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Help with conditional expectation question

I have problem with exercise, I didn't solve.

Let $X$ and $Y$ be i.i.d. random variables with $E(X)$ defined. Show that

$$E(X|X+Y)=E(Y|X+Y)= \frac{X+Y}{2}$$ (a.s.)

Thanks very much for your help.

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marked as duplicate by Nate Eldredge, Davide Giraudo, leonbloy, cardinal, t.b. May 1 '12 at 17:37

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I dislike the notation. $X$ and $Y$ on the right term are neither random variables nor given values (what is given is just their sum) –  leonbloy May 1 '12 at 15:14
    
@leonbloy : They are random variables. Suppose the conditional expected value of the random variable $U$ given the event $V=v$, where $V$ is a random variable, is some function $g(v)$ of $v$. Then $\mathbb{E}(U\mid V=v)=g(v)$. Then one defines $\mathbb{E}(U\mid V)$ to be the random variable $g(V)$. This is perfectly standard. –  Michael Hardy May 1 '12 at 15:33
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2 Answers 2

The first part of the equation:

$$E(X|X+Y) = E(Y|X+Y)$$

is true by symmetry. They are independent and identical.

Now, what is $E(X|X+Y)+E(Y|X+Y)$?

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Hint:

  • Using the linearity of conditional expectation (and an other property), show that $E(X\mid X+Y)+E(Y\mid X+Y)=X+Y$.
  • Show that $E(X\mid X+Y)=E(Y\mid X+Y)$ by the following argument. Take $B$ a set in the $\sigma$-algebra generated by $X+Y$ ($B=(X+Y)^{-1}(B')$ for some $B'$), then write $$\int_B X \, dP=\int_{\Bbb R}\int_{\mathbb R}x\chi_{x+y\in B'} \, dP_X(x)\, dP_Y(y),$$ then use independence and a substitution.
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+1 for this and the answer by Thomas Andrews. I think the latter is better. This answer relies on Kolmogorov's way of making certain things logically precise. But if Thomas Andrews' answer didn't work within Kolmogorov's system, we would reject Kolmogorov's system rather than rejecting Thomas Andrews' answer. –  Michael Hardy May 1 '12 at 15:13
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Replace $B=(X+Y)(B')$ by $B=(X+Y)^{-1}(B')$ and $dP_XdP_Y$ by $dP_X(x)dP_Y(y)$. –  Did May 8 '12 at 11:13
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