Example 4. A discrete random variable $X$ is given by the distribution law:

| $X$ | 1 | 2 |
| :---: | :---: | :---: |
| $P$ | 0.4 | 0.6 |

Find the first, second, and third central moments of the random variable $X$.
Solution. According to the definition (see formula 2.6.I), we first find the initial moments using formula (2.6.2):

$$
\begin{gathered}
v_{1}=M(X)=1 \cdot 0.4 + 2 \cdot 0.6 = 1.6, \quad(a=1.6) \\
v_{2}=M\left(X^{2}\right)=1^{2} \cdot 0.4 + 2^{2} \cdot 0.6 = 2.8 \\
v_{3}=M\left(X^{3}\right)=1^{3} \cdot 0.4 + 2^{3} \cdot 0.6 = 5.2
\end{gathered}
$$

We will now compute the central moments. It was previously proven that $\mu_{1}=$. і (see example 2, $\left.\mu_{1}=M((X-a))=0\right)$. According to formulas (2.6.12) and (2.6.13):

$$
\begin{gathered}
\mu_{2}=v_{2}-v_{1}^{2}=2.8-(1.6)^{2}=0.24 \\
\mu_{3}=v_{3}-3 v_{1} v_{2}+2 v_{1}^{3}=5.2-3 \cdot 1.6 \cdot 2.8+2 \cdot(1.6)^{3}=-0.48
\end{gathered}
$$

Remark. The values of $\mu_{2}$ and $\mu_{3}$ can also be found directly, according to formulas (2.6.8) and (2.6.9):

$$
\mu_{2}=M(X-a)^{2}=(1-1.6)^{2} \cdot 0.4 + (2-1.6)^{2} \cdot 0.6 = 0.36 \cdot 0.4 + 0.16 \cdot 0.6 = 0.24
$$

$$
\mu_{3}=M(X-a)^{3}=(1-1.6)^{3} \cdot 0.4 + (2-1.6)^{3} \cdot 0.6 = (-0.216) \cdot 0.4 + 0.064 \cdot 0.6 = -0.48
$$