Small-Sample Results for the Kaplan-Meier Estimator

研究问题: 在比例风险模型下，如何获得 Kaplan–Meier 生存函数估计量的小样本精确矩，并据此分析偏差及方差近似？
方法与思路: 作者在比例风险模型下推导估计量任意正阶矩的精确表达式，利用该结果研究偏差并比较精确方差与渐近方差。
材料陈述的贡献: 摘要称得到正阶矩的精确表达式，使比例风险条件下的偏差分析以及精确方差和渐近方差的比较成为可能。
阅读关注点: 应核对比例风险条件具体涉及哪些分布、删失机制与观测独立性要求，并检查矩表达式如何用于小样本偏差和方差计算。
阅读边界: 仅据 OpenAlex 索引摘要，稿件版本未注明；摘要未说明比例风险关系的具体对象、删失条件或比较结果，不能推及所有生存数据或推断刊本变化。

GPT-6.1 Sol · 高 · 2026-10-03

原文摘要

Whereas much is known about the asymptotic properties of the Kaplan-Meier (1958) estimator (KME) of a survival function, exact results for small samples have been difficult to obtain. In this article, we obtain an exact expression for the αth moment (α > 0) of the KME under a model of proportional hazards. This enables us, under proportional hazards, to (a) study the bias of the KME, and (b) compare the exact variance of the KME to its asymptotic variance.

摘要来源：OpenAlex 索引摘要（稿件版本未注明）

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