For zero-mean r.v. $X$ with var. $\sigma^2$, want to show $E[e^{2X}]\leq e^{2\sigma^2}$.

Question

Let $X$ have zero mean, $E[X]=0$ and finite variance $E[X^2]=\sigma^2<\infty$. I'm trying to show $$ E[e^{2X}] \leq e^{2\sigma^2}. $$ I started out with this related question, but I hadn't quite phrased it correctly. So this is the update. Several points worth noting. You can show that $$ E[e^{X}] \leq E[X^2]^{1/2}\,E[e^{2X}]^{1/2} + 1. $$ To see this, note that for $x\in\mathbb{R}$, $$ e^x-1 \leq |x|\,e^x, $$ so that $$ E[e^X-1] \leq E[|X|\,e^X]. $$ The zero-mean assumption implies $0\leq E[e^X-1]$, so Cauchy-Schwarz gives $$ E[e^X-1] \leq E[X^2]^{1/2}\,E[e^{2X}]^{1/2}. $$ Move $E[-1]=-1$ to the right side to finish the derivation. If you replace $X$ by $nX$, you get something that starts to look recursive, $$ E[e^{nX}] \leq E[(nX)^2]^{1/2}\,E[e^{2nX}]^{1/2} + 1. $$ The recursion suggests (to me) a power series lurking. Unfortunately, it gets recursive in a really ugly way because of the square roots. If there's a way to do it, I suspect there's a nicer way to show it than through the recursion. If $X$ is Gaussian, then the statement is true, which I discovered by looking at the sub-Gaussian literature. I'd hoped some of their techniques would help, but so far they haven't. I'd prefer to avoid a Gaussian or sub-Gaussian assumption.

Answer 1

Counterexample:

Say, $\sigma=1$. Take $X_N$ to be the random variable taking values in $\left\{-\frac{1}{N},N\right\}$ for $N>1$ with $\text{Prob}\left(X_N=-\frac{1}{N}\right)=\frac{N^2}{N^2+1}$ and $\text{Prob}\left(X_N=N\right)=\frac{1}{N^2+1}$. Obviously, $E\left[X_N\right]=0$ and $\text{Var}\left(X_N\right)=1=\sigma^2$.

Now, $E\left[X_N^n\right]=\frac{N^2\cdot(-1/N)^n+1\cdot N^n}{1+N^2}$. If $n\geq 3$, then $$E\left[X_N^n\right]\geq\frac{N^n-1}{N^2+1}>N^{n-2}-N^{n-3}\,.$$ Thence, for $t>0$, $$E\big[\exp\left(tX_N\right)\big] > \sum_{n\geq 3}\,\frac{t^n}{n!}\left(N^{n-2}-N^{n-3}\right)=\frac{N-1}{N^3}\left(\exp(Nt)-1-Nt-\frac{N^2t^2}{2}\right)\,.$$ That is, $E\big[\exp\left(tX_N\right)\big] \in \Omega\left(\frac{1}{N^2}\exp(Nt)\right)$ for large $N$, implying that $E\big[\exp\left(tX_N\right)\big]\to \infty$ as $N\to\infty$. So, the bound $$E\big[\exp\left(2X_N\right)\big] \leq \exp\left(2\sigma^2\right)=\exp(2)$$ cannot hold for all $N>1$.