Determine joint density of expressions of two variables

Question

I have two identically distributed and independent variables, $X$ and $Y$, on $(0, 1)$. I also have $U = X + Y$ and $V = X / Y$. The problem is to computer the joint density of $U$ and $V$. How do I go about doing this?

Answer 1

HINT

Assuming $X,Y$ are continuous on $(0,1)$.

Given $U,V$ you have $X = UV/(V+1)$ and $Y = U/(V+1)$.

Transform $\mathbb{P}[(U,V) \in A]$ to $\mathbb{P}(X,Y) \in B]$

UPDATE

So you have $$ F_{UV}(u,v) = \mathbb{P}[U < u, V<v] = \mathbb{P}[X+Y < u, X<vY] = \iint_A f_{XY}(x,y)dxdy $$ where $A$ is the region bounded by $0 < x,y < 1, x+y<u, x<vy$, and since $X,Y$ are independent, $f_{XY}(x,y) = f_X(x) f_Y(y)$. Can you complete the problem now?

UPDATE 2

Your region is a triangle, bounded by the $y$-axis between the origin and $(0,u)$ from the left, the line $x+y = u$ from the top and the line $x = vy$ from the bottom. Hence the bounds of integration would be $x/v < y < u-x$ with $0 < x < uv/(v+1)$.

Let the pdf and cdf of $X$ and $Y$ be denoted by $g$ and $G$. Thus you have $$ \begin{split} F_{UV}(u,v) &= \int_{x=0}^{x=uv/(v+1)} \int_{y=x/v}^{y=u-x} g(x) g(y) dy dx \\ &= \int_{x=0}^{x=uv/(v+1)} \left(G(u-x)- G\left( \frac{x}{v} \right)\right) g(x) dx \end{split} $$

If you know, for example, that $X,Y$ are uniform, then on $(0,1)$ we have $g(x) = 1$ and $G(x) = x$, so this integral becomes $$ \begin{split} F_{UV}(u,v) &= \int_{x=0}^{x=uv/(v+1)} \left((u-x)- \left( \frac{x}{v} \right)\right) 1 dx \\ &= \frac{v u^2}{v+1} - \left(\frac{uv}{v+1}\right)^2 \frac{v+1}{2v} \\ &= \frac{2vu^2 - u^2 v}{2(v+1)} \\ &= \frac{vu^2}{2(v+1)} \end{split} $$ and $f_{UV}(uv)$ is the mixed partial of that as I wrote before.