1665. It is known that the density of the distribution of a random variable $X$ has the form

$$
p(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{x^{2}}{2 \sigma^{2}}}
$$

Find the density of the distribution of the random variable $Y=X^{2}$.
Solution. First, let's find the distribution function of the random variable $Y$. Since it does not take negative values, then $F(y)=P(Y0$. Then

$$
\begin{aligned}
F(y) & =P(Y0
\end{aligned}
$$

Therefore, the desired density function is given by the formula

$$
\begin{aligned}
\varphi(y)=F^{\prime}(y) & =\left\{\begin{array}{cc}
0 & \text { for } y \leqslant 0, \\
\frac{2}{\sigma \sqrt{2 \pi}} \cdot e^{-\frac{y}{2 \sigma^{2}}} \cdot \frac{1}{2 \sqrt{y}} \text { for } y>0
\end{array}=\right. \\
& =\left\{\begin{array}{cc}
0 \quad \text { for } y \leqslant 0, \\
\frac{1}{\sigma \sqrt{2 \pi y}} \cdot e^{-\frac{y}{2 \sigma^{2}}} & \text { for } y>0 .
\end{array}\right.
\end{aligned}
$$