Example 4. Find the probability density $p(x, y)$ of a two-dimensional random variable $(X, Y)$ given the known distribution function

$$
F(x, y)=\left(\frac{1}{\pi} \operatorname{arctg} x+\frac{1}{2}\right) \cdot\left(\frac{1}{\pi} \operatorname{arctg} y+\frac{1}{2}\right)
$$
## Solution. Since

$$
\begin{gathered}
F_{x}^{\prime}(x, y)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}} \cdot\left(\frac{1}{\pi} \operatorname{arctg} y+\frac{1}{2}\right) \\
F_{x y}^{\prime \prime}(x, y)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}} \cdot \frac{1}{\pi} \cdot \frac{1}{1+y^{2}}
\end{gathered}
$$

then by formula (2.8.11) we obtain

$$
p(x, y)=\frac{1}{\pi^{2}\left(1+x^{2}\right)\left(1+y^{2}\right)}
$$