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

$$
f(x, y)=\frac{1}{\pi} \mathrm{e}^{-(1 / 2)\left(x^{2}+2 x y+3 y^{2}\right)}
$$

Find: a) the marginal density functions of the components; b) the conditional density functions of the components.
Solution. a) Let's find the density of the distribution of the component $X$:

$$
f_{1}(x)=\int_{-\infty}^{\infty} f(x, y) \mathrm{d} y=\frac{1}{\pi} \int_{-\infty}^{\infty} \mathrm{e}^{-(1 / 2)\left(x^{2}+2 x y+5 y^{2}\right)} \mathrm{d} y
$$

We factor out the multiplier $\mathrm{e}^{-x^{2} / 2}$, which does not depend on the integration variable $y$, and complete the remaining exponent to a perfect square; then

$$
\begin{gathered}
f_{1}(x)=(1 / \pi) \cdot \mathrm{e}^{-x^{2} / 2} \cdot \mathrm{e}^{x^{2} / 10} \cdot \sqrt{2 / 5} \int_{-\infty}^{\infty} \mathrm{e}^{-(\sqrt{5 / 2} y+\sqrt{2 / 5} x)^{2}} \times \\
\times d(\sqrt{5 / 2} y+\sqrt{2 / 5} x)
\end{gathered}
$$

Considering that the Poisson integral $\int_{-\infty}^{\infty} \mathrm{e}^{-u^{2}} \mathrm{~d} u=\sqrt{\pi}$, we finally obtain the density of the distribution of the component $X$:

$$
f_{1}(x)=\sqrt{2 /(5 \pi)} e^{-0.4 x^{2}}
$$

Similarly, we find the density of the distribution of the component $Y$:

$$
f_{2}(y)=\sqrt{2 / \pi} e^{-2 y^{2}}
$$

b) Let's find the conditional densities of the distributions of the components. Performing elementary calculations, we get:

$$
\begin{aligned}
& \varphi(x \mid y)=\frac{f(x, y)}{f_{2}(y)}=\frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-0.5(x+y)^{2}} \\
& \psi(y \mid x)=\frac{f(x, y)}{f_{1}(x)}=\frac{\sqrt{5}}{\sqrt{2 \pi}} \mathrm{e}^{-0.1(x+\sqrt{5} y)^{2}}
\end{aligned}
$$