Calculate $E[X|X+Y=z]$ for independent and identically distributed $X$ and $Y$.

Question

Let $X,Y$ be independent and indentically distributed random variables taking values in $(0,\infty)$ and $z>0$. I want to calculate $E[X\mid X+Y=z]$. My intuition tells me, that its value should be $\frac z2$. My idea was to prove this using the equation \begin{align*} z &=E[X+Y\mid X+Y=z] \\ &=E[X|X+Y=z]+E[Y\mid X+Y=z] \\ &=2E[X\mid X+Y=z], \end{align*} which holds if $X$ and $Y$ conditioned on $X+Y=z$ are also identically distributed.

This is easily verifiable for absolutely continuous $X$ and $Y$ with pdfs $f_X$ and $f_Y$, since we can calculate \begin{align*} f_{X,X+Y}(x,y) &=f_{X,Y}(x,y-x) \\ &=f_{X}(x)f_Y(y-x) \\ &=f_Y(x)f_X(y-x) \\ &=f_{Y,X}(x,y-x) \\ &=f_{Y,X+Y}(x,y), \end{align*} which implies $f_{X\mid X+Y}(x\mid z)=f_{Y\mid X+Y}(x\mid z)$

How can I show this for general $X$ and $Y$? Or can you give me another idea on how to calculate the conditional expectation? Many thanks in advance.

Answer 1

By definition: $$\int E[X\mid X+Y]I_A(X+Y)dP_{\Omega}=\int XI_A(X+Y)dP_{\Omega}\tag{1}$$ But: $\hspace{40pt}\int E[X\mid X+Y]I_A(X+Y)dP_{\Omega}=\int E[X\mid x+y] I_A(x+y)dP_{X\circ Y}$

And: $\hspace{80pt}\int XI_A(X+Y)dP_{\Omega}=\int xI_A(x+y))dP_{X\circ Y}$

Therefore, for (1) to be valid, it is enough that: $$\int E[X\mid x+y ]I_A(x+y)dP_{X\circ Y}=\int xI_A(x+y))dP_{X\circ Y}$$ But this last equation, and therefore $E[X\mid x+y]$, only depends on the distribution of $(X,Y)$ which is the same as that of $(Y,X)$.

Answer 2

Let $W_1 = E(X\mid Z)$ and $W_2 = E(Y\mid Z)$. Then for any $Z$ measurable set $A$ it is the case that $$ EW_1I(Z\in A) = EXI(Z\in A) = EYI(Z\in A) = EW_2I(Z\in A) $$ The first equality and the last equality are by the definition of conditional expectation. The middle equality is true since $X$ and $Y$ are i.i.d.

It follows that $W_1 = W_2$ a.s.

Hence $$ X+Y=E(X+Y\mid Z) = E(X\mid Z) + E(Y\mid Z) = 2E(X\mid Z) $$

Therefore $$ E(X\mid Z) = E(Y\mid Z) = \frac{X+Y}{2} $$