Suppose you're a shopkeeper in the business of selling Items. An "Item" is a thing whose only property is that the quantity that can be bought by a purchaser must be a positive integer; all Items are identical.

A book whose identity I do not recall (I'll post it here maybe tomorrow?) seems to hold that the way to model the number $Y$ of Items that your customers will buy today is as a negative binomial distribution: \begin{align} & \Pr(Y=n) = \binom{-r}{\phantom{+}n} (-q)^n p^r \text{ for } n=0,1,2,3,\ldots \\[8pt] \text{where } & 0<p<1,\, q = 1-p,\, r>0, \\[8pt] \text{and } & \binom {-r}{\phantom{+}n} = \frac{\overbrace{(-r)(-r-1)(-r-2)\cdots (-r-n+1)}^{\text{$n$ factors}}}{n!}. \end{align} Supposing the number $N$ of purchasers today has a Poisson distribution (which makes sense for well known reasons under some plausible assumptions) that would mean the number of Items purchased by each purchaser has a logarithmic series distribution: $$ \Pr(X=x) = \frac 1 {-\log(1-u)}\cdot \frac{u^x} x \text{ for } x = 1,2,3,\ldots $$ where $0<u<1$ and $\dfrac1{-\log(1-u)} \cdot \dfrac u{1-u}$ is the average number of Items purchased by each purchaser.

(To prove this, use moment-generating functions. I tried to do it by more straightforward combinatorial reasoning and it's a mess.) (Postscript some days later: ok, I now see that there's at least one other fairly convenient way to prove this.)

My question is: Is there some "intuitive" argument for the logarithmic series distribution to occur here that is independent of any knowledge that the negative binomial distribution occurs here, so that one can deduce from that intuitive argument that the negative binomial distribution should occur here? Similar to the well known "intuitive" arguments that would show the Poisson distribution should occur above? ("Intuitive" and "intuition" are words that mathematicians use too much, without examining the various different meanings for which they use it. This practice exhibits Augean stability.)

  • 1
    $\begingroup$ Since the first use of this series was motivated by curve-fitting rather than intuitions, I doubt there’s much good intuition to be had. $\endgroup$
    – Matt F.
    Dec 7, 2022 at 3:46
  • $\begingroup$ My brother, can I ask a personal question? Why are you suspended on Mathematics SE? $\endgroup$ Jan 7 at 12:02
  • $\begingroup$ @SineoftheTime I was suspended until late March by Xander Henderson for what he called disruptive posting. I have never posted with intent to disrupt, nor am I aware that anything I have posted has had that effect, nor was Xander Henderson willing to tell me which among my postings he objected to, nor was one of the community managers willing to answer that. After nine months, another community manager finally gave in and told me which one. She said the posting expressed a refusal to accept a community consensus about deletion of questions, but in fact the posting was$\,\ldots\qquad$ $\endgroup$ Jan 9 at 4:03
  • $\begingroup$ $\ldots\,$asking questions whose answers I do not know, seeking information. I think that's pretty typical of how Stackexchange operates. $\qquad$ $\endgroup$ Jan 9 at 4:03
  • $\begingroup$ @MichaelHardy Well, that's sad. I feel sorry for you, expecially because suspend without giving a reason seems bossy and an abuse of power. Hope you'll be back soon and stronger :) $\endgroup$ Jan 9 at 13:13

2 Answers 2


$\newcommand\la\lambda\newcommand\La\Lambda\newcommand{\Ga}{\Gamma}\newcommand{\R}{\mathbb R}$Some preliminaries: To make it explicit that the distribution of $Y$ depends on the parameter $r$, let $Y_r:=Y$, so that \begin{equation*} \Pr(Y_r=j) = \binom{-r}j (-q)^j p^r \tag{1}\label{1} \end{equation*} for $j=0,1,\dots$. We know that the number $N$ of purchasers has the Poisson distribution with some parameter $\mu$.

We have \begin{equation*} Y_r=\sum_{i=1}^N X_i, \end{equation*} where $X_i$ is the number of Items bought by the $i$th purchaser. Suppose that the purchasers act independently and are statistically interchangeable, so that the $X_i$'s are iid. We also suppose that the number $N$ of purchasers is independent of the numbers $X_i$ of Items bought by purchasers. It follows that the distribution of $Y_r$ is a compound Poisson distribution: \begin{equation*} P_{Y_r}=\sum_{k=0}^\infty \Pr(N=k)\,P_X^{*k} =\sum_{k=0}^\infty \frac{\mu^k}{k!}\,e^{-\mu}\,P_X^{*k}, \tag{2}\label{2} \end{equation*} where $X:=X_1$, $P_Z$ denotes the distribution of a random variable (r.v.) $Z$, and $Q^{*k}:=\underbrace{Q*\cdots*Q}_{k}$.

We have to plausibly/"intuitively" recover the distribution $P_X$ of $X$ given \eqref{2}. In view of the infinite divisibility of the negative binomial distribution of $Y_r$ and by \eqref{2}, we have \begin{equation*} \lim_{n\to\infty}P_{Y_{r/n}}^{*n}=P_{Y_r} =\lim_{n\to\infty}\big(P_0+\tfrac\mu n\,(P_X-P_0)\big)^{*n}. \end{equation*} So, it is plausible that for large $n$ \begin{equation*} (1-\tfrac\mu n)P_0+\tfrac\mu n\,P_X=P_0+\frac\mu n\,(P_X-P_0) \approx P_{Y_{r/n}}. \end{equation*} Therefore and because $X\ge1$, we see that $P_X$ is the limit as $n\to\infty$ of the conditional distribution of $Y_{r/n}$ given $Y_{r/n}\ne0$. So, for integers $j\ge1$, in view of \eqref{1}, \begin{equation*} \Pr(X=j)=\lim_{s\downarrow0}\frac{\Pr(Y_s=j)}{1-\Pr(Y_s=0)} \\ =\lim_{s\downarrow0}\frac{(j-1+s)(j-2+s)\cdots(1+s)s\; q^j p^s/j!}{1-p^s} =\frac{q^j/j}{-\ln p}, \end{equation*} as desired.


Answer to the changed question

In the initial version of the question, the negative binomial distribution of $Y_r$ appeared to be assumed, and the question appeared to be asking for a more intuitive derivation of the logarithmic series distribution for $X$ based on that "negative binomial" assumption.

After the above answer was posted, the OP changed the question, which now calls for a derivation of the logarithmic series distribution for $X$ without assuming the negative binomial distribution of $Y_r$.

Let us try to answer the changed question as well. Here I will be following lines of Fisher's paper (pp. 54--58).

Let $Z$ denote the random number of books purchased by a potential buyer (not necessarily an actual purchaser). Assume that $Z$ depends on some "hidden" positive r.v. $\La$ so that the conditional distribution of $Z$ given $\La$ is the Poisson distribution with mean $\La$: \begin{equation*} \Pr(Z=z|\La)=\frac{\La^z}{z!}\,e^{-\La}\,1(z=0,1,\dots). \tag{10}\label{10} \end{equation*} In turn, assume that $\La$ has the gamma distribution with positive parameters $a$ and $b$, so that \begin{equation*} \Pr(\La\in d\la)=\frac1{\Ga(a)b^a}\,\la^{a-1}e^{-\la/b}\,1(\la>0)\,d\la. \tag{20}\label{20} \end{equation*} Then \begin{align} \Pr(Z=z)&=\int_\R\Pr(\La\in d\la)\Pr(Z=z|\La) \notag \\ &=\frac{\Ga(a+z)}{\Ga(a)}\frac1{z!}\,\frac{1}{b^a (1+1/b)^{a+z}}\,1(z=0,1,\dots) \notag \\ &=\frac{a(a+1)\cdots(a+z-1)}{z!}\,p^a(1-p)^z\,1(z=0,1,\dots), \tag{30}\label{30} \end{align} where \begin{equation*} p:=\frac1{1+b}. \end{equation*}

The "Poisson" assumption \eqref{10} seems natural.

Unfortunately, just as in the mentioned paper by Fisher, I cannot offer a reason to model the mixing distribution of $\La$ as a gamma distribution, except that then it is easy to get \eqref{30}. One may note here, though, that the gamma distribution is the conjugate prior distribution with respect to the Poisson family of distribution -- a fact prized by Bayesians.

Anyhow, once this modeling is accepted, we next note that the expected number of purchases by a potential buyer is $EZ=EE(Z|\La)=E\La=ab$, which may be reasonably assumed small. If $b>0$ is fixed (that is, if $p=\frac1{1+b}\in(0,1)$ is fixed), then it follows that $a>0$ is small.

So, the distribution of $Z$ will be close to the Dirac distribution supported on the singleton set $\{0\}$. However, since we are interested in those potential buyers who are actually purchasers, we consider the conditional distribution of $Z$ given $Z>0$. In view of \eqref{30}, the probability mass function (pmf) -- say $f_{a,p}$ -- of the just mentioned conditional distribution is given by the formula \begin{align*} & f_{a,p}(z)=\frac{\Pr(Z=z)\,1(z>0)}{1-\Pr(Z=0)} \\[6pt] = {} & \frac{a(a+1)\cdots(a+z-1)}{z!}\,\frac{p^a(1-p)^z}{1-p^a} \, 1(z=1,2,\dots). \end{align*} Since $a$ is small, it is natural to consider the limit \begin{equation*} f_{0+,p}(z):=\lim_{a\downarrow0}f_{a,p}(z) =\frac1{-\ln p}\frac{(1-p)^z}z\,1(z=1,2,\dots). \end{equation*} So, $f_{0+,p}$ is the pmf of the logarithmic series distribution, as desired.

  • 1
    $\begingroup$ This seems to address the wrong question. The question is NOT how to prove the assertion that I already made in the question, nor how to do it in a more "intuitive" way; rather the question is why one should expect the number of items bought by the $i$th purchaser to have a logarithmic series distribution (so that we can then deduce that the amount bought each day has a negative binomial distribution), rather than taking that as given at the outset, as you appear to do. $\endgroup$ Nov 29, 2022 at 22:00
  • $\begingroup$ I've edited the question to clarify. $\endgroup$ Nov 29, 2022 at 22:07
  • 1
    $\begingroup$ @MichaelHardy : I think your question, as it was stated and as it could be naturally understood (with the negative binomial and Poisson distributions as givens), was answered. However, your added "clarification" does turn the question upside-down. $\endgroup$ Nov 29, 2022 at 22:55
  • $\begingroup$ @MichaelHardy : Your changed question has now been answered as well. $\endgroup$ Dec 5, 2022 at 22:25
  • 1
    $\begingroup$ Although this is not exactly the kind of answer I had in mind, it shows that my thinking about this particular compound Poisson process had not advanced very far and makes it clear that the logarithmic series distribution is a sort of infinitesimal generator of this process and shows in what direction I can move forward in thinking about this further. $\endgroup$ Dec 8, 2022 at 21:21

Q: Is there some "intuitive" argument for the logarithmic series distribution to occur as a model for the number of items bought?

To aid intuition I might offer the following: If I treat the number $n$ of items as a continuous variable $\nu$, then the logarithmic series distribution $P(n)\propto u^n/n$ describes a truncated log-normal distribution, $P(\nu)\propto e^{-\alpha\ln^2(\nu/\beta)}/\nu$. The distribution is truncated by restricting $\nu\geq 1$ (since customers who do not buy any item are not counted), and the exponent is linearized in $\nu$ to obtain a distribution of the form $P(\nu)\propto u^\nu/\nu$.

The plot shows an example (data points = logarithmic series distribution for $u=0.99$; curve = truncated log-normal distribution with $\alpha=0.058$, $\beta=2$).

The log-normal distribution generically appears when the random variable is determined by events that enter multiplicatively rather than additively. I can imagine that this applies to the number of items purchased by a given customer. For example, if it's food I would take into account the number of social events I would be hosting at my home, multiplied by the number of guests, times their appetite.

It seems this mechanism of multiplicative fluctuations is operative in a great variety of settings. Perhaps someone with data skills could test the following

Conjecture: The number of answers each user has posted on MO in the previous year follows a logarithmic series distribution.

  • 1
    $\begingroup$ @MartinSleziak, can you show us how to run the query that gets the data for Carlo’s question? $\endgroup$
    – Matt F.
    Dec 9, 2022 at 4:43
  • $\begingroup$ @MattF. Better to ask this in the MO Editors lounge chat room, but it's not yet clear what you mean. $\endgroup$
    – David Roberts
    Dec 10, 2022 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.