What type of distribution would rand()/rand() produce? If rand() is a function which produces a linearly distributed random number over a range not containing zero, then what type of distribution would rand() / rand() produce?
I know it would center at 1, and there would be a few extreme values very close to zero or very large.
Does this type of random distribution have a name?
 A: If your interval is $(a,b)$, then look at the square $U=(a,b)\times(a,b)$.  If $\frac{b}{a}\geq z\geq 1$, then the subspace of $U$ such that $x/y>z$ is just a lower right triangle of $U$  going from $(az,a)$ to $(b,a)$ to $(b,\frac{b}z)$. This has area equal to $\frac{(b-az)(\frac{b}z - a)}2 = \frac{(b-az)^2}{2z}$.  So if $g(z)=p(x/y>z) = \frac{(b-az)^2}{2z(b-a)^2}$, then the distribution should be $-g'(z)$ for $z>1$.  
This gives a distribution of $\frac{\frac{b^2}{z^2}-a^2}{2(b-a)^2}$ for $z\geq 1$.
For $\frac{a}{b}<z\leq 1$, we do a similar computation to show that:
$$p(\frac{x}{y}<z) = \frac{(bz-a)^2}{2z(b-a)^2}$$
The derivative gives us the distribution for these $z$, so the value is here is:
$$\frac{b^2-\frac{a^2}{z^2}}{2(b-a)^2}$$
So the result is:
$$\frac{b^2-\frac{a^2}{z^2}}{2(b-a)^2}\text {when }\frac{a}b \leq z\leq 1$$
and
$$\frac{\frac{b^2}{z^2}-a^2}{2(b-a)^2}\text{ when }1\leq z \leq \frac{b}a$$
A: Let $U_1$ be a uniform random variable with domain $[a,b]$ for $b>a>0$, and let $U_2$ be another identically distributed independent uniform variable.
Let $X= \frac{U_1}{U_2}$. It is clear that $\frac{b}{a} \geqslant X \geqslant \frac{a}{b}$ with probability 1. Let $x$ be from such an interval. Then
$$ \begin{eqnarray}
 F(x) &=&  \mathbb{P}\left( X \leqslant x\right) = \mathbb{P}\left( U_1 \leqslant U_2 x \right) = \mathbb{E}\left( \min\left(1, \max\left(0,  \frac{x U_2 - a}{b-a}  \right) \right)  \right) \\
 & = &
  \int_0^1 \min\left(1, \max\left(0,  \frac{x (a+(b-a) u) - a}{b-a}  \right) \right) \mathrm{d} u \\
  &=& \int_0^1 \min\left(1, \max\left(0,  u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\
  \end{eqnarray}
$$
Consider now two cases, $\frac{b}{a} \geqslant x \geqslant 1$ and $1 > x \geqslant \frac{a}{b}$.


*

*$\frac{b}{a} \geqslant x \geqslant 1$, which implies $u \cdot x + \frac{a}{b-a} (x-1) > 0$:


$$ 
\begin{eqnarray}
   F(x) &=& \int_0^1 \min\left(1, \max\left(0,  u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\ 
   &=& \int_0^1 \min\left(1, u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
   &=& \int_0^{u^\ast} \left( u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u + \int_{u^\ast}^1 \mathrm{d} u \\
   &=& \left(\frac{u^\ast}{2} x - \frac{b - a x}{ x(b-a)}\right) u^\ast + 1 \\
   &=& 1 - \frac{1}{2 x} \left( \frac{b-a x}{b-a} \right)^2 
\end{eqnarray}
$$
where $u^\ast$ solves $u x + \frac{a}{b-a} (x-1) = 1 $, i.e. $u^\ast = \frac{b-a x}{x (b-a)}$.


*

*$1 > x \geqslant \frac{a}{b}$ implies $u \cdot x + \frac{a}{b-a} (x-1) < 1$:
$$
\begin{eqnarray}
   F(x) &=& \int_0^1 \min\left(1, \max\left(0,  u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\ 
  &=& \int_0^1 \max\left(0,  u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
  &=& \int_{u_\ast}^1 \left(u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
  &=& x \left( \frac{1}{2}- \frac{u_\ast^2}{2} \right) + \frac{a}{b-a} (x-1) \left( 1- u_\ast \right) \\
  &=& \frac{ (a- b x)^2}{2 (b-a)^2 x}
\end{eqnarray}
$$
where $u_\ast$ solves $ u \cdot x + \frac{a}{b-a} (x-1)  = 0$, i.e. $u_\ast = 
\frac{a}{b-a} \frac{1-x}{x}$.


Given $F(x)$ computed above, the probability density follows by differentiation.
