Stable distributions are invariant under convolutions. What sub-families $F$ of the stable distributions are also closed under multiplication? In the sense that if $f\in F$ and $g\in F $, then the product probability density function, $f \cdot g$ (up to a normalization constant) also belongs to $F$?

Note: I substantially changed the content of this question. But the idea is essentially the same, and now it is much simpler. I only had a partial answer, so I think it's okay.

  • $\begingroup$ If the domain is bounded the mean and variance (indeed all the moments) must be finite. How confident are you that any known distributions exist that satisfy all the conditions? $\endgroup$ – Glen_b Jun 25 '14 at 2:40
  • $\begingroup$ @Glen_b If it is possible to prove that no distribution exists with all of these conditions, I'll accept an answer with that proof. $\endgroup$ – becko Jun 25 '14 at 13:33
  • $\begingroup$ What precisely is "the" bounded uniform distribution in (5)? Is it one distribution (and if so, what are its parameters), or is it a family of uniform distributions (and if so, what family is it)? $\endgroup$ – whuber Jun 25 '14 at 18:06
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    $\begingroup$ (1) By "sub-family" do you mean of the stable distributions? (2a) If so, then given that the product of Gaussians obviously is another Gaussian, you have an immediate answer in the positive. (2b) If not, then there are myriad answers. Start with any family $\mathcal F$ of continuous distributions with everywhere positive density. The smallest family that contains $\mathcal F$ and is closed under renormalized products of density functions does the job. You can compute these explicitly when $\mathcal F$ has just one element. $\endgroup$ – whuber Nov 7 '16 at 15:51
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    $\begingroup$ @whuber Yes, I mean a sub-family of the stable distributions. You are right, a Gaussian satisfies my criteria. I was actually looking for other examples, but I forgot to mention that. Are there any other distributions that also satisfy my criteria? I'll update the question, thanks for helping me make it clearer. $\endgroup$ – becko Nov 7 '16 at 16:38

A "stable distribution" is a particular kind of location-scale family of distributions. The class of stable distributions is parameterized by two real numbers, the stability $\alpha\in(0,2]$ and skewness $\beta\in[-1,1]$.

A result quoted in the Wikipedia article resolves this question about closure under products of density functions. When $f$ is the density of a stable distribution with $\alpha \lt 2$, then asymptotically

$$f(x) \sim |x|^{-(1+\alpha)} g(\operatorname{sgn}(x), \alpha, \beta)$$

for an explicitly given function $g$ whose details do not matter. (In particular, $g$ will be nonzero either for all positive $x$ or all negative $x$ or both.) The product of any two such densities therefore will be asymptotically proportional to $|x|^{-2(1+\alpha)}$ in at least one tail. Since $2(1+\alpha)\ne 1+\alpha$, this product (after renormalization) cannot correspond to any distribution in the same stable family.

(Indeed, because $3(1+\alpha) \ne 1+\alpha^\prime$ for any possible $\alpha^\prime\in(0,2]$, the product of any three such density functions cannot even be the density function of any stable distribution. That destroys any hope of extending the idea of product closure from a single stable distribution to a set of stable distributions.)

The only remaining possibility is $\alpha=2$. These are the Normal distributions, with densities proportional to $\exp(-(x-\mu)^2/(2\sigma^2))$ for the location and scale parameters $\mu$ and $\sigma$. It is straightforward to check that a product of two such expressions is of the same form (because the sum of two quadratic forms in $x$ is another quadratic form in $x$).

The unique answer, then, is that the Normal distribution family is the only product-of-density-closed stable distribution.

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    $\begingroup$ Cool! This is then a nice way to define a normal distribution, as the unique stable and closed under products. Thanks $\endgroup$ – becko Nov 7 '16 at 19:01

I know this is a partial answer and I'm not an expert, but this might help: if one of two unimodal pdfs is log-concave, then their convolution is unimodal. Due to Ibragimov (1956), via these notes. Apparently, if both are log-concave, then the convolution is also log-concave.

As far as product closure, the only "clean" result I know of for product distributions is the limit theorem described in this math.se answer.

How about a truncated version of these? The bounded uniform distribution is a limiting case of its shape parameter, and as far as I'm aware they're unimodal and log-concave so they have unimodal, log-concave convolutions. I have no clue about their products . When I have more time later this week I could try and run some simulations to see if I get log-concave products of truncated error distributions. Maybe Govindarajulu (1966) would help.

I'm not sure what the policy on crossposting is, but it seems like the math.se people might be able to help you as well. Out of curiosity, are you trying to build an algebraic structure out of probability distributions?

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    $\begingroup$ The policy on crossposting is contained in the very first page of the help. It says 'please don't crosspost'. We should choose the one best site for our question. A question can be migrated if necessary. If part of a question is better suited to a different site the question should be asked as two separate questions (which can be linked). $\endgroup$ – Glen_b Jun 25 '14 at 21:28
  • $\begingroup$ The result on preserving log-concavity is also described on the Wikipedia page: en.wikipedia.org/wiki/… $\endgroup$ – Thomas Ahle Mar 24 at 17:00

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