How to prove $\mathrm{E}(e^{tX}) \le \exp(t^2\mathrm{Var}(X))$ where $|X|\le 1$ and $\mathrm{E}(X)=0$? Let $X$ be a random variable with $|X|\le1$ and $\mathrm{E}(X)=0$. Then for any $-1\le t\le1$, prove that $$\mathrm{E}(e^{tX}) \le \exp(t^2\mathrm{Var}(X)).$$
Maybe it is about Jensen inequality but I only get $\mathrm{E}(e^{tX}) \ge e^{t\mathrm{E}(X)} = 1$. And I used Taylor expansion but doesn't work.
 A: Applying Taylor expansion we have $$e^{tx} = 1+tx+\frac12t^2x^2+\sum_{i=3}^\infty\frac{t^ix^i}{i!}$$
$$\exp(t^2E(X^2)) = 1+t^2E(X^2)+\sum_{i=2}^\infty \frac{t^{2i}E^{i}(X^2)}{i!}$$
Since $|t|\leq 1$ and $|x|\leq 1$, we know that $t^kx^k \leq t^2x^2$ for $k\geq 2$, therefore\begin{align*}
e^{tx} &\leq 1+tx+\frac12 t^2x^2 + \sum_{i=3}^\infty\frac{t^2x^2}{i!}\\
&\leq 1+tx+\frac12t^2x^2+t^2x^2\sum_{i=3}^\infty \frac{1}{3^{i-2}}\\
&= 1+tx+t^2x^2
\end{align*}
Thus $$E(e^{tX}) \leq 1+tE(X)+t^2E(X^2) = 1+t^2E(X^2)\leq \exp(t^2E(X^2)) = \exp(t^2Var(X))$$
A: If we do the Taylor development for both sides (noting that $E(X)=0$), we get:
$$1+\frac{t^2}2E(X^2)+\frac{t^3}6E(X^3)+\frac{t^4}{24}E(X^4)+\cdots\leq1+t^2E(X^2)+\frac{t^4}2E(X^2)^2\cdots$$
We can rewrite this as:
$$\sum_{i=1}^\infty \frac{t^{2i}}{(2i)!}E(X^{2i})\left(1 + \frac{t}{(2i+1)}\frac{E(X^{2i+1})}{E(X^{2i})}\right) \leq \sum_{i=1}^\infty \frac{t^{2i}}{i!}E(X^{2})^i$$
Given that $|X|\leq1$ then $E(X^{2i+1}) \leq E(X^{2i})$. [Note: This is a change over the initial post.]
So, with $|t| \leq 1$:
$$\frac{t}{(2i+1)}\frac{E(X^{2i+1})}{E(X^{2i})} \leq 1$$
And:
$$\sum_{i=1}^\infty 2\frac{t^{2i}}{(2i)!}E(X^{2i}) \leq \sum_{i=1}^\infty \frac{t^{2i}}{i!}E(X^{2})^i$$
We also have that $E(X^2)^i \leq E(X^{2i})$ (Jensen) so the sum becomes:
$$\sum_{i=1}^\infty 2\frac{t^{2i}}{(2i)!}E(X^{2i}) \leq \sum_{i=1}^\infty \frac{t^{2i}}{i!}E(X^{2})^i  \leq \sum_{i=1}^\infty \frac{t^{2i}}{i!}E(X^{2i})$$
And this should prove your point.
