$E[e^x]=e^{E[x]}$ holds?

Question

$E$ is the expectation and $x$ is the random variable. If it holds, how to prove? If it fails, how to calculate $E[e^x]$ in a simplified way?

Answer 1

By Jensen's inequality we have that $$e^{E[X]}\leq E[e^X]$$

We can have equality, say for example if $X$ is constant. We can also have $E[e^X]-e^{E[X]}$ as large as we want. For example let $X$ be $-M$ with probability $1/2$ and $M$ with probability $1/2$. Then $E[e^X]-e^{E[X]}=\cosh(M)-1$ which can be made as big as you want.

Answer 2

Jensen's Inequality states that if $X$ is a random variable and $\phi$ is a convex function, then $$\phi(\mathbb{E}[X]) \le \mathbb{E}[\phi(X)].$$ Furthermore, if $\phi$ is strictly convex, then equality can only hold if $X$ is a constant random variable.

In particular, if you let $\phi(x) = e^x$, you get that $e^{\mathbb{E}[X]} \le \mathbb{E}[e^X]$. Since $e^x$ is strictly convex, equality will only hold if $X$ is a constant random variable.

The general formula for calculating $\mathbb{E}[\phi(X)]$ is $\displaystyle\int_{-\infty}^{\infty}\phi(x)f(x)\,dx$ where $f(x)$ is the probability density function of $X$. So if you know the p.d.f. of $X$, then you can try to calculate $\mathbb{E}[e^X] = \displaystyle\int_{-\infty}^{\infty}e^xf(x)\,dx$.