Erratum to: Hennessy J, Dasgupta T, Miratrix L, Pattanayak C, Sarkar P. A conditional randomization test to account for covariate imbalance in randomized experiments. J Causal Inference 2016;4(1):61–80 (https://doi.org/10.1515/jci-2015-0018). There was an error in Hennessy et al. [1] and we are very grateful to Peng Ding for pointing it out. Proposition 1, restated below, is incorrect, our proof being a mis-application of a result from Rosenbaum [2]. Proposition 1. Let X denote a categorical covariate with J levels, observed after a two-armed randomized experiment is conducted with N units. Let Nj denote the observed number of units that belong to stratum j, and let NTj and NCj denote the number of units assigned to treatment and control respectively, in stratum j, such thatNTj+NCj = Nj, and ∑J j=1 Nj = N. Then the conditional randomization test using the simple difference test statistic 4̂sd = Ȳobs T – Ȳobs C and the balance function (NT1, ..., NTJ) is equivalent to the conditional randomization test using the composite test statistic 4̂ps = ∑J j=1 Nj N 4̂sd,j, where 4̂sd,j denotes the simple difference test statistic for the jth stratum. We can show the proposition is not true by a simple counterexample. In order for the two conditional tests to be equivalent, they must yield the same p-values. Consider the situation where N = 5, X = (1, 1, 1, 2, 2), w = (1, 0, 0, 1, 0), and yobs = (1.13, 0.49, –0.31, 0.98, 1.68). In this case, 4̂sd = 0.435 and 4̂ps = 0.344. To find the p-values for the conditional test, we consider the values of the test statistics across all 6 alternative randomizations where NT1 = 1 and NT2 = 1. To calculate the p-values, we find the proportion of test statistics as or more extreme than the observed. For 4̂sd = 0.435, there are three test statistics as large or larger (0.435, 0.485, 1.018), so the 2-sided p-value is 2 ⋅ 3/6 = 1. For 4̂ps = 0.344, there are two test statistics as large or larger (0.344, 0.904), so the 2-sided p-value is 2 ⋅ 2/6 = 2/3. Since the p-values do not agree, the tests are not equivalent. The incorrect proof in Appendix A mis-applied a result from Rosenbaum [2] that showed that in a linear regression of the response on the treatment indicator and covariates, a conditional randomization test based on the treatment indicator coefficient is equivalent to the conditional randomization test based on the simple difference test statistic. This proof rests on the fact that the columns corresponding to the covariates are fixed across randomizations. While our 4̂ps does equal a regression coefficient, that regression includes interactions between the treatment indicator and the covariates. However, these interaction terms are not

中文翻译：

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勘误表：协变量不平衡的条件随机化检验

勘误表：Hennessy J、Dasgupta T、Miratrix L、Pattanayak C、Sarkar P. 用于解释随机实验中协变量不平衡的条件随机化检验。J 因果推断 2016；4(1):61–80 (https://doi.org/10.1515/jci-2015-0018)。轩尼诗等人有错误。[1] 非常感谢彭丁的指出。下面重申的命题 1 是不正确的，我们的证明是对 Rosenbaum [2] 结果的错误应用。命题 1. 让 X 表示具有 J 水平的分类协变量，这是在用 N 单位进行双臂随机实验后观察到的。设 Nj 表示观测到的属于第 j 层的单元数，设 NTj 和 NCj 分别表示分配给处理和控制的单元数，在第 j 层，使得 NTj+NCj = Nj，且 ∑J j=1 Nj = N。那么使用简单差异检验统计量 4̂sd = Ȳobs T – Ȳobs C 和平衡函数 (NT1, ..., NTJ) 的条件随机化检验等价于使用复合检验统计量 4̂ps = ∑J j=1 的条件随机化检验Nj N 4̂sd,j，其中4̂sd,j 表示第j层的简单差分检验统计量。我们可以通过一个简单的反例证明这个命题是不正确的。为了使两个条件测试等效，它们必须产生相同的 p 值。考虑 N = 5, X = (1, 1, 1, 2, 2), w = (1, 0, 0, 1, 0) 和 yobs = (1.13, 0.49, –0.31, 0.98, 1.68) 的情况）。在这种情况下，4̂sd = 0.435 和 4̂ps = 0.344。为了找到条件检验的 p 值，我们考虑了所有 6 个替代随机化的检验统计值，其中 NT1 = 1 和 NT2 = 1。要计算 p 值，我们发现测试统计的比例与观察到的一样或更极端。对于 4̂sd = 0.435，有三个或更大的检验统计量 (0.435, 0.485, 1.018)，因此两侧 p 值是 2 ⋅ 3/6 = 1。对于 4̂ps = 0.344，有两个检验统计量如下大或更大 (0.344, 0.904)，因此 2 边 p 值为 2 ⋅ 2/6 = 2/3。由于 p 值不一致，因此测试不等效。附录 A 中的错误证明错误地应用了 Rosenbaum [2] 的结果，该结果表明，在响应对治疗指标和协变量的线性回归中，基于治疗指标系数的条件随机化检验等效于条件随机化检验基于简单差异检验统计量。这个证明基于这样一个事实，即对应于协变量的列在随机化中是固定的。虽然我们的 4̂ps 确实等于回归系数，但该回归包括治疗指标和协变量之间的相互作用。但是，这些交互项不是