# 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:

1. $E(X|X+Y)=(X+Y)/2$
2. $E(X|XY)=\sqrt{XY}$

My thoughts:

1. 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!

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The first one has been asked several times here, search for it. – leonbloy May 14 '12 at 19:19

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.

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As $X$ and $Y$ are nonnegative, can't we just apply the first result to $\log(X)$ and $\log(Y)$ for the second question? – leonbloy May 14 '12 at 19:29
$E(\log X\mid \log X + \log Y) = (\log X + \log Y)/2$. So $\exp E(\log X\mid \log X + \log Y) = \sqrt{XY\,{}}$, and hence $\exp E(\log X\mid XY) = \sqrt{XY\,{}}$. But generally $\exp E(\log V)$ is not the same as $E(V)$. – Michael Hardy May 14 '12 at 19:44
of course, you're right – leonbloy May 14 '12 at 19:45
@MichaelHardy Unfortunately, we can't move the Y inside as it may not be $\sigma(XY)$-measurable. – Ben Derrett May 17 '12 at 10:24

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.

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