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This is an exercise of Shiryaev's Probability on page 233:

Let $\xi$ and $\eta$ be independent identically distributed random variables with $E\xi$ defined. Show that $$E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2} \text{(a.s.)}$$

How can I say that for any $A\in\sigma(\xi+\eta)$ we have $E\xi I_A=E\eta I_A$. Thank you!

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1 Answer 1

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The distributions of $(\xi,\eta)$ and $(\xi',\eta')=(\eta,\xi)$ coincide hence, for every measurable bounded $u$, $$ \mathbb E(\xi u(\xi+\eta))=\mathbb E(\xi' u(\xi'+\eta'))=\mathbb E(\eta u(\xi+\eta)). $$ In particular, if $A$ is in $\sigma(\xi+\eta)$, there exists a Borel subset $B$ such that $A=[\xi+\eta\in B]$ hence, for $u=\mathbf 1_B$, the identity above yields $\mathbb E(\xi\mathbf 1_A)=\mathbb E(\eta\mathbf 1_A)$.

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I don't know the relation between $u$ and $\mathbf 1_A$. Can you say a little more? Thank you! –  Danielsen Dec 3 '12 at 10:06
    
See Edit. $ $ $ $ –  Did Dec 3 '12 at 10:24
    
Thank you very much! –  Danielsen Dec 3 '12 at 10:28

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