Maximise expectation of exponential given mean and variance The problem is as follows:
Suppose that $X$ is a random variable with $\mathbb{E}X=0$, $\mathbb{D}X = \sigma^2$ and having a finite support: $P(|X|\leq a)=1$. What is the maximum possible value of $\mathbb{E}e^{X}$ and on what RV is this value achieved?
My thought was to consider the value of the characteristic function of such an RV at the point $-i$:
\begin{gather*}
\mathbb{E}e^X=\varphi(-i)=\sum_{k=0}^{\infty}\frac{\mathbb{E}X^k}{k!}=1+\frac{\sigma^2}{2}+\sum_{k=3}^{\infty}\frac{\mathbb{E}X^k}{k!}
\end{gather*}
which may have worked if not for the condition of boundedness of support, which I have no idea how to incorporate here.
Of course I'm not asking for a complete solution but any ideas/approaches are highly welcome.
 A: First of all, @DilipSarwate , @cardinal , thank you very much for your helpful advice.
I tried to carry this idea of putting as much mass at $a$ to a solution and it turns out that you can actually do even better as you proposed. Namely, you can put as much as $\frac{\sigma^2}{\sigma^2+a^2}$ there while still satisfying the constraints. 
I would also write a formal proof that you can't do better which came out pretty clumsy so please bear with its lack of constructiveness - after all, this problem is about guessing the right answer, and I've already wasted my weekend on this problem so I won't let it take away more of my life making this proof better.
Consider such RV $X_0$ that
\begin{gather*}
  X_0=\begin{cases}
   a,\,\,P=\frac{\sigma^2}{\sigma^2+a^2},\\
   -\frac{\sigma^2}{a},\,\,P=\frac{a^2}{\sigma^2+a^2}.
  \end{cases}
 \end{gather*}
It is straightforward to make sure that it satisfies all the constraints. Now, for this RV
\begin{gather*}
\mathbb{E}e^{X_0}=\frac{e^a\sigma^2+e^{-\frac{\sigma^2}{a}}a^2}{\sigma^2+a^2}.
\end{gather*}
Let us now prove that we can't do better. It is fairly easy to see that there exists a quadratic function $f(x):=\alpha x^2+\beta x+\gamma$ that satisfies the following constraints (three linear equations for its coefficients):
    \begin{gather*}
  f(a)=e^a,\\
  f(-\sigma^2/a)=e^{-\frac{\sigma^2}{a}},\\
  f^\prime(-\sigma^2/a)=(e^{x})^\prime|_{x=-\sigma^2/a}=e^{-\frac{\sigma^2}{a}}\,\,(\text{common tangent}).
 \end{gather*}
For example, simply solving the equations we obtain:
    \begin{gather*}
  \alpha=\frac{e^aa^2-e^{-\frac{\sigma^2}{a}}a(a+a^2+\sigma^2)}{(\sigma^2+a^2)^2},\\
  \beta=e^{-\frac{\sigma^2}{a}}+\frac{2\sigma^2}{a}\alpha,\\
  \gamma=\mathbb{E}X_0-\alpha\sigma^2.
 \end{gather*}
Putting our strengths together for the last struggle, we can see through simple analysis of derivatives that
    \begin{gather*}
  \forall x\in [-a;a]\,\,f(x)\geq e^x.
 \end{gather*}
Thus, for any X satisfying the constraints
    \begin{gather*}
  \mathbb{E}e^{X}\leq\mathbb{E}f(X)=\alpha\mathbb{E}X^2+\gamma=\alpha\sigma^2+\gamma=\frac{e^a\sigma^2+e^{-\frac{\sigma^2}{a}}a^2}{\sigma^2+a^2}=\mathbb{E}e^{X_0}.
 \end{gather*}
Case closed.
