Lower bounds on the MGF for a mean zero random variable with variance $\sigma^2$

Question

Let $X$ be mean-zero with variance $\sigma^2$. Is there a lower bound on the MGF for $X$ (or even simpler, $E e^X$) in terms of $\sigma^2$: $E[e^X] \ge f(\sigma^2)$?

What about the general case where we are given the first $k$ moments?

Answer 1

I don't think so. Consider for now the case $\sigma^2=1$; the claim is that the only lower bound you can get is the trivial $\mathbb{E}[e^X] \geq 1$. To see why, consider any $n\geq 1$, and the random variable $X_n$ defined by $$X_n = \begin{cases} \frac{1}{n} & \text{ with probability } 1-\frac{1}{n^2+1}\\ -n & \text{ with probability } \frac{1}{n^2+1}\\ \end{cases} $$ Then $\mathbb{E}[X_n] = 0$, $\operatorname{Var}[X_n] = 1$, but $$ \mathbb{E}[e^{X_n}] = e^{1/n} \left(1-\frac{1}{n^2+1}\right) + \frac{e^{-n}}{n^2+1} = 1+\frac{1}{n} + o\!\left(\frac{1}{n}\right) $$ As $n\to \infty$, this gets arbitrarily close to $1$.

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For general $\sigma>0$, consider $Y_n :=\sigma X_n$. The same computations will give $\mathbb{E}[Y_n] = 0$, $\operatorname{Var}[Y_n] = \sigma^2$, but $$ \mathbb{E}[e^{Y_n}] = e^{\sigma/n} \left(1-\frac{1}{n^2+1}\right) + \frac{e^{-\sigma n}}{n^2+1} \leq e^{\sigma/n} + \frac{1}{n^2+1} = 1+\frac{\sigma}{n} + o\!\left(\frac{1}{n}\right) $$