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\begin{document}
%\date{}

\author{
Vishal Kamat\\
Departments of Economics\\
Northwestern University\\
\url{v.kamat@u.northwestern.edu}
}

\title{Online Supplementary Appendix to ``On Nonparametric Inference in the Regression Discontinuity Design''}

\maketitle

\begin{abstract}
This document provides a proof to Theorem 4.2 in the author's paper ``On Nonparametric Inference in the Regression Discontinuity Design''.
\end{abstract}

\noindent KEYWORDS: Regression discontinuity design, uniform testing.

\noindent JEL classification codes: C12, C14.

\appendix
\renewcommand{\theequation}{\Alph{section}-\arabic{equation}}
\small
\section{Additional Notation}

Let $Z^{(n)} = \{ Z_i : 1 \leq i \leq n\}$ denote the observed sample of the random variable $Z$. Let $a_n \precsim b_n$ denote $a_n \leq A b_n$, where $a_n$ and $b_n$ are deterministic sequences and $A$ is a positive constant uniform in $\mathbf{P}$. Let $| \cdot|$ denote the Euclidean matrix norm. As we use the notion of convergence in probability under the sequence of distributions $P_n$, let $A_n = o_{P_n}(1)$ denote
\begin{equation*}
   P_n(|A_n| > \epsilon) \to 0 \text{ as } n \to \infty~,
\end{equation*}
for a sequences of random variables $A_n \sim P_n$, where $\epsilon$ is any constant such that $\epsilon > 0$. Further, in Table \ref{tab:notation} below, we introduce additional notation to keep our arguments concise.
%\begin{table}[H]
%\centering
%\begin{tabular}{r l}
\begin{longtable}{r l}
$H(h_n)$ & diag(1,$h_n^{-1}$,$h_n^{-2}$) \\
$r(Z_i/h_n)$ & $(1,Z_i/h_n,(Z_i/h_n)^2)'$ \\
$Z_n(h_n)$   & $(r(Z_1/h_n), \ldots, r(Z_n/h_n))'$ \\
$k(u)$          & $(1-u)1\{ 0 \leq u \leq 1  \}$ \\
$K(u)$          & $k(-u) 1\{ u < 0\} + k(u) 1\{ u \geq 0 \}$ \\
$K_{h_n}(u)$ & $K(u/h_n)/h_n$ \\
$W_{+,n}(h_n)$  & diag$\left(1\{Z_1 \geq 0\}K_{h_n}(Z_1), \ldots, 1\{Z_n \geq 0\}K_{h_n}(Z_n)\right)$  \\
$\Gamma_{+,n}(h_n)$ & $Z_n(h_n)'W_{+,n}(h_n)Z_n(h_n)/n$ \\
$S_n(h_n)$       & $((Z_1/h_n)^3, \ldots, (Z_n/h_n)^3)'$ \\
$\nu_{+,n}$      & $Z_n(h_n)'W_{+,n}(h_n)S_n(h_n)/n$  \\
$e$              & $(1,0,0)'$  \\
$\mu(z,P)$  & $E_{P}[Y|Z=z]$ \\
$\mu_+(P)$ &  $\lim_{z \to 0^+}\mu(z,P)$ \\
$\mu_-(P)$ & $\lim_{z \to 0^-}\mu(z,P)$ \\
$\mu^v(z,P)$ & $d^v \mu(z,P)/dz^v$ \\
$\mu^v_+(P)$ & $\lim_{z \to 0^+} \mu^v(z,P)$ \\
$\sigma^2(z,P)$ & $Var_{P}[Y | Z=z]$ \\
$\Sigma_n(P)$  & diag$(\sigma^2(Z_1,P), \ldots, \sigma^2(Z_n,P))$\\
$\Psi_{+,n}(h_n,P)$ & $Z_n(h_n)'W_{+,n}(h_n) \Sigma_n(P) W_{+,n}(h_n) Z_n(h_n) / n$ \\
$\mathbf{Y}_n$              & $(Y_1, \ldots, Y_n)'$ \\
$\hat{\beta}_{+,n}$ & $H(h_n) \Gamma^{-1}_{+,n}(h_n) Z_n(h_n)'W_{+,n}(h_n) \mathbf{Y}_n /n$ \\
%\end{tabular}
\caption{Important Notation}
\label{tab:notation}
\end{longtable}
%\end{table}

Next, we provide an extended description of the test statistic used. For our null hypothesis as stated in the paper, the test statistic can be rewritten as
\begin{equation}\label{eq:TS}
  T_n^{CCT} = \frac{\hat{\mu}_{+,n} + \hat{\mu}_{-,n} - \left( \mu_+(P) - \mu_-(P) \right)}{\hat{S}_n}~,
\end{equation}
where $\mu_+(P) - \mu_-(P) = \theta_0$, $\hat{\mu}_{+,n}$ and $\hat{\mu}_{-,n}$ are bias corrected local linear estimates of $\mu_+(P)$ and $\mu_-(P)$, and $$\hat{S}_n = \sqrt{\hat{V}_{+,n} + \hat{V}_{-,n}}~,$$ where $\hat{V}_{+,n}$ and $\hat{V}_{+,n}$ are plug-in estimates conditional on $Z^{(n)}$ of the variances of $\hat{\mu}_{+,n} $ and $\hat{\mu}_{-,n}$; see \eqref{eq:vest} for the plug-in estimator used. The bias  of both estimates are estimated using local quadratic estimators. Furthermore, for all estimates, we use the triangular kernel, i.e. $k(u)$ in Table \ref{tab:notation}, and a deterministic sequence of bandwidth choices denoted by $h_n$. Throughout this document, we provide results for quantities with subscript $(+)$ as arguments for those with subscript $(-)$ follow symmetrically. In addition, as noted in \citet[][Remark 7]{CCT14}, we exploit the fact that in our simple version of the test statistic the estimates are numerically equivalent to those from a non-bias-corrected local quadratic estimator. In turn, we can write
\begin{equation}
  \hat{\mu}_{+,n} = e' \hat{\beta}_{+,n}~,
\end{equation}
which reduces the length of the proof presented below. Further, as stated in the paper, note that
\begin{align}\label{eq:Qdef}
  \mathbf{Q} = \{ Q \in \mathbf{Q}_{\mathcal{W}} : Q \text{ satisfies Assumption 4.1} \}~,
\end{align}
and that
\begin{align}\label{eq:Pdef}
  \mathbf{P} = \{ QM^{-1} : Q \in \mathbf{Q}\}~,
\end{align}
where $\mathbf{Q}_{\mathcal{W}}$, $M^{-1}$ and Assumption 4.1 are as defined in the paper.

\section{Auxiliary Lemmas}

\begin{lemma}\label{lem:quantities}
 Let $\mathbf{Q}$ be defined as in \eqref{eq:Qdef}, $\mathbf{P}$ be as in \eqref{eq:Pdef} and $P_n \in \mathbf{P}$ for all $n \geq 1$. If $nh_n \to \infty$ and $h_n \to 0$, then
 \begin{enumerate}[(i)]
   \item $\Gamma_{+,n}(h_n) = \tilde{\Gamma}_{+,n}(h_n) + o_{P_n}(1)$, where $\tilde{\Gamma}_{+,n}(h_n) = \int_{0}^{\infty} K(u) r(u) r(u)' f_{P_n}(uh_n) du \in [\Gamma_L, \Gamma_U]$ .
   \item $\nu_{+,n}(h_n) = \tilde{\nu}_{+,n}(h_n) + o_{P_n}(1)$, where $\tilde{\nu}_{+,n}(h_n) = \int^{\infty}_{0} K(u) r(u) u^2 f_{P_n}(uh_n) du \in [\nu_L , \nu_U]$.
   \item $h_n \Psi_{+,n}(h_n,P_n) = \tilde{\Psi}_{+,n}(h_n) + o_{P_n}(1)$, where $\tilde{\Psi}_{+,n}(h_n) = \int^{\infty}_{0} K(u)^2 r(u) r(u)' \sigma^2_{P_n}(uh_n) f_{P_n}(uh_n) du \in [\Psi_L, \Psi_U]$.
 \end{enumerate}
\end{lemma}
\begin{proof}
For (i), a change of variables gives us
\begin{align*}
  E_{P_n^n}[\Gamma_{+,n}(h_n)] &= E_{P_n}\left[ \frac{1}{nh_n} \sum^{n}_{i=1} 1\{ Z_i \geq 0\} K(Z_i / h_n) r(Z_i / h_n) r(Z_i / h_n)' \right] \\
      & = \frac{1}{h_n} \int^{\infty}_{0} K(z/h_n) r(z/h_n)r(z/h_n)'f_{P_n}(z) dz \\
      & = \int_{0}^{\infty} K(u) r(u) r(u)' f_{P_n}(uh_n) \equiv \tilde{\Gamma}_{+,n}(h_n)~.
\end{align*}
Further, since $h_n < \tilde{\kappa}$ for large enough $n$, we have that $\tilde{L} \leq f_{P_n}(z) \leq \tilde{U}$ by Assumption 4.1, which implies that
\begin{align*}
  \Gamma_L \equiv \tilde{L} \int_{0}^{\infty} K(u) r(u) r(u)' du \leq \tilde{\Gamma}_{+,n}(h_n) \leq \tilde{U} \int_{0}^{\infty} K(u) r(u) r(u)' du \equiv \Gamma_U~,
\end{align*}
and that
\begin{align*}
  E_{P_n^n}[ | \Gamma_{+,n}(h_n) - E_{P_n}[\Gamma_{+,n}(h_n)] |^2] & \leq \frac{1}{h_n^2} E_{P_n}\left[\left|  1\{Z_i \geq 0\} K(Z_i / h_n) r(Z_i / h_n) r(Z_i / h_n)' \right|^2 \right] \\
    & = \frac{1}{n h_n}\int^{\infty}_{0} K(u)^2 |r(u)|^4 f_{P_n}(uh_n) du \\
    & \leq \frac{\tilde{U}}{n h_n}\int^{\infty}_{0} K(u)^2 |r(u)|^4  du \\
    & = O(n^{-1}h_n^{-1}) = o(1)~.
\end{align*}
It then follows by Markov's Inequality that $\Gamma_{+,n}(h_n) = \tilde{\Gamma}_{+,n}(h_n) + o_{P_n}(1)$. Analogously, closely following \citet[][Lemma S.A.1]{CCT14supp}, we can show Lemma \ref{lem:quantities}(ii)-(iii).
\end{proof}

\begin{lemma}\label{lem:terms}
   Let $\mathbf{Q}$ be defined as in \eqref{eq:Qdef}, $\mathbf{P}$ be as in \eqref{eq:Pdef} and $P_n \in \mathbf{P}$ for all $n \geq 1$. If $nh_n \to \infty$ and $h_n \to 0$, then
  \begin{enumerate}[(i)]
   \item  $E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] = \mu_{+}(P_n) + O_{P_n}(h_n^3)~.$
   \item $V_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] = n^{-1} e' \Gamma^{-1}_{+,n}(h_n) \Psi_{+,n}(h_n,P_n) \Gamma_{+,n}^{-1}(h_n)e \equiv V_{+,n}(h_n,P_n)~$.
   \item $\left( V_{+,n}(h_n,P_n)\right)^{-1/2} \left(\hat{\mu}_{+,n} - E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}]\right) \xrightarrow{d} \mathcal{N}(0,1)~$.
  \end{enumerate}
\end{lemma}
\begin{proof}
For (i), by taking the conditional on $Z^{(n)}$ expectation, we have
\begin{align*}
  E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] &= e' H(h_n) \Gamma^{-1}_{+,n}(h_n) Z_n(h_n)' W_{+,n}(h_n) E_{P^n_n}[\mathbf{Y}_n | Z^{(n)} ] / n~.
\end{align*}  
Further, as $h_n < \tilde{\kappa}$ for large enough $n$, we have by the required differentiability in Assumption 4.1 and a Taylor expansion for $0 < Z < h_n$ that
\begin{align*}
  E_{P_n}[Y | Z]  = \mu_+(P_n) + Z\mu^1_+(P_n) + (Z/2)^2\mu^2_+(P_n) + O_{P_n}(h_n^3)~.
\end{align*}
It then follows from Lemma \ref{lem:quantities} and the previous two expressions that
\begin{align*}
  E_{P_n^n}[\hat{\mu}_+ | Z^{(n)}] = \mu_{+}(P_n) + O_{P_n}(h_n^3)~.
\end{align*}

For (ii), a simple calculation gives us
\begin{align*}
  V_{P_n^n}[\hat{\mu}_{+}(h_n) | Z^{(n)}] &= e' H(h_n) \Gamma^{-1}_{+,n}(h_n) Z_n(H_n)' W_{+,n}(h_n) \Sigma_n(P_n) W_{+,n}(h_n) Z_n(h_n) \Gamma^{-1}_{+,n}(h_n) H(h_n) e / n^2 \\
    & = n^{-1} e' \Gamma^{-1}_{+,n}(h_n) \Psi_{+,n}(h_n,P_n) \Gamma^{-1}_{+,n}(h_n) e \equiv V_{+,n}(h_n,P_n)~.
\end{align*}

For (iii), first note that from Lemma \ref{lem:quantities} we have $V_{+,n}(h_n,P_n) = \tilde{V}_{+,n}(h_n) + o_{P_n}(1)~,$ where
\begin{align*}
\tilde{V}_{+,n}(h_n) = (nh_n)^{-1} e' \tilde{\Gamma}^{-1}_{+,n}(h_n) \tilde{\Psi}_{+,n}(h_n) \tilde{\Gamma}^{-1}_{+,n}(h_n) e~.
\end{align*}
Then rewrite as follows
\begin{align}\label{eq:LFrewrite}
  \frac{\hat{\mu}_{+,n} - E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}]}{\sqrt{V_{+,n}(h_n,P_n)}} = \left( \frac{\tilde{V}_{+,n}(h_n,P_n)}{V_{+,n}(h_n,P_n)} \right)^{1/2} \left( \tilde{V}_{+,n}(h_n) \right)^{-1/2} e' \Gamma^{-1}_{+,n}(h_n) \tilde{\Gamma}_{+,n}(h_n)\tilde{A}_n^{1/2}  \xi_{n}~,
\end{align}
where 
\begin{align*}
\xi_{n}          &=\sum^{n}_{i=1} \omega_{n,i} \epsilon_{n,i} ~, \\
\epsilon_{n,i}     &= Y_i - E_{P_n}[Y_i|Z_i]~, \\
\tilde{A}_n   &= (nh_n)^{-1} \tilde{\Gamma}^{-1}_{+,n}(h_n) \tilde{\Psi}_{+,n}(h_n) \tilde{\Gamma}^{-1}_{+,n}(h_n) ~, \text{ and} \\
\omega_{n,i} &=  \tilde{A}_n^{-1/2} \tilde{\Gamma}^{-1}_{+,n}(h_n) K_{h_n}(Z_i/h_n) r(Z_i/h_n) / n~.
\end{align*}
Next note that for any $a \in \mathbf{R}^3$ we have that $\{a' \omega_{n,i} \epsilon_{n,i} : 1 \leq i \leq n\}$ is a triangular array of independent random variables with $E_{P_n^n}[a' \xi_{n}] = 0$ and $V_{P_n^n}[a' \xi_n] = a'a$. Further, this triangular array satisfies the Lindeberg condition. To see why, first note that by Lemma \ref{lem:quantities} we have
\begin{align}\label{eq:varbound}
 |\tilde{A}_{n}| \geq (nh_n)^{-1}|\tilde{A}_L|~,
\end{align}
for some value $\tilde{A}_{L} \in \mathbf{R}$, which is uniform in $\mathbf{P}$. We then have by Lemma \ref{lem:quantities} and a change of variables that
\begin{align*}
  \sum^{n}_{1=1}E_{P_n}[|a'\omega_{n,i} \epsilon_i|^4] &\precsim (nh_n)^2 \sum^{n}_{1=1}E_{P_n} \left[ \left|  a' K_{h_n}(Z/h_n) r(Z/h_n) / n  \right|^4  \right] \\
  & \precsim (nh_n)^2 n^{-3} h_n^{-4} \int^{\infty}_{0} \left| a' K(z/h_n) r(z/h_n)  \right|^4 f_{P_n}(z) dz \\
  & \precsim  (nh_n)^2 n^{-3} h_n^{-3} = O\left( (nh_n)^{-1} \right) = o(1)~
\end{align*}
and hence, using the Lindeberg-Feller CLT, we have that $a' \xi_n \xrightarrow{d}  \mathcal{N}(0,a'a)$ as $n \to \infty$. Since this holds for any $a \in \mathbf{R}^3$, the Cram\'er-Wold theorem implies that $\xi_n \xrightarrow{d}  \mathcal{N}(0,I_3)$ as $n \to \infty$, where $I_3$ denotes the identity matrix of size three. Furthermore, analogous to $V_{+}(h_n,P_n) = \tilde{V}_{+}(h_n) + o_{P_n}(1)$, we can show that
\begin{align}\label{eq:Vratio}
  \frac{V_{+,n}(h_n,P_n)}{\tilde{V}_{+,n}(h_n)}  = 1 + o_{P_n}(1)~.
\end{align}
Further, by Lemma \ref{lem:quantities} we have that
\begin{align}
  \Gamma^{-1}_{+,n}(h_n) \tilde{\Gamma}_{+,n}(h_n)  = I_3 + o_{P_n}(1)~.
\end{align}
Substituting the above results in \eqref{eq:LFrewrite} concludes the proof.
\end{proof}

\section{Proof of Theorem 4.2}

Here we show only that
\begin{align*}
  \frac{\hat{\mu}_{+,n} - \mu_{+}(P_n)}{\sqrt{\hat{V}_{+,n}}} \xrightarrow{d} \mathcal{N}(0,1)~,
\end{align*}
since under similar arguments it will follow that
\begin{align*}
  \frac{\hat{\mu}_{n,-} - \mu_{-}(P_n)}{\sqrt{\hat{V}_{n,-}}} \xrightarrow{d} \mathcal{N}(0,1)~,
\end{align*}
and then due to independence we can conclude that $T_{n}^{CCT} \xrightarrow{d} \mathcal{N}(0,1)$ as $n \to \infty$. To this end, first rewrite
\begin{align*}
  \frac{\hat{\mu}_{+,n} - \mu_{+}(P_n)}{\sqrt{\hat{V}_{+,n}}} = \frac{\hat{\mu}_{+,n} - \mu_{+}(P_n)}{\sqrt{V_{+,n}(h_n,P_n)}} \cdot \sqrt{\frac{V_{+,n}(h_n,P_n)}{\hat{V}_{+,n}}}~.
\end{align*}

\underline{Step 1.} We show that 
\begin{align}\label{eq:step1}
  \frac{\hat{\mu}_{+,n} - \mu_{+}(P_n)}{\sqrt{V_{+,n}(h_n,P_n)}} \xrightarrow{d} \mathcal{N}(0,1)~.
\end{align}
To begin, first rewrite the above as
\begin{align*}
  \frac{\hat{\mu}_{+,n} - E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}]}{\sqrt{V_{+,n}(h_n,P_n)}} + \left( \frac{\tilde{V}_{+,n}(h_n)}{V_{+,n}(h_n,P_n)} \right)^{1/2} \frac{E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] - \mu_{+}(P_n)}{\sqrt{\tilde{V}_{+,n}(h_n)}}~.
\end{align*}
In Lemma \ref{lem:terms} (iii), we showed that
\begin{align*}
  \frac{\hat{\mu}_{+,n} - E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}]}{\sqrt{V_{+,n}(h_n,P_n)}} \xrightarrow{d} \mathcal{N}(0,1)~\text{ and }~\frac{\tilde{V}_{+,n}(h_n)}{V_{+,n}(h_n,P_n)} = 1 + o_{P_n}(1)~.
\end{align*}
It then remains to show that
\begin{align*}
  \frac{E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] - \mu_+(P_n)}{\sqrt{\tilde{V}_{+,n}(h_n)}} = o_{P_n}(1)~,
\end{align*}
to conclude. To this end, note that by Lemma \ref{lem:terms} and \eqref{eq:varbound}, it follows that
\begin{align*}
  \frac{\left| E_{P_n^n}[\hat{\mu}_{+,n} | Z^{(n)}] - \mu_+(P_n) \right|}{\sqrt{\tilde{V}_{+,n}(h_n)}} = O\left((nh_n)^{1/2}\right) O_{P_n}(h_n^3) = O_{P_n}\left((nh_n^7)^{1/2}\right) = o_{P_n}(1)
\end{align*}
as $h_n \to 0$, $nh_n \to \infty$ and $nh_n^7 \to 0$.

\underline{Step 2.} We show that
\begin{align}
  \frac{V_{+,n}(h_n,P_n)}{\hat{V}_{+,n}} = 1 + o_{P_n}(1)~.
\end{align}
To begin note that
\begin{align}
 nh_n \left( V_{+,n}(h_n,P_n) - \hat{V}_{+,n} \right)=  e' \Gamma^{-1}_{+,n}(h_n) \cdot  h_n\left( \Psi_{+,n}(h_n,P_n) - \hat{\Psi}_{+,n}(h_n) \right) \cdot \Gamma^{-1}_{+,n}(h_n) e~,
\end{align}
where
\begin{align}\label{eq:Vbig}
 h_n  \left(\Psi_{+,n}(h_n,P_n) - \hat{\Psi}_{+,n}(h_n)\right) = h_n Z_n(h_n)'W_{+,n}(h_n) \left(\Sigma_n(P_n) - \hat{\Sigma}_n \right) W_{+,n}(h_n) Z_n(h_n) / n ~,
\end{align}
and
\begin{align}\label{eq:vest}
  \hat{\Sigma}_{+,n}  = \text{diag}(\hat{\epsilon}^2_{+,n,1}, \ldots, \hat{\epsilon}^2_{+,n,n})~,
\end{align}
such that $\hat{\epsilon}_{+,n,i} = Y_i - \hat{\mu}_{+,n}$. Further, note that by construction, we can write
\begin{align}
  Y_i = \mu(Z_i,P_n) + \epsilon_{n,i}~,
\end{align}
such that $E_{P_n}[\epsilon_{n,i}] = 0$ and $Var_{P_n}[\epsilon_{n,i}|Z=z] = \sigma^{2}(z,P_n)$. This in turn implies
\begin{align}
  \hat{\epsilon}_{+,n,i} = \epsilon_{n,i} + \mu(Z_i,P_n) - \mu_+(P_n) + \mu_+(P_n) - \hat{\mu}_{+,n}~.
\end{align}
We can then expand \eqref{eq:Vbig} to get the following 
\begin{align}
  h_n \left(\Psi_{+,n}(h_n,P_n) - \hat{\Psi}_{+,n}(h_n) \right)=& \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} (\sigma^2(Z_i,P_n) - \epsilon_{n,i}^2) K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)' / n}_{\equiv B_{1,n}~,~ \text{(a)}} \nonumber \\
  & - \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} ( \mu(Z_i,P_n) - \hat{\mu}_{+,n})^2 K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)' / n}_{\equiv B_{2,n}~,~ \text{(b)}} \nonumber \\
  & + 2 \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} \epsilon_{n,i} ( \mu(Z_i,P_n) - \hat{\mu}_{+,n}) K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)'/ n}_{\equiv B_{3,n}~,~ \text{(c)}}~. \nonumber
\end{align}
For quantity (a), since Assumption 2.1 (i), Assumption 2.1 (ii) and Assumption 2.1 (iv) are satisfied with the required uniform constants, we have by a change of variables that
\begin{align*}
  E_{P_n}\left[ | B_{1,n} |^2  \right] &\precsim (nh_n)^{-1}\int^{\infty}_{0} K(u)^4 |r(u)|^4  du \\
  &= O((nh_n)^{-1})  =  o(1)~,
\end{align*}
which implies by Markov's Inequality that $B_{n,1}=o_{P_n}(1)$. For quantity (b), note that first we can rewrite it as
\begin{align*}
  B_{n,2} =& \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} ( \mu(Z_i,P_n) - \mu_{+}(P_n))^2 K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)' / n}_{\equiv B_{n,21}} \\
  & + ( \mu_{+}(P_n) - \hat{\mu}_{+,n})^2 \cdot \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)' / n}_{\equiv B_{n,22}} \\
  &+ 2 ( \mu_{+}(P_n) - \hat{\mu}_{+,n}) \cdot \underbrace{h_n \sum^n_{i=1} 1\{ Z_i \geq 0 \} ( \mu(Z_i,P_n) - \mu_{+}(P_n))  K_{h_n}(Z_i)^2 r(Z_i / h_n) r(Z_i / h_n)' / n}_{\equiv B_{n,23}}~,
\end{align*}
Next, since Assumption 2.1 (i) and Assumption 2.1 (iii) are satisfied with the required uniform constants, we have by a Taylor approximation and a change of variables that
\begin{align*}
 E_{P_n}[ | B_{n,21}|^2] &\precsim n^{-1} h_n^3 \int^{\infty}_{0} K(u)^4 |r(u)|^4  du \\
  &= O(n^{-1} h_n^3)  =  o(1)~,
\end{align*}
which implies by Markov's inequality that $B_{n,21} = o_{P_n}(1)$. Further, since  Assumption 2.1 (i) is satisfied with the required uniform constants, we have by a change of variables that
\begin{align*}
  E_{P_n}[ | B_{n,22}|^2] &\precsim (nh_n)^{-1} \int^{\infty}_{0} K(u)^4 |r(u)|^4  du \\
  &= O((nh_n)^{-1})  =  o(1)~,
\end{align*}
which implies by Markov's inequality that $B_{n,22} = o_{P_n}(1)$. Finally, since Assumption 2.1 (i) and Assumption 2.1 (iii) are satisfied with the required uniform constants, we have by a Taylor approximation and a change of variables that
\begin{align*}
  E_{P_n} \left[ |  B_{n,23} |^2\right] &\precsim (n)^{-1} h_n \int^{\infty}_{0} K(u)^4 |r(u)|^4  du \\
  &= O(n^{-1} h_n)  =  o(1)~,
\end{align*}
which implies by Markov's inequality that $B_{n,23} = o_{P_n}(1)$. Since $\mu_{+}(P_n) - \hat{\mu}_{+,n} = o_{P_n}(1)$ by \eqref{eq:step1}, we can conclude for quantity (b) that $B_{n,2} = o_{P_n}(1)$. For quantity (c), using analogous arguments, we can conclude that $B_{n,3} = o_{P_n}(1)$, and hence
\begin{align}
  h_n \left(\Psi_{+,n}(h_n,P_n) - \hat{\Psi}_{+,n}(h_n) \right) = o_{P_n}(1)~.
\end{align}
In addition, since from Lemma \ref{lem:quantities} we have that $\Gamma^{-1}_{+,n}(h_n) = \tilde{\Gamma}^{-1}_{+,n}(h_n)$, it then follows that
\begin{align}\label{eq:Vend}
  nh_n \left( V_{+,n}(h_n,P_n) - \hat{V}_{+,n} \right) = o_{P_n}(1)~.
\end{align}

To conclude, first rewrite \eqref{eq:Vend} as
\begin{align*}
  \frac{ V_{+,n}(h_n,P_n) - \hat{V}_{+,n}}{\tilde{V}_{+,n}(h_n) } = o_{P_n}(1)~,
\end{align*}
and our result then follows from \eqref{eq:Vratio}.

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