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I would like to see some indications on how to approach this: Let $\xi$ and $\eta$ be independent, identically distributed random variables, with $E\xi$ defined (that is, $\xi$ is semi-integrable). Show that $E(\xi | \xi + \eta)=E(\eta | \xi + \eta ) = (\xi + \eta) / 2$ almost surely.

david
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  • Nice question, very intuitive result it seems. The first part follows by symmetry of $\xi$ and $\eta$. Thinking about the second part. – gt6989b May 31 '13 at 03:41

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This seems intuitive. The first equality $\mathbb{E}[\xi|\xi + \eta] = \mathbb{E}[\eta|\xi + \eta]$ follows by symmetry in $\xi$ and $\eta$. Let $x = \mathbb{E}[\xi|\xi + \eta]$.

Note that

$$ 2x = x + x = \mathbb{E}[\xi|\xi + \eta] + \mathbb{E}[\eta|\xi + \eta] = \mathbb{E}[\xi + \eta|\xi + \eta] = \xi + \eta. $$ Therefore, $$x = \frac{\xi+\eta}{2}$$ as desired.

gt6989b
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  • The first equality E[ξ|ξ+η]=E[η|ξ+η] follows by symmetry< Be careful: $E[\xi|\xi+\eta]=\xi$ is also a "symmetric" statement.

     – fedja May 31 '13 at 03:47
  • @fedja I'm sorry, I didn't follow that - what is symmetric about it? It clearly prefers $\xi$ to $\eta$, i.e. if I replace $\xi$ by $\eta$ and vice versa, I get a different statement. The one we made is indeed symmetric - such a transformation results in itself. – gt6989b May 31 '13 at 03:50
  • @fedja From the point of the problem, $\xi$ and $\eta$ are indistinguishable, it is symmetric in them... – gt6989b May 31 '13 at 03:51
  • OK, I just thought you meant "trivial notational symmetry" (i.e., you get the same if you swap the variables in the statement and then swap their notation). Apparently you meant a deeper thing, in which case you are right but it would be nice to clarify exactly what symmetry you referred to :). – fedja May 31 '13 at 04:02