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On the limiting variance of matching estimators
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\title{\LARGE On the limiting variance of matching estimators}
\author{
Songliang Chen\thanks{Department of Mathematical Sciences, Tsinghua University, Beijing, China. E-mail: \tt{[email removed]}} ~~and~ Fang Han\thanks{Department of Statistics, University of Washington, Seattle, WA 98195, USA; e-mail: {\tt [email removed]}}}
\date{\today}
\maketitle
\begin{abstract}
This paper examines the limiting variance of nearest neighbor matching estimators for average treatment effects with a fixed number of matches. We present, for the first time, a closed-form expression for this limit. Here the key is the establishment of the limiting second moment of the catchment area’s volume, which resolves a question of Abadie and Imbens. At the core of our approach is a new universality theorem on the measures of high-order Voronoi cells, extending a result by Devroye, Gy\"{o}rfi, Lugosi, and Walk.
\end{abstract}
{\bf Keywords:} matching estimators, nearest neighbors, Voronoi cells, stochastic geometry
\section{Introduction}
Consider estimating the population average treatment effect (ATE),
\[
\tau := {\mathrm E}\{Y(1)-Y(0)\},
\]
based on $n$ independent observations of $(Y(W),X,W)$, where $X\in{\mathbb R}^d$ represents some pre-treatment covariates, $W\in\{0,1\}$ indicates the treatment status, and $(Y(0),Y(1))\in{\mathbb R}^2$ are two potential outcomes \citep{neyman1923applications,rubin1974estimating} referring to being treated or not.
For estimating $\tau$, the nearest neighbor (NN) matching estimator is intuitive and widely used \citep{stuart2010matching,imbens2024causal}. It imputes the missing potential outcomes by averaging the $M$ within-match outcomes in the opposite treatment group. In a landmark paper, Abadie and Imbens \citep[Theorem 4]{abadie2006large} proved the following central limit theorem for the {\it normalized} NN matching estimator, $\hat\tau_M$, with $M$ assumed to be fixed:
\[
V_{M}^{-1/2}\cdot \sqrt{n}(\hat\tau_M-B_{M}-\tau) \text{ converges in distribution to }\cN(0,1).
\]
Here $B_M=B_{M,n}$ represents the bias term and $V_M=V_{M,n}$ stands for a {\it random} approximation to the variance of $\sqrt{n}\hat\tau_M$.
While techniques to correct the bias $B_{M}$ have been mature \citep{abadie2011bias}, still little is known about the random term $V_{M}$, whose stochastic behavior is ``difficult to work out'' \citep[Page 248]{abadie2006large}. The following theorem, presented informally here and to be rigorously stated in Section \ref{sec:theory}, settles the limit of ${\mathrm E} [V_{M}]$ and is our central result.
\begin{theorem}[Main theorem, informal]\label{thm:informal} Under certain regularity conditions and assuming $M$ to be fixed, the following hold true.
\begin{enumerate}[label=(\roman*)]
\item The limit of ${\mathrm E}[V_{M}]$ exists and admits a closed form $\sigma_{M,d}^2$ to be introduced in Theorem \ref{thm4.2} ahead.
\item $\sqrt{n}(\hat\tau_M-B_M-\tau)$ converges in distribution to $\cN(0,\sigma_{M,d}^2)$.
\end{enumerate}
\end{theorem}
\subsection{Related literature}
This paper contributes to a growing body of literature on large-sample theory for NN matching estimators. This line of research was pioneered by Abadie and Imbens in their series of seminal papers \citep{abadie2006large,abadie2008failure,abadie2011bias,abadie2012martingale,abadie2016matching} and has been extended by subsequent works, including \cite{lin2022regression}, \cite{lin2023estimation}, \cite{lu2023flexible}, \cite{huo2023adaptation}, \cite{cattaneo2023rosenbaum}, \cite{demirkaya2024optimal}, \cite{he2024propensity}, \cite{holzmann2024multivariate}, \cite{li2024matching}, \cite{ulloa2024propensity}, and \cite{lin2024consistency}, among many others. In particular, \cite{lin2023estimation} calculated the limiting variance of $\hat\tau_M$ as $M$ increases to infinity with $n$, and showed that the limit matches the corresponding semiparametric efficiency lower bound \citep{hahn1998role}. In this regard, the current paper extends their theory to the fixed-$M$ asymptotic regime.
For deriving the form of $\sigma_{M,d}^2$, a crucial step is to calculate the limiting second moment of the volume of a stochastic object, which Abadie and Imbens termed the ``catchment area'' \citep[Page 260]{abadie2006large}. As pointed out in \citet[footnote 9]{abadie2006large} and \citet[Remark 2.1]{lin2023estimation}, this problem is closely related to a stochastic geometry problem of calculating the volume of Voronoi cells \citep{voronoi1908nouvelles}, and reduces to it when $M=1$. In this regard, our work also extends a result of Devroye, Gy\"{o}rfi, Lugosi, and Walk \citep[Theorem 1]{devroye2017measure} --- they calculated the limiting second moment of Voronoi's cells --- to the case when $M>1$.
More broadly speaking, this paper constitutes to the literature of statistical theory for NN-based methods \citep{MR3445317}, which aim to estimate a certain functional of the data generating distribution using a data-based NN graph. In this respect, the current paper is particularly related to a line of research on the asymptotic distribution-free properties of NN functionals \citep{MR532236,MR682809,shi2021ac,han2022azadkia}, to which this paper contributes a new distribution-free theorem (Theorem \ref{assump3.1} ahead, with two densities therein set to be identical).
\subsection{Notation and set-up}
For any positive integer $n$, denote $[n]=\{1,2,\ldots,n\}$. Let $\{(Y_i(0),Y_i(1),X_i,W_i)\}_{i\in[n]}$ be $n$ realizations of $(Y(0),Y(1),X,W)$. Write $Y=Y(W)$ and denote the support of $X$ by $\cX\subset {\mathbb R}^d$. Let $\{(Y_i=Y_i(W_i),X_i,W_i)\}_{i\in[n]}$ be the observation and
\[
n_1:=\sum_{i=1}^nW_i~~~{\rm and}~~~n_0:=\sum_{i=1}^n(1-W_i)
\]
be the sizes of treated and control groups, respectively. Following \cite{abadie2006large}, we define the conditional average treatment effect
\[
\tau(x)={\mathrm E}[Y(1)-Y(0) \,|\, X=x]
\]
and introduce, for any $x\in\cX$ and $w\in \{0,1\}$, the following two conditional moments,
\[
\mu_w(x)={\mathrm E}[Y\,|\, X=x, W=w]~~~{\rm and}~~~\sigma_w^2(x)={\rm Var}(Y\,|\, X=x,W=w).
\]
It is immediate that $\tau(x)=\mu_1(x)-\mu_0(x)$ if $(Y(0),Y(1))$ is independent of $W$ conditional on $X=x$ and $W$ is not degenerate given $X=x$. Furthermore, if $(Y(0),Y(1))$ is independent of $W$ conditional on almost all $X=x$ and ${\mathrm P}(W=1\,|\, X)$ is bounded away from 0 and 1 almost surely, then
\[
\tau={\mathrm E}[\tau(X)]={\mathrm E}[\mu_1(X)-\mu_0(X)].
\]
Lastly, for any $i\in[n]$, let $\varepsilon_i:=Y_i-\mu_{W_i}(X_i)$ be the residual and, for any $x\in\cX$, let $e(x):={\mathrm P}(W=1\,|\, X=x)$ be the propensity score \citep{rosenbaum1983central}.
\section{Nearest neighbor matching}
Following \cite{abadie2006large}, for any given number of matches $M$, we consider the following matching-based estimator of the ATE,
\[
\hat{\tau}_M = \frac{1}{n}\sum_{i = 1}^n\Big\{\hat{Y}_i(1) - \hat{Y}_i(0)\Big\},
\]
where for each $i\in[n]$, we define
\[
\hat{Y}_i(1)= \begin{cases}
Y_i, & \mbox{ if } W_i=1,\\
M^{-1}\sum_{j \in \cJ_M(i)}Y_j, & \mbox{ if } W_i=0
\end{cases}
~~~{\rm and}~~~
\hat{Y}_i(0)= \begin{cases}
Y_i, & \mbox{ if } W_i=0,\\
M^{-1}\sum_{j \in \cJ_M(i)}Y_j, & \mbox{ if } W_i=1.
\end{cases}
\]
Here $\cJ_M(i)$ indexes all $M$ NNs, based on the Euclidean metric $\norm{\cdot}$, in the opposite treated group of subject $i$. In other words, $\cJ_M(i)$ indexes all $j \in [n]$ such that $W_j = 1 - W_i$ and
\[
\sum_{k = 1, W_k = 1 - W_i}^n{\mathds 1}\Big\{\Big\| X_k - X_i\Big\| \leq \Big\|X_j - X_i\Big\| \Big\} \leq M.
\]
Let $K_M(i)$ denote the times the subject $i$ is used as a match, i.e.,
\[
K_M(i) := \sum_{j = 1, W_j = 1 - W_i}^n{\mathds 1}\Big\{i \in \cJ_M(j)\Big\}.
\]
The estimator $\hat{\tau}_M$ can also be written as
\[
\hat{\tau}_M = \frac{1}{n}\sum_{i = 1}^n(2W_i -1)\Big\{1 + \frac{K_M(i)}{M}\Big\}Y_i,
\]
which, as correctly pointed out in \citet[Chapter 15.3.2]{ding2024first} and rigorously formalized in \cite{lin2023estimation}, resembles inverse probability weighted estimators.
\section{Theory}\label{sec:theory}
Our main theorem, Theorem \ref{thm:informal}, is proved under the following four sets of assumptions.
\begin{assumption} \label{assump4.1}
$\{(Y_i(0),Y_i(1),X_i,W_i)\}_{i\in[n]}$ are independently drawn from $(Y(0),Y(1),X,W)$.
\end{assumption}
\begin{assumption} \label{assump4.2} For almost all $x\in\cX$,
\begin{itemize}
\item [(i)] $(Y(0),Y(1))$ is independent of $W$ conditional on $X=x$;
\item [(ii)] there exists some constant $\eta > 0$ such that $\eta < e(x) < 1 - \eta$.
\end{itemize}
\end{assumption}
\begin{assumption} \label{assump4.3}
\begin{itemize}
\item [(i)] The support of $X$ is compact and convex;
\item [(ii)] the random vector $X$ has Lebesgue density $f_X$ and there exist some constants $\ell, u\in (0,\infty)$ such that $\ell \leq \inf_{x\in\cX}f_X(x)\leq \sup_{x\in\cX}f_X(x) \leq u$.
\end{itemize}
\end{assumption}
\begin{assumption}\label{assump4.4}
For $w = 0, 1$,
\begin{itemize}
\item [(i)] $\mu_w(x)$ and $\sigma_w^2(x)$ are Lipschitz in $\cX$;
\item [(ii)] ${\mathrm E}[Y^4 \,|\, X = x, W = w]$ is uniformly bounded in $\cX$;
\item [(iii)] There exist some constants $\overline{\sigma}^2, \underline{\sigma}^2 > 0$ such that $\underline{\sigma}^2 < \sigma_w^2(x) < \overline{\sigma}^2$ for all $x\in\cX$.
\end{itemize}
\end{assumption}
The above four are identical to Assumptions 1-4 in \cite{abadie2006large}.
In \cite{abadie2006large}, it is shown that $\hat{\tau}_M$ has the following decomposition
\[
\hat{\tau}_M - \tau = \overline{\tau(X)} + E_M + B_M - \tau,
\]
with
\[
\overline{\tau(X)} := \frac{1}{n}\sum_{i = 1}^n\Big\{\mu_1(X_i) - \mu_0(X_i)\Big\},~~
E_M := \frac{1}{n}\sum_{i = 1}^n(2W_i - 1)\Big\{1 + \frac{K_M(i)}{M}\Big\}\varepsilon_i,
\]
and
\[
B_M = \frac{1}{n}\sum_{i = 1}^n(2W_i - 1)\Big[\frac{1}{M}\sum_{j \in \cJ_M(i)}\Big\{\mu_{1 - W_i}(X_i) - \mu_{1- W_i}(X_j)\Big\}\Big].
\]
Defining
\[
V^E = \frac{1}{n}\sum_{i = 1}^n\Big\{1 + \frac{K_M(i)}{M}\Big\}^2\sigma_{W_i}^2(X_i),~~
V^{\tau(X)} = {\mathrm E}[(\tau(X) - \tau)^2],~~{\rm and}~~V_M:=V^E+V^{\tau(X)},
\]
we summarize the asymptotic properties of $\hat{\tau}_M$, derived in \cite{abadie2006large} and \cite{abadie2011bias}, as follows.
\begin{theorem}[\cite{abadie2006large, abadie2011bias}] \label{thm4.1}
\begin{itemize}
\item[(i)]
Assume Assumption \ref{assump4.1}-\ref{assump4.4} hold. We then have
\[
V_M^{-1/2}\cdot \sqrt{n}(\hat{\tau}_M - B_M - \tau) \text{ converges in distribution to } \cN(0, 1).
\]
\item[(ii)] Under additional smoothness conditions on $\mu_w$'s \citep[Theorem 2]{abadie2011bias}, there exists a bias-correction statistic $\hat B_M$ such that $\sqrt{n}(\hat B_M-B_M)$ converges in probability to 0 and
\[
V_M^{-1/2}\cdot \sqrt{n}(\hat{\tau}_M^{\rm bc} - \tau) \text{ converges in distribution to } \cN(0, 1),
\]
with $\hat{\tau}_M^{\rm bc}:=\hat\tau_M-\hat B_M$.
\end{itemize}
\end{theorem}
It is evident that their theory focuses on the normalized $\hat\tau_M$, whose limiting variance relies on the asymptotic behavior of ${\mathrm E}[V^E]$. In \cite{abadie2006large}, the limit of ${\mathrm E}[V^E]$ is derived only in the special case of $d = 1$; see, for example, Theorem 5 therein.
Using our results on high-order Voronoi cells to be detailed in Section \ref{voronoi} ahead, we derive, for the first time, the limiting variance of $\hat\tau_M$ for a general dimension $d\geq 1$. This result is formalized in the following theorem, based on an additional continuity assumption on the density functions.
\begin{assumption}\label{assump4.5} For $w=0,1$, the Lebesgue density function of $X \,|\, W=w$ is continuous.
\end{assumption}
\begin{theorem}[Main theorem] \label{thm4.2} Assume Assumptions \ref{assump4.1}-\ref{assump4.5} hold.
\begin{itemize}
\item[(i)] We have
\[
\sqrt{n}(\hat{\tau}_M - B_M - \tau) \text{ converges in distribution to } \cN(0, \sigma^2_{M,d}),
\]
where $\sigma_{M,d}^2$ takes the form
\begin{align*}
&\sigma^2_{M,d}:=\lim_{n\to\infty} {\mathrm E}[V_{M}] \\
&=V^{\tau(X)} + \frac{1}{M^2}{\mathrm E}\Big[\sigma_1^2(X)\Big\{\frac{\alpha(M, d)}{e(X)} + \Big(\alpha(M, d) - M^2 - M\Big)e(X) + \Big(2M^2 + M - 2\alpha(M, d)\Big)\Big\}\Big] \\
&+ \frac{1}{M^2}{\mathrm E}\Big[\sigma_0^2(X)\Big\{\frac{\alpha(M, d)}{1 - e(X)} + \Big(\alpha(M, d) - M^2 - M)(1 - e(X)\Big) + \Big(2M^2 + M - 2\alpha(M, d)\Big)\Big\}\Big],
\end{align*}
with $\alpha(M,d)$, defined in Equation \eqref{eq:alphaMd} ahead, being a {\it distribution-free} constant only depending on $M$ and $d$.
\item[(ii)] Assume further the same conditions as in \citet[Theorem 2]{abadie2011bias}, the same bias-corrected estimator $\hat\tau_{M}^{bc}$ satisfies
\[
\sqrt{n}(\hat{\tau}_M^{\rm bc} - \tau) \text{ converges in distribution to } \cN(0, \sigma^2_{M,d}).
\]
\end{itemize}
\end{theorem}
\section{Measures of high-order Voronoi cells} \label{voronoi}
The key in the proof of Theorem \ref{thm4.2} is to quantify the volume of an NN-related stochastic object, which itself deserves a separate section to discuss. To this end, similar to the presentation of \citet[Section 2]{lin2023estimation}, let $\nu_0$ and $\nu_1$ be two laws on ${\mathbb R}^d$ with Lebesgue densities $f_0$ and $f_1$, respectively. Let $X_1, \dots, X_{n}$ be $n$ independent draws from $\nu_0$.
\begin{definition}[$M$-th NN map] Let $\cX_M(\cdot): {\mathbb R}^d \to \{X_i\}_{i = 1}^{n}$ be the map that returns the input $z$'s $M$-th NN in $\{X_i\}_{i = 1}^{n}$, that is, the value $x \in \{X_i\}_{i = 1}^{n}$ such that
\[
\sum_{i = 1}^{n}{\mathds 1}\Big\{\Big\| X_i - z\Big\| \leq \Big\| x - z\Big\| \Big\} = M.
\]
\end{definition}
\begin{definition}[Catchment area] \label{def3.1} Let $\cA_M(\cdot): \bR^d \to \cB(\bR^d)$ be the map from $\bR^d$ to the class of all Borel sets in $(\bR^d,\|\cdot\|)$ such that
\[
\cA_M(x) = \cA_M\Big(x; \{X_i\}_{i = 1}^{n}\Big) := \Big\{z \in \bR^d: \Big\| z - x\Big\| \leq \Big\| \cX_M(x) - x\Big\| \Big\}.
\]
\end{definition}\label{def3.2}
As noted in \citet[Remark 2.1]{lin2023estimation}, the sets $\{\cA_1(X_i)\}_{i\in[n]}$ are almost surely disjoint, partition ${\mathbb R}^d$ into $n$ polygons, and correspond exactly to the Voronoi cells introduced by \cite{voronoi1908nouvelles}. Accordingly, in this paper, we refer to the general sets $\{\cA_M(X_i)\}_{i\in[n]}$, for some $M\geq 1$, as {\it high-order Voronoi cells}. However, it should be noted that for $M>1$, $\{\cA_M(X_i)\}_{i\in[n]}$ are generally not disjoint. Additionally, they differ from the $K$-th order Voronoi cell/tessellation commonly defined in stochastic geometry; cf. \citet[Chapter 3.2]{boots1999spatial}.
Let $B(x,r)$ be the closed ball in $({\mathbb R},\|\cdot\|)$ of center $x\in{\mathbb R}^d$ and radius $r>0$. Following \cite{devroye2017measure}, introduce $D$ to be a random vector uniformly distributed in $B(0,1)$. Define $\bar{1} = (1, 0, \dots, 0) \in \bR^d$ and let $\bar{B} = B(\bar{1}, 1) \cup B(D, \Vert D \Vert)$. Define the random variable $T$ as
\[
T = \frac{\lambda(\bar{B})}{\lambda(B(0,1))},
\]
with $\lambda$ representing the Lebesgue measure on $\bR^d$. Lastly, introduce
\[
\alpha(d) := {\mathrm E}\Big[\frac{2}{T^2}\Big].
\]
The following is then a generalization of \citet[Theorem 1]{devroye2017measure}, which focuses on the case of $\nu_1(\cA_1(x))$ with $\nu_0=\nu_1$, to the measures of high-order Voronoi cells.
\begin{assumption}\label{assump3.1} Assume $\nu_1$ is absolutely continuous with regard to $\nu_0$ so that the Lebesgue densities $f_0$ and $f_1$ share a common support $\cX$ that is compact. Furthermore, assume $f_0$ and $f_1$ to be continuous in $\cX$.
\end{assumption}
\begin{theorem} \label{thm3.1}
\begin{itemize}
\item[(i)] Under Assumption \ref{assump3.1}, for $\nu_0$-almost all $x$ we have
\begin{align}
\lim_{n\to\infty} n{\mathrm E}\Big[\nu_1(\cA_M(x)) \mid X_1 = x\Big] &= M\cdot \frac{f_1(x)}{f_0(x)}, \label{eq:thm3.1-1}\\
\lim_{n\to\infty}n^2{\mathrm E}\Big[\nu_1(\cA_M(x))^2 \mid X_1 = x\Big] &= \alpha(M,d)\cdot \left(\frac{f_1(x)}{f_0(x)}\right)^2, \label{eq:thm3.1-2}
\end{align}
where we define
\begin{align}\label{eq:alphaMd}
\alpha(M, d) := \alpha(d)\sum_{i + k \leq M - 1, j +k \leq M - 1}\frac{c_{ijk}(d)(i + j + k + 1)!}{i!j!k!},
\end{align}
with $c_{ijk}(d)$ introduced in Lemma \ref{lem5.2} ahead being distribution-free constant depending only on $i, j, k,$ and $d$.
\item[(ii)] For $d = 1$, we have $\alpha(M, 1) = M(2M+1)/2$. For $M=1$, we have $\alpha(1,d)=\alpha(d)$.
\end{itemize}
\end{theorem}
In the special case where $\nu_0=\nu_1$, Theorem \ref{thm3.1} demonstrates that both $ n{\mathrm E}[\nu_0(\cA_M(x)) \mid X_1 = x]$ and $n^2{\mathrm E}[\nu_0(\cA_M(x))^2 \mid X_1 = x]$ have distribution-free limits that only depend on the NN number, $M$, and the dimension, $d$. While the closed form of $\alpha(M,d)$ is generally unavailable unless either $d$ or $M$ is 1, these values can be computed numerically. The results are presented in Tables \ref{tab:1} and \ref{tab:2}.
\begin{table}[H]\label{tab:1}
\caption{Values of $\alpha(d)$}
\centering
\begin{tabular}{cccccccccccc}
\hline
& d= 1 & $d = 2$ & $d = 3$ &$d = 4$ & $d = 5$ & $d = 6$ & $d = 7$ & $d = 8$ & $d = 9$ & $d = 10$ &\\
\hline
$\alpha(d)$& 1.50& 1.28 & 1.18 & 1.12& 1.08 & 1.06& 1.04 & 1.03 & 1.02 & 1.02 &\\
\hline
\end{tabular}
\label{tab:1}
\end{table}
\begin{table}[H]\
\caption{Values of $\alpha(M,d)$}
\centering
\begin{tabular}{ccccccccccc}
\hline
$\alpha(M, d) $& $d = 2$ & $d = 3$ &$d = 4$ & $d = 5$ & $d = 6$ & $d = 7$ & $d = 8$ & $d = 9$ & $d = 10$ &\\
\hline
$M = 2$& 4.57 & 4.37 & 4.26 & 4.18 & 4.13 & 4.09 & 4.07 & 4.05 & 4.07 & \\
\hline
$M = 3$& 9.86 & 9.57 & 9.40 & 9.28 & 9.21 & 9.14 & 9.10 & 9.08 & 9.17&\\
\hline
$M = 4$& 17.15 & 16.77 & 16.54 & 16.38 & 16.28 & 16.19 & 16.15 & 16.10 & 16.31 & \\
\hline
$M = 5$& 26.45 & 25.97 & 25.67 & 25.48 & 25.36 & 25.23 & 25.18 & 25.13 & 25.49&\\
\hline
$M = 6$& 37.74 & 37.16 & 36.83 & 36.58 & 36.45 & 36.27 & 36.22 & 36.16 & 36.71 &\\
\hline
$M = 7$& 51.03 & 50.34 & 49.98 & 49.68 & 49.53 & 49.32 & 49.25 & 49.19 & 49.97 &\\
\hline
$M = 8$& 66.32 & 65.54 & 65.12 & 64.79 & 64.62 & 64.36 & 64.29 & 64.22 & 65.27&\\
\hline
$M = 9$& 83.60 & 82.73 & 82.27 & 81.89 & 81.71 & 81.40 & 81.33 & 81.25 & 82.62 &\\
\hline
$M = 10$& 102.89 & 101.92 & 101.43 & 100.99 & 100.81 & 100.43 & 100.37 & 100.28 & 102.00 &\\
\hline
\end{tabular}
\label{tab:2}
\end{table}
\section{Proofs of the main results}
\subsection{Proof of Theorem \ref{thm3.1}}
Let $Z_1, Z_2$ be two independent copies of $Z \sim \nu_1$. Define
\begin{align*}
&V_1 = \nu_0(B(Z_1, \Vert Z_1 - x \Vert)),~~ V_2 = \nu_0(B(Z_2,\Vert Z_2 - x \Vert)),\\
~~ {\rm and}~~ &V = \nu_0\big(B(Z_1, \Vert Z_1 - x \Vert) \cup B(Z_2, \Vert Z_2 - x \Vert)\big).
\end{align*}
We first introduce the following lemma.
\begin{lemma}\label{lem5.1}
Denote the Lebesgue density functions of $V_1, V$ by $f_{V_1}, f_V$. We then have
\[
\begin{aligned}
\lim_{t \to 0+} f_{V_1}(t) &= \frac{f_1(x)}{f_0(x)} ~~~{\rm and}~~~\lim_{t \to 0+} \frac{f_V(t)}{t} &= \alpha(d)\Big\{\frac{f_1(x)}{f_0(x)}\Big\}^2.
\end{aligned}
\]
\end{lemma}
\noindent\textbf{Step 1.} We first prove \eqref{eq:thm3.1-1}. Observe that
\[
\begin{aligned}
&\quad {\mathrm E}\Big[\nu_1(\cA_M(x)) \mid X_1 = x\Big]\\
&= {\mathrm P}\Big(Z_1 \in \cA_M(x) \mid X_1 = x\Big)\\
&= {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert \cX_M(Z_1) - Z_1 \Vert \mid X_1 = x\Big) \\
& = {\mathrm P}\Big(\text{at most } M - 1 \text{ indices } i \in \{2, \dots, n\} \text{ satisfy } \Vert x - Z_1\Vert > \Vert X_i - Z_1\Vert \mid X_1 = x\Big)\\
& = \sum_{0 \leq k \leq M - 1}{\mathrm P}\Big(\text{there exist exactly } k \text{ indices } i \in \{2, \dots n\} \text{ such that } \Vert x - Z_1\Vert > \Vert X_i - Z_1\Vert \mid X_1 = x\Big)\\
& = \sum_{0 \leq k \leq M - 1} \binom{n - 1}{k} {\mathrm P}\Big(\Vert x - Z_1\Vert > \Vert X_2 - Z_1 \Vert\Big)^k \cdot {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert X_2 - Z_1 \Vert\Big)^{n - 1 - k}\\
& = \sum_{0 \leq k \leq M - 1} \binom{n - 1}{k} {\mathrm P}\Big(X_2 \in B_{Z_1, \Vert Z_1 - x\Vert}\Big)^k {\mathrm P}\Big(X_2 \notin B_{Z_1, \Vert Z_1 - x\Vert}\Big)^{n - k - 1}\\
& = \sum_{0 \leq k \leq M - 1} {\mathrm E}\Big[\binom{n - 1}{k} V_1^k(1 - V_1)^{n - 1 - k}\Big].
\end{aligned}
\]
For each $0 \leq k \leq M - 1$ and all $0 < \delta < 1$, we have
\[
n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}{\mathds 1}\{V_1 \geq \delta\}\Big] \leq n\binom{n - 1}{k}(1 - \delta)^{n - 1 - k}.
\]
Since $n\binom{n - 1}{k}$ is a polynomial of $n$ and $(1 - \delta)^n$ decays to 0 exponentially fast, for any fixed $0 < \delta < 1$, we then have
\[
n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}{\mathds 1}\{V_1 \geq \delta\}\Big] \to 0, \quad \text{ as } n \to \infty.
\]
By Lemma \ref{lem5.1}, for any $\varepsilon > 0$, there exists $\delta\in(0,1)$ such that for any $v_1 \in [0, \delta]$, the density $f_{V_1}(v_1)$ satisfies
\[
(1 - \varepsilon)\frac{f_1(x)}{f_0(x)} \leq f_{V_1}(v_1) \leq (1 + \varepsilon)\frac{f_1(x)}{f_0(x)}.
\]
Thus
\begin{align*}
&\quad n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}{\mathds 1}\{V_1 < \delta\}\Big] \\
&= n\binom{n - 1}{k}\int_0^{\delta}v_1^k(1 - v_1)^{n - 1 - k}f_{V_1}(v_1) \mathrm{d} v_1 \\
& \leq (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \int_0^\delta v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \\
& = (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \left(\int_0^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 - \int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \right) \\
& = (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \frac{\Gamma(k + 1)\Gamma(n - k)}{\Gamma(n + 1)} - (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k}\int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \\
& = (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} - (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k}\int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 .
\end{align*}
Since
\[
n\binom{n - 1}{k}\int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \leq n\binom{n - 1}{k}(1 - \delta)^{n - 1 - k} \to 0, \quad \text{as } n \to \infty,
\]
we obtain
\begin{align}\label{eq:5.1}
\limsup_{n \to \infty} n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}\Big] \leq (1 + \varepsilon)\frac{f_1(x)}{f_0(x)}.
\end{align}
On the other hand, we can similarly derive the lower bound since
\begin{align*}
&\quad n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}{\mathds 1}\{V_1 < \delta\}\Big] \\
& \geq (1 - \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \int_0^\delta v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \\
& = (1 - \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \left(\int_0^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 - \int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \right) \\
& = (1 - \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k} \frac{\Gamma(k + 1)\Gamma(n - k)}{\Gamma(n + 1)} - (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k}\int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 \\
& = (1 - \varepsilon)\frac{f_1(x)}{f_0(x)} - (1 + \varepsilon)\frac{f_1(x)}{f_0(x)} n\binom{n - 1}{k}\int_\delta^1 v_1^k(1 - v_1)^{n - 1- k} \mathrm{d} v_1 .
\end{align*}
Thus
\begin{align}\label{eq:5.2}
\liminf_{n \to \infty} n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}\Big] \geq (1 - \varepsilon)\frac{f_1(x)}{f_0(x)}.
\end{align}
Since $\varepsilon$ is taken arbitrarily, combining \eqref{eq:5.1} and \eqref{eq:5.2}, we obtain
\[
\lim_{n \to \infty} n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}\Big] = \frac{f_1(x)}{f_0(x)}.
\]
Therefore we have
\[
\lim_{n \to \infty}n{\mathrm E}\Big[\nu_1(\cA_M(x))\mid X_1 = x\Big] = \lim_{n \to \infty} \sum_{0 \leq k \leq M-1}n{\mathrm E}\Big[\binom{n - 1}{k}V_1^k(1 - V_1)^{n - 1 - k}\Big] = M\frac{f_1(x)}{f_0(x)},
\]
which completes the proof of \eqref{eq:thm3.1-1}.\\
\noindent\textbf{Step 2.} We next prove \eqref{eq:thm3.1-2}. Similar to the previous argument, we observe
\begin{align*}
&\quad {\mathrm E}\Big[\nu_1(\cA_M(x))^2 \mid X_1 = x\Big]\\
&= {\mathrm P}\Big(Z_1 \in \cA_M(x), Z_2 \in \cA_M(x) \mid X_1 = x\Big)\\
&= {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert \cX_M(Z_1) - Z_1 \Vert, \Vert x - Z_2\Vert \leq \Vert \cX_M(Z_2) - Z_2 \Vert \mid X_1 = x\Big) .
\end{align*}
Define the following four random sets
\begin{align*}
& A = \{l \in \{2, \dots, n\}: \Vert x - Z_1\Vert \leq \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert \leq \Vert X_l - Z_2\Vert \}, \\
& B = \{l \in \{2, \dots, n\}: \Vert x - Z_1\Vert \leq \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert > \Vert X_l - Z_2\Vert \}, \\
& C = \{l \in \{2, \dots, n\}: \Vert x - Z_1\Vert > \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert \leq \Vert X_l - Z_2\Vert \}, \\
& D = \{l \in \{2, \dots, n\}: \Vert x - Z_1\Vert > \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert > \Vert X_l - Z_2\Vert \}.
\end{align*}
Given $Z_1, Z_2$, let $p_A, p_B, p_C, p_D$ be the probabilities of a given index $l$ belonging to $A, B, C, D$, respectively. It then holds true that
\begin{align*}
p_A &:= {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert \leq \Vert X_l - Z_2\Vert \mid Z_1, Z_2\Big) \\
& = \nu_0 \Big(B(Z_1, \Vert Z_1 - x\Vert)^c \cap B(Z_2, \Vert Z_2 - x\Vert)^c\Big)\\
& = \nu_0\Big((B(Z_1, \Vert Z_1 - x\Vert) \cup B(Z_2, \Vert Z_2 - x\Vert))^c\Big)\\
& = 1 - V,\\
p_B &:= {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert > \Vert X_l - Z_2\Vert \mid Z_1, Z_2\Big) \\
& = \nu_0 \Big(B(Z_1, \Vert Z_1 - x\Vert)^c \cap B(Z_2, \Vert Z_2 - x\Vert)\Big) \\
& = V - V_1,\\
p_C &:= {\mathrm P}\Big(\Vert x - Z_1\Vert > \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert \leq \Vert X_l - Z_2\Vert \mid Z_1, Z_2\Big) \\
& = \nu_0 \Big(B(Z_1, \Vert Z_1 - x\Vert) \cap B(Z_2, \Vert Z_2 - x\Vert)^c\Big) \\
& = V - V_2,\\
p_D & := {\mathrm P}\Big(\Vert x - Z_1\Vert > \Vert X_l - Z_1\Vert, \Vert x - Z_2\Vert > \Vert X_l - Z_2\Vert \mid Z_1, Z_2\Big) \\
& = \nu_0 \Big(B(Z_1, \Vert Z_1 - x\Vert)^c \cap B(Z_2, \Vert Z_2 - x\Vert)^c\Big)\\
& = V_1 + V_2 - V.
\end{align*}
Observe that $A, B, C, D$ are pairwisely disjoint, $|A| + |B| + |C| + |D| = n - 1$, and we have
\begin{align*}
B\cup D &= \Big\{l \in \{2, \dots, n\}: \Vert x - Z_1\Vert > \Vert X_l - Z_1\Vert \Big\},\\
C \cup D &= \Big\{l \in \{2, \dots, n\}: \Vert x - Z_2\Vert > \Vert X_l - Z_2\Vert \Big\}.
\end{align*}
We then have
\[
\begin{aligned}
& \quad {\mathrm P}\Big(\Vert x - Z_1\Vert \leq \Vert \cX_M(Z_1) - Z_1 \Vert, \Vert x - Z_2\Vert \leq \Vert \cX_M(Z_2) - Z_2 \Vert \mid X_1 = x\Big) \\
& = {\mathrm P}\Big(|B \cup D| \leq M - 1, |C \cup D| \leq M - 1 \mid X_1 = x\Big) \\
& = {\mathrm P}\Big(|B| + |D| \leq M - 1, |C| + |D| \leq M - 1\Big) \\
& = \sum_{i + k , j + k \leq M - 1}{\mathrm E}\Big[{\mathrm P}(|D| = k, |B| = i, |C| = j \mid Z_1, Z_2)\Big] \\
& = \sum_{i + k , j + k \leq M - 1}\binom{n - 1}{k}\binom{n - 1 - k}{i}\binom{n - 1 - k - i}{j}{\mathrm E}\Big[p_A^{n - 1 - i - j - k}p_B^ip_C^jp_D^k\Big] \\
& = \sum_{i + k , j + k \leq M - 1} \frac{(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}\Big[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k}\Big].
\end{aligned}
\]
Define
\[
\begin{aligned}
&P_{ijk}(v, v_1, v_2) = \Big(1 - \frac{v_1}{v}\Big)^i\Big(1 - \frac{v_2}{v}\Big)^j\Big(\frac{v_1 + v_2}{v} - 1\Big)^k,\\
&Q_{ijk}(v) = {\mathrm E}\Big[\Big(1 - \frac{V_1}{V}\Big)^i\Big(1 - \frac{V_2}{V}\Big)^j\Big(\frac{V_1 + V_2}{V} - 1\Big)^k\mid V = v\Big] = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid V = v\Big] .
\end{aligned}
\]
We then have
\begin{align*}
& \quad {\mathrm E}\Big[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k}\Big] \\
& = {\mathrm E}\Big[{\mathrm E}[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k \mid V](1 - V)^{n - 1 - i - j - k}\Big] \\
& = {\mathrm E}\Big[Q_{ijk}(V)V^{i + j + k}(1 - V)^{n - 1 - i - j - k}\Big].
\end{align*}
\begin{lemma}[Distribution-free limits] \label{lem5.2} It holds true that
\[
\lim_{v \to 0+}Q_{ijk}(v) = c_{ijk}(d),
\]
where $c_{ijk}(d)$ is a positive constant depending only on the indices $i, j, k$ and dimension $d$. Specifically, when the dimension $d = 1 $, this constant can be explicitly calculated as
\[
c_{ijk}(1) = \begin{cases}
1, & \mbox{ if } i = j = k = 0,\\
\frac{1}{3}\frac{i!j!}{(i + j + 1)!}, & \mbox{ if } i, j \neq 0, k = 0,\\
\frac{1}{3}\frac{i!k!}{(i + k + 1)!}, & \mbox{ if } i, k \neq 0, j = 0,\\
\frac{1}{3}\frac{j!k!}{(j + k + 1)!}, & \mbox{ if } j, k \neq 0, i = 0,\\
\frac{2}{3}\frac{1}{k + 1}, & \mbox{ if } k \neq 0, i = j = 0,\\
\frac{2}{3}\frac{1}{j + 1}, & \mbox{ if } j \neq 0, i = k = 0,\\
\frac{2}{3}\frac{1}{i + 1}, & \mbox{ if } i \neq 0, j = k = 0,\\
0, & \mbox{ if } i, j, k \neq 0.
\end{cases}
\]
\end{lemma}
Back to the proof, Lemmas \ref{lem5.1} and \ref{lem5.2} combined imply that, for any $\varepsilon > 0$, there exists some $\delta > 0$ such that for any $0 < v < \delta$, we have
\[
(1 - \varepsilon)\alpha(d)\Big\{\frac{f_1(x)}{f_0(x)}\Big\}^2 \leq f_V(v) \leq (1 + \varepsilon)\alpha(d)\Big\{\frac{f_1(x)}{f_0(x)}\Big\}^2,
\]
\[
(1 - \varepsilon)c_{ijk}(d) \leq Q_{ijk}(v) \leq (1 + \varepsilon)c_{ijk}(d).
\]
Similar to the argument in {\bf Step 1}, we now have
\begin{align*}
& \quad n^2\frac{(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}\Big[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k} {\mathds 1} \{V \leq \delta\}\Big] \\
& = n^2\frac{(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}\Big[Q_{ijk}(V)V^{i + j + k}(1 - V)^{n - 1 - i - j - k} {\mathds 1} \{V \leq \delta \}\Big]\\
& = n^2\frac{(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} \int_0^\delta Q_{ijk}(v)v^{i + j + k}(1 - v)^{n - 1 - i - j - k} \mathrm{d} v \\
& \leq (1 + \varepsilon)^2 c_{ijk}(d) \alpha(d) \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} \int_0^\delta v^{i + j + k + 1}(1 - v)^{n - 1 - i - j - k} \mathrm{d} v \\
& = (1 + \varepsilon)^2 c_{ijk}(d) \alpha(d) \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} (\int_0^1 - \int_\delta^1) v^{i + j + k + 1}(1 - v)^{n - 1 - i - j - k} \mathrm{d} v \\
& = (1 + \varepsilon)^2 c_{ijk}(d) \alpha(d) \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} \frac{\Gamma(i + j + k + 2)\Gamma(n - i - j - k)}{\Gamma(n + 2)} \\
& \quad - (1 + \varepsilon)^2 c_{ijk}(d) \alpha(d) \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} \int_\delta^1 v^{i + j + k + 1}(1 - v)^{n - 1 - i - j - k} \mathrm{d} v\\
& = (1 + \varepsilon)^2 \alpha(d) \frac{n}{n + 1}\frac{c_{ijk}(d)(i + j + k + 1)!}{i!j!k!} \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\\
& \quad - (1 + \varepsilon)^2 c_{ijk}(d) \alpha(d) \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} \int_\delta^1 v^{i + j + k + 1}(1 - v)^{n - 1 - i - j - k} \mathrm{d} v.
\end{align*}
It is ready to check that the second term in the last display converges to $0$ as $n \to \infty$. We thus have
\begin{align*}
&\limsup_{n \to \infty}\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k}]\\
\leq& (1 + \varepsilon)^2\alpha(d)\frac{c_{ijk}(d)(i + j + k + 1)!}{i!j!k!}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\end{align*}
Similarly, we can obtain
\[
\begin{aligned}
&\liminf_{n \to \infty}\frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k}]\\
\geq& (1 - \varepsilon)^2\alpha(d)\frac{c_{ijk}(d)(i + j + k + 1)!}{i!j!k!}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\end{aligned}
\]
Combining the above two bounds, it then holds true that
\begin{align*}
& \quad \lim_{n \to \infty} n^2{\mathrm E}[\nu_1(\cA_M(x))^2 \mid X_1 = x] \\
& = \lim_{n \to \infty}\sum_{i + k , j + k \leq M - 1} \frac{n^2(n - 1)!}{i!j!k!(n - 1 - i - j - k)!} {\mathrm E}\Big[(V - V_1)^i(V - V_2)^j(V_1 + V_2 - V)^k(1 - V)^{n - 1 - i - j - k}\Big] \\
& = \sum_{i +k, j + k \leq M- 1}\alpha(d)\frac{c_{ijk}(d)(i + j + k + 1)!}{i!j!k!}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2 \\
& = \alpha(M, d)\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\end{align*}
~\\
\textbf{Step 3.} We close the proof by calculating the explicit values of $\alpha(M,d)$ when either $M$ or $d$ is 1. The results for $M=1$ is \citet[Theorem 1]{devroye2017measure}. It remains to prove
\[
\alpha(M,1) = M(2M + 1)/2.
\]
The previous arguments yield
\[
\alpha(M, 1) = \alpha(1)\sum_{i + k \leq M - 1, j + k \leq M - 1}\frac{c_{ijk}(1)(i + j + k + 1)!}{i!j!k!}.
\]
By the form of $c_{ijk}(1)$ in Lemma \ref{lem5.2}, we have
\[
\frac{c_{ijk}(1)(i + j + k + 1)!}{i!j!k!}= \begin{cases}
1, & \mbox{ if } i = j = k = 0,\\
\frac{1}{3}, & \mbox{ if } \text{only one of }i, j, k \text{ is equal to }0,\\
\frac{2}{3}, & \mbox{ if } \text{only one of }i, j, k \text{ is not equal to } 0 ,\\
0, & \mbox{ if } i,j,k \neq 0.
\end{cases}
\]
It can be shown that, under the restriction $i + k \leq M - 1, j + k \leq M - 1$, there are $2M^2 - 5M + 3$ combinations of $(i, j, k)$ satisfying only one of $i, j, k$ is equal to 0; $3M - 3$ combinations satisfying only one of $i, j, k$ is not equal to 0; and one combination satisfying $(i, j, k) = (0, 0, 0)$. Adding together, we have
\[
\sum_{i + k \leq M - 1, j + k \leq M - 1}\frac{c_{ijk}(1)(i + j + k + 1)!}{i!j!k!} = \frac{1}{3}(2M^2 - 5M + 3) + \frac{2}{3}(3M - 3) + 1 = \frac{1}{3}(2M^2 + M).
\]
Since $\alpha(1) = 3/2$, we obtain
\[
\alpha(M,1) = \frac{3}{2}\cdot \frac{1}{3}(2M^2 + M) = \frac{M(2M + 1)}{2},
\]
and thus completes the proof.
\subsection{Proof of Theorem \ref{thm4.2}}
Think of $\nu_0, \nu_1$ as the conditional distributions of $X \mid W = 0$, $X \mid W = 1$. Let $f_0(\cdot), f_1(\cdot)$ be the corresponding Lebesgue density functions.
By Assumption \ref{assump4.5}, we have both $f_0(\cdot)$ and $f_1(\cdot)$ to be continuous in $\cX$.
~\\
{\bf Step 1.} We first prove the following convergence result for ${\mathrm E}[V^E]$:
\begin{align*}
&\lim_{n\to\infty}{\mathrm E}[V^E] \\
=& \frac{1}{M^2}{\mathrm E}\Big[\sigma_1^2(X)\Big(\frac{\alpha(M, d)}{e(X)} + (\alpha(M, d) - M^2 - M)e(X) + (2M^2 + M - 2\alpha(M, d))\Big)\Big] \\
&+ \frac{1}{M^2}{\mathrm E}\Big[\sigma_0^2(X)\Big(\frac{\alpha(M, d)}{1 - e(X)} + (\alpha(M, d) - M^2 - M)(1 - e(X)) + (2M^2 + M - 2\alpha(M, d))\Big)\Big].
\end{align*}
Recall that
\[
V^E = \frac{1}{n}\sum_{i = 1}^n \Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_{W_i}^2(X_i).
\]
Let $p = {\mathrm P}(W_i = 1)$. Since $(X_i, W_i)_{i = 1}^n$ are independent and identically distributed, we have
\begin{align*}
{\mathrm E}[V^E] =& {\mathrm E}\Big[\Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_{W_i}^2(X_i)\Big] \\
=& {\mathrm E}\Big[\Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_1^2(X_i) \mid W_i = 1\Big]p + {\mathrm E}\Big[\Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_0^2(X_i) \mid W_i = 0\Big](1 - p).
\end{align*}
Fix the treatment indicators $\mathbf{W}=(W_1,\ldots,W_n)$ and covariates in the control group $\{X_j\}_{j:W_j = 0}$.
It is then immediate that
\[
K_M(i) \mid \mathbf{W}, \{X_j\}_{j:W_j = 0}, W_i = 0 \sim \text{Binomial}\Big(n_1, \nu_1(\cA_M(X_i))\Big).
\]
We accordingly have
\begin{align*}
&{\mathrm E}\Big[K_M(i) \mid \mathbf{W}, \{X_j\}_{j:W_j = 0}, W_i = 0\Big] = n_1\nu_1(\cA_M(X_i)),\\
&{\mathrm E}\Big[K_M(i)^2 \mid \mathbf{W}, \{X_j\}_{j:W_j = 0}, W_i = 0\Big] = n_1\nu_1(\cA_M(X_i)) + n_1(n_1 - 1)\nu_1(\cA_M(X_i))^2.
\end{align*}
We first introduce the following lemma.
\begin{lemma} \label{lem5.3} We have both
${\mathrm E}[n_0 \nu_1(\cA_M(x)) \mid W_i = 0]$ and ${\mathrm E}[n_0^2\nu_1(\cA_M(x))^2 \mid W_i = 0]$ are uniformly bounded for all $n$.
\end{lemma}
Back to ${\mathrm E}[V^E]$, we have
\begin{align*}
& {\mathrm E}\Big[\left(1 + \frac{K_M(i)}{M}\right)^2\sigma_0^2(X_i) \mid W_i = 0\Big] \\
=& {\mathrm E}\Big[\left(1 + \frac{2}{M}n_1\nu_1(\cA_M(X_i)) + \frac{1}{M^2}(n_1\nu_1(\cA_M(X_i)) + n_1(n_1 - 1)\nu_1(\cA_M(X_i))^2\right)\sigma_0^2(X_i)\mid W_i = 0\Big] \\
=& {\mathrm E}\Big[\left(1 + \frac{n_1}{n_0}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)n_0\nu_1(\cA_M(X_i)) + \frac{n_1(n_1 - 1)}{n_0^2}\frac{1}{M^2}n_0^2\nu_1(\cA_M(X_i))^2\right)\sigma_0^2(X_i)\mid W_i = 0\Big].
\end{align*}
By the law of large numbers, we have
\[
n_1/n_0 \xrightarrow{a.s.} p/(1 - p).
\]
By Lemma \ref{lem5.3} and Assumption \ref{assump4.4},
\[
{\mathrm E}\Big[n_0 \nu_1(\cA_M(x)) \mid W_i = 0\Big],~ {\mathrm E}\Big[n_0^2\nu_1(\cA_M(x))^2 \mid W_i = 0\Big],~ {\rm and}~ \sigma_w^2(x) \text{ are all uniformly bounded.}
\]
Adding together yields
\[
\begin{aligned}
& {\mathrm E}\Big[\Big(1 + \frac{n_1}{n_0}(\frac{2}{M}+ \frac{1}{M^2})n_0\nu_1(\cA_M(X_i)) + \frac{n_1(n_1 - 1)}{n_0^2}\frac{1}{M^2}n_0^2\nu_1(\cA_M(X_i))^2\Big)\sigma_0^2(X_i)\mid W_i = 0\Big] \\
& = {\mathrm E}\Big[\Big(1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)n_0\nu_1(\cA_M(X_i)) + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}n_0^2\nu_1(\cA_M(X_i))^2\Big)\sigma_0^2(X_i)\mid W_i = 0\Big] + o(1).
\end{aligned}
\]
Let
\[
F_n(x) = {\mathrm E}\Big[\Big(1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)n_0\nu_1(\cA_M(X_i)) + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}n_0^2\nu_1(\cA_M(X_i))^2\Big)\mid X_i = x, W_i = 0\Big].
\]
Theorem \ref{thm3.1} then implies, for almost all $x \in \cX$,
\[
F_n(x) \to F(x) = 1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)M\cdot\frac{f_1(x)}{f_0(x)} + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}\cdot\alpha(M,d)\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\]
Since $f_0, f_1$ are both supported on a compact set $\cX$ and are continuous, $F(\cdot)$ is uniformly bounded. Also, since $F_n(\cdot)$'s are continuous, and the pointwise convergence on compact set implies uniform convergence, thus by dominated convergence theorem we obtain
\begin{align*}
&{\mathrm E}\Big[\Big(1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)n_0\nu_1(\cA_M(X_i)) + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}n_0^2\nu_1(\cA_M(X_i))^2\Big)\sigma_0^2(X_i)\mid W_i = 0\Big] \\
=& {\mathrm E}\Big[F_n(X_i)\sigma_0^2(X_i) \mid W_i = 0\Big]\\
=& {\mathrm E}\Big[F(X_i) \sigma_0^2(X_i) \mid W_i = 0\Big] + o(1) \\
=& {\mathrm E}\Big[\Big(1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)M\frac{f_1(X_i)}{f_0(X_i)} + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}\alpha(M,d)\Big(\frac{f_1(X_i)}{f_0(X_i)}\Big)^2\Big)\sigma_0^2(X_i)\mid W_i = 0\Big] + o(1).
\end{align*}
Furthermore, we have
\[
{\mathrm E}[\sigma_0^2(X_i) \mid W_i = 0](1 - p) = {\mathrm E}\Big[\frac{1 - W_i}{1 - p}\sigma_0^2(X_i)\Big](1 - p) = {\mathrm E}[\sigma_0^2(X_i)(1 - W_i)] = {\mathrm E}[\sigma_0^2(X_i)(1 - e(X_i))].
\]
Note that
\[
\frac{f_1(X_i)}{f_0(X_i)} = \frac{e(X_i)/p}{(1 - e(X_i))/(1 - p)}.
\]
We accordingly have
\begin{align*}
&{\mathrm E}\Big[\frac{p}{1 - p}\Big(\frac{2}{M} + \frac{1}{M^2}\Big)M\frac{f_1(X_i)}{f_0(X_i)}\sigma_0^2(X_i) \mid W_i = 0\Big](1 - p)\\
= & {\mathrm E}\Big[\frac{p}{1 - p}\Big(\frac{2}{M} + \frac{1}{M^2}\Big)M\frac{f_1(X_i)}{f_0(X_i)}\sigma_0^2(X_i) (1 - W_i)\Big]\\
= & {\mathrm E}\Big[\frac{p}{1 - p}\Big(\frac{2}{M} + \frac{1}{M^2}\Big)M\frac{f_1(X_i)}{f_0(X_i)}\sigma_0^2(X_i) (1 - e(X_i))\Big]\\
= & {\mathrm E}\Big[\Big(\frac{2}{M} + \frac{1}{M^2}\Big)M\frac{e(X_i)}{1 - e(X_i)}\sigma_0^2(X_i) (1 - e(X_i))\Big] \\
= & {\mathrm E}\Big[(2 + \frac{1}{M})e(X_i)\sigma_0^2(X_i)\Big],
\end{align*}
and similarly,
\begin{align*}
&{\mathrm E}\Big[\frac{p^2}{(1 - p)^2}\frac{1}{M^2}\alpha(M,d)\Big(\frac{f_1(X_i)}{f_0(X_i)}\Big)^2 \sigma_0^2(X_i)\mid W_i = 0\Big](1 - p) \\
= & {\mathrm E}\Big[\frac{p^2}{(1 - p)^2}\frac{1}{M^2}\alpha(M,d)\Big(\frac{f_1(X_i)}{f_0(X_i)}\Big)^2 \sigma_0^2(X_i)(1 - W_i)\Big] \\
= & {\mathrm E}\Big[\frac{p^2}{(1 - p)^2}\frac{1}{M^2}\alpha(M,d)\Big(\frac{f_1(X_i)}{f_0(X_i)}\Big)^2 \sigma_0^2(X_i)(1 - e(X_i))\Big] \\
= & {\mathrm E}\Big[\frac{\alpha(M,d)}{M^2}\frac{e(X_i)^2}{(1 - e(X_i))^2}\sigma_0^2(X_i)(1 - e(X_i))\Big]\\
= & {\mathrm E}\Big[\frac{\alpha(M,d)}{M^2}\frac{e(X_i)^2}{1 - e(X_i)}\sigma_0^2(X_i)\Big].
\end{align*}
Combining the above two identities yields
\begin{align*}
& {\mathrm E}\Big[\Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_0^2(X_i) \mid W_i = 0\Big](1 - p)\\
= &{\mathrm E}\Big[\Big(1 + \frac{p}{1- p}\Big(\frac{2}{M}+ \frac{1}{M^2}\Big)M\frac{f_1(X_i)}{f_0(X_i)} + \frac{p^2}{(1 - p)^2}\frac{1}{M^2}\alpha(M,d)\Big(\frac{f_1(X_i)}{f_0(X_i)}\Big)^2\Big)\sigma_0^2(X_i)\mid W_i = 0\Big] + o(1) \\
= & \frac{1}{M^2}{\mathrm E}\Big[\sigma_0^2(X)\Big(\frac{\alpha(M, d)}{1 - e(X)} + (\alpha(M, d) - M^2 - M)(1 - e(X)) + (2M^2 + M - 2\alpha(M, d))\Big)\Big] + o(1).
\end{align*}
Similarly, we have
\begin{align*}
& {\mathrm E}\Big[\Big(1 + \frac{K_M(i)}{M}\Big)^2\sigma_1^2(X_i) \mid W_i = 1\Big]p\\
= & \frac{1}{M^2}{\mathrm E}\Big[\sigma_1^2(X)\Big(\frac{\alpha(M, d)}{e(X)} + (\alpha(M, d) - M^2 - M)e(X) + (2M^2 + M - 2\alpha(M, d))\Big)\Big] + o(1).
\end{align*}
Combine these two yields the convergence result for ${\mathrm E}[V^E]$.
~\\
{\bf Step 2.} In \citet[Page 40]{abadie2002simple}, it is proved that the variance of $V^E$ converges to 0, and thus
\[
V^E -{\mathrm E}[V^E]\xrightarrow{\enskip {\mathrm P} \enskip} 0.
\]
Employing Theorem \ref{thm4.1} and Slutsky's theorem then completes the proof.
\section{Proofs of the rest results}
\subsection{Proof of Lemma \ref{lem5.1}}
The first and second parts correspond to the proofs of Equation (2.2) and the fourth identity in Part (ii) of Theorem 2.1 in \cite{devroye2017measure}, respectively. The proofs then only take minor modifications with regard to the change of measures, and are accordingly omitted.
\subsection{Proof of Lemma \ref{lem5.2}}
\textbf{Step 1}: We first consider the case of dimension $d \geq 2$.
Denote the conditional density functions of $V \mid V_1, V_2$ and $V_1, V_2 \mid V$ by $f_{V \mid V_1, V_2}(\cdot \mid \cdot, \cdot)$ and $f_{V_1, V_2 \mid V_2}(\cdot ,\cdot \mid \cdot)$, respectively. Let
\[
D(v) = \Big\{(v_1, v_2) \in \bR^2: 0 \leq v_1, v_2 \leq v, v_1 + v_2 \geq v\Big\}
\]
be the support of $V_1, V_2 \mid V = v$. Thus we have
\begin{align*}
Q_{ijk}(v) &= {\mathrm E}\Big[P_{ijk}(v, V_1, V_2) \mid V = v\Big] \\
& = \int_{D(v)}P_{ijk}(v,v_1,v_2) f_{V_1, V_2\mid v}(v_1, v_2\mid v) \mathrm{d} v_1 \mathrm{d} v_2 \\
& = \int_{D(v)}P_{ijk}(v,v_1,v_2) \frac{f_{V \mid V_1, V_2}(v\mid v_1, v_2) f_{V_1, V_2}(v_1, v_2)}{f_V(v)} \mathrm{d} v_1 \mathrm{d} v_2 \\
& = \int_{D(1)}P_{ijk}(v, vv_1, vv_2)\frac{f_{V \mid V_1, V_2}(v\mid vv_1, vv_2) f_{V_1, V_2}(vv_1, vv_2)}{f_V(v)}v^2 \mathrm{d} v_1 \mathrm{d} v_2\\
& = \int_{D(1)}P_{ijk}(1, v_1, v_2)(vf_{V \mid V_1, V_2}(v\mid vv_1, vv_2))\frac{f_{V_1, V_2}(vv_1, vv_2)}{f_V(v)/v} \mathrm{d} v_1 \mathrm{d} v_2.
\end{align*}
By Lemma \ref{lem5.1}, for fixed $v_1, v_2$, as $v \to 0+$, we have
\[
f_{V_1, V_2}(vv_1, vv_2) = f_{V_1}(vv_1)f_{V_2}(vv_2) \to \Big(\frac{f_1(x)}{f_0(x)}\Big)^2
~~~{\rm and}~~~
\frac{f_V(v)}{v} \to \alpha(d)\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\]
It suffice to consider the convergence of $vf_{V \mid V_1, V_2}(v\mid vv_1, vv_2)$.
Introduce
\[
R_1 = \Vert Z_1 - x\Vert,~~ \theta_1 =\frac{Z_1 - x}{\Vert Z_1 - x\Vert},~~ R_2 = \Vert Z_2 - x\Vert,~~ \theta_2 = \frac{Z_2 - x}{\Vert Z_2 - x \Vert}.
\]
Note that the volume of $B(Z_1, \Vert Z_1 - x\Vert) \cup B(Z_2, \Vert Z_2 - x\Vert) $ is uniquely determined by $R_1, R_2$, and the directions $\theta_1, \theta_2$. We thus denote this volume by
\[
\lambda\Big(B(Z_1, \Vert Z_1 - x\Vert) \cup B(Z_2, \Vert Z_2 - x\Vert)\Big) =: S(R_1, R_2, \theta_1, \theta_2).
\]
Then for any $k > 0$, we have
\[
S(kR_1, kR_2, \theta_1, \theta_2) = k^dS(R_1, R_2, \theta_1, \theta_2).
\]
By the argument in Lemma \ref{lem5.1}, for any $\varepsilon > 0$, there exists $\delta > 0$ such that if $V < \delta $ holds, then
\[
(1 - \varepsilon)f_0(x) \leq \frac{\nu_0(B(Z_1, \Vert Z_1 - x\Vert) \cup B(Z_2, \Vert Z_2 - x\Vert))}{\lambda(B(Z_1, \Vert Z_1 - x\Vert) \cup B(Z_2, \Vert Z_2 - x\Vert))} = \frac{V}{S(R_1, R_2, \theta_1, \theta_2)} \leq (1 + \varepsilon)f_0(x).
\]
Similarly, letting $A(r) = c_d r^d$ denote the Lebesgue measure of a ball with radius $r$ in $\bR^d$, we have
\[
(1 - \varepsilon)f_0(x) \leq \frac{\nu_0(B(Z_1, \Vert Z_1 - x\Vert) )}{\lambda(B(Z_1, \Vert Z_1 - x\Vert) )} = \frac{V_1}{A(R_1)} \leq (1 + \varepsilon)f_0(x),
\]
\[
(1 - \varepsilon)f_0(x) \leq \frac{\nu_0(B(Z_2, \Vert Z_2 - x\Vert) )}{\lambda(B(Z_2, \Vert Z_2 - x\Vert) )} = \frac{V_2}{A(R_2)} \leq (1 + \varepsilon)f_0(x).
\]
For any fixed $v_1, v_2, t > 0$ and $ 0 < v < \delta$, consider the conditional probability ${\mathrm P}(V \leq vt \mid V_1 = vv_1, V_2 = vv_2)$. We have
\begin{align*}
&\quad {\mathrm P}(V \leq vt \mid V_1 = vv_1, V_2 = vv_2) \\
& \leq {\mathrm P}\Big( (1 - \varepsilon)f_0(x)S(R_1, R_2, \theta_1, \theta_2) \leq vt \mid V_1 = vv_1, V_2 = vv_2\Big) \\
& \leq {\mathrm P}\Big( (1 - \varepsilon)f_0(x) S\Big(\Big(\frac{V_1}{c_d(1 + \varepsilon)f_0(x)}\Big)^{1/d}, \Big(\frac{V_2}{c_d(1 + \varepsilon)f_0(x)}\Big)^{1/d}, \theta_1, \theta_2 \Big) \leq vt \mid V_1 = vv_1, V_2 = vv_2\Big) \\
& = {\mathrm P}\Big(\frac{1 - \varepsilon}{1 + \varepsilon}v c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \theta_1, \theta_2) \leq vt \mid V_1 = vv_1, V_2 = vv_2\Big) \\
& = {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \theta_1, \theta_2) \leq \frac{1 + \varepsilon}{1 - \varepsilon}t \mid V_1 = vv_1, V_2 = vv_2\Big).
\end{align*}
This shows
\begin{align}\label{eq:6.1}
{\mathrm P}(V \leq vt \mid V_1 = vv_1, V_2 = vv_2) \leq {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \theta_1, \theta_2) \leq \frac{1 + \varepsilon}{1 - \varepsilon}t \mid V_1 = vv_1, V_2 = vv_2\Big).
\end{align}
Similarly, we have
\begin{align}\label{eq:6.2}
{\mathrm P}(V \leq vt \mid V_1 = vv_1, V_2 = vv_2) \geq {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \theta_1, \theta_2) \leq \frac{1 - \varepsilon}{1 + \varepsilon}t \mid V_1 = vv_1, V_2 = vv_2\Big).
\end{align}
We then consider the limiting distribution of $\theta_1 \mid V_1 = vv_1$. Define $\bS^{d-1}$ to be the unit sphere in $({\mathbb R}^d,\|\cdot\|)$. For any measurable set $A \subset \bS^{d-1}$ and $0 < b < \delta$, we have
\begin{align*}
{\mathrm P}(\theta_1 \in A \mid V_1 \leq b) &= \frac{{\mathrm P}(\theta_1 \in A, V_1 \leq b)}{{\mathrm P}(V_1 \leq b)} \\
& \leq \frac{{\mathrm P}(\theta_1 \in A, R_1 \leq (\frac{b}{c_d(1 - \varepsilon)f_0(x)})^{1/d} )}{{\mathrm P}(R_1 \leq (\frac{b}{c_d(1 + \varepsilon)f_0(x)})^{1/d} )}\\
& = \frac{\int_{\{(z - x)/\Vert z - x \Vert \in A,\Vert z - x \Vert \leq (\frac{b}{c_d(1 - \varepsilon)f_0(x)})^{1/d}\}}f_1(z) \mathrm{d} z}{ \int_{\{\Vert z - x \Vert \leq (\frac{b}{c_d(1 + \varepsilon)f_0(x)})^{1/d}\}} f_1(z) \mathrm{d} z} \\
& \leq \frac{(1 + \varepsilon)f_1(x)}{(1 - \varepsilon)f_1(x)} \frac{\lambda(\{(z - x)/\Vert z - x \Vert \in A,\Vert z - x \Vert \leq (\frac{b}{c_d(1 - \varepsilon)f_0(x)})^{1/d}\})}{ \lambda \{\Vert z - x \Vert \leq (\frac{b}{c_d(1 + \varepsilon)f_0(x)})^{1/d}\}}\\
& = \frac{(1 + \varepsilon)f_1(x)}{(1 - \varepsilon)f_1(x)} \frac{\mu_{d}(A)b/c_d(1 - \varepsilon)f_0(x)}{b/c_d(1 + \varepsilon)f_0(x) }\\
& = \frac{(1 + \varepsilon)^2}{(1 - \varepsilon)^2}\mu_{d}(A).
\end{align*}
Here $\mu_d$ is the normalized Hausdorff measure on sphere $\bS^{d - 1}$ such that $\mu_d({\bS^{d - 1}}) = 1$.
Similarly, we have
\[
{\mathrm P}(\theta_1 \in A \mid V_1 \leq b) \geq \frac{(1 - \varepsilon)^2}{(1 + \varepsilon)^2}\mu_{d}(A).
\]
Therefore we obtain
\[
\lim_{b \to 0+}{\mathrm P}(\theta_1 \in A \mid V_1 \leq b) = \mu_d(A).
\]
By L'Hopital's rule, we have
\[
\lim_{b \to 0+}\frac{{\mathrm P}(\theta_1 \in A, V_1 \leq b)}{{\mathrm P}(V_1 \leq b)} = \lim_{b \to 0+}\frac{\frac{\partial}{\partial b}{\mathrm P}(\theta_1 \in A, V_1 \leq b)}{\frac{\mathrm{d}}{\mathrm{d} b}{\mathrm P}(V_1 \leq b)} = \lim_{b \to 0+}{\mathrm P}( \theta_1 \in A \mid V_1 = b) = \mu_d(A).
\]
Since the above holds for any measurable set $A \subset \bS^{d - 1}$, then as $v \to 0+$, we have
\begin{align}\label{eq:6.3}
\theta_1 \mid V_1 = vv_1 \xrightarrow{\enskip d \enskip} \mathrm{Unif}(\bS^{d-1}).
\end{align}
Let $\tilde{\theta}_1$ and $\tilde{\theta}_2$ be two independent copies of $\mathrm{Unif}(\bS^{d-1})$. Then combining with \eqref{eq:6.1} we have
\[
\begin{aligned}
& \quad \limsup_{v \to 0+}{\mathrm P}(V \leq vt \mid V_1 = vv_1, V_2 = vv_2) \\
& \leq \limsup_{v \to 0+}{\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \theta_1, \theta_2) \leq \frac{1 + \varepsilon}{1 - \varepsilon}t \mid V_1 = vv_1, V_2 = vv_2\Big) \\
& \leq {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \tilde{\theta}_1, \tilde{\theta}_2) \leq \frac{1 + \varepsilon}{1 - \varepsilon}t\Big).
\end{aligned}
\]
Similarly, combining with \eqref{eq:6.2} we have
\[
\liminf_{v \to 0+}{\mathrm P}\Big(V \leq vt \mid V_1 = vv_1, V_2 = vv_2\Big) \geq {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \tilde{\theta}_1, \tilde{\theta}_2 )\leq \frac{1 - \varepsilon}{1 + \varepsilon}t\Big).
\]
Since the above holds for arbitrary $\varepsilon > 0$, we then obtain
\[
\lim_{v \to 0+}{\mathrm P}\Big(\frac{V}{v}\leq t \mid V_1 = vv_1, V_2 = vv_2\Big) = {\mathrm P}\Big(c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \tilde{\theta}_1, \tilde{\theta}_2 )\leq t\Big),
\]
for all $t > 0$. Therefore as $v \to 0+$, the condition distribution converges:
\[
\frac{V}{v} \mid V_1 = vv_1, V_2 = vv_2 \xrightarrow{\enskip d \enskip} c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \tilde{\theta}_1, \tilde{\theta}_2 ).
\]
Denote the density of $c_d^{-1}S(v_1^{1/d}, v_2^{1/d}, \tilde{\theta}_1, \tilde{\theta}_2 )$ by $f_{S, v_1, v_2}(t)$. By linear transform of density functions, we have
\[
\lim_{v \to 0+}vf_{V \mid V_1, V_2}(v \mid vv_1, vv_2) = \lim_{v \to 0+}f_{\frac{V}{v} \mid V_1, V_2}(1 \mid vv_1, vv_2) = f_{S, v_1, v_2}(1).
\]
For any sequence ${v_n}$ converges to 0, define
\begin{align*}
&g_n(v_1, v_2) = P_{ijk}(1, v_1, v_2)(v_nf_{V \mid V_1, V_2}(v_n\mid v_nv_1, v_nv_2))\frac{f_{V_1, V_2}(v_nv_1, v_nv_2)}{f_V(v_n)/v_n}\\
{\rm and}~~~&g(v_1, v_2) = \frac{P_{ijk}(1, v_1, v_2) f_{S, v_1, v_2}(1)}{\alpha(d)}.
\end{align*}
Notice that $Q_{ijk}(v) \leq 1$ is uniformly bounded. Therefore
\[
Q_{ijk}(v_n) = \int_{D(1)}g_n \leq 1.
\]
Since $g_n(v_1, v_2) \to g(v_1, v_2)$, invoking Fatou's Lemma yields
\[
\int_{D(1)}g \leq \liminf_{n \to \infty}\int_{D(1)}g_n \leq 1.
\]
Thus $g$ is integrable on $D(1)$. Define $D(1, \delta)$ to be
\[
D(1, \delta) = \Big\{(v_1, v_2) \in \bR^2: \delta \leq v_1, v_2 \leq 1 - \delta, v_1 + v_2 \geq 1 + \delta\Big\};
\]
specifically, $D(1) = D(1, 0)$. Then for any $\varepsilon > 0$, there exists some $\delta > 0$ such that
\begin{align}\label{eq:6.4}
\int_{D(1, \delta)} g \geq \int_{D(1)}g - \varepsilon ~~~{\rm and}~~~
\int_{D(1, \delta)} g_n \geq \int_{D(1)}g_n - \varepsilon.
\end{align}
Notice $g(v_1, v_2) < \infty$ for every $(v_1, v_2 ) \in D(1, \delta)$ and $D(1, \delta)$ is a compact subset on $\bR^2$. By the continuity of $g(v_1, v_2)$, we have $g(v_1, v_2)$ is uniformly bounded on $D(1, \delta)$. By dominated convergence theorem on this compact set, we then have
\[
\lim_{n \to \infty} \int_{D(1, \delta)}g_n = \int_{D(1,\delta)}g.
\]
Combining \eqref{eq:6.4} we obtain
\[
\int_{D(1)}g \geq \lim_{n \to \infty}\int_{D(1)}g_n - 2\varepsilon.
\]
Thus we obtain
\[
\lim_{n \to \infty}\int_{D(1)}g_n = \int_{D(1)}g.
\]
This shows
\[
\begin{aligned}
\lim_{v \to 0+}Q_{ijk}(v) &= \lim_{v \to 0+}\int_{D(1)}P_{ijk}(1, v_1, v_2)(vf_{V \mid V_1, V_2}(v\mid vv_1, vv_2))\frac{f_{V_1, V_2}(vv_1, vv_2)}{(f_V(v)/v)} \mathrm{d} v_1 \mathrm{d} v_2 \\
&= \int_{D(1)}\frac{P_{ijk}(1, v_1, v_2) f_{S, v_1, v_2}(1)}{\alpha(d)} \mathrm{d} v_1 \mathrm{d} v_2 \\
&=: c_{ijk}(d).
\end{aligned}
\]
It can be seen that the constant $c_{ijk}(d)$ does not depend on the distribution of $\nu_1, \nu_0$.
~\\
\textbf{Step 2}: Next we consider the case of $d = 1$ and give the exact value of $c_{ijk}(1)$.
Following the same notation as in \textbf{Step 1}, define
\[
R_1 = \Vert Z_1 - x\Vert,~~ \theta_1 =\frac{Z_1 - x}{\Vert Z_1 - x\Vert},~~ R_2 = \Vert Z_2 - x\Vert,~~ \theta_2 = \frac{Z_2 - x}{\Vert Z_2 - x \Vert}.
\]
Since $d = 1$, then $Z_1, Z_2 \in \bR$ and $\theta_1, \theta_2 \in \{1, -1\}$. Therefore $V$ can be directly expressed by $V_1, V_2$ as follows
\[
V = \begin{cases}
V_1 + V_2, & \mbox{ if } \theta_1 = -\theta_2,\\
\max\{V_1, V_2\}, & \mbox{ if } \theta_1 = \theta_2.
\end{cases}
\]
Obviously we have
\[
(V - V_1)(V - V_2)(V_1 + V_2 - V) = 0.
\]
Thus when $i, j, k > 0$, we have
\[
Q_{ijk}(v) = 0.
\]
Observe that
\[
\begin{aligned}
Q_{ijk}(v) & = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid V = v\Big] \\
& = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2 \mid V = v)\\
&~~~+ {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = -\theta_2, V = v\Big]\cdot{\mathrm P}(\theta_1 = -\theta_2 \mid V = v),
\end{aligned}
\]
and
\[
{\mathrm P}(\theta_1 = -\theta_2 \mid V \leq v) = \frac{{\mathrm P}(\theta_1 = - \theta_2, V \leq v)}{{\mathrm P}(V \leq v)} = \frac{{\mathrm P}(\theta_1 = - \theta_2, V_1 + V_2 \leq v)}{{\mathrm P}(V \leq v)}.
\]
By \eqref{eq:6.3} in \textbf{Step 1}, we have $\theta_i \mid V_i = v$ converges in distribution to $\mathrm{Unif}(\bS^{d-1}) = \mathrm{Unif}\{1, - 1\}$. Letting $\tilde{\theta}_1, \tilde{\theta}_2$ be independent $\mathrm{Unif}\{1, - 1\}$, we have
\[
\lim_{v \to 0+}{\mathrm P}(\theta_1 = - \theta_2\mid V_1 + V_2 \leq v) = {\mathrm P}(\tilde{\theta}_1 = -\tilde{\theta}_2) = 1/2.
\]
Accordingly, it holds true that
\[
{\mathrm P}(\theta_1 = -\theta_2 \mid V \leq v) = \frac{{\mathrm P}(\theta_1 = - \theta_2, V_1 + V_2 \leq v)}{{\mathrm P}(V \leq v)} = \frac{{\mathrm P}(V_1 + V_2 \leq v)}{2{\mathrm P}(V \leq v)}.
\]
From Lemma \ref{lem5.1}, we have
\[
\lim_{v \to 0+}\frac{{\mathrm P}(V \leq v)}{v^2} = \frac{\alpha(1)}{2}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2= \frac{3}{4}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\]
We also have
\[
\lim_{v \to 0+} f_{V_1}(v) = \frac{f_1(x)}{f_0(x)} .
\]
Combining the above identities with the fact that $V_1, V_2$ are independent and have the same distribution, we obtain
\[
\begin{aligned}
\lim_{v \to 0+}\frac{{\mathrm P}(V_1 + V_2 \leq v)}{v^2} & = \lim_{v \to 0+}\int_{0 \leq x + y \leq v}\frac{f_{V_1}(x)f_{V_1}(y)}{v^2}\mathrm{d} x\mathrm{d} y = \lim_{v \to 0+} \int_{0 \leq x + y \leq 1}f_{V_1}(vx)f_{V_1}(vy) \mathrm{d} x\mathrm{d} y \\
& = \Big(\frac{f_1(x)}{f_0(x)}\Big)^2\int_{0 \leq x + y \leq 1}\mathrm{d} x\mathrm{d} y = \frac{1}{2}\Big(\frac{f_1(x)}{f_0(x)}\Big)^2.
\end{aligned}
\]
Therefore
\begin{align}\label{eq:6.5}
\lim_{v\to 0+} {\mathrm P}(\theta_1 = -\theta_2 \mid V = v) = \lim_{v\to 0+} {\mathrm P}(\theta_1 = -\theta_2 \mid V \leq v) = \lim_{v \to 0+}\frac{{\mathrm P}(V_1 + V_2 \leq v)/v^2}{2{\mathrm P}(V \leq v)/v^2} = \frac{1}{3},
\end{align}
and similarly we obtain
\begin{align}\label{eq:6.6}
\lim_{v\to 0+} {\mathrm P}(\theta_1 = \theta_2 \mid V = v) = \frac{2}{3}.
\end{align}
Recall that when $i, j, k > 0$, we have $Q_{ijk}(v) = 0$. We then consider the following two cases for $c_{ijk}(1)$.
~\\
\textbf{Case 1}: Only one of $i, j, k$ is $0$. Suppose $k = 0$ and $i,j > 0$. Then $\theta_1 = \theta_2$ implies $P_{ijk}(V, V_1, V_2) = 0$, and thus
\[
\begin{aligned}
Q_{ijk}(v) &= {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = -\theta_2, V = v\Big]\cdot {\mathrm P}(\theta_1 = -\theta_2 \mid V = v) \\
& = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = -\theta_2, V_1 + V_2 = v\Big]\cdot{\mathrm P}(\theta_1 = -\theta_2 \mid V = v) \\
& = {\mathrm E}\Big[V_1^j(v - V_1)^i/v^{i + j}\mid V_1 + V_2 = v, \theta_1 = -\theta_2\Big]\cdot{\mathrm P}(\theta_1 = -\theta_2 \mid V = v). \\
\end{aligned}
\]
Since the density functions of $V_1, V_2$ converge to a constant around 0, thus the condistional distribution $(V_i/v) \mid V_i \leq v, \theta_i$ converges in disrtibution to $\mathrm{Unif}[0, 1]$. Letting $U_1, U_2$ be independent $\mathrm{Unif}[0, 1]$, then given $V_1 + V_2 = v, \theta_1 = - \theta_2$, we have
\[
\frac{V_1}{v} \mid V_1 + V_2 = v, \theta_1 = - \theta_2 \xrightarrow{\enskip d \enskip} U_1 \mid U_1 + U_2 = 1 \sim \mathrm{Unif}[0, 1], \quad \text{as}~~ v \to 0+.
\]
Therefore
\[
\lim_{v \to 0+} {\mathrm E}\Big[V_1^j(1 - V_1)^i/v^{i + j}\mid V_1 + V_2 = v, \theta_1 = -\theta_2\Big] = \int_0^1x^j(1 - x)^i \mathrm{d} x = \frac{i!j!}{(i + j + 1)!}.
\]
Combining the above with \eqref{eq:6.5}, we have
\[
c_{ijk}(1) = \lim_{v \to 0+}Q_{ijk}(v) = \frac{1}{3}\cdot\frac{i!j!}{(i + j + 1)!}.
\]
Next suppose $j = 0$, $i, k > 0$. Then if either $\theta_1 = - \theta_2$ or $\theta_1 = \theta_2, V_1 > V_2$, it holds true that $P_{ijk}(V, V_1, V_2) = 0$, and thus
\[
\begin{aligned}
Q_{ijk}(v) &= {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V_1 \leq V_2, V = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2, V_1 \leq V_2 \mid V = v) \\
& = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V_1 \leq V_2 = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2, V_1 \leq V_2 \mid V = v) \\
& = {\mathrm E}\Big[(v - V_1)^iV_1^k/v^{i + k}\mid \theta_1 = \theta_2, V_1 \leq V_2 = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2, V_1 \leq V_2 \mid V = v). \\
\end{aligned}
\]
Similarly, we have
\[
\frac{V_1}{v} \mid V_1 \leq V_2 = v, \theta_1 = \theta_2 \xrightarrow{\enskip d \enskip} U_1 \mid U_1 \leq U_2 = 1 \sim \mathrm{Unif}[0, 1], \quad \text{as} ~~v \to 0+.
\]
And by symmetry, we have
\[
{\mathrm P}(\theta_1 = \theta_2, V_1 \leq V_2 \mid V = v) = \frac{1}{2}{\mathrm P}(\theta_1 = \theta_2 \mid V = v).
\]
Then by \eqref{eq:6.6} we obtain
\[
c_{ijk}(1) = \lim_{v \to 0+}Q_{ijk}(v) = \frac{i!k!}{(i + k + 1)!}\cdot\frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}\frac{i!k!}{(i + k + 1)!}.
\]
Since $i, j$ are symmtric, then the case when $i = 0$, $j, k > 0$ is similar.
\textbf{Case 2}: Only one of $i, j, k$ is not $0$. Suppose $i = j = 0$, $k > 0$. Then $\theta_1 = -\theta_2$ implies $P_{ijk}(V,V_1, V_2) = 0$, and thus
\begin{align*}
Q_{ijk}(v) &= {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V = v\Big]\cdot{\mathrm P}(\theta_1 = \theta_2 \mid V = v) \\
& = {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, \max\{V_1, V_2\} = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2 \mid V = v).
\end{align*}
By symmetry, we have
\[
{\mathrm P}(V_1 \leq V_2 = v\mid \theta_1 = \theta_2) = {\mathrm P}(V_2 \leq V_1 = v\mid \theta_1 = \theta_2).
\]
It then implies
\begin{align*}
\quad {\mathrm E}[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, \max\{V_1, V_2\} = v] &= \frac{{\mathrm E}[P_{ijk}(V, V_1, V_2) {\mathds 1} \{\max\{V_1, V_2\} = v\} \mid \theta_1 = \theta_2]}{{\mathrm P}(\max\{V_1, V_2\} = v\mid \theta_1 = \theta_2)}\\
& = \frac{2{\mathrm E}[P_{ijk}(V, V_1, V_2) {\mathds 1} \{V_1 \leq V_2 = v\} \mid \theta_1 = \theta_2]}{2{\mathrm P}(V_1 \leq V_2 = v\mid \theta_1 = \theta_2)} \\
& = {\mathrm E}[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V_1 \leq V_2, V_2 = v] \\
& = {\mathrm E}[V_1^k/v^k\mid \theta_1 = \theta_2, V_1 \leq V_2 = v].
\end{align*}
We then have
\[
c_{ijk}(1) = \lim_{v \to 0+}Q_{ijk}(v) = \frac{2}{3}\cdot\frac{1}{k + 1}.
\]
Suppose $j = k = 0$, $i > 0$. Then $\theta_1 = \theta_2, V_1 > V_2$ implies $P_{ijk}(V, V_1, V_2) = 0$. Thus
\[
\begin{aligned}
Q_{ijk}(v) =& {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = -\theta_2, V = v\Big]\cdot {\mathrm P}(\theta_1 = -\theta_2 \mid V = v)\\
&+ {\mathrm E}\Big[P_{ijk}(V, V_1, V_2) \mid \theta_1 = \theta_2, V_1 \leq V_2 = v\Big]\cdot{\mathrm P}(\theta_1 = \theta_2, V_1 \leq V_2 \mid V = v) \\
=& {\mathrm E}\Big[(v - V_1)^i/v^i \mid \theta_1 = -\theta_2, V_1 + V_2 = v\Big]\cdot {\mathrm P}(\theta_1 = -\theta_2 \mid V = v) \\
&+ {\mathrm E}\Big[(v - V_1)^i/v^i \mid \theta_1 = \theta_2, V_1 \leq V_2 = v\Big]\cdot {\mathrm P}(\theta_1 = \theta_2, V_1\leq V_2 \mid V = v).
\end{aligned}
\]
Combining the above with results in \textbf{Case 1}, we obtain
\[
c_{ijk}(1) = \lim_{v \to 0+}Q_{ijk}(v) = \frac{1}{3}\cdot\frac{1}{i + 1} + \frac{1}{3}\cdot\frac{1}{i + 1} = \frac{2}{3}\cdot\frac{1}{i + 1}.
\]
By symmetry, a similar result holds for $i = k = 0$, $j > 0$.
Combining all results in \textbf{Case 1} and \textbf{Case 2} completes the proof.
\subsection{Proof of Lemma \ref{lem5.3}}
By the proof of Theorem \ref{thm3.1}, we have
\[
{\mathrm E}\Big[n_0 \nu_1(\cA_M(x)) \mid W_i = 0\Big] = {\mathrm E}\Big[\frac{n_0}{n_1}n_1\nu_1(\cA_M(x)) \mid W_i = 0\Big] = {\mathrm E}\Big[\frac{n_0}{n_1}K_M(i) \mid W_i = 0\Big].
\]
Therefore we have
\[
{\mathrm E}\Big[\frac{n_0}{n_1}K_M(i) \mid W_i = 0\Big] \leq {\mathrm E}\Big[\frac{n_0^2}{n_1^2} \mid W_i = 0\Big] {\mathrm E}\Big[K_M(i)^2 \mid W_i = 0\Big].
\]
By Lemma 3 in \cite{abadie2006large} we have ${\mathrm E}[K_M(i)^q]$ is uniformly bounded in $n$ for all $q > 0$. Thus we obtain ${\mathrm E}\Big[n_0 \nu_1(\cA_M(X_i)) \mid W_i = 0\Big] $ is uniformly bounded in $n$.
Also, by Lemma S.3 in \cite{abadie2016matching}, we have
\[
{\mathrm E}\Big[\Big(\frac{n}{n_1}\Big)^r\Big] \leq C_r,
\]
for some constant $C_r$ depending only on $r$. Thus we obtain ${\mathrm E}\Big[n_0 \nu_1(\cA_M(X_i)) \mid W_i = 0\Big]$ is uniformly bounded in $n$.
Similarly, we have
\[
\begin{aligned}
{\mathrm E}\Big[n_0^2\nu_1(\cA_M(x))^2 \mid W_i = 0\Big] &= {\mathrm E}\Big[\frac{n_0^2}{n_1(n_1 -1)}(K_M(i)^2 - K_M(i)) \mid W_i = 0\Big] \\
& \leq {\mathrm E}\Big[\frac{n_0^4}{n_1^2(n_1 -1)^2} \mid W_i = 0\Big] {\mathrm E}\Big[K_M(i)^4 \mid W_i = 0\Big].
\end{aligned}
\]
By the same reason above we have ${\mathrm E}\Big[n_0^2\nu_1(\cA_M(x))^2 \mid W_i = 0\Big]$ is uniformly bounded in $n$.
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}