$\hat{\tau}_{\textnormal{tdim}}$,
transformed difference-in-means estimator; $\hat{\tau}_{\textnormal{str}}$,
stratified transformed difference-in-means estimator;
SR, simple randomization; STR, stratified randomization; MIN, Pocock and
Simon’s minimization;
SD, standard deviation; SE, standard error; CP, coverage probability.

Table 11: Simulation results under Model 3 for $n=500$ and $\pi=1/2$

|  | SR | STR | MIN
---|---|---|---|---
|  | Bias | SD | SE | CP | Bias | SD | SE | CP | Bias | SD | SE | CP
$\tau$ = 0 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | 0.000 | 0.126 | 0.120 | 0.940 | -0.006 | 0.100 | 0.102 | 0.957 | 0.003 | 0.115 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | -0.001 | 0.105 | 0.102 | 0.946 | -0.006 | 0.100 | 0.102 | 0.957 | 0.000 | 0.101 | 0.102 | 0.956
| $\hat{\tau}_{\textnormal{md}}$ | -0.002 | 0.154 | 0.161 | 0.955 | -0.005 | 0.135 | 0.145 | 0.964 | 0.002 | 0.144 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.004 | 0.136 | 0.147 | 0.964 | -0.005 | 0.136 | 0.145 | 0.964 | 0.000 | 0.134 | 0.146 | 0.965
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.002 | 0.137 | 0.125 | 0.925 | -0.002 | 0.111 | 0.114 | 0.951 | 0.006 | 0.121 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | 0.001 | 0.119 | 0.114 | 0.939 | -0.002 | 0.111 | 0.114 | 0.950 | 0.004 | 0.112 | 0.114 | 0.956
Laplace tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | -0.006 | 0.148 | 0.149 | 0.942 | 0.001 | 0.136 | 0.135 | 0.942 | 0.006 | 0.141 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | -0.007 | 0.133 | 0.135 | 0.942 | 0.001 | 0.136 | 0.135 | 0.944 | 0.003 | 0.129 | 0.135 | 0.967
| $\hat{\tau}_{\textnormal{md}}$ | 0.002 | 0.160 | 0.156 | 0.941 | 0.002 | 0.140 | 0.143 | 0.945 | 0.000 | 0.148 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.002 | 0.144 | 0.144 | 0.954 | 0.002 | 0.140 | 0.143 | 0.945 | -0.002 | 0.137 | 0.144 | 0.953
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.000 | 0.145 | 0.143 | 0.953 | -0.001 | 0.127 | 0.131 | 0.945 | 0.006 | 0.132 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | -0.001 | 0.129 | 0.131 | 0.954 | -0.001 | 0.127 | 0.131 | 0.945 | 0.004 | 0.121 | 0.131 | 0.965
Cauchy tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | 84.118 | 2821.234 | 109.258 | 0.980 | -98.881 | 2988.736 | 106.743 | 0.987 | -101.940 | 3099.181 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | 93.859 | 3130.406 | 108.938 | 0.978 | -98.910 | 2990.158 | 106.743 | 0.987 | -93.180 | 2821.739 | 104.067 | 0.978
| $\hat{\tau}_{\textnormal{md}}$ | 0.006 | 0.196 | 0.213 | 0.975 | 0.001 | 0.182 | 0.200 | 0.974 | 0.009 | 0.194 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.004 | 0.181 | 0.202 | 0.972 | 0.001 | 0.182 | 0.200 | 0.972 | 0.006 | 0.184 | 0.201 | 0.967
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.005 | 0.188 | 0.198 | 0.963 | -0.001 | 0.177 | 0.181 | 0.958 | 0.001 | 0.186 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | 0.003 | 0.174 | 0.186 | 0.967 | -0.001 | 0.177 | 0.181 | 0.960 | -0.001 | 0.175 | 0.182 | 0.963
$\tau$ = 1 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | 0.000 | 0.123 | 0.120 | 0.944 | 0.002 | 0.103 | 0.102 | 0.946 | -0.004 | 0.109 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | -0.002 | 0.104 | 0.102 | 0.949 | 0.002 | 0.103 | 0.102 | 0.945 | -0.004 | 0.095 | 0.102 | 0.964
| $\hat{\tau}_{\textnormal{md}}$ | 0.003 | 0.151 | 0.160 | 0.964 | 0.001 | 0.134 | 0.146 | 0.961 | 0.000 | 0.134 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.001 | 0.137 | 0.146 | 0.961 | 0.001 | 0.134 | 0.146 | 0.961 | 0.001 | 0.126 | 0.146 | 0.971
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.004 | 0.131 | 0.125 | 0.941 | 0.003 | 0.115 | 0.114 | 0.943 | -0.002 | 0.117 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | 0.003 | 0.117 | 0.114 | 0.939 | 0.003 | 0.115 | 0.114 | 0.943 | -0.002 | 0.107 | 0.114 | 0.962
Laplace tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | 0.004 | 0.155 | 0.149 | 0.944 | 0.000 | 0.133 | 0.135 | 0.956 | -0.004 | 0.141 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | 0.002 | 0.140 | 0.135 | 0.950 | 0.000 | 0.133 | 0.135 | 0.957 | -0.005 | 0.128 | 0.135 | 0.957
| $\hat{\tau}_{\textnormal{md}}$ | 0.005 | 0.162 | 0.157 | 0.938 | 0.002 | 0.145 | 0.144 | 0.945 | -0.002 | 0.156 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.003 | 0.149 | 0.145 | 0.940 | 0.002 | 0.145 | 0.144 | 0.945 | -0.003 | 0.145 | 0.144 | 0.947
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.004 | 0.145 | 0.144 | 0.948 | 0.000 | 0.129 | 0.131 | 0.946 | 0.003 | 0.139 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | 0.003 | 0.131 | 0.132 | 0.950 | 0.001 | 0.129 | 0.131 | 0.946 | 0.002 | 0.128 | 0.131 | 0.948
Cauchy tail | $\hat{\tau}_{\textnormal{naive-dim}}$ | 1.387 | 83.917 | 9.725 | 0.976 | 3.544 | 87.378 | 9.797 | 0.978 | 3.096 | 85.248 | / | /
| $\hat{\tau}_{\textnormal{str-dim}}$ | 1.518 | 84.781 | 9.694 | 0.972 | 3.552 | 87.676 | 9.797 | 0.980 | 3.205 | 84.577 | 9.710 | 0.980
| $\hat{\tau}_{\textnormal{md}}$ | 0.004 | 0.207 | 0.211 | 0.957 | 0.002 | 0.190 | 0.200 | 0.962 | -0.014 | 0.198 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.003 | 0.195 | 0.201 | 0.951 | 0.002 | 0.190 | 0.200 | 0.962 | -0.016 | 0.189 | 0.202 | 0.962
| $\hat{\tau}_{\textnormal{tdim}}$ | 0.003 | 0.195 | 0.196 | 0.942 | 0.006 | 0.180 | 0.181 | 0.949 | -0.010 | 0.187 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | 0.003 | 0.183 | 0.184 | 0.944 | 0.007 | 0.180 | 0.181 | 0.950 | -0.011 | 0.179 | 0.183 | 0.945

  * Note: $\hat{\tau}_{\textnormal{naive-dim}}$, standard difference-in-means estimator; $\hat{\tau}_{\textnormal{str-dim}}$, stratified difference-in-means esti-
mator; $\hat{\tau}_{\textnormal{md}}$, difference-in-medians estimator;
$\hat{\tau}_{\textnormal{wt-md}}$, difference-in-weighted-medians estimator;
$\hat{\tau}_{\textnormal{tdim}}$,
transformed difference-in-means estimator; $\hat{\tau}_{\textnormal{str}}$,
stratified transformed difference-in-means estimator;
SR, simple randomization; STR, stratified randomization; MIN, Pocock and
Simon’s minimization;
SD, standard deviation; SE, standard error; CP, coverage probability.

###  C.3 Initial estimators

In this section, we evaluate the influence of the initial estimator on the
final treatment effect estimators, adopting the same settings as in the main
text. Tables 12-14 present the results for the weighted average difference-in-
medians estimator $\hat{\tau}_{\textnormal{str-md}}$, the difference-in-
weighted-medians estimator $\hat{\tau}_{\textnormal{wt-md}}$, and the
stratified transformed difference-in-means estimator
$\hat{\tau}_{\textnormal{str}}$ when using each of the first two as the
initial estimator. To distinguish them, we denote
$\hat{\tau}_{\textnormal{str}}$ as $\hat{\tau}_{\textnormal{str},1}$ and
$\hat{\tau}_{\textnormal{str},2}$ when using $\hat{\tau}_{\textnormal{str-
md}}$ and $\hat{\tau}_{\textnormal{wt-md}}$ as initial estimators,
respectively. To the best of our knowledge, the asymptotic properties of
$\hat{\tau}_{\textnormal{str-md}}$ under covariate-adaptive randomization have
not yet been studied, so we do not report inference-related results for it. We
find that $\hat{\tau}_{\textnormal{str}}$ exhibits similar performance when
using these two different initial estimators, with the differences being
smaller than the Monte Carlo error. Thus, $\hat{\tau}_{\textnormal{str}}$
appears to be robust to the choice of the initial estimator. For the two
initial estimators themselves, $\hat{\tau}_{\textnormal{str-md}}$ demonstrates
comparable or smaller variance compared to $\hat{\tau}_{\textnormal{wt-md}}$
in all cases. Notably, under a Cauchy tail distribution,
$\hat{\tau}_{\textnormal{str-md}}$ has a variance close to that of
$\hat{\tau}_{\textnormal{str}}$. Further investigation reveals that the error
in density estimation contributes significantly to the variance of
$\hat{\tau}_{\textnormal{str}}$. When the sample size used for density
estimation exceeds, for example, 350, $\hat{\tau}_{\textnormal{str}}$ begins
to exhibit significantly smaller variance.

Table 12: Simulation results under Model 1 for different initial estimators

|  | SR | STR | MIN
---|---|---|---|---
|  | Bias | SD | SE | CP | Bias | SD | SE | CP | Bias | SD | SE | CP
$\tau$ = 0 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.005 | 0.090 | / | / | -0.003 | 0.090 | / | / | -0.004 | 0.086 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.005 | 0.095 | 0.106 | 0.961 | -0.004 | 0.094 | 0.106 | 0.962 | -0.003 | 0.094 | 0.106 | 0.960
| $\hat{\tau}_{\textnormal{str},1}$ | 0.005 | 0.079 | 0.082 | 0.950 | -0.005 | 0.078 | 0.081 | 0.964 | -0.003 | 0.080 | 0.082 | 0.948
| $\hat{\tau}_{\textnormal{str},2}$ | 0.005 | 0.079 | 0.082 | 0.945 | -0.005 | 0.078 | 0.081 | 0.965 | -0.003 | 0.080 | 0.082 | 0.949
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.003 | 0.092 | / | / | -0.005 | 0.093 | / | / | 0.001 | 0.090 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.007 | 0.107 | 0.109 | 0.948 | -0.003 | 0.107 | 0.109 | 0.959 | 0.000 | 0.108 | 0.109 | 0.945
| $\hat{\tau}_{\textnormal{str},1}$ | 0.006 | 0.096 | 0.092 | 0.932 | -0.005 | 0.093 | 0.092 | 0.942 | 0.001 | 0.093 | 0.092 | 0.954
| $\hat{\tau}_{\textnormal{str},2}$ | 0.006 | 0.095 | 0.092 | 0.942 | -0.006 | 0.092 | 0.092 | 0.943 | 0.001 | 0.091 | 0.092 | 0.958
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.001 | 0.113 | / | / | 0.007 | 0.121 | / | / | 0.005 | 0.112 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.001 | 0.138 | 0.138 | 0.942 | 0.005 | 0.135 | 0.138 | 0.946 | 0.004 | 0.130 | 0.138 | 0.965
| $\hat{\tau}_{\textnormal{str},1}$ | 0.002 | 0.118 | 0.118 | 0.948 | 0.003 | 0.118 | 0.118 | 0.943 | 0.006 | 0.116 | 0.118 | 0.958
| $\hat{\tau}_{\textnormal{str},2}$ | 0.001 | 0.118 | 0.118 | 0.949 | 0.003 | 0.118 | 0.118 | 0.945 | 0.005 | 0.116 | 0.118 | 0.956
$\tau$ = 1 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.003 | 0.087 | / | / | 0.009 | 0.087 | / | / | -0.003 | 0.083 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.001 | 0.095 | 0.106 | 0.968 | 0.006 | 0.092 | 0.106 | 0.977 | 0.001 | 0.091 | 0.106 | 0.974
| $\hat{\tau}_{\textnormal{str},1}$ | 0.003 | 0.077 | 0.082 | 0.963 | 0.006 | 0.075 | 0.081 | 0.964 | -0.002 | 0.075 | 0.082 | 0.970
| $\hat{\tau}_{\textnormal{str},2}$ | 0.003 | 0.078 | 0.082 | 0.968 | 0.007 | 0.075 | 0.081 | 0.960 | -0.002 | 0.075 | 0.082 | 0.967
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.090 | / | / | 0.004 | 0.089 | / | / | 0.003 | 0.083 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.003 | 0.100 | 0.109 | 0.972 | 0.004 | 0.108 | 0.109 | 0.946 | -0.002 | 0.100 | 0.109 | 0.965
| $\hat{\tau}_{\textnormal{str},1}$ | 0.002 | 0.092 | 0.092 | 0.941 | 0.004 | 0.094 | 0.092 | 0.946 | -0.001 | 0.089 | 0.092 | 0.962
| $\hat{\tau}_{\textnormal{str},2}$ | 0.002 | 0.091 | 0.092 | 0.944 | 0.004 | 0.092 | 0.092 | 0.945 | -0.001 | 0.088 | 0.092 | 0.965
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.117 | / | / | -0.008 | 0.119 | / | / | -0.006 | 0.114 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.002 | 0.138 | 0.138 | 0.948 | -0.004 | 0.143 | 0.138 | 0.938 | -0.007 | 0.138 | 0.138 | 0.948
| $\hat{\tau}_{\textnormal{str},1}$ | -0.002 | 0.119 | 0.118 | 0.944 | -0.007 | 0.121 | 0.118 | 0.934 | -0.006 | 0.119 | 0.118 | 0.944
| $\hat{\tau}_{\textnormal{str},2}$ | -0.002 | 0.119 | 0.118 | 0.944 | -0.007 | 0.122 | 0.118 | 0.931 | -0.006 | 0.120 | 0.118 | 0.942

  * Note: $\hat{\tau}_{\textnormal{str-md}}$, stratified difference-in-medians estimator; $\hat{\tau}_{\textnormal{wt-md}}$, difference-in-weighted-medians esti-
mator; $\hat{\tau}_{\textnormal{str},1}$, stratified transformed difference-
in-means estimator with $\hat{\tau}_{\textnormal{str-md}}$ as initial
estimator;
$\hat{\tau}_{\textnormal{str},2}$, stratified transformed difference-in-means
estimator with $\hat{\tau}_{\textnormal{wt-md}}$ as initial estimator; SR,
simple
randomization; STR, stratified randomization; MIN, Pocock and Simon’s
minimization; SD, standard
deviation; SE, standard error; CP, coverage probability.

Table 13: Simulation results under Model 2 for different initial estimators

|  | SR | STR | MIN
---|---|---|---|---
|  | Bias | SD | SE | CP | Bias | SD | SE | CP | Bias | SD | SE | CP
$\tau$ = 0 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.090 | / | / | -0.003 | 0.090 | / | / | 0.001 | 0.087 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.004 | 0.094 | 0.100 | 0.965 | -0.002 | 0.090 | 0.100 | 0.969 | -0.003 | 0.088 | 0.100 | 0.961
| $\hat{\tau}_{\textnormal{str},1}$ | 0.001 | 0.079 | 0.079 | 0.944 | -0.003 | 0.076 | 0.078 | 0.951 | -0.001 | 0.078 | 0.079 | 0.948
| $\hat{\tau}_{\textnormal{str},2}$ | 0.001 | 0.079 | 0.079 | 0.943 | -0.003 | 0.077 | 0.078 | 0.954 | 0.000 | 0.078 | 0.079 | 0.945
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.090 | / | / | 0.002 | 0.092 | / | / | 0.005 | 0.088 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.003 | 0.097 | 0.098 | 0.951 | 0.000 | 0.101 | 0.098 | 0.939 | 0.002 | 0.096 | 0.098 | 0.944
| $\hat{\tau}_{\textnormal{str},1}$ | 0.004 | 0.091 | 0.090 | 0.949 | 0.002 | 0.093 | 0.090 | 0.947 | 0.003 | 0.089 | 0.090 | 0.946
| $\hat{\tau}_{\textnormal{str},2}$ | 0.004 | 0.090 | 0.090 | 0.951 | 0.001 | 0.092 | 0.090 | 0.949 | 0.002 | 0.088 | 0.090 | 0.951
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.005 | 0.118 | / | / | -0.003 | 0.117 | / | / | 0.001 | 0.119 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.003 | 0.129 | 0.131 | 0.946 | 0.000 | 0.127 | 0.132 | 0.956 | -0.001 | 0.130 | 0.132 | 0.954
| $\hat{\tau}_{\textnormal{str},1}$ | 0.005 | 0.116 | 0.117 | 0.946 | 0.001 | 0.116 | 0.118 | 0.951 | 0.001 | 0.118 | 0.118 | 0.951
| $\hat{\tau}_{\textnormal{str},2}$ | 0.005 | 0.116 | 0.117 | 0.946 | 0.000 | 0.115 | 0.118 | 0.955 | 0.000 | 0.118 | 0.118 | 0.949
$\tau$ = 1 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.005 | 0.091 | / | / | 0.001 | 0.090 | / | / | 0.003 | 0.088 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.003 | 0.091 | 0.101 | 0.963 | -0.004 | 0.092 | 0.101 | 0.969 | 0.003 | 0.089 | 0.101 | 0.973
| $\hat{\tau}_{\textnormal{str},1}$ | -0.002 | 0.079 | 0.079 | 0.947 | -0.001 | 0.079 | 0.079 | 0.931 | 0.001 | 0.078 | 0.079 | 0.955
| $\hat{\tau}_{\textnormal{str},2}$ | -0.002 | 0.079 | 0.079 | 0.945 | -0.001 | 0.079 | 0.079 | 0.929 | 0.002 | 0.078 | 0.079 | 0.957
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.002 | 0.092 | / | / | -0.003 | 0.093 | / | / | -0.001 | 0.087 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.000 | 0.098 | 0.098 | 0.938 | -0.006 | 0.099 | 0.098 | 0.945 | 0.000 | 0.099 | 0.098 | 0.936
| $\hat{\tau}_{\textnormal{str},1}$ | 0.000 | 0.091 | 0.089 | 0.949 | -0.003 | 0.090 | 0.089 | 0.955 | 0.000 | 0.090 | 0.089 | 0.945
| $\hat{\tau}_{\textnormal{str},2}$ | 0.000 | 0.090 | 0.089 | 0.947 | -0.003 | 0.089 | 0.089 | 0.955 | 0.000 | 0.090 | 0.089 | 0.953
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.001 | 0.120 | / | / | 0.002 | 0.114 | / | / | -0.004 | 0.121 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.004 | 0.133 | 0.131 | 0.935 | 0.002 | 0.127 | 0.131 | 0.953 | -0.001 | 0.131 | 0.131 | 0.952
| $\hat{\tau}_{\textnormal{str},1}$ | 0.002 | 0.118 | 0.117 | 0.940 | 0.001 | 0.116 | 0.118 | 0.946 | -0.002 | 0.122 | 0.118 | 0.948
| $\hat{\tau}_{\textnormal{str},2}$ | 0.001 | 0.118 | 0.117 | 0.939 | 0.000 | 0.115 | 0.118 | 0.947 | -0.003 | 0.122 | 0.118 | 0.945

  * Note: $\hat{\tau}_{\textnormal{str-md}}$, stratified difference-in-medians estimator; $\hat{\tau}_{\textnormal{wt-md}}$, difference-in-weighted-medians esti-
mator; $\hat{\tau}_{\textnormal{str},1}$, stratified transformed difference-
in-means estimator with $\hat{\tau}_{\textnormal{str-md}}$ as initial
estimator;
$\hat{\tau}_{\textnormal{str},2}$, stratified transformed difference-in-means
estimator with $\hat{\tau}_{\textnormal{wt-md}}$ as initial estimator; SR,
simple
randomization; STR, stratified randomization; MIN, Pocock and Simon’s
minimization; SD, standard
deviation; SE, standard error; CP, coverage probability.

Table 14: Simulation results under Model 3 for different initial estimators

|  | SR | STR | MIN
---|---|---|---|---
|  | Bias | SD | SE | CP | Bias | SD | SE | CP | Bias | SD | SE | CP
$\tau$ = 0 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.002 | 0.095 | / | / | 0.001 | 0.089 | / | / | -0.003 | 0.088 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.002 | 0.097 | 0.105 | 0.970 | 0.003 | 0.095 | 0.105 | 0.968 | 0.001 | 0.090 | 0.105 | 0.980
| $\hat{\tau}_{\textnormal{str},1}$ | 0.000 | 0.082 | 0.082 | 0.946 | 0.002 | 0.079 | 0.081 | 0.964 | -0.001 | 0.077 | 0.081 | 0.957
| $\hat{\tau}_{\textnormal{str},2}$ | 0.000 | 0.081 | 0.082 | 0.949 | 0.002 | 0.079 | 0.081 | 0.958 | -0.002 | 0.077 | 0.082 | 0.951
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.001 | 0.096 | / | / | 0.005 | 0.093 | / | / | -0.006 | 0.094 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.000 | 0.105 | 0.104 | 0.945 | 0.002 | 0.103 | 0.104 | 0.956 | -0.001 | 0.102 | 0.104 | 0.953
| $\hat{\tau}_{\textnormal{str},1}$ | 0.001 | 0.095 | 0.093 | 0.943 | 0.003 | 0.093 | 0.093 | 0.948 | 0.000 | 0.092 | 0.093 | 0.950
| $\hat{\tau}_{\textnormal{str},2}$ | 0.000 | 0.093 | 0.093 | 0.948 | 0.003 | 0.092 | 0.093 | 0.951 | -0.001 | 0.091 | 0.093 | 0.957
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.004 | 0.128 | / | / | 0.000 | 0.118 | / | / | 0.000 | 0.119 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.005 | 0.137 | 0.137 | 0.948 | -0.004 | 0.133 | 0.138 | 0.950 | -0.001 | 0.131 | 0.138 | 0.958
| $\hat{\tau}_{\textnormal{str},1}$ | 0.003 | 0.125 | 0.120 | 0.932 | -0.002 | 0.116 | 0.120 | 0.963 | -0.002 | 0.116 | 0.120 | 0.951
| $\hat{\tau}_{\textnormal{str},2}$ | 0.003 | 0.126 | 0.120 | 0.929 | -0.002 | 0.116 | 0.120 | 0.961 | -0.002 | 0.117 | 0.120 | 0.950
$\tau$ = 1 |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.093 | / | / | 0.005 | 0.093 | / | / | 0.003 | 0.091 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.001 | 0.094 | 0.105 | 0.961 | 0.003 | 0.099 | 0.105 | 0.951 | 0.007 | 0.095 | 0.105 | 0.964
| $\hat{\tau}_{\textnormal{str},1}$ | 0.000 | 0.079 | 0.081 | 0.949 | 0.005 | 0.082 | 0.081 | 0.945 | 0.005 | 0.078 | 0.082 | 0.956
| $\hat{\tau}_{\textnormal{str},2}$ | 0.000 | 0.079 | 0.081 | 0.948 | 0.004 | 0.082 | 0.081 | 0.945 | 0.005 | 0.077 | 0.082 | 0.957
Laplace tail | $\hat{\tau}_{\textnormal{str-md}}$ | 0.000 | 0.090 | / | / | -0.001 | 0.092 | / | / | -0.005 | 0.093 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | 0.002 | 0.101 | 0.104 | 0.950 | -0.002 | 0.103 | 0.104 | 0.948 | 0.001 | 0.103 | 0.104 | 0.949
| $\hat{\tau}_{\textnormal{str},1}$ | 0.003 | 0.093 | 0.093 | 0.949 | -0.003 | 0.090 | 0.093 | 0.947 | 0.002 | 0.094 | 0.093 | 0.951
| $\hat{\tau}_{\textnormal{str},2}$ | 0.002 | 0.092 | 0.093 | 0.954 | -0.003 | 0.089 | 0.093 | 0.949 | 0.001 | 0.092 | 0.093 | 0.950
Cauchy tail | $\hat{\tau}_{\textnormal{str-md}}$ | -0.005 | 0.121 | / | / | 0.009 | 0.122 | / | / | 0.004 | 0.123 | / | /
| $\hat{\tau}_{\textnormal{wt-md}}$ | -0.004 | 0.134 | 0.137 | 0.951 | 0.009 | 0.134 | 0.136 | 0.956 | 0.000 | 0.137 | 0.137 | 0.949
| $\hat{\tau}_{\textnormal{str},1}$ | -0.003 | 0.119 | 0.119 | 0.954 | 0.006 | 0.119 | 0.120 | 0.950 | 0.001 | 0.126 | 0.119 | 0.932
| $\hat{\tau}_{\textnormal{str},2}$ | -0.003 | 0.119 | 0.119 | 0.954 | 0.006 | 0.119 | 0.120 | 0.952 | 0.001 | 0.126 | 0.119 | 0.933

  * Note: $\hat{\tau}_{\textnormal{str-md}}$, stratified difference-in-medians estimator; $\hat{\tau}_{\textnormal{wt-md}}$, difference-in-weighted-medians esti-
mator; $\hat{\tau}_{\textnormal{str},1}$, stratified transformed difference-
in-means estimator with $\hat{\tau}_{\textnormal{str-md}}$ as initial
estimator;
$\hat{\tau}_{\textnormal{str},2}$, stratified transformed difference-in-means
estimator with $\hat{\tau}_{\textnormal{wt-md}}$ as initial estimator; SR,
simple
randomization; STR, stratified randomization; MIN, Pocock and Simon’s
minimization; SD, standard
deviation; SE, standard error; CP, coverage probability.

###  C.4 Splitting techniques

We investigate the influence of sample splitting techniques within model 1 for
$n=1000$. The results for the other two models are similar so we omit them. We
assess two splitting approaches: (i) simple random splitting devoid of stratum
and treatment information, akin to the method in Athey et al., (2023) (SR
splitting), and (ii) splitting the sample within each stratum and treatment
arm, as outlined in Section 3 (CAR splitting). Table 15 shows the results.
Both splitting techniques exhibit comparable performance, with neither method
dominating the other.

Table 15: Simulation results under Model 1 for different splitting techniques

|  |  | SR | STR | MIN
---|---|---|---|---|---
|  |  | Bias | SD | SE | CP | Bias | SD | SE | CP | Bias | SD | SE | CP
$\tau$ = 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | 0.005 | 0.090 | 0.091 | 0.941 | -0.005 | 0.078 | 0.081 | 0.963 | -0.002 | 0.089 | / | /
|  | SR splitting | 0.000 | 0.090 | 0.091 | 0.949 | -0.002 | 0.077 | 0.081 | 0.966 | 0.003 | 0.085 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | 0.005 | 0.079 | 0.082 | 0.945 | -0.005 | 0.078 | 0.081 | 0.965 | -0.003 | 0.080 | 0.082 | 0.949
|  | SR splitting | 0.002 | 0.079 | 0.081 | 0.952 | -0.002 | 0.077 | 0.081 | 0.967 | 0.003 | 0.076 | 0.082 | 0.962
Laplace tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | 0.006 | 0.107 | 0.103 | 0.932 | -0.006 | 0.092 | 0.092 | 0.943 | 0.002 | 0.101 | / | /
|  | SR splitting | -0.003 | 0.107 | 0.102 | 0.933 | -0.001 | 0.091 | 0.092 | 0.957 | 0.000 | 0.100 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | 0.006 | 0.095 | 0.092 | 0.942 | -0.006 | 0.092 | 0.092 | 0.943 | 0.001 | 0.091 | 0.092 | 0.958
|  | SR splitting | -0.001 | 0.095 | 0.092 | 0.943 | -0.001 | 0.091 | 0.092 | 0.957 | 0.000 | 0.091 | 0.092 | 0.950
Cauchy tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | 0.002 | 0.128 | 0.128 | 0.940 | 0.003 | 0.118 | 0.118 | 0.946 | 0.006 | 0.124 | / | /
|  | SR splitting | -0.003 | 0.126 | 0.128 | 0.950 | -0.004 | 0.115 | 0.118 | 0.950 | -0.001 | 0.123 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | 0.001 | 0.118 | 0.118 | 0.949 | 0.003 | 0.118 | 0.118 | 0.945 | 0.005 | 0.116 | 0.118 | 0.956
|  | SR splitting | -0.001 | 0.117 | 0.118 | 0.948 | -0.004 | 0.115 | 0.118 | 0.948 | -0.002 | 0.116 | 0.118 | 0.962
$\tau$ = 1 |  |  |  |  |  |  |  |  |  |  |  |  |  |
Normal tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | 0.003 | 0.091 | 0.091 | 0.948 | 0.007 | 0.075 | 0.081 | 0.961 | -0.002 | 0.084 | / | /
|  | SR splitting | 0.002 | 0.089 | 0.091 | 0.945 | -0.001 | 0.078 | 0.081 | 0.960 | -0.004 | 0.086 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | 0.003 | 0.078 | 0.082 | 0.968 | 0.007 | 0.075 | 0.081 | 0.960 | -0.002 | 0.075 | 0.082 | 0.967
|  | SR splitting | 0.000 | 0.078 | 0.082 | 0.955 | -0.001 | 0.078 | 0.081 | 0.959 | -0.004 | 0.076 | 0.082 | 0.968
Laplace tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | 0.002 | 0.105 | 0.103 | 0.937 | 0.004 | 0.092 | 0.092 | 0.945 | -0.001 | 0.097 | / | /
|  | SR splitting | 0.002 | 0.102 | 0.103 | 0.950 | -0.002 | 0.089 | 0.092 | 0.959 | -0.003 | 0.096 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | 0.002 | 0.091 | 0.092 | 0.944 | 0.004 | 0.092 | 0.092 | 0.945 | -0.001 | 0.088 | 0.092 | 0.965
|  | SR splitting | 0.000 | 0.089 | 0.092 | 0.954 | -0.002 | 0.089 | 0.092 | 0.958 | -0.003 | 0.088 | 0.092 | 0.964
Cauchy tail | $\hat{\tau}_{\textnormal{tdim}}$ | CAR splitting | -0.002 | 0.129 | 0.128 | 0.946 | -0.007 | 0.122 | 0.118 | 0.933 | -0.006 | 0.126 | / | /
|  | SR splitting | 0.003 | 0.132 | 0.128 | 0.945 | 0.001 | 0.119 | 0.119 | 0.951 | -0.003 | 0.128 | / | /
| $\hat{\tau}_{\textnormal{str}}$ | CAR splitting | -0.002 | 0.119 | 0.118 | 0.944 | -0.007 | 0.122 | 0.118 | 0.931 | -0.006 | 0.120 | 0.118 | 0.942
|  | SR splitting | 0.001 | 0.122 | 0.119 | 0.941 | 0.001 | 0.119 | 0.119 | 0.951 | -0.004 | 0.120 | 0.118 | 0.950

  * Note: $\hat{\tau}_{\textnormal{tdim}}$, transformed difference-in-means estimator; $\hat{\tau}_{\textnormal{str}}$, stratified transformed difference-in-
means estimator; SR, simple randomization; STR, stratified randomization; MIN,
Pocock and
Simon’s minimization; SD, standard deviation; SE, standard error; CP, coverage
probability.

##  Appendix D Proof of main results

Let $S^{(n)}=\\{S_{1},\ldots,S_{n}\\}$, $A^{(n)}=\\{A_{1},\ldots,A_{n}\\}$,
$U^{(n)}=\\{I_{1\in\mathcal{L}_{1}},\ldots,I_{n\in\mathcal{L}_{1}}\\}$,
$Y^{(n)}=\\{Y_{1},\ldots,Y_{n}\\}$,
$D^{(\mathcal{L}_{j})}=\\{Y_{i},A_{i},S_{i}\\}_{i\in\mathcal{L}_{j}}$, for
$j=1,2$. Consider a measurable function $f(Y^{(n)},A^{(n)},S^{(n)},U^{(n)})$
such as $\hat{f}_{0(1)}^{{}^{\prime}}/\hat{f}_{0(1)}$. Given the independence
of the splitting indicators $I_{i\in\mathcal{L}_{1}}$’s from
$\\{Y^{(n)},A^{(n)},S^{(n)}\\}$, and that Assumptions 1–2 hold, using the
technique of Bugni et al., (2018) for deriving the law of large numbers and
central limit theory, we can conclude that conditional on
$(S^{(n)},A^{(n)},U^{(n)})$, $f(Y^{(n)},A^{(n)},S^{(n)},U^{(n)})$ has the same
distribution as $f(Y^{{}^{\prime}(n)},A^{(n)},S^{(n)},U^{(n)})$, where
$Y^{{}^{\prime}(n)}=\\{Y_{1}^{{}^{\prime}},\ldots,Y_{n}^{{}^{\prime}}\\}$ with
$Y_{i}{{}^{\prime}}=A_{i}Y_{i}{{}^{\prime}}(1)+(1-A_{i})Y_{i}^{{}^{\prime}}(0)$.
Furthermore, $\\{Y_{i}^{{}^{\prime}}\\}_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=a}$
are i.i.d. following the same distribution as $\\{Y_{i}(a)\mid S_{i}=k\\}$ and
are independent of $(S^{(n)},A^{(n)},U^{(n)})$. For simplicity, we will say
that conditional on $(S^{(n)},A^{(n)},U^{(n)})$,
$\\{Y_{i}\\}_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=a}$ are i.i.d. following the
same distribution as $\\{Y_{i}(a)\mid S_{i}=k\\}$. This technique will be
repeatedly used for convenience.

###  D.1 Proof of Theorem 1

We need to show that

$\sup_{y\in\mathbb{R}}\bigg{\\{}\bigg{|}\frac{\hat{f}_{0(j)}^{\prime}}{\hat{f}_{0(j)}}\bigg{|}(y)\bigg{\\}}=o_{P}(n^{1/2}),$
(2)
$\int\left(\frac{\hat{f}_{0(j)}^{\prime}}{\hat{f}_{0(j)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}(y)dF_{0}(y)=o_{P}(1),$
(3)

and for $m$ such that $m/n$ converges to a positive constant, with an i.i.d.
sample $\\{Y_{1}^{\prime},...,Y_{m}^{\prime}\\}$ drawn from the density
$f_{[k]0}$ that is independent of $\hat{f}_{0(j)}^{\prime}/\hat{f}_{0(j)}$,
and any ${\delta}_{n}=O_{P}\left(n^{-1/2}\right)$, as $n\rightarrow\infty$, we
have

$\frac{1}{m}\sum_{i=1}^{m}\left|\left(\frac{\hat{f}_{0(j)}^{\prime}}{\hat{f}_{0(j)}}\right)^{\prime}\right|\left(Y_{i}^{\prime}+{\delta}_{n}\right)=O_{P}(1),$
(4)
$\frac{1}{m}\sum_{i=1}^{m}\bigg{(}\frac{\hat{f}_{0(j)}^{\prime}}{\hat{f}_{0(j)}}\bigg{)}^{\prime}\left(Y_{i}^{\prime}+{\delta}_{n}\right)=\frac{1}{m}\sum_{i=1}^{m}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{\prime}\left(Y_{i}^{\prime}\right)+o_{P}(1).$
(5)

###### Proof.

By symmetry, we only need to prove the results for $j=1$. We will prove
equations (2)–(5) in steps (I)–(IV), respectively.

(I) We prove equation (2). By definition and Assumption 6, it is easy to see
that

$\sup_{y\in\mathbb{R}}\bigg{\\{}\bigg{|}\frac{\hat{f}_{0(j)}^{\prime}}{\hat{f}_{0(j)}}\bigg{|}(y)\bigg{\\}}\leq
c_{n}=o(\sigma_{n}^{-1})=o_{P}(n^{1/2}).$

(II) We prove equation (3). Let $\mathcal{D}$ be the truncation field defined
as:

$\mathcal{D}=\\{y\in\mathbb{R}:\hat{f}_{\sigma_{n}}(y)\geq d_{n},\ |y|\leq
e_{n},\ |\hat{f}_{\sigma_{n}}^{\prime}(y)|\leq c_{n}\hat{f}_{\sigma_{n}}(y),\
|\hat{f}_{\sigma_{n}}^{\prime\prime}(y)|\leq b_{n}\hat{f}_{\sigma_{n}}(y)\\}.$

Firstly, we show that

$\int_{\\{y:f_{0}(y)>0\\}}\left\\{\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}(y)I_{\mathcal{D}}(y)-\frac{f_{\sigma_{n}}^{\prime}}{f_{\sigma_{n}}}(y)\right\\}^{2}f_{\sigma_{n}}(y)dy\xrightarrow{p}0,$
(6)
$\int_{\\{y:f_{0}(y)>0\\}}\left\\{\frac{\hat{f}_{\sigma_{n}}^{\prime\prime}}{\hat{f}_{\sigma_{n}}}(y)I_{\mathcal{D}}(y)-\frac{f_{\sigma_{n}}^{\prime\prime}}{f_{\sigma_{n}}}(y)\right\\}^{2}f_{\sigma_{n}}(y)dy\xrightarrow{p}0,$
(7)

where $f_{\sigma_{n}}(y)$ is the convolution of $f_{0}$ and
$\phi_{\sigma_{n}}$, defined as
$f_{\sigma_{n}}(y)=\int_{-\infty}^{\infty}f_{0}(x)\phi_{\sigma_{n}}(y-x)dx$.
Bickel, (1982) proved similar results for i.i.d. samples. We will generalize
it to the samples under covariate-adaptive randomization with considerations
of dependent treatment assignments.

We partition the integral in equation (6) into two components: one over
$\mathcal{D}$ and the other over $\mathcal{D}^{C}$, the complement of
$\mathcal{D}$. Let’s denote these two components as $T_{1}$ and $T_{2}$. Note
that $T_{1}\geq 0$ and $T_{2}\geq 0$.

By the dominated convergence theorem, it suffices for $T_{1}\xrightarrow{p}0$
to show that $P(T_{1}>\epsilon\mid S^{(n)},A^{(n)},U^{(n)})\xrightarrow{p}0$
for any $\epsilon>0$. By the Markov inequality,

$\displaystyle P(T_{1}>\epsilon\mid
S^{(n)},A^{(n)},U^{(n)})\leq\epsilon^{-1}E\\{T_{1}\mid
S^{(n)},A^{(n)},U^{(n)}\\}.$

On $\mathcal{D}$, we have

$\displaystyle E\\{T_{1}\mid S^{(n)},A^{(n)},U^{(n)}\\}\leq
2\int_{\mathcal{D}}f_{\sigma_{n}}^{-1}(y)E\Big{[}\big{\\{}\hat{f}_{\sigma_{n}}^{\prime}(y)-f_{\sigma_{n}}^{\prime}(y)\big{\\}}^{2}\mid
S^{(n)},A^{(n)},U^{(n)}\Big{]}dy$
$\displaystyle\quad\quad\quad+2\int_{\mathcal{D}}c_{n}^{2}f_{\sigma_{n}}^{-1}(y)E\Big{[}\big{\\{}\hat{f}_{\sigma_{n}}(y)-f_{\sigma_{n}}(y)\big{\\}}^{2}\mid
S^{(n)},A^{(n)},U^{(n)}\Big{]}dy.$ (8)

Let the superscript $(l)$ denote the $l$-th order derivative with respect to
$y$. Define
$f_{\sigma_{n}[k]}(y)=\int_{-\infty}^{\infty}f_{[k]0}(x)\phi_{\sigma_{n}}(y-x)dx.$
Then, we have

$\displaystyle\operatorname{Var}\\{\hat{f}_{\sigma_{n}}^{(l)}(y)\mid
S^{(n)},A^{(n)},U^{(n)}\\}$ (9) $\displaystyle=$
$\displaystyle\operatorname{Var}\left(\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0(1)}}\cdot\frac{1}{n_{[k]0(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\phi^{(l)}_{i}\mid
S^{(n)},A^{(n)},U^{(n)}\right)$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{K}\left(\frac{n_{[k]0(1)}}{n_{0(1)}}\right)^{2}\operatorname{Var}\left(\frac{1}{n_{[k]0(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\phi^{(l)}_{i}\mid
S^{(n)},A^{(n)},U^{(n)}\right)$ $\displaystyle\leq$
$\displaystyle\sum_{k=1}^{K}\left(\frac{n_{[k]0(1)}}{n_{0(1)}}\right)^{2}\cdot\kappa_{l}^{SR}\sigma_{n}^{-2l-1}n_{[k]0(1)}^{-1}f_{\sigma_{n}[k]}(y)$
$\displaystyle=$
$\displaystyle\kappa_{l}^{SR}\sigma_{n}^{-2l-1}\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0(1)}^{2}}f_{\sigma_{n}[k]}(y),\quad
l=0,1,2,$

where $\phi_{i}^{(l)}=\phi_{\sigma_{n}}^{(l)}(y-Y_{i})$. The second equality
holds because, conditional on $S^{(n)},A^{(n)}$ and $U^{(n)}$, the
$\phi_{i}^{(l)}$’s from different strata are independent (refer to the
discussion at the start of Section D). The inequality in the third line
results from applying the results in the proof of Bickel, (1982, Lemma 6.1) to
each stratum.

By a similar argument, we have

$\displaystyle E\\{\hat{f}_{\sigma_{n}}^{(l)}(y)\mid
S^{(n)},A^{(n)},U^{(n)}\\}$
$\displaystyle=\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0(1)}}E\left(\frac{1}{n_{[k]0(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\phi^{(l)}_{i}\mid
S^{(n)},A^{(n)},U^{(n)}\right)$
$\displaystyle=\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0(1)}}f_{\sigma_{n}[k]}^{(l)}(y).$

Since
$f_{\sigma_{n}}^{(l)}(y)=\sum_{k=1}^{K}p_{[k]}f_{\sigma_{n}[k]}^{(l)}(y)$,
then

$\Big{|}E\\{\hat{f}_{\sigma_{n}}^{(l)}(y)\mid
S^{(n)},A^{(n)},U^{(n)}\\}-f_{\sigma_{n}}^{(l)}(y)\Big{|}\leq\frac{\Delta_{n}}{\min_{k=1,\ldots,K}p_{[k]}}f_{\sigma_{n}}^{(l)}(y).$

Let $\Pi$ denote the ratio of a circle’s circumference to its diameter. A
straightforward calculation yields

$\displaystyle f_{\sigma_{n}[k]}(y)=$
$\displaystyle\int_{-\infty}^{\infty}f_{[k]0}(x)\phi_{\sigma_{n}}(y-x)dx\leq\int_{-\infty}^{\infty}(\sqrt{2\Pi}\sigma_{n})^{-1}f_{[k]0}(x)dx=(\sqrt{2\Pi}\sigma_{n})^{-1},$
$\displaystyle\big{\\{}f_{\sigma_{n}}(y)\big{\\}}^{2}=$ $\displaystyle
f_{\sigma_{n}}(y)\sum_{k=1}^{K}p_{[k]}f_{\sigma_{n}[k]}(y)\leq(\sqrt{2\Pi}\sigma_{n})^{-1}f_{\sigma_{n}}(y),$
$\displaystyle\big{\\{}f_{\sigma_{n}[k]}^{\prime}(y)\big{\\}}^{2}=$
$\displaystyle\bigg{(}\int_{-\infty}^{\infty}f_{[k]0}(x)\phi_{\sigma_{n}}^{\prime}(y-x)dx\bigg{)}^{2}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\phi_{\sigma_{n}}^{\prime}(y-u)\phi_{\sigma_{n}}^{\prime}(y-v)$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\sigma_{n}^{-4}(y-u)(y-v)\phi_{\sigma_{n}}(y-u)\phi_{\sigma_{n}}(y-v)$
$\displaystyle\leq$
$\displaystyle\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\frac{(y-u)^{2}+(y-v)^{2}}{2}\phi_{\sigma_{n}}(y-u)\phi_{\sigma_{n}}(y-v)$
$\displaystyle=$
$\displaystyle\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)(y-u)^{2}\phi_{\sigma_{n}}(y-u)du\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(y-v)f_{[k]0}(v)dv$
$\displaystyle\leq$
$\displaystyle{\sigma_{n}^{-2}}\int_{-\infty}^{\infty}2e^{-1}(\sqrt{2\Pi}\sigma_{n})^{-1}f_{[k]0}(u)du\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(y-v)f_{[k]0}(v)dv$
$\displaystyle=$ $\displaystyle
2(\sqrt{2\Pi}e)^{-1}{\sigma_{n}^{-3}}f_{\sigma_{n}[k]}(y),$
$\displaystyle\big{\\{}f_{\sigma_{n}[k]}^{\prime\prime}(y)\big{\\}}^{2}=$
$\displaystyle\bigg{(}\int_{-\infty}^{\infty}f_{[k]0}(x)\phi_{\sigma_{n}}^{\prime\prime}(y-x)dx\bigg{)}^{2}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\phi_{\sigma_{n}}^{\prime\prime}(y-u)\phi_{\sigma_{n}}^{\prime\prime}(y-v)$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv$
$\displaystyle\cdot\sigma_{n}^{-4}\bigg{\\{}\frac{(y-u)^{2}}{\sigma_{n}^{2}}-1\bigg{\\}}\bigg{\\{}\frac{(y-v)^{2}}{\sigma_{n}^{2}}-1\bigg{\\}}\phi_{\sigma_{n}}(y-u)\phi_{\sigma_{n}}(y-v)$
$\displaystyle\leq$
$\displaystyle\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\frac{(y-u)^{4}+(y-v)^{4}}{2\sigma_{n}^{4}}\phi_{\sigma_{n}}(y-u)\phi_{\sigma_{n}}(y-v)$
$\displaystyle+\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)du\int_{-\infty}^{\infty}f_{[k]0}(v)dv\cdot\phi_{\sigma_{n}}(y-u)\phi_{\sigma_{n}}(y-v)$
$\displaystyle=$
$\displaystyle\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)\cdot(y-u)^{4}/\sigma_{n}^{4}\cdot\phi_{\sigma_{n}}(y-u)du\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(y-v)f_{[k]0}(v)dv$
$\displaystyle+\sigma_{n}^{-4}\int_{-\infty}^{\infty}f_{[k]0}(u)\phi_{\sigma_{n}}(y-u)du\int_{-\infty}^{\infty}f_{[k]0}(v)\phi_{\sigma_{n}}(y-v)dv$
$\displaystyle\leq$
$\displaystyle{\sigma_{n}^{-4}}\int_{-\infty}^{\infty}16e^{-2}(\sqrt{2\Pi}\sigma_{n})^{-1}f_{[k]0}(u)du\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(y-v)f_{[k]0}(v)dv+\sigma_{n}^{-4}f_{\sigma_{n}[k]}^{2}(y)$
$\displaystyle=$ $\displaystyle
16(\sqrt{2\Pi}e^{2})^{-1}{\sigma_{n}^{-5}}f_{\sigma_{n}[k]}(y)+(\sqrt{2\Pi})^{-1}\sigma_{n}^{-5}f_{\sigma_{n}[k]}(y).$

Then, applying the Cauchy–Schwarz inequality and noting that
$\sum_{k=1}^{K}p_{[k]}=1$,

$\displaystyle\big{\\{}f_{\sigma_{n}}^{\prime}(y)\big{\\}}^{2}$
$\displaystyle=$
$\displaystyle\Big{\\{}\sum_{k=1}^{K}p_{[k]}f_{\sigma_{n}[k]}^{\prime}(y)\Big{\\}}^{2}\leq\sum_{k=1}^{K}p_{[k]}\big{\\{}f_{\sigma_{n}[k]}^{\prime}(y)\big{\\}}^{2}\leq
2(\sqrt{2\Pi}e)^{-1}{\sigma_{n}^{-3}}f_{\sigma_{n}}(y),$
$\displaystyle\big{\\{}f_{\sigma_{n}}^{\prime\prime}(y)\big{\\}}^{2}$
$\displaystyle=$
$\displaystyle\Big{\\{}\sum_{k=1}^{K}p_{[k]}f_{\sigma_{n}[k]}^{\prime\prime}(y)\Big{\\}}^{2}\leq\sum_{k=1}^{K}p_{[k]}\big{\\{}f_{\sigma_{n}[k]}^{\prime\prime}(y)\big{\\}}^{2}\leq(16+e^{2})(\sqrt{2\Pi}e^{2})^{-1}{\sigma_{n}^{-5}}f_{\sigma_{n}}(y).$

Therefore,

$\displaystyle\Big{|}E\\{\hat{f}_{\sigma_{n}}^{(l)}(y)\mid
S^{(n)},A^{(n)},U^{(n)}\\}-f_{\sigma_{n}}^{(l)}(y)\Big{|}^{2}\leq\frac{\Delta_{n}^{2}}{\min_{k=1,\ldots,K}p_{[k]}^{2}}\frac{4}{\sqrt{2\Pi}}\sigma_{n}^{-2l-1}f_{\sigma_{n}}(y),\
l=0,1,2.$ (10)

Let
$C_{l}=\max\\{\kappa_{l}^{SR},4/\sqrt{2\Pi}\\}/\min_{k=1,\ldots,K}p_{[k]}^{2}$
and $\Lambda_{n}=\max_{k=1,\ldots,K}n_{[k]0(1)}/n_{0(1)}$. Then $C_{l}$ is a
constant independent of $n$ and
$\Lambda_{n}\xrightarrow{p}\max_{k=1,\ldots,K}p_{[k]}$. Combining (9) and
(10), we have

$E\Big{[}\big{\\{}\hat{f}_{\sigma_{n}}^{(l)}(y)-f_{\sigma_{n}}^{(l)}(y)\big{\\}}^{2}\mid
S^{(n)},A^{(n)},U^{(n)}\Big{]}\leq
C_{l}\Big{\\{}\frac{\Lambda_{n}}{n_{0(1)}}+\Delta_{n}^{2}\Big{\\}}\sigma_{n}^{-2l-1}f_{\sigma_{n}}(y).$

Substituting it into equation (D.1), we have

$E\\{T_{1}\mid S^{(n)},A^{(n)},U^{(n)}\\}\leq
2C_{1}\Big{\\{}\frac{\Lambda_{n}}{n_{0(1)}}+\Delta_{n}^{2}\Big{\\}}\frac{e_{n}}{\sigma_{n}^{3}}+2C_{0}\Big{\\{}\frac{\Lambda_{n}}{n_{0(1)}}+\Delta_{n}^{2}\Big{\\}}\frac{c_{n}^{2}e_{n}}{\sigma_{n}}.$

By Assumption 6, we have $E\\{T_{1}\mid
S^{(n)},A^{(n)},U^{(n)}\\}\xrightarrow{p}0$. Therefore
$T_{1}\xrightarrow{p}0$.

Similarly, let $T_{1}^{\prime}$ and $T_{2}^{\prime}$ denote the integrals in
equation (7) over $\mathcal{D}$ and $\mathcal{D}^{C}$, respectively. Then, we
have

$E\\{T_{1}^{\prime}\mid S^{(n)},A^{(n)},U^{(n)}\\}\leq
2C_{2}\Big{\\{}\frac{\Lambda_{n}}{n_{0(1)}}+\Delta_{n}^{2}\Big{\\}}\frac{e_{n}}{\sigma_{n}^{5}}+2C_{0}\Big{\\{}\frac{\Lambda_{n}}{n_{0(1)}}+\Delta_{n}^{2}\Big{\\}}\frac{b_{n}^{2}e_{n}}{\sigma_{n}}.$

By Assumption 6, we have $E\\{T_{1}^{\prime}\mid
S^{(n)},A^{(n)},U^{(n)}\\}\xrightarrow{p}0$. Therefore
$T_{1}^{\prime}\xrightarrow{p}0$.

On $\mathcal{D}^{C}$, we have

$\displaystyle E(T_{2})\leq$ $\displaystyle\int\frac{f_{\sigma_{n}}^{\prime
2}}{f_{\sigma_{n}}}(y)\bigg{[}P\left\\{\left|\hat{f}_{\sigma_{n}}^{\prime}(y)\right|>c_{n}\hat{f}_{\sigma_{n}}(y)\right\\}+P\left\\{\left|\hat{f}_{\sigma_{n}}^{\prime\prime}(y)\right|>b_{n}\hat{f}_{\sigma_{n}}(y)\right\\}$
$\displaystyle+P\left\\{\hat{f}_{\sigma_{n}}(y)<d_{n},f_{0}(y)>0\right\\}+I\left(|y|>e_{n}\right)\bigg{]}dy,$
$\displaystyle E(T_{2}^{\prime})\leq$
$\displaystyle\int\frac{f_{\sigma_{n}}^{\prime\prime
2}}{f_{\sigma_{n}}}(y)\bigg{[}P\left\\{\left|\hat{f}_{\sigma_{n}}^{\prime}(y)\right|>c_{n}\hat{f}_{\sigma_{n}}(y)\right\\}+P\left\\{\left|\hat{f}_{\sigma_{n}}^{\prime\prime}(y)\right|>b_{n}\hat{f}_{\sigma_{n}}(y)\right\\}$
$\displaystyle+P\left\\{\hat{f}_{\sigma_{n}}(y)<d_{n},f_{0}(y)>0\right\\}+I\left(|y|>e_{n}\right)\bigg{]}dy.$

Both of them converge to zero if

$\displaystyle\hat{f}_{\sigma_{n}}(y)\xrightarrow{p}f_{0}(y),$
$\displaystyle\forall y,\quad\text{if}\quad n\sigma_{n}\rightarrow\infty,$
(11)
$\displaystyle\hat{f}_{\sigma_{n}}^{\prime}(y)\xrightarrow{p}f_{0}^{\prime}(y),$
$\displaystyle a.e.\quad y,\quad\text{if}\quad
n\sigma_{n}^{3}\rightarrow\infty,$ (12)
$\displaystyle\hat{f}_{\sigma_{n}}^{\prime\prime}(y)\xrightarrow{p}f_{0}^{\prime\prime}(y),$
$\displaystyle a.e.\quad y,\quad\text{if}\quad
n\sigma_{n}^{5}\rightarrow\infty,$ (13)
$\displaystyle\int\frac{f_{\sigma_{n}}^{\prime 2}}{f_{\sigma_{n}}}(y)dy$
$\displaystyle\leq$ $\displaystyle\int\frac{f_{0}^{\prime
2}}{f_{0}}(y)dy,\quad\forall n,$ (14)
$\displaystyle\int\frac{f_{\sigma_{n}}^{\prime\prime 2}}{f_{\sigma_{n}}}(y)dy$
$\displaystyle\leq$ $\displaystyle\int\frac{f_{0}^{\prime\prime
2}}{f_{0}}(y)dy,\quad\forall n;$ (15)

see Bickel, (1982).

For deriving (11), conditional on $S^{(n)},A^{(n)}$ and $U^{(n)}$, we have

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\phi_{\sigma}(y-Y_{i})\xrightarrow{p}f_{[k]0}(y),\quad
n\sigma_{n}\rightarrow\infty.$

Thus, conditional on $S^{(n)},A^{(n)}$ and $U^{(n)}$,
$\hat{f}_{\sigma_{n}}(y)-\sum_{k=1}^{K}(n_{[k]0(1)}/n_{0(1)})f_{[k]0}(y)\xrightarrow{p}0$,
implying unconditional convergence in probability. Moreover, since
$p_{n[k](1)}=n_{[k]0(1)}/n_{0(1)}\xrightarrow{p}p_{[k]}$, we have
$\hat{f}_{\sigma_{n}}(y)\xrightarrow{p}f_{0}(y)$. Similarly, we can derive
(12) and (13). Since (14) is not related to $Y_{i}$, it holds by the proof in
Bickel, (1982). Finally, applying the Cauchy–Schwarz inequality gives

$\frac{f_{\sigma_{n}}^{\prime\prime
2}(y)}{f_{\sigma_{n}}(y)}=\frac{\big{\\{}\int
f_{0}^{\prime\prime}(y-x)\phi_{\sigma_{n}}(x)dx\big{\\}}^{2}}{\int
f_{0}(y-x)\phi_{\sigma_{n}}(x)dx}\leq\int\frac{f_{0}^{\prime\prime
2}(y-x)}{f_{0}(y-x)}\phi_{\sigma_{n}}(x)dx,\quad\forall y.$

Integrating the above inequality with respect to $y$, we have

$\displaystyle\int_{-\infty}^{\infty}\frac{f_{\sigma_{n}}^{\prime\prime
2}(y)}{f_{\sigma_{n}}(y)}dy$ $\displaystyle\leq$
$\displaystyle\int_{-\infty}^{\infty}\frac{f_{0}^{\prime\prime
2}(y-x)}{f_{0}(y-x)}dy\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(x)dx$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\frac{f_{0}^{\prime\prime
2}(z)}{f_{0}(z)}dz\int_{-\infty}^{\infty}\phi_{\sigma_{n}}(x)dx=\int_{-\infty}^{\infty}\frac{f_{0}^{\prime\prime
2}(z)}{f_{0}(z)}dz,$

that is, (15) holds.

Therefore, we obtain $E(T_{2})\to 0$ and $E(T_{2}^{\prime})\to 0$, implying
$T_{2}\xrightarrow{p}0$ and $T_{2}^{\prime}\xrightarrow{p}0$. Consequently,
equations (6) and (7) hold.

Moreover, by (11)–(15), we have, as $n\rightarrow\infty$,

$\int_{\\{y:f_{0}(y)>0\\}}\left\\{\frac{f_{\sigma_{n}}^{\prime}}{\sqrt{f_{\sigma_{n}}}}(y)-\frac{f_{0}^{\prime}}{\sqrt{f_{0}}}(y)\right\\}^{2}dy\rightarrow
0,$ (16)

and

$\int_{\\{y:f_{0}(y)>0\\}}\left\\{\frac{f_{\sigma_{n}}^{\prime\prime}}{\sqrt{f_{\sigma_{n}}}}(y)-\frac{f_{0}^{\prime\prime}}{\sqrt{f_{0}}}(y)\right\\}^{2}dy\rightarrow
0.$

By Lemmas 6.3 in Bickel, (1982), as $n\rightarrow\infty$, the following term
is bounded,

$\int_{\\{y:f_{0}(y)>0\\}}\sigma_{n}^{-2}\left(\sqrt{f_{\sigma_{n}}(y)}-\sqrt{f_{0}(y)}\right)^{2}dy.$

Since $|\hat{f}_{\sigma_{n}}^{\prime}/\hat{f}_{\sigma_{n}}|(y)\leq
c_{n}=o(\sigma_{n}^{-1})$,
$|\hat{f}_{\sigma_{n}}^{\prime\prime}/\hat{f}_{\sigma_{n}}|(y)\leq
b_{n}=o(\sigma_{n}^{-1})$, then we have

$\int_{\\{y:f_{0}(y)>0\\}}\bigg{(}\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}\bigg{)}^{2}(y)\left(\sqrt{f_{\sigma_{n}}(y)}-\sqrt{f_{0}(y)}\right)^{2}dy\xrightarrow{p}0,$
(17)
$\int_{\\{y:f_{0}(y)>0\\}}\bigg{(}\frac{\hat{f}_{\sigma_{n}}^{\prime\prime}}{\hat{f}_{\sigma_{n}}}\bigg{)}^{2}(y)\left(\sqrt{f_{\sigma_{n}}(y)}-\sqrt{f_{0}(y)}\right)^{2}dy\xrightarrow{p}0.$

Finally, we bound the $L_{2}$-loss of $g_{n}(y)$ by

$\displaystyle\int\left\\{\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}(y)I_{\mathcal{D}}(y)-\frac{f_{0}^{\prime}}{f_{0}}(y)\right\\}^{2}f_{0}(y)dy$
$\displaystyle\leq
3\left[\int_{\\{y:f_{0}(y)>0\\}}\left\\{\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}(y)-\bigg{(}\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}\bigg{)}\bigg{(}\frac{f_{\sigma_{n}}}{f_{0}}\bigg{)}^{1/2}(y)\right\\}^{2}I_{\mathcal{D}}(y)f_{0}(y)dy\right.$
$\displaystyle\quad+\int_{\\{y:f_{0}(y)>0\\}}\left\\{\bigg{(}\frac{\hat{f}_{\sigma_{n}}^{\prime}}{\hat{f}_{\sigma_{n}}}\bigg{)}\bigg{(}\frac{f_{\sigma_{n}}}{f_{0}}\bigg{)}^{1/2}(y)I_{\mathcal{D}}(y)-\bigg{(}\frac{f_{\sigma_{n}}^{\prime}}{f_{\sigma_{n}}}\bigg{)}\bigg{(}\frac{f_{\sigma_{n}}}{f_{0}}\bigg{)}^{1/2}(y)\right\\}^{2}f_{0}(y)dy$
$\displaystyle\left.\quad+\int_{\\{y:f_{0}(y)>0\\}}\left\\{\bigg{(}\frac{f_{\sigma_{n}}^{\prime}}{f_{\sigma_{n}}}\bigg{)}\bigg{(}\frac{f_{\sigma_{n}}}{f_{0}}\bigg{)}^{1/2}(y)-\frac{f_{0}^{\prime}}{f_{0}}(y)\right\\}^{2}f_{0}(y)dy\right],$

where the first term tends to zero in probability by (17), the second term
tends to zero in probability by (6), and the last term tends to zero in
probability by (16). Thus, equation (3) holds.

Similarly, we can show that

$\int\left\\{\frac{\hat{f}_{\sigma_{n}}^{\prime\prime}}{\hat{f}_{\sigma_{n}}}(y)I_{\mathcal{D}}(y)-\frac{f_{0}^{\prime\prime}}{f_{0}}(y)\right\\}^{2}f_{0}(y)dy=o_{P}(1).$

(III) We prove equation (4). We slightly abuse notation by using
$(\hat{f}_{0(1)}^{\prime\prime}/\hat{f}_{0(1)})(y)$ to denote
$(\hat{f}_{\sigma_{n}}^{\prime\prime}/\hat{f}_{\sigma_{n}})(y)\cdot
I_{\mathcal{D}}(y)$. Leveraging the Cauchy–Schwarz inequality, we have

$\int\bigg{|}\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}-\frac{f_{0}^{\prime\prime}}{f_{0}}\bigg{|}(y)dF_{0}(y)=o_{P}(1).$

For a specific stratum $k$, since
$f_{[k]0}(y)\leq(\min_{k=1,...,K}p_{[k]})^{-1}f_{0}(y)$, then we have

$\int\bigg{|}\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}-\frac{f_{0}^{\prime\prime}}{f_{0}}\bigg{|}(y)dF_{[k]0}(y)=o_{P}(1).$
(18)

Next, by Cauchy–Schwarz inequality, we have

$\displaystyle\left(\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\bigg{|}(y)dF_{0}(y)\right)^{2}$
$\displaystyle\leq\int\left(\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}(y)dF_{0}(y)\cdot\int\left(\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}+\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}(y)dF_{0}(y).$

The first integration on the right side of the above inequality is $o_{P}(1)$
by equation (3). Moreover, the second integration is $O_{P}(1)$ due to
equation (3) and

$I(f_{0})=\int\left(\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}(y)dF_{0}(y)<\infty.$

Thus,

$\left(\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\bigg{|}(y)dF_{0}(y)\right)^{2}=o_{P}(1).$

Since $f_{[k]0}(y)\leq(\min_{k=1,...,K}p_{[k]})^{-1}f_{0}(y)$, then we have

$\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\bigg{|}(y)dF_{[k]0}(y)=o_{P}(1).$
(19)

Combining (18) and (19), by the triangle inequality, we have

$\displaystyle\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}(y)dF_{[k]0}(y)=$
$\displaystyle\int\bigg{|}\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}-\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}\bigg{|}(y)dF_{[k]0}(y)$
$\displaystyle\leq$
$\displaystyle\int\bigg{|}\frac{f_{0}^{\prime\prime}}{f_{0}}\bigg{|}(y)dF_{[k]0}(y)+\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}(y)dF_{[k]0}(y)+o_{P}(1).$

Noting that

$\sum_{k=1}^{K}p_{[k]}\cdot\int\bigg{|}\frac{f_{0}^{\prime\prime}}{f_{0}}\bigg{|}(y)dF_{[k]0}(y)=\int|f_{0}^{\prime\prime}(y)|dy<\infty,$
$\sum_{k=1}^{K}p_{[k]}\cdot\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}(y)dF_{[k]0}(y)=\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}(y)dF_{0}(y)=I(f_{0})<\infty,$

we have

$\int|(\hat{f}_{0(1)}^{\prime}/\hat{f}_{0(1)})^{\prime}|(y)dF_{[k]0}(y)\leq(\min_{k=1,...,K}p_{[k]})^{-1}\big{\\{}\int|f_{0}^{\prime\prime}(y)|dy+I(f_{0})\big{\\}}+o_{P}(1).$

By the argument in Bickel, (1982), we establish (4) by substituting
${\delta}_{n}$ with $\tilde{\delta}_{n}=t_{n}n^{-1/2}\to 0$, where $t_{n}$ is
an arbitrary convergent deterministic sequence.

By Assumption 7, we have

$\displaystyle\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}(y+\tilde{\delta}_{n})f_{[k]0}(y)dy=$
$\displaystyle\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}(y)f_{[k]0}(y-\tilde{\delta}_{n})dy$
(20) $\displaystyle=$
$\displaystyle\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}(y)\cdot
f_{[k]0}(y)\big{\\{}1+o(1)\big{\\}}dy$ $\displaystyle\leq$
$\displaystyle\big{(}\min_{k=1,...,K}p_{[k]}\big{)}^{-1}\cdot\bigg{\\{}\int|f_{0}^{\prime\prime}(y)|dy+I(f_{0})\bigg{\\}}+o_{P}(1).$

If $\\{Y_{i}^{\prime}\\}_{i=1,...,m}$ are i.i.d. samples drawn from the
density $f_{[k]0}$ and are independent of
$\hat{f}_{0(1)}^{\prime}/\hat{f}_{0(1)}$, by the weak law of large numbers, we
have

$\displaystyle\frac{1}{m}\sum_{i=1}^{m}\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}\left(Y_{i}^{\prime}+\tilde{\delta}_{n}\right)\mid\hat{f}_{0(1)}$
$\displaystyle=\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}(y+\tilde{\delta}_{n})f_{[k]0}(y)dy+o_{P}(1)$
$\displaystyle\leq\big{(}\min_{k=1,...,K}p_{[k]}\big{)}^{-1}\cdot\bigg{\\{}\int|f_{0}^{\prime\prime}(y)|dy+I(f_{0})\bigg{\\}}+o_{P}(1).$

Therefore,

$\frac{1}{m}\sum_{i=1}^{m}\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\bigg{|}\left(Y_{i}^{\prime}+\tilde{\delta}_{n}\right)=O_{P}(1).$

(IV) We prove equation (5). Similar to (III), if
$\\{Y_{i}^{\prime}\\}_{i=1,...,m}$ are i.i.d. samples drawn from the density
$f_{[k]0}$ and are independent of $\hat{f}_{0(1)}^{\prime}/\hat{f}_{0(1)}$, by
the weak law of large numbers, we have

$\displaystyle\frac{1}{m}\sum_{i=1}^{m}\bigg{\\{}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}(Y_{i}^{\prime}+\tilde{\delta}_{n})-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{\prime}(Y_{i}^{\prime})\bigg{\\}}\mid\hat{f}_{0(1)}$
$\displaystyle=$
$\displaystyle\int\bigg{\\{}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}(y+\tilde{\delta}_{n})-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{\prime}(y)\bigg{\\}}f_{[k]0}(y)dy+o_{P}(1)$
$\displaystyle=$
$\displaystyle\int\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}(y)f_{[k]0}(y-\tilde{\delta}_{n})dy-\int\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}(y)f_{[k]0}(y-\tilde{\delta}_{n})dy-\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{\prime}(y)f_{[k]0}(y)dy+o_{P}(1)$
$\displaystyle\leq$
$\displaystyle\int\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}(y)f_{[k]0}(y)dy-\int\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}(y)f_{[k]0}(y)dy+o(1)\cdot\bigg{\\{}\int\bigg{|}\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}\bigg{|}(y)f_{[k]0}(y)dy$
$\displaystyle+\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}\bigg{|}(y)f_{[k]0}(y)dy\bigg{\\}}-\bigg{\\{}\int\frac{f_{0}^{\prime\prime}}{f_{0}}(y)f_{[k]0}(y)dy-\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}(y)f_{[k]0}(y)dy\bigg{\\}}+o_{P}(1)$
$\displaystyle\leq$
$\displaystyle\int\bigg{|}\frac{\hat{f}_{0(1)}^{\prime\prime}}{\hat{f}_{0(1)}}-\frac{f_{0}^{\prime\prime}}{f_{0}}\bigg{|}(y)f_{[k]0}(y)dy+\int\bigg{|}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\bigg{|}(y)f_{[k]0}(y)dy+o_{P}(1)$
$\displaystyle=$ $\displaystyle o_{P}(1).$

Therefore,

$\frac{1}{m}\sum_{i=1}^{m}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{\prime}\left(Y_{i}^{\prime}+\tilde{\delta}_{n}\right)=\frac{1}{m}\sum_{i=1}^{m}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{\prime}\left(Y_{i}^{\prime}\right)+o_{P}(1).$

∎

###  D.2 Proof of Proposition 1

###### Proof.

Proposition 1 directly follows from Zhang and Zheng, (2020, Theorem 3.1 and
Theorem 3.2), and therefore, we omit its proof. ∎

###  D.3 Proof of Proposition 2

###### Proof.

Let $\mathcal{Y}$ be a set of real numbers or random variables, and
$\textnormal{median}(\mathcal{Y})$ denotes the median of its elements.
Condition on $(S^{(n)},A^{(n)})$, and employing the results under simple
randomization (Athey et al.,; 2023, Theorem A.1), we have

$\sqrt{n_{[k]}}\big{\\{}\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=1})-\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=0})-\tau_{[k]}\big{\\}}\mid(S^{(n)},A^{(n)})\xrightarrow{d}N(0,\sigma_{[k],\textnormal{median}}^{2}),$

where
$\sigma_{[k],\textnormal{median}}^{2}=\\{4\pi(1-\pi)f_{[k]0}^{2}(F_{[k]0}^{-1}(1/2))\\}^{-1}$.

As $f_{[k]0}(F_{[k]0}^{-1}(1/2))>0$ and $n_{[k]}/n\xrightarrow{P}p_{[k]}$, it
follows that

$\sqrt{n}\\{\textnormal{median}(\\{Y_{i}\\}_{S_{i}=k,A_{i}=1})-\textnormal{median}(\\{Y_{i}\\}_{S_{i}=k,A_{i}=0})-\tau_{[k]}\\}=O_{P}(1).$

By weighting the above term with the stratum proportion $p_{n[k]}$ and and
summing over $k$, we have

$\sqrt{n}\bigg{[}\sum_{k=1}^{K}p_{n[k]}\left\\{\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=1})-\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=0})\right\\}-\sum_{k=1}^{K}p_{n[k]}\tau_{[k]}\bigg{]}=O_{P}(1).$

Given $\sqrt{n}(p_{n[k]}-p_{[k]})=O_{P}(1)$ and
$\sum_{k=1}^{K}p_{[k]}\tau_{[k]}=\tau$, we have

$\sqrt{n}\bigg{[}\sum_{k=1}^{K}p_{n[k]}\left\\{\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=1})-\textnormal{median}(\left\\{Y_{i}\right\\}_{S_{i}=k,A_{i}=0})\right\\}-\tau\bigg{]}=O_{P}(1).$

Therefore, $\sqrt{n}(\hat{\tau}_{\textnormal{str-md}}-\tau)=O_{P}(1).$

∎

###  D.4 Proof of Theorem 2

Before proving the result, we introduce two useful lemmas whose proofs are
provided in Section E. Note that Lemma 1 below relies on the less stringent
Assumption 3 rather than the more restrictive Assumption 4.

######  Lemma 1.

Under Assumptions 1–3 and 5, $\hat{I}(f_{0})=I\left(f_{0}\right)+o_{P}(1).$

######  Remark 8.

Lemma 1 requires weaker conditions compared to Theorem 2. Even in cases where
the designs do not satisfy Assumption 4 (e.g., Pocock and Simon’s
minimization), the consistency of $\hat{I}(f_{0})$ can still be preserved as
long as the design satisfies Assumption 3.

######  Lemma 2.

Under Assumptions 1, 2, 4 and 5, we have

$\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)=\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)+o_{P}(1),$
$\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{2}}\psi_{\hat{f}_{0(1)}}(A_{i},Y_{i};\tau)=\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)+o_{P}(1).$

###### Proof of Theorem 2.

By definition and the first order Taylor expansion, we have

$\displaystyle\hat{\tau}_{\textnormal{tdim}}-\tau$
$\displaystyle=\tilde{\tau}-\tau+\frac{1}{n}\left(\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tilde{\tau}\right)+\sum_{i\in\mathcal{L}_{2}}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\tilde{\tau}\right)\right)$
$\displaystyle=\tilde{\tau}-\tau+\frac{1}{n}\left(\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tau\right)+\sum_{i\in\mathcal{L}_{2}}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\tau\right)\right)$
$\displaystyle\quad+\frac{1}{n}(\tilde{\tau}-\tau)\left(\sum_{i\in\mathcal{L}_{1}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{i\in\mathcal{L}_{2}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\bar{\tau}\right)\right),$

where $\bar{\tau}$ is between $\tilde{\tau}$ and $\tau$. By Lemma 2, we have

$\displaystyle\hat{\tau}_{\textnormal{tdim}}-\tau$ $\displaystyle=$
$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\psi_{f_{0}}(A_{i},Y_{i};\tau)+(\tilde{\tau}-\tau)\bigg{[}1+\frac{1}{n}\Big{\\{}\sum_{i\in\mathcal{L}_{1}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)$
$\displaystyle+\sum_{i\in\mathcal{L}_{2}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\bar{\tau}\right)\Big{\\}}\bigg{]}+o_{P}\Big{(}\frac{1}{\sqrt{n}}\Big{)}$
$\displaystyle=$ $\displaystyle
B_{1}+(\tilde{\tau}-\tau)(1+B_{2})+o_{P}\Big{(}\frac{1}{\sqrt{n}}\Big{)},$

where

$\displaystyle B_{1}=$
$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right),$
$\displaystyle B_{2}=$
$\displaystyle\frac{1}{n}\left(\sum_{i\in\mathcal{L}_{1}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{i\in\mathcal{L}_{2}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\bar{\tau}\right)\right).$

Recall that $Z_{i}(a)=-I(f_{a})^{-1}\cdot(f_{a}^{\prime}/f_{a})(Y_{i}(a))$,
$a=0,1$, and abbreviate $Z_{i}(A_{i})=A_{i}Z_{i}(1)+(1-A_{i})Z_{i}(0)$ to
$Z_{i}$. Since $f_{1}(y)=f_{0}(y-\tau)$, then $I(f_{1})=I(f_{0})$. Moreover,

$\bigg{(}\int_{-\infty}^{\infty}|f_{0}^{\prime}(y)|dy\bigg{)}^{2}\leq\int_{-\infty}^{\infty}\frac{\big{(}f_{0}^{\prime}\big{)}^{2}}{f_{0}}(y)dy\cdot\int_{-\infty}^{\infty}f_{0}(y)dy=I(f_{0})\cdot
1<\infty.$

This implies $\int_{-\infty}^{\infty}|f_{0}^{\prime}(y)|dy<\infty$, concluding
that $f_{0}(\infty)$ and $f_{0}(-\infty)$ both exist and equal 0.
Consequently,

$-I(f_{0})\cdot
E\\{Z_{i}(0)\\}=\int_{-\infty}^{\infty}\frac{\big{(}f_{0}^{\prime}\big{)}}{f_{0}}(y)dF_{0}(y)=\int_{-\infty}^{\infty}f_{0}^{\prime}(y)dy=f_{0}(\infty)-f_{0}(-\infty)=0.$

Therefore, $E\\{Z_{i}(0)\\}=0$. Similarly, $E\\{Z_{i}(1)\\}=0$.

For $Y_{i}$’s satisfying Assumptions 1, 2 and 4, it can be easily shown that
these assumptions remain valid when $Y_{i}$ is replaced by $Z_{i}$. Moreover,
the transformed outcomes $Z_{i}(a)$’s have finite second-order moment with
$E\\{Z_{i}(a)\\}^{2}=[{I(f_{0})}]^{-1}$, and

$\displaystyle B_{1}$
$\displaystyle=-\frac{1}{n\pi}\sum_{i=1}^{n}A_{i}\frac{1}{I\left(f_{0}\right)}\cdot\frac{f_{0}^{\prime}}{f_{0}}(Y_{i}-\tau)+\frac{1}{n(1-\pi)}\sum_{i=1}^{n}(1-A_{i})\frac{1}{I\left(f_{0}\right)}\cdot\frac{f_{0}^{\prime}}{f_{0}}(Y_{i})$
$\displaystyle=\frac{1}{n\pi}\sum_{i=1}^{n}A_{i}Z_{i}-\frac{1}{n(1-\pi)}\sum_{i=1}^{n}(1-A_{i})Z_{i}.$

Applying Bugni et al., (2018, Theorem 4.1) to $Z_{i}(a)$, we have
$n_{1}^{-1/2}\sum_{i=1}^{n}A_{i}Z_{i}=O_{P}(1)$, and
$n_{0}^{-1/2}\sum_{i=1}^{n}(1-A_{i})Z_{i}=O_{P}(1)$. Alongside
$n_{1}/n\xrightarrow{p}\pi$ and $n_{0}/n\xrightarrow{p}(1-\pi)$, we have

$B_{1}=\frac{1}{n_{1}}\sum_{i=1}^{n}A_{i}Z_{i}-\frac{1}{n_{0}}\sum_{i=1}^{n}(1-A_{i})Z_{i}+o_{P}\Big{(}\frac{1}{\sqrt{n}}\Big{)}.$

By Theorem 4.1 in Bugni et al., (2018), we have

$\sqrt{n}B_{1}\xrightarrow{d}N(0,V_{Z}^{2}+V_{H}^{2}+V_{A}^{2}),$

where

$V_{Z}^{2}=\frac{\textnormal{Var}\\{\tilde{Z}_{i}(1)\\}}{\pi}+\frac{\textnormal{Var}\\{\tilde{Z}_{i}(0)\\}}{1-\pi},\quad
V_{H}^{2}=E\\{\check{Z}_{i}(1)-\check{Z}_{i}(0)\\}^{2},\quad
V_{A}^{2}=E\Big{[}q_{[S_{i}]}\Big{\\{}\frac{\check{Z}_{i}(1)}{\pi}+\frac{\check{Z}_{i}(0)}{1-\pi}\Big{\\}}^{2}\Big{]},$

with $\tilde{Z}_{i}(a)=Z_{i}(a)-E\\{Z_{i}(a)\mid S_{i}\\}$ and
$\check{Z}_{i}(a)=E\\{Z_{i}(a)\mid S_{i}\\}-E\\{Z_{i}(a)\\}$, $a=0,1$.

Moreover, since $\sqrt{n}(\tilde{\tau}-\tau)=O_{P}(1)$ and
$n_{[k]1(1)}/n\xrightarrow{P}\pi p_{[k]}/2$, then applying Assumption 5(iii)
for $\\{Y_{i}\\}_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=a}$ conditional on
$\\{S^{(n)},A^{(n)},U^{(n)}\\}$, we have

$\displaystyle\frac{1}{n_{[k]1(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\left(\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\right)^{\prime}\left(Y_{i}-\tau+(\tau-\bar{\tau})\right)=\frac{1}{n_{[k]1(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\left(\frac{{f_{0}}^{\prime}}{f_{0}}\right)^{\prime}\left(Y_{i}-\tau\right)+o_{P}(1).$

A similar result holds for
$\\{Y_{i}\\}_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=a}$. Summing over $k$ with
weights $n_{[k]1(1)}/n$, we have

$\displaystyle B_{2}$
$\displaystyle=\frac{1}{n}\left(\sum_{i\in\mathcal{L}_{1}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{i\in\mathcal{L}_{2}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\bar{\tau}\right)\right)$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\frac{\partial}{\partial\tau}\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right)+o_{P}(1)$
$\displaystyle=-\frac{1}{nI(f_{0})}\sum_{i=1}^{n}\bigg{\\{}\frac{A_{i}}{\pi}\cdot\frac{\partial}{\partial\tau}\frac{{f_{0}}^{\prime}}{f_{0}}\left(Y_{i}-\tau\right)-\frac{1-A_{i}}{\pi}\cdot\frac{\partial}{\partial\tau}\frac{{f_{0}}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}+o_{P}(1).$
$\displaystyle=\frac{1}{nI(f_{0})}\sum_{i=1}^{n}\frac{A_{i}}{\pi}\left(\frac{{f_{0}}^{\prime}}{f_{0}}\right)^{\prime}\left(Y_{i}-\tau\right)+o_{P}(1).$

Since $\int_{-\infty}^{\infty}|f_{0}^{\prime\prime}(y)|dy<\infty$, then

$E\left(\frac{{f_{0}}^{\prime}}{f_{0}}\right)^{\prime}\left(Y_{i}(1)-\tau\right)=\int\left({f_{0}}^{\prime}/f_{0}\right)^{\prime}(y-\tau)dF_{1}(y)=\int\left({f_{0}}^{\prime}/f_{0}\right)^{\prime}(y)dF_{0}(y)=-I(f_{0})<\infty.$

Additionally, $\int_{-\infty}^{\infty}|f_{0}^{\prime\prime}(y)|dy<\infty$ also
indicates that $|({f_{0}}^{\prime}/f_{0})^{\prime}(Y_{i}(1)-\tau)|$ has a
finite first-order moment. By the strong law of large numbers under covariate-
adaptive randomization (Bugni et al.,; 2018, Lemma B.3), we have

$\frac{1}{n_{1}}\sum_{i=1}^{n}A_{i}\left(\frac{{f_{0}}^{\prime}}{f_{0}}\right)^{\prime}\left(Y_{i}-\tau\right)=-\pi
I(f_{0})+o_{P}(1).$

Together with $n_{1}/n\xrightarrow{p}\pi$, we have $B_{2}+1=o_{P}(1)$.

Since $\sqrt{n}(\tilde{\tau}-\tau)=O_{P}(1)$, then by Slutsky’s theorem, we
have

$\sqrt{n}(\hat{\tau}_{\textnormal{tdim}}-\tau)=\sqrt{n}B_{1}+o_{P}(1)\xrightarrow{d}N(0,V_{Z}^{2}+V_{H}^{2}+V_{A}^{2}).$

∎

###  D.5 Proof of Corollary 1

###### Proof.

Recall that
$V_{A}^{2}=E[q_{[S_{i}]}\\{\check{Z}_{i}(1)/\pi+\check{Z}_{i}(0)/(1-\pi)\\}^{2}]$,
where $\check{Z}_{i}(a)=E\\{Z_{i}(a)\mid S_{i}\\}-E\\{Z_{i}(a)\\}$, $a=0,1$.
When $q_{[k]}=0$ for $k=1,...,K$, we have $V_{A}^{2}=0$. By Theorem 2, we have

$\sqrt{n}(\hat{\tau}_{\textnormal{tdim}}-\tau)\stackrel{{\scriptstyle
d}}{{\rightarrow}}N(0,V_{Z}^{2}+V_{H}^{2}).$

∎

###  D.6 Proof of Theorem 3

###### Proof.

By Lemma 1, we have $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$, which implies that
$\\{\pi(1-\pi)\hat{I}(f_{0})\\}^{-1}$ is a consistent estimator of
$\\{\pi(1-\pi)I(f_{0})\\}^{-1}$. Note that $E\\{Z_{i}(a)\\}=0$ and
$\textnormal{Var}\\{Z_{i}(a)\\}=E\\{Z_{i}(a)\\}^{2}=I(f_{0})^{-1}$ for
$a=0,1$. Thus, we can decompose $\\{\pi(1-\pi)I(f_{0})\\}^{-1}$ as follows:

$\displaystyle\\{\pi(1-\pi)I(f_{0})\\}^{-1}$ $\displaystyle=$
$\displaystyle\frac{\textnormal{Var}\\{Z_{i}(1)\\}}{\pi}+\frac{\textnormal{Var}\\{Z_{i}(0)\\}}{1-\pi}$
$\displaystyle=$ $\displaystyle\frac{\textnormal{Var}\\{E(Z_{i}(1)\mid
S_{i})\\}}{\pi}+\frac{\textnormal{Var}\\{E(Z_{i}(0)\mid
S_{i})\\}}{1-\pi}+\frac{E[\textnormal{Var}\\{Z_{i}(1)\mid
S_{i}\\}]}{\pi}+\frac{E[\textnormal{Var}\\{Z_{i}(0)\mid S_{i}\\}]}{1-\pi}.$

Since
$\textnormal{Var}\\{\tilde{Z}_{i}(a)\\}=\textnormal{Var}[Z_{i}(a)-E\\{Z_{i}(a)\mid
S_{i}\\}]=E[\textnormal{Var}\\{Z_{i}(a)\mid S_{i}\\}],\ a=0,1,$ then
$V_{Z}^{2}=E[\textnormal{Var}\\{Z_{i}(1)\mid
S_{i}\\}]/\pi+E[\textnormal{Var}\\{Z_{i}(0)\mid S_{i}\\}]/(1-\pi)$. Thus,

$\\{\pi(1-\pi)I(f_{0})\\}^{-1}=V_{Z}^{2}+\frac{\textnormal{Var}\\{E(Z_{i}(1)\mid
S_{i})\\}}{\pi}+\frac{\textnormal{Var}\\{E(Z_{i}(0)\mid
S_{i})\\}}{1-\pi}=V_{Z}^{2}+\frac{E\\{\check{Z}_{i}(1)\\}^{2}}{\pi}+\frac{E\\{\check{Z}_{i}(0)\\}^{2}}{1-\pi}.$

A straightforward calculation yields

$\frac{\check{Z}_{i}^{2}(1)}{\pi}+\frac{\check{Z}_{i}^{2}(0)}{1-\pi}=\\{\check{Z}_{i}(1)-\check{Z}_{i}(0)\\}^{2}+\pi(1-\pi)\Big{\\{}\frac{\check{Z}_{i}(1)}{\pi}+\frac{\check{Z}_{i}(0)}{1-\pi}\Big{\\}}^{2},$

which implies

$\frac{E\\{\check{Z}_{i}(1)\\}^{2}}{\pi}+\frac{E\\{\check{Z}_{i}(0)\\}^{2}}{1-\pi}=V_{H}^{2}+E\Big{[}\pi(1-\pi)\Big{\\{}\frac{\check{Z}_{i}(1)}{\pi}+\frac{\check{Z}_{i}(0)}{1-\pi}\Big{\\}}^{2}\Big{]}.$

Thus,

$\\{\pi(1-\pi)I(f_{0})\\}^{-1}=V_{Z}^{2}+V_{H}^{2}+E\Big{[}\pi(1-\pi)\Big{\\{}\frac{\check{Z}_{i}(1)}{\pi}+\frac{\check{Z}_{i}(0)}{1-\pi}\Big{\\}}^{2}\Big{]}.$

Recall that $\sigma_{\textnormal{tdim}}^{2}=V_{Z}^{2}+V_{H}^{2}+V_{A}^{2}$,
where

$V_{A}^{2}=E\Big{[}q_{[S_{i}]}\Big{\\{}\frac{\check{Z}_{i}(1)}{\pi}+\frac{\check{Z}_{i}(0)}{1-\pi}\Big{\\}}^{2}\Big{]}.$

Since $0\leq q_{[k]}\leq\pi(1-\pi)$ for $k=1,...,K$, we have
$\sigma_{\textnormal{tdim}}^{2}\leq\\{\pi(1-\pi)I(f_{0})\\}^{-1}$ and the
equality holds if $q_{[k]}=\pi(1-\pi)$ for $k=1,\ldots,K$. ∎

###  D.7 Proof of Theorem 4

We need Lemma 3 below to prove Theorem 4. The proof of Lemma 3 is provided in
Section E. Let $\bar{Z}_{a}=n_{a}^{-1}\sum_{i:A_{i}=a}Z_{i}$ and
$\bar{Z}_{[k]a}=n_{[k]a}^{-1}\sum_{i:S_{i}=k,A_{i}=a}Z_{i}$ for $a=0,1$.

######  Lemma 3.

Under Assumptions 1–3, and 5, we have
$\bar{\hat{Z}}_{[k]a}-\bar{Z}_{[k]a}=o_{P}(1)$ and
$n_{[k]a}^{-1}\sum_{i:S_{i}=k,A_{i}=a}(\hat{Z}_{i}^{2}-Z_{i}^{2})=o_{P}(1)$
for $a=0,1$.

###### Proof of Theorem 4.

Recall that $Z_{i}(a)=-I(f_{a})^{-1}\cdot(f_{a}^{\prime}/f_{a})(Y_{i}(a))$
($a=0,1$) has finite second-order moment $I(f_{0})^{-1}$, and Assumptions 1–4
hold if we replace $Y_{i}$ with $Z_{i}$. Define

$\displaystyle\tilde{V}_{Z}^{2}$
$\displaystyle=\frac{1}{\pi}\sum_{k=1}^{K}p_{n[k]}\Big{\\{}\frac{1}{n_{[k]1}}\sum_{i:S_{i}=k}A_{i}(Z_{i}-\bar{Z}_{[k]1})^{2}\Big{\\}}+\frac{1}{1-\pi}\sum_{k=1}^{K}p_{n[k]}\Big{\\{}\frac{1}{n_{[k]0}}\sum_{i:S_{i}=k}(1-A_{i})(Z_{i}-\bar{Z}_{[k]0})^{2}\Big{\\}},$
$\displaystyle\tilde{V}_{H}^{2}$
$\displaystyle=\sum_{k=1}^{K}p_{n[k]}\left\\{(\bar{Z}_{[k]1}-\bar{Z}_{1})-(\bar{Z}_{[k]0}-\bar{Z}_{0})\right\\}^{2},$
$\displaystyle\tilde{V}_{A}^{2}$
$\displaystyle=\sum_{k=1}^{K}p_{n[k]}q_{[k]}\Big{\\{}\frac{(\bar{Z}_{[k]1}-\bar{Z}_{1})}{\pi}+\frac{(\bar{Z}_{[k]0}-\bar{Z}_{0})}{1-\pi}\Big{\\}}^{2}.$

Applying Proposition 1 in Ma et al., (2022) to $Z_{i}(a)$, we have
$\tilde{V}_{A}^{2}-V_{A}^{2}\xrightarrow{p}0$ under Assumptions 1, 2, and 4.
Applying Theorems 3.2 and 3.3 in Bugni et al., (2019) to $Z_{i}(a)$, we have
$\tilde{V}_{Z}^{2}-V_{Z}^{2}\xrightarrow{p}0$ and
$\tilde{V}_{H}^{2}-V_{H}^{2}\xrightarrow{p}0$ under Assumptions 1–3. The only
difference between $(\hat{V}_{Z}^{2},\hat{V}_{H}^{2},\hat{V}_{A}^{2})$ and
$(\tilde{V}_{Z}^{2},\tilde{V}_{H}^{2},\tilde{V}_{A}^{2})$ is the substitution
of $\hat{Z}_{i}$ with $Z_{i}$.

Lemma 3 implies that

$\displaystyle\hat{V}_{Z}^{2}=$
$\displaystyle\frac{1}{\pi}\sum_{k=1}^{K}p_{n[k]}\Big{\\{}\frac{1}{n_{[k]1}}\sum_{S_{i}=k,A_{i}=1}(Z_{i}-\bar{Z}_{[k]1})^{2}+o_{P}(1)\Big{\\}}$
$\displaystyle+\frac{1}{1-\pi}\sum_{k=1}^{K}p_{n[k]}\Big{\\{}\frac{1}{n_{[k]0}}\sum_{S_{i}=k,A_{i}=0}(Z_{i}-\bar{Z}_{[k]0})^{2}+o_{P}(1)\Big{\\}}=\tilde{V}_{Z}^{2}+o_{P}(1),$
$\displaystyle\hat{V}_{H}^{2}=$
$\displaystyle\sum_{k=1}^{K}p_{n[k]}\left\\{(\bar{Z}_{[k]1}-\bar{Z}_{1})-(\bar{Z}_{[k]0}-\bar{Z}_{0})+o_{P}(1)\right\\}^{2}=\tilde{V}_{H}^{2}+o_{P}(1),$
$\displaystyle\hat{V}_{A}^{2}=$
$\displaystyle\sum_{k=1}^{K}p_{n[k]}q_{[k]}\Big{\\{}\frac{(\bar{Z}_{[k]1}-\bar{Z}_{1})+o_{P}(1)}{\pi}+\frac{(\bar{Z}_{[k]0}-\bar{Z}_{0})+o_{P}(1)}{1-\pi}\Big{\\}}^{2}=\tilde{V}_{A}^{2}+o_{P}(1).$

Here, the $o_{P}(1)$ term can be moved outside because $\tilde{V}_{Z}^{2}$,
$\tilde{V}_{H}^{2}$ and $\tilde{V}_{A}^{2}$ are all bounded in probability.
Therefore, $\hat{V}_{Z}^{2}$, $\hat{V}_{H}^{2}$, $\hat{V}_{A}^{2}$ are all
consistent estimators, which implies that
$\hat{\sigma}^{2}_{\textnormal{tdim}}=\hat{V}_{Z}^{2}+\hat{V}_{H}^{2}+\hat{V}_{A}^{2}$
converges in probability to
$\sigma^{2}_{\textnormal{tdim}}={V}_{Z}^{2}+{V}_{H}^{2}+{V}_{A}^{2}$.

∎

###  D.8 Proof of Theorem 5

Before proving the result, we introduce a lemma, with its proof provided in
Section E.

######  Lemma 4.

Under Assumptions 1–3 and 5, we have

$\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)=\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)+o_{P}(1),$
$\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{2},S_{i}=k}\psi_{\hat{f}_{0(1)}}^{(k)}(A_{i},Y_{i};\tau)=\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{2},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)+o_{P}(1).$

###### Proof of Theorem 5.

By definition and the first order Taylor expansion, we have

$\displaystyle\hat{\tau}_{\textnormal{str}}-\tau$
$\displaystyle=\tilde{\tau}-\tau+\sum_{k=1}^{K}\frac{1}{n}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\tilde{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\tilde{\tau}\right)\right)$
$\displaystyle=\tilde{\tau}-\tau+\sum_{k=1}^{K}\frac{1}{n}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\tau\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\tau\right)\right)$
$\displaystyle\quad+\sum_{k=1}^{K}\frac{1}{n}(\tilde{\tau}-\tau)\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)\right),$

where $\bar{\tau}$ is between $\tilde{\tau}$ and $\tau$.

By Lemma 4, we have

$\displaystyle\hat{\tau}_{\textnormal{str}}-\tau$ $\displaystyle=$
$\displaystyle\frac{1}{n}\sum_{k=1}^{K}\sum_{\begin{subarray}{c}S_{i}=k\end{subarray}}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)+o_{P}\Big{(}\frac{1}{\sqrt{n}}\Big{)}$
$\displaystyle+(\tilde{\tau}-\tau)\left[1+\frac{1}{n}\sum_{k=1}^{K}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)\right)\right]$
$\displaystyle=$ $\displaystyle
B_{3}+(\tilde{\tau}-\tau)(1+B_{4})+o_{P}\Big{(}\frac{1}{\sqrt{n}}\Big{)},$

where

$\displaystyle B_{3}=$
$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau),$
$\displaystyle B_{4}=$
$\displaystyle\frac{1}{n}\sum_{k=1}^{K}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)\right).$

Recall that $Z_{i}(a)=-I(f_{a})^{-1}\cdot(f_{a}^{\prime}/f_{a})(Y_{i}(a))$,
$a=0,1$. For $Y_{i}$’s satisfying Assumption 1–3, it is straightforward to
verify that these assumptions still hold if we replace $Y_{i}$ with $Z_{i}$.
By definition,

$\displaystyle B_{3}$
$\displaystyle=-\frac{1}{n}\sum_{k=1}^{K}\sum_{i:S_{i}=k}\frac{A_{i}}{\pi_{n[k]}}\cdot\frac{1}{I\left(f_{0}\right)}\cdot\frac{f_{0}^{\prime}}{f_{0}}(Y_{i}-\tau)+\frac{1}{n}\sum_{k=1}^{K}\sum_{i:S_{i}=k}\frac{1-A_{i}}{1-\pi_{n[k]}}\cdot\frac{1}{I\left(f_{0}\right)}\cdot\frac{f_{0}^{\prime}}{f_{0}}(Y_{i})$
$\displaystyle=\frac{1}{n}\sum_{k=1}^{K}\sum_{i:S_{i}=k}\frac{A_{i}Z_{i}}{\pi_{n[k]}}-\frac{1}{n}\sum_{k=1}^{K}\sum_{i:S_{i}=k}\frac{(1-A_{i})Z_{i}}{1-\pi_{n[k]}},$

which is the fully saturated linear regression estimator proposed by Bugni et
al., (2019), applied to $Z_{i}$. Recall that
$E\\{Z_{i}(1)\\}=E\\{Z_{i}(0)\\}=0$. By Theorem 3.1 in Bugni et al., (2019),
we have

$\sqrt{n}B_{3}\stackrel{{\scriptstyle
d}}{{\rightarrow}}N(0,V_{Z}^{2}+V_{H}^{2}).$

Next, we show that $B_{4}=-1+o_{P}(1)$. By the proof of Theorem 2, we have
$B_{2}=-1+o_{P}(1)$ (note that this conclusion still holds under the weaker
Assumption 3). Moreover,

$\displaystyle B_{4}-B_{2}=$
$\displaystyle\frac{1}{n}\sum_{k=1}^{K}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)\right)$
$\displaystyle-\frac{1}{n}\sum_{k=1}^{K}\left(\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)+\sum_{\begin{subarray}{c}i\in\mathcal{L}_{2},S_{i}=k\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(1)}}\left(A_{i},Y_{i};\bar{\tau}\right)\right).$

By symmetry, we only need to show that

$\frac{1}{n}\sum_{k=1}^{K}\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}^{(k)}\left(A_{i},Y_{i};\bar{\tau}\right)-\frac{1}{n}\sum_{k=1}^{K}\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0\end{subarray}}\frac{\partial}{\partial\tau}\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\bar{\tau}\right)=o_{P}(1),$

which is implied by, for any ${\delta}_{n}=O_{P}(n^{-1/2})$,

$\frac{1}{n}\sum_{k=1}^{K}\Bigg{[}p_{n[k]}\bigg{(}\frac{1}{1-\pi_{n[k]}}-\frac{1}{1-\pi}\bigg{)}\cdot\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0\end{subarray}}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{\prime}\left(Y_{i}+{\delta}_{n}\right)\Bigg{]}=o_{P}(1).$
(21)

By Assumption 5(iii), we have
$n^{-1}\sum_{\begin{subarray}{c}i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0\end{subarray}}(\hat{f}_{0(2)}^{\prime}/\hat{f}_{0(2)})^{\prime}(Y_{i}+{\delta}_{n})$
is bounded in probability (conditional on
$(S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})})$,
$\\{Y_{i}\\}_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}$ are i.i.d.). Recall that
$\pi_{n[k]}\xrightarrow{p}\pi$, we have (21) holds. Thus $B_{4}=-1+o_{P}(1)$.

Note that $\sqrt{n}(\tilde{\tau}-\tau)=O_{P}(n^{-1/2})$, we have
$(\tilde{\tau}-\tau)(1+B_{4})=o_{P}(n^{-1/2})$. By Slutsky’s theorem,

$\sqrt{n}(\hat{\tau}_{\textnormal{str}}-\tau)\stackrel{{\scriptstyle
d}}{{\rightarrow}}N(0,V_{Z}^{2}+V_{H}^{2}).$

By the proof of Theorem 4, $\hat{V}_{Z}^{2}$ and $\hat{V}_{H}^{2}$ are
consistent estimators of $V_{Z}^{2}$ and $V_{H}^{2}$ under Assumptions 1–3.
Thus,
$\hat{\sigma}^{2}_{\textnormal{str}}=\hat{V}_{Z}^{2}+\hat{V}_{H}^{2}\xrightarrow{p}\sigma^{2}_{\textnormal{str}}=V_{Z}^{2}+V_{H}^{2}$
under Assumptions 1–3. ∎

##  Appendix E Proof of lemmas

###  E.1 Proof of Lemma 1

###### Proof.

By definition,

$\hat{I}(f_{0})=\frac{1}{n_{0}}\bigg{[}\sum_{i\in\mathcal{L}_{1},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})\bigg{\\}}^{2}+\sum_{i\in\mathcal{L}_{2},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i})\bigg{\\}}^{2}\bigg{]}.$

It holds that

$\displaystyle\frac{1}{n_{0}}\sum_{i\in\mathcal{L}_{1},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})\bigg{\\}}^{2}-\frac{I(f_{0})}{2}$
$\displaystyle=$
$\displaystyle\left[\frac{1}{n_{0}}\sum_{i\in\mathcal{L}_{1},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})\bigg{\\}}^{2}-\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0}}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y)\right]$
$\displaystyle+\sum_{k=1}^{K}\left(\frac{n_{[k]0(1)}}{n_{0}}-\frac{p_{[k]}}{2}\right)\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y)$
$\displaystyle+\left[\sum_{k=1}^{K}\frac{p_{[k]}}{2}\int\left\\{\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\right\\}(y)dF_{[k]0}(y)\right]$
$\displaystyle:=$ $\displaystyle B_{1}+B_{2}+B_{3}.$

For the first term $B_{1}$, since the splitting is independent of
$S^{(n)},A^{(n)},Y^{(n)}(a)=\\{Y_{1}(a),\ldots,Y_{n}(a)\\}$, for $a=0,1$, we
have

$E\left[\frac{1}{n_{0}}\sum_{i\in\mathcal{L}_{1},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})\bigg{\\}}^{2}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\right]=\sum_{k=1}^{K}\frac{n_{[k]0(1)}}{n_{0}}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y).$

By the triangle inequality, we have

$\displaystyle\left|\bigg{\\{}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y)\bigg{\\}}^{1/2}-\bigg{\\{}\int\left(\frac{{f}_{0}^{\prime}}{{f}_{0}}\right)^{2}(y)dF_{[k]0}(y)\bigg{\\}}^{1/2}\right|^{2}$
$\displaystyle\leq$
$\displaystyle\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{{f}_{0}^{\prime}}{{f}_{0}}\right)^{2}(y)dF_{[k]0}(y)$
$\displaystyle\leq$
$\displaystyle\frac{1}{\min_{k=1,\ldots,K}p_{[k]}}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{{f}_{0}^{\prime}}{{f}_{0}}\right)^{2}(y)dF_{0}(y)=o_{P}(1),$

where the second inequality is due to
$f_{0}(y)=\sum_{k=1}^{K}p_{[k]}f_{[k]0}(y)\geq(\min_{k^{\prime}=1,\ldots,K}p_{[k^{\prime}]})\cdot
f_{[k]0}(y),$ for all $k=1,\ldots,K,$ and the last equality is due to
Assumption 5(iii). Combined with
$\\{\int({f}_{0}^{\prime}/{f}_{0})^{2}(y)dF_{[k]0}(y)\\}^{1/2}=O_{P}(1)$, we
have

$\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y)=\int\left(\frac{{f}_{0}^{\prime}}{{f}_{0}}\right)^{2}(y)dF_{[k]0}(y)+o_{P}(1).$
(22)

Therefore, by the weak law of large numbers, we have $\\{B_{1}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\\}=o_{P}(1)$, which implies
$B_{1}=o_{P}(1)$.

For the second term $B_{2}$, by (22) and $I(f_{0})<\infty$, we have

$\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)^{2}(y)dF_{[k]0}(y)=O_{P}(1).$

Since ${n_{[k]0(1)}}/{n_{0}}-{p_{[k]}}/{2}=o_{P}(1)$, then $B_{2}=o_{P}(1)$.

For the third term $B_{3}$, by Assumption 5(iii) and the fact that
$F_{0}(y)=\sum_{k=1}^{K}p_{[k]}F_{[k]0}(y)$, we have

$B_{3}=2^{-1}\int\left\\{\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}-\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\right\\}(y)dF_{0}(y)=o_{P}(1).$
(23)

Combining them together, we have

$\frac{1}{n_{0}}\sum_{i\in\mathcal{L}_{1},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})\bigg{\\}}^{2}=\frac{I(f_{0})}{2}+o_{P}(1).$

Similarly,

$\frac{1}{n_{0}}\sum_{i\in\mathcal{L}_{2},A_{i}=0}\bigg{\\{}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i})\bigg{\\}}^{2}=\frac{I(f_{0})}{2}+o_{P}(1).$

Therefore, $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$.

∎

###  E.2 Proof of Lemma 2

###### Proof.

We will prove the result for the first claim ($i\in\mathcal{L}_{1}$) only, as
the proof for the second claim is similar. By Lemma 1, we have
$\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$. Given that $0<I(f_{0})<\infty$, it
suffices to show that

$\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)=o_{P}(1).$
(24)

Note that

$\displaystyle\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)$
$\displaystyle=$
$\displaystyle\left[\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)-I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)\right]$
$\displaystyle+\left[I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)\right]$
$\displaystyle:=$ $\displaystyle B_{1}+B_{2}.$

It suffices for (24) to show that $B_{1}=o_{P}(1)$ and $B_{2}=o_{P}(1)$.

For the first term $B_{1}$, by definition, we have

$\hat{I}(f_{0})\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)=\sum_{i\in\mathcal{L}_{1},A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)\left(Y_{i}-\tau\right)-\sum_{i\in\mathcal{L}_{1},A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\right)\left(Y_{i}\right),$
$I(f_{0})\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)=\sum_{i\in\mathcal{L}_{1},A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{{f}_{0(2)}^{\prime}}{{f}_{0(2)}}\right)\left(Y_{i}-\tau\right)-\sum_{i\in\mathcal{L}_{1},A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{{f}_{0(2)}^{\prime}}{{f}_{0(2)}}\right)\left(Y_{i}\right).$

Thus,

$\displaystyle\hat{I}(f_{0})\sum_{i\in\mathcal{L}_{1}}\psi_{\hat{f}_{0(2)}}(A_{i},Y_{i};\tau)-I(f_{0})\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)$
$\displaystyle=$
$\displaystyle\sum_{i\in\mathcal{L}_{1},A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{{f}_{0(2)}^{\prime}}{{f}_{0(2)}}\right)\left(Y_{i}-\tau\right)-\sum_{i\in\mathcal{L}_{1},A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{{f}_{0(2)}^{\prime}}{{f}_{0(2)}}\right)\left(Y_{i}\right).$

Then, by Assumptions 1–2, we have

$\displaystyle\textnormal{Var}\bigg{(}\bigg{\\{}\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\left(\hat{I}\left(f_{0}\right)\cdot\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tau\right)-I\left(f_{0}\right)\cdot\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right)\right)\bigg{\\}}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ (25) $\displaystyle=$
$\displaystyle\frac{1}{n}\textnormal{Var}\bigg{(}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)-\sum_{i\in\mathcal{L}_{1},A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle\mid S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{n}\textnormal{Var}\bigg{(}\bigg{\\{}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)$
$\displaystyle-\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\bigg{\\}}^{2}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ $\displaystyle\leq$
$\displaystyle\frac{2K}{n}\sum_{k=1}^{K}\Bigg{[}\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$
$\displaystyle+\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}\Bigg{]}.$

Moreover,

$\displaystyle\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{1}{1-\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ (26) $\displaystyle=$
$\displaystyle\frac{1}{(1-\pi)^{2}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\textnormal{Var}\bigg{\\{}\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{\\}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{(1-\pi)^{2}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}E\bigg{\\{}\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{\\}}$ $\displaystyle=$
$\displaystyle\frac{n_{[k]0(1)}}{(1-\pi)^{2}}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}\left(y\right)dF_{[k]0}(y)$
$\displaystyle=$ $\displaystyle o_{P}(n),$

where the last equality is due to
$n_{[k]0(1)}/n\xrightarrow{p}p_{[k]}(1-\pi)/2$ and

$\displaystyle\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}\left(y\right)dF_{[k]0}(y)\leq\frac{1}{\min_{k=1,\ldots,K}p_{[k]}}\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}\left(y\right)dF_{0}(y)=o_{P}(1).$
(27)

Similarly,

$\displaystyle\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{1}{\pi}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}=o_{P}(n),$ (28)

By substituting (26) and (28) into (25), we have

$\displaystyle\textnormal{Var}\bigg{(}\bigg{\\{}\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\left(\hat{I}\left(f_{0}\right)\cdot\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tau\right)-I\left(f_{0}\right)\cdot\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right)\right)\bigg{\\}}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ (29) $\displaystyle=$
$\displaystyle o_{P}(1).$

Next, we show that the conditional bias also converges to zero in probability.
Specifically, we will show that

$\displaystyle
E\bigg{(}\bigg{\\{}\sum_{i\in\mathcal{L}_{1}}\left(\hat{I}\left(f_{0}\right)\cdot\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tau\right)-I\left(f_{0}\right)\cdot\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right)\right)\bigg{\\}}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ (30) $\displaystyle=$
$\displaystyle\sum_{k=1}^{K}\bigg{(}\frac{n_{[k]1(1)}}{n\pi}-\frac{n_{[k]0(1)}}{n(1-\pi)}\bigg{)}\cdot\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(y\right)dF_{[k]0}(y)=o_{P}\bigg{(}\frac{1}{\sqrt{n}}\bigg{)}.$

Under Assumptions 1, 2, and 4, by Bugni et al., (2019, Lemma C.2), we have
$n_{[k]1}/n-p_{[k]}\pi=O_{P}(n^{-1/2})$. Since $(n_{[k]1}-1)/2\leq
n_{[k]1(1)}\leq n_{[k]1}/2$, we have
$n_{[k]1(1)}/(n\pi)-p_{[k]}/2=O_{P}(n^{-1/2})$. Similarly,
$n_{[k]0(1)}/\\{n(1-\pi)\\}-p_{[k]}/2=O_{P}(n^{-1/2})$. Moreover, by (27) and
the Cauchy–Schwarz inequality, we have
$\int(\hat{f}_{0(2)}^{\prime}/\hat{f}_{0(2)}-f_{0}^{\prime}/f_{0})\left(y\right)dF_{[k]0}(y)=o_{P}(1).$
Therefore, (30) holds.

Combining (29) and (30), we have

$\displaystyle
E\bigg{(}\bigg{\\{}\frac{1}{\sqrt{n}}\sum_{i\in\mathcal{L}_{1}}\left(\hat{I}\left(f_{0}\right)\cdot\psi_{\hat{f}_{0(2)}}\left(A_{i},Y_{i};\tau\right)-I\left(f_{0}\right)\cdot\psi_{f_{0}}\left(A_{i},Y_{i};\tau\right)\right)\bigg{\\}}^{2}\mid
S^{(n)},A^{(n)},aU^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ $\displaystyle=$
$\displaystyle o_{P}(1).$

Therefore, conditional on $S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}$,
$B_{1}$ converges to zero in probability, which implies $B_{1}=o_{P}(1)$
unconditionally.

For the second term $B_{2}$, in the proof of Theorem 2, we have shown that
$n^{-1/2}\sum_{i=1}^{n}\psi_{f_{0}}(A_{i},Y_{i};\tau)$ is asymptotically
normal with zero mean and finite variance. Since the splitting is independent
of $S^{(n)},A^{(n)},Y^{(n)}(a)$, we have
$n^{-1/2}\sum_{i\in\mathcal{L}_{1}}\psi_{f_{0}}(A_{i},Y_{i};\tau)=O_{P}(1)$.
By Lemma 1, we have $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$. Therefore,
$B_{2}=o_{P}(1)$.

∎

###  E.3 Proof of Lemma 3

###### Proof.

Recall that $\hat{Z}_{i}$ is defined as follows:
$\hat{Z}_{i}=-\hat{I}(f_{0})^{-1}(\hat{f}_{0(3-j)}^{\prime}/\hat{f}_{0(3-j)})(Y_{i}),$
for $i\in\mathcal{L}_{j},\ A_{i}=0$, $j=1,2$, and
$\hat{Z}_{i}=-\hat{I}(f_{0})^{-1}(\hat{f}_{0(3-j)}^{\prime}/\hat{f}_{0(3-j)})(Y_{i}-\tilde{\tau}),$
for $i\in\mathcal{L}_{j},\ A_{i}=1$, $j=1,2$. Recall that
$\bar{\hat{Z}}_{[k]a}=\sum_{S_{i}=k,A_{i}=a}\hat{Z}_{i}/n_{[k]a}$ is the
stratum-specific sample mean of $\hat{Z}_{i}$ in stratum $k$ under treatment
arm $a$ and $\bar{\hat{Z}}_{a}=\sum_{i:A_{i}=a}\hat{Z}_{i}/n_{a}$ is the
sample mean of $\hat{Z}_{i}$ under treatment arm $a$, $a=0,1$, $k=1,\ldots,K$.

(I) We show that $\bar{\hat{Z}}_{[k]0}-\bar{Z}_{[k]0}=o_{P}(1)$. By
definition, we have

$\displaystyle\bar{\hat{Z}}_{[k]0}-\bar{Z}_{[k]0}$ $\displaystyle=$
$\displaystyle
n_{[k]0}^{-1}\cdot\hat{I}(f_{0})^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i})+\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i})\bigg{\\}}$
$\displaystyle-n_{[k]0}^{-1}\cdot
I(f_{0})^{-1}\sum_{i:S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}(Y_{i})$
$\displaystyle=$ $\displaystyle\frac{n_{[k]0(1)}}{n_{[k]0}}\cdot
n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle+\frac{n_{[k]0(2)}}{n_{[k]0}}\cdot
n_{[k]0(2)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}.$

Since $n_{[k]0(1)}/n_{[k]0}\xrightarrow{p}1/2$ and
$n_{[k]0(2)}/n_{[k]0}\xrightarrow{p}1/2$, it suffices for
$\bar{\hat{Z}}_{[k]0}-\bar{Z}_{[k]0}=o_{P}(1)$ to show that

$n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}=o_{P}(1),$
$n_{[k]0(2)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}=o_{P}(1).$

We will prove the first claim only, as the proof of the second claim is
similar. We make the following decomposition:

$\displaystyle
n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle=n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\left(Y_{i}\right)-\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle\quad+n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)-I(f_{0})^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle:=B_{1}+B_{2}.$

Next, we show that $B_{1}=o_{P}(1)$ and $B_{2}=o_{P}(1)$. By Lemma 1,
$\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$. Thus, $\hat{I}(f_{0})^{-1}=O_{P}(1)$. It
suffices for $B_{1}=o_{P}(1)$ to show that

$\displaystyle
n_{[k]0(1)}^{-1}E\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\left|\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0(2)}^{\prime}}{f_{0(2)}}\right|\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{\\}}=o_{P}(1).$

It holds that

$\displaystyle
n_{[k]0(1)}^{-1}E\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\left|\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0(2)}^{\prime}}{f_{0(2)}}\right|\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{\\}}$ $\displaystyle=$
$\displaystyle\int\left|\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right|\left(y\right)dF_{[k]0}(y)$
$\displaystyle\leq$
$\displaystyle\left\\{\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)^{2}\left(y\right)dF_{[k]0}(y)\right\\}^{1/2}=o_{P}(1),$

where the last equality is due to (27).

For $B_{2}=o_{P}(1)$, it suffices to show that

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)=O_{P}(1).$

By applying the Cauchy–Schwarz inequality and using
$f_{[k]0}(y)\leq\min_{k=1,\ldots,K}p_{[k]}^{-1}f_{0}(y)$, we have

$E\bigg{|}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{|}=\int\bigg{|}\frac{f_{0}^{\prime}}{f_{0}}\left(y\right)\bigg{|}dF_{[k]0}(y)\leq\frac{1}{\min_{k=1,\ldots,K}p_{[k]}}\bigg{[}\int\bigg{\\{}\frac{f_{0}^{\prime}}{f_{0}}\left(y\right)\bigg{\\}}^{2}dF_{0}(y)\bigg{]}^{1/2}<\infty.$

Then, we have

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)}=E\bigg{\\{}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}+o_{P}(1).$

Thus,

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)=E\bigg{\\{}\frac{f_{0}^{\prime}}{f_{0}}\left(Y_{i}\right)\bigg{\\}}+o_{P}(1).$

(II) We show that $\bar{\hat{Z}}_{[k]1}-\bar{Z}_{[k]1}=o_{P}(1)$. Similar to
the proof of (I), we have

$\displaystyle
n_{[k]1}^{-1}\cdot\hat{I}(f_{0})^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i}-\tau)+\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i}-\tau)\bigg{\\}}$
$\displaystyle-n_{[k]1}^{-1}\cdot
I(f_{0})^{-1}\sum_{S_{i}=k,A_{i}=1}\frac{\hat{f}_{0}}{f_{0}}(Y_{i}-\tau)=o_{P}(1).$

The difference arises from our lack of knowledge about $\tau$, leading us to
substitute $\tilde{\tau}$ for $\tau$ in the definition of
$\bar{\hat{Z}}{[k]1}$. Considering this substitution,
$\bar{\hat{Z}}{[k]1}-\bar{Z}_{[k]1}=o_{P}(1)$ is implied by

$\frac{n_{[k]1(1)}}{n_{[k]1}}\cdot
n_{[k]1(1)}^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i}-\tilde{\tau})-\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}(Y_{i}-\tau)\bigg{\\}}=o_{P}(1).$
$\frac{n_{[k]1(2)}}{n_{[k]1}}\cdot
n_{[k]1(2)}^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i}-\tilde{\tau})-\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}(Y_{i}-\tau)\bigg{\\}}=o_{P}(1).$

We will prove the first claim only, as the proof of the second claim is
similar. Note that $n_{[k]1(1)}/n_{[k]1}\xrightarrow{P}1/2$. By the first-
order Taylor expansion, the first claim is implied by

$\frac{1}{n_{[k]1(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}(\tau-\tilde{\tau})\cdot\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{\prime}\big{(}(Y_{i}-\tau)+(\tau-\bar{\tau})\big{)}=o_{P}(1),$

where $\bar{\tau}$ is between $\tilde{\tau}$ and $\tau$.

Recall that $\sqrt{n}(\tilde{\tau}-\tau)=O_{P}(1)$, thus
$\tau-\bar{\tau}=O_{P}(n^{-1/2})=O_{P}(n_{[k]1(1)}^{-1/2})$, and it suffices
to show that

$\frac{1}{n_{[k]1(1)}}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{\prime}\big{(}Y_{i}^{*}+(\tau-\bar{\tau})\big{)}=O_{P}(1),$

where $Y_{i}^{*}=Y_{i}-\tau\sim F_{0[k]}(y)$ for
$i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1$. The above statement follows from
Assumption 5(iii).

(III) We show that $\bar{\hat{Z}}_{a}-\bar{Z}_{a}=o_{P}(1)$. By definition,
$\bar{\hat{Z}}_{a}=\sum_{i:A_{i}=a}\hat{Z}_{i}/n_{a}=\sum_{k=1}^{K}(n_{[k]a}/n_{a})\bar{\hat{Z}}_{[k]a}$.
Then

$\bar{\hat{Z}}_{a}-\bar{Z}_{a}=\sum_{k=1}^{K}(n_{[k]a}/n_{a})(\bar{\hat{Z}}_{[k]a}-\bar{Z}_{[k]a})=o_{P}(1),$

where the last equality is due to $n_{[k]a}/n_{a}\xrightarrow{p}p_{[k]}$ and
$\bar{\hat{Z}}_{[k]a}-\bar{Z}_{[k]a}=o_{P}(1)$.

(IV) We show that
$n_{[k]0}^{-1}\sum_{S_{i}=k,A_{i}=0}(\hat{Z}_{i}^{2}-Z_{i}^{2})=o_{P}(1)$,
which is implied by

$n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}\left(Y_{i}\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\bigg{\\}}=o_{P}(1),$
$n_{[k]0(2)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}\left(Y_{i}\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\bigg{\\}}=o_{P}(1).$

We will prove the first equation ($i\in\mathcal{L}_{1}$) only, as the proof
for the second equation ($i\in\mathcal{L}_{2}$) is similar.

Similar to (I), we make the following decomposition:

$\displaystyle
n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}\left(Y_{i}\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle=n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}\left(Y_{i}\right)-\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle\quad+n_{[k]0(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\bigg{\\}}$
$\displaystyle:=B_{3}+B_{4}.$

By (22), we have
$n_{[k]0(1)}^{-1}[\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\\{(\hat{f}_{0(2)}^{\prime}/\hat{f}_{0(2)})^{2}-(f_{0}^{\prime}/f_{0})^{2}\\}(Y_{i})]=o_{P}(1)$.
By Lemma 1, $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$, which implies
$\hat{I}(f_{0})^{-2}=O_{P}(1)$. Therefore, $B_{3}=o_{P}(1)$.

It suffices for $B_{4}=o_{P}(1)$ to show that

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)=O_{P}(1).$

Note that

$E\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)=\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(y\right)dF_{[k]0}(y)\leq\frac{1}{\min_{k=1,\ldots,K}p_{[k]}}\int\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(y\right)dF_{0}(y)<\infty.$

Then, we have

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)}=E\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)+o_{P}(1).$

Thus,

$n_{[k]0(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)=E\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}\right)+o_{P}(1)=O_{P}(1).$

(V) We show that
$n_{[k]1}^{-1}\sum_{S_{i}=k,A_{i}=1}(\hat{Z}_{i}^{2}-Z_{i}^{2})=o_{P}(1)$.
Similar to the proof of (IV), we have

$n_{[k]1(1)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}\left(Y_{i}-\tau\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}-\tau\right)\bigg{\\}}=o_{P}(1),$
$n_{[k]0(2)}^{-1}\bigg{\\{}\hat{I}(f_{0})^{-2}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{1(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}\left(Y_{i}-\tau\right)-I(f_{0})^{-2}\sum_{i\in\mathcal{L}_{2},S_{i}=k,A_{i}=1}\bigg{(}\frac{f_{0}^{\prime}}{f_{0}}\bigg{)}^{2}\left(Y_{i}-\tau\right)\bigg{\\}}=o_{P}(1).$

The difference arises from our lack of knowledge about $\tau$, leading us to
substitute $\tilde{\tau}$ for $\tau$ in the definition of
$\bar{\hat{Z}}{[k]1}$. Considering this substitution,
$\bar{\hat{Z}}_{[k]1}-\bar{Z}_{[k]1}=o_{P}(1)$ is implied by

$n_{[k]1(1)}^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}(Y_{i}-\tilde{\tau})-\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{2}(Y_{i}-\tau)\bigg{\\}}=o_{P}(1),$
$n_{[k]1(2)}^{-1}\bigg{\\{}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}(Y_{i}-\tilde{\tau})-\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(1)}^{\prime}}{\hat{f}_{0(1)}}\bigg{)}^{2}(Y_{i}-\tau)\bigg{\\}}=o_{P}(1).$

As before we prove the first equation ($i\in\mathcal{L}_{1}$) only, as the
proof for the second equation ($i\in\mathcal{L}_{2}$) is similar.

By the first-order Taylor expansion, it suffices to show that

$n_{[k]1(1)}^{-1}\cdot(\tilde{\tau}-\tau)\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}(Y_{i}-\bar{\tau})\cdot\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{\prime}(Y_{i}-\bar{\tau})=o_{P}(1),$
(31)

where $\bar{\tau}$ is between $\tilde{\tau}$ and $\tau$.

By Assumption 5(iii), we have

$n_{[k]1(1)}^{-1}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\bigg{|}\bigg{(}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{)}^{\prime}\bigg{|}(Y_{i}-\bar{\tau})=O_{P}(1).$

Note that for any $i\in\mathcal{L}_{1}$ with $S_{i}=k$ and $A_{i}=1$,

$\bigg{|}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{|}(Y_{i}-\bar{\tau})\leq\sup_{y\in\mathbb{R}}\bigg{\\{}\bigg{|}\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}\bigg{|}(y)\bigg{\\}}=o_{P}(n^{1/2}).$

Combining with $\tilde{\tau}-\tau=O_{P}(n^{-1/2})$, we obtain (31).

∎

###  E.4 Proof of Lemma 4

###### Proof.

We adopt a similar approach to the proof of Lemma 2. As before, we will only
prove the result for $i\in\mathcal{L}_{1}$.

By Lemma 1, we have $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$. Since
$0<I(f_{0})<\infty$, then it suffices to show that

$\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)=o_{P}(1).$
(32)

Firstly, we have

$\displaystyle\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)$
(33) $\displaystyle=$
$\displaystyle\left[\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\right]$
$\displaystyle+\left[I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)-\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\right]$
$\displaystyle:=$ $\displaystyle B_{1}+B_{2}.$

It suffices for (32) to show that $B_{1}=o_{P}(1)$ and $B_{2}=o_{P}(1)$.

In the proof of Theorem 5, we have shown that
$n^{-1/2}\sum_{k=1}^{K}\sum_{i:S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)$
converges in distribution to a normal distribution with zero mean and finite
variance. Since the splitting is independent of $S^{(n)},A^{(n)},Y^{(n)}(a)$,
then
$n^{-1/2}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)=O_{P}(1)$.
By Lemma 1, we have $\hat{I}(f_{0})=I(f_{0})+o_{P}(1)$. Therefore,
$B_{2}=o_{P}(1)$.

Next, we show that $B_{1}=o_{P}(1)$, which is implied by

$\displaystyle\textnormal{Var}\bigg{(}\bigg{\\{}\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\bigg{\\}}$
$\displaystyle\quad\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}=o_{P}(1)$ (34)

and

$\displaystyle
E\bigg{(}\bigg{\\{}\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\bigg{\\}}$
$\displaystyle\quad\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}=o_{P}(1).$ (35)

To prove (E.4), we have

$\displaystyle\textnormal{Var}\bigg{(}\bigg{\\{}\hat{I}(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-I(f_{0})\cdot\frac{1}{\sqrt{n}}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\bigg{\\}}$
$\displaystyle\mid S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{n}\textnormal{Var}\bigg{(}\bigg{\\{}\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\frac{1}{\pi_{n[k]}}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)$
$\displaystyle-\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\frac{1}{1-\pi_{n[k]}}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\bigg{\\}}\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$ $\displaystyle\leq$
$\displaystyle\frac{2K}{n}\sum_{k=1}^{K}\Bigg{[}\frac{1}{\pi_{n[k]}^{2}}\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$
$\displaystyle+\frac{1}{(1-\pi_{n[k]})^{2}}\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}\Bigg{]}.$

We have shown in (26) and (28) that

$\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=0}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}=o_{P}(n),$
$\textnormal{Var}\bigg{(}\sum_{i\in\mathcal{L}_{1},S_{i}=k,A_{i}=1}\cdot\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(Y_{i}-\tau\right)\mid
S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}=o_{P}(n).$

Since $\pi_{n[k]}\xrightarrow{p}\pi$, then equation (E.4) holds.

To prove (E.4), we have

$\displaystyle
E\bigg{(}\bigg{\\{}\hat{I}(f_{0})\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{\hat{f}_{0(2)}}^{(k)}(A_{i},Y_{i};\tau)-I(f_{0})\sum_{k=1}^{K}\sum_{i\in\mathcal{L}_{1},S_{i}=k}\psi_{f_{0}}^{(k)}(A_{i},Y_{i};\tau)\bigg{\\}}$
$\displaystyle\quad\mid S^{(n)},A^{(n)},U^{(n)},D^{(\mathcal{L}_{2})}\bigg{)}$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{K}\bigg{(}\frac{n_{[k]1(1)}}{\pi_{n[k]}}-\frac{n_{[k]0(1)}}{(1-\pi_{n[k]})}\bigg{)}\cdot\int\left(\frac{\hat{f}_{0(2)}^{\prime}}{\hat{f}_{0(2)}}-\frac{f_{0}^{\prime}}{f_{0}}\right)\left(y\right)dF_{[k]0}(y).$

Note that
$n_{[k]1(1)}/(n\pi_{n[k]})=(n_{[k]1(1)}/n_{[k]1})(n_{[k]}/n)=p_{[k]}/2+O_{P}(n^{-1/2})$,
and similarly, $n_{[k]0(1)}/\\{n(1-\pi_{n[k]})\\}=p_{[k]}/2+O_{P}(n^{-1/2})$.
Thus, $n_{[k]1(1)}/\pi_{n[k]}-n_{[k]0(1)}/(1-\pi_{n[k]})=O_{P}(n^{1/2})$. We
have shown by (27) and the Cauchy–Schwarz inequality that,
$\int(\hat{f}_{0(2)}^{\prime}/\hat{f}_{0(2)}-f_{0}^{\prime}/f_{0})\left(y\right)dF_{[k]0}(y)=o_{P}(1).$
Therefore, equation (E.4) holds.

∎