Conditional expectation $E[X\mid X+Y]$ Let $X$ and $Y$ be independent, identically distributed integrable random variables. Show that $$E[X\mid X+Y]=\frac{X+Y}{2}$$
 A: $$
X+Y = \operatorname{E}(X+Y \mid X+Y) = \operatorname{E}(X\mid X+Y) + \operatorname{E} (Y\mid X+Y).
$$
If the two conditional expectations on the right are equal to each other (and symmetry shows that they are), then each must be equal to half of the expression on the right.
Postscript on symmetry: It was given that $X,Y$ are identically distributed and independent.  That implies $(X,Y)$ has the same distribution as $(Y,X)$, a considerably weaker statement.  That weaker statement is enough to get the conclusion.  But assuming identical distribution of $X$ and $Y$ is weaker than identical distribution of $(X,Y)$ and $(Y,X)$, and the former is not strong enough to get the conclusion.  The following example demonstrates this:
$$
(X,Y) = \begin{cases} (0,2) & \text{with probability }1/3, \\ (1,0) & \text{with probability }1/3, \\ (2,1) & \text{with probability }1/3. \end{cases}
$$
For this joint distribution,


*

*$X$ and $Y$ are both uniformly distributed on $\{0,1,2\}$; and

*the distributions of $(X,Y)$ and $(Y,X)$ are not the same; and

*$\operatorname{E}(X \mid X+Y=2) = 2 \ne 1 = (X+Y)/2$.

A: As Michael suggested, the problem can be reduced to show that $E[X\mid X + Y] = E[Y\mid X + Y]$. Intuitively, this is quite obvious by symmetry. To show it rigorously, lets' denote $E[X \mid X + Y]$ by $X'$ and $E[Y \mid X + Y] = Y'$. Our goal is to show $X' = Y'$. By the definition of conditional expectation, for every $H \in \sigma(X + Y)$, we have
$$\int_H X dP = \int_H X' dP, \quad \int_H Y dP = \int_H Y' dP. \tag{1}$$
Since $X$ and $Y$ are i.i.d., for the general element $H = \{X + Y \leq z\} \in \sigma(X + Y)$, we have 
$$\int_H X dP = \int_H Y dP = \int_{-\infty}^\infty uF(z - u)dF(u).$$
Together with $(1)$, we derive that
$$\int_H (X' - Y') dP = \int_H (Y' - X') dP = 0, \quad \forall H \in \sigma(X + Y).\tag{2}$$
Consider the set $H_1 = [\omega: X'(\omega) < Y'(\omega)]$, since both $X'$ and $Y'$ are measurable $\sigma(X + Y)$, so is $X' - Y'$, hence $H_1 \in \sigma(X + Y)$. By $(2)$, we thus conclude
$$\int_{H_1} (Y' - X') dP = 0. \tag{3}$$
Since $Y' - X'$ is everywhere positive on $H_1$, $(3)$ implies that $P[H_1] = 0$. Similarly, you can show that $P[X' > Y'] = 0$. Therefore $P[X' = Y'] = 1$. 
