# Showing for a variable $X$ with mean and variance 1 that $\mathbb{E}\left(e^{X}\right)\geq e$

Let $X$ be a discrete non-negative variable with $\mathbb{E}\left(X\right)=Var\left(X\right)=1$. I'd like to show that necessarily $\mathbb{E}\left(e^{X}\right)\geq e$.

My initial intuition was to use Chernoff, that gives for any $t,a>0$ $$\mathbb{P}\left(X\geq a\right)\leq M_{X}\left(t\right)e^{-ta}$$ So with $t=a=1$ we have $\mathbb{E}\left(e^{X}\right)\geq e\cdot \mathbb{P}\left(X\geq 1\right)$ which isn't much help since obviously $\mathbb{P}\left(X\geq 1\right)<1$.

The only relevant knowledge that the mean and variance provide as far as I can see are that $$M_{X}'\left(0\right)=1$$ and $$M_{X}''\left(0\right)=2$$ but I fail to see how this helps me.

## 1 Answer

For $x \ge 0$ we have $e^x \ge \frac{1}{2} x^2 + x + 1$, so $$E e^X \ge \frac{1}{2} E[X^2] + E[X] + 1 = 3.$$

EDIT: As pointed out by Lord Shark, Jensen's inequality also works. I actually prefer this to my original solution, since it does not require the variance assumption on $X$.

$$E e^X \ge e^{E[X]} = e.$$