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arXiv 2210.02027math.PRmath.CO
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伯努利时钟:通过循环卷积得到的伯努利多项式的概率与组合解释

The Bernoulli clock: probabilistic and combinatorial interpretations of the Bernoulli polynomials by circular convolution

Yassine El Maazouz, Jim Pitman

AI总结:

本文通过循环卷积给出了阶乘归一化伯努利多项式的概率与组合解释,提出了“伯努利时钟”模型,并证明其对称性是伯努利多项式对称性的组合类比。

摘要:

已知阶乘归一化的伯努利多项式 $b_n(x) = B_n(x)/n!$ 可由以下条件刻画:$b_0(x) = 1$,且对于 $n >0$,$b_n(x)$ 是 $b_{n-1}(x)$ 的原函数,并满足条件 $\int_0^1 b_n(x) dx = 0$。我们提供了一个相关的刻画:$b_1(x) = x - 1/2$,且对于 $n >0$,$(-1)^{n-1} b_n(x)$ 是 $b_1(x)$ 与其自身的 $n$ 重循环卷积。等价地,$1 - 2^n b_n(x)$ 是 $x \in (0,1)$ 处的概率密度,对应于 $n$ 个独立随机变量之和的小数部分,其中每个随机变量在 $x \in (0,1)$ 处具有 beta$(1,2)$ 概率密度 $2(1-x)$。该结果具有一种新颖的组合类比,即“伯努利时钟”:用一个包含 $2 n$ 小时的时钟,将多重集 $\{1,1, 2,2, \ldots, n,n\}$ 的均匀随机排列标记在小时刻度上,这意味着从 $2 n$ 个小时中均匀随机地选取两个不同的小时并标记为 $1$,然后从剩下的 $2 n - 2$ 个小时中均匀随机地选取两个不同的小时并标记为 $2$,以此类推。从第 $0 = 2n$ 小时开始,顺时针移动到第一个标记为 $1$ 的小时,继续顺时针移动到第一个标记为 $2$ 的小时,以此类推,在伯努利时钟上持续顺时针移动,直到遇到两个标记为 $n$ 的小时中的第一个,此时停在 $1$ 到 $2n$ 之间的一个随机小时 $I_n$ 上。我们证明了,对于每个正整数 $n$,事件 $( I_n = 1)$ 的概率为 $(1 - 2^n b_n(0))/(2n)$,其中 $n! b_n(0) = B_n(0)$ 是第 $n$ 个伯努利数。对于 $ 1 \le k \le 2 n$,差值 $δ_n(k):= 1/(2n) - P( I_n = k)$ 是关于 $k$ 的多项式函数,具有令人惊讶的对称性 $δ_n( 2 n + 1 - k) = (-1)^n δ_n(k)$,这是伯努利多项式众所周知的对称性 $b_n(1-x) = (-1)^n b_n(x)$ 的组合类比。

英文摘要:

The factorially normalized Bernoulli polynomials $b_n(x) = B_n(x)/n!$ are known to be characterized by $b_0(x) = 1$ and $b_n(x)$ for $n >0$ is the antiderivative of $b_{n-1}(x)$ subject to $\int_0^1 b_n(x) dx = 0$. We offer a related characterization: $b_1(x) = x - 1/2$ and $(-1)^{n-1} b_n(x)$ for $n >0$ is the $n$-fold circular convolution of $b_1(x)$ with itself. Equivalently, $1 - 2^n b_n(x)$ is the probability density at $x \in (0,1)$ of the fractional part of a sum of $n$ independent random variables, each with the beta$(1,2)$ probability density $2(1-x)$ at $x \in (0,1)$. This result has a novel combinatorial analog, the {\em Bernoulli clock}: mark the hours of a $2 n$ hour clock by a uniform random permutation of the multiset $\{1,1, 2,2, \ldots, n,n\}$, meaning pick two different hours uniformly at random from the $2 n$ hours and mark them $1$, then pick two different hours uniformly at random from the remaining $2 n - 2$ hours and mark them $2$, and so on. Starting from hour $0 = 2n$, move clockwise to the first hour marked $1$, continue clockwise to the first hour marked $2$, and so on, continuing clockwise around the Bernoulli clock until the first of the two hours marked $n$ is encountered, at a random hour $I_n$ between $1$ and $2n$. We show that for each positive integer $n$, the event $( I_n = 1)$ has probability $(1 - 2^n b_n(0))/(2n)$, where $n! b_n(0) = B_n(0)$ is the $n$th Bernoulli number. For $ 1 \le k \le 2 n$, the difference $δ_n(k):= 1/(2n) - ¶( I_n = k)$ is a polynomial function of $k$ with the surprising symmetry $δ_n( 2 n + 1 - k) = (-1)^n δ_n(k)$, which is a combinatorial analog of the well known symmetry of Bernoulli polynomials $b_n(1-x) = (-1)^n b_n(x)$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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