I am having trouble finding the range of $u$ and $v$ given that $x$ and $y$ satisfy
$$0<x<y<1, \quad u=x-y, v=x+y$$
I am working on finding joint distributions and generally have very difficult time getting an "acceptable" range of $u$ and $v$. (I apologize in advance for the lack of proper words for I am not sure how to describe this situation).
So far, I do not have trouble getting
$$x = \frac{u+v}{2} \quad \text{and} \quad y = -\frac{u-v}{2}$$
but I do not know what to do with the expression
$$0 < \frac{u+v}{2} < -\frac{u-v}{2} < 1$$
Here are two things that I would like to have advice in.
1), How would we know if we are able to "split" the inequality so that $u$ and $v$ does not depend on each other?
Like the original $x$ and $y$ are dependent on each other. How can we decide from the inequality that it would look like, for example, $$u<0 \quad \text{and} \quad 1<v$$
instead of the mixed expression that can be derived from above such as
$$0 < u+v < v-u < 2$$?
2), When the question is "what is the pdf?", my understanding is that you must also present the support of that distribution. So, when do I know that the support is written in a satisfactory way?
In my case, is there any reason to believe that
$$0<u+v < v-u < 2$$
would be good enough?
As I understand your problem, you want to know where in the plane the points $(x-y,x+y)$ will fall. Here’s how you want to look at the question.
You have correctly described the region of the plane that the points $(x,y)$ fill out, it’s the triangle whose corners are the origin, I’ll call it $\Bbb O$, the point $(1,1)$, which I’ll call $\Bbb D$, and the point $(0,1)$, which I’ll call $\Bbb T$.
Now, ignoring the inequalities for a moment, you’re sending a general point $(x,y)$ in the plane to the point $(u,v)=(x-y,x+y)$. This is a linear transformation of the plane (as two-dimensional vector space). But you don’t need to have much background there, to see that your triangle $\triangle\Bbb O\Bbb D\Bbb T$ will be mapped by this transformation to another triangle $\triangle\Bbb O'\Bbb D'\Bbb T'$. Here you have \begin{align} \Bbb O=(0,0)&\mapsto\Bbb O'=(0,0)\\ \Bbb D=(1,1)&\mapsto\Bbb D'=(0,2)\\ \Bbb T=(0,1)&\mapsto\Bbb T'=(-1,1)\,, \end{align} in other words, the new triangle has vertices $(0,0)$, $(0,2)$, and $(-1,1)$.
And that’s your answer. You can check several sample points like $x=1/3,y=2/3$ get transformed to points within the new triangle, I won’t do that work for you. You can also check that the three sides of the original triangle such as $\{(0,y):0\le y\le1\}$ get transformed to sides of the new triangle.
The moral of the story? Always try to find a geometric interpretation of the problem.