I came across this and this answers explaning how to compute joint probability distribution by using differential forms, which I think it is an elegant method to deal with this kind of problem. I have tried to apply this approach to the following problem:
Let $U,V$ be normally distributed random variables and $W$ be uniformly distributed over $(0,1)$.
Let also two other continuous dependent variables $X$ and $Y$ given by
$$X=U+W ~~~~ \text{and} ~~~~ Y=V+W.$$
I want to compute the joint probability distribution $f_{XY}(x,y)$ from $f_{UV}(u,v)$ and $f_W(w)$.
My thoughts:
First, I compute the wedge product $$dx \wedge dy = (du+dw)\wedge(dv+dw) = du\wedge dv+du\wedge dw+dw\wedge dv$$
However, when I write the probability element $$ f_{XY}(x,y)|dx \wedge dy|, $$
I have no idea how to proceed from this
$$|dx \wedge dy| = |du\wedge dv+du\wedge dw+dw\wedge dv|. $$
I appreciate any help!
EDIT:
I think I have got the solution but I am not sure if it is correct. I created a new variable $Z=W$. This allows me to write the differential element $|dx\wedge dy \wedge dz|$ as a three-form $g(u,v,w)|du\wedge dv \wedge dw|.$ Thus,
$$ dx\,dy\,dz=|dx\wedge dy \wedge dz| =|(du+dw)\wedge (dv+dw)\wedge dw| = |du\wedge dv\wedge dw|=du\,dv\,dw.$$
Therefore, the joint probability element $f_{XYZ}(x,y,z)\,dx\,dy\,dz$ is given by
$$f_{XYZ}(x,y,z)\,dx\,dy\,dz = f_{UVW}(u,v,w)\,du\,dv\,dw \implies f_{XYZ}(x,y,z) = f_{UVW}(x-z,y-z,z) = f_{UV}(x-z,y-z)f_{W}(z),$$ and
$$f_{XY}(x,y) = \int_{-\infty}^{\infty}f_{XYZ}(x,y,z)\,dz = \int_{-\infty}^{\infty}f_{UV}(x-z,y-z)f_{W}(z)\,dz = \int_{0}^{1}f_{UV}(x-z,y-z)\,dz.$$
Is my solution correct?
Thanks in advance!