If $X,Y,Z$ are independent random variables having identical density functions If $X,Y,Z$ are independent random variables having identical density functions $f(x) = e^{-x} $ , $0 < x < \infty$ , derive the joint probability density of the vector $(U,V,W)$ , where
\begin{align}
U &= X + Y \\
V &= X + Z \\
W &= Y + Z
\end{align}
I'm not sure on how to do this but, if were to find the joint probability density... I would do moment generate function for each $U,$ $V$ and $W$  and then multiple those. 
 A: One idea is to use the change of variables theorem. Consider the random vector $(X,Y,Z)$ and notice that:
$$f_{(X,Y,Z)}(x,y,z) = e^{-(x+y+z)} \mathbf{1}_{\mathbb{R}_+^3}(x,y,z).$$
Define the transformation 
$$T:(x,y,z) \mapsto (x + y, x + z, y + z)$$ which is linear and possesses the inverse
$$S:(u,v,w) \mapsto \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right).$$
By the change of variables theorem, the random vector $(U,V,W) = T(X,Y,Z)$ has density
$$f_{(U,V,W)}(u,v,w) = f_{(X,Y,Z)}(S(u,v,w)) |\det \mathbf{D}S(u,v,w)|.$$
Now, $\det \mathbf{D}S(u,v,w) = \det S = -\dfrac{1}{2},$ so
$\begin{align}
f_{(U,V,W)}(u,v,w) &=\dfrac{1}{2} f_{(X,Y,Z)}\left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\
&= \dfrac{1}{2} e^{-\frac{u+v+w}{2}}  \mathbf{1}_{\mathbb{R}_+^3} \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\
&= \dfrac{1}{2} e^{-\frac{u+v+w}{2}}  \mathbf{1}_{\mathrm{G}}(u,v,w),
\end{align}$
where $\mathrm{G}$ is the set of $(u,v,w)$ such that the three relations $u + v \geq w,$ $u + w \geq v$ and $v + w \geq u$ hold. (Naturally, these three relations imply $u,v,w \geq 0$).
