While re-dipping into the immortal Feller Vol I (An Introduction to Probability Theory and its Applications - 3rd Edn.), in Chapter IV (Combination of Events), I came across the following enticing statement:
As Von Mises has shown, the probability of finding exactly m cells with k-tuple occupancy can again be approximated by the Poisson distribution, but this time $\lambda$ must be defined as $$\lambda=n\frac{e^{-r/n}}{k!}\left(\frac{r}{n}\right)^{k}$$
For context, in the preceding, Feller demonstrates that in the case $k=0$, with $\lambda:=ne^{-r/n}$, with $r$ and $n$ increasing such that $\lambda$ is contained in some finite interval $0<a<\lambda<b$, we have that the probability that there exist $m$ empty cells, $p_{m}(r,n)$, has the following property:$$p_{m}(r,n)-e^{-\lambda}\frac{\lambda^m}{m!}\rightarrow 0 \tag{*}$$ i.e. that $p_{m}(r,n)$ has a $Poisson(\lambda)$ limit under the given conditions.
I thought I'd have a go at proving the $k$-tuple generalisation as an exercise, which Feller asserts requires only 'minor changes' to the argument used for proving $(*)$. My question is simply whether the proof attempt is correct?! Hopefully this is not superfluous, and if essentially correct, of interest to someone. (Health warning: been a long time since have done any mathematics, just 'having a go'...)
We have $r$ distinguishable balls and $n$ distinguishable cells, so that under a Maxwell-Boltzmann assumption there are $n^r$ distinguishable $r$-tuples possible (though balls are distinguishable, different balls in the same cell are order - agnostic, i.e. this is not akin to a 'flags on poles' problem).
Let $$\begin{align*}p_{m}(r,n,k)&:=\mathbb{P}(\mathrm{ there \space are \space} m \mathrm{\space cells \space with\space} k-\mathrm{tuple \space occupancy}) \\ p_i&:=\mathbb{P}( \mathrm{cell} \space i \space \mathrm{has}\space k-\mathrm{tuple \space occupancy})\end{align*}$$ then $$\begin{align*} p_i&= {r\choose k}\frac{(n-1)^{r-k}}{n^{r}}={r\choose k}\frac{1}{n^{k}}(1-\frac{1}{n})^{r-k}\\ \end{align*}$$
Here we choose the $k$ balls for the $i^{th}$ cell, then distribute the residual $r-k$ balls in the remaining $n-1$ cells, arriving at the familiar binomial expression. Generalising this, we have that: $$p_{m}(r,n,k) = {n\choose m} {r \choose {\underbrace{k,...,k}_{m\space\mathrm{times}},r-mk}}\frac{1}{n^{mk}}(1-\frac{m}{n})^{r-mk}p_{0}(r,n-m,k)\tag{1}$$ Where $p_{0}(r,n-m,k)$ is the probability that none of the remaining $n-m$ cells have $k$-fold occupancy. The heuristic for this formula is first to choose the $m$ 'special' cells, then distribute the $r$ balls among the cells, so that the $m$ designated cells get $k$ balls apiece, with the residual $r-mk$ balls allocated to 'the rest'. The final probability factor $p_{0}(r,n-m,k)$ discards those arrangements where there is a cell in the $n-m$ non-designates that has $k$ balls.
Now $$\begin{align*} p_{0}(r,n-m,k) &= \mathbb{P}(\mathrm{none \space of \space the \space} n-m \mathrm{\space cells \space have \space }k \mathrm{\space balls \space}) \\&=1-\mathbb{P}\left(\bigcup_{i=1}^{n-m}\{\mathrm{cell} \space i \space \mathrm{has}\space k-\mathrm{balls} \}\right) \\ &=1+\sum_{\nu=1}^{n-m}(-1)^{\nu}S_{\nu} =\sum_{\nu=0}^{n-m}(-1)^{\nu}S_{\nu} \tag{2}\end{align*}$$ where $S_{\nu}=\sum_{i_1<..<i_{\nu}}p_{i_1,...,i_{\nu}}$, or the sum over all $\nu$-tuples of distinct cells such that cell $i_1$,...,cell $i_{\nu}$ all contain $k$ balls. The last line follows from inclusion-exclusion. Using reasoning similar to that used to write down $(1)$ above, we see that $$S_{\nu}={n \choose \nu}{r \choose {\underbrace{k,...,k}_{\nu \space\mathrm{times}},r-\nu k}}\frac{1}{n^{\nu k}}(1-\frac{\nu}{n})^{r-\nu k}$$ So that in fact $$\begin{align*}p_{m}(r,n,k) &= S_m p_{0}(r,n-m,k)\tag{3} \\ \end{align*}$$
To proceed, here's the definition of $\lambda$ again: $\lambda=n\frac{e^{-r/n}}{k!}\left(\frac{r}{n}\right)^{k}$
Claim 1: $S_\nu-\frac{\lambda^\nu}{\nu}\rightarrow 0\tag{4}$
First we re-write $S_{\nu}$ as
$$\begin{align*} S_{\nu} &= \frac{(n)_{\nu}}{\nu!} \frac{(r)_{\nu k}}{(k!)^{\nu}}\frac{1}{n^{\nu k}}(1-\frac{\nu}{n})^{r- \nu k} \end{align*}$$
Where $(n)_{\nu}$ for example denotes the truncated or falling factorial with $\nu$ terms starting at $n$. Now clearly $(n-\nu)^{\nu}<(n)_{\nu}<n^{\nu}$, and similarly $(r-\nu k)^{\nu k}<(r)_{\nu k}<r^{\nu k}$, which we substitute into the preceding expression to start to estimate $\nu!S_\nu$:
$$\begin{align*} (n-\nu)^\nu \frac{(r-\nu k)^{\nu k}}{(k!)^\nu}\left(\frac{1}{n^k}\right)^\nu\left(1-\frac{\nu}{n} \right)^{r-\nu k} &= \frac{(r-\nu k)^{\nu k}}{(k!)^\nu}\left(\frac{1}{n^k}\right)^\nu\frac{(n-\nu)^{r-\nu k + \nu}}{n^{r-\nu k + \nu}}\cdot n^\nu \\ &= \left\{\frac{n}{k!}\frac{(r-\nu k)^k}{n^k}\right\}^\nu\left(1-\frac{\nu}{n} \right)^{r-\nu k + \nu}< \nu! S_\nu \\ \text{On the other hand} \space \space \space \nu!S_{\nu} &< \left\{ \frac{n}{k!} \left(\frac{r}{n}\right)^k \right\}^\nu\left(1-\frac{\nu}{n} \right)^{r-\nu k} \end{align*} $$
Let $\alpha:=\frac{n}{k!} \left(\frac{r}{n}\right)^k$ to simplify notation, so that $\lambda = \alpha e^{-r/n}$. Further, let $t:=\nu / n$. For $t \in (0,1)$ we have that $ \frac{1}{1-t} = \sum_{i=0}^\infty t^i $, whence $t<-\mathrm{log}(1-t)<\frac{t}{1-t}$ on integrating. We then have that $e^{-\frac{t}{1-t}}<1-t< e^{-t}$, or $e^{-\frac{\nu}{n-\nu}}<1-\frac{\nu}{n}<e^{-\nu/n}$. Combining this with the preceding estimates, we then have:
$$\begin{align*} \left\{ \frac{n}{k!}\left(\frac{r-\nu k}{n} \right)^k e^{-\frac{r-\nu k + \nu}{n-\nu}} \right\}^\nu &= \left\{ \frac{n}{k!}\left(\frac{r-\nu k}{n} \right)^k \right\}^\nu \left(e^{-\frac{\nu}{n-\nu}}\right) ^{r-\nu k + \nu} \\ &< \left\{ \frac{n}{k!}\left(\frac{r-\nu k}{n} \right)^k \right\}^\nu \left(1-\frac{\nu}{n}\right) ^{r-\nu k + \nu} <\nu!S_\nu \\ \text{a similar argument shows } \space \space \space \nu!S_\nu &<\left\{ \frac{n}{k!}\left(\frac{r}{n} \right)^k e^{-\frac{r-\nu k}{n}} \right\}^\nu \end{align*} $$
With admittedly some hand-waving, hopefully this is enough to claim that for fixed $\nu, k$, the two arms of the inequalities $ \left\{ \frac{n}{k!}\left(\frac{r-\nu k}{n} \right)^k e^{-\frac{r-\nu k + \nu}{n-\nu}} \right\}^\nu<\nu!S_\nu<\left\{ \frac{n}{k!}\left(\frac{r}{n} \right)^k e^{-\frac{r-\nu k}{n}} \right\}^\nu$ tend in ratio to unity as $r,n$ get large, so that $\nu! S_\nu$ tends arbitrarily close to $\lambda^\nu$ to establish the claim.
Claim 2: $\space \space \space \space \space \space e^{-\lambda}-p_0(r,n-m,k)\rightarrow 0$
This just follows from $(2)$ above and that $e^{-\lambda}-p_0(r,n-m,k) = \sum_{\nu=0}^\infty(-1)^\nu \left(\frac{\lambda^\nu}{\nu!}-S_\nu\right) \rightarrow 0$ via Claim 1.
Putting the results of Claims $1/2$ into $(3)$ we have that $$p_m(r,n,k)-e^{-\lambda}\frac{\lambda^m}{m!}\rightarrow 0 \tag*{$\square$}$$
The above is just hopefully a correct extension of the $k=0$ case demonstrated in the book with a generalised combinatorial start point and ensuing manipulations to get to the limit. Thanks!