430. The joint density function of a continuous two-dimensional random variable $(X, Y)$ is given by:

$$
f(x, y)= \begin{cases}4 x y e^{-x^{2}-y^{2}} & (x>0, y>0) \\ 0 & (x<0 \text { or } y<0) .\end{cases}
$$

Find: a) the expectations; b) the variances of the components $X$ and $Y$.
Solution. a) First, let's find the density of the distribution of the component $X$:

$$
f_{1}(x)=\int_{0}^{\infty} f(x, y) d y=4 x e^{-x^{2}} \int_{0}^{\infty} y e^{-y^{2}} d y=2 x e^{-x^{2}} \quad(x>0)
$$

Similarly, we get

$$
f_{2}(y)=2 y e^{-y^{2}} \quad(y>0)
$$

Let's find the expected value of the component $X$:

$$
M(X)=\int_{0}^{\infty} x f_{1}(x) \mathrm{d} x=\int_{0}^{\infty} x \cdot\left(2 x e^{-x^{2}} \mathrm{~d} x\right)
$$

Integrating by parts and considering that the Poisson integral $\int_{0}^{\infty} \mathrm{e}^{-x^{2}} \mathrm{~d} x=$ $=\sqrt{\pi} / 2$, we get $M(X)=\sqrt{\pi} / 2$. It is clear that $M(Y)=\sqrt{\pi} / 2$.

b) Let's find the variance of $X$:

$$
\begin{aligned}
& D(X)=\int_{0}^{\infty} x^{2} f_{1}(x) \mathrm{d} x-\left[\left.M(X)\right|^{2}=\right. \\
= & \int_{0}^{\infty} x^{2}\left(2 x \mathrm{e}^{-x^{2}} \mathrm{~d} x\right)-(\sqrt{\pi} / 2)^{2}=1-\pi / 4
\end{aligned}
$$

It is clear that $D(Y)=1-\pi / 4$.