Problem
Let $X, Y \sim Uniform(0, 1)$ be independent. Find the PDF for $X - Y$ and $X / Y$.
Solution
Part 1: PDF of $X - Y$
$X, Y \sim Uniform(0,1)$ means
$f_{X,Y}(x,y) = \begin{cases} 1 & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$
our transformation function $g(x, y) = x - y$.
$F_Z(z) = P(Z \leq z) = P(g(x,y) \leq z) = P(\{x,y\}: g(x,y) \leq z)$
Let's call $P(\{x,y\}: g(x,y) \leq z)$ the region $A_z$.
$F_Z(z) = \int\int_{A_z} f(x,y) dy,dx$
We can see that $x$ and $y$ go from 0 to 1, $z$ will range from -1 to 1, but how to we reason about $A_z$? A diagram helps:
as $x$ and $y$ go from 0 to 1, we can imagine $z$ sweeping down from the upper left to the bottom right. There are two regions we need to consider: $-1 \leq z < 0$ in the top left and $0 \leq z < 1$ in the bottom right.
For $-1 \leq z < 0$ $A_z$ is the triangle with vertices $(0, -z), (0, 1), (z+1, 1)$ with area $\frac{1}{2}(z+1)^2$
For $0 \leq z < 1$ $A_z$ is everything in the unit square minus triangle with vertices $(z, 0), (1, 0), (1, 1-z)$ with area $1 - \frac{1}{2}(1-z)^2$
(it really helps to draw this out)
So
$F_Z(z) = \begin{cases} 0 & z < -1 \\ \frac{1}{2}(z+1)^2 & -1 \leq z < 0 \\ 1 - \frac{1}{2}(1-z)^2 & 0 < z \leq 1 \\ 0 & z \geq 1 \end{cases} $
differentiating we get:
$f_Z(z)= \begin{cases} 1 + z & -1 \leq z < 0 \\ 1 - z & 0 < z \leq 1 \\ 0 & \text{otherwise} \end{cases} $
This PDF is known as the triangle distribution:
