If I recall correctly, for $X$, $Y$ normally distributed, the ratio $X/Y$ is Cauchy-distributed. This is sort of like a power law, but isn't quite. So:

Are there any simple distributions for two RVs $X$, $Y$ s.t the ratio is really power-law distributed (at least in some regime)?

  • $\begingroup$ I like the question and upvoted it. But is there any motivation for the question? $\endgroup$ – Srivatsan Oct 21 '11 at 15:02
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    $\begingroup$ I have what appears to be such a data set (corroborated by MLE but I haven't done anything exhaustive). The fits appear excellent. However, it's a small enough data set that I can't easily ID plausible distributions for the ratio (these seem noisier when considered separately). $\endgroup$ – S Huntsman Oct 21 '11 at 18:31
  • $\begingroup$ @MikeWierzbicki See a more recent discussion here which indicates that the ratio of an exponential to a gamma is related to, but not quite, Pareto as Wikipedia claims $\endgroup$ – Dilip Sarwate Oct 26 '11 at 22:26

As noted here, Wikipedia's answer to the question posed by the OP is not quite correct. If $X$ is a Gamma random variable with parameters $(n,1)$ and $Y$ is an exponential random variable with parameter $1$ and $X$ and $Y$ are independent random variables, then $Y/X$ is not a Pareto random variable, but $Z = Y/X + 1$ is:
$$P\{Z > z\} = P\{Y/X + 1 > z\} = z^{-n}~ \text{for}~ z > 1.$$ A common scale parameter $\lambda$ could be included in both variables, but since we are looking at ratios, the scale parameter cancels out, and it is convenient to set $\lambda = 1$.

Suppose that $X$ denotes the time of the $n$-th arrival after $t = 0$ in a Poisson process with arrival rate $1$. Then $X$ is a Gamma random variable with parameters $(n,1)$. Let $Y$ denote the additional waiting time for the $(n+1)$-th arrival. Then, $Y$ is an exponential random variable with parameter $1$ and is independent of $X$. Thus, we have the situation described in the previous paragraph. But notice that the $(n+1)$-th arrival time is just $W = Y + X$, and $Z = Y/X + 1 = (Y+X)/X = W/X$ is thus the ratio of the $(n+1)$-th and $n$-th arrival times, and is a Pareto random variable. Naturally $Z > 1$.

In summary, 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 $a > 1$.

  • $\begingroup$ +1 Nicely done. I didn't bother with fact-checking as my answer was meant more of a 'point-in-the-right-direction' rather than a 'here's-the-solution.' More wise advice from Dilip: fact-check your sources. $\endgroup$ – user13888 Oct 27 '11 at 3:57

According to Wikipedia, if $X\sim\mathrm{Exponential}(\lambda)$ and $Y\sim\Gamma (n, \frac{1}{\lambda})$, then $\frac{X}{Y}\sim$Pareto$(1,n)$.

NOTE: Wikipedia is not quite right on this one. See Dilip's response.

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    $\begingroup$ Just a note of caution about notation: $E[X] = \frac{1}{\lambda}$ but $E[Y] = \frac{n}{\lambda}$, that is, in Wikipedia's notation, $X \sim \Gamma(1, \frac{1}{\lambda})$ while many other authors would say $X \sim \Gamma(1, \lambda)$ $\endgroup$ – Dilip Sarwate Oct 22 '11 at 2:08
  • $\begingroup$ Wise advice. Always check what notation is being used. $\endgroup$ – user13888 Oct 22 '11 at 3:13

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