Example 3. Express the central moments of the second, third, and fourth orders through the initial moments.
Solution. Taking into account the properties of the expectation, the definitions of moments, and assuming \( M(X) = a \), we obtain

\[
\begin{aligned}
& \mu_{2}=M(X-a)^{2}=M\left(X^{2}-2 a X+a^{2}\right)=M\left(X^{2}\right)-2 a M(X)+M\left(a^{2}\right)= \\
& =M\left(X^{2}\right)-2 a^{2}+a^{2}=M\left(X^{2}\right)-a^{2}=v_{2}-v_{1}^{2} \\
& \quad \mu_{2}=v_{2}-v_{1}^{2}, \text{ or } D(X)=v_{2}-v_{1}^{2}
\end{aligned}
\]

\[
\begin{aligned}
& \mu_{3}=M(X-a)^{3}=M\left(X^{3}-3 a X^{2}+3 a^{2} X-a^{3}\right)=M\left(X^{3}\right)-3 a M\left(X^{2}\right)+ \\
& +3 a^{2} M(X)-M\left(a^{3}\right)=M\left(X^{3}\right)-3 a M\left(X^{2}\right)+3 a^{2} M(X)-a^{3}= \\
& =v_{3}-3 v_{1} v_{2}+3 v_{1}^{3}-v_{1}^{3}=v_{3}-3 v_{1} v_{2}+2 v_{1}^{3} \\
& \mu_{3}=v_{3}-3 v_{1} v_{2}+2 v_{1}^{3} ; \\
& \mu_{4}=M(X-a)^{4}=M\left(X^{4}-4 a X^{3}+6 a^{2} X^{2}-4 a^{3} X+a^{4}\right)= \\
& =M\left(X^{4}\right)-M\left(4 a X^{3}\right)+M\left(6 a^{2} X^{2}\right)-M\left(4 a^{3} X\right)+M\left(a^{4}\right)= \\
& =M\left(X^{4}\right)-4 a M\left(X^{3}\right)+6 a^{2} M\left(X^{2}\right)-4 a^{3} M(X)+M\left(a^{4}\right)= \\
& =M\left(X^{4}\right)-4 a M\left(X^{3}\right)+6 a^{2} M\left(X^{2}\right)-4 a^{3} M(X)+a^{4}= \\
& =v_{4}-4 v_{1} v_{3}+6 v_{1}^{2} \cdot v_{2}-4 v_{1}^{3} \cdot v_{1}+v_{1}^{4} \\
& \mu_{4}=v_{4}-4 v_{1} v_{3}+6 v_{1}^{2} \cdot v_{2}-3 v_{1}^{4}
\end{aligned}
\]

Note that formula (2.6.12) is equivalent to formula (2.5.4).