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.)



Why should the logarithmic series distribution model the number of "Items" bought?
$\newcommand\la\lambda\newcommand\La\Lambda\newcommand{\Ga}{\Gamma}\newcommand{\R}{\mathbb R}$**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](https://en.wikipedia.org/wiki/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.


$$\mbox{\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*}$$


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](https://www.jstor.org/stable/1411#metadata_info_tab_contents) (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\mid\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\mid\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](https://en.wikipedia.org/wiki/Conjugate_prior#Table_of_conjugate_distributions) 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 $EZ=EE(Z\mid\La)=E\La=ab$, 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.


