Joint distribution of $U = X + Y$ and $V = X - Y$ I have two independent continuous random variables, $X$ and $Y$, which are uniformly distributed over the interval $[0,1]$. From this I have two further random variables, $U$ and $V$, which are defined as $U = X + Y$, and $V = X - Y$.
I am trying to figure out the density function for the joint probability distribution of $(U,V)$ but am struggling. I have calculated the density functions of U and V on their own, but do not think it makes sense to simply multiply these together, as I do not think U and V are independent. Is this assumption correct?
My university lecturer suggested sketching out the range of values over which $(U,V)$ is defined, which seems to suggest he is trying to lead me towards a more intuitive solution, but I would appreciate any explanation (analytical or otherwise) which would help me understand how to solve problems such as these.
 A: Let $g(x, y) = (x+y, x-y)$. Then $(U, V) = g(X, Y)$.
You know that $$f_{UV}(u, v) = f_{XY}(h(u, v))|J_h(u,v)|$$
where $h = g^{-1}$, $J_h$ is the jacobian matrix of $h$
So $h(u, v) = (\frac{u+v}2, \frac{u-v}2)$ $|J_h| = \frac 12$, $f_{XY}(x, y) = f_X(x)f_Y(y)$ (because $X$ and $Y$ are independent)
Therefore $$f_{XY}(x,y) = \begin{cases} 1 & \text{on the set $[0, 1] \times [0, 1]$} \\ 0  & \text{otherwise} \end{cases}$$
hence $f_{UV}(u, v) = \frac 12$ only when $h(u, v) \in [0, 1] \times [0, 1]$. That is, only on the set $A = g([0, 1] \times [0, 1])$ which is a square with vertices on $(0, 0), (1, 1), (1, -1), (2, 0)$
So finally 
$$f_{UV}(u, v) = \begin{cases} \frac 12 & \text{on the set $A$} \\ 0  & \text{otherwise} \end{cases}$$
This also tells you that no, $U$ and $V$ are not independent, because the support of their joint density is not a rectangle whose sides are parallel to the axis
A: $F\left(u,v\right)=P\left\{ U\leq u\wedge V\leq v\right\} =P\left\{ X+Y\leq u\wedge X-Y\leq v\right\} =\iint_{A}dxdy$
where $A=\left\{ \left(x,y\right)\in\left[0,1\right]^{2}\mid x+y\leq u\wedge x-y\leq v\right\} $.
If you have that then find density $f\left(u,v\right)=\left(d^{2}/dudv\right)F\left(u,v\right)$
A: $X = \left(U + V\right)/2\,,\quad Y = \left(U - V\right)/2$
\begin{align}
\color{#ff0000}{\large{\cal P}\left(U,V\right)}
&=
{\partial^{2} \over \partial U\,\partial V}
\int_{-\infty}^{V}{\rm d}V'\int_{-\infty}^{U}{\rm d}U'\,{\cal P}\left(U',V'\right)
\\[3mm]&=
{\partial^{2} \over \partial U\,\partial V}
\int_{0}^{1}{\rm d}X\int_{0}^{1}{\rm d}Y\,
\Theta\left(U - X - Y\right)\Theta\left(V - X + Y\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\int_{0}^{1}{\rm d}Y\,
\delta\left(U - X - Y\right)\,\delta\left(V - X + Y\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\,\delta\left(U - X - \left[X - V\right]\right)
\int_{0}^{1}{\rm d}Y\,\delta\left(Y - \left[X - V\right]\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\,\delta\left(U + V - 2X\right)\
\Theta\left(X - V\right)\ \Theta\left(1 - \left[X - V\right]\right)
\\[3mm]&=
\Theta\left({U + V \over 2} - V\right)\Theta\left(1 + V - {U + V \over 2}\right)
{ 1 \over 2}\int_{0}^{1}{\rm d}X\,\delta\left(X - {U + V \over 2}\right)
\\[3mm]&=
\color{#ff0000}{\large%
{1 \over 2}\,\Theta\left(U - V\right)\ \Theta\left(2 -\left[U - V\right]\right)\
\Theta\left(U + V\right)\ \Theta\left(2 - \left[U + V\right]\right)}
\end{align}
This expression 'define the region':
\begin{eqnarray*}
0 < & U - V < 2
\\
0 < & U + V < 2
\end{eqnarray*}
