\documentclass[11pt]{article} \usepackage{amsfonts,amsmath,amssymb,amsthm} \usepackage[pdftex]{graphicx} \usepackage[hmargin=1in,vmargin=1in]{geometry} \usepackage{natbib} \usepackage[colorlinks,citecolor=blue]{hyperref} \usepackage{enumerate} \usepackage{caption} \usepackage[finnish, british]{babel} \usepackage{capt-of} \usepackage{hyperref} \usepackage[flushleft]{threeparttable} \usepackage{longtable} \usepackage{float} \bibliographystyle{ims} \def\qed{\rule{2mm}{2mm}} \def\indep{\perp \!\!\! \perp} \parskip = 1.5ex plus 0.5 ex minus0.2 ex \let\footnote=\endnote \def\enotesize{\normalsize} %------------ for Plots --------------- \usepackage{tikz,graphicx,pgfplots,pgfkeys,subcaption} %The subcaption package is used in place of "subfig" %\pgfplotsset{compat=1.8} %backwards compatibility \newenvironment{customlegend}[1][]{ \begingroup \csname pgfplots@init@cleared@structures\endcsname \pgfplotsset{#1} }{ \csname pgfplots@createlegend\endcsname \endgroup } \def\addlegendimage{\csname pgfplots@addlegendimage\endcsname} %----------- Line Numbers ------------- \usepackage[pagewise,mathlines]{lineno} %\linenumbers \synctex=1 \mathchardef\dash="2D %-------------------------------------- \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{definition}{Definition}[section] \newtheorem{condition}{Condition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{assumption}{Assumption}[section] \theoremstyle{definition} \newtheorem{example}{Example}[section] \newtheorem{remark}{Remark}[section] \usepackage{etoolbox} % This package goes here to add \qed to remarks automatically. \AtEndEnvironment{remark}{~\qed}% \AtEndEnvironment{example}{~\qed}% \DeclareMathOperator*{\var}{var} \renewcommand{\baselinestretch}{1.25} \renewcommand{\thesection}{A.\arabic{section}} \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}. \clearpage %\small \bibliography{referencessupp} \nocite{*} \clearpage \end{document}