Let be $X$ and $Y$ random variables with joint probability function $f(x,y) = 4xy$ with $0<x<1$, $x<y<1$ and $-1<x<0$, $x<y<0$.
I want to find $\mathbb{E}(X+Y \mid Y-X)$.
I tried to do a rotation of coordinates, $u=\frac{1}{\sqrt{2}}(x+y)$ and $v=\frac{1}{\sqrt{2}}(-x+y)$. However, the new function $f(u,v)$ fails with the condition:
$$\int\int_{R}f(u,v)dudv = 1,$$
where $R$ is the same region: $0<x<1$, $x<y<1$ and $-1<x<0$, $x<y<0$.
What is the correct form to solve this problem? I'm really confused.
If $u=\frac{1}{\sqrt{2}}(x+y)$, $v=\frac{1}{\sqrt{2}}(y-x)$ is a rotation, then $$ (x,y)=(0,0) \mapsto (u,v)=(0,0), $$ $$ (x,y)=(1,1) \mapsto (u,v)=(\sqrt{2},0), $$ $$ (x,y)=(0,1) \mapsto (u,v)=\left(\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right), $$ $$ (x,y)=(-1,-1) \mapsto (u,v)=(-\sqrt{2},0), $$ $$ (x,y)=(-1,0) \mapsto (u,v)=\left(-\tfrac1{\sqrt{2}},\tfrac1{\sqrt{2}}\right). $$
And the pdf $f_{U,V}(u,v)=2u^2-2v^2$ is a valid pdf inside this region. Also it is evident that $$ \mathbb E[X+Y\mid Y-X] = \mathbb E[\sqrt{2}U\mid \sqrt{2}V]=\sqrt{2}\mathop{\mathbb E}[U\mid V] = 0 $$ since the distribution of $U$ is symmetric for any fixed $V$.