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Nov 15, 2019 at 10:47 comment added Mateusz Kwaśnicki @bursneh: You can preview some pages on Google Books. Chapter 8 on GNBC contains various "mixture-type" results, if the multiplication theorem is true for GNBCs, it should follow easily from one of them, I suppose.
Nov 15, 2019 at 10:30 comment added bursneh @MateuszKwaśnicki This is excellent news. I don't have access to the book right now so I'll take your word for it. I'm going to see if I can work through a proof of the multiplicity based on the definition of GNBCs. If I'm successful, I'll post it as an answer.
Nov 15, 2019 at 9:40 comment added Mateusz Kwaśnicki @bursneh: Lesson learned: GGC is closed under multiplication, see Theorem 6.2.1 in Bondesson's book. Then a natural guess would be that "generalised negative binomial convolutions" share similar properties in the discrete setting. To my surprise, the name I just made up is in use, and a whole chapter of Bondesson's book is devoted precisely to this subject. Unfortunately, I did not find a variant of the multiplication theorem for GNBC there.
Nov 14, 2019 at 20:41 comment added Mateusz Kwaśnicki @bursneh: Is the product of two exponential distributions a GGC? Obviously, it is completely monotone, but I see no reason why it should be a GGC. Have you checked that?
Nov 14, 2019 at 18:02 comment added bursneh There's some literature in the case that $S \subset \mathbb{R}^n$ on this topic. As far as I understand, generalised gamma convolutions (GGC) may satisfy the above conditions; they are infinitely divisible, however, I'm still searching for a reference which proves that the product of two independent GGC random variables is still GGC. Despite this, I'm primarily interested in the case in which $S$ has countable support.
Nov 14, 2019 at 17:25 comment added user44143 It looks like no to me: Wikipedia has a list of distributions with nice products (en.wikipedia.org/wiki/Product_distribution) and a list of distributions closed under sums (en.wikipedia.org/wiki/Stable_distribution), and they don't seem have any non-trivial intersection.
Nov 14, 2019 at 16:32 history edited bursneh CC BY-SA 4.0
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Nov 14, 2019 at 14:47 comment added Mark Schultz-Wu Random variables in this hypothesized family (only the special case that they're normal), although the general case may be possible. Moreover if you wanted to gain intuition by looking at the pdf of any mentioned distribution, this may be more difficult to do for Meijer-G functions.
Nov 14, 2019 at 14:45 comment added Mark Schultz-Wu This isn't a full answer, but you can phrase the pdf of the product of two (independent) normal distributions in terms of Bessel Functions, which can then be phrased in terms of Meijer-G functions. The standard exponential can phrased in terms of Meijer-G functions as well (and Meijer-G functions satisfy a certain convolution theorem). By identifying a suitable parametrization of Meijer-G functions you likely can get a family of probability distributions. This doesn't yet answer how to compute the product of arbitrary ...
Nov 14, 2019 at 12:09 comment added bursneh @user44191 This is precisely the type of situation I wished to exclude. In your case, for each element in the support set, there is a corresponding parameter. I've updated the question to hopefully be more clear about this.
Nov 14, 2019 at 12:05 history edited bursneh CC BY-SA 4.0
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Nov 14, 2019 at 11:59 history edited bursneh CC BY-SA 4.0
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Nov 14, 2019 at 6:57 comment added user44191 Can't the entire class of random variables on $\mathbb{Z}$ be represented by a countable number of parameters (say, $P(0), P(1|\text{not}0), P(-1|\text{not}0, \text{not}1), \dots$)? I'm not sure I understand your bijection condition.
Nov 14, 2019 at 2:35 review Close votes
Nov 18, 2019 at 3:05
Nov 13, 2019 at 20:52 history asked bursneh CC BY-SA 4.0