arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~

泊松-二项分布点概率中距离为2为何特殊

Why Distance Two Is Exceptional for Poisson-Binomial Point Probabilities

Igor Kleiner

arXiv 2610.10664首次发表:更新:

发表机构

Technion – Israel Institute of Technology; Holon Institute of Technology (HIT)(以色列理工学院; 霍隆理工学院)

机构由 AI 辅助整理,请以论文原文为准。

AI 中文总结

该研究针对泊松-二项分布点概率的阈值问题,发现距离2是唯一非相邻例外,推导了其阈值的显式表达式,证明结合固定均值约简与离散优化,还对实根多项式的非相邻系数反转给出严格限制。

AI 中文摘要

设S为独立伯努利随机变量的有限和,研究均值需多大时,S=k+r处的概率质量可严格超过S=k处的概率质量。对于阈值T_{k,r}:=inf{ES:P(S=k+r)>P(S=k)},相邻情况对应经典Darroch边界k+1/(k+2),而间隙r≥3时阈值均为k+1。距离2是唯一的非相邻例外:T_{k,2}=k+Δ_k,其中Δ_k及其阶梯优化器显式表达,且Δ_k=1/2+1/√(2k)-1/(4k)+O(k^{-3/2})。证明结合经典的固定均值约简为移位二项分布律,以及精确离散优化。该结果还对非负系数实根多项式的非相邻系数反转给出严格限制。

英文摘要

Let $S$ be a finite sum of independent Bernoulli random variables. We ask how large the mean must be before the atom at $k+r$ can strictly exceed the atom at $k$. For the threshold $T_{k,r}:=\inf\{\mathbb{E}S:\mathbb{P}(S=k+r)>\mathbb{P}(S=k)\}$, the adjacent case recovers the classical Darroch boundary $k+1/(k+2)$, while every gap $r\ge3$ has threshold $k+1$. Distance two is the unique nonadjacent exception: $T_{k,2}=k+Δ_k$, where $Δ_k$ and its staircase optimizer are explicit and $Δ_k=1/2+1/\sqrt{2k}-1/(4k)+O(k^{-3/2})$. The proof combines the classical fixed-mean reduction to shifted binomial laws with an exact discrete optimization. The result also yields sharp restrictions on nonadjacent coefficient reversals in real-rooted polynomials with nonnegative coefficients.

Comments11 pages

论文原文

arXiv 摘要页 · PDF 原文 · HTML 原文

↑