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I don't really know how to start proving this question.

Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.

Show that $E(\xi|\xi+\eta)$=$E(\eta|\xi+\eta)=\frac{\xi+\eta}{2}$

Does anyone here have any idea for starting this question?

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As this question has no explicit mention of stochastic processes, I replaced the tag with more appropriate ones. – Sasha Nov 3 '11 at 14:54
Thomas Andrews' answer to a re-posting of essentially this same question is simple and elegant. – Michael Hardy May 1 '12 at 15:35

2 Answers

up vote 6 down vote

There's a subtle point here, which bothered me the first time I saw this problem.

Henry's answer has the essential idea, which is to use symmetry. Didier Piau's comment points out that the symmetry comes from the fact that $(\xi, \eta)$ and $(\eta, \xi)$ are identically distributed. But, straight from the definition of conditional expectation, it isn't clear that symmetry in the joint distributions is enough to get the result. I ended up having to prove the following lemma:

Lemma. Let $X,Y$ be random variables. There is a measurable function $f$ such that $E[X|Y] = f(Y)$ a.s. Moreover, if $(X', Y')$ is identically distributed to $(X,Y)$, then $E[X' | Y'] = f(Y')$ a.s. for the same function $f$.

Also, to address the point in kkk's comment: Just knowing that $\xi, \eta$ are identically distributed is not sufficient. Here is a counterexample. Let $\Omega = \{a,b,c\}$ have three outcomes, each with probability $1/3$ (and $\mathcal{F} = 2^\Omega$). Let $X(a) = 0$, $X(b)=1$, $X(c)=2$; and $Y(a)=1$, $Y(b)=2$, $Y(c)=0$. Thus $X$ is uniformly distributed on $\{0,1,2\}$, and $Y = X + 1 \bmod 2$, so $Y$ is also uniformly distributed on $\{0,1,2\}$.

Now we have $(X+Y)(a) = 1$, $(X+Y)(b)=3$, $(X+Y)(c)=2$. So $X+Y$ is a 1-1 function on $\Omega$ and thus $\sigma(X+Y) = \mathcal{F}$, so both $X,Y$ are $\sigma(X+Y)$-measurable. Thus $E[X|X+Y]=X$, $E[Y|X+Y]=Y$. However, $X$, $Y$, and $\frac{X+Y}{2}$ are all different.

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I like this lemma, and the example too! – Byron Schmuland Nov 3 '11 at 14:34
This lemma IS a straight consequence of the definition of conditional expectation (and I remember seeing it in textbooks under headings like Conditional expectations depend only on the distributions). But it is nice to see it stated separately. – Did Nov 3 '11 at 16:47
Now things seem to be complicated for me when it comes to measure theory.So Byron, according to that lemma, are you saying, we still could stick to proof u gave before, if there exists such function? – kkk Nov 4 '11 at 6:54
Sorry, the proof henry presents – kkk Nov 4 '11 at 6:56
up vote 3 down vote

$E(\xi\mid \xi+\eta)=E(\eta\mid \xi+\eta)$ since they are identically distributed. (Independent does not matter here.)

So $2E(\xi\mid \xi+\eta)=2E(\eta\mid \xi+\eta) = E(\xi\mid \xi+\eta)+E(\eta\mid \xi+\eta) =E(\xi+\eta\mid \xi+\eta) = \xi+\eta$ since the sum $\xi+\eta$ is fixed.

Now divide by two.

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Qualification: $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)$ because $(\xi,\eta)$ and $(\eta,\xi)$ are identically distributed. The condition that $\xi$ and $\eta$ are identically distributed is not enough. – Did Nov 3 '11 at 11:36
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Thank you so much, i was thinking in a more complicated way.But Didier Might be right, Even if $\xi$ and $\eta$ are identically distirbuted,would that be enough to show the conditional expecation. are the same? – kkk Nov 3 '11 at 11:37
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Might be right is absolutely sweet. – Did Nov 3 '11 at 12:03

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