Let $X$ and $Y$ be independent random variables following $U(-1,1)$. Find the joint CDF of $U=X-Y$ and $V=X$.
I found the Jacobian of the transformation to be equal to $1$, and $f_Xf_Y=\frac{1}{4}$. I'm confused regarding the final solution.
Is $\displaystyle P(U\leq s,V\leq t) = \int_{-1}^{t}\int_{-2}^{s}\frac{1}{8}du dv$? Somehow this seems wrong to me.
If you are using the transformation formula, you should have got, for the joint density
$$f_{U,V}(u,v)=\frac{f_{X,Y}(x,y)}{\left |\frac{\partial(U,V) }{\partial(X,Y)}\right|}=\frac{1}{4}$$
but this is not the end of the story, you need to get the support of the transformed variables. To write $-2<U<2$ and $-1<V<1$ is not totally right: both inequalities are true, but they don't give you the support, because (say) you cannot have simultaneously $V=0.9$ and $U=1.9$.
(You can also guess that something is wrong in that the integral of the density over the support must be one).
The correct way is to note that if we allow for $V=X$ its full range $-1<V<1$ then we must put that dependence into the other variable: $U=Y-X=Y-V$ , hence the range of $U$ is $(-1-V, 1-V)$
Then the support is $-1<V<1$ and $-1-V< U <1-V$ which corresponds to a parallelogram.
(Notice BTW that two variables with uniform joint density over a -straight- rectangular support are independent - which is the case for $X,Y$, but it's not -it should not- for $U,V$ )
Sanity check:
$$ \int f_{U,V}= \int_{-1}^1 \int_{-1-V}^{1-V} \frac{1}{4} dU dV= \frac{1}{4} \int_{-1}^1 2 dV= 1$$