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In my work on discrete urn problems, I have run across a family of discrete distributions with an unusual form, and I am trying to find out if this family of distributions is in the statistical literature. To do this, I would like to know if it has a name. Let $S(n,k,r)$ denote the non-central Stirling numbers of the second kind, and let $S(n,k)$ denote the (central) Stirling numbers of the second kind. It is well-known that these two types of numbers are related by:

$$S(n,k,r) = \sum_{t=k}^n {n \choose t} r^{n-t} S(t,k).$$

Thus, we can form a valid family of discrete distributions with probability mass function:

$$p(t|n,k,r) = {n \choose t} \frac{r^{n-t} \cdot S(t,k)}{S(n,k,r)} \quad \quad \quad \text{for } t = k,...,n.$$

I have encountered this distribution in some analysis I am doing on urn problems, and I have managed to derive some of its properties. I am trying to identify if this distribution has been examined in the academic literature, which is hard to do, since I don't know its name.


My Question: Does this distributional family have a name? Is there any existing academic literature on this distribution that anyone can point me to?

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    $\begingroup$ Don't know if it's much direct use but you might at least locate some clues in S. B. Nandi and S. K. Dutta (1986), "Some Discrete Distributions Involving Stirling Numbers", Sankhyā: The Indian Journal of Statistics, Series B Vol. 48, No. 3 (Dec.), pp. 301-314. Perhaps a bit more of a long shot: Keener, R; Rothman, E; Starr, N. (1987), "Distributions on Partitions." Ann. Statist. 15, no. 4, 1466--1481. doi:10.1214/aos/1176350604. projecteuclid.org/euclid.aos/1176350604 $\endgroup$ – Glen_b -Reinstate Monica Jul 30 '19 at 8:24
  • $\begingroup$ You could try some numerical examples in the form numerator/denominator and try the oeis.org $\endgroup$ – kjetil b halvorsen Jul 30 '19 at 11:09
  • $\begingroup$ Trying to reverse engineer the urn problem I get to the following. When drawing $n$ balls and placing them in one of the $k$ urns with $(1-r)/k$ probability. Then $p(t \vert n,k,r)$ is the probability that $t$ balls are in the urn, conditional that each urn is non-empty. Is this your case as well? $\endgroup$ – Sextus Empiricus Aug 5 '19 at 15:41
  • $\begingroup$ @Martijn: That's roughly how I got it. In my case, it came up in the extended occupancy problem, as the distribution of the effective number of balls conditional on the occupancy number ---i.e., essentially the same way as you have specified it. Very impressive reverse-engineering! $\endgroup$ – Reinstate Monica Aug 6 '19 at 1:20
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1. Relation with minimal covers

This answer is not a distribution but a related set of numbers.

For the cases with $r = 2^{k}-k-1$ you can express the probability $p(t \vert n,k,r)$ by using numbers relating to minimal covers of a set with cardinality $n$ ( from 'Minimal covers of finite sets' by T.Hearne and C.Wagner):

$$p(t \vert n,k,r=2^{k}-k-1) = \frac{M(n,k,t)}{M^\star(n,k)}$$

  • Here $M(n,k,j)$ is the number of $k$ element minimal covers of a $n$-set, such that $j$ points are uniquely covered (where $j\geq k$ since each element of a minimal cover must contain at least one unique member). $$M(n,k,j) = {{n}\choose{j}}\left( 2^{k}-k-1\right)^{n-j}S(j,k)$$
  • And $M^\star(n,k)$ is the number of $k$ element minimal covers of a $n$-set. $$M^\star(n,k) = \sum_{j=k}^n M(n,k,j)$$

2. Relation conditional urn and balls problem

This answer is not a solution of a related distribution but it is how I imagined the problem could be when I was trying to reverse engineer what kind of urn problem is underlying the formula. I believe it may be helpful for others to visualize the problem.

This is how I got to think of the related urn problem: First consider that the expression ${{n}\choose{t}}S(t,k)$ is the number of ways to select $t$ balls from a total of $n$ balls and put them into $k$ identical urns such that each urn is non-empty (this term also occurs also in the referenced article of Hearne and Wagner). Then, the difference with the expression in the question is the term $r^{n-t}$, and this resembles the number of ways to distribute the remaining $n-t$ balls among $r$ urns. So this leads to the following urn-problem.

Put $n$ balls into $k+r$ urns. Let $T$ be the number of balls in the first $k$ urns, let $E$ be the event that each of these $k$ urns is non-empty. Then $$P(T=t \vert E) = \frac{ {{n}\choose{t}}r^{n-t}S(t,k) }{\sum_{j=k}^n{{n}\choose{j}}r^{n-j}S(j,k)} $$

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  • $\begingroup$ This is very helpful (+1). In my case I am dealing with a problem where $r$ can be any positive real number (not just an integer), which extends the distribution beyond what you have looked at here. But this is still very useful. $\endgroup$ – Reinstate Monica Aug 29 '19 at 23:49

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