Proof of some statements involving conditional expectation Let $(\Omega, \mathcal{F}, \mu)$ be a probability space. Let $X$ and $Y$ be i.i.d. nonnegative random variables. Show if the following is true:


*

*$E(X|X+Y)=(X+Y)/2$

*$E(X|XY)=\sqrt{XY}$


My thoughts:


*

*Since $\sigma(X)$ is equal to $\sigma(Y)$ is equal to $\sigma(X+Y)$ and because of $\mathcal{G}\subset\mathcal{F}$ it follows that $E(X|\mathcal{G})=X$. 


$$\int E(X|X+Y)d\mu) = \int E(X|X)d\mu) = \int E(X|Y)d\mu) = \left(\int E(X|Y)d\mu)+\int E(X|X)d\mu)\right)/2 = (X+Y)/2 $$
Is this correct so far? For the second point I am lacking an idea how to proof that. Any inspiration is welcome. Thanks!
 A: The second statement is false. Suppose $X$ and $Y$ are independent Bernoulli random variables, each taking the value $1$ with probability half and $0$ otherwise. Let $A$ be the event that $XY=0$. 
$$\begin{align}
\mathbb{E}[\sqrt{XY}\mathbb{1}_A]=0
\end{align}$$
Note that $\mathbb{1}_A=1-XY$.
$$\begin{align}
\mathbb{E}[X\mathbb{1}_A]&=\mathbb{E}[X(1-XY)]\\
&=\mathbb{E}[X]-\mathbb{E}[X^2]\mathbb{E}[Y]\\
&= \frac{1}{2}-\frac{1}{4}\\
&= \frac{1}{4}\\
&\neq \mathbb{E}[\sqrt{XY}\mathbb{1}_A]
\end{align}$$
This also aligns with our intuition in this case. Suppose I flip two coins and tell you that one is tails (event $A$). Given this information, there's some chance that the first coin is heads ($X=1$), so we know that $\mathbb{E}[X|XY]$ should be non-zero.
A: Your first question has been addressed here on two recent occasions.  Only your second question prevents this from being an exact duplicate.
Notice that $E(X\mid X+Y) = E(Y\mid X+Y)$ by symmetry, and their sum is $E(X+Y\mid X+Y)=X+Y$.  If the sum of two numbers is $X+Y$ and the two numbers are the same number, then what number is it?
Now $E(X\mid XY)=E(Y\mid XY)$.  Their product is $E(X\mid XY)E(Y\mid XY)$.  This can be considered to be $(\text{constant}\cdot E(Y\mid XY)$, since the first factor is a function of $XY$, and we're conditioning on $XY$ so functions of $XY$ are "constant".  Therefore the product is $E(\text{constant}\cdot Y\mid XY)= E( E(X\mid XY)\cdot Y\mid XY)$. [LATER EDIT: The next clause seems not to be generally true, so this argument is at best incomplete.] Now the "$Y$" can be moved inside $E(X\mid XY)$ for the same reason that we just applied,[end of clause] and we get $E(E(XY\mid XY)\mid XY)= E(XY\mid XY)= XY$.  So this is a bit more involved than the situation with sums, but it gives you the result.
