>>612
D:x+y<1 |XY|=1-x-y
¬D:x+y>1 |XY|=x+y-1
|XY|=|x+y-1|
E(|XY|)=∬|x+y-1|dxdy=1/3
P(|XY|>t)=(1-t)^2
ρ(t)=(d/dt)P(|XY|<t)=2(1-t)