A joint probability distribution $f(x,y)=2\mathrm{I}_{(0<x<y<1)}$ is given, and I need to calculate $$\mathrm{Var}[Y+X-E(X|Y)]$$ I can separate it into $$\mathrm{Var}[Y]+\mathrm{Var}[X-E(X|Y)]$$ since $$\mathrm{Cov}(Y,X-E(X|Y))=0$$, but how do I get $\mathrm{Var}[X-E(X|Y)]$?
It seems $E(X|Y)=Y/2$, if that being said, is it okay to simply do like $$\mathrm{Var}[X-E(X|Y)]=\mathrm{Var}(X-Y/2)=\mathrm{Var}(X)+\mathrm{Var}(Y)/4-\mathrm{Cov}(X,Y)$$ ?