Suppose that $X$ and $Y$ are independent and $X\sim E(1)=\Gamma(1,1)$ and $Y\sim\Gamma(2,1)$. Then I am asked to find $E[X\mid X+Y]$ by using the following result:
Let $(U,V)$ be an $n+m$ dimensional random vector with density $(u,v)\mapsto f(u,v)$ with respect to $\lambda_{n+m}$. Put for $u\in\mathbb{R}^n$ and $v\in\mathbb{R}^m$ f_V(v)=\int_{\mathbb{R}^n} f(u,v)\lambda_n (du)\quad \text{and} \quad f_{U\mid V}(u\mid v)=\frac{f(u,v)}{f_V(v)}1_{\{0
Then for every Borel function $\psi: \mathbb{R}^{n+m}\to\mathbb{R}$ with E[|\psi(U,Y)|]<\infty we have that $ E[\psi(U,V)\mid V]=\varphi(V) \quad\text{a.s.}, $ where $ \varphi(v)=\int_{\mathbb{R}^n} \psi(u,v) f_{U\mid V}(u\mid v)\lambda_n (d v). $
So I was thinking that I would use this result with $U=X$ and $V=X+Y$ and $\psi(x,y)=x$. Then we are clearly in the scope of this result. Since $X$ and $Y$ are independent we also have that $X+Y\sim\Gamma(3,1)$, and hence $f_V$ is just a Gamma-density.
Now my question is, how do I go by finding the joint density $f$ in the easiest way? I have tried looking at probabilities $P(X\leq a,X+Y\leq b)$ but without any luck (it got very messy). An additional question is: Is it possible to obtain the conditional expectation in other ways than using this result?
Thanks in advance.