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A recent question 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?


I suspect that this result is quite well known in the theory of Poisson processes but I don't have a reference for it.

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Actually it is $Y/X + 1$ that follows $\mathrm{Pareto}(1,n)$. – Sasha Oct 26 '11 at 20:49
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+\infty)$ and not $(1,+\infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$. – Did Oct 26 '11 at 20:54
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question? – Dilip Sarwate Oct 27 '11 at 3:36
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[X\leqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$). – Did Oct 27 '11 at 7:56

1 Answer 1

up vote 4 down vote accepted

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 $\mathrm{Pareto}(1,n)$

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+1 Thanks for confirming my calculations. – Dilip Sarwate Oct 27 '11 at 3:37
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $X\sim\operatorname{Exp}(\alpha)$ and $Y\sim\Gamma(n,\beta)$). – M.B.M. Oct 3 '12 at 5:11
@M.B.M. This is just the conditional tower law: $$ \mathbb{E}\left(\exp(i t Y/X) \right) = \mathbb{E}\left(\mathbb{E}\left(\exp(i t Y/X) | X \right)\right) = \mathbb{E}\left( \phi_Y\left(t/X \right)\right) $$ where $\phi_Y(t)$ is the characteristic function of $Y$ random variable. – Sasha Oct 3 '12 at 5:24

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