Example 3. Find the mathematical expectation of a discrete random variable distributed according to the Poisson law.
Solution. Based on the table defining the distribution law and formula (2.4.6), we find

$$
\begin{gathered}
M(X)=0 \cdot e^{-a}+1 \cdot \frac{a}{1!} e^{-a}+2 \cdot \frac{a^{2}}{2!} e^{-a}+\ldots+k \frac{a^{k}}{k!} e^{-a}+\ldots= \\
=\sum_{k=1}^{\infty} \frac{a^{k}}{(k-1)!} e^{-a}=a e^{-a} \sum_{k=1}^{\infty} \frac{a^{k-1}}{(k-1)!}=a e^{-a} e^{a}=a \\
M(X)=a
\end{gathered}
$$

Thus, the mathematical expectation of a random variable distributed according to the Poisson law is equal to the number $a$ - the parameter of this distribution. Example 4. Find the variance of a discrete random variable having a Poisson distribution.

Solution. We will compute the variance using formula (2.5.4). First, we need to find the mathematical expectation of the square of this variable:

$$
\begin{aligned}
M\left(X^{2}\right)= & 0^{2} \cdot e^{-a}+1^{2} \cdot \frac{a}{1!} e^{-a}+2^{2} \cdot \frac{a^{2}}{2!} e^{-a}+\ldots+k^{2} \frac{a^{k}}{k!} e^{-a}+\ldots= \\
& =1 \cdot \frac{a}{0!} e^{-a}+2 \cdot \frac{a^{2}}{1!} e^{-a}+\ldots+k \frac{a^{k}}{(k-1)!} e^{-a}+\ldots
\end{aligned}
$$

Setting $k-1=m$, we get

$$
\begin{gathered}
M\left(X^{2}\right)=\sum_{k=1}^{\infty} k \frac{a^{k}}{(k-1)!} e^{-a}=\sum_{m=0}^{\infty}(m+1) \frac{a^{m+1}}{m!} e^{-a}= \\
=\sum_{m=0}^{\infty} m \frac{a^{m+1}}{m!} e^{-a}+\sum_{m=0}^{\infty} \frac{a^{m+1}}{m!} e^{-a}= \\
=a^{2} e^{-a} \sum_{m=1}^{\infty} \frac{a^{m-1}}{(m-1)!}+a e^{-a} \sum_{m=0}^{\infty} \frac{a^{m}}{m!}=a^{2} e^{-a} e^{a}+a e^{-a} e^{a}=a^{2}+a .
\end{gathered}
$$

Using formula (2.5.4), we find the variance:

$$
D(X)=M\left(X^{2}\right)-(M(X))^{2}=a^{2}+a-a^{2}=a
$$

Thus,

$$
D(X)=a
$$

i.e., the variance of a random variable distributed according to the Poisson law is equal to the number $a$ - the parameter of this distribution.