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Possible Duplicate:
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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    @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.2012-05-01

2 Answers 2

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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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    See Nate Eldredge's counter-example in https://math.stackexchange.com/questions/78546/conditional-expectation-for-a-sum-of-iid-random-variables-e-xi-mid-xi-eta-e/2385794#2385794.2017-08-08
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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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    Replace $B=(X+Y)(B')$ by $B=(X+Y)^{-1}(B')$ and $dP_XdP_Y$ by $dP_X(x)dP_Y(y)$.2012-05-08