A recent question (Are there any (pairs of) simple distributions that give rise to a power law ratio?) asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if $X$ is a Gamma random variable with parameters $(n, 1)$ and $Y$ is an exponential random variable with parameter $1$, then $Y/X$ is a Pareto random variable with parameters $(1, n)$. Presumably $X$ and $Y$ need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what $(1, n)$ means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in $(1,\infty)$ and its complementary CDF decays away as $z^{-n}$.
My question is: what is the intuitive explanation for the ratio $Y/X$ to have value $1$ or more? It would seem that all positive values should occur, and indeed the event
$\{Y < X\}$ should have large probability since the Gamma random variable has larger mean than the exponential random variable.
I did work out the complementary CDF of $Y/X$ and got $(1+z)^{-n}$ for $z > 0$ which is not quite what Wikipedia claims.
Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that
for $z > 0$, $P\{Y/X > z\} = (1+z)^{-n}$ which of course implies that
$$P\{Y/X + 1 > z\} = P\{Y/X > z-1\} = (1 + z – 1)^{-n} = z^{-n}~ \text{for}~ z > 1$$
and thus it is $Y/X + 1$ which is a Pareto random variable, not $Y/X$ as claimed by
Wikipedia. This leads to a simple answer to
a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio?
If $W$ and $X$ are the $(n+1)$-th and $n$-th arrival times in a (homogeneous) Poisson process, then $W/X$ is a Pareto $(1,n)$ random variable: $P\{W/X > a\} = a^{-n}$ for $a > 1$.
I suspect that this result is quite well known in the theory
of Poisson processes but I don't have a reference for it.
Best Answer
As I claimed in the comments, $Y/X + 1$ follows the $\mathrm{Pareto}(1,n)$, not $Y/X$.
$$ \phi_{Y/X}(t)= \mathbb{E}\left(\exp\left(i t \frac{Y}{X}\right)\right) = \mathbb{E}\left( \phi_Y\left( \frac{t}{X} \right) \right) = \mathbb{E}\left( \frac{X}{X-i t} \right) $$ Writing the last expectation explicitly: $$ \begin{eqnarray} \phi_{Y/X}(t) &=& \frac{1}{(n-1)!} \int_0^\infty \frac{x^n}{x - i t} \mathrm{e}^{-x} \mathrm{d} x = \frac{1}{(n-1)!} \int_0^\infty \mathrm{d} u \int_0^\infty x^n \mathrm{e}^{-x} \mathrm{e}^{-u(x - i t)} \mathrm{d} x \\ &=& \frac{1}{(n-1)!} \int_0^\infty \mathrm{e}^{i t u} n! (1+u)^{-1-n} \, \mathrm{d} u = \mathrm{e}^{-i t} \int_0^\infty \mathrm{e}^{i t (u+1)} \, \frac{n}{(1+u)^{n+1}} \mathrm{d} u \\ &=& \mathrm{e}^{-i t} \phi_{\mathrm{Pareto}(1,n)}(t) \end{eqnarray} $$ This proves that $Y/X \stackrel{d}{=} Z - 1$, where $Z$ follows $\mathrm{Pareto}(1,n)$