Joint PDF of two random variables and their sum What is the joint PDF of two uniformly distributed random variables and their sum?
 A: I will try to address the question you posed in the comments, namely:

Given 3 independent random variables $U$, $V$ and $W$ uniformly distributed on $(0,1)$, find the joint probability distribution function of $X=U+V$ and $Y=U+W$.

Gives $0<x<2$ and $0<y<2$, we first compute $F_{X,Y}(x,y)$:
$$\begin{eqnarray}
 F_{X,Y}(x,y) &=& \mathbb{P}\left(X \leqslant x, Y \leqslant y\right) = \mathbb{P}\left(U+V \leqslant x, U+W \leqslant y\right) = \mathbb{E}\left(\mathbb{P}\left(U+V \leqslant x, U+W \leqslant y|U\right)\right)\\
  &=& \mathbb{E}\left(F_V(x-U)F_W(y-U)\right) = \mathbb{E}\left(\min(1, \max(x-U,0))\min(1, \max(y-U,0))\right)
\end{eqnarray}
$$
The pdf, being a derivative of the cdf, will then read:
$$
    f_{X,Y}(x,y) = \frac{\partial}{\partial x}
\frac{\partial}{\partial y}  F_{X,Y}(x,y) = \mathbb{E}\left( [1+U>x>U] [1+U > y>U] \right) = \mathbb{P}\left(U < x < 1+U \land U < y<1+U\right)
$$
The evaluation of the latter expectation is straightforward but tedious, so I asked Mathematica for help:

A: 
The problem I'm trying to tackle is really finding the marginal pdf p(e,f) given that a,b,c are uniform and iid RVs while e,f are a+b and a+c respectively. 

Call $X=U+V$ and $Y=U+W$ for $(U,V,W)$ i.i.d. uniform on $(0,1)$. The common density of $U$, $V$ and $W$ is $\mathbf 1_{(0,1)}$ hence the density of $(X,Y)$ is the function $f$ defined by
$$
f(x,y)=\int_0^1\mathbf 1_{(0,1)}(x-u)\,\mathbf 1_{(0,1)}(y-u)\,\mathrm du.
$$
Introducing $s=\min(x,y)$ and $t=\max(x,y)$, this is
$f(x,y)=s$ if $0\lt s,t\lt1$ and $f(x,y)=s-t+1$ if $1\lt t\lt2$ and $t\lt1+s$.
