Given two independent exponential distributed random variables $X,Y$ what is the condintional expectation $E[X| X+Y]$.
This is my exercise and I don't really see how to solve it. I know that I need the shared density function $f_{X|X+Y}$ then I get:
$$ E[X|X+Y=z] = \frac{1}{\int f_{X|X+Y}(x,y)dx} \int x f_{X|X+Y}(x,y)\text{d} x $$ The the wanted expectation is just $$ E[X|X+Y] = \frac{1}{\int f_{X|X+Y}(x,Y)d x}\int x f_{X|X+Y}(x,Y) d x $$
But I don't get the shared density function.
That is a good start, but you are not quite there. Replace that $Y$ with $Z-X$ and you obtain:
$\begin{align} \mathbf{\text{Let:}}\quad Z & = X+Y \\[2ex] \mathsf E(X\mid X+Y) & = \int_0^Z x f_{X\mid Y}(x\mid Z-x)\operatorname d x \\[1ex] & = \dfrac{ \displaystyle\int_0^Z x f_X(x)f_Y(Z-x) \operatorname d x }{ \displaystyle \int_0^Z f_X(x)f_Y(Z-x)\operatorname d x } \end{align}$
Can you take it from here?