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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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    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)2012-05-01
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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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    Although it is true, it is not by simple symmetry. For example, if we replace the sigma algebra $\sigma(X+Y)$ by $\mathcal{F}$, then it is false. Therefore, it is true because we can prove it.2017-08-07
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    @DannyPak-KeungChan Umm, what does this have to do with $\sigma$ algebras? It is true by symmetry. You can obviously prove it, but the proof is essentially a symmetry proof.2017-08-07
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    Here, $\mathcal{F}$ denotes the whole $\sigma$-algebra of the probability space $(\Omega,\mathcal{F},P)$. Note that the expression $E(X\mid\mathcal{F})=E(Y\mid\mathcal{F})$ is symmetric in $X$ and $Y$. However, in general, it is false. That $E(X\mid X+Y)=E(Y\mid X+Y)$ does not follow from symmetry. For example, if $X$ and $Y$ are not independent, the expression $E(X\mid X+Y)=E(Y\mid X+Y)$ is still symmetric in $X$ and $Y$. However, without independence, the equality does not hold in general. Your argument using ``symmetry'' cannot explain this.2017-08-08
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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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    +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.2012-05-01
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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