What's the relation between $\mathbf E(X)$ and $\mathbf E(e^X)$?

Question

Given $\mathbf E(X)=0$ and $-1\le X\le 1$, show that $\mathbf E(\text{exp}(\sqrt{2}X))\le e^{\sqrt{2}}-\sqrt{2}$.

It seems that Jensen's inequality will help, but I have no idea.

Thanks in advance.

Answer 1

Hint: Since $-1 \le X \le 1$, we have $X^k \le 1$, and thus, $E[X^k] \le 1$ for all integers $k \ge 2$. Also, you are given $E[X] = 0$.

Now, use $e^{X\sqrt{2}} = \displaystyle\sum_{k = 0}^{\infty}\dfrac{(\sqrt{2})^k}{k!}X^k$ along with linearity of expectation and see what you get.

Answer 2

The following is well-known:-

If $X$ is a random variable(real valued and to speak very formally $X$ is an integrable/measurable real valued random variable) and $\varphi$ is a convex function then $\varphi(E(X))\leq E(\varphi(X))$

Now the exponential function is convex. This answer the question you asked in the title of your question. But it'll not help you prove $E(e^{\sqrt{2}X})<3$.