F (x, y) = 2 (0

Let Z=X+Y, then p (z) = p (z) = p (z ≤ z) = p (x+y ≤ z) = ∫ ∫ 2ddxdy, (the integral area is the intersection of a straight line Y=-X+Z and the triangle area).

=∫2d xdy(s = z- 1)+∫( 1，z- 1)dx∫(z-x，0)2dy-∫2d xdy(s = 1/2)

= 4z-z 2-3 In the above formula, (1, z- 1) indicates the upper and lower limit of the integral.

So: the distribution function f (z) = 4z-z 2-3 (1

So: probability density: f (z) = 4-2z (1

E(Z)=∫z(4-2z)dz=4/3，e(z^2)=∫z^2(4-2z)dz= 1 1/6

d(z)=e(x^2)-e(x)^2= 1/ 18

I don't know if the answer is right, just for reference.

This problem is a bit complicated for the integration of ordinary * * * regions, and the others are basically the process of finding expectations and variances according to definitions.