answer
That is valid, but you may want to expand on the
since that's the crux of the argument.