Is this a known formula? $ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)$

Question

I try to quantify a partition, are there any known indicators/caracteristic numbers? Something which came to my mind was $$ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right), $$

with the following condition

$$ \sum_{k=0}^n a_k = N .$$

Is this a known formula? I'd like to have an indicator which tells me if the partition is well spread or concentrated on some few numbers. I hope my question is clear as I know not a lot about partitions. Thank you for your help.

Edit: If $a_0=N$ and all others $a_k$ are 0 this formula gives 0. If all $a_k$ are 1 and $n=N$ tends to infinity this formula goes to $\frac{1}{e}$. So I am wondering too if $\frac{1}{e}$ is the upper bound for a finite $N$ for all partitions.

Edit2: Thanks a lot for different proofs that $\frac{1}{e}$ is the upper bound. I still like to know if someone knows something more about this formula. If someone has an interessing fact, that would be nice.

Answer 1

Since $$1-x\leq e^{-x},$$ if $x\geq 0,$ the upper bound you state: $$ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)\leq e^{-a_1/N} \times \cdots\times e^{-a_N/N} =e^{-1}, $$ holds for any finite partition.

Your measure can be looked at as the $n^{th}$ power of the geometric mean of the relative sizes of the complements of partition atoms.

Answer 2

Let $f(x) = -\log(1-x)$ for $x \in [0,1)$. Since $f''(x) = \frac{1}{(1-x)^2} > 0$. $f(x)$ is convex over $[0,1)$.
By Jensen's inequality, for any $a_0,\ldots, a_n \ge 0$ such that $\sum_{i=0}^n a_i = N$, we have

$$-\sum_{i=0}^n\log\left(1 - \frac{a_i}{N}\right) \ge -(n+1)\log\left(1-\frac{1}{n+1}\right)$$ In the Taylor expansion of $f(x) = \sum\limits_{k=1}^\infty \frac{x^k}{k}$, all coefficients are positive. This leads to

$$-(n+1)\log\left(1-\frac{1}{n+1}\right) = (n+1)\sum_{k=1}^\infty \frac{(n+1)^{-k}}{k} > (n+1)\sum_{k=1}^1 \frac{(n+1)^{-k}}{k} = 1$$ As a result,

$$\prod_{i=0}^n\left(1 - \frac{a_i}{N}\right) \le \left(1-\frac{1}{n+1}\right)^{n+1} < e^{-1}$$ In short, $e^{-1}$ is indeed the upper limit for a finite $N$ for all partitions.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\bracks{\prod_{k = 0}^{n}\pars{1 - {a_{k} \over N}}} \bracks{z^{N}}z^{\sum_{j = 0}^{n}a_{j}}}} = \bracks{z^{N}}\prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}} \\[1cm] &\ln\pars{\prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}}} = \sum_{k = 0}^{n}\bracks{\ln\pars{1 - {a_{k} \over N}} + a_{k}\ln\pars{z}} \\[5mm]\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,& \bracks{\ln\pars{z} - {1 \over N}}\sum_{k = 0}^{n}a_{k} \\[1cm] &\ \prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, z^{\sum_{j = 0}^{n}a_{j}}\expo{-\pars{\sum_{j = 0}^{n}a_{j}}/N} = z^{N}\expo{-1} \\[1cm] &\bbox[10px,#ffd]{\ds{\bracks{\prod_{k = 0}^{n}\pars{1 - {a_{k} \over N}}} \bracks{z^{N}}z^{\sum_{j = 0}^{n}a_{j}}}} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \bbx{\expo{-1}} \end{align}