Let $U=X+Y$, $V=X-Y$, while $X,Y\sim U[0,1]$ and independent. Prove or disprove.. 
Let $U=X+Y$, $V=X-Y$, while $X,Y\sim U[0,1]$ and independent. Prove or disprove: 
$(U,V)$ has a uniform distribution on some area in the plane. 
$U$ and $V+1$ are distributed the same (sorry if the translation is bad, would be happy to know how it's usually written). 
$U,V$ are independent. 
$U,V$ are (uncoordinated - not sure of the translation), but what it means is $Cov(U,V)=0$

My work: 

*

*For first statement:

Intuitively this is true, but I wanted to find the CDF: 
$F_{U,V}(u,v)=P(X+Y \le u, X-Y \le v)=P(Y \le u-X)P(X \le v+Y)=(u-X)(v+Y)$ whenever $u,v\le 1$.
I'm confused if what I did is correct and would love to hear feedback. 


*For second statement:
$P(V+1 \le v)=P(V \le v-1)=0$
$P(U \le u) = P(X+Y \le u)=P(X \le u-Y)=$.. I'm a little stuck here, what does it mean that $X$ is less than $u-Y$ since $Y$ could be anything, this is giving me some problems.


*For third statement:  
I need to  either prove that $F_UF_V=F_{U,V}$ or disprove it.
My intuition says that they're dependent, since they both depend on $X,Y$.
$F_U(u)F_V(v)=P(X+Y \le u , X-Y \le v) $, again I'm struggling with calculating these, How do I reach $X,Y$ or stuff that I know how to deal with, without complicating myself? 


*The last one was not hard, all I've done is $Cov(X+Y,X-Y)=Var(X)-Var(Y)=0$.
Any help and feedback is really appreciated, thanks in advance.
 A: 

*

*Intuitively this is true, but I wanted to find the CDF:


I'd suggest working with the pdf rather than the CDF.
Use the Jacobian change-of-variables transformation: $$f_{U,V}(u,v)=\begin{Vmatrix}\tfrac{\partial (u+v)/2}{\partial u}&\tfrac{\partial (u+v)/2}{\partial v}\\\tfrac{\partial (u-v)/2}{\partial u}&\tfrac{\partial (u-v)/2}{\partial v}\end{Vmatrix} f_{X,Y}(\tfrac {u+v}2,\tfrac {u-v}2)$$
This is much easier to evaluate and work with.
$$\phantom{f_{U,V}(u,v)=\tfrac 12(\mathbf 1_{0\leqslant u\lt 1}\mathbf 1_{-u\leqslant v\leqslant u}+\mathbf 1_{1\leqslant u\leqslant 2}\mathbf 1_{u-2\leqslant v\leqslant 2-u})}$$
A: To prove or disprove the first statement, I will directly find the joint density of $(U, V)$ using Jacobian transformation.
If $ \ U = X + Y, V = X - Y, \ \ X = \cfrac{U+V}{2}, Y = \cfrac{U-V}{2}$
$|J| = \cfrac{1}{2}$
As $f_{XY} (x, y) = 1$,
$f_{UV}(u, v) = f_{XY} \left(\cfrac{u+v}{2}, \cfrac{u-v}{2}\right) |J| = \cfrac{1}{2}$
Now as $ \ 0 \leq x \leq 1, 0 \leq y \leq 1$, $(U,V)$ is uniformly distributed over region $R$ defined below,
i) $-u \leq v \leq u \ $ for $ \ 0 \leq u \leq 1 \ $ and,
ii) $u - 2 \leq v \leq 2 - u \ $ for $ \ 1 \leq u \leq 2$
The above also shows that $U, V$ are not independent.
