$\displaystyle\qquad+\mathbb{E}\mathinner{\biggl{[}\gamma(S_{i})\mathinner{\Bigl{(}\mathbb{E}\mathinner{\bigl{[}(1-\pi)\varpi_{i}(1)+\pi\varpi_{i}(0)|S_{i}\bigr{]}}\Bigr{)}}^{2}\biggr{]}}+\pi^{2}(1-\pi)^{2}\mathbb{E}\mathinner{\Bigl{(}\mathbb{E}\mathinner{\bigl{[}\varpi_{i}(1)-\varpi_{i}(0)|S_{i}\bigr{]}}\Bigr{)}}^{2},$

the last convergence in distribution is by a similar argument in the proof of
Bugni et al. (2018, Lemma B.2) and the fact that

$\displaystyle\mathbb{E}(\varpi_{i}(1)-\varpi_{i}(0))=\lambda_{1}\mathbb{E}(Y_{i}(D_{i}(1))-Y_{i}(D_{i}(1))-(D_{i}(1)-D_{i}(0))\tau)=0.$

Thus (E.4) implies both
$\frac{1}{\sqrt{n}}\sum_{i\in[n]}(A_{i}-\pi)(Y_{i}-D_{i}\tau)$ and
$\frac{1}{\sqrt{n}}\sum_{i\in[n]}\mathbb{X}_{i}(A_{i}-\pi)$ are $O_{p}(1)$. In
addition, let
$\hat{\lambda}=\left(\frac{1}{n}\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\right)^{-1}\frac{1}{n}\sum_{i\in[n]}\mathbb{X}_{i}(Y_{i}-D_{i}\tau)$.
We can show

$\displaystyle\hat{\lambda}\overset{p}{\longrightarrow}\lambda^{*}\mathrel{\mathop{\mathchar
58\relax}}=\left(\mathbb{E}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\right)^{-1}\mathbb{E}\mathbb{X}_{i}\left[\pi(Y_{i}(D_{i}(1))-D_{i}(1)\tau)+(1-\pi)(Y_{i}(D_{i}(0))-D_{i}(0)\tau)\right].$

Therefore, by letting $\lambda_{1}=1$ and $\lambda_{2}=\lambda^{*}$, we have

$\displaystyle\sqrt{n}(\hat{\tau}_{TSLS}-\tau)\rightsquigarrow\mathcal{N}(0,\sigma_{TSLS}^{2}),$

where

$\displaystyle\sigma_{TSLS}^{2}$
$\displaystyle=\frac{\sigma_{TSLS,0}^{2}+\sigma_{TSLS,1}^{2}+\sigma_{TSLS,2}^{2}+\sigma_{TSLS,3}^{2}}{(\mathbb{E}(D_{i}(1)-D_{i}(0)))^{2}},$
$\displaystyle\sigma_{TSLS,0}^{2}$
$\displaystyle=\frac{\mathbb{E}\mathinner{\bigl{[}Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}-\mathbb{E}[Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}|S_{i}]\bigr{]}}^{2}}{1-\pi},$
$\displaystyle\sigma_{TSLS,1}^{2}$
$\displaystyle=\frac{\mathbb{E}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}-\mathbb{E}[Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}|S_{i}]\bigr{]}}^{2}}{\pi},$
$\displaystyle\sigma_{TSLS,2}^{2}$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\mathinner{\bigl{[}Y(D(1))-Y(D(0))-(D(1)-D(0))\tau|S_{i}\bigr{]}}\right]^{2},$
$\displaystyle\sigma_{TSLS,3}^{2}$
$\displaystyle=\mathbb{E}\left\\{\gamma(S_{i})\mathinner{\biggl{(}\mathbb{E}\mathinner{\biggl{[}\frac{Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}}{\pi}+\frac{Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*}}{1-\pi}\bigg{|}S_{i}\biggr{]}}\biggr{)}}^{2}\right\\}.$

To see the second result, we note that
$\bar{X}_{i}=(D_{i},\mathbb{X}_{i}^{\top})^{\top}$ and
$\bar{Z}_{i}=(A_{i},\mathbb{X}_{i}^{\top})^{\top}$. Denote
$\breve{Z}_{i}=(\tilde{A}_{i},\mathbb{X}_{i}^{\top})^{\top}$. Then, we have

$\displaystyle
e_{1}^{\top}[S_{\bar{X},\bar{Z}}S_{\bar{Z},\bar{Z}}^{-1}S_{\bar{Z},\bar{X}}]^{-1}$
$\displaystyle=[S_{\bar{X},\breve{Z}}S_{\breve{Z},\breve{Z}}^{-1}S_{\breve{Z},\bar{X}}]^{-1}$
$\displaystyle=e_{1}^{\top}\biggl{\\{}\begin{pmatrix}\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n&\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n\\\
0&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}\begin{pmatrix}\sum_{i\in[n]}\tilde{A}_{i}^{2}/n&0\\\
0&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}^{-1}$
$\displaystyle\times\begin{pmatrix}\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n&0\\\
\sum_{i\in[n]}D_{i}\mathbb{X}_{i}/n&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}\biggr{\\}}^{-1}$
$\displaystyle=e_{1}^{\top}\begin{pmatrix}\frac{(\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n)^{2}}{(\sum_{i\in[n]}\tilde{A}_{i}^{2}/n)}+\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n\left[\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\right]^{-1}\sum_{i\in[n]}D_{i}\mathbb{X}_{i}/n&\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n\\\
\sum_{i\in[n]}D_{i}\mathbb{X}_{i}/n&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}^{-1}$
$\displaystyle\stackrel{{\scriptstyle
p}}{{\longrightarrow}}[\pi(1-\pi)]^{-1}(\mathbb{E}(D(1)-D(0)))^{-2}\begin{pmatrix}1&-\gamma_{D}^{\top}\end{pmatrix}.$

and

$\displaystyle S_{\bar{X},\bar{Z}}S_{\bar{Z},\bar{Z}}\bar{Z}_{i}$
$\displaystyle=S_{\bar{X},\breve{Z}}S_{\breve{Z},\breve{Z}}\breve{Z}_{i}$
$\displaystyle=\begin{pmatrix}\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n&\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n\\\
0&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}\begin{pmatrix}\sum_{i\in[n]}\tilde{A}_{i}^{2}/n&0\\\
0&\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n\end{pmatrix}^{-1}\begin{pmatrix}\tilde{A}_{i}\\\
\mathbb{X}_{i}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}(\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n)(\sum_{i\in[n]}\tilde{A}_{i}^{2}/n)^{-1}&(\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n)(\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n)^{-1}\\\
0&I\end{pmatrix}\begin{pmatrix}\tilde{A}_{i}\\\ \mathbb{X}_{i}\end{pmatrix},$

where
$\gamma_{D}=(\mathbb{E}\mathbb{X}_{i}\mathbb{X}_{i}^{\top})^{-1}\mathbb{E}(\mathbb{X}_{i}(\pi
D_{i}(1)+(1-\pi)D_{i}(0)))$. Further note that

$\displaystyle\begin{pmatrix}(\sum_{i\in[n]}D_{i}\tilde{A}_{i}/n)(\sum_{i\in[n]}\tilde{A}_{i}^{2}/n)^{-1}&(\sum_{i\in[n]}D_{i}\mathbb{X}_{i}^{\top}/n)(\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n)^{-1}\\\
0&I\end{pmatrix}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\begin{pmatrix}\mathbb{E}(D(1)-D(0))&\gamma_{D}^{\top}\\\
0&I\end{pmatrix}$

and

$\displaystyle\hat{\lambda}_{TSLS}\equiv\begin{pmatrix}\hat{\alpha}_{1,TSLS}\\\
\vdots\\\ \hat{\alpha}_{S,TSLS}\\\ \hat{\theta}_{TSLS}\end{pmatrix}$
$\displaystyle=(\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n)^{-1}(\sum_{i\in[n]}\mathbb{X}_{i}(Y_{i}-D_{i}\hat{\tau}_{TSLS})/n)$
$\displaystyle=\hat{\lambda}+(\sum_{i\in[n]}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}/n)^{-1}(\sum_{i\in[n]}\mathbb{X}_{i}D_{i}/n)(\tau-\hat{\tau}_{TSLS})\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\lambda^{*}.$

Then, we have

$\displaystyle\hat{e}_{i}=e_{i}-D_{i}(\hat{\tau}_{TSLS}-\tau)-\mathbb{X}_{i}^{\top}(\hat{\lambda}_{TSLS}-\lambda^{*}),$

where $e_{i}=Y_{i}-D_{i}\tau-\mathbb{X}_{i}^{\top}\lambda^{*}$. In addition,
as shown above, we have
$\tilde{A}_{i}=A_{i}-\pi-\mathbb{X}_{i}^{\top}(\hat{\theta}-\theta)$ and
$\hat{\theta}\stackrel{{\scriptstyle p}}{{\longrightarrow}}\theta$. This
implies,

$\displaystyle\frac{1}{n}\sum_{i\in[n]}\hat{e}_{i}^{2}\begin{pmatrix}\tilde{A}_{i}^{2}&\tilde{A}_{i}\mathbb{X}_{i}^{\top}\\\
\tilde{A}_{i}\mathbb{X}_{i}&\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle=\frac{1}{n}\sum_{i\in[n]}e_{i}^{2}\begin{pmatrix}(A_{i}-\pi)^{2}&(A_{i}-\pi)\mathbb{X}_{i}^{\top}\\\
(A_{i}-\pi)\mathbb{X}_{i}&\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}+o_{P}(1)$
$\displaystyle=\frac{1}{n}\sum_{i\in[n]}\left[A_{i}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}+(1-A_{i})(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\right]$
$\displaystyle\times\begin{pmatrix}(A_{i}-\pi)^{2}&(A_{i}-\pi)\mathbb{X}_{i}^{\top}\\\
(A_{i}-\pi)\mathbb{X}_{i}&\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}+o_{P}(1)$
$\displaystyle=\frac{1}{n}\sum_{i\in[n]}A_{i}\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle+\frac{1}{n}\sum_{i\in[n]}(1-A_{i})\begin{pmatrix}\pi^{2}(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}+o_{P}(1).$

For the first term on the RHS of the above display, we have

$\displaystyle\frac{1}{n}\sum_{i\in[n]}A_{i}\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle=\frac{1}{n}\sum_{i\in[n]}A_{i}\biggl{\\{}\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle-\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]\biggr{\\}}$
$\displaystyle+\frac{1}{n}\sum_{i\in[n]}(A_{i}-\pi)\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]$
$\displaystyle+\frac{1}{n}\sum_{i\in[n]}\pi\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]$
$\displaystyle\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\pi\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\right]$
$\displaystyle\equiv\Omega_{1}.$

To see the convergence in probability in the above display, we note that

$\displaystyle
A_{i}\biggl{\\{}\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle-\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]\biggr{\\}}$

is independent and conditionally mean zero given $(A^{(n)},S^{(n)})$.
Therefore, by the conditional Chebyshev’s inequality, we have

$\displaystyle\frac{1}{n}\sum_{i\in[n]}A_{i}\biggl{\\{}\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle-\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]\biggr{\\}}$
$\displaystyle=o_{P}(1).$

Also, by Assumption 2, we have

$\displaystyle\frac{1}{n}\sum_{i\in[n]}(A_{i}-\pi)\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]$
$\displaystyle=o_{P}(1).$

Last, by the usual Law of Large numbers for i.i.d. data, we have

$\displaystyle\frac{1}{n}\sum_{i\in[n]}\pi\mathbb{E}\left[\begin{pmatrix}(1-\pi)^{2}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
(1-\pi)(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\bigg{|}S_{i}\right]$
$\displaystyle\stackrel{{\scriptstyle p}}{{\longrightarrow}}\Omega_{1}.$

Similarly, we have

$\displaystyle\frac{1}{n}\sum_{i\in[n]}(1-A_{i})\begin{pmatrix}\pi^{2}(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}$
$\displaystyle\stackrel{{\scriptstyle
p}}{{\longrightarrow}}(1-\pi)\mathbb{E}\left[\begin{pmatrix}\pi^{2}(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}&-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}^{\top}\\\
-\pi(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}&(Y_{i}(D_{i}(0))-D_{i}(0)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}\mathbb{X}_{i}\mathbb{X}_{i}^{\top}\end{pmatrix}\right]$
$\displaystyle\equiv\Omega_{0}.$

Consequently, we have

$\displaystyle\hat{\sigma}_{TSLS,naive}^{2}$
$\displaystyle\stackrel{{\scriptstyle
p}}{{\longrightarrow}}\frac{\begin{pmatrix}1&-\gamma_{D}^{\top}\end{pmatrix}\begin{pmatrix}\mathbb{E}(D(1)-D(0))&\gamma_{D}^{\top}\\\
0&I\end{pmatrix}(\Omega_{1}+\Omega_{0})\begin{pmatrix}\mathbb{E}(D(1)-D(0))&\gamma_{D}^{\top}\\\
0&I\end{pmatrix}^{\top}\begin{pmatrix}1\\\
-\gamma_{D}\end{pmatrix}}{[\pi(1-\pi)]^{2}[\mathbb{E}(D(1)-D(0))]^{4}}$
$\displaystyle=\frac{\pi^{-1}\mathbb{E}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}+(1-\pi)^{-1}\mathbb{E}(Y_{i}(D_{i}(1))-D_{i}(1)\tau-\mathbb{X}_{i}^{\top}\lambda^{*})^{2}}{[\mathbb{E}(D(1)-D(0))]^{2}}$
$\displaystyle=\frac{\sigma_{TSLS,0}^{2}+\sigma_{TSLS,1}^{2}+\sigma_{TSLS,2}^{2}+\tilde{\sigma}_{TSLS,3}^{2}}{(\mathbb{E}(D_{i}(1)-D_{i}(0)))^{2}}.$

For the last result, by the proof Theorem 3.1 with
$\overline{\mu}^{b}(a,s,x)=0$ for $a=0,1$ and $b=D,Y$ and $\pi(s)=\pi$, we
have

$\displaystyle\sigma_{NA}^{2}$ $\displaystyle=\frac{\sum_{s\in
S}\frac{p(s)}{\pi}Var(Y(D(1))-\tau D(1)|S=s)+\sum_{s\in
S}\frac{p(s)}{1-\pi}Var(Y(D(0))-\tau D(0)|S=s)}{\mathbb{P}(D(1)>D(0))^{2}}$
$\displaystyle+\frac{Var(\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\tau\left(\mathbb{E}[D_{i}(1)-D_{i}(0)|S_{i}]\right))}{\mathbb{P}(D(1)>D(0))^{2}}$
$\displaystyle=\frac{\mathbb{E}\frac{1}{\pi}Var(Y(D(1))-\tau
D(1)|S)+\frac{1}{1-\pi}Var(Y(D(0))-\tau
D(0)|S)}{\mathbb{P}(D(1)>D(0))^{2}}+\frac{\sigma_{TSLS,2}^{2}}{\mathbb{P}(D(1)>D(0))^{2}}.$

Then, we have $\sigma_{NA}^{2}<\sigma_{TSLS}^{2}$ if and only if

$\mathbb{E}\left[\frac{1}{\pi}Var(Y(D(1))-\tau
D(1)|S)+\frac{1}{1-\pi}Var(Y(D(0))-\tau
D(0)|S)\right]<\sigma_{TSLS,0}^{2}+\sigma_{TSLS,1}^{2}+\sigma_{TSLS,3}^{2},$

which is equivalent to

$\displaystyle
2\left[\frac{\mathbb{E}cov(Y_{i}(D_{i}(1))-D_{i}(1)\tau,\mathbb{X}_{i}^{\top}\lambda^{*}|S)}{\pi}+\frac{\mathbb{E}cov(Y_{i}(D_{i}(0))-D_{i}(0)\tau,\mathbb{X}_{i}^{\top}\lambda^{*}|S)}{1-\pi}\right]\leq\frac{\mathbb{E}Var(\mathbb{X}_{i}^{\top}\lambda^{*}|S)}{\pi(1-\pi)}+\sigma_{TSLS,3}^{2}.$

##  Appendix F Proof of Theorem 3.1

Let

$\displaystyle G$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}\mathinner{\Bigl{[}\mathinner{\bigl{(}Y(1)-Y(0)\bigr{)}}\mathinner{\bigl{(}D(1)-D(0)\bigr{)}}\Bigr{]}},$
$\displaystyle H$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}\mathinner{\bigl{[}D(1)-D(0)\bigr{]}},$
$\displaystyle\hat{G}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{n}\sum_{i\in[n]}\left[\frac{A_{i}(Y_{i}-\hat{\mu}^{Y}(1,S_{i},X_{i}))}{\hat{\pi}(S_{i})}-\frac{(1-A_{i})(Y_{i}-\hat{\mu}^{Y}(0,S_{i},X_{i}))}{1-\hat{\pi}(S_{i})}+\hat{\mu}^{Y}(1,S_{i},X_{i})-\hat{\mu}^{Y}(0,S_{i},X_{i})\right],$
$\displaystyle\hat{H}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{n}\sum_{i\in[n]}\left[\frac{A_{i}(D_{i}-\hat{\mu}^{D}(1,S_{i},X_{i}))}{\hat{\pi}(S_{i})}-\frac{(1-A_{i})(D_{i}-\hat{\mu}^{D}(0,S_{i},X_{i}))}{1-\hat{\pi}(S_{i})}+\hat{\mu}^{D}(1,S_{i},X_{i})-\hat{\mu}^{D}(0,S_{i},X_{i})\right].$

Then, we have

$\displaystyle\sqrt{n}(\hat{\tau}-\tau)$
$\displaystyle=\sqrt{n}\mathinner{\biggl{(}\frac{\hat{G}}{\hat{H}}-\frac{G}{H}\biggr{)}}$
$\displaystyle=\frac{1}{\hat{H}}\sqrt{n}(\hat{G}-G)-\frac{G}{\hat{H}H}\sqrt{n}(\hat{H}-H)$
$\displaystyle=\frac{1}{\hat{H}}\mathinner{\biggl{[}\sqrt{n}(\hat{G}-G)-\tau\sqrt{n}(\hat{H}-H)\biggr{]}}.$
(F.1)

Next, we divide the proof into three steps. In the first step, we obtain the
linear expansion of $\sqrt{n}(\hat{G}-G)$. Based on the same argument, we can
obtain the linear expansion of $\sqrt{n}(\hat{H}-H)$. In the second step, we
obtain the linear expansion of $\sqrt{n}(\hat{\tau}-\tau)$ and then prove the
asymptotic normality. In the third step, we show the consistency of
$\hat{\sigma}$. The second result in the Theorem is obvious given the
semiparametric efficiency bound derived in Theorem 4.1.

Step 1. We have

$\displaystyle\sqrt{n}(\hat{G}-G)$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i\in[n]}\biggl{[}\frac{A_{i}(Y_{i}-\hat{\mu}^{Y}(1,S_{i},X_{i}))}{\hat{\pi}(S_{i})}-\frac{(1-A_{i})(Y_{i}-\hat{\mu}^{Y}(0,S_{i},X_{i}))}{1-\hat{\pi}(S_{i})}$
$\displaystyle+\hat{\mu}^{Y}(1,S_{i},X_{i})-\hat{\mu}^{Y}(0,S_{i},X_{i})\biggr{]}-\sqrt{n}G$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{[}\hat{\mu}^{Y}(1,S_{i},X_{i})-\frac{A_{i}\hat{\mu}^{Y}(1,S_{i},X_{i})}{\hat{\pi}(S_{i})}\biggr{]}}$
$\displaystyle+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{[}\frac{(1-A_{i})\hat{\mu}^{Y}(0,S_{i},X_{i})}{1-\hat{\pi}(S_{i})}-\hat{\mu}^{Y}(0,S_{i},X_{i})\biggr{]}}$
$\displaystyle+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}Y_{i}}{\hat{\pi}(S_{i})}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{(1-A_{i})Y_{i}}{1-\hat{\pi}(S_{i})}-\sqrt{n}G$
$\displaystyle=\mathrel{\mathop{\mathchar 58\relax}}R_{n,1}+R_{n,2}+R_{n,3},$

where

$\displaystyle R_{n,1}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{[}\hat{\mu}^{Y}(1,S_{i},X_{i})-\frac{A_{i}\hat{\mu}^{Y}(1,S_{i},X_{i})}{\hat{\pi}(S_{i})}\biggr{]}},$
$\displaystyle R_{n,2}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{[}\frac{(1-A_{i})\hat{\mu}^{Y}(0,S_{i},X_{i})}{1-\hat{\pi}(S_{i})}-\hat{\mu}^{Y}(0,S_{i},X_{i})\biggr{]}},$
$\displaystyle R_{n,3}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}Y_{i}}{\hat{\pi}(S_{i})}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{(1-A_{i})Y_{i}}{1-\hat{\pi}(S_{i})}-\sqrt{n}G.$

Lemma P.1 shows that

$\displaystyle R_{n,1}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\pi(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1),$
$\displaystyle R_{n,2}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}\frac{1}{1-\pi(S_{i})}-1\biggr{)}}(1-A_{i})\tilde{\mu}^{Y}(0,S_{i},X_{i})-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(0,S_{i},X_{i})+o_{p}(1),$
$\displaystyle R_{n,3}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(S_{i})}\tilde{W}_{i}A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\pi(S_{i})}\tilde{Z}_{i}+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}.$

This implies

$\displaystyle\sqrt{n}(\hat{G}-G)$
$\displaystyle=\biggl{\\{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{\mu}^{Y}(1,S_{i},X_{i})-\tilde{\mu}^{Y}(0,S_{i},X_{i})+\frac{\tilde{W}_{i}}{\pi(S_{i})}\right]A_{i}$
$\displaystyle+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{\mu}^{Y}(0,S_{i},X_{i})+\tilde{\mu}^{Y}(1,S_{i},X_{i})-\frac{\tilde{Z}_{i}}{1-\pi(S_{i})}\right](1-A_{i})\biggr{\\}}$
$\displaystyle+\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}\right\\}+o_{p}(1).$
(F.2)

Similarly, we can show that

$\displaystyle\sqrt{n}(\hat{H}-H)$
$\displaystyle=\biggl{\\{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{\mu}^{D}(1,S_{i},X_{i})-\tilde{\mu}^{D}(0,S_{i},X_{i})+\frac{\tilde{D}_{i}(1)}{\pi(S_{i})}\right]A_{i}$
$\displaystyle+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{\mu}^{D}(0,S_{i},X_{i})+\tilde{\mu}^{D}(1,S_{i},X_{i})-\frac{\tilde{D}_{i}(0)}{1-\pi(S_{i})}\right](1-A_{i})\biggr{\\}}$
$\displaystyle+\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[D_{i}(1)-D_{i}(0)|S_{i}]-\mathbb{E}[D_{i}(1)-D_{i}(0)]\bigr{)}}\right\\}+o_{p}(1),$
(F.3)

where $\tilde{D}_{i}(a)=D_{i}(a)-\mathbb{E}(D_{i}(a)|S_{i})$ for $a=0,1$ and
$\tilde{\mu}^{D}(0,s,X_{i})=\overline{\mu}^{D}(0,s,X_{i})-\mathbb{E}(\overline{\mu}^{D}(0,S_{i},X_{i})|S_{i}=s).$

Combining (F.1), (F.2), and (F.3), we obtain the linear expansion for
$\hat{\tau}$ as

$\displaystyle\sqrt{n}(\hat{\tau}-\tau)$
$\displaystyle=\frac{1}{\hat{H}}\left[\sqrt{n}(\hat{G}-G)-\tau\sqrt{n}(\hat{H}-H)\right]$
$\displaystyle=\frac{1}{\hat{H}}\left[\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i}+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\right]+o_{p}(1),$

where $\mathcal{D}_{i}=\\{Y_{i}(1),Y_{i}(0),D_{i}(1),D_{i}(0),X_{i}\\}$,

$\displaystyle\Xi_{1}(\mathcal{D}_{i},S_{i})$
$\displaystyle=\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{\mu}^{Y}(1,S_{i},X_{i})-\tilde{\mu}^{Y}(0,S_{i},X_{i})+\frac{\tilde{W}_{i}}{\pi(S_{i})}\right]$
$\displaystyle-\tau\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{\mu}^{D}(1,S_{i},X_{i})-\tilde{\mu}^{D}(0,S_{i},X_{i})+\frac{\tilde{D}_{i}(1)}{\pi(S_{i})}\right],$
$\displaystyle\Xi_{0}(\mathcal{D}_{i},S_{i})$
$\displaystyle=\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{\mu}^{Y}(0,S_{i},X_{i})+\tilde{\mu}^{Y}(1,S_{i},X_{i})-\frac{\tilde{Z}_{i}}{1-\pi(S_{i})}\right]$
$\displaystyle-\tau\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{\mu}^{D}(0,S_{i},X_{i})+\tilde{\mu}^{D}(1,S_{i},X_{i})-\frac{\tilde{D}_{i}(0)}{1-\pi(S_{i})}\right],$
$\displaystyle\Xi_{2}(S_{i})$
$\displaystyle=\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}-\tau\left[\mathbb{E}[D_{i}(1)-D_{i}(0)|S_{i}]-\mathbb{E}[D_{i}(1)-D_{i}(0)]\right].$

Step 2. Lemma P.2 implies that

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i}\rightsquigarrow\mathcal{N}(0,\sigma_{1}^{2}),\quad\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i})\rightsquigarrow\mathcal{N}(0,\sigma_{0}^{2}),\quad\text{and}$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\rightsquigarrow\mathcal{N}(0,\sigma_{2}^{2}),$

and the three terms are asymptotically independent, where

$\displaystyle\sigma_{1}^{2}=\mathbb{E}\pi(S_{i})\Xi_{1}^{2}(\mathcal{D}_{i},S_{i}),\quad\sigma_{0}^{2}=\mathbb{E}(1-\pi(S_{i}))\Xi_{0}^{2}(\mathcal{D}_{i},S_{i}),\quad\text{and}\quad\sigma_{2}^{2}=\mathbb{E}\Xi_{2}^{2}(S_{i}).$

This further implies $\hat{H}\overset{p}{\longrightarrow}H$ and

$\displaystyle\sqrt{n}(\hat{\tau}-\tau)\rightsquigarrow\mathcal{N}\left(0,\frac{\sigma_{1}^{2}+\sigma_{0}^{2}+\sigma_{2}^{2}}{H^{2}}\right),$

Step 3. We aim to show the consistency of $\hat{\sigma}^{2}$. First note that

$\frac{1}{n}\sum_{i=1}^{n}\Xi_{H,i}=\hat{H}\overset{p}{\longrightarrow}H=\mathbb{E}(D_{i}(1)-D_{i}(0)).$

In addition, Lemma P.3 shows.

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}A_{i}\hat{\Xi}_{1}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{1}^{2},\quad\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})\hat{\Xi}_{0}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{0}^{2},\quad\text{and}\quad\frac{1}{n}\sum_{i=1}^{n}\hat{\Xi}_{2}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{2}^{2}.$

This implies $\hat{\sigma}^{2}\overset{p}{\longrightarrow}\sigma^{2}$.

##  Appendix G Proof of Theorem 4.1

Without loss of generality, we assume
$A_{i}=\phi_{i}(\\{S_{i}\\}_{i\in[n]},U)$, where $\phi_{i}(\cdot)$ is a
deterministic function and $U$ is a random variable (vector) with density
$P_{U}(\cdot)$ and is independent of everything else in the data. Further
denote $\mathcal{Y}_{i}(a)=\\{Y_{i}(D_{i}(a)),D_{i}(a),X_{i}\\}$. We consider
parametric submodels indexed by a generic parameter $\theta$. The likelihoods
of $S_{i}$ evaluated at $s$ and $\mathcal{Y}_{i}(a)$ given $S_{i}=s$ evaluated
at $\overline{y}$ are written as $f_{S}(s;\theta)$ and
$f_{\mathcal{Y}(a)|S}(\overline{y}|s;\theta)$ for $a=0,1$, respectively. The
density of $U$ does not depend on $\theta$. Let
$\theta_{n}=\theta^{*}+h/\sqrt{n}$, where $\theta^{*}$ indexes the true
underlying DGP.

By Assumption 1, the joint likelihood of
$\\{Y_{i},X_{i},S_{i},A_{i}\\}_{i\in[n]}$ under $\theta$ can be written as

$\displaystyle
P_{U}(u)\Pi_{i\in[n]}\left[f_{S}(s_{i};\theta)\Pi_{a=0,1}f_{\mathcal{Y}(a)|S}(\tilde{y}_{i}(a)|s_{i};\theta)^{1\\{\phi_{i}(s_{1},\cdots,s_{n},u)=a\\}}\right]$

where $(x_{i},y_{i}(d_{i}(a)),d_{i}(a),u,s_{i})$ are the realizations
$(X_{i},Y_{i}(D_{i}(a)),D_{i}(a),U,S_{i})$ for $i\in[n]$ and
$\tilde{y}_{i}(a)=\\{y_{i}(d_{i}(a)),d_{i}(a),x_{i}\\}$. We make the following
regularity assumptions with respect to the submodel.

######  Assumption 10.

  1. (i)

Suppose $f_{S}(s;\theta)$ and $f_{\mathcal{Y}(a)|S}(\overline{y}|s;\theta)$
for $a=0,1$ are differentiable in quadratic mean at $\theta^{*}$ with score
functions $g_{s}(S_{i})$ and $g_{a}(\mathcal{Y}_{i}(a)|S_{i})$ for $a=0,1$,
respectively, such that

$\displaystyle\dot{f}_{\mathcal{Y}(a)|S}(\overline{y}|s;\theta)=\frac{\partial\log(f_{\mathcal{Y}(a)|S}(\overline{y}|s;\theta))}{\partial\theta},\quad\dot{f}_{S}(s;\theta)=\frac{\partial\log(f_{S}(s;\theta))}{\partial\theta},$
$\displaystyle\dot{f}_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}(a)|S_{i};\theta^{*})=g_{a}(\mathcal{Y}_{i}(a)|S_{i}),\quad\text{and}\quad\dot{f}_{S}(S_{i};\theta^{*})=g_{s}(S_{i}).$

  2. (ii)

Suppose $\dot{f}_{\mathcal{Y}(a)|S}(\overline{y}|s;\theta)$ and
$\dot{f}_{S}(s;\theta)$ are continuous at $\theta^{*}$ so that there exist a
sequence $t_{n}=o(1)$ and a function $L_{a}(\mathcal{Y}_{i}(a),S_{i})$ such
that

$\displaystyle|\dot{f}_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}(a)|S_{i};\theta^{*}+h/\sqrt{n})-g_{a}(\mathcal{Y}_{i}(a)|S_{i})|+|\dot{f}_{S}(S_{i};\theta^{*}+h/\sqrt{n})-g_{s}(S_{i})|\leq
t_{n}L_{a}(\mathcal{Y}_{i}(a),S_{i})$

and $\mathbb{E}|Y_{i}(D_{i}(a))L_{a}(\mathcal{Y}_{i}(a),S_{i})|<\infty$ for
$a=0,1$.

  3. (iii)

Suppose there exists a constant $C>0$ such that

$\displaystyle\max_{s\in\mathcal{S}}\mathbb{E}\mathinner{\Bigl{[}\left.\mathinner{\\!\bigl{\lvert}\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})g_{1}(\mathcal{Y}_{i}(1)|S_{i})\bigr{\rvert}}+\mathinner{\\!\bigl{\lvert}\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i})\bigr{\rvert}}\right|S_{i}=s\Bigr{]}}$
$\displaystyle\leq C$
$\displaystyle\max_{s\in\mathcal{S}}\mathbb{E}\mathinner{\Bigl{[}\left.\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})^{2}+\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})^{2}\right|S_{i}=s\Bigr{]}}$
$\displaystyle\leq C,$

where $\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})$,
$\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})$ and $\underline{\Xi}_{2}(S_{i})$
are defined as $\Xi_{1}(\mathcal{D}_{i},S_{i})$,
$\Xi_{0}(\mathcal{D}_{i},S_{i})$ and $\Xi_{2}(S_{i})$ in (3.7)–(3.9),
respectively, with the researcher-specified working model
$\overline{\mu}^{b}(a,s,x)$ equal to the true specification $\mu^{b}(a,s,x)$
for all $(a,b,s,x)\in\\{0,1\\}\times\\{D,Y\\}\times\mathcal{SX}$.

We denote
$\tau(\theta)=\mathbb{E}_{\theta}(Y_{i}(1)-Y_{i}(0)|D_{i}(1)>D_{i}(0))$, where
$\mathbb{E}_{\theta}(\cdot)$ means the expectation is taken with the
parametric submodel indexed by $\theta$. We further denote
$\mathbb{E}(\cdot)=\mathbb{E}_{\theta^{*}}(\cdot)$, which is the expectation
with respect to the true DGP.

Proof of Theorem 4.1. Following the same argument in Armstrong (2022), in
order to show the semiparametric efficiency bound, we only need to show (1)
local asymptotic normality of the log likelihood ratio for the parametric
submodel with tangent set of the form

$\displaystyle\mathbb{T}=\begin{pmatrix}&\Psi(\mathcal{D}_{i},S_{i},A_{i})=g_{s}(S_{i})+A_{i}g_{1}(\mathcal{Y}_{i}(1)|S_{i})+(1-A_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i})\mathrel{\mathop{\mathchar
58\relax}}\\\
&\mathbb{E}\mathinner{\bigl{[}g_{s}^{2}(S_{i})+\sum_{a=0,1}g_{a}^{2}(\mathcal{Y}_{i}(a)|S_{i})\bigr{]}}<\infty,\mathbb{E}g_{s}(S_{i})=0,\mathbb{E}(g_{a}(\mathcal{Y}_{i}(a)|S_{i})|S_{i})=0,\\\
&\mathbb{E}(g_{1}(\mathcal{Y}_{i}(1)|S_{i})|X_{i},S_{i})=\mathbb{E}(g_{0}(\mathcal{Y}_{i}(0)|S_{i})|X_{i},S_{i})\end{pmatrix}.$
(G.1)

and (2)
$\sqrt{n}(\tau(\theta^{*}+h/\sqrt{n})-\tau(\theta^{*}))=\langle\tilde{\Psi},\Psi\rangle_{\bar{\mathbb{P}}}h+o(1)$,
where $\tilde{\Psi}(\mathcal{D}_{i},S_{i},A_{i})$ is the efficient score
defined as

$\displaystyle\tilde{\Psi}(\mathcal{D}_{i},S_{i},A_{i})=\mathinner{\bigl{[}\underline{\Xi}_{2}(S_{i})+A_{i}\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})+(1-A_{i})\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})\bigr{]}}/\mathbb{E}[D_{i}(1)-D_{i}(0)]$
(G.2)

and
$\langle\tilde{\Psi},\Psi\rangle_{\bar{\mathbb{P}}}=\frac{1}{n}\sum_{i\in[n]}\mathbb{E}\tilde{\Psi}(\mathcal{D}_{i},S_{i},A_{i})\Psi(\mathcal{D}_{i},S_{i},A_{i})$
is the inner product w.r.t. measure
$\bar{\mathbb{P}}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{n}\sum_{i\in[n]}\mathbb{P}_{i}$. We establish these two
results in two steps.

##### Step 1.

Denote $\theta_{n}=\theta^{*}+h/\sqrt{n}$ where $\theta^{*}$ is fixed and
$\mathbb{P}_{n,h}$ as the joint distribution of
$\\{Y_{i},X_{i},S_{i},A_{i}\\}_{i\in[n]}$ under $\theta_{n}$. The log
likelihood ratio for $\theta_{n}$ against $\theta^{*}$ is given by

$\displaystyle\ell_{n,h}=\sum_{i\in[n]}\tilde{\ell}_{s}(S_{i};\theta_{n})+\sum_{a=0,1}\sum_{i\in[n]}1\\{A_{i}=a\\}\tilde{\ell}_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}|S_{i};\theta_{n}),$

where $\mathcal{Y}_{i}=(Y_{i},D_{i},X_{i})$,
$\tilde{\ell}_{s}(S_{i};\theta_{n})=\log\left(\frac{f_{S}(S_{i};\theta_{n})}{f_{S}(S_{i};\theta^{*})}\right)$,
and
$\tilde{\ell}_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}|S_{i};\theta_{n})=\log\left(\frac{f_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}|S_{i};\theta_{n})}{f_{\mathcal{Y}(a)|S}(\mathcal{Y}_{i}|S_{i};\theta^{*})}\right)$
for $a=0,1$. Then, Armstrong (2022, Corollary 3.1) shows $\ell_{n,h}$
converges in distribution to a
$\mathcal{N}(-h^{\prime}\tilde{I}^{*}h/2,h^{\prime}\tilde{I}^{*}h)$ law under
$\theta^{*}$ where $\tilde{I}^{*}$ is the limit of

$\displaystyle\mathbb{E}_{\theta^{*}}g_{s}^{2}(S_{i})+\frac{1}{n}\sum_{i\in[n]}\sum_{a=0,1}1\\{A_{i}=a\\}\mathbb{E}_{\theta^{*}}\left[g_{a}^{2}(\mathcal{Y}_{i}(a)|S_{i})|S_{i}\right].$

and the score for this parametric submodel can be written as

$\displaystyle\Psi(\mathcal{D}_{i},S_{i},A_{i})=g_{s}(S_{i})+A_{i}g_{1}(\mathcal{Y}_{i}(1)|S_{i})+(1-A_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i}).$
(G.3)

We note that by definition, we have

$\displaystyle\mathbb{E}g_{s}(S_{i})=0\quad\text{and}\quad\mathbb{E}(g_{a}(\mathcal{Y}_{i}(a)|S_{i})|S_{i})=0.$

In addition, we have the equality that, for an arbitrary function $h(\cdot)$
of $X$ such that $\mathbb{E}h^{2}(X)<\infty$,

$\displaystyle\mathbb{E}_{\theta}(h(X)|S)=\int_{x}h(x)f_{X|S}(x|S;\theta)\text{d}x$
$\displaystyle=\int_{x}h(x)\mathinner{\biggl{[}\int_{y(d(a)),d(a)}f_{Y(D(a)),D(a)|X,S}(y(d(a)),d(a)|x,S;\theta)\text{d}y(d(a))\text{d}d(a)\biggr{]}}f_{X|S}(x|S;\theta)\text{d}x$
$\displaystyle=\int_{y(d(a)),d(a),x}h(x)f_{\mathcal{Y}(a)|S}(y(d(a)),d(a),x|S;\theta)\text{d}y(d(a))\text{d}d(a)\text{d}x$
(G.4)

for $a=0,1$, where $f_{\mathcal{Y}(a)|S}(y(d(a)),d(a),x|s;\theta)$ is the
joint likelihood of $(Y(D(a)),D(a),X)$ given $S$ for $a=0,1$. We note that,
for $a=0,1$,

$\displaystyle\frac{\partial
f_{\mathcal{Y}(a)|S}(y(d(a)),d(a),x|S;\theta^{*})}{\partial\theta}$
$\displaystyle=f_{\mathcal{Y}(a)|S}(y(d(a)),d(a),x|S;\theta^{*})g_{a}(\mathcal{Y}(a)|S).$

Therefore, taking derivatives of $\theta$ in (G.4) and evaluating the
derivatives at $\theta^{*}$, we have

$\displaystyle\mathbb{E}\mathinner{\bigl{[}h(X)g_{1}(\mathcal{Y}(1)|S)|S\bigr{]}}=\mathbb{E}\mathinner{\bigl{[}h(X)g_{0}(\mathcal{Y}(0)|S)|S\bigr{]}},$

which implies
$\mathbb{E}\mathinner{\bigl{[}g_{1}(\mathcal{Y}(1)|S)-g_{0}(\mathcal{Y}(0)|S)|X,S\bigr{]}}=0$.
Therefore, the tangent set can be written in (G.1).

##### Step 2.

We have

$\displaystyle\tau(\theta)=\frac{\mathbb{E}_{\theta}(Y_{i}(D_{i}(1))-Y_{i}(D_{i}(0)))}{\mathbb{E}_{\theta}(D_{i}(1)-D_{i}(0))}.$

By the mean-value theorem, we have

$\displaystyle\tau(\theta^{*}+h/\sqrt{n})-\tau(\theta^{*})$
$\displaystyle=\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\tilde{\theta}}\frac{h}{\sqrt{n}}$
$\displaystyle=\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}\frac{h}{\sqrt{n}}+\left[\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\tilde{\theta}}-\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}\right]\frac{h}{\sqrt{n}}.$

Let
$G(\theta)=\mathbb{E}_{\theta}\mathinner{\bigl{[}Y(D(1))-Y(D(0))\bigr{]}}$,
$H(\theta)=\mathbb{E}_{\theta}\mathinner{\bigl{[}D(1)-D(0)\bigr{]}}$,
$G=G(\theta^{*})$, and $H=H(\theta^{*})$. Note that
$\tau(\theta)=G(\theta)/H(\theta)$ and $\tau=G/H$. Then, we have

$\displaystyle\frac{\partial G(\theta)}{\partial\theta}$
$\displaystyle=\mathbb{E}_{\theta}\mathinner{\bigl{[}Y(D(1))(\dot{f}_{\mathcal{Y}(1)|S}(\mathcal{Y}(1)|S;\theta)+\dot{f}_{S}(S;\theta)\bigr{]}}-\mathbb{E}_{\theta}\mathinner{\bigl{[}Y(D(0))(\dot{f}_{\mathcal{Y}(0)|S}(\mathcal{Y}(0)|S;\theta)+\dot{f}_{S}(S;\theta))\bigr{]}}$
$\displaystyle\frac{\partial H(\theta)}{\partial\theta}$
$\displaystyle=\mathbb{E}_{\theta}\mathinner{\bigl{[}D(1)(\dot{f}_{\mathcal{Y}(1)|S}(\mathcal{Y}(1)|S;\theta)+\dot{f}_{S}(S;\theta))\bigr{]}}-\mathbb{E}_{\theta}\mathinner{\bigl{[}D(0)(\dot{f}_{\mathcal{Y}(0)|S}(\mathcal{Y}(0)|S;\theta)+\dot{f}_{S}(S;\theta))\bigr{]}}.$

Therefore by Assumption 10 we can find a constant $L$ such that

$\displaystyle\biggl{|}\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\tilde{\theta}}-\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}\biggr{|}$
$\displaystyle=\biggl{|}\frac{H(\tilde{\theta})\frac{\partial
G(\tilde{\theta})}{\partial\theta}-G(\tilde{\theta})\frac{\partial
H(\tilde{\theta})}{\partial\theta}}{H^{2}(\tilde{\theta})}-\frac{H(\theta^{*})\frac{\partial
G(\theta^{*})}{\partial\theta}-G(\theta^{*})\frac{\partial
H(\theta^{*})}{\partial\theta}}{H^{2}(\theta^{*})}\biggr{|}$
$\displaystyle\leq t_{n}L.$

This implies

$\displaystyle\sqrt{n}(\tau(\theta^{*}+h/\sqrt{n})-\tau(\theta^{*}))=\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}h+o(1).$
(G.5)

In addition, following the calculation by Frölich (2007), we have

$\displaystyle\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}=\frac{\left[\frac{\partial
G(\theta)}{\partial\theta}-\tau\frac{\partial
H(\theta)}{\partial\theta}\right]\bigg{|}_{\theta=\theta^{*}}}{H}$
$\displaystyle=\frac{\mathbb{E}\left[(Y(D(1))-\tau
D(1))(g_{1}(\mathcal{Y}(1)|S)+g_{s}(S))\right]}{H}-\frac{\mathbb{E}\left[(Y(D(0))-\tau
D(0))(g_{0}(\mathcal{Y}(0)|S)+g_{s}(S))\right]}{H},$

where for notation simplicity, we write $\mathbb{E}_{\theta^{*}}$ as
$\mathbb{E}$. Let

$\displaystyle\Gamma(X,S)$
$\displaystyle=\biggl{[}\pi(S)(\mathbb{E}(Z|X,S)-\mathbb{E}(Z|S))+\left(1-\pi(S)\right)(\mathbb{E}(W|X,S)-\mathbb{E}(W|S))$
$\displaystyle-\tau\left(\pi(S)(\mathbb{E}(D(0)|X,S)-\mathbb{E}(D(0)|S))+\left(1-\pi(S)\right)(\mathbb{E}(D(1)|X,S)-\mathbb{E}(D(1)|S))\right)\biggr{]}.$

Then, we have

$\displaystyle Y(D(1))-\tau
D(1)=\pi(S)\underline{\Xi}_{1}(\mathcal{D},S)+\mathbb{E}(W-\tau
D(1)|S)+\Gamma(X,S),$ $\displaystyle Y(D(0))-\tau
D(0)=-(1-\pi(S))\underline{\Xi}_{0}(\mathcal{D},S)+\mathbb{E}(Z-\tau
D(0)|S)+\Gamma(X,S).$

This implies

$\displaystyle\mathbb{E}(Y(D(1))-\tau D(1))(g_{1}(\mathcal{Y}(1)|S)+g_{s}(S))$
$\displaystyle=\mathbb{E}\pi(S)\underline{\Xi}_{1}(\mathcal{D},S)g_{1}(\mathcal{Y}(1)|S)+\mathbb{E}\Gamma(X,S)g_{1}(\mathcal{Y}(1)|S)+\mathbb{E}(\mathbb{E}(W-\tau
D(1)|S)g_{s}(S)),$ $\displaystyle\mathbb{E}(Y(D(0))-\tau
D(0))(g_{0}(\mathcal{Y}(0)|S)+g_{s}(S))$
$\displaystyle=-\mathbb{E}(1-\pi(S))\Xi_{0}(\mathcal{D},S)g_{0}(\mathcal{Y}(0)|S)+\mathbb{E}\Gamma(X,S)g_{0}(\mathcal{Y}(0)|S)+\mathbb{E}(\mathbb{E}(Z-\tau
D(0)|S)g_{s}(S)),$

where we have used $\mathbb{E}[\underline{\Xi}_{a}(\mathcal{D},S)|S]=0$,
$\mathbb{E}[\Gamma(X,S)|S]=0$ and $\mathbb{E}[g_{a}(\mathcal{Y}(a)|S)|S]=0$
for $a=0,1$. Then

$\displaystyle\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}$
$\displaystyle=\frac{\mathbb{E}\pi(S)\underline{\Xi}_{1}(\mathcal{D},S)g_{1}(\mathcal{Y}(1)|S)}{H}+\frac{\mathbb{E}(1-\pi(S))\underline{\Xi}_{0}(\mathcal{D},S)g_{0}(\mathcal{Y}(0)|S)}{H}+\frac{\mathbb{E}g_{s}(S)\underline{\Xi}_{2}(S)}{H}$
$\displaystyle+\frac{\mathbb{E}\Gamma(X,S)(g_{1}(\mathcal{Y}(1)|S)-g_{0}(\mathcal{Y}(0)|S))}{H}$
$\displaystyle=\frac{\mathbb{E}\pi(S)\underline{\Xi}_{1}(\mathcal{D},S)g_{1}(\mathcal{Y}(1)|S)}{H}+\frac{\mathbb{E}(1-\pi(S))\underline{\Xi}_{0}(\mathcal{D},S)g_{0}(\mathcal{Y}(0)|S)}{H}+\frac{\mathbb{E}g_{s}(S)\underline{\Xi}_{2}(S)}{H}.$
(G.6)

where the last equality is due to (G.1).

On the other hand, we note that

$\displaystyle\langle\tilde{\Psi},\Psi\rangle_{\bar{\mathbb{P}}}=\frac{1}{n}\sum_{i\in[n]}\left[\frac{\mathbb{E}\mathinner{\bigl{[}g_{s}(S_{i})\underline{\Xi}_{2}(S_{i})\bigr{]}}}{H}+\frac{\mathbb{E}\mathinner{\bigl{[}A_{i}\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})g_{1}(\mathcal{Y}_{i}(1)|S_{i})\bigr{]}}}{H}+\frac{\mathbb{E}\mathinner{\bigl{[}(1-A_{i})\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i})\bigr{]}}}{H}\right]$
$\displaystyle=\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}+\frac{1}{n}\sum_{i\in[n]}\left[\frac{\mathbb{E}\mathinner{\bigl{[}(A_{i}-\pi(S_{i}))\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})g_{1}(\mathcal{Y}_{i}(1)|S_{i})\bigr{]}}}{H}-\frac{\mathbb{E}\mathinner{\bigl{[}(A_{i}-\pi(S_{i}))\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i})\bigr{]}}}{H}\right].$

In addition, by Assumption 10, we have, for some constant $C>0$, that

$\displaystyle\left|\frac{1}{n}\sum_{i\in[n]}\left[\frac{\mathbb{E}(A_{i}-\pi(S_{i}))\underline{\Xi}_{1}(\mathcal{D}_{i},S_{i})g_{1}(\mathcal{Y}_{i}(1)|S_{i})}{H}-\frac{\mathbb{E}(A_{i}-\pi(S_{i}))\underline{\Xi}_{0}(\mathcal{D}_{i},S_{i})g_{0}(\mathcal{Y}_{i}(0)|S_{i})}{H}\right]\right|$
$\displaystyle\leq\frac{C}{n}\sum_{s\in\mathcal{S}}\mathbb{E}|B_{n}(s)|=o(1),$

where the inequality is by law of iterated expectation and Assumption 10(iii)
and the last equality is due to $\mathbb{E}|B_{n}(s)|/n=o(1)$.171717Since
$|B_{n}(s)/n|\leq 1$, $\\{B_{n}(s)/n\\}$ is uniformly integrable. Then from
$B_{n}(s)/n=o_{p}(1)$, we have $\mathbb{E}|B_{n}(s)|/n=o(1)$. This implies

$\displaystyle\langle\tilde{\Psi},\Psi\rangle_{\bar{\mathbb{P}}}=\frac{\partial\tau(\theta)}{\partial\theta}\bigg{|}_{\theta=\theta^{*}}+o(1).$
(G.7)

Combining (G.5), (G.6) and (G.7), we obtained the desired result for Step 2.
Last, it is obvious from the previous calculation that

$\displaystyle\langle\tilde{\Psi},\tilde{\Psi}\rangle_{\bar{\mathbb{P}}}\rightarrow\underline{\sigma}^{2}.$

##  Appendix H Proof of Theorem 5.1

The proof is divided into two steps. In the first step, we show Assumption
3(i). In the second step, we establish Assumptions 3(ii) and 3(iii).

Step 1. Recall

$\displaystyle\Delta^{Y}(a,s,X_{i})=\hat{\mu}^{Y}(a,s,X_{i})-\overline{\mu}^{Y}(a,s,X_{i})=\Lambda_{a,s}^{Y}(X_{i},\hat{\theta}_{a,s})-\Lambda_{a,s}^{Y}(X_{i},\theta_{a,s}),$

and $\\{X_{i}^{s}\\}_{i\in[n]}$ is generated independently from the
distribution of $X_{i}$ given $S_{i}=s$, and so is independent of
$\\{A_{i},S_{i}\\}_{i\in[n]}$. Let
$M_{a,s}(\theta_{1},\theta_{2})\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}[\Lambda_{a,s}^{Y}(X_{i},\theta_{1})-\Lambda_{a,s}^{Y}(X_{i},\theta_{2})|S_{i}=s]=\mathbb{E}[\Lambda_{a,s}^{Y}(X_{i}^{s},\theta_{1})-\Lambda_{a,s}^{Y}(X_{i}^{s},\theta_{2})]$.
We have

$\displaystyle\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\Delta^{Y}(a,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{Y}(a,s,X_{i})}{n_{0}(s)}\biggr{|}$
$\displaystyle\leq\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}+\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{0}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{0}(s)}\biggr{\rvert}}$
$\displaystyle=o_{p}(n^{-1/2}).$ (H.1)

To see the last equality, we note that, for any $\varepsilon>0$, with
probability approaching one (w.p.a.1), we have

$\displaystyle\max_{s\in\mathcal{S}}||\hat{\theta}_{a,s}-\theta_{a,s}||_{2}\leq\varepsilon.$

Therefore, on the event
$\mathcal{A}_{n}(\varepsilon)\mathrel{\mathop{\mathchar
58\relax}}=\\{\max_{s\in\mathcal{S}}||\hat{\theta}_{a,s}-\theta_{a,s}||_{2}\leq\varepsilon,\min_{s\in\mathcal{S}}n_{1}(s)\geq\varepsilon
n\\}$ we have

$\displaystyle\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\biggl{|}\\{A_{i},S_{i}\\}_{i\in[n]}$
$\displaystyle\overset{d}{=}\mathinner{\\!\biggl{\lvert}\frac{\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}[\Delta^{Y}(a,s,X_{i}^{s})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\biggl{|}\\{A_{i},S_{i}\\}_{i\in[n]}\leq||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}\biggl{|}\\{A_{i},S_{i}\\}_{i\in[n]},$

where

$\displaystyle\mathcal{F}=\\{\Lambda_{a,s}^{Y}(X_{i}^{s},\theta_{1})-\Lambda_{a,s}^{Y}(X_{i}^{s},\theta_{2})-M_{a,s}(\theta_{1},\theta_{2})\mathrel{\mathop{\mathchar
58\relax}}||\theta_{1}-\theta_{2}||_{2}\leq\varepsilon\\}.$

Therefore, for any $\delta>0$ we have

$\displaystyle\mathbb{P}\mathinner{\biggl{(}\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\geq\delta
n^{-1/2}\biggr{)}}$
$\displaystyle\leq\mathbb{P}\mathinner{\biggl{(}\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\geq\delta
n^{-1/2},\mathcal{A}_{n}(\varepsilon)\biggr{)}}+\mathbb{P}(\mathcal{A}_{n}^{c}(\varepsilon))$
$\displaystyle\leq\mathbb{E}\mathinner{\biggl{[}\mathbb{P}\mathinner{\biggl{(}\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\geq\delta
n^{-1/2},\mathcal{A}_{n}(\varepsilon)\biggl{|}\\{A_{i},S_{i}\\}_{i\in[n]}\biggr{)}}\biggr{]}}+\mathbb{P}(\mathcal{A}_{n}^{c}(\varepsilon))$
$\displaystyle\leq\sum_{s\in\mathcal{S}}\mathbb{E}\left[\mathbb{P}\left(||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}\geq\delta
n^{-1/2}\biggl{|}\\{A_{i},S_{i}\\}_{i\in[n]}\right)1\\{n_{1}(s)\geq
n\varepsilon\\}\right]+\mathbb{P}(\mathcal{A}_{n}^{c}(\varepsilon))$
$\displaystyle\leq\sum_{s\in\mathcal{S}}\mathbb{E}\left\\{\frac{n^{1/2}\mathbb{E}\left[||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\right]1\\{n_{1}(s)\geq
n\varepsilon\\}}{\delta}\right\\}+\mathbb{P}(\mathcal{A}_{n}^{c}(\varepsilon)).$

By Assumption 4, $\mathcal{F}$ is a VC-class with a fixed VC index and
envelope $L_{i}$ such that
$\mathbb{E}(L_{i}^{q}|\\{A_{i},S_{i}\\}_{i\in[n]})\leq C<\infty$. This implies
$\mathbb{E}\max_{i\in[n_{1}(s)]}L_{i}^{2}\leq Cn_{1}^{2/q}(s)$. In addition,

$\displaystyle\sup_{f\in\mathcal{F}}\mathbb{P}f^{2}\leq\mathbb{E}L_{i}^{2}(\theta_{1}-\theta_{2})^{2}\leq
C\varepsilon^{2}.$

Invoke Chernozhukov et al. (2014, Corollary 5.1) with $A$ and $\nu$ being
fixed constants, and $\sigma^{2}$, $F$, $M$ being $C\varepsilon^{2}$, $L$,
$\max_{1\leq i\leq n_{1}(s)}L_{i}$, respectively, in our setting. We have

$\displaystyle
n^{1/2}\mathbb{E}\left[||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\right]1\\{n_{1}(s)\geq
n\varepsilon\\}$ $\displaystyle\leq
C\mathinner{\biggl{(}\sqrt{\frac{n}{n_{1}(s)}\varepsilon^{2}\log(1/\varepsilon)}+n^{1/2}n_{1}^{1/q-1}(s)\log(1/\varepsilon)\biggr{)}}1\\{n_{1}(s)\geq
n\varepsilon\\}$ $\displaystyle\leq
C(\varepsilon^{1/2}\log^{1/2}(1/\varepsilon)+n^{1/q-1/2}\varepsilon^{1/q-1}\log(1/\varepsilon)).$

Therefore,

$\displaystyle\mathbb{E}\left\\{\frac{n^{1/2}\mathbb{E}\mathinner{\bigl{[}||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\bigr{]}}1\\{n_{1}(s)\geq
n\varepsilon\\}}{\delta}\right\\}\leq
C\mathbb{E}\left(\varepsilon^{1/2}\log^{1/2}(1/\varepsilon)+n^{1/q-1/2}\varepsilon^{1/q-1}\log(1/\varepsilon)\right)/\delta.$

By letting $n\rightarrow\infty$ followed by $\varepsilon\rightarrow 0$, we
have

$\displaystyle\lim_{n\rightarrow\infty}\mathbb{P}\mathinner{\biggl{(}\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}\geq\delta
n^{-1/2}\biggr{)}}=0,$

Therefore,

$\displaystyle\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{1}(s)}\biggr{\rvert}}=o_{p}(n^{-1/2}).$

For the same reason, we have

$\displaystyle\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{0}(s)}[\Delta^{Y}(a,s,X_{i})-M_{a,s}(\hat{\theta}_{a,s},\theta_{a,s})]}{n_{0}(s)}\biggr{\rvert}}=o_{p}(n^{-1/2}),$

and (H.1) holds.

Step 2. We have

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\Delta^{Y,2}(a,S_{i},X_{i})=\frac{1}{n}\sum_{i=1}^{n}\sum_{s\in\mathcal{S}}1\\{S_{i}=s\\}(\Lambda_{a,s}^{Y}(X_{i},\hat{\theta}_{a,s})-\Lambda_{a,s}^{Y}(X_{i},\theta_{a,s}))^{2}$
$\displaystyle\leq\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}L_{i}^{2}\biggr{)}}C\max_{s\in\mathcal{S}}||\hat{\theta}_{a,s}-\theta_{a,s}||_{2}^{2}=o_{p}(1).$

This verifies Assumption 3(ii). Assumption 3(iii) holds by Assumption 4(ii).

##  Appendix I Proof of Theorem 5.2

Let

$\displaystyle\nu^{Y}(a,S_{i},X_{i})=\mathbb{E}(Y_{i}(D_{i}(a))|S_{i},X_{i})-\mathbb{E}(Y_{i}(D_{i}(a))|S_{i})\quad\text{and}$
$\displaystyle\nu^{D}(a,S_{i},X_{i})=\mathbb{E}(D_{i}(a)|S_{i},X_{i})-\mathbb{E}(D_{i}(a)|S_{i}).$
(I.1)

Also recall that $W_{i}=Y_{i}(D_{i}(1))$, $Z_{i}=Y_{i}(D_{i}(0))$,
$\mu^{Y}(a,S_{i},X_{i})=\mathbb{E}(Y_{i}(D_{i}(a))|S_{i},X_{i})$. Then, we
have

$\displaystyle\mathbb{E}\pi(S_{i})\Xi_{1}^{2}(\mathcal{D}_{i},S_{i})$
$\displaystyle=\mathbb{E}\left\\{\frac{\left(W_{i}-\mu^{Y}(1,S_{i},X_{i})-\tau(D_{i}(1)-\mu^{D}(1,S_{i},X_{i}))\right)^{2}}{\pi(S_{i})}\right\\}$
$\displaystyle+\mathbb{E}\biggl{\\{}\pi(S_{i})\biggl{[}\frac{\nu^{Y}(1,S_{i},X_{i})-\tilde{\mu}^{Y}(1,S_{i},X_{i})-\tau(\nu^{D}(1,S_{i},X_{i})-\tilde{\mu}^{D}(1,S_{i},X_{i}))}{\pi(S_{i})}$
$\displaystyle+\tilde{\mu}^{Y}(1,S_{i},X_{i})-\tilde{\mu}^{Y}(0,S_{i},X_{i})-\tau(\tilde{\mu}^{D}(1,S_{i},X_{i})-\tilde{\mu}^{D}(0,S_{i},X_{i}))\biggr{]}^{2}\biggr{\\}}.$

Similarly, we have

$\displaystyle\mathbb{E}(1-\pi(S_{i}))\Xi_{0}^{2}(\mathcal{D}_{i},S_{i})$
$\displaystyle=\mathbb{E}\left\\{\frac{\left(Z_{i}-\mu^{Y}(0,S_{i},X_{i})-\tau(D_{i}(0)-\mu^{D}(0,S_{i},X_{i}))\right)^{2}}{1-\pi(S_{i})}\right\\}$
$\displaystyle+\mathbb{E}\biggl{\\{}(1-\pi(S_{i}))\biggl{[}\frac{\nu^{Y}(0,S_{i},X_{i})-\tilde{\mu}^{Y}(0,S_{i},X_{i})-\tau(\nu^{D}(0,S_{i},X_{i})-\tilde{\mu}^{D}(0,S_{i},X_{i}))}{1-\pi(S_{i})}$
$\displaystyle-\left(\tilde{\mu}^{Y}(1,S_{i},X_{i})-\tilde{\mu}^{Y}(0,S_{i},X_{i})-\tau(\tilde{\mu}^{D}(1,S_{i},X_{i})-\tilde{\mu}^{D}(0,S_{i},X_{i}))\right)\biggr{]}^{2}\biggr{\\}}.$

Last, we have

$\displaystyle\mathbb{E}\Xi_{2}^{2}(S_{i})$
$\displaystyle=\mathbb{E}(\mu^{Y}(1,S_{i},X_{i})-\mu^{Y}(0,S_{i},X_{i})-\tau(\mu^{D}(1,S_{i},X_{i})-\mu^{D}(0,S_{i},X_{i})))^{2}$
$\displaystyle-\mathbb{E}(\nu^{Y}(1,S_{i},X_{i})-\nu^{Y}(0,S_{i},X_{i})-\tau(\nu^{D}(1,S_{i},X_{i})-\nu^{D}(0,S_{i},X_{i})))^{2}$

Let

$\displaystyle\sigma_{\ast}^{2}$
$\displaystyle=(\mathbb{P}(D_{i}(1)>D_{i}(0)))^{-2}\biggl{\\{}\mathbb{E}\mathinner{\biggl{[}\frac{\mathinner{\bigl{(}W_{i}-\mu^{Y}(1,S_{i},X_{i})-\tau(D_{i}(1)-\mu^{D}(1,S_{i},X_{i}))\bigr{)}}^{2}}{\pi(S_{i})}\biggr{]}}$
$\displaystyle+\mathbb{E}\mathinner{\biggl{[}\frac{\mathinner{\bigl{(}Z_{i}-\mu^{Y}(0,S_{i},X_{i})-\tau[D_{i}(0)-\mu^{D}(0,S_{i},X_{i})]\bigr{)}}^{2}}{1-\pi(S_{i})}\biggr{]}}$
$\displaystyle+\mathbb{E}\mathinner{\Bigl{(}\mu^{Y}(1,S_{i},X_{i})-\mu^{Y}(0,S_{i},X_{i})-\tau[\mu^{D}(1,S_{i},X_{i})-\mu^{D}(0,S_{i},X_{i})]\Bigr{)}}^{2}\biggr{\\}},$

which does not depend on the working models
$\overline{\mu}^{b}(a,S_{i},X_{i})$ for $a=0,1$ and $b=D,Y$. Then, we have

$\displaystyle\sigma^{2}((t_{a,s},b_{a,s})_{a=0,1,s\in\mathcal{S}})=\frac{\sigma_{*}^{2}+V((t_{a,s},b_{a,s})_{a=0,1,s\in\mathcal{S}})}{\mathbb{P}(D_{i}(1)>D_{i}(0))^{2}},$

where $\sigma_{*}^{2}$ does not depend on
$(t_{a,s},b_{a,s})_{a=0,1,s\in\mathcal{S}}$ and

$\displaystyle V((t_{a,s},b_{a,s})_{a=0,1,s\in\mathcal{S}})$
$\displaystyle=\mathbb{E}\mathinner{\biggl{(}\sqrt{\frac{\pi(S_{i})}{1-\pi(S_{i})}}A_{0}(S_{i},X_{i})+\sqrt{\frac{1-\pi(S_{i})}{\pi(S_{i})}}A_{1}(S_{i},X_{i})\biggr{)}}^{2}$
$\displaystyle=\sum_{s\in\mathcal{S}}p(s)\mathbb{E}\mathinner{\biggl{[}\mathinner{\biggl{(}\sqrt{\frac{\pi(s)}{1-\pi(s)}}A_{0}(s,X_{i})+\sqrt{\frac{1-\pi(s)}{\pi(s)}}A_{1}(s,X_{i})\biggr{)}}^{2}\biggl{|}S_{i}=s\biggr{]}}$

where for $a=0,1$,

$\displaystyle A_{a}(s,x)$
$\displaystyle=\nu^{Y}(a,s,x)-\tilde{\mu}^{Y}(a,s,x)-\tau(\nu^{D}(a,s,x)-\tilde{\mu}^{D}(a,s,x))$
$\displaystyle=(\nu^{Y}(a,s,x)-\tau\nu^{D}(a,s,x))-\tilde{\Psi}_{i,s}^{\top}(t_{a,s}-\tau
b_{a,s}),$

and $(\tilde{\mu}^{Y}(a,s,x),\tilde{\mu}^{D}(a,s,x))$ and
$(\nu^{Y}(a,s,x),\nu^{D}(a,s,x))$ are defined in (3.5) and (I.1),
respectively. Specifically, we have

$\displaystyle\tilde{\mu}^{Y}(a,s,x)=\tilde{\Psi}_{i,s}^{\top}t_{a,s},\quad\tilde{\mu}^{D}(a,s,x)=\tilde{\Psi}_{i,s}^{\top}b_{a,s},\quad\text{and}\quad\tilde{\Psi}_{i,s}=\Psi_{i,s}-\mathbb{E}(\Psi_{i,s}|S_{i}=s).$

In order to minimize $V((t_{a,s},b_{a,s})_{a=0,1,s\in\mathcal{S}})$, it
suffices to minimize

$\mathbb{E}\mathinner{\biggl{[}\mathinner{\biggl{(}\sqrt{\frac{\pi(s)}{1-\pi(s)}}A_{0}(s,X_{i})+\sqrt{\frac{1-\pi(s)}{\pi(s)}}A_{1}(s,X_{i})\biggr{)}}^{2}\biggl{|}S_{i}=s\biggr{]}}$

for each $s\in\mathcal{S}$. In addition, we have

$\displaystyle\mathbb{E}\mathinner{\biggl{[}\mathinner{\biggl{(}\sqrt{\frac{\pi(s)}{1-\pi(s)}}A_{0}(s,X_{i})+\sqrt{\frac{1-\pi(s)}{\pi(s)}}A_{1}(s,X_{i})\biggr{)}}^{2}\biggl{|}S_{i}=s\biggr{]}}=\mathbb{E}\left((\overline{y}_{i,s}-\tilde{\Psi}_{i,s}^{\top}\gamma_{s})^{2}\biggl{|}S_{i}=s\right),$

where

$\displaystyle\overline{y}_{i,s}=\sqrt{\frac{1-\pi(s)}{\pi(s)}}(\nu^{Y}(1,s,X_{i})-\tau\nu^{D}(1,s,X_{i}))+\sqrt{\frac{\pi(s)}{1-\pi(s)}}(\nu^{Y}(0,s,X_{i})-\tau\nu^{D}(0,s,X_{i}))$

and

$\displaystyle\gamma_{s}=\sqrt{\frac{1-\pi(s)}{\pi(s)}}(t_{1,s}-\tau
b_{1,s})+\sqrt{\frac{\pi(s)}{1-\pi(s)}}(t_{0,s}-\tau b_{0,s}).$

By solving the first order condition, we find that

$\displaystyle\Theta^{*}$
$\displaystyle=\begin{pmatrix}&(\theta_{a,s}^{*},\beta_{a,s}^{*})_{a=0,1,s\in\mathcal{S}}\mathrel{\mathop{\mathchar
58\relax}}\\\
&\sqrt{\frac{1-\pi(s)}{\pi(s)}}(\theta_{1,s}^{*}-\tau\beta_{1,s}^{*})+\sqrt{\frac{\pi(s)}{1-\pi(s)}}(\theta_{0,s}^{*}-\tau\beta_{0,s}^{*})=\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)^{-1}\mathbb{E}(\tilde{\Psi}_{i,s}\overline{y}_{i,s}|S_{i}=s)\end{pmatrix}$
$\displaystyle=\begin{pmatrix}&(\theta_{a,s}^{*},\beta_{a,s}^{*})_{a=0,1,s\in\mathcal{S}}\mathrel{\mathop{\mathchar
58\relax}}\\\
&\sqrt{\frac{1-\pi(s)}{\pi(s)}}(\theta_{1,s}^{*}-\tau\beta_{1,s}^{*})+\sqrt{\frac{\pi(s)}{1-\pi(s)}}(\theta_{0,s}^{*}-\tau\beta_{0,s}^{*})\\\
&=\sqrt{\frac{1-\pi(s)}{\pi(s)}}(\theta_{1,s}^{L}-\tau\beta_{1,s}^{L})+\sqrt{\frac{\pi(s)}{1-\pi(s)}}(\theta_{0,s}^{L}-\tau\beta_{0,s}^{L}).\end{pmatrix},$

where

$\displaystyle\theta_{a,s}^{L}$
$\displaystyle=[\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)]^{-1}[\mathbb{E}(\tilde{\Psi}_{i,s}\nu^{Y}(a,s,X_{i})|S_{i}=s)]$
$\displaystyle=[\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)]^{-1}[\mathbb{E}(\tilde{\Psi}_{i,s}\mathbb{E}(Y_{i}(D_{i}(a))|S_{i},X_{i})|S_{i}=s)]$
$\displaystyle=[\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)]^{-1}[\mathbb{E}(\tilde{\Psi}_{i,s}Y_{i}(D_{i}(a))|S_{i}=s)].$

Similarly, we have

$\displaystyle\beta_{a,s}^{L}$
$\displaystyle=[\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)]^{-1}[\mathbb{E}(\tilde{\Psi}_{i,s}D_{i}(a)|S_{i}=s)].$

This concludes the proof.

##  Appendix J Proof of Theorem 5.3

In order to verify Assumption 3, by Theorem 5.1, it suffices to show that
$\hat{\theta}_{a,s}^{L}\overset{p}{\longrightarrow}\theta_{a,s}^{L}$ and
$\hat{\beta}_{a,s}^{L}\overset{p}{\longrightarrow}\beta_{a,s}^{L}$. We focus
on the former with $a=1$. Let $\\{W_{i}^{s},X_{i}^{s}\\}_{i\in[n]}$ be
generated independently from the joint distribution of
$(Y_{i}(D_{i}(1)),X_{i})$ given $S_{i}=s$ and denote
$\Psi_{i,s}^{s}=\Psi_{s}(X_{i}^{s})$,
$\tilde{\Psi}_{i,s}^{s}=\Psi_{s}(X_{i}^{s})-\mathbb{E}\Psi_{s}(X_{i}^{s})$,
$\dot{\Psi}_{i,1,s}^{s}=\Psi_{s}(X_{i}^{s})-\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Psi_{s}(X_{i}^{s})$,
and
$\dot{\Psi}_{i,0,s}^{s}=\Psi_{s}(X_{i}^{s})-\frac{1}{n_{0}(s)}\sum_{i=N(s)+n_{1}(s)+1}^{N(s)+n(s)}\Psi_{s}(X_{i}^{s})$.
Then, we have

$\displaystyle\hat{\theta}_{1,s}^{L}\overset{d}{=}\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\dot{\Psi}_{i,1,s}^{s}\dot{\Psi}_{i,1,s}^{s,\top}\biggr{)}}^{-1}\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\dot{\Psi}_{i,1,s}^{s}W_{i}^{s}\biggr{)}}.$

As
$\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Psi_{i,s}^{s}\overset{p}{\longrightarrow}\mathbb{E}\Psi_{i,s}^{s}=\mathbb{E}(\Psi_{s}(X_{i}^{s}))=\mathbb{E}(\Psi_{s}(X_{i})|S_{i}=s)$
by the standard LLN, we have

$\displaystyle\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\dot{\Psi}_{i,1,s}^{s}\dot{\Psi}_{i,1,s}^{s,\top}\biggr{)}}=\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\Psi}_{i,s}^{s}\tilde{\Psi}_{i,s}^{s,\top}\biggr{)}}+o_{p}(1),$
$\displaystyle\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\dot{\Psi}_{i,1,s}^{s}W_{i}^{s}\biggr{)}}=\mathinner{\biggl{(}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\Psi}_{i,s}^{s}W_{i}^{s}\biggr{)}}+o_{p}(1).$

In addition, by the standard LLN,

$\displaystyle\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\Psi}_{i,s}^{s}\tilde{\Psi}_{i,s}^{s,\top}\overset{p}{\longrightarrow}\mathbb{E}\tilde{\Psi}_{i,s}^{s}\tilde{\Psi}_{i,s}^{s,\top}=\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s),$
$\displaystyle\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\Psi}_{i,s}^{s}W_{i}^{s}\overset{p}{\longrightarrow}\mathbb{E}\tilde{\Psi}_{i,s}^{s}W_{i}^{s}=\mathbb{E}(\tilde{\Psi}_{i,s}Y_{i}(D_{i}(1))|S_{i}=s).$

Last, Assumption 5 implies
$\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)$ is
invertible, this means

$\displaystyle\hat{\theta}_{1,s}^{L}\overset{p}{\longrightarrow}\left[\mathbb{E}(\tilde{\Psi}_{i,s}\tilde{\Psi}_{i,s}^{\top}|S_{i}=s)\right]^{-1}\mathbb{E}(\tilde{\Psi}_{i,s}Y_{i}(D_{i}(1))|S_{i}=s)=\theta_{1,s}^{L}.$

Similarly, we can show that
$\hat{\theta}_{0,s}^{L}\overset{p}{\longrightarrow}\theta_{0,s}^{L}$ and
$\hat{\beta}_{a,s}^{L}\overset{p}{\longrightarrow}\beta_{a,s}^{L}$ for $a=0,1$
and $s\in\mathcal{S}$. Therefore, Assumption 3 holds, and thus, all the
results in Theorem 3.1 hold for $\hat{\tau}_{L}$. Then, the optimality result
in the second half of Theorem 5.3 is a direct consequence of Theorem 5.2.

Last, we compare the asymptotic variances of TSLS estimator and the estimator
with the optimal linear adjustment with $\pi(s)=\pi$ for $s\in\mathcal{S}$ and
$\Psi_{i,s}=X_{i}$. In this special case, we first note that the asymptotic
variance of the estimator with the optimal linear adjustment is

$\displaystyle\frac{\sigma_{1}^{2}+\sigma_{0}^{2}+\sigma_{2}^{2}}{[\mathbb{E}(D(1)-D(0))]^{2}},$

where

$\displaystyle\sigma_{0}^{2}$
$\displaystyle=\mathbb{E}(1-\pi)\Xi_{0}^{2}(\mathcal{D}_{i},S_{i})$
$\displaystyle\Xi_{0}(\mathcal{D}_{i},S_{i})$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\left[\left(\frac{1}{1-\pi}-1\right)\tilde{X}_{i}^{\top}\theta_{0s}+\tilde{X}_{i}^{\top}\theta_{1s}-\frac{\tilde{Z}_{i}}{1-\pi}\right]-\tau\left[\left(\frac{1}{1-\pi}-1\right)\tilde{X}_{i}^{\top}\beta_{0s}+\tilde{X}_{i}^{\top}\beta_{1s}-\frac{\tilde{D}_{i}(0)}{1-\pi}\right]$
$\displaystyle\sigma_{1}^{2}$
$\displaystyle=\mathbb{E}\pi\Xi_{1}^{2}(\mathcal{D}_{i},S_{i}),$
$\displaystyle\Xi_{1}(\mathcal{D}_{i},S_{i})$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\left[\left(1-\frac{1}{\pi}\right)\tilde{X}_{i}^{\top}\theta_{1s}-\tilde{X}_{i}^{\top}\theta_{0s}+\frac{\tilde{W}_{i}}{\pi}\right]-\tau\left[\left(1-\frac{1}{\pi}\right)\tilde{X}_{i}^{\top}\beta_{1s}-\tilde{X}_{i}^{\top}\beta_{0s}+\frac{\tilde{D}_{i}(1)}{\pi}\right],$
$\displaystyle\sigma_{2}^{2}$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\mathinner{\bigl{[}Y(D(1))-Y(D(0))-(D(1)-D(0))\tau|S_{i}\bigr{]}}\right]^{2},$

with

$\displaystyle\theta_{as}$
$\displaystyle=\left[\mathbb{E}(\tilde{X}_{is}\tilde{X}_{is}^{\top}|S_{i}=s)\right]^{-1}\left[\mathbb{E}(\tilde{X}_{is}Y_{i}(D_{i}(a))|S_{i}=s)\right]\quad\text{and}$
$\displaystyle\beta_{as}$
$\displaystyle=\left[\mathbb{E}(\tilde{X}_{is}\tilde{X}_{is}^{\top}|S_{i}=s)\right]^{-1}\left[\mathbb{E}(\tilde{X}_{is}D_{i}(a)|S_{i}=s)\right],\quad
a=0,1.$ (J.1)

Observe that $\sigma_{TSLS,1}^{2}+\sigma_{TSLS,0}^{2}$ can also be written as

$\displaystyle\mathbb{E}\left[\pi(S_{i})\Xi_{1}(\mathcal{D}_{i},S_{i})^{2}+(1-\pi(S_{i}))\Xi_{0}(\mathcal{D}_{i},S_{i})^{2}\right],$
(J.2)

where

$\displaystyle\Xi_{1}(\mathcal{D}_{i},S_{i})$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{X}_{i}^{\top}\theta_{1s}-\tilde{X}_{i}^{\top}\theta_{0s}+\frac{\tilde{W}_{i}}{\pi(S_{i})}\right]$
$\displaystyle-\tau\left[\left(1-\frac{1}{\pi(S_{i})}\right)\tilde{X}_{i}^{\top}\beta_{1s}-\tilde{X}_{i}^{\top}\beta_{0s}+\frac{\tilde{D}_{i}(1)}{\pi(S_{i})}\right],$
$\displaystyle\Xi_{0}(\mathcal{D}_{i},S_{i})$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{X}_{i}^{\top}\theta_{0s}+\tilde{X}_{i}^{\top}\theta_{1s}-\frac{\tilde{Z}_{i}}{1-\pi(S_{i})}\right]$
$\displaystyle-\tau\left[\left(\frac{1}{1-\pi(S_{i})}-1\right)\tilde{X}_{i}^{\top}\beta_{0s}+\tilde{X}_{i}^{\top}\beta_{1s}-\frac{\tilde{D}_{i}(0)}{1-\pi(S_{i})}\right],$

with $\theta_{1s}=\theta_{0s}$, $\beta_{1s}=\beta_{0s}$ and
$\theta_{1s}-\tau\beta_{1s}=\lambda_{x}^{*}$, where $\lambda_{x}^{*}$ is the
first $d_{x}$ coefficients of $\lambda^{*}$ defined in Theorem 2.1 where
$d_{x}$ is the dimension of $X_{i}$. By Theorem 5.2, we achieve the optimal
linear adjustment when $\theta_{a,s}$ and $\beta_{a,s}$ satisfy (J.1), which
implies

$\displaystyle\sigma_{1}^{2}+\sigma_{0}^{2}\leq\sigma_{TSLS,1}^{2}+\sigma_{TSLS,0}^{2}.$

In addition, we have $\sigma_{2}^{2}=\sigma_{TSLS,2}^{2}$ and
$0\leq\sigma_{TSLS,3}^{2}$, which implies the desired result.

##  Appendix K Proof of Theorem 5.4

Let $\\{D_{i}^{s}(1),X_{i}^{s}\\}_{i\in[n]}$ be generated independently from
the joint distribution of $(D_{i}(1),X_{i})$ given $S_{i}=s$,
$\Psi_{i,s}^{s}=\Psi_{s}(X_{i}^{s})$, and
$\mathring{\Psi}_{i,s}^{s}=(1,\Psi_{i,s}^{s,\top})^{\top}$. Then, we have,
pointwise in $b$,

$\displaystyle\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}\left[D_{i}\log(\lambda(\mathring{\Psi}_{i,s}^{\top}b))+(1-D_{i})\log(1-\lambda(\mathring{\Psi}_{i,s}^{\top}b))\right]$
$\displaystyle\overset{d}{=}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\left[D_{i}^{s}(1)\log(\lambda(\mathring{\Psi}_{i,s}^{s,\top}b))+(1-D_{i}^{s}(1))\log(1-\lambda(\mathring{\Psi}_{i,s}^{s,\top}b))\right]$
$\displaystyle\overset{p}{\longrightarrow}\mathbb{E}\left[D_{i}^{s}(1)\log(\lambda(\mathring{\Psi}_{i,s}^{s,\top}b))+(1-D_{i}^{s}(1))\log(1-\lambda(\mathring{\Psi}_{i,s}^{s,\top}b))\right]$
$\displaystyle=\mathbb{E}\left[D_{i}(1)\log(\lambda(\mathring{\Psi}_{i,s}^{\top}b))+(1-D_{i}(1))\log(1-\lambda(\mathring{\Psi}_{i,s}^{\top}b))|S_{i}=s\right].$

As the logistic likelihood function is concave in $b$, the pointwise
convergence in $b$ implies uniform convergence, i.e.,

$\displaystyle\sup_{b}\biggl{|}$ $\displaystyle\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}\left[D_{i}\log(\lambda(\mathring{\Psi}_{i,s}^{\top}b))+(1-D_{i})\log(1-\lambda(\mathring{\Psi}_{i,s}^{\top}b))\right]$
$\displaystyle-\mathbb{E}\left[D_{i}(1)\log(\lambda(\mathring{\Psi}_{i,s}^{\top}b))+(1-D_{i}(1))\log(1-\lambda(\mathring{\Psi}_{i,s}^{\top}b))|S_{i}=s\right]\biggr{|}\overset{p}{\longrightarrow}0.$

Then, by the standard proof for the extremum estimation, we have
$\hat{\beta}_{a,s}^{MLE}\overset{p}{\longrightarrow}\beta_{a,s}^{MLE}$.
Similarly, we can show that
$\hat{\theta}_{a,s}^{OLS}\overset{p}{\longrightarrow}\theta_{a,s}^{OLS}$. The
verifies Assumption 4(i). Assumptions 4(ii) and 4(iii) follow from Assumption
6(ii). Then, the desired results hold due to Theorem 5.1.

##  Appendix L Proof of Theorem 5.5

We note that the adjustments proposed in Theorem 5.5 are still parametric.
Specifically, we have

$\displaystyle\overline{\mu}^{Y}(a,s,X_{i})=\Lambda_{a,s}^{Y}(X_{i},\\{\beta_{1,s}^{MLE},\beta_{0,s}^{MLE},\theta_{a,s}^{F}\\}),$
$\displaystyle\overline{\mu}^{D}(a,s,X_{i})=\Lambda_{a,s}^{D}(X_{i},\\{\beta_{1,s}^{MLE},\beta_{0,s}^{MLE},\beta_{a,s}^{F}\\}),$
$\displaystyle\hat{\mu}^{Y}(a,s,X_{i})=\Lambda_{a,s}^{Y}(X_{i},\\{\hat{\beta}_{1,s}^{MLE},\hat{\beta}_{0,s}^{MLE},\hat{\theta}_{a,s}^{F}\\}),\quad\text{and}$
$\displaystyle\hat{\mu}^{D}(a,s,X_{i})=\Lambda_{a,s}^{D}(X_{i},\\{\hat{\beta}_{1,s}^{MLE},\hat{\beta}_{0,s}^{MLE},\hat{\beta}_{a,s}^{F}\\}),$

where

$\displaystyle\Lambda_{a,s}^{Y}(X_{i},\\{b_{1},b_{0},t_{a}^{*}\\})=\begin{pmatrix}\Psi_{i,s}^{\top}\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}b_{1})\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}b_{0})\end{pmatrix}^{\top}t_{a}^{*}\quad\text{and}\quad\Lambda_{a,s}^{D}(X_{i},\\{b_{1},b_{0},b_{a}^{*}\\})=\begin{pmatrix}\Psi_{i,s}^{\top}\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}b_{1})\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}b_{0})\end{pmatrix}^{\top}b_{a}^{*}.$

Therefore, in view of Theorem 5.1, to verify Assumption 3, it suffices to show
that $\hat{\theta}_{a,s}^{F}\overset{p}{\longrightarrow}\theta_{a,s}^{F}$ and
$\hat{\beta}_{a,s}^{F}\overset{p}{\longrightarrow}\beta_{a,s}^{F}$, as we have
already shown the consistency of $\hat{\beta}_{a,s}^{MLE}$ in the proof of
Theorem 5.4. We focus on $\hat{\theta}_{a,s}^{F}$. Define
$\dot{\Phi}_{i,a,s}\mathrel{\mathop{\mathchar
58\relax}}=\Phi_{i,s}-\frac{1}{n_{a}(s)}\sum_{i\in I_{a}(s)}\Phi_{i,s}$, where

$\displaystyle\Phi_{i,s}=\begin{pmatrix}\Psi_{i,s}\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{1,s}^{MLE})\\\
\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{0,s}^{MLE})\end{pmatrix}.$

We first show that

$\displaystyle\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\breve{\Phi}_{i,a,s}\breve{\Phi}_{i,a,s}^{\top}=\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}\dot{\Phi}_{i,a,s}^{\top}+o_{p}(1).$ (L.1)

Let $v,u\in\Re^{d_{\Psi}+2}$ be two arbitrary vectors such that
$||u||_{2}=||v||_{2}=1$. Then, we have

$\displaystyle\mathinner{\\!\biggl{\lvert}v^{\top}\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\left(\breve{\Phi}_{i,a,s}\breve{\Phi}_{i,a,s}^{\top}-\dot{\Phi}_{i,a,s}\dot{\Phi}_{i,a,s}^{\top}\right)\biggr{]}}u\biggr{\rvert}}$
$\displaystyle=\mathinner{\\!\biggl{\lvert}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\left[(v^{\top}\breve{\Phi}_{i,a,s})(u^{\top}\breve{\Phi}_{i,a,s})-(v^{\top}\dot{\Phi}_{i,a,s})(u^{\top}\dot{\Phi}_{i,a,s})\right]\biggr{\rvert}}$
$\displaystyle=\mathinner{\\!\biggl{\lvert}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\left[v^{\top}(\breve{\Phi}_{i,a,s}-\dot{\Phi}_{i,a,s})(u^{\top}\breve{\Phi}_{i,a,s})+(v^{\top}\dot{\Phi}_{i,a,s})u^{\top}(\breve{\Phi}_{i,a,s}-\dot{\Phi}_{i,a,s})\right]\biggr{\rvert}}$
$\displaystyle\leq\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}||\breve{\Phi}_{i,a,s}-\dot{\Phi}_{i,a,s}||_{2}(||\breve{\Phi}_{i,a,s}||_{2}+||\dot{\Phi}_{i,a,s}||_{2})$
(L.2)

where the first inequality is due to Cauchy-Schwarz inequality. We now show
(L.2) is $o_{p}(1)$. First note that

$\displaystyle||\breve{\Phi}_{i,a,s}-\dot{\Phi}_{i,a,s}||_{2}\leq\sum_{a^{\prime}=0,1}\|B_{a^{\prime}}\|_{2}$

where

$\displaystyle B_{a^{\prime}}\mathrel{\mathop{\mathchar
58\relax}}=\lambda(\mathring{\Psi}_{i,s}^{\top}\hat{\beta}_{a^{\prime},s}^{MLE})-\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{a^{\prime},s}^{MLE})-\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,s}^{\top}\hat{\beta}_{a^{\prime},s}^{MLE})-\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{a^{\prime},s}^{MLE})\bigr{]}}.$

Note that

$\displaystyle\lambda(\mathring{\Psi}_{i,s}^{\top}\hat{\beta}_{a^{\prime},s}^{MLE})-\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{a^{\prime},s}^{MLE})$
$\displaystyle=\frac{\partial\lambda(\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE})}{\partial\beta_{a^{\prime},s}}(\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE})$
$\displaystyle\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\lambda(\mathring{\Psi}_{i,s}^{\top}\hat{\beta}_{a^{\prime},s}^{MLE})-\lambda(\mathring{\Psi}_{i,s}^{\top}\beta_{a^{\prime},s}^{MLE})$
$\displaystyle=\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\frac{\partial\lambda(\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE})}{\partial\beta_{a^{\prime},s}}\biggr{]}}(\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE})$

where $\tilde{\beta}_{a^{\prime},s}^{MLE}$ is a mid-point of
$\hat{\beta}_{a^{\prime},s}^{MLE}$ and $\beta_{a^{\prime},s}^{MLE}$. Hence

$\displaystyle\|B_{a^{\prime}}\|_{2}=\mathinner{\\!\biggl{\lVert}\frac{\partial\lambda(\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE})}{\partial\beta_{a^{\prime},s}}-\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\frac{\partial\lambda(\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE})}{\partial\beta_{a^{\prime},s}}\biggr{\rVert}}_{2}\|\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE}\|_{2}.$

Since $\partial\lambda(u)/\partial u\leq 1$,

$\displaystyle\mathinner{\\!\biggl{\lVert}\frac{\partial\lambda(\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE})}{\partial\beta_{a^{\prime},s}}\biggr{\rVert}}_{2}=\mathinner{\\!\biggl{\lVert}\left.\frac{\partial\lambda(u)}{\partial
u}\right|_{u=\mathring{\Psi}_{i,s}^{\top}\tilde{\beta}_{a^{\prime},s}^{MLE}}\cdot\mathring{\Psi}_{i,s}^{\top}\biggr{\rVert}}_{2}\leq\|\mathring{\Psi}_{i,s}\|_{2}.$

Thus,

$\displaystyle\|B_{a^{\prime}}\|_{2}$
$\displaystyle\leq\mathinner{\biggl{(}\|\mathring{\Psi}_{i,s}\|_{2}+\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\|\mathring{\Psi}_{i,s}\|_{2}\biggr{)}}\|\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE}\|_{2}$
$\displaystyle\leq\mathinner{\biggl{(}2+\|\Psi_{i,s}\|_{2}+\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\|\Psi_{i,s}\|_{2}\biggr{)}}\|\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE}\|_{2},$
$\displaystyle||\breve{\Phi}_{i,a,s}-\dot{\Phi}_{i,a,s}||_{2}$
$\displaystyle\leq\mathinner{\biggl{(}2+\|\Psi_{i,s}\|_{2}+\frac{1}{n_{a}(s)}\cdot\sum_{i\in
I_{a}(s)}\|\Psi_{i,s}\|_{2}\biggr{)}}\sum_{a^{\prime}=0,1}\|\hat{\beta}_{a^{\prime},s}^{MLE}-\beta_{a^{\prime},s}^{MLE}\|_{2}.$
(L.3)

Moreover, we can show

$\displaystyle||\breve{\Phi}_{i,a,s}||_{2}+||\dot{\Phi}_{i,a,s}||_{2}\leq
2\mathinner{\biggl{(}4+||\Psi_{i,s}||_{2}+\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\|\Psi_{i,s}\|_{2}\biggr{)}}.$ (L.4)

Substituting (L.3), (L.4) and the fact that
$||\hat{\beta}_{a,s}^{MLE}-\beta_{a,s}^{MLE}||_{2}=o_{p}(1)$ into (L.2), we
show that (L.2) is $o_{p}(1)$. As it holds for arbitrary $u,v$, it implies
(L.1). Similarly, we can show that

$\displaystyle\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\breve{\Phi}_{i,a,s}Y_{i}=\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}Y_{i}+o_{p}(1).$ (L.5)

Following the same argument in the proof of Theorem 5.3, we can show that

$\displaystyle\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}\dot{\Phi}_{i,a,s}^{\top}\biggr{]}}^{-1}\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}Y_{i}\biggr{]}}\overset{p}{\longrightarrow}\theta_{a,s}^{F}.$

In addition, by Assumption 7, with probability approaching one, there exists a
constant $c>0$ such that

$\displaystyle\lambda_{\min}\mathinner{\biggl{(}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}\dot{\Phi}_{i,a,s}^{\top}\biggr{)}}\geq c.$ (L.6)

Combining (L.1), (L.5), and (L.6), we can show that

$\displaystyle\hat{\theta}_{a,s}^{F}$
$\displaystyle=\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\breve{\Phi}_{i,a,s}\breve{\Phi}_{i,a,s}^{\top}\biggr{]}}^{-1}\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\breve{\Phi}_{i,a,s}Y_{i}\biggr{]}}$
$\displaystyle=\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}\dot{\Phi}_{i,a,s}^{\top}\biggr{]}}^{-1}\mathinner{\biggl{[}\frac{1}{n_{a}(s)}\sum_{i\in
I_{a}(s)}\dot{\Phi}_{i,a,s}Y_{i}\biggr{]}}+o_{p}(1)\overset{p}{\longrightarrow}\theta_{a,s}^{F}.$

Similarly, we have
$\hat{\beta}_{a,s}^{F}\overset{p}{\longrightarrow}\beta_{a,s}^{F}$, which
implies all the results in Theorem 3.1 hold for $\hat{\tau}_{F}$. The
optimality result in the second half of the theorem is a direct consequence of
Theorem 5.2.

##  Appendix M Proof of Theorem 5.6

We focus on verifying Assumption 3 for $\hat{\mu}^{D}(a,s,X_{i})$. The proof
for $\hat{\mu}^{Y}(a,s,X_{i})$ is similar and hence omitted. Following the
proof of Theorem 5.4, we note that, for each $a=0,1$ and $s\in\mathcal{S}$,
the data in cell $I_{a}(s)$, denoted $\\{D_{i}^{s}(a),X_{i}^{s}\\}_{i\in[n]}$,
can be viewed as i.i.d. following the joint distribution of $(D_{i}(a),X_{i})$
given $S_{i}=s$ conditionally on $\\{A_{i},S_{i}\\}_{i\in[n]}$. Then following
the standard logistic sieve regression in Hirano et al. (2003), we have

$\displaystyle\max_{a=0,1,s\in\mathcal{S}}||\hat{\beta}_{a,s}^{NP}-\beta_{a,s}^{NP}||_{2}=O_{p}\left(\sqrt{h_{n}/n_{a}(s)}\right).$

Then we have

$\displaystyle\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\Delta^{D}(a,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{D}(a,s,X_{i})}{n_{0}(s)}\biggr{|}$
$\displaystyle\leq\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{NP})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})\bigr{)}}}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{NP})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})\bigr{)}}}{n_{0}(s)}\biggr{|}$
$\displaystyle\qquad+\mathinner{\\!\biggl{\lvert}\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}\mathinner{\Bigl{(}R^{D}(a,s,X_{i})-\mathbb{E}[R^{D}(a,s,X_{i})|S_{i}=s]\Bigr{)}}\biggr{\rvert}}$
$\displaystyle\qquad+\mathinner{\\!\biggl{\lvert}\frac{1}{n_{0}(s)}\sum_{i\in
I_{0}(s)}\mathinner{\Bigl{(}R^{D}(a,s,X_{i})-\mathbb{E}[R^{D}(a,s,X_{i})|S_{i}=s]\Bigr{)}}\biggr{\rvert}}=\mathrel{\mathop{\mathchar
58\relax}}I+II+III.$ (M.1)

To bound term $I$ in (M.1), we define
$M_{a,s}(\beta_{1},\beta_{2})\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{2})|S_{i}=s\bigr{]}}=\mathbb{E}[\lambda(\mathring{\Psi}_{i,n}^{s,\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{s,\top}\beta_{2})]$,
where $\mathring{\Psi}_{i,n}^{s}=\mathring{\Psi}(X_{i}^{s})$. Then we have

$\displaystyle I$ $\displaystyle\leq\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{NP})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})-M_{a,s}(\hat{\beta}_{a,s}^{NP},\beta_{a,s}^{NP})\bigr{]}}}{n_{1}(s)}\biggr{|}$
$\displaystyle\qquad+\biggl{|}\frac{\sum_{i\in
I_{0}(s)}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{NP})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})-M_{a,s}(\hat{\beta}_{a,s}^{NP},\beta_{a,s}^{NP})\bigr{]}}}{n_{0}(s)}\biggr{|}=\mathrel{\mathop{\mathchar
58\relax}}I_{1}+I_{2}.$

Following the argument in the proof of Theorem 5.1, in order to show
$I_{1}=o_{p}(n^{-1/2})$, we only need to show

$n^{1/2}\mathbb{E}\mathinner{\bigl{[}||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\bigr{]}}1\\{n_{1}(s)\geq
n\varepsilon,n_{0}(s)\geq n\varepsilon\\}=o(1),$

where $\varepsilon$ is an arbitrary but fixed constant, and

$\mathcal{F}\mathrel{\mathop{\mathchar
58\relax}}=\mathinner{\Bigl{\\{}\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})\mathrel{\mathop{\mathchar
58\relax}}\beta_{1}\in\Re^{h_{n}},||\beta_{1}-\beta_{a,s}^{NP}||_{2}\leq
C\sqrt{h_{n}/n_{a}(s)}\Bigr{\\}}},$

for some constant $C>0$. Furthermore, we note that $\mathcal{F}$ has a bounded
envelope, is of the VC-type with VC-index upper bounded by $Ch_{n}$,181818See
van der Vaart and Wellner (1996, Section 2.6.5) for the calculation of the VC
index. and has

$\displaystyle\sup_{f\in\mathcal{F}}\mathbb{E}\mathinner{\bigl{[}f^{2}|\\{A_{i},S_{i}\\}_{i\in[n]}\bigr{]}}\leq\frac{Ch_{n}}{n_{a}(s)}.$

Invoking Chernozhukov et al. (2014, Corollary 5.1) with $A$ being a constant,
$\nu=Ch_{n}$, $\sigma^{2}=Ch_{n}/n_{a}(s)$, and $F$ and $M$ being $2h_{n}$, we
have

$\displaystyle
n^{1/2}\mathbb{E}\left[||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\right]1\\{n_{1}(s)\geq
n\varepsilon,n_{0}(s)\geq n\varepsilon\\}$ $\displaystyle\leq
C\sqrt{\frac{n}{n_{1}(s)}}\mathinner{\biggl{(}\sqrt{\frac{h_{n}^{2}\log
n}{n_{a}(s)}}+\frac{h_{n}\log n}{\sqrt{n_{1}(s)}}\biggr{)}}1\\{n_{1}(s)\geq
n\varepsilon,n_{0}(s)\geq n\varepsilon\\}$ $\displaystyle\leq
C\sqrt{\frac{1}{\varepsilon}}\mathinner{\biggl{(}\sqrt{\frac{h_{n}^{2}\log
n}{n\varepsilon}}+\frac{h\log n}{\sqrt{n\varepsilon}}\biggr{)}}\rightarrow 0,$

as $n\to\infty$.

Similarly, we can show $I_{2}=o_{p}(n^{-1/2})$. In addition, we note that

$\displaystyle
II\overset{d}{=}\mathinner{\\!\biggl{\lvert}\frac{1}{n_{1}(s)}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\mathinner{\Bigl{(}R^{D}(a,s,X_{i}^{s})-\mathbb{E}[R^{D}(a,s,X_{i}^{s})]\Bigr{)}}\biggr{\rvert}}=o_{p}(n^{-1/2})$

by the Chebyshev’s inequality as
$\mathbb{E}R^{D,2}(a,s,X_{i}^{s})=\mathbb{E}[R^{D,2}(a,s,X_{i})|S_{i}=s]=o(1)$
by Assumption 8(ii). Similarly we have $III=o_{p}(n^{-1/2})$. Combining the
bounds of $I$, $II$, $III$ with (M.1), we have

$\displaystyle\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\Delta^{D}(a,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{D}(a,s,X_{i})}{n_{0}(s)}\biggr{|}=o_{p}(n^{-1/2}),$

which verifies Assumption 3(i).

To verify Assumption 3(ii), we note that

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\Delta^{D,2}(a,s,X_{i})\leq\frac{2}{n}\sum_{i=1}^{n}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{NP})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{NP})\bigr{)}}^{2}+\frac{2}{n}\sum_{i=1}^{n}R^{D,2}(a,S_{i},X_{i})$
$\displaystyle\leq\frac{2}{n}\sum_{i=1}^{n}\|\mathring{\Psi}_{i,n}\|_{2}^{2}\|\hat{\beta}_{a,s}^{NP}-\beta_{a,s}^{NP}\|_{2}^{2}+\frac{2}{n}\sum_{i=1}^{n}R^{D,2}(a,S_{i},X_{i})$
$\displaystyle=\frac{2}{n}\sum_{i=1}^{n}\|\mathring{\Psi}_{i,n}\|_{2}^{2}\|\hat{\beta}_{a,s}^{NP}-\beta_{a,s}^{NP}\|_{2}^{2}+o_{p}(1)\leq
2\max_{i}\|\mathring{\Psi}_{i,n}\|_{2}^{2}\max_{s}\|\hat{\beta}_{a,s}^{NP}-\beta_{a,s}^{NP}\|_{2}^{2}+o_{p}(1)$
$\displaystyle=O_{p}\mathinner{\bigl{(}\zeta^{2}(h_{n})h_{n}/n_{a}(s)\bigr{)}}+o_{p}(1)=o_{p}(1)$

where the first equality is due to Assumption 8(ii), and the second equality
is due to Assumption 8(iv).

Last, Assumption 3(iii) is implied by Assumption 1(vi) via Jensen’s
inequality.

##  Appendix N Proof of Theorem 5.7

We focus on verifying Assumption 3 for $\hat{\mu}^{D}(a,s,X_{i})$. The proof
for $\hat{\mu}^{Y}(a,s,X_{i})$ is similar and hence omitted. Following the
proof of Theorem 5.4, we note that, for each $a=0,1$ and $s\in\mathcal{S}$,
the data in cell $I_{a}(s)$, denoted $\\{D_{i}^{s}(a),X_{i}^{s}\\}_{i\in[n]}$,
can be viewed as i.i.d. following the joint distribution of $(D_{i}(a),X_{i})$
given $S_{i}=s$ conditionally on $\\{A_{i},S_{i}\\}_{i\in[n]}$. Then following
the standard logistic Lasso regression in Belloni et al. (2017), we have

$\displaystyle\max_{a=0,1,s\in\mathcal{S}}||\hat{\beta}_{a,s}^{R}-\beta_{a,s}^{R}||_{2}=O_{p}\left(\sqrt{h_{n}\log
p_{n}/n_{a}(s)}\right)\quad\text{and}\quad\max_{a=0,1,s\in\mathcal{S}}||\hat{\beta}_{a,s}^{R}||_{0}=O_{p}(h_{n}).$

Then, we have

$\displaystyle\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\Delta^{D}(a,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{D}(a,s,X_{i})}{n_{0}(s)}\biggr{|}$
$\displaystyle\leq\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\bigr{)}}}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\bigr{)}}}{n_{0}(s)}\biggr{|}$
$\displaystyle\qquad+\mathinner{\\!\biggl{\lvert}\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}\mathinner{\Bigl{(}R^{D}(a,s,X_{i})-\mathbb{E}[R^{D}(a,s,X_{i})|S_{i}=s]\Bigr{)}}\biggr{\rvert}}$
$\displaystyle\qquad+\mathinner{\\!\biggl{\lvert}\frac{1}{n_{0}(s)}\sum_{i\in
I_{0}(s)}\mathinner{\Bigl{(}R^{D}(a,s,X_{i})-\mathbb{E}[R^{D}(a,s,X_{i})|S_{i}=s]\Bigr{)}}\biggr{\rvert}}\mathrel{\mathop{\mathchar
58\relax}}=I+II+III.$ (N.1)

To bound term $I$ in (M.1), we define
$M_{a,s}(\beta_{1},\beta_{2})\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{2})|S_{i}=s\bigr{]}}=\mathbb{E}[\lambda(\mathring{\Psi}_{i,n}^{s,\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{s,\top}\beta_{2})]$,
where $\mathring{\Psi}_{i,n}^{s}=\mathring{\Psi}(X_{i}^{s})$. Then we have

$\displaystyle I$ $\displaystyle\leq\biggl{|}\frac{\sum_{i\in
I_{1}(s)}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})-M_{a,s}(\hat{\beta}_{a,s}^{R},\beta_{a,s}^{R})\bigr{]}}}{n_{1}(s)}\biggr{|}$
$\displaystyle\qquad+\biggl{|}\frac{\sum_{i\in
I_{0}(s)}\mathinner{\bigl{[}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})-M_{a,s}(\hat{\beta}_{a,s}^{R},\beta_{a,s}^{R})\bigr{]}}}{n_{0}(s)}\biggr{|}=\mathrel{\mathop{\mathchar
58\relax}}I_{1}+I_{2}.$

Following the argument in the proof of Theorems 5.1 and 5.6, in order to show
$I_{1}=o_{p}(n^{-1/2})$, we only need to show

$n^{1/2}\mathbb{E}\mathinner{\bigl{[}||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\bigr{]}}1\\{n_{1}(s)\geq
n\varepsilon,n_{0}(s)\geq n\varepsilon\\}=o(1),$

where $\varepsilon$ is an arbitrary but fixed constant, and

$\mathcal{F}\mathrel{\mathop{\mathchar
58\relax}}=\mathinner{\Bigl{\\{}\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{1})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\mathrel{\mathop{\mathchar
58\relax}}\beta_{1}\in\Re^{h_{n}},||\beta_{1}-\beta_{a,s}^{R}||_{2}\leq
C\sqrt{h_{n}\log(p_{n})/n_{a}(s)},||\beta_{1}||_{0}\leq Ch_{n}\Bigr{\\}}},$

for some constant $C>0$. Furthermore, we note that $\mathcal{F}$ has a bounded
envelope and

$\displaystyle\sup_{Q}N(\mathcal{F},e_{Q},\varepsilon||F||_{Q,2})\leq\left(\frac{c_{1}p_{n}}{\varepsilon}\right)^{c_{2}h_{n}},$

where $c_{1},c_{2}$ are two fixed constants, $N(\cdot)$ is the covering
number, $e_{Q}(f,g)=\sqrt{Q|f-g|^{2}}$, and the supremum is taken over all
discrete probability measures $Q$. Last, we have

$\displaystyle\sup_{f\in\mathcal{F}}\mathbb{E}\mathinner{\bigl{[}f^{2}|\\{A_{i},S_{i}\\}_{i\in[n]}\bigr{]}}\leq\frac{Ch_{n}\log
p_{n}}{n_{a}(s)}.$

Invoking Chernozhukov et al. (2014, Corollary 5.1) with $A=Cp_{n}$,
$\nu=Ch_{n}$, $\sigma^{2}=Ch_{n}\log(p_{n})/n_{a}(s)$, and $F$ and $M$ being
$2$, we have

$\displaystyle
n^{1/2}\mathbb{E}\left[||\mathbb{P}_{n_{1}(s)}-\mathbb{P}||_{\mathcal{F}}|\\{A_{i},S_{i}\\}_{i\in[n]}\right]1\\{n_{1}(s)\geq
n\varepsilon,n_{0}(s)\geq n\varepsilon\\}$ $\displaystyle\leq
C\sqrt{\frac{n}{n_{1}(s)}}\left(\sqrt{h_{n}\frac{h_{n}\log
p_{n}}{n_{a}(s)}\log\mathinner{\biggl{(}\frac{p_{n}}{\sqrt{\frac{h_{n}\log
p_{n}}{n_{a}(s)}}}\biggr{)}}}+\frac{h_{n}}{\sqrt{n_{1}(s)}}\log\mathinner{\biggl{(}\frac{p_{n}}{\sqrt{\frac{h_{n}\log
p_{n}}{n_{a}(s)}}}\biggr{)}}\right)1\\{n_{1}(s)\geq n\varepsilon,n_{0}(s)\geq
n\varepsilon\\}$ $\displaystyle\leq
C\left(\sqrt{\frac{n}{n_{1}(s)}}\right)\left(\frac{h_{n}\log(p_{n})}{\sqrt{n_{1}(s)\wedge
n_{0}(s)}}\right)1\\{n_{1}(s)\geq n\varepsilon,n_{0}(s)\geq
n\varepsilon\\}\rightarrow 0.$

The bounds for $I_{2}$, $II$ and $III$ can be established following the same
argument as in the proof of Theorem 5.6. We omit the detail for brevity. This
leads to Assumption 3(i).

To verify Assumption 3(ii), we note that

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\Delta^{D,2}(a,s,X_{i})\leq\frac{2}{n}\sum_{i=1}^{n}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\bigr{)}}^{2}+\frac{2}{n}\sum_{i=1}^{n}R^{D,2}(a,S_{i},X_{i})$
$\displaystyle=\frac{2}{n}\sum_{i=1}^{n}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\bigr{)}}^{2}+o_{p}(1)=o_{p}(1),$

where the first equality is due to Assumption 9(iii) and the second equality
is by Assumption 9(vi) and the fact that

$\displaystyle\frac{2}{n}\sum_{i=1}^{n}\mathinner{\bigl{(}\lambda(\mathring{\Psi}_{i,n}^{\top}\hat{\beta}_{a,s}^{R})-\lambda(\mathring{\Psi}_{i,n}^{\top}\beta_{a,s}^{R})\bigr{)}}^{2}\lesssim\frac{(\hat{\beta}_{a,s}^{R}-\beta_{a,s}^{R})^{\top}}{n}\sum_{i=1}^{n}\mathring{\Psi}_{i,n}\mathring{\Psi}_{i,n}^{\top}(\hat{\beta}_{a,s}^{R}-\beta_{a,s}^{R})\lesssim||\hat{\beta}_{a,s}^{R}-\beta_{a,s}^{R}||_{2}^{2}=o_{p}(1),$

where the first probability inequality is due to the fact that
$\lambda(\cdot)$ is Lipschitz continuous with Lipschitz constant 1. Last,
Assumption 3(iii) is implied by Assumption 1(vi) via Jensen’s inequality.

##  Appendix O Proof of Theorem B.1

Some part of the proof of part (i) is due to Ansel et al. (2018) while some
part of the proof is original. Let $U_{i}\mathrel{\mathop{\mathchar
58\relax}}=(1,X_{i}^{\top})^{\top}$ and
$\hat{\lambda}_{as}\mathrel{\mathop{\mathchar
58\relax}}=(\hat{\gamma}_{as}^{b},\hat{\nu}_{as}^{b,\top})^{\top}$ for $a=0,1$
and $b=Y,D$. Consider $\hat{\lambda}_{0s}^{D}$ as an example; note that

$\hat{\lambda}_{0s}^{D}=\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}U_{i}^{\top}\biggr{)}}^{-1}\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}D_{i}.$

Consider the denominator of $\hat{\lambda}_{0s}^{D}$:

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}U_{i}^{\top}=\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}U_{i}U_{i}^{\top}+\frac{1}{n}\sum_{i=1}^{n}(1-\pi(s))1\\{S_{i}=s\\}U_{i}U_{i}^{\top}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{(}U_{i}U_{i}^{\top}-\mathbb{E}[U_{i}U_{i}^{\top}|S_{i}]\bigr{)}}+\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathbb{E}[U_{i}U_{i}^{\top}|S_{i}]$
$\displaystyle\qquad+\frac{1}{n}\sum_{i=1}^{n}(1-\pi(s))1\\{S_{i}=s\\}U_{i}U_{i}^{\top}.$
(O.1)

Consider the first term of (O.1). Note that

$\displaystyle\mathbb{E}\mathinner{\biggl{[}\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{(}U_{i}U_{i}^{\top}-\mathbb{E}[U_{i}U_{i}^{\top}|S_{i}]\bigr{)}}|A^{(n)},S^{(n)}\biggr{]}}=0.$

Invoking the conditional Chebyshev’s inequality, we have, for any $a>0$,
$1\leq k,\ell\leq\text{dim}(U_{i})$,

$\displaystyle\mathbb{P}\mathinner{\biggl{(}\mathinner{\\!\biggl{\lvert}\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}\biggr{\rvert}}\geq
a|A^{(n)},S^{(n)}\biggr{)}}$
$\displaystyle\leq\frac{1}{a^{2}}\operatorname*{var}\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}|A^{(n)},S^{(n)}\biggr{)}}$
$\displaystyle=\frac{\sum_{i,j\in[n]}(\pi(s)-A_{i})(\pi(s)-A_{j})1\\{S_{i}=s\\}1\\{S_{j}=s\\}\mathbb{E}\mathinner{\Bigl{[}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}\mathinner{\bigl{(}U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]\bigr{)}}|A^{(n)},S^{(n)}\Bigr{]}}}{a^{2}n^{2}}$
$\displaystyle=\frac{\sum_{i\in[n]}(\pi(s)-A_{i})^{2}1\\{S_{i}=s\\}\mathbb{E}\mathinner{\Bigl{[}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}^{2}|A^{(n)},S^{(n)}\Bigr{]}}}{a^{2}n^{2}}$
$\displaystyle\leq\frac{\sum_{i\in[n]}(\pi(s)-A_{i})^{2}1\\{S_{i}=s\\}\mathbb{E}\mathinner{\Bigl{[}U_{ik}^{2}U_{i\ell}^{2}|S_{i}\Bigr{]}}}{a^{2}n^{2}}\leq\frac{\sum_{i\in[n]}\mathbb{E}\mathinner{\Bigl{[}U_{ik}^{2}U_{i\ell}^{2}|S_{i}=s\Bigr{]}}}{a^{2}n^{2}}=o(1)$
(O.2)

where the second equality is due to

$\displaystyle\mathbb{E}\mathinner{\Bigl{[}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}\mathinner{\bigl{(}U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]\bigr{)}}|A^{(n)},S^{(n)}\Bigr{]}}$
$\displaystyle=\mathbb{E}\mathinner{\Bigl{[}\mathinner{\bigl{(}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]\bigr{)}}\mathinner{\bigl{(}U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]\bigr{)}}|S^{(n)}\Bigr{]}}$
$\displaystyle=\mathbb{E}\mathinner{\Bigl{[}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]|S^{(n)}\Bigr{]}}\mathbb{E}\mathinner{\Bigl{[}U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]|S^{(n)}\Bigr{]}}$
$\displaystyle=\mathbb{E}\mathinner{\Bigl{[}U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]|S_{i}\Bigr{]}}\mathbb{E}\mathinner{\Bigl{[}U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]|S_{j}\Bigr{]}}=0$

for $i\neq j$, where the second equality is due to that
$U_{ik}U_{i\ell}-\mathbb{E}[U_{ik}U_{i\ell}|S_{i}]$ and
$U_{jk}U_{j\ell}-\mathbb{E}[U_{jk}U_{j\ell}|S_{j}]$ are independent
conditional on $S^{(n)}$. From (O.2), we deduce that the first term of (O.1)
is $o_{p}(1)$. Consider the second term of (O.1).

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}\mathbb{E}[U_{i}U_{i}^{\top}|S_{i}]=\mathbb{E}[UU^{\top}|S=s]\frac{1}{n}\sum_{i=1}^{n}(\pi(s)-A_{i})1\\{S_{i}=s\\}$
$\displaystyle=\mathbb{E}[UU^{\top}|S=s]\frac{1}{n}B_{n}(s)=o_{p}(1).$

Consider the third term of (O.1).

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(1-\pi(s))1\\{S_{i}=s\\}U_{i}U_{i}^{\top}=(1-\pi(s))\frac{n(s)}{n}\frac{1}{n(s)}\sum_{i=1}^{n}1\\{S_{i}=s\\}U_{i}U_{i}^{\top}$
$\displaystyle\qquad\xrightarrow{p}(1-\pi(s))p(s)\mathbb{E}[UU^{\top}|S=s].$

We hence have

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}U_{i}^{\top}\xrightarrow{p}(1-\pi(s))p(s)\mathbb{E}[UU^{\top}|S=s].$

Similarly, we have

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}D_{i}\xrightarrow{p}(1-\pi(s))\hat{p}(s)\mathbb{E}[UD(0)|S=s]$
$\displaystyle\hat{\lambda}_{0s}^{D}\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}[UD(0)|S=s]$
$\displaystyle\hat{\lambda}_{1s}^{D}\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}[UD(1)|S=s].$

Thus, we have

$\displaystyle\sum_{s\in\mathcal{S}}\hat{p}(s)(\hat{\gamma}_{1s}^{D}-\hat{\gamma}_{0s}^{D}+(\hat{\nu}_{1s}^{D}-\hat{\nu}_{0s}^{D})^{\top}\bar{X}_{s})=\sum_{s\in\mathcal{S}}(\hat{\lambda}_{1s}^{D}-\hat{\lambda}_{0s}^{D})^{\top}\left(\begin{array}[c]{c}\frac{1}{n}\sum_{i\in[n]}1\\{S_{i}=s\\}\\\
\frac{1}{n}\sum_{i\in[n]}X_{i}1\\{S_{i}=s\\}\end{array}\right)$
$\displaystyle\qquad=\sum_{s\in\mathcal{S}}\frac{n(s)}{n}\frac{1}{n(s)}\sum_{i\in[n]}1\\{S_{i}=s\\}U_{i}^{\top}(\hat{\lambda}_{1s}^{D}-\hat{\lambda}_{0s}^{D})$
$\displaystyle\qquad\xrightarrow{p}\sum_{s\in\mathcal{S}}p(s)\mathbb{E}[U^{\top}|S=s]\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}\mathinner{\Bigl{[}U\mathinner{\bigl{(}D(1)-D(0)\bigr{)}}|S=s\Bigr{]}}$
$\displaystyle\qquad=\sum_{s\in\mathcal{S}}p(s)\mathbb{E}\mathinner{\bigl{[}D(1)-D(0)|S=s\bigr{]}}=\mathbb{E}\mathinner{\bigl{[}D(1)-D(0)\bigr{]}}$

where the second last equality is due to
$\mathbb{E}[U^{\top}|S=s]\mathinner{\bigl{(}\mathbb{E}[UU^{\top}|S=s]\bigr{)}}^{-1}=(1,0,\ldots,0)$
(Ansel et al. (2018) p290). Thus, the denominator of
$\sqrt{n}(\hat{\tau}_{S}-\tau)$ converges in probability to
$\mathbb{E}[D(1)-D(0)]$.

We now consider the numerator of $\sqrt{n}(\hat{\tau}_{S}-\tau)$. Relying on a
similar argument, we have

$\displaystyle\hat{\lambda}_{1s}^{Y}=\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}U_{i}^{\top}\biggr{)}}^{-1}\frac{1}{n}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}Y_{i}(D_{i}(1))$
$\displaystyle\qquad\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}[UY(D(1))|S=s]$
$\displaystyle\hat{\lambda}_{0s}^{Y}=\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}U_{i}^{\top}\biggr{)}}^{-1}\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}U_{i}Y_{i}(D_{i}(0))$
$\displaystyle\qquad\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}[UY(D(0))|S=s]$
$\displaystyle\hat{\eta}_{1s}\mathrel{\mathop{\mathchar
58\relax}}=\hat{\lambda}_{1s}^{Y}-\tau\hat{\lambda}_{1s}^{D}\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}\mathinner{\Bigl{[}U\mathinner{\bigl{[}Y(D(1))-\tau
D(1)\bigr{]}}|S=s\Bigr{]}}=\mathrel{\mathop{\mathchar 58\relax}}\eta_{1s}$
$\displaystyle\hat{\eta}_{0s}\mathrel{\mathop{\mathchar
58\relax}}=\hat{\lambda}_{0s}^{Y}-\tau\hat{\lambda}_{0s}^{D}\xrightarrow{p}\mathinner{\Bigl{(}\mathbb{E}[UU^{\top}|S=s]\Bigr{)}}^{-1}\mathbb{E}\mathinner{\Bigl{[}U\mathinner{\bigl{[}Y(D(0))-\tau
D(0)\bigr{]}}|S=s\Bigr{]}}=\mathrel{\mathop{\mathchar 58\relax}}\eta_{0s}.$

The numerator of $\sqrt{n}(\hat{\tau}_{S}-\tau)$ could be written as

$\displaystyle\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)(\hat{\gamma}_{1s}^{Y}-\hat{\gamma}_{0s}^{Y}+(\hat{\nu}_{1s}^{Y}-\hat{\nu}_{0s}^{Y})^{\top}\bar{X}_{s})-\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)(\hat{\gamma}_{1s}^{D}-\hat{\gamma}_{0s}^{D}+(\hat{\nu}_{1s}^{D}-\hat{\nu}_{0s}^{D})^{\top}\bar{X}_{s})\tau$
$\displaystyle=\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\frac{1}{n(s)}\sum_{i\in[n]}1\\{S_{i}=s\\}U_{i}^{\top}\mathinner{\bigl{[}\hat{\lambda}_{1s}^{Y}-\tau\hat{\lambda}_{1s}^{D}-(\hat{\lambda}_{0s}^{Y}-\tau\hat{\lambda}_{0s}^{D})\bigr{]}}$
$\displaystyle=\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}(\hat{\eta}_{1s}-\eta_{1s})-\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}(\hat{\eta}_{0s}-\eta_{0s})+\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}(\eta_{1s}-\eta_{0s})$
(O.3)

where $\bar{U}_{s}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{n(s)}\sum_{i\in[n]}1\\{S_{i}=s\\}U_{i}\xrightarrow{p}\mathbb{E}[U|S=s]$.
Consider the first term of (O.3).

$\displaystyle\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}(\hat{\eta}_{1s}-\eta_{1s})$
$\displaystyle=\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}\mathinner{\biggl{(}\frac{1}{n}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}U_{i}^{\top}\biggr{)}}^{-1}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}$
$\displaystyle=\sum_{s\in\mathcal{S}}\hat{p}(s)\mathbb{E}[U^{\top}|S=s]\mathinner{\biggl{(}\pi(s)\hat{p}(s)\mathbb{E}[UU^{\top}|S=s]\biggr{)}}^{-1}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}+o_{p}(1)$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\pi(s)}\mathbb{E}[U^{\top}|S=s]\mathinner{\biggl{(}\mathbb{E}[UU^{\top}|S=s]\biggr{)}}^{-1}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}+o_{p}(1)$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}+o_{p}(1)$ (O.4)

where the second equality is based on that

$n^{-1/2}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}U_{i}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}=O_{p}(1),$

which is implied by the asymptotic normality of (O.7), which we will prove
shortly, and the last equality is due to
$\mathbb{E}[U^{\top}|S=s]\mathinner{\bigl{(}\mathbb{E}[UU^{\top}|S=s]\bigr{)}}^{-1}=(1,0,\ldots,0)$
(Ansel et al. (2018) p290). Likewise, the second term of (O.3)

$\displaystyle\sqrt{n}\sum_{s\in\mathcal{S}}\hat{p}(s)\bar{U}_{s}^{\top}(\hat{\eta}_{0s}-\eta_{0s})$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{1-\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{[}Y_{i}(D_{i}(0))-\tau
D_{i}(0)-U_{i}^{\top}\eta_{0s}\bigr{]}}+o_{p}(1).$ (O.5)

Note that

$\displaystyle\eta_{as}=\left(\begin{array}[c]{c}\mathbb{E}\mathinner{\bigl{[}Y(D(a))-\tau
D(a)|S=s\bigr{]}}-\mathbb{E}[X^{\top}\nu_{as}^{YD}|S=s]\\\
\nu_{as}^{YD}\end{array}\right)$

for $a=0,1$ via the Frisch-Waugh Theorem. Hence

$\displaystyle U_{i}^{\top}\eta_{as}=\mathbb{E}\mathinner{\bigl{[}Y(D(a))-\tau
D(a)|S=s\bigr{]}}+X_{i}^{\top}\nu_{as}^{YD}-\mathbb{E}[X^{\top}\nu_{as}^{YD}|S=s].$
(O.6)

Substituting (O.4), (O.5) and (O.6) into (O.3), we could write the numerator
of $\sqrt{n}(\hat{\tau}_{S}-\tau)$ as

$\displaystyle\sum_{s\in\mathcal{S}}\frac{1}{\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}\mathinner{\bigl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-U_{i}^{\top}\eta_{1s}\bigr{]}}$
$\displaystyle\qquad-\sum_{s\in\mathcal{S}}\frac{1}{1-\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{[}Y_{i}(D_{i}(0))-\tau
D_{i}(0)-U_{i}^{\top}\eta_{0s}\bigr{]}}$ (O.7)
$\displaystyle\qquad+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i\in[n]}1\\{S_{i}=s\\}U_{i}^{\top}(\eta_{1s}-\eta_{0s})+o_{p}(1)$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}\mathinner{\biggl{[}Y_{i}(D_{i}(1))-\tau
D_{i}(1)-\mathbb{E}\mathinner{\bigl{[}Y(D(1))-\tau
D(1)|S=s\bigr{]}}-\mathinner{\bigl{(}X_{i}^{\top}\nu_{1s}^{YD}-\mathbb{E}[X^{\top}\nu_{1s}^{YD}|S=s]\bigr{)}}\biggr{]}}$
$\displaystyle\quad-\sum_{s\in\mathcal{S}}\frac{1}{1-\pi(s)}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\biggl{[}Y_{i}(D_{i}(0))-\tau
D_{i}(0)-\mathbb{E}\mathinner{\bigl{[}Y(D(0))-\tau
D(0)|S=s\bigr{]}}-\mathinner{\bigl{(}X_{i}^{\top}\nu_{0s}^{YD}-\mathbb{E}[X^{\top}\nu_{0s}^{YD}|S=s]\bigr{)}}\biggr{]}}$
$\displaystyle\quad+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i\in[n]}1\\{S_{i}=s\\}\mathbb{E}\mathinner{\Bigl{[}Y(D(1))-Y(D(0))-\tau(D(1)-D(0))|S=s\Bigr{]}}$
$\displaystyle\quad+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i\in[n]}1\\{S_{i}=s\\}\mathinner{\Bigl{(}X_{i}^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})-\mathbb{E}[X^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})|S=s]\Bigr{)}}+o_{p}(1)$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}\mathinner{\biggl{[}\frac{Y_{i}(D_{i}(1))-\tau
D_{i}(1)-X_{i}^{\top}\nu_{1s}^{YD}-\mathbb{E}\mathinner{\bigl{[}Y(D(1))-\tau
D(1)-X^{\top}\nu_{1s}^{YD}|S=s\bigr{]}}}{\pi(s)}\biggr{]}}$
$\displaystyle\quad+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i\in[n]}A_{i}1\\{S_{i}=s\\}\mathinner{\Bigl{(}X_{i}^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})-\mathbb{E}[X^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})|S=s]\Bigr{)}}$
$\displaystyle\quad-\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\biggl{[}\frac{Y_{i}(D_{i}(0))-\tau
D_{i}(0)-X_{i}^{\top}\nu_{0s}^{YD}-\mathbb{E}\mathinner{\bigl{[}Y(D(0))-\tau
D(0)-X^{\top}\nu_{0s}^{YD}|S=s\bigr{]}}}{1-\pi(s)}\biggr{]}}$
$\displaystyle\quad+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i\in[n]}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\Bigl{(}X_{i}^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})-\mathbb{E}[X^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})|S=s]\Bigr{)}}$
$\displaystyle\quad+\frac{1}{\sqrt{n}}\sum_{i\in[n]}\mathbb{E}\mathinner{\Bigl{[}Y(D(1))-Y(D(0))-\tau(D(1)-D(0))|S\Bigr{]}}+o_{p}(1).$
(O.8)

Define

$\displaystyle\rho_{is}(1)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{Y_{i}(D_{i}(1))-\tau
D_{i}(1)-X_{i}^{\top}\nu_{1s}^{YD}}{\pi(s)}+X_{i}^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})$
$\displaystyle\rho_{is}(0)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{Y_{i}(D_{i}(0))-\tau
D_{i}(0)-X_{i}^{\top}\nu_{0s}^{YD}}{1-\pi(s)}-X_{i}^{\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD}).$

Then the first four terms of (O.8) could be written compactly as

$\displaystyle R_{n,1}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}\mathinner{\bigl{[}\rho_{is}(1)-\mathbb{E}[\rho_{is}(1)|S_{i}=s]\bigr{]}}$
$\displaystyle\qquad-\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})1\\{S_{i}=s\\}\mathinner{\bigl{[}\rho_{is}(0)-\mathbb{E}[\rho_{is}(0)|S_{i}=s]\bigr{]}}.$

Define $R_{n,2}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{n}}\sum_{i\in[n]}\mathbb{E}\mathinner{\bigl{[}Y(D(1))-Y(D(0))-\tau(D(1)-D(0))|S\bigr{]}}$.
To establish the asymptotic distribution of (O.8), we first argue that

$\displaystyle(R_{n,1},R_{n,2})\overset{d}{=}(R_{n,1}^{*},R_{n,2})+o_{p}(1)$

for a random variable $R_{n,1}^{*}$ that satisfies
$R_{n,1}^{*}\perp\\!\\!\\!\perp R_{n,2}$. Conditional on
$\\{S^{(n)},A^{(n)}\\}$, the distribution of $R_{n,1}$ is the same as the
distribution of the same quantity where units are ordered by strata and then
ordered by $A_{i}=1$ first and $A_{i}=0$ second within strata. To this end,
define $N(s)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{i=1}^{n}1\\{S_{i}<s\\}$ and $F(s)\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{P}(S_{i}<s)$. Furthermore, independently for each
$s\in\mathcal{S}$ and independently of $\\{S^{(n)},A^{(n)}\\}$, let
$\mathinner{\bigl{\\{}Y_{i}(1)^{s},Y_{i}(0)^{s},D_{i}(1)^{s},D_{i}(0)^{s},X_{i}^{s}\mathrel{\mathop{\mathchar
58\relax}}1\leq i\leq n\bigr{\\}}}$ be i.i.d. over $i$ with distribution equal
to that of $(Y(1),Y(0),D(1),D(0),X)|S=s$. Define

$\displaystyle\tilde{\rho}_{is}(a)\mathrel{\mathop{\mathchar
58\relax}}=\rho_{is}(a)-\mathbb{E}[\rho_{is}(a)|S_{i}=s],\qquad\tilde{\rho}_{is}^{s}(a)\mathrel{\mathop{\mathchar
58\relax}}=\rho_{is}^{s}(a)-\mathbb{E}[\rho_{is}^{s}(a)|S_{i}=s],$

where

$\displaystyle\rho_{is}^{s}(1)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{Y_{i}^{s}(D_{i}^{s}(1))-\tau
D_{i}^{s}(1)-X_{i}^{s,\top}\nu_{1s}^{YD}}{\pi(s)}+X_{i}^{s,\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD})$
$\displaystyle\rho_{is}^{s}(0)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{Y_{i}^{s}(D_{i}^{s}(0))-\tau
D_{i}^{s}(0)-X_{i}^{s,\top}\nu_{0s}^{YD}}{1-\pi(s)}-X_{i}^{s,\top}(\nu_{1s}^{YD}-\nu_{0s}^{YD}).$

Then we have

$\displaystyle R_{n,1}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}1\\{S_{i}=s\\}\mathinner{\Bigl{[}A_{i}\tilde{\rho}_{is}(1)-(1-A_{i})\tilde{\rho}_{is}(0)\Bigr{]}}.$

Define

$\displaystyle\tilde{R}_{n,1}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\sum_{s\in\mathcal{S}}\mathinner{\biggl{[}\frac{1}{\sqrt{n}}\sum_{i=n\frac{N(s)}{n}+1}^{n\mathinner{\bigl{(}\frac{N(s)}{n}+\frac{n_{1}(s)}{n}\bigr{)}}}\tilde{\rho}_{is}^{s}(1)-\frac{1}{\sqrt{n}}\sum_{n\mathinner{\bigl{(}\frac{N(s)}{n}+\frac{n_{1}(s)}{n}\bigr{)}}+1}^{n\mathinner{\bigl{(}\frac{N(s)}{n}+\frac{n(s)}{n}\bigr{)}}}\tilde{\rho}_{is}^{s}(0)\biggr{]}}$
$\displaystyle R_{n,1}^{*}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\sum_{s\in\mathcal{S}}\mathinner{\biggl{[}\frac{1}{\sqrt{n}}\sum_{i=\lfloor
nF(s)\rfloor+1}^{\lfloor
n(F(s)+\pi(s)p(s))\rfloor}\tilde{\rho}_{is}^{s}(1)-\frac{1}{\sqrt{n}}\sum_{i=\lfloor
n(F(s)+\pi(s)p(s))\rfloor+1}^{\lfloor
n(F(s)+p(s))\rfloor}\tilde{\rho}_{is}^{s}(0)\biggr{]}}.$

Thus $R_{n,1}|S^{(n)},A^{(n)}\overset{d}{=}\tilde{R}_{n,1}|S^{(n)},A^{(n)}$
(and as a by-product $R_{n,1}\overset{d}{=}\tilde{R}_{n,1}$). Since $R_{n,2}$
is a function of $\\{S^{(n)},A^{(n)}\\}$, we have, arguing along the line of a
joint distribution being the product of a conditional distribution and a
marginal distribution,
$(R_{n,1},R_{n,2})\overset{d}{=}(\tilde{R}_{n,1},R_{n,2})$. Define the
following partial sum process

$\displaystyle g_{n}(u)\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{n}}\sum_{i=1}^{\lfloor
nu\rfloor}\tilde{\rho}^{s}_{is}(1).$

Under our assumptions, $g_{n}(u)$ converges weakly to a suitably scaled
Brownian motion. Next, by elementary properties of Brownian motion, we have
that

$\displaystyle
g_{n}\mathinner{\bigl{(}F(s)+\pi(s)p(s)\bigr{)}}-g_{n}\mathinner{\bigl{(}F(s)\bigr{)}}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=\lfloor nF(s)\rfloor+1}^{\lfloor
n(F(s)+\pi(s)p(s))\rfloor}\tilde{\rho}^{s}_{is}(1)\rightsquigarrow\mathcal{N}\mathinner{\biggl{(}0,\pi(s)p(s)\operatorname*{var}\mathinner{\bigl{(}\rho_{s}(1)|S=s\bigr{)}}\biggr{)}}.$
(O.9)

Furthermore, since

$\displaystyle\mathinner{\biggl{(}\frac{N(s)}{n},\frac{n_{1}(s)}{n}\biggr{)}}\xrightarrow{p}\mathinner{\bigl{(}F(s),\pi(s)p(s)\bigr{)}},$

it follows that

$\displaystyle
g_{n}\mathinner{\biggl{(}\frac{N(s)+n_{1}(s)}{n}\biggr{)}}-g_{n}\mathinner{\biggl{(}\frac{N(s)}{n}\biggr{)}}-\mathinner{\biggl{[}g_{n}\mathinner{\bigl{(}F(s)+\pi(s)p(s)\bigr{)}}-g_{n}\mathinner{\bigl{(}F(s)\bigr{)}}\biggr{]}}\xrightarrow{p}0$
(O.10)

where the convergence follows from the stochastic equicontinuity of the
partial sum process. Using (O.9) and (O.10), we have:

$\displaystyle(R_{n,1},R_{n,2})$
$\displaystyle\overset{d}{=}(\tilde{R}_{n,1},R_{n,2})=(R^{*}_{n,1},R_{n,2})+o_{p}(1)$
(O.11) $\displaystyle R^{*}_{n,1}$
$\displaystyle\rightsquigarrow\mathcal{N}\mathinner{\biggl{(}0,\sum_{s\in\mathcal{S}}\mathinner{\biggl{[}\pi(s)p(s)\operatorname*{var}\mathinner{\bigl{(}\rho_{s}(1)|S=s\bigr{)}}+\mathinner{\bigl{[}1-\pi(s)\bigr{]}}p(s)\operatorname*{var}\mathinner{\bigl{(}\rho_{s}(0)|S=s\bigr{)}}\biggr{]}}\biggr{)}}$
$\displaystyle=\mathcal{N}\mathinner{\biggl{(}0,\mathbb{E}\mathinner{\Bigl{[}\pi(S)\mathinner{\bigl{(}\rho_{S}(1)-\mathbb{E}[\rho_{S}(1)|S]\bigr{)}}^{2}+(1-\pi(S))\mathinner{\bigl{(}\rho_{S}(0)-\mathbb{E}[\rho_{S}(0)|S]\bigr{)}}^{2}\Bigr{]}}\biggr{)}}$
$\displaystyle=\mathrel{\mathop{\mathchar 58\relax}}\zeta_{1}$

where the convergence in distribution is due to an analogous argument for
$\tilde{\rho}^{s}_{is}(0)$ and the independence of
$\mathinner{\bigl{\\{}Y_{i}(1)^{s},Y_{i}(0)^{s},D_{i}(1)^{s},D_{i}(0)^{s},X_{i}^{s}\mathrel{\mathop{\mathchar
58\relax}}1\leq i\leq n,s\in\mathcal{S}\bigr{\\}}}$ across both $i$ and $s$.
Moreover, since $R_{n,1}^{*}$ is a function of
$\mathinner{\bigl{\\{}Y_{i}(1)^{s},Y_{i}(0)^{s},D_{i}(1)^{s},D_{i}(0)^{s},X_{i}^{s}\mathrel{\mathop{\mathchar
58\relax}}1\leq i\leq n,s\in\mathcal{S}\bigr{\\}}}$ $\perp\\!\\!\\!\perp
S^{(n)},A^{(n)}$, and $R_{n,2}$ is a function of $\\{S^{(n)},A^{(n)}\\}$, we
see that $R_{n,1}^{*}\perp\\!\\!\\!\perp R_{n,2}$. Thus (O.11) implies

$\displaystyle(R_{n,1},R_{n,2})\overset{d}{=}(R^{*}_{n,1},R_{n,2})+o_{p}(1)\rightsquigarrow\mathinner{\bigl{(}\zeta_{1},\zeta_{2}\bigr{)}}$

where $\zeta_{1}$ and $\zeta_{2}$ are independent, with

$\displaystyle\zeta_{2}\mathrel{\mathop{\mathchar
58\relax}}=\mathcal{N}\mathinner{\biggl{(}0,\mathbb{E}\mathinner{\biggl{[}\mathinner{\Bigl{(}\mathbb{E}\mathinner{\bigl{[}Y(D(1))-Y(D(0))-\tau(D(1)-D(0))|S\bigr{]}}\Bigr{)}}^{2}\biggr{]}}\biggr{)}}.$

We hence show that the asymptotic distribution of the numerator of
$\sqrt{n}(\hat{\tau}_{S}-\tau)$ is $\zeta_{1}+\zeta_{2}$. This completes the
proof of part (i). The proof of part (ii), available upon request, is omitted
in the interest of space as it is quite similar to that of part (ii) of
Theorem 3.1.

##  Appendix P Technical Lemmas Used in the Proof of Theorem 3.1

######  Lemma P.1.

Suppose assumptions in Theorem 3.1 hold. Then, we have

$\displaystyle R_{n,1}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\pi(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1),$
$\displaystyle R_{n,2}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}\frac{1}{1-\pi(S_{i})}-1\biggr{)}}(1-A_{i})\tilde{\mu}^{Y}(0,S_{i},X_{i})-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(0,S_{i},X_{i})+o_{p}(1),$
$\displaystyle R_{n,3}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(S_{i})}\tilde{W}_{i}A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\pi(S_{i})}\tilde{Z}_{i}+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}+o_{p}(1),$

where for $a=0,1$,

$\displaystyle\tilde{\mu}^{Y}(a,S_{i},X_{i})\mathrel{\mathop{\mathchar
58\relax}}=\overline{\mu}^{Y}(a,S_{i},X_{i})-\overline{\mu}^{Y}(a,S_{i}),\quad\overline{\mu}^{Y}(a,S_{i})\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{E}\mathinner{\bigl{[}\overline{\mu}^{Y}(a,S_{i},X_{i})|S_{i}\bigr{]}},$
$\displaystyle W_{i}\mathrel{\mathop{\mathchar
58\relax}}=Y_{i}(1)D_{i}(1)+Y_{i}(0)(1-D_{i}(1)),\quad
Z_{i}\mathrel{\mathop{\mathchar
58\relax}}=Y_{i}(1)D_{i}(0)+Y_{i}(0)(1-D_{i}(0)),$
$\displaystyle\tilde{W}_{i}\mathrel{\mathop{\mathchar
58\relax}}=W_{i}-\mathbb{E}[W_{i}|S_{i}],\quad\text{and}\quad\tilde{Z}_{i}\mathrel{\mathop{\mathchar
58\relax}}=Z_{i}-\mathbb{E}[Z_{i}|S_{i}].$

###### Proof.

We have

$\displaystyle R_{n,1}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{[}\hat{\mu}^{Y}(1,S_{i},X_{i})-\frac{A_{i}\hat{\mu}^{Y}(1,S_{i},X_{i})}{\hat{\pi}(S_{i})}\biggr{]}}$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(S_{i})}{\hat{\pi}(S_{i})}\hat{\mu}^{Y}(1,S_{i},X_{i})$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(S_{i})}{\hat{\pi}(S_{i})}\mathinner{\bigl{[}\hat{\mu}^{Y}(1,S_{i},X_{i})-\overline{\mu}^{Y}(1,S_{i},X_{i})+\overline{\mu}^{Y}(1,S_{i},X_{i})\bigr{]}}$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(S_{i})}{\hat{\pi}(S_{i})}\Delta^{Y}(1,S_{i},X_{i})-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}}{\hat{\pi}(S_{i})}\overline{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\overline{\mu}^{Y}(1,S_{i},X_{i})$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(S_{i})}{\hat{\pi}(S_{i})}\Delta^{Y}(1,S_{i},X_{i})-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}}{\hat{\pi}(S_{i})}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\tilde{\mu}^{Y}(1,S_{i},X_{i}),$
(P.1)

where the last equality is due to

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}}{\hat{\pi}(S_{i})}\overline{\mu}^{Y}(1,S_{i})$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\overline{\mu}^{Y}(1,S_{i}).$

Consider the first term of (P.1).

$\displaystyle\mathinner{\\!\biggl{\lvert}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(S_{i})}{\hat{\pi}(S_{i})}\Delta^{Y}(1,S_{i},X_{i})\biggr{\rvert}}=\mathinner{\\!\biggl{\lvert}\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=1}^{n}\frac{A_{i}-\hat{\pi}(s)}{\hat{\pi}(s)}\Delta^{Y}(1,s,X_{i})1\\{S_{i}=s\\}\biggr{\rvert}}$
$\displaystyle=\frac{1}{\sqrt{n}}\mathinner{\\!\biggl{\lvert}\sum_{s\in\mathcal{S}}\frac{1}{\hat{\pi}(s)}\sum_{i=1}^{n}A_{i}\Delta^{Y}(1,s,X_{i})1\\{S_{i}=s\\}-\sum_{s\in\mathcal{S}}\sum_{i=1}^{n}\Delta^{Y}(1,s,X_{i})1\\{S_{i}=s\\}\biggr{\rvert}}$
$\displaystyle=\frac{1}{\sqrt{n}}\mathinner{\\!\biggl{\lvert}\sum_{s\in\mathcal{S}}\sum_{i\in
I_{1}(s)}\Delta^{Y}(1,s,X_{i})\frac{n(s)}{n_{1}(s)}-\sum_{s\in\mathcal{S}}\sum_{i\in
I_{0}(s)\cup I_{1}(s)}\Delta^{Y}(1,s,X_{i})\biggr{\rvert}}$
$\displaystyle=\frac{1}{\sqrt{n}}\mathinner{\\!\biggl{\lvert}\sum_{s\in\mathcal{S}}\sum_{i\in
I_{1}(s)}\Delta^{Y}(1,s,X_{i})\frac{n_{0}(s)}{n_{1}(s)}-\sum_{s\in\mathcal{S}}\sum_{i\in
I_{0}(s)}\Delta^{Y}(1,s,X_{i})\biggr{\rvert}}$
$\displaystyle=\frac{1}{\sqrt{n}}\mathinner{\\!\biggl{\lvert}\sum_{s\in\mathcal{S}}n_{0}(s)\mathinner{\biggl{[}\frac{\sum_{i\in
I_{1}(s)}\Delta^{Y}(1,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{Y}(1,s,X_{i})}{n_{0}(s)}\biggr{]}}\biggr{\rvert}}$
$\displaystyle\leq\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}n_{0}(s)\mathinner{\\!\biggl{\lvert}\frac{\sum_{i\in
I_{1}(s)}\Delta^{Y}(1,s,X_{i})}{n_{1}(s)}-\frac{\sum_{i\in
I_{0}(s)}\Delta^{Y}(1,s,X_{i})}{n_{0}(s)}\biggr{\rvert}}=o_{p}(1)$

where the last equality is due to Assumption 3. Thus

$\displaystyle R_{n,1}$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}}{\hat{\pi}(S_{i})}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1)$
$\displaystyle=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}}{\hat{\pi}(S_{i})}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1)$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\hat{\pi}(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1).$
(P.2)

In addition, we note that

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\hat{\pi}(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\pi(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})$
$\displaystyle+\sum_{s\in\mathcal{S}}\mathinner{\biggl{(}\frac{1}{\pi(s)}-\frac{1}{\hat{\pi}(s)}\biggr{)}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(1,s,X_{i})1\\{S_{i}=s\\}.$

Note that under Assumption 1(i), conditional on $\\{S^{(n)},A^{(n)}\\}$, the
distribution of

$\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(1,s,X_{i})1\\{S_{i}=s\\}$

is the same as the distribution of the same quantity where units are ordered
by strata and then ordered by $A_{i}=1$ first and $A_{i}=0$ second within
strata. To this end, define $N(s)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{i=1}^{n}1\\{S_{i}<s\\}$ and $F(s)\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{P}(S_{i}<s)$. Furthermore, independently for each
$s\in\mathcal{S}$ and independently of $\\{S^{(n)},A^{(n)}\\}$, let
$\mathinner{\bigl{\\{}X_{i}^{s}\mathrel{\mathop{\mathchar 58\relax}}1\leq
i\leq n\bigr{\\}}}$ be i.i.d with marginal distribution equal to the
distribution of $X_{i}|S=s$. Define

$\displaystyle\tilde{\mu}^{b}(a,s,X_{i}^{s})$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\overline{\mu}^{b}(a,s,X_{i}^{s})-\mathbb{E}\mathinner{\bigl{[}\overline{\mu}^{b}(a,s,X_{i}^{s})|S_{i}=s\bigr{]}}$

Then, we have, for $s\in\mathcal{S}$,

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(1,s,X_{i})1\\{S_{i}=s\\}\stackrel{{\scriptstyle
d}}{{=}}\frac{1}{\sqrt{n}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\mu}^{Y}(1,s,X_{i}^{s}).$

In addition, we have

$\displaystyle\mathbb{E}\left[\left(\frac{1}{\sqrt{n}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\mu}^{Y}(1,s,X_{i}^{s})\right)^{2}\biggl{|}S^{(n)},A^{(n)}\right]$
$\displaystyle=\frac{n_{1}(s)}{n}\mathbb{E}\mathinner{\bigl{[}\tilde{\mu}^{Y,2}(a,s,X_{i}^{s})|S^{(n)}\bigr{]}}$
$\displaystyle\leq\frac{n_{1}(s)}{n}E\left[\overline{\mu}^{Y,2}(a,s,X_{i})|S_{i}=s\right]=O_{p}(1),$

which implies

$\displaystyle\max_{s\in\mathcal{S}}\biggl{|}\frac{1}{\sqrt{n}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\tilde{\mu}^{Y}(1,s,X_{i}^{s})\biggr{|}=O_{p}(1).$

Combining this with the facts that
$\max_{s\in\mathcal{S}}|\hat{\pi}(s)-\pi(s)|=o_{p}(1)$ and
$\min_{s\in\mathcal{S}}\pi(s)>c>0$ for some constant $c$, we have

$\displaystyle\sum_{s\in\mathcal{S}}\mathinner{\biggl{(}\frac{1}{\pi(s)}-\frac{1}{\hat{\pi}(s)}\biggr{)}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}A_{i}\tilde{\mu}^{Y}(1,s,X_{i})1\\{S_{i}=s\\}=o_{p}(1)$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\hat{\pi}(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\pi(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1).$

Therefore, we have

$\displaystyle R_{n,1}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\biggl{(}1-\frac{1}{\pi(S_{i})}\biggr{)}}A_{i}\tilde{\mu}^{Y}(1,S_{i},X_{i})+\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(1-A_{i})\tilde{\mu}^{Y}(1,S_{i},X_{i})+o_{p}(1).$

The linear expansion of $R_{n,2}$ can be established in the same manner. For
$R_{n,3}$, note that

$\displaystyle Y_{i}$
$\displaystyle=Y_{i}(1)\mathinner{\bigl{[}D_{i}(1)A_{i}+D_{i}(0)(1-A_{i})\bigr{]}}+Y_{i}(0)\mathinner{\bigl{[}1-D_{i}(1)A_{i}-D_{i}(0)(1-A_{i})\bigr{]}}$
$\displaystyle=\mathinner{\bigl{[}Y_{i}(1)D_{i}(1)-Y_{i}(0)D_{i}(1)\bigr{]}}A_{i}+\mathinner{\bigl{[}Y_{i}(1)D_{i}(0)-Y_{i}(0)D_{i}(0)\bigr{]}}(1-A_{i})+Y_{i}(0).$

Then

$\displaystyle A_{i}Y_{i}$
$\displaystyle=\mathinner{\bigl{[}Y_{i}(1)D_{i}(1)+Y_{i}(0)(1-D_{i}(1))\bigr{]}}A_{i},$
$\displaystyle(1-A_{i})Y_{i}$
$\displaystyle=\mathinner{\bigl{[}Y_{i}(1)D_{i}(0)+Y_{i}(0)(1-D_{i}(0))\bigr{]}}(1-A_{i}),$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}Y_{i}}{\hat{\pi}(S_{i})}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\mathinner{\bigl{[}Y_{i}(1)D_{i}(1)+Y_{i}(0)(1-D_{i}(1))\bigr{]}}A_{i}=\mathrel{\mathop{\mathchar
58\relax}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}W_{i}A_{i},$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{(1-A_{i})Y_{i}}{1-\hat{\pi}(S_{i})}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\mathinner{\bigl{[}Y_{i}(1)D_{i}(0)+Y_{i}(0)(1-D_{i}(0))\bigr{]}}(1-A_{i})}{1-\hat{\pi}(S_{i})}=\mathrel{\mathop{\mathchar
58\relax}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{Z_{i}(1-A_{i})}{1-\hat{\pi}(S_{i})}.$

Thus we have

$\displaystyle R_{n,3}$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{A_{i}Y_{i}}{\hat{\pi}(S_{i})}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{(1-A_{i})Y_{i}}{1-\hat{\pi}(S_{i})}-\sqrt{n}G$
$\displaystyle=\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\tilde{W}_{i}A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\hat{\pi}(S_{i})}\tilde{Z}_{i}\right\\}$
$\displaystyle+\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\hat{\pi}(S_{i})}\mathbb{E}[Z_{i}|S_{i}]-\sqrt{n}G\right\\}.$
(P.3)

We now consider the second term on the RHS of (P.3). First note that

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\hat{\pi}(S_{i})-\pi(S_{i})}{\hat{\pi}(S_{i})\pi(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i},$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(s)}\mathbb{E}[W_{i}|S_{i}=s]A_{i}1\\{S_{i}=s\\}$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\mathbb{E}[W_{i}|S_{i}=s]}{\pi(s)}(A_{i}-\pi(s))1\\{S_{i}=s\\}+\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(s)}\mathbb{E}[W_{i}|S_{i}=s]\pi(s)1\\{S_{i}=s\\}$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\pi(s)\sqrt{n}}\sum_{i=1}^{n}(A_{i}-\pi(s))1\\{S_{i}=s\\}+\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\sqrt{n}}\sum_{i=1}^{n}1\\{S_{i}=s\\}$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\pi(s)\sqrt{n}}B_{n}(s)+\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\sqrt{n}}n(s),$
(P.4)

and

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\hat{\pi}(S_{i})-\pi(S_{i})}{\hat{\pi}(S_{i})\pi(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{\hat{\pi}(s)-\pi(s)}{\hat{\pi}(s)\pi(s)}\mathbb{E}[W_{i}|S_{i}=s]A_{i}1\\{S_{i}=s\\}$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{B_{n}(s)}{n(s)\hat{\pi}(s)\pi(s)}\mathbb{E}[W_{i}|S_{i}=s]A_{i}1\\{S_{i}=s\\}$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{B_{n}(s)\mathbb{E}[W|S=s]}{\sqrt{n}n(s)\hat{\pi}(s)\pi(s)}\sum_{i=1}^{n}A_{i}1\\{S_{i}=s\\}=\sum_{s\in\mathcal{S}}\frac{B_{n}(s)\mathbb{E}[W|S=s]}{\sqrt{n}n(s)\hat{\pi}(s)\pi(s)}n_{1}(s)$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{B_{n}(s)\mathbb{E}[W|S=s]}{\sqrt{n}\pi(s)}.$

Therefore, we have

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}=\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\sqrt{n}}n(s).$

Similarly, we have

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\hat{\pi}(S_{i})}\mathbb{E}[Z_{i}|S_{i}]=\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[Z|S=s]}{\sqrt{n}}n(s)$

Then, we have

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\mathbb{E}[W_{i}|S_{i}]A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\hat{\pi}(S_{i})}\mathbb{E}[Z_{i}|S_{i}]-\sqrt{n}G$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[W|S=s]}{\sqrt{n}}n(s)-\sum_{s\in\mathcal{S}}\frac{\mathbb{E}[Z|S=s]}{\sqrt{n}}n(s)-\sqrt{n}G$
$\displaystyle=\sum_{s\in\mathcal{S}}\sqrt{n}\mathinner{\biggl{(}\frac{n(s)}{n}-p(s)\biggr{)}}\mathbb{E}[W-Z|S=s]+\sum_{s\in\mathcal{S}}\sqrt{n}p(s)\mathbb{E}[W-Z|S=s]-\sqrt{n}G$
$\displaystyle=\sum_{s\in\mathcal{S}}\sqrt{n}\mathinner{\biggl{(}\frac{n(s)}{n}-p(s)\biggr{)}}\mathbb{E}[W-Z|S=s]+\sqrt{n}\mathbb{E}[W-Z]-\sqrt{n}G$
$\displaystyle=\sum_{s\in\mathcal{S}}\frac{n(s)}{\sqrt{n}}\mathbb{E}[W-Z|S=s]-\sqrt{n}\mathbb{E}[W-Z]$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=1}^{n}\mathinner{\Bigl{(}1\\{S_{i}=s\\}\mathbb{E}[W_{i}-Z_{i}|S_{i}=s]\Bigr{)}}-\sqrt{n}\mathbb{E}[W-Z]$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\sqrt{n}\mathbb{E}[W-Z]$
$\displaystyle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}.$
(P.5)

Combining (P.3) and (P.5), we have

$\displaystyle R_{n,3}$
$\displaystyle=\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\hat{\pi}(S_{i})}\tilde{W}_{i}A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\hat{\pi}(S_{i})}\tilde{Z}_{i}\right\\}+\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}\right\\}$
$\displaystyle=\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\pi(S_{i})}\tilde{W}_{i}A_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1-A_{i}}{1-\pi(S_{i})}\tilde{Z}_{i}\right\\}$
$\displaystyle+\left\\{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{\bigl{(}\mathbb{E}[W_{i}-Z_{i}|S_{i}]-\mathbb{E}[W_{i}-Z_{i}]\bigr{)}}\right\\}+o_{p}(1),$

where the second equality holds because

$\displaystyle\left(\frac{1}{\pi(s)}-\frac{1}{\hat{\pi}(s)}\right)\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\tilde{W}_{i}A_{i}1\\{S_{i}=s\\}=o_{p}(1)\quad\text{and}$
$\displaystyle\left(\frac{1}{\pi(s)}-\frac{1}{\hat{\pi}(s)}\right)\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\tilde{Z}_{i}(1-A_{i})1\\{S_{i}=s\\}=o_{p}(1)$

due to the same argument used in the proofs of $R_{n,1}$. ∎

######  Lemma P.2.

Under the assumptions in Theorem 3.1, we have

$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i}\rightsquigarrow\mathcal{N}\left(0,\mathbb{E}\pi(S_{i})\Xi_{1}^{2}(\mathcal{D}_{i},S_{i})\right),$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i})\rightsquigarrow\mathcal{N}\left(0,\mathbb{E}(1-\pi(S_{i}))\Xi_{0}^{2}(\mathcal{D}_{i},S_{i})\right),\quad\text{and}$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\rightsquigarrow\mathcal{N}(0,\mathbb{E}\Xi_{2}^{2}(S_{i})),$

and the three terms are asymptotically independent.

###### Proof.

Note that under Assumption 1(i), conditional on $\\{S^{(n)},A^{(n)}\\}$, the
distribution of

$\left(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i},~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i})\right)$

is the same as the distribution of the same quantity where units are ordered
by strata and then ordered by $A_{i}=1$ first and $A_{i}=0$ second within
strata. To this end, define $N(s)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{i=1}^{n}1\\{S_{i}<s\\}$ and $F(s)\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{P}(S_{i}<s)$. Furthermore, independently for each
$s\in\mathcal{S}$ and independently of $\\{S^{(n)},A^{(n)}\\}$, let
$\mathinner{\bigl{\\{}\mathcal{D}_{i}^{s}\mathrel{\mathop{\mathchar
58\relax}}1\leq i\leq n\bigr{\\}}}$ be i.i.d with marginal distribution equal
to the distribution of $\mathcal{D}|S=s$. Then, we have

$\displaystyle\left(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i},~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i})\right)\biggl{|}S^{(n)},A^{(n)}$
$\displaystyle\stackrel{{\scriptstyle
d}}{{=}}\left(\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Xi_{1}(\mathcal{D}_{i}^{s},s),~{}\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{N(s)+n_{1}(s)+1}^{N(s)+n(s)}\Xi_{0}(\mathcal{D}_{i}^{s},s)\right)\biggl{|}S^{(n)},A^{(n)}.$

In addition, since $\Xi_{2}(S_{i})$ is a function of $\\{S^{(n)},A^{(n)}\\}$,
we have, arguing along the line of a joint distribution being the product of a
conditional distribution and a marginal distribution,

$\displaystyle\left(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{1}(\mathcal{D}_{i},S_{i})A_{i},~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{0}(\mathcal{D}_{i},S_{i})(1-A_{i}),~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\right)$
$\displaystyle\stackrel{{\scriptstyle
d}}{{=}}\left(\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Xi_{1}(\mathcal{D}_{i}^{s},s),~{}\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{N(s)+n_{1}(s)+1}^{N(s)+n(s)}\Xi_{0}(\mathcal{D}_{i}^{s},s),~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\right).$

Define $\Gamma_{a,n}(u,s)=\frac{1}{\sqrt{n}}\sum_{i=1}^{\lfloor
un\rfloor}\Xi_{a}(\mathcal{D}_{i}^{s},s)$ for $a=0,1,s\in\mathcal{S}$. We have

$\displaystyle\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Xi_{1}(\mathcal{D}_{i}^{s},s)=\sum_{s\in\mathcal{S}}\left[\Gamma_{1,n}\left(\frac{N(s)+n_{1}(s)}{n},s\right)-\Gamma_{1,n}\left(\frac{N(s)}{n},s\right)\right],$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{N(s)+n_{1}(s)+1}^{N(s)+n(s)}\Xi_{0}(\mathcal{D}_{i}^{s},s)=\sum_{s\in\mathcal{S}}\left[\Gamma_{0,n}\left(\frac{N(s)+n(s)}{n},s\right)-\Gamma_{0,n}\left(\frac{N(s)+n_{1}(s)}{n},s\right)\right].$

In addition, the partial sum process (w.r.t. $u\in[0,1]$) is stochastic
equicontinuous and

$\displaystyle\mathinner{\biggl{(}\frac{N(s)}{n},\frac{n_{1}(s)}{n}\biggr{)}}\xrightarrow{p}\mathinner{\bigl{(}F(s),\pi(s)p(s)\bigr{)}}.$

Therefore,

$\displaystyle\left(\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{i=N(s)+1}^{N(s)+n_{1}(s)}\Xi_{1}(\mathcal{D}_{i}^{s},s),~{}\frac{1}{\sqrt{n}}\sum_{s\in\mathcal{S}}\sum_{N(s)+n_{1}(s)+1}^{N(s)+n(s)}\Xi_{0}(\mathcal{D}_{i}^{s},s),~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\right)$
$\displaystyle=\begin{pmatrix}&\sum_{s\in\mathcal{S}}\left[\Gamma_{1,n}\left(F(s)+p(s)\pi(s),s\right)-\Gamma_{1,n}\left(F(s),s\right)\right],\\\
&\sum_{s\in\mathcal{S}}\left[\Gamma_{0,n}\left(F(s)+p(s),s\right)-\Gamma_{0,n}\left(F(s)+\pi(s)p(s),s\right)\right],\\\
&~{}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\end{pmatrix}+o_{p}(1)$

and by construction,

$\displaystyle\sum_{s\in\mathcal{S}}\left[\Gamma_{1,n}\left(F(s)+p(s)\pi(s),s\right)-\Gamma_{1,n}\left(F(s),s\right)\right],$
$\displaystyle\sum_{s\in\mathcal{S}}\left[\Gamma_{0,n}\left(F(s)+p(s),s\right)-\Gamma_{0,n}\left(F(s)+p(s)\pi(s),s\right)\right],$
$\displaystyle\text{and}\quad\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})$

are independent. Last, we have

$\displaystyle\sum_{s\in\mathcal{S}}\left[\Gamma_{1,n}\left(F(s)+p(s)\pi(s),s\right)-\Gamma_{1,n}\left(F(s),s\right)\right]\rightsquigarrow\mathcal{N}\left(0,\mathbb{E}\pi(S_{i})\Xi_{1}^{2}(\mathcal{D}_{i},S_{i})\right)$
$\displaystyle\sum_{s\in\mathcal{S}}\left[\Gamma_{0,n}\left(F(s)+p(s),s\right)-\Gamma_{0,n}\left(F(s)+p(s)\pi(s),s\right)\right]\rightsquigarrow\mathcal{N}\left(0,\mathbb{E}(1-\pi(S_{i}))\Xi_{0}^{2}(\mathcal{D}_{i},S_{i})\right)$
$\displaystyle\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\Xi_{2}(S_{i})\rightsquigarrow\mathcal{N}\left(0,\mathbb{E}\Xi_{2}^{2}(S_{i})\right).$

This implies the desired result. ∎

######  Lemma P.3.

Suppose assumptions in Theorem 3.1 hold. Then,

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}A_{i}\hat{\Xi}_{1}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{1}^{2},\quad\frac{1}{n}\sum_{i=1}^{n}(1-A_{i})\hat{\Xi}_{0}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{0}^{2},\quad\text{and}\quad\frac{1}{n}\sum_{i=1}^{n}\hat{\Xi}_{2}^{2}(\mathcal{D}_{i},S_{i})\overset{p}{\longrightarrow}\sigma_{2}^{2}.$

###### Proof.

To derive the limit of
$\frac{1}{n}\sum_{i=1}^{n}A_{i}\hat{\Xi}_{1}^{2}(\mathcal{D}_{i},S_{i})$, we
first define

$\displaystyle\tilde{\Xi}_{1}^{*}(\mathcal{D}_{i},s)$
$\displaystyle=\left[\left(1-\frac{1}{\pi(s)}\right)\overline{\mu}^{Y}(1,s,X_{i})-\overline{\mu}^{Y}(0,s,X_{i})+\frac{Y_{i}}{\pi(s)}\right]$
$\displaystyle-\tau\left[\left(1-\frac{1}{\pi(s)}\right)\overline{\mu}^{D}(1,s,X_{i})-\overline{\mu}^{D}(0,s,X_{i})+\frac{D_{i}}{\pi(s)}\right]\quad\text{and}$
$\displaystyle\breve{\Xi}_{1}(\mathcal{D}_{i},s)$
$\displaystyle=\left[\left(1-\frac{1}{\hat{\pi}(s)}\right)\overline{\mu}^{Y}(1,s,X_{i})-\overline{\mu}^{Y}(0,s,X_{i})+\frac{Y_{i}}{\hat{\pi}(s)}\right]$
$\displaystyle-\hat{\tau}\left[\left(1-\frac{1}{\hat{\pi}(s)}\right)\overline{\mu}^{D}(1,s,X_{i})-\overline{\mu}^{D}(0,s,X_{i})+\frac{D_{i}}{\hat{\pi}(s)}\right]$

Then, we have

$\displaystyle\left[\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}(\tilde{\Xi}_{1}^{*}(\mathcal{D}_{i},s)-\tilde{\Xi}_{1}(\mathcal{D}_{i},s))^{2}\right]^{1/2}$
$\displaystyle\leq\left[\frac{1}{n_{1}(s)}\sum_{i\in
I_{1}(s)}(\tilde{\Xi}_{1}^{*}(\mathcal{D}_{i},s)-\breve{\Xi}_{1}(\mathcal{D}_{i},s))^{2}\right]^{1/2}+\left[\frac{1}{n_{1}(s)}\sum_{i\in