Example 1. The random variable $\xi$ is distributed according to the normal law with parameters $a$ and $\sigma^{2}$. From the sample $x_{1}, \ldots, x_{n}$ of values of $\xi$, the empirical moments $M_{1}^{*}=\bar{x}=2.3$ and $M_{2}^{*}=\overline{x^{2}}=8.7$ have been determined. Using these moments, find the parameters $a$ and $\sigma^{2}$.
## Solution.

1. We calculate the theoretical moments $M \xi^{k}$ for $k=1,2$, expressing them in terms of the unknown parameters $a$ and $\sigma$. Since $\xi$ is normally distributed with parameters $a$ and $\sigma^{2}$, we have

$$
\begin{aligned}
& M \xi=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{-\infty}^{+\infty} x e^{-(x-a)^{2} /\left(2 \sigma^{2}\right)} d x=a \\
& M \xi^{2}=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \int_{-\infty}^{+\infty} x^{2} e^{-(x-a)^{2} /\left(2 \sigma^{2}\right)} d x=a^{2}+\sigma^{2}
\end{aligned}
$$

(In this case, $a_{1}=a, a_{2}=\sigma^{2}, m_{1}\left(a_{1}, a_{2}\right)=a_{1}, m_{2}\left(a_{1}, a_{2}\right)=$ $\left.=a_{1}^{2}+a_{2}.\right)$

2. By equating the found theoretical moments $M \xi, M \xi^{2}$ to the given empirical moments $M_{1}^{*}=2.3$ and $M_{2}^{*}=8.7$, we obtain a system of 2 equations

$$
\left\{\begin{array}{l}
a=2.3 \\
a^{2}+\sigma^{2}=8.7
\end{array}\right.
$$

3. We solve the system of equations (2) and obtain $a=2.3, \sigma^{2}=3.41$.

Answer. $\quad a=2.3, \quad \sigma^{2}=3.41$.