49. Along with the generating function $G(s)$, in many cases the concept of the moment generating function is also useful:

$$
M(s)=\mathrm{E} e^{s X}
$$

(assuming that $s$ are such that $\mathrm{E} e^{s X} < \infty$). If the moment generating function $M(s)$ is finite in a neighborhood of zero, i.e., if there exists a $\delta > 0$ such that $M(s) < \infty$ for $|s| < \delta$, then the derivatives $M^{(n)}(s)$ exist for $|s| < \delta$. If $X$ is a Poisson random variable with parameter $\lambda > 0$, then the moment generating function has the form $M(s)=e^{-\lambda\left(1-e^{s}\right)}$ for all $s \in \mathbb{R}$.

(d) Provide an example of two dependent random variables $X$ and $Y$ for which the moment generating function $M_{X+Y}(s)=\mathrm{E} e^{s(X+Y)}$ is the product of the moment generating functions $M_{X}(s)=\mathrm{E} e^{s X}$ and $M_{Y}(s)=\mathrm{E} e^{s Y}$.
Solution. (a) Let us assume for definiteness that for some $a>0$ the relation holds

$$
\left(\mathrm{E} e^{a|X|} \leqslant\right) \mathrm{E} e^{a X}+\mathrm{E} e^{-a X}0} x^{n} e^{-(a-|s|) x} \cdot \mathrm{E} e^{a|X|}0, p+q+r=1$. Let us define the distribution of ( $X, Y$ ) using the table:

| $X \backslash Y$ | 0 | 1 | 2 |  |
| :---: | :---: | :---: | :---: | :---: |
| 0 | $p^{2}$ | $p q+\varepsilon$ | $p r-\varepsilon$ | $p$ |
| 1 | $q p-\varepsilon$ | $q^{2}$ | $q r+\varepsilon$ | $q$ |
| 2 | $r p+\varepsilon$ | $r q-\varepsilon$ | $r^{2}$ | $r$ |
|  | $p$ | $q$ | $r$ |  |

Here $\varepsilon$ is an arbitrary positive number for which all expressions in the table are non-negative. The $X, Y$ thus defined are dependent and identically distributed, and

$$
\begin{aligned}
& M_{X+Y}(s)=p^{2}+2 p q e^{s}+\left(2 p r+q^{2}\right) e^{2 s}+2 q r e^{3 s}+r^{2} e^{4 s}= \\
&=\left(p+q e^{s}+r e^{2 s}\right)^{2}=M_{X}(s) M_{Y}(s)
\end{aligned}
$$