I have some preliminary findings that are a little strange.

I have a couple of related data sets that are small (~150) positive integer samples and that seem rather strange. Because they're small, I don't take their histograms seriously, so I've performed Gaussian kernel density estimation on both of them. (While I haven't been that careful about bandwidth selection, I've at least taken this into account, and doubt it is a significant issue in the present context--but I could be wrong.)

The weird thing is this: in both cases, I get a remarkably good-looking fit to a Cauchy distribution (I've done this by hand, no MLE stuff in this instance*, but I've corroborated the fits by looking at normal and log-log plots) which is centered at a positive number (i.e., the KDE basically looks like this). But as I've said, the samples are (and must be) positive integers, so the support of the distribution is the same. On the other hand, the Cauchy distribution fits very well on the positive region.

The only thing that I can imagine producing something like this besides a mistake on my part is that these samples might admit some kind of interpretation of sums of IIDRVs and that leads to the apparent Cauchy-ness due to the generalized CLT.

What might cause this? Explanations for mistakes are particularly welcome. However, please note my doubts about the kernel bandwidth being an issue.

*However, I've tried MLE fits to some other common distributions (e.g., gamma) and these are terrible, especially in comparison.

  • $\begingroup$ People are voting as offtopic, probably because they want the question to be migrated to Statistical Analysis. I am not sure if there is a consensus about questions of this kind being offtopic. $\endgroup$
    – Rasmus
    Oct 25, 2011 at 15:32
  • 1
    $\begingroup$ At root, this is a question about whether or not the generalized CLT has anything to say about what I'm seeing. To me, this is a mathematical question. If I were primarily concerned about bandwidth selection or was sure I'd made a mistake, I would agree with the votes for "off-topic", but that isn't the motivation at all. I expect a decent answer will be mathematical in character. $\endgroup$
    – S Huntsman
    Oct 25, 2011 at 16:00
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    $\begingroup$ There is a large area of overlap between stats.stackexchange.com and math.stackexchange.com. Somethings clearly belong here and not there. If one were arguing about which is the right way to model the Behrens-Fisher problem, maybe you could say that's not even a mathematical question, but even that might fit in here because mathematical matters bear upon it. $\endgroup$ Oct 25, 2011 at 16:27
  • $\begingroup$ In what sense is the Cauchy a good fit? Just eyeballing it, or something more formal? $\endgroup$
    – guy
    Oct 26, 2011 at 17:45

1 Answer 1


A simple way to produce heavy-tailed distributions is to consider (thin-tailed) distributions with a random parameter. Assume for instance that $N$ has a geometric distribution with parameter $A$ and that $A$ has a Beta distribution, that is, $$ \mathrm P(A\in dx)=\frac1{\mathrm B(a,b)}x^{a-1}(1-x)^{b-1}\cdot[0\leqslant x\leqslant 1]\cdot\mathrm dx, $$ and, for every nonnegative integer $n$, $$ \mathrm P(N\geqslant n\mid A)=A^n. $$ Then, $$ \mathrm P(N\geqslant n)=\frac1{\mathrm B(a,b)}\int\limits_0^1x^{n+a-1}(1-x)^{b-1}\mathrm dx=\frac{\mathrm B(a+n,b)}{\mathrm B(a,b)}=\frac{\Gamma(a+b)\Gamma(a+n)}{\Gamma(a)\Gamma(a+b+n)}, $$ hence $\mathrm P(N\geqslant n)$ is equivalent to a multiple of $n^{-b}$ when $n\to\infty$.

To get a tail equivalent to $n^{-1}$ as in a (discrete) Cauchy distribution, choose $b=1$, that is, $$ \mathrm P(A\in dx)=ax^{a-1}\cdot[0\leqslant x\leqslant 1]\cdot\mathrm dx, $$ in other words, assume that $A=U^{1/a}$ where $U$ is uniformly distributed on $(0,1)$.

  • $\begingroup$ Thanks, this is interesting. However, it's not just the tail that is a good fit (and I have just posted to MO about this at mathoverflow.net/questions/79623), it's the whole domain of the data (incl. near 0). Your answer does however remind me that discrete stable distributions can be obtained by combining (IIRC) a Poissonian number of Poisson RVs. $\endgroup$
    – S Huntsman
    Oct 31, 2011 at 22:08
  • $\begingroup$ No, the sum of a Poisson number of Poisson is thin-tailed. The sum of a Poisson number of so-called Sibuya distributions is stable, see here if only for the references. $\endgroup$
    – Did
    Oct 31, 2011 at 22:44
  • $\begingroup$ Ah, you caught my mistake, thanks. I should have said Sibuya: this is implicit in my MO question. $\endgroup$
    – S Huntsman
    Nov 1, 2011 at 4:20

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