383. A random variable $X$ is uniformly distributed in the interval $(-\pi / 2, \pi / 2)$. Find the density function $g(y)$ of the random variable $Y=\sin X$.
Solution. Let's find the density function $f(x)$ of the random variable $X$. The variable $X$ is uniformly distributed in the interval $(-\pi / 2, \pi / 2)$, so in this interval

$$
f(x)=\frac{1}{\pi / 2-(-\pi / 2)}=\frac{1}{\pi}
$$

outside the considered interval, $f(x)=0$.

The function $y=\sin x$ is monotonic in the interval $(-\pi / 2, \pi / 2)$, hence it has an inverse function $x=\psi(y)=$ $=\arcsin y$ in this interval. Let's find the derivative $\psi^{\prime}(y)$:

$$
\psi^{\prime}(y)=1 / \sqrt{1-y^{2}}
$$

We will find the desired density function using the formula

$$
g(y)=f[\psi(y)] \cdot\left|\psi^{\prime}(y)\right|
$$

Considering that $f(x)=1 / \pi \quad$ (thus, $\quad f[\psi(y)]=1 / \pi) \quad$ and $\left|\boldsymbol{\Psi}^{\prime}(y)\right|=1 / \sqrt{1-y^{2}}$, we get

$$
g(y)=1 /\left(\pi \sqrt{1-y^{2}}\right)
$$

Since $y=\sin x$, and $-\pi / 2<x<\pi / 2$, then $-1<y<1$. Therefore, in the interval $(-1,1)$ we have $g(y)=1 /\left(\pi \sqrt{1-y^{2}}\right)$; outside this interval, $g(y)=0$.

Control:

$$
\begin{gathered}
\int_{-1}^{1} g(y) \mathrm{d} y=\frac{1}{\pi} \int_{-1}^{1} \frac{\mathrm{d} y}{\sqrt{1-y^{2}}}=\frac{2}{\pi} \int_{0}^{1} \frac{\mathrm{d} y}{\sqrt{1-y^{2}}}=\left.\frac{2}{\pi} \operatorname{arosin} y\right|_{0} ^{1}= \\
=2 / \pi \cdot \pi / 2=1
\end{gathered}
$$