Calculate $E(X|X+Y)$ where $X,Y$ have expotential distribution with parameter 1

Question

Calculate $E(X|X+Y)$ where $X,Y$ have expotential distribution with parameter 1. I calculated distribution of $X+Y$, which if I am not mistaken is $\frac{e^{-x-y}}{x+y}$, where $x+y>=0$, but I am not sure how to calculate the joint distribution of $ (X,X+Y)$ , since $X$ and $X+Y$ are clearly dependent. Or maybe there is a simplier way of calculating this expected value?

Answer 1

The given density function for $(X,Y)$ is $f_{X,Y}(x,y) =e^{-x-y}1_{x,y>0}$. By applying change of variables formula to $(X,Y)\mapsto (X,Z)$ where $Z=X+Y$, we obtain the p.d.f. of $(X,Z)$ as $$ f_{X,Z}(x,z) = e^{-z}1_{x>0, z-x>0}. $$ From this, we can marginalize $f_Z(z)=\int f_{X,Z}(x,z)\ dx =\int_0^z e^{-z}1_{z>0}\ dx=ze^{-z}1_{z>0}$ and find that $$ f_{X|Z}(x|z) = \frac{f_{X,Z}(x,z)}{f_Z(z) }=\frac{1}{z}1_{x>0, z-x>0}=\frac1{z}1_{0<x<z}. $$ This shows $X|Z$ is uniformly distributed on $[0,z]$, hence we get $$ \Bbb E[X|X+Y]=\Bbb E[X|Z] =\frac{Z}2=\frac{X+Y}2. $$

Intuitively, since $X$ and $Y$ have identical distribution, we would have $$ \Bbb E[X|X+Y=z]=\Bbb E[Y|X+Y=z]=\frac12\Bbb E[X+Y|X+Y=z]=\frac{z}2. $$

Answer 2

Just use the fact that $X$ and $Y$ are iid. We have, because $X+Y$ is $\sigma(X+Y)$-measurable:

$$\mathbb{E}[X|X+Y] +\mathbb{E}[Y|X+Y] = \mathbb{E}[X+Y | X+Y ] = X+Y. $$

But by the iid assumption,

$$ \mathbb{E}[X|X+Y] =\mathbb{E}[Y|X+Y] . $$

Putting the above together then easily gives $\mathbb{E}[X|X+Y] =\frac{1}{2}(X+Y).$