$Y = EX\mid G$ then $E (X-Y)Y = 0$

Question

I am asked to prove

If $0 \leq X \ {\color{blue} {\leq C}}$ (i.e. bounded) and $Y = EX\mid G$, then $E(X-Y)Y = 0$.

I managed to prove the statement, since $$E(X-Y)Y = E[E[(X-Y)Y\mid G]] = E[YE[X-Y\mid G]] = E[Y(EX\mid G - Y)] = 0.$$ Here I used the law of iterated expectation, taking out what is known and the definition of $Y$. Here is where my question comes in though: why is boundedness necessary? Am I missing something?

Answer 1

Your argument works fine as long as $EX^{2}<\infty$. It is not enough to assume that $E|X|<\infty$ because even though $Y$ is well defined $EXY$ may not exist. $0 \leq X \leq C$ is too strong an assumption.

Answer 2

I do not understand what you mean by $G-Y$ (it seems that $G$ is a $\sigma$-algebra). However, $E[X-Y\mid G]=E[X\mid G]-E[Y\mid G]=Y-E[Y\mid G]=0$ since $Y$ is $G$-measurable.

We actually only need $X$ to be square integrable in order to be sure that $XY$ and $Y^2$ have a finite expectation.