207. Find the mathematical expectation of a discrete random variable $X$ distributed according to the Poisson law:

$$
\begin{array}{cccccc}
X & 0 & 1 & 2 & k & k \\
p & \mathrm{e}^{-\lambda} & \frac{\lambda \mathrm{e}^{-\lambda}}{1!} & \frac{\lambda^{2} \mathrm{e}^{-\lambda}}{2!} \cdots \frac{\lambda^{k} \mathrm{e}^{-\lambda}}{k!} \cdots
\end{array}
$$
Solution. By the definition of the expectation for the case when the number of possible values of $\boldsymbol{X}$ is a countable set,

$$
M(X)=\sum_{k=0}^{\infty} k \cdot \frac{\lambda k_{\mathrm{e}}-\lambda}{k!}
$$

Considering that the first term of the sum is zero when $k=0$, we take the smallest value of $k$ to be one:

$$
M(X)=\sum_{k=1}^{\infty} k \cdot \frac{\lambda^{k} \mathrm{e}^{-\lambda}}{k(k-1)!}=\lambda \mathrm{e}^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!}
$$

By setting $k-1=m$, we get

$$
M(X)=\lambda e^{-\lambda} \sum_{m=0}^{\infty} \frac{\lambda m}{m!}
$$

Taking into account that

$$
\sum_{m=0}^{\infty} \frac{\lambda^{m}}{m!}=\mathrm{e}^{\lambda}, \text { we finally have }
$$

$$
M(X)=\lambda \cdot \mathrm{e}^{-\lambda \cdot \mathrm{e}^{\lambda}}=\lambda
$$

Thus,

$$
M(X)=\lambda
$$

i.e., the expectation of the Poisson distribution is equal to the parameter of this distribution $\lambda$.