I have a random vector $X,Y$ (with respect to the Lebesgue measure):
$$f_{X,Y}=1/2 \mathbb I _{|x|+|y|\leq1}$$
And I want to show that $X+Y$ and $X-Y$ are independent. One way to show it, is to show that:
$$f_{X+Y}f_{X-Y}=f_{X+Y,X-Y}$$
I understand that I can calculate the densities for $X+Y$, $X-Y$ with the usual convolution formula. Although maybe there is the better way... in any case how do I get the joint density?
Notice that $|x|+|y| \leq 1$ if and only if $-1 \leq x+y \leq 1$ and $-1 \leq x-y \leq 1$, and so,
$$ \mathbf{1}_{\{|x|+|y|\leq 1\}} = \mathbf{1}_{\{x+y \in [-1,1]\}} \mathbf{1}_{\{x-y \in [-1,1]\}}. $$
This already tells that the joint density of $X+Y$ and $X-Y$ must split. Indeed, the change of variables $(u, v) = (x+y, x-y)$ shows
\begin{align*} f_{X,Y}(x,y)\, dxdy &= \frac{1}{4} \mathbf{1}_{\{u\in [-1,1]\}} \mathbf{1}_{\{v\in [-1,1]\}} \, dudv. \end{align*}
Therefore the joint density of $X+Y$ and $X-Y$ splits and they are independent.