For defined in the condition of random values:
The two -dimensional law of the distribution of the random vector XI = (XI1; XI2), one -dimensional laws of the distribution of the components, the vector of the average and the covalization matrix;
The random values of the XI1 and XI2 are evenly distributed in the quadrangle with peaks (2, 2), (1, -2), (-2, -2), (-2, 2).
In my decision, I divided the trapezoid into a rectangle and a triangle.
Please check my solution. I checked everything numerically using Wolfram Mathematica. Now I need to make sure my idea and its application are correct.
$S=4\cdot4-4\cdot1\cdot\frac12=14;$
$f_{\xi}(x;y)=\frac{1}{S}$ if $(x;y) \in D, 0$ otherwise$;$
$f\xi_{11}=\int_{-2}^{2}\frac{1}{S}\,dy=\frac{2}{7};$
$f\xi_{12}=\int_{4\cdot x-6}^{2}\frac{1}{S}\,dy=\frac{4-2\cdot x}{7};$
$f\xi_{2}=\int_{-2}^{\frac{y+6}{4}}\frac{1}{S}\,dx=\frac{y+14}{56};$
$E\xi_2=\int_{-2}^{2}y\cdot f\xi_{2}\,dy=\frac{2}{21};$
$E\xi_1=\int_{-2}^{1}x\cdot f\xi_{11}\,dx+\int_{1}^{2}x\cdot f\xi_{12}\,dx=-\frac{5}{21};$
$E\xi_2^2=\int_{-2}^{2}y^2\cdot f\xi_{2}\,dy=\frac{4}{3};$
$E\xi_1^2=\int_{-2}^{1}x^2\cdot f\xi_{11}\,dx+\int_{1}^{2}x^2\cdot f\xi_{12}\,dx=-\frac{47}{42};$
$D\xi_1=E\xi_1^2-(E\xi_1)^2=\frac{937}{882};$
$D\xi_2=E\xi_2^2-(E\xi_2)^2=\frac{584}{441};$
$cov(\xi_1;\xi_2)=\int_{-2}^{2}(y-E\xi_2)\int_{-2}^{\frac{y+6}{4}}(x-E\xi_1)\,dx\,dy=\frac{146}{63};$
$\sum_{\xi}=((D\xi_1;cov(\xi_1;\xi_2));(cov(\xi_1;\xi_2);D\xi_2)).$
Link to the drawing: https://i.stack.imgur.com/QDFVN.png
