Conditional expectation for a sum of iid random variables: $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$ 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
$E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$
Does anyone here have any idea for starting this question?

I state in full and prove in details.
Proposition: Let $(\Omega,\mathcal{F},P$) be a probability space. Let $X,Y$ be
i.i.d. random variables with $E\left[|X|\right]<\infty$. Let $\mathcal{G}=\sigma(X+Y)$.
Then $\operatorname{E} \left[X \mid \mathcal{G}\right] = \operatorname{E} \left[Y \mid \mathcal{G}\right]=\frac{1}{2}(X+Y)$.
Proof: Let $\mu_{XY}$ be the joint distribution measure on $\mathbb{R}^{2}$
induced by $(X,Y)$. That is, $\mu_{XY}(B)=P\left(\left\{ \omega \mid (X(\omega), Y(\omega)) \in B\right\} \right)$.
Let $\mu_X$ and $\mu_Y$ be the distribution measures on $\mathbb{R}$
induced by $X$ and $Y$ respectively. Since $X$ and $Y$ are independent,
we have $\mu_{XY}=\mu_X\times\mu_Y$. Moreover, since $X$ and
$Y$ are identically distributed, $\mu_X=\mu_Y$. We denote $\mu=\mu_X=\mu_Y$.
Let $A\in\mathcal{G}$ be arbitrary. There exists a Borel set $B\subseteq\mathbb{R}$
such that $A=(X+Y)^{-1}(B)$. Hence $1_{A}(\omega)=1_{B}(X(\omega)+Y(\omega))$
for any $\omega\in\Omega$.
We have 
\begin{align}
& \int_A \operatorname{E}\left[X\mid\mathcal{G}\right]\,dP = \int_A X\,dP=\int 1_B(X+Y)X \, dP = \int 1_B(x+y)x\,d\mu_{XY}(x,y) \\[10pt]
= {} & \iint1_{B}(x+y)x\,d\mu_{X}(x) \, d\mu_Y(y) = \iint 1_B(x+y)x \, d\mu(x) \, d\mu(y).
\end{align}
By the same argument, 
$$
\int_A \operatorname{E}\left[Y\mid\mathcal{G}\right]\,dP=\iint1_{B}(x+y)y \, d\mu(x) \, d\mu(y).
$$
Now it is clear that 
$$
\int_A \operatorname{E}\left[X\mid\mathcal{G}\right]\,dP=\int_A \operatorname{E} \left[Y\mid\mathcal{G}\right] \,dP
$$
and hence $\operatorname{E} \left[X \mid \mathcal{G}\right] = \operatorname{E}\left[Y \mid \mathcal{G}\right]$.
Lastly, $\operatorname{E}\left[X+Y\mid\mathcal{G}\right]=X+Y$. It follows that $\operatorname{E}\left[X\mid\mathcal{G}\right]=\operatorname{E} \left[Y \mid \mathcal{G} \right]=\frac 1 2 (X+Y)$.
