Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
The first one has been asked several times here, search for it. – leonbloy May 14 '12 at 19:19
up vote 0 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.