Assumption-lean falsification tests of rate double-robustness of double-machine-learning estimators
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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
- 研究不对干扰函数施加降低复杂度的限制时，能否检验速率双重稳健性，从而证伪 Wald 区间有效性的论据。
- 方法与思路
- 作者针对偏差的柯西—施瓦茨上界是否为小于根号样本量倒数的阶构造检验，并讨论可检验性与功效限制。
- 材料陈述的贡献
- 摘要称检验在部分备择下具有非平凡功效，可否定速率双重稳健性的论据，但任何此类弱假设检验都无法一致。
- 阅读关注点
- 应核对可检测备择、检验统计量及尺寸控制，特别区分否定区间有效性的论据与直接证明区间覆盖失效。
- 阅读边界
- 这是 arXiv 预印本，同行评审及期刊接收未核实；摘要明确强调不拒绝不支持原假设，且未给出检验具体条件。
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 (ii) 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., 2018). A DML estimator $\widehatψ_{1}$ of $ψ$ depends on estimates $\widehat{p} (x)$ and $\widehat{b} (x)$ 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 $o (n^{- 1/2})$. 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 $\widehatψ_{1}$. 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 $H_{0}$: "rate double-robustness holds", with non-trivial power against certain alternatives. If $H_{0}$ is rejected, we will have falsified her justification. However, no assumption-lean test of $H_{0}$, including ours, can be a consistent test. Thus, the failure of our test to reject is not meaningful evidence in favor of $H_{0}$.