Bounding the expectation of a function of a zero-mean random variable

Question

I have a random variable $X$ with mean zero, $E[X]=0$, and finite second moment, $E[X^2]=\sigma^2<\infty$. I'm wondering if it's possible to show the following bound: $$ E[(e^{X/2}-1)^2] \leq c\sigma^2, $$ for some $c>0$. Is zero-mean sufficient? Or maybe I need more assumptions on the tails of $X$'s density or the size of $\sigma^2$. A couple things I've tried:

1. Intuitively, $(e^{x/2}-1)^2\approx (x/2)^2$ for $x$ near zero, which gives me hope. But I want something more rigorous.
2. It is not true that $(e^{x/2}-1)^2\leq cx^2$ for fixed $c$ and all $x\in\mathbb{R}$. There's a crossover point at some positive $x$. However, that crossover increases as $c$ increases.
3. I tried playing with the moment generating function $M(t) = E[e^{tX}]$, and I found (by Jensen's inequality) that $$ E[e^{tX}-1] \geq 0,\quad t\in\mathbb{R}. $$ But this seemed to bound things in the wrong direction.

I'd greatly appreciate any help.

EDIT

Thanks for the helpful discussion so far. I think it has helped me rephrase the question. It should be: given zero-mean $X$ with pdf $f(x)$ and second moment $\sigma^2<\infty$, find $c$ (that may depend on $X$) such that $$ E[(e^{X/2}-1)^2] \leq c\sigma^2. $$

Answer 1

I figured out something that worked for what I needed, which was slightly different from the question I'd originally posed. It was based on this inequality: $$ e^x-1 \leq |x|\,e^x. $$ Thus, $$ e^X - 1 \leq |X|\,e^X, $$ almost surely, which implies $$ E[e^X-1] \leq E[|X|\,e^X]. $$ With $E[X]=0$ and $e^x$ convex, Jensen's inequality implies $$ 1 = e^0 = e^{E[X]} \leq E[e^X]\quad \Longrightarrow\quad 0 \leq E[e^X]-1 = E[e^X-1]. $$ Therefore $$ 0 \leq (E[e^X-1])^2 \leq (E[|X|\,e^X])^2. $$ By Cauchy-Schwarz, $$ (E[|X|\,e^X])^2 \leq E[X^2]\,E[(e^{X})^2] = \sigma^2\,E[e^{2X}], $$ which finishes the derivation of what I ended up needing. Sorry for any confusion with the initial formulation.

Answer 2

You should have a feeling that this couldn't possibly be true. Ignoring the mean zero condition, consider $X$ a constant. The left side increases exponentially, but the right quadratically. Now considering the mean zero condition again, if we just take $X=+x$ and $X=-x$ with both probabilities $1/2$, then $E[(e^{X/2}-1)^2]$ still scales exponentially, whereas $\sigma^2$ scales quadratically.

What goes wrong here is that $e^x$ blows up when $x$ blows up, so the taylor approximation is no good. If we instead look at $e^{ix}$, then using taylor estimates we find that the taylor remainder $$|R_n(x)| \leq \min \left(2|x|^n/n!, |x|^{n+1}/(n+1)!\right).$$ Thus $$ E[|e^{iX}-1|^2] = E[|R_0(X)|^2] \leq E[\min(2,X)^2] \leq \min(4,\sigma^2). $$ You may want to play with these kinds of results.