Does $\mathbb{E}e^X=e^{\mathbb{E}X}$?

Question

Does $\mathbb{E}e^X=e^{\mathbb{E}X}$ hold? Can we just Taylor expand and use linearity?

Answer 1

Counterexample: Let $X$ be standard normal. Then the expectation of $X$ is $0$. The expectation of $e^X$ is $\sqrt e$, which is not equal to $e^0 = 1$.

Linearity doesn't help us get all the way. Sure, if all is well-behaved, we may have $$ \Bbb E(e^X) = \Bbb E\left(1 + X +\frac{X^2}2 + \frac{X^3}6 + \cdots\right) = \Bbb E(1) + \Bbb E(X) + \frac{\Bbb E(X^2)}2 + \frac{\Bbb E(X^3)}6 + \cdots $$ but in general $\Bbb E(X^n)\neq (\Bbb EX)^n$, so you can't easily make this into $$ e^{\Bbb EX} = \Bbb E(1) + \Bbb E(X) + \frac{(\Bbb EX)^2}2 + \frac{(\Bbb EX)^3}6 + \cdots $$ Of course, the failure of transforming any single term of the power series the way you want doesn't automatically mean that the expressions are unequal in value. But it makes it implausible. And the above counterexample shows that they are indeed unequal, at least some of the time.

Answer 2

$e^{EX} \leq Ee^{X}$ and equality cannot hold unless $X$ is almost surely constant.

Take any non-constant random variable (like the outcome of a coin toss) to get a counter-example.

When you use Taylor expansion and use linearity the problem you face is $EX^{n} =(EX)^{n}$ is not true.

Answer 3

The AM-GM or Jensen'sInequality is: $$\frac{e^{x_1}+e^{x_2}}{2} \ge e^{\frac{x_1+x_2}{2}}, ~x_1,x_2 \in R.$$ Equailty holds only if $x_1=x_2.$ This is nothing but $$E (e^{X}) \ge e^{E x}.$$ so the equality will hold only if the variables are equal and so equality will be trivial here.