Suppose $\displaystyle f_{X|Y}(x|y)=\frac{1}{y}e^{-\frac{x}{y}}$ and $\displaystyle f_Y(y)=e^{-y}$ where $\displaystyle 0<x,y<\infty$
Why is it obvious that $\displaystyle E(X|Y)=Y$ and $\displaystyle Var(X|Y)=Y^2$ ??
Suppose $\displaystyle f_{X|Y}(x|y)=\frac{1}{y}e^{-\frac{x}{y}}$ and $\displaystyle f_Y(y)=e^{-y}$ where $\displaystyle 0<x,y<\infty$
Why is it obvious that $\displaystyle E(X|Y)=Y$ and $\displaystyle Var(X|Y)=Y^2$ ??