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经典离散分布的平均绝对偏差:坍缩恒等式、完全渐近展开与包络级数

The mean absolute deviation of the classical discrete distributions: collapse identities, complete asymptotic expansions, and enveloping series

Neven Elezović

arXiv 2608.06232首次发表:更新:

AI 中文总结

本文针对四种经典离散分布,证明其平均绝对偏差的坍缩恒等式,推导相关渐近展开并揭示级数包络特性,扩展了配套论文的二项分布展开。

AI 中文摘要

对于二项分布、泊松分布、负二项分布和超几何分布这四种经典离散律,其关于均值的平均绝对偏差会坍缩为单个点质量。我们给出这些恒等式的一个通用望远镜证明,并通过大小偏差解释所得闭式。随后,我们推导了泊松分布(λ→∞)、负二项分布(r→∞,p固定)和超几何分布(N→∞,边际成固定比例)的完全渐近展开,扩展了配套论文中的二项分布展开。系数以伯努利多项式闭式形式给出,精确携带均值的格位移;在整数均值处,这些展开简化为符号交替的奇级数,且单个Binet核论证表明,这些级数包络归一化平均绝对偏差的对数,即连续部分和对其进行夹逼。

英文摘要

For each of the four classical discrete laws --- binomial, Poisson, negative binomial and hypergeometric --- the mean absolute deviation about the mean collapses to a single point mass. We give a common telescoping proof of these identities and interpret the resulting closed forms by size biasing. We then derive complete asymptotic expansions for the Poisson ($λ\to\infty$), negative binomial ($r\to\infty$, $p$ fixed) and hypergeometric ($N\to\infty$, margins in fixed proportion) cases, extending the binomial expansion from the companion papers. The coefficients are given in closed Bernoulli-polynomial form and carry the lattice displacement of the mean exactly. At integer means the expansions reduce to sign-alternating odd series, and a single Binet-kernel argument shows that these series envelop the logarithm of the normalised mean absolute deviation: successive partial sums bracket it.

Comments23 pages

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