Example 7. Calculate the central moments of a random variable distributed according to the normal law.
Solution. According to the definition (see formula (2.6.7)) and formula (3.5.1), we obtain

$$
\mu_{k}=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty}(x-\dot{a})^{k} e^{-(x-a)^{2} / 2 \sigma^{2}} d x
$$

We transition to a new variable $t$ using the formula $t=\frac{x-a}{\sigma}$, then $x=a+\sigma t, d x=\sigma d t$, and

$$
\mu_{k}=\frac{\sigma^{k}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t^{k} e^{-t^{2} / 2} d t
$$

We consider separately the cases when $k$ is an odd number and $k$ is an even number. Let $k=2 l+1$ be an odd number, then

$$
\mu_{2 l+1}=\frac{\sigma^{2 l+1}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t^{2 l+1} e^{-t^{2} / 2} d t=\frac{\sigma^{2 l+1}}{\sqrt{2 \pi}}\left(\int_{-\infty}^{0} t^{2 l+1} e^{-t^{2} / 2} d t+\int_{0}^{+\infty} t^{2 l+1} e^{-t^{2} / 2} d t\right)
$$

The substitution of the variable $z=t^{2}$ in both integrals leads to the equality

$$
\begin{aligned}
& \mu_{2 l+1}=\frac{\sigma^{2 l+1}}{\sqrt{2 \pi}}\left[\frac{1}{2} \int_{-\infty}^{0} z^{\prime} e^{-z / 2} d z+\frac{1}{2} \int_{0}^{+\infty} z^{\prime} e^{-z / 2} d z\right]= \\
& =\frac{\sigma^{2 l+1}}{\sqrt{2 \pi}}\left[-\frac{1}{2} \int_{0}^{+\infty} z^{\prime} e^{-z / 2} d z+\frac{1}{2} \int_{0}^{+\infty} z^{\prime} e^{-z / 2} d z\right]=0
\end{aligned}
$$

Thus, all central moments of odd orders of a random variable distributed according to the normal law are equal to zero.

If $k=2 l$ is an even number, then

$$
\mu_{2 l}=\frac{\sigma^{2 l}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t^{2 l} e^{-t^{2} / 2} d t=\frac{2 \sigma^{2 l}}{\sqrt{2 \pi}} \int_{0}^{+\infty} t^{2 l} e^{-t^{2} / 2} d t
$$

since in this case the integrand function is even.

Applying the method of integration by parts, we obtain

$$
\int_{0}^{+\infty} t^{2 l} e^{-t^{2} / 2} d t=(2 l-1) \int_{0}^{+\infty} t^{2 l-2} e^{-t^{2} / 2} d t
$$

Using this formula, we sequentially find:

$$
\begin{gathered}
\mu_{2 l}=\frac{2 \sigma^{2 l}}{\sqrt{2 \pi}}(2 l-1) \int_{0}^{+\infty} t^{2 l-2} e^{-t^{2} / 2} d t= \\
=\sqrt{\frac{2}{\pi}} \sigma^{2 l}(2 l-1)(2 l-3) \cdot \int_{0}^{+\infty} e^{2 l-4} e^{-t^{2} / 2} d t=\ldots= \\
=\sqrt{\frac{2}{\pi}} \sigma^{2 l}(2 l-1)(2 l-3) \ldots 3 \cdot 1 \cdot \int_{0}^{+\infty} e^{-t^{2} / 2} d t= \\
=\sigma^{2 l}(2 l-1)(2 l-3) \ldots 3 \cdot 1 \cdot \sqrt{\frac{2}{\pi}} \cdot \frac{\sqrt{2 \pi}}{2}=1 \cdot 3 \ldots(2 l-1) \sigma^{2 l}
\end{gathered}
$$

i.e.

$$
\mu_{2 l}=1 \cdot 3 \ldots(2 l-1) \sigma^{2 l}
$$

In particular, the central moment of the fourth order of a random variable distributed according to the normal law, $\mu_{4}=1 \cdot 3 \cdot \sigma^{4}=3 \sigma^{4}$.