I'm struggling with this exercise:
Suppose the density of $(X,Y)'$ is given by $f(x,y)=\frac{1}{2\pi}\exp({-\frac{1}{2}(x^2-2xy+2y^2)})$. Determine the conditional distributions, particularly the conditional expectations and the conditional variances.
I saw the solutions and it's said that $EX=EY=0$. It may be silly but can't get why. Moreover, the solutions show $\Lambda^{-1}$ but how can i find it?