TITLE: Uniform distribution joint $\to$ marginal
QUESTION [0 upvotes]: Let vector $(X,Y)$ have a uniform distribution on the set $N = \{ (x,y): x<1,y<1,1<x+y\}$. Determine distribution $X-Y$. 
So far I've thought of this:
\begin{align}
P[X | Y=y] &\sim U(0,1-x) \space \forall y \in (0,1)\\
P[Y|X=x] &\sim U(0,1-y) \space \forall x \in (0,1)\\
\end{align}
but honestly I don't know how to go about it.
Any hints?

REPLY [1 votes]: There is actually a fully "visual" solution to this. 
The couple $(X,Y)$ is uniformly distributed on the triangle $T$ with vertices $(1,0)$, $(1,1)$ and $(0,1)$. Thus $-1\leqslant X-Y\leqslant1$ with full probability. 
Let $-1\leqslant z\leqslant1$. The intersection of the triangle $T$ with the line $x-y=z$ is a segment of length $1-|z|$. 
(This segment starts from the boundary $x+y=1$ of $T$ at the point $(\frac12(1+z),\frac12(1-z))$ and ends at the point $(1,1-z)$ on the boundary $x=1$ if $z\gt0$ and at the point $(1+z,1)$ on the boundary $y=1$ if $z\lt0$.)
Thus the density of $X-Y$ is proportional to the function $f:z\mapsto(1-|z|)\mathbf 1_{|z|\leqslant1}$. Since the integral of $f$ is $1$, $f$ is the density of $X-Y$.
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