1645. A random variable that takes the value $m$ with probability

$$
P_{m}(\lambda)=\frac{\lambda^{m} e^{-\lambda}}{m!}(m=0,1, \ldots, m, \ldots ; \lambda>0)
$$

is said to be distributed according to the Poisson distribution. Find the mathematical expectation and variance of this random variable.
Solution. $\quad E X=\sum_{m=0}^{\infty} m P_{m}(\lambda)=\sum_{m=0}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}$ (the first term is omitted, as it equals zero). In the obtained sum, we replace $m$ with $k+1$ and get

$E X=\sum_{m=1}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}=\lambda e^{-\lambda} \sum_{m=1}^{\infty} \frac{\lambda^{m-1}}{(m-1)!}=\lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}=\lambda e^{-i} e^{\lambda}=\lambda$, since $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots$

Knowing $E X$, by property 4 of variance: $D X=E X^{2}-(E X)^{2}$, we can determine $D X$. For this, it remains to find $E X^{2}$:

$$
E X^{2}=\sum_{m=0}^{\infty} m^{2} \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} m^{2} \frac{\lambda^{m} e^{-\lambda}}{m!}=\sum_{m=1}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}
$$

Again, let $m-1=k$. Then

$$
\begin{gathered}
E X^{2}=\sum_{m=1}^{\infty} m \frac{\lambda^{m} e^{-\lambda}}{(m-1)!}=\sum_{k=0}^{\infty}(k+1) \frac{\lambda^{k+1} e^{-\lambda}}{k!}=\lambda \sum_{k=0}^{\infty} k \frac{\lambda^{k} e^{-\lambda}}{k!}+ \\
+\lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}=\lambda E X+\lambda=\lambda^{2}+\lambda
\end{gathered}
$$

From this

$$
D X=\lambda^{2}+\lambda-\lambda^{2}=\lambda
$$