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AI 根据所提供的论文材料归类:摘要针对一般双重稳健泛函和双重机器学习理论,未明确限定因果效应或其他所列应用方向。

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关注这篇论文的作者
Lin Liu
Rajarshi Mukherjee
James M. Robins

双重机器学习估计量速率双重稳健性的弱假设证伪检验

研究问题
在不对干扰函数施加降低复杂度的假设时,能否检验双重机器学习置信区间的有效性,或至少证伪其依赖的速率双重稳健性依据?
方法与思路
作者转而检验偏差的 Cauchy–Schwarz 上界是否小于根号样本量倒数的阶,构造弱假设下的有效检验,而非直接检验置信区间有效性。
材料陈述的贡献
摘要称,这些检验对某些备择具有非平凡功效,但任何此类弱假设检验都不能一致;拒绝可证伪论证依据,不拒绝不能支持原假设。
阅读关注点
应核对速率双重稳健性与实际偏差条件的区别、检验统计量构造、能够发现的备择,以及不可能性结论和检验有效性的适用范围。
阅读边界
所给 RePEc 出版社摘要未说明具体构造与功效数值;拒绝论证依据不等于证明区间失效,不拒绝也非有效性证据,其他版本变化未知。

原文摘要

The class of doubly robust (DR) functionals studied by Rotnitzky et al. (2021) is of central importance in economics and biostatistics. It strictly includes both (i) the class of mean-square continuous functionals that can be written as an expectation of an affine functional of a conditional expectation studied by Chernozhukov et al. (2022b) and the class of functionals studied by Robins et al. (2008). The present state-of-the-art estimators for DR functionals ψ are double-machine-learning (DML) estimators (Chernozhukov et al., 2018a). A DML estimator ψ̂1 of ψ depends on estimates p̂(x) and b̂(x) of a pair of nuisance functions p(x) and b(x), and is said to satisfy “rate double-robustness” if the Cauchy–Schwarz upper bound of its bias is o(n−1/2). Rate double-robustness implies that the bias is o(n−1/2), but the converse is false. Were it achievable, our scientific goal would have been to construct valid, assumption-lean (i.e. no complexity-reducing assumptions on b or p) tests of the validity of a nominal (1−α) Wald confidence interval (CI) centered at ψ̂1. But this would require a test of the bias to be o(n−1/2), which can be shown not to exist. We therefore adopt the less ambitious goal of falsifying, when possible, an analyst’s justification for her claim that the reported (1−α) Wald CI is valid. In many instances, an analyst justifies her claim by imposing complexity-reducing assumptions on b and p to ensure “rate double-robustness”. Here we exhibit valid, assumption-lean tests of H0: “rate double-robustness holds”, with non-trivial power against certain alternatives. If H0 is rejected, we will have falsified her justification. However, no assumption-lean test of H0, including ours, can be a consistent test. Thus, the failure of our test to reject is not meaningful evidence in favor of H0.

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