60. (See [34].) Let $\mathrm{P}(|X| \leqslant b)=1, \mathrm{E} X=0$ and $\mathrm{D} X=\sigma^{2}$ for some $\sigma, b>0$. What is the maximum possible value of $\mathrm{E} e^{X}$?
Solution. We will show that the maximum possible value of $\mathrm{E} e^{X}$ is achieved on a Bernoulli random variable $X_{*}$, one of whose values is $b$ (and, consequently, the other value is $-\sigma^{2} / b$, since $\mathrm{E} X_{*}=0$ and $\left.\mathrm{E} X_{*}^{2}=\mathrm{D} X_{*}=\sigma^{2}\right)$. In other words,

$$
\mathrm{E} e^{X} \leqslant \mathrm{E} e^{X_{*}}=\frac{e^{b} \sigma^{2}+e^{-\sigma^{2} / b} b^{2}}{\sigma^{2}+b^{2}}
$$

The parabola $f(x)=\alpha x^{2}+\beta x+\gamma$, which coincides with $e^{x}$ at points $x=$ $=-\sigma^{2} / b \in[-b, b]$ (see problem II.6.59) and $x=b$, and also has a common tangent with $e^{x}$ at the point $x=-\sigma^{2} / b$, lies above $e^{x}$ on $[-b, b) \backslash\left\{-\sigma^{2} / b\right\}$. From this, the desired estimate $\mathrm{E} e^{X} \leqslant \mathrm{E} f(X)=\mathrm{E} e^{X_{*}}$ follows.