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\begin{document}

\title{Supplementary Material on \textquotedblleft Adaptive Nonparametric
Regression with Conditional Heteroskedasticity\textquotedblright }
\author{Sainan Jin$^{a}$, Liangjun Su$^{a},$ Zhijie Xiao$^{b}$ \\
%EndAName
$^{a\text{ }}$\textit{School of Economics, Singapore Management University }%
\\
$^{b}$ \textit{Department of Economics,} \textit{Boston College}}
\date{}
\maketitle

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This appendix provides proofs for all technical lemmas in the above paper.\ 

\section{Proofs of the Technical Lemmas}

To facilitate the proof, we define an $N\times N$ matrix $M_{n}(x)$ and $%
N\times 1$ vectors $\Psi _{s,n}(x)$ ($s=1,2$) as:%
\begin{equation}
M_{n}\left( x\right) \equiv \left[ 
\begin{array}{cccc}
M_{n,0,0}(x) & M_{n,0,1}(x) & ... & M_{n,0,p}(x) \\ 
M_{n,1,0}(x) & M_{n,1,1}(x) & ... & M_{n,1,p}(x) \\ 
\vdots & \vdots & \ddots & \vdots \\ 
M_{n,p,0}(x) & M_{n,p,1}(x) & ... & M_{t,p,p}(x)%
\end{array}%
\right] ,\text{ }\Psi _{s,n}(x)\equiv \left[ 
\begin{array}{c}
\Psi _{s,n,0}(x) \\ 
\Psi _{s,n,1}(x) \\ 
\vdots \\ 
\Psi _{s,n,p}(x)%
\end{array}%
\right] ,  \label{A1}
\end{equation}%
where $M_{n,\left\vert j\right\vert ,\left\vert k\right\vert }(x)$ is an $%
N_{\left\vert j\right\vert }\times N_{\left\vert k\right\vert }$ submatrix
with the $\left( l,r\right) $ element given by 
\begin{equation*}
\left[ M_{n,\left\vert j\right\vert ,\left\vert k\right\vert }\left(
x\right) \right] _{l,r}\equiv \frac{1}{nh^{d}}\sum_{i=1}^{n}\left( \frac{%
X_{i}-x}{h_{1}}\right) ^{\phi _{\left\vert j\right\vert }(l)+\phi
_{\left\vert k\right\vert }(r)}K\left( \frac{X_{i}-x}{h_{1}}\right) ,
\end{equation*}%
$\Psi _{1,n,\left\vert j\right\vert }(x)$ is an $N_{\left\vert j\right\vert
}\times 1$ subvector whose $r$-th element is given by%
\begin{equation*}
\left[ \Psi _{1,n,\left\vert j\right\vert }(x)\right] _{r}\equiv \frac{1}{%
nh^{d}}\sum_{i=1}^{n}\left( \frac{X_{i}-x}{h_{1}}\right) ^{\phi _{\left\vert
j\right\vert }(r)}K\left( \frac{X_{i}-x}{h_{1}}\right) Y_{i},
\end{equation*}%
and $\Psi _{2,n,\left\vert j\right\vert }(x)$ is an $N_{\left\vert
j\right\vert }\times 1$ subvector whose $r$-th element is given by%
\begin{equation*}
\left[ \Psi _{2,n,\left\vert j\right\vert }(x)\right] _{r}\equiv \frac{1}{%
nh^{d}}\sum_{i=1}^{n}\left( \frac{X_{i}-x}{h_{1}}\right) ^{\phi _{\left\vert
j\right\vert }(r)}K\left( \frac{X_{i}-x}{h_{1}}\right) u_{i}^{2}.
\end{equation*}%
Define $\widetilde{\Psi }_{2,n}(x)$ analogously as $\Psi _{2,n}(x)$ with $%
u_{i}^{2}$ being replaced by $\tilde{u}_{i}^{2},$ where $\tilde{u}_{i}\equiv
Y_{i}-\tilde{m}\left( X_{i}\right) .$ The $p$-th order local polynomial
estimates of $m\left( x\right) $ and $\sigma ^{2}\left( x\right) $ are given
respectively by 
\begin{equation*}
\tilde{m}\left( x\right) =e_{1}^{^{\top }}M_{n}^{-1}\left( x\right) \Psi
_{1,n}(x)\text{ and }\tilde{\sigma}^{2}\left( x\right) =e_{1}^{^{\top
}}M_{n}^{-1}\left( x\right) \widetilde{\Psi }_{2,n}(x)\text{.}
\end{equation*}

For $s=1,2,$ let%
\begin{equation*}
U_{s,n}\left( x\right) \equiv \left[ 
\begin{array}{c}
U_{s,n,0}(x) \\ 
U_{s,n,1}(x) \\ 
: \\ 
U_{1,n,p}(x)%
\end{array}%
\right] ,\text{ }B_{s,n}\left( x\right) \equiv \left[ 
\begin{array}{c}
B_{s,n,0}(x) \\ 
B_{s,n,1}(x) \\ 
: \\ 
B_{s,n,p}(x)%
\end{array}%
\right] ,\text{ }
\end{equation*}%
where $U_{s,n,l}\left( x\right) $ and $B_{s,n,l}\left( x\right) $ are
defined analogously as $\Psi _{s,n,l}(x)$ so that $U_{s,n,\left\vert
j\right\vert }\left( x\right) $ and $B_{s,n,\left\vert j\right\vert }\left(
x\right) $ are $N_{\left\vert j\right\vert }\times 1$ subvectors whose $r$%
-th elements are given by%
\begin{eqnarray*}
\left[ U_{s,n,\left\vert j\right\vert }\left( x\right) \right] _{r} &=&\frac{%
1}{nh_{1}^{d}}\sum_{i=1}^{n}\left( \frac{X_{i}-x}{h_{1}}\right) ^{\phi
_{\left\vert j\right\vert }(r)}K\left( \frac{X_{i}-x}{h_{1}}\right) u_{s,i},
\\
\left[ B_{s,n,\left\vert j\right\vert }\left( x\right) \right] _{r} &=&\frac{%
1}{nh_{1}^{d}}\sum_{i=1}^{n}\left( \frac{X_{i}-x}{h_{1}}\right) ^{\phi
_{\left\vert j\right\vert }(r)}K\left( \frac{X_{i}-x}{h_{1}}\right) \Delta
_{s,i}\left( x\right) ,
\end{eqnarray*}%
where $u_{1,i}\equiv u_{i},$ $u_{2,i}\equiv
u_{i}^{2}-E(u_{i}^{2}|X_{i})=\sigma ^{2}\left( X_{i}\right) (\varepsilon
_{i}^{2}-1),$ and $\Delta _{s,i}\left( x\right) \equiv \beta _{s}\left(
X_{i}\right) -\sum_{0\leq \left\vert \mathbf{j}\right\vert \leq p}\beta _{s%
\mathbf{j}}\left( x\right) $ $\times \left( X_{i}-x\right) ^{\mathbf{j}}.$
We further define $\tilde{U}_{2,n}\left( x\right) $ analogously as $%
U_{2,n}\left( x\right) $ but with $u_{2,i}$ being replaced by $\tilde{u}%
_{2,i}\equiv \tilde{u}_{i}^{2}-E\left( u_{i}^{2}|X_{i}\right) .$ Then%
\begin{eqnarray}
\tilde{m}(x)-m\left( x\right) &=&e_{1}^{^{\top }}M_{n}^{-1}(x)U_{1,n}\left(
x\right) +e_{1}^{^{\top }}M_{n}^{-1}(x)B_{1,n}\left( x\right) ,\text{ and }
\label{A2} \\
\tilde{\sigma}^{2}(x)-\sigma ^{2}\left( x\right) &=&e_{1}^{^{\top
}}M_{n}^{-1}(x)\tilde{U}_{2,n}\left( x\right) +e_{1}^{^{\top
}}M_{n}^{-1}(x)B_{2,n}\left( x\right) .  \notag
\end{eqnarray}%
By Masry 1996(a), we can readily show that 
\begin{equation}
\tilde{m}(x)-m\left( x\right) =e_{1}^{^{\top }}\left[ f_{X}\left( x\right) M%
\right] ^{-1}\frac{1}{n}\sum_{i=1}^{n}K_{h_{1}}\left( x-X_{i}\right) \mathbf{%
Z}_{i}u_{i}+h_{1}^{p+1}e_{1}^{^{\top }}M^{-1}B\mathbf{m}^{\left( p+1\right)
}\left( x\right) +o_{p}(h_{1}^{p+1})  \label{a1}
\end{equation}%
uniformly in $x.$ Furthermore, 
\begin{equation}
\underset{x\in \mathcal{X}}{\sup }\left\vert M_{n}(x)-f_{X}\left( x\right)
M\right\vert =O_{p}\left( \upsilon _{0n}\right) \text{ and }\underset{x\in 
\mathcal{X}}{\sup }\left\vert \tilde{m}(x)-m\left( x\right) \right\vert
=O_{p}\left( \upsilon _{1n}\right) .  \label{a2}
\end{equation}

The following lemma studies the asymptotic property of the local polynomial
estimator $\tilde{\sigma}^{2}(x)$ of $\sigma ^{2}(x).$

\begin{lemma}
\label{LemC1}Suppose Assumptions A1-A5 hold. Then $\tilde{\sigma}%
^{2}(x)-\sigma ^{2}(x)=e_{1}^{^{\top }}M_{n}^{-1}(x)U_{2,n}\left( x\right)
+e_{1}^{^{\top }}M_{n}^{-1}(x)$ $\times B_{2,n}\left( x\right) +O_{p}\left(
(\upsilon _{0n}+\upsilon _{1n})\upsilon _{1n}\right) $ uniformly in $x.$
\end{lemma}

\noindent \textbf{Proof of Lemma \ref{LemC1}. }Let $K^{\ast }\left(
X_{i},x\right) \equiv e_{1}^{^{\top }}M_{n}^{-1}(x)K\left( \left(
X_{i}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}_{i}.$ Then $\tilde{\sigma}%
^{2}(x)=(nh_{1}^{d})^{-1}\sum_{i=1}^{n}K^{\ast }$ $(X_{i},x)\tilde{u}%
_{i}^{2}.$ It follows from $M_{n}^{-1}(x)M_{n}(x)=I_{N}$ that%
\begin{equation*}
\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) =\frac{1}{%
nh_{1}^{d}}\sum_{i=1}^{n}e_{1}^{^{\top }}M_{n}^{-1}(x)K\left( \left(
X_{i}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}_{i}=1,
\end{equation*}%
and%
\begin{equation*}
\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) \left(
X_{i}-x\right) ^{\mathbf{j}}=\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}e_{1}^{^{\top
}}M_{n}^{-1}(x)K\left( \left( X_{i}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}%
_{i}\left( X_{i}-x\right) ^{\mathbf{j}}=0,
\end{equation*}%
for $1\leq \left\vert \mathbf{j}\right\vert \leq p.$ Consequently, 
\begin{equation*}
\tilde{\sigma}^{2}(x)-\sigma ^{2}(x)=e_{1}^{^{\top }}M_{n}^{-1}(x)\widetilde{%
\Psi }_{2,n}(x)=\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left(
X_{i},x\right) \left\{ \tilde{u}_{i}^{2}-\overline{\sigma }^{2}\left(
x,X_{i}\right) \right\} ,
\end{equation*}%
where $\overline{\sigma }^{2}\left( x,X_{i}\right) \equiv \sum_{0\leq
\left\vert \mathbf{j}\right\vert \leq p}\left( D^{(\mathbf{j})}\sigma
^{2}\right) (x)\left( X_{i}-x\right) ^{\mathbf{j}}.$ Noting that $\tilde{u}%
_{i}^{2}=[Y_{i}-\tilde{m}(X_{i})]^{2}$ $=[\sigma (X_{i})\varepsilon
_{i}+m\left( X_{i}\right) $ $-\tilde{m}(X_{i})]^{2}$ $=\sigma
^{2}(X_{i})\varepsilon _{i}^{2}+2\sigma (X_{i})\varepsilon _{i}[m\left(
X_{i}\right) -\tilde{m}(X_{i})]+[m\left( X_{i}\right) -\tilde{m}(X_{i})]^{2},
$ we have 
\begin{eqnarray*}
\tilde{\sigma}^{2}(x)-\sigma ^{2}(x) &=&\frac{1}{nh_{1}^{d}}%
\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) \left\{ \sigma ^{2}(X_{i})-%
\overline{\sigma }^{2}\left( x,X_{i}\right) \right\}  \\
&&+\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) \sigma
^{2}(X_{i})\left( \varepsilon _{i}^{2}-1\right)  \\
&&+\frac{2}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) \sigma
(X_{i})\varepsilon _{i}\left[ m\left( X_{i}\right) -\tilde{m}(X_{i})\right] 
\\
&&+\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right) \left[
m\left( X_{i}\right) -\tilde{m}(X_{i})\right] ^{2} \\
&\equiv &A_{1}\left( x\right) +A_{2}\left( x\right) +2A_{3}\left( x\right)
+A_{4}\left( x\right) ,\text{ say.}
\end{eqnarray*}

Noting that $\Delta _{2,i}\left( x\right) =\sigma ^{2}(X_{i})-\overline{%
\sigma }^{2}\left( x,X_{i}\right) ,$ we have $A_{1}\left( x\right)
=e_{1}^{^{\top }}M_{n}^{-1}(x)B_{2,n}\left( x\right) .$ In addition $%
A_{2}\left( x\right) =e_{1}^{^{\top }}M_{n}^{-1}(x)U_{2,n}\left( x\right) $
by the definition of $u_{2,i},$ and $\sup_{x\in \mathcal{X}}\left\vert
A_{4}\left( x\right) \right\vert =\upsilon _{1n}^{2}$ by (\ref{a2}). For $%
A_{3}\left( x\right) ,$ write $-A_{3}\left( x\right) =A_{31}\left( x\right)
+A_{32}\left( x\right) ,$ where 
\begin{eqnarray*}
A_{31}\left( x\right) &\equiv &\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast
}\left( X_{i},x\right) u_{i}e_{1}^{^{\top }}M_{n}^{-1}(X_{i})U_{1,n}\left(
X_{i}\right) ,\text{ and } \\
A_{32}\left( x\right) &\equiv &\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast
}\left( X_{i},x\right) u_{i}e_{1}^{^{\top }}M_{n}^{-1}(X_{i})B_{1,n}\left(
X_{i}\right) .
\end{eqnarray*}%
Note that%
\begin{eqnarray*}
A_{31}\left( x\right) &=&\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left(
X_{i},x\right) u_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{i}\right) \right]
^{-1}U_{1,n}\left( X_{i}\right) \\
&&-\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right)
u_{i}e_{1}^{^{\top }}\left\{ M_{n}\left( x\right) ^{-1}-\left[ Mf_{X}\left(
X_{i}\right) \right] ^{-1}\right\} U_{1,n}\left( X_{i}\right) \\
&\equiv &A_{31,1}\left( x\right) -A_{31,2}\left( x\right) ,\text{ say.}
\end{eqnarray*}%
We dispose $A_{31,2}\left( x\right) $ first. By (\ref{a2}), the facts that $%
\sup_{x\in \mathcal{X}}\left\Vert U_{1,n}\left( x\right) \right\Vert
=O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n})$ and $\sup_{x\in \mathcal{X}}\frac{%
1}{nh_{1}^{d}}\sum_{i=1}^{n}\left\vert K^{\ast }\left( X_{i},x\right)
u_{i}\right\vert =O_{p}\left( 1\right) $, we have 
\begin{eqnarray*}
\underset{x\in \mathcal{X}}{\sup }\left\vert A_{31,2}\left( x\right)
\right\vert &\leq &\underset{x\in \mathcal{X}}{\sup }\left\Vert M_{n}\left(
x\right) ^{-1}-\left[ Mf_{X}\left( X_{i}\right) \right] ^{-1}\right\Vert 
\underset{x\in \mathcal{X}}{\sup }\left\Vert U_{1,n}\left( x\right)
\right\Vert \underset{x\in \mathcal{X}}{\sup }\frac{1}{nh_{1}^{d}}%
\sum_{i=1}^{n}\left\vert K^{\ast }\left( X_{i},x\right) u_{i}\right\vert \\
&=&O_{p}(\upsilon _{0n})O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
)O_{p}(1)=O_{p}(\upsilon _{0n}n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}).
\end{eqnarray*}%
Using $U_{1,n}\left( x\right) =\frac{1}{nh_{1}^{d}}\sum_{j=1}^{n}K\left(
\left( X_{j}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}_{j}u_{j}$ and $%
K^{\ast }\left( X_{i},x\right) =e_{1}^{^{\top }}M_{n}^{-1}(x)K\left( \left(
X_{i}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}_{i},$ we have%
\begin{eqnarray*}
A_{31,1}\left( x\right) &=&\frac{1}{n^{2}h_{1}^{2d}}e_{1}^{^{\top
}}M_{n}^{-1}(x)\sum_{i=1}^{n}\sum_{j=1}^{n}K\left( \left( X_{i}-x\right)
/h_{1}\right) \mathbf{\tilde{Z}}_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left(
X_{i}\right) \right] ^{-1}K\left( \left( X_{j}-X_{j}\right) /h_{1}\right) 
\mathbf{\tilde{Z}}_{j}u_{i}u_{j} \\
&=&\frac{1}{n^{2}h_{1}^{2d}}e_{1}^{^{\top }}M_{n}^{-1}(x)\sum_{1\leq i\neq
j\leq n}K\left( \left( X_{i}-x\right) /h_{1}\right) \mathbf{\tilde{Z}}%
_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{i}\right) \right] ^{-1}K\left(
\left( X_{j}-X_{j}\right) /h_{1}\right) \mathbf{\tilde{Z}}_{j}u_{i}u_{j} \\
&&+\frac{1}{n^{2}h_{1}^{2d}}e_{1}^{^{\top
}}M_{n}^{-1}(x)\sum_{i=1}^{n}K\left( \left( X_{i}-x\right) /h_{1}\right) 
\mathbf{\tilde{Z}}_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{i}\right) %
\right] ^{-1}K\left( 0\right) \mathbf{\tilde{Z}}_{i}u_{i}^{2} \\
&\equiv &A_{31,1a}\left( x\right) +A_{31,1b}\left( x\right) ,\text{ say.}
\end{eqnarray*}%
Let $\varsigma _{ij}\left( x\right) \equiv \{e_{1}^{^{\top }}\left[
Mf_{X}\left( x\right) \right] ^{-1}K\left( \left( X_{i}-x\right)
/h_{1}\right) \mathbf{\tilde{Z}}_{i}\}\{e_{1}^{^{\top }}\left[ Mf_{X}\left(
X_{i}\right) \right] ^{-1}K\left( \left( X_{j}-X_{i}\right) /h_{1}\right) 
\mathbf{\tilde{Z}}_{j}\}u_{i}u_{j}.$ Then by (\ref{a2}), $A_{31,1a}\left(
x\right) =[1+O_{p}\left( \upsilon _{0n}\right) ]\bar{A}_{31,1a}\left(
x\right) ,$ where 
\begin{equation*}
\bar{A}_{31,1a}\left( x\right) =\frac{1}{n^{2}h_{1}^{2d}}\sum_{1\leq i\neq
j\leq n}\varsigma _{ij}\left( x\right) .
\end{equation*}%
is a second order degenerate $U$-statistic. We can readily show that $\bar{A}%
_{31,1a}\left( x\right) =O_{p}\left( n^{-1}h_{1}^{-d}\right) $ for each $x$
by Chebyshev inequality. By using Bickel's (1975) standard chaining
argument, we can show $\sup_{x\in \mathcal{X}}\left\vert \bar{A}%
_{31,1a}\left( x\right) \right\vert =O_{p}\left( n^{-1}h_{1}^{-d}\log
n\right) .$ For $A_{31,1b}\left( x\right) ,$ we have 
\begin{eqnarray*}
\sup_{x\in \mathcal{X}}\left\vert A_{31,1b}\left( x\right) \right\vert &\leq
&\frac{1}{nh_{1}^{d}}\sup_{x\in \mathcal{X}}\left\Vert
M_{n}^{-1}(x)\right\Vert \sup_{x\in \mathcal{X}}\left\Vert \frac{1}{%
nh_{1}^{d}}\sum_{i=1}^{n}K\left( \left( X_{i}-x\right) /h_{1}\right) \mathbf{%
\tilde{Z}}_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{i}\right) \right]
^{-1}K\left( 0\right) \mathbf{\tilde{Z}}_{i}u_{i}^{2}\right\Vert \\
&=&O_{p}\left( n^{-1}h_{1}^{-d}\right) O_{p}\left( 1\right) O_{p}\left(
1\right) =O_{p}\left( n^{-1}h_{1}^{-d}\right) .
\end{eqnarray*}%
It follows that $\sup_{x\in \mathcal{X}}\left\vert A_{31,1}\left( x\right)
\right\vert =O_{p}\left( n^{-1}h_{1}^{-d}\log n\right) $. Consequently, we
have shown that $\sup_{x\in \mathcal{X}}\left\vert A_{31}\left( x\right)
\right\vert =O_{p}(\upsilon _{0n}n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}).$

Note that%
\begin{eqnarray*}
A_{32}\left( x\right) &=&\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left(
X_{i},x\right) u_{i}e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{i}\right) \right]
^{-1}B_{1,n}\left( X_{i}\right) \\
&&-\frac{1}{nh_{1}^{d}}\sum_{i=1}^{n}K^{\ast }\left( X_{i},x\right)
u_{i}e_{1}^{^{\top }}\left\{ M_{n}\left( x\right) ^{-1}-\left[ Mf_{X}\left(
X_{i}\right) \right] ^{-1}\right\} B_{1,n}\left( X_{i}\right) \\
&\equiv &A_{32,1}\left( x\right) -A_{32,2}\left( x\right) ,\text{ say.}
\end{eqnarray*}%
As in the study of $A_{31}\left( x\right) ,$ using (\ref{a2}) and the fact
that $\sup_{x\in \mathcal{X}}\left\vert B_{1,n}\left( x\right) \right\vert
=O_{p}(h_{1}^{p+1})$ we can readily show that $\sup_{x\in \mathcal{X}%
}\left\vert A_{32,2}\left( x\right) \right\vert =O_{p}(\upsilon
_{0n}h_{1}^{p+1})$ and that $\sup_{x\in \mathcal{X}}\left\vert
A_{32,2}\left( x\right) \right\vert =O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
h_{1}^{p+1}).$ Hence $\sup_{x\in \mathcal{X}}\left\vert A_{32}\left(
x\right) \right\vert =O_{p}(\upsilon _{0n}h_{1}^{p+1}).$ Consequently, $%
\sup_{x\in \mathcal{X}}\left\vert A_{3}\left( x\right) \right\vert
=O_{p}(\upsilon _{0n}\upsilon _{1n}).$ This completes the proof. ${\tiny %
\blacksquare }$\medskip

\noindent \textbf{Remark C.1.} Using the notation defined in the proof of
Lemma \ref{LemC1}, we can also show that $A_{1}\left( x\right) $ $%
=h_{1}^{p+1}e_{1}^{^{\top }}M^{-1}B\mathbf{\sigma }^{2}{}^{\left( p+1\right)
}\left( x\right) +o_{p}(h_{1}^{p+1}),$ and $\sqrt{nh_{1}^{d}}A_{2}\left(
x\right) \overset{d}{\rightarrow }N(0,\left( \sigma ^{4}(x)/f_{X}\left(
x\right) \right) E\left( \varepsilon _{1}^{2}-1\right) ^{2}$ $e_{1}^{^{\top
}}M^{-1}\Gamma M^{-1}e_{1}^{^{\top }}).$ By standard results on local
polynomial estimators, Lemma A.1 implies%
\begin{equation}
\underset{x\in \mathcal{X}}{\sup }\left\vert \tilde{\sigma}^{2}(x)-\sigma
^{2}(x)\right\vert =O_{p}\left( \upsilon _{1n}\right) ,  \label{a3}
\end{equation}%
where $\upsilon _{1n}$ is the rate we can obtain even if the conditional
mean function $m\left( x\right) $ is known. \medskip 

Let $\delta _{i}$ and $v_{ri}\left( x\right) $ be as defined in Appendix A.
To prove Lemmas A.1-A.2, we will frequently use the facts that 
\begin{eqnarray}
\delta _{i} &=&O_{p}\left( h^{p+1}\right) \text{ uniformly on the set }%
\left\{ K_{ix}>0\right\} ,  \label{fact1} \\
v_{ri}\left( x\right)  &=&O_{p}\left(
(h_{1}^{p+1}+n^{-1/2}h_{1}^{-d/2})\left( 1+\left( h/h_{1}\right) ^{p}\right)
\right) \text{ on the set }\left\{ K_{ix}>0\right\} ,\text{ }r=1,2,
\label{fact2} \\
\underset{\left\{ K_{ix}>0\right\} }{\max }\left\vert v_{ri}\left( x\right)
\right\vert  &=&O_{p}\left( \upsilon _{2n}\right) ,\text{ }r=1,2.
\label{fact3}
\end{eqnarray}%
To facilitate the asymptotic analysis, we also define the kernel density and
derivative estimator based on the unobserved errors $\left\{ \varepsilon
_{j}\right\} $:%
\begin{equation*}
\overline{f}_{i}\left( e_{i}\right) =\frac{1}{nh_{0}}\sum_{j\neq
i}k_{0}\left( \frac{e_{i}-\varepsilon _{j}}{h_{0}}\right) ,\text{ and }%
\overline{f}_{i}^{\left( s\right) }\left( e_{i}\right) =\frac{1}{nh_{0}^{1+s}%
}\sum_{j\neq i}k_{0}^{\left( s\right) }\left( \frac{e_{i}-\varepsilon _{j}}{%
h_{0}}\right) \text{ for }s=1,2,3.
\end{equation*}%
We will need the result in the following lemma which is adopted from Hansen
(2008).

\begin{lemma}
\label{LemC2}Let $\varepsilon _{i},$ $i=1,...,n,$ be IID. Assume that $%
\left( i\right) $ the PDF of $\varepsilon _{i},$ $f\left( \cdot \right) ,$
is uniformly bounded, and the $\left( p+1\right) $th derivative of $%
f^{\left( s\right) }\left( \varepsilon \right) $ is uniformly continuous; $%
\left( ii\right) $ there exists $q>0$ such that $\sup_{\varepsilon
}\left\vert \varepsilon \right\vert ^{q}f\left( \varepsilon \right) <\infty $
and $|k_{0}^{\left( s\right) }\left( e\right) |\leq C\left\vert e\right\vert
^{-q}$ for $\left\vert e\right\vert $ large; $\left( iii\right) $ $%
k_{0}\left( \cdot \right) $ is a ($p+1$)th order kernel and $\int \left\vert
e\right\vert ^{p+s+1}\left\vert k_{0}\left( e\right) \right\vert de<\infty $%
; $\left( iv\right) $ $h_{0}\rightarrow 0$ and $nh_{0}^{1+2s}/\log
n\rightarrow \infty $ as $n\rightarrow \infty .$ Then%
\begin{equation*}
\max_{1\leq i\leq n}\left\vert \overline{f}_{i}^{\left( s\right) }\left( 
\bar{\varepsilon}_{i}\right) -f^{\left( s\right) }\left( \bar{\varepsilon}%
_{i}\right) \right\vert =O_{p}(h_{0}^{p+1}+n^{-1/2}h_{0}^{-1/2-s}\sqrt{\log n%
}).
\end{equation*}
\end{lemma}

\noindent \textbf{Proof of Lemma \ref{LemC2}. }The above result is
essentially a special case of Theorem 6 in Hansen (2008) who allows for
strong mixing processes. For an IID\ sequence, the parameters $\beta $ and $%
\theta $ in Hansen (2008) correspond to $\infty $ and one, respectively.
Another noticeable difference is that Hansen considers the usual kernel
estimates whereas we consider the leave-one-out kernel estimates here. The
difference between these two kernel estimates is uniformly $%
(nh_{0}^{1+s})^{-1}k_{0}^{\left( s\right) }\left( 0\right) ,$ which is $%
o(n^{-1/2}h_{0}^{-1/2-s}\sqrt{\log n})$ under condition $\left( iv\right) $
and thus does not contribute to the uniform convergence rate of $\overline{f}%
_{i}^{\left( s\right) }\left( \bar{\varepsilon}_{i}\right) -f^{\left(
s\right) }\left( \bar{\varepsilon}_{i}\right) $ to 0. ${\tiny \blacksquare }$%
\medskip \medskip 

\noindent \textbf{Proof of Lemma A.1. }We only prove the lemma with $s=0$ as
the other cases can be treated analogously. Write $\tilde{f}_{i}\left( 
\overrightarrow{\varepsilon }_{i}\right) -f\left( \bar{\varepsilon}%
_{i}\right) $ $=[\overline{f}\left( \bar{\varepsilon}_{i}\right) -f\left( 
\bar{\varepsilon}_{i}\right) ]+[\tilde{f}_{i}\left( \overrightarrow{%
\varepsilon }_{i}\right) -\overline{f}\left( \bar{\varepsilon}_{i}\right) ].$
Noting that $k_{0}$ is a ($p+1$)-th order kernel with compact support by
Assumption A6, the conditions on the kernel in Lemma \ref{LemC2} are
satisfied. One can readily check that the other conditions in that lemma are
also satisfied under Assumptions A1, A2, and A7. So we can apply Lemma \ref%
{LemC2} to obtain $\max_{1\leq i\leq n}\left\vert \overline{f}_{i}\left( 
\bar{\varepsilon}_{i}\right) -f\left( \bar{\varepsilon}_{i}\right)
\right\vert =O_{p}(h_{0}^{p+1}+n^{-1/2}h_{0}^{-1/2}\sqrt{\log n}).$ Let 
\begin{eqnarray}
r_{1ij} &\equiv &\frac{\bar{\varepsilon}_{i}\left[ \varphi \left(
P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{1/2}-\varphi (P_{i}(%
\mathbf{\tilde{\beta}}_{2}))^{1/2}\right] }{\varphi (P_{i}(\mathbf{\tilde{%
\beta}}_{2}))^{1/2}}-\frac{v_{1i}\left( x\right) }{\varphi (P_{i}(\mathbf{%
\tilde{\beta}}_{2}))^{1/2}}+\frac{\tilde{m}\left( X_{j}\right) -m\left(
X_{j}\right) }{\sigma \left( X_{j}\right) }  \notag \\
&&+\left[ \varepsilon _{j}+\frac{m\left( X_{j}\right) -\tilde{m}\left(
X_{j}\right) }{\sigma \left( X_{j}\right) }\right] \frac{\tilde{\sigma}%
\left( X_{j}\right) -\sigma \left( X_{j}\right) }{\tilde{\sigma}\left(
X_{j}\right) }.  \label{r1ij1}
\end{eqnarray}%
Then%
\begin{equation}
\overrightarrow{\varepsilon }_{i}-\tilde{\varepsilon}_{j}=\left( \bar{%
\varepsilon}_{i}-\varepsilon _{j}\right) +r_{1ij}.  \label{r1ij2}
\end{equation}%
By a first order Taylor expansion with an integral remainder, we have 
\begin{eqnarray}
\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) -\overline{f}%
\left( \bar{\varepsilon}_{i}\right)  &=&\frac{1}{nh_{0}}\sum_{j\neq i}\left[
k_{0}\left( \frac{\overrightarrow{\varepsilon }_{i}-\tilde{\varepsilon}_{j}}{%
h_{0}}\right) -k_{0}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{%
h_{0}}\right) \right]   \notag \\
&=&\frac{-1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{v_{1i}\left( x\right) 
}{\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{1/2}}  \notag \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \bar{\varepsilon}_{i}\left[
\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{1/2}-\varphi (P_{i}(%
\mathbf{\tilde{\beta}}_{2}))^{1/2}\right] \varphi (P_{i}(\mathbf{\tilde{\beta%
}}_{2}))^{-1/2}  \notag \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{\tilde{m}\left(
X_{j}\right) -m\left( X_{j}\right) }{\sigma \left( X_{j}\right) }  \notag \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \left[ \varepsilon _{j}+%
\frac{m\left( X_{j}\right) -\tilde{m}\left( X_{j}\right) }{\sigma \left(
X_{j}\right) }\right] \frac{\tilde{\sigma}\left( X_{j}\right) -\sigma \left(
X_{j}\right) }{\tilde{\sigma}\left( X_{j}\right) }  \notag \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}\int_{0}^{1}\left[ k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}+wr_{1ij}}{h_{0}}\right)
-k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}%
\right) \right] dwr_{1ij}  \notag \\
&\equiv &-B_{1i}\left( x\right) +B_{2i}\left( x\right) +B_{3i}\left(
x\right) +B_{4i}\left( x\right) +B_{5i}\left( x\right) ,\text{ say.}
\label{A1_5x}
\end{eqnarray}%
We will establish the uniform probability order for $B_{ji}\left( x\right) ,$
$j=1,2,...,5,$ in order.

For $B_{1i}\left( x\right) ,$ we apply Lemma \ref{LemC2} to obtain that,
uniformly in $i,$ 
\begin{equation}
\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) =f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) +O_{p}\left( n^{-1/2}h_{0}^{-3/2}\sqrt{\log n}%
+h_{0}^{p+1}\right) .  \label{A4}
\end{equation}%
Then by (\ref{fact3}) and the uniform boundedness of $f^{\prime }\left(
\varepsilon \right) $, we have 
\begin{equation}
\max_{\left\{ K_{ix}>0\right\} }\left\vert B_{1i}\left( x\right) \right\vert
=O_{p}\left( \upsilon _{2n}\right) .  \label{A1x}
\end{equation}%
Similarly, by (\ref{A4}), (\ref{fact3}), and the uniform boundedness of $%
f^{\prime }\left( \varepsilon \right) \varepsilon ,$ we have 
\begin{equation}
\max_{\left\{ K_{ix}>0\right\} }\left\vert B_{2i}\left( x\right) \right\vert
=O_{p}\left( \upsilon _{2n}\right) .  \label{A2x}
\end{equation}%
Expanding $M_{n}^{-1}(x)$ around its probability limit $\left[ Mf_{X}\left(
x\right) \right] ^{-1},$ we have%
\begin{eqnarray*}
B_{3i}\left( x\right) &=&\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \sigma
^{-1}\left( X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) %
\right] ^{-1}U_{1,n}\left( X_{j}\right) \\
&&-\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\overline{%
\varepsilon }_{i}-\varepsilon _{j}}{h_{0}}\right) \sigma ^{-1}\left(
X_{j}\right) e_{1}^{^{\top }}\alpha _{n}\left( X_{j}\right) M_{n}^{-1}\left(
X_{j}\right) U_{1,n}\left( X_{j}\right) \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \sigma ^{-1}\left(
X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right]
^{-1}B_{1,n}\left( X_{j}\right) \\
&&-\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \sigma ^{-1}\left(
X_{j}\right) e_{1}^{^{\top }}\alpha _{n}\left( X_{j}\right) M_{n}^{-1}\left(
X_{j}\right) B_{1,n}\left( X_{j}\right) \\
&\equiv &B_{31i}\left( x\right) -B_{32i}\left( x\right) +B_{33i}\left(
x\right) -B_{34i}\left( x\right) ,
\end{eqnarray*}%
where $\alpha _{n}\left( x\right) \equiv \left[ Mf_{X}\left( x\right) \right]
^{-1}[M_{n}\left( x\right) -Mf_{X}\left( x\right) ].$ Write 
\begin{eqnarray*}
B_{31i}\left( x\right) &=&\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \sigma
^{-1}\left( X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) %
\right] ^{-1}U_{1,n}\left( X_{j}\right) \\
&=&\frac{1}{nh_{0}^{2}}\sum_{j\neq i}E_{j}\left[ k_{0}^{\prime }\left( \frac{%
\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \right] \sigma
^{-1}\left( X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) %
\right] ^{-1}U_{1,n}\left( X_{j}\right) \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}\left\{ k_{0}^{\prime }\left( \frac{%
\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) -E_{j}\left[
k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}%
\right) \right] \right\} \sigma ^{-1}\left( X_{j}\right) e_{1}^{^{\top }}%
\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}U_{1,n}\left( X_{j}\right) \\
&&\equiv B_{31i,1}\left( x\right) +B_{31i,2}\left( x\right) ,\text{ say.}
\end{eqnarray*}

For $B_{31i,1}\left( x\right) ,$ we have%
\begin{eqnarray*}
\max_{1\leq i\leq n}\left\vert B_{31i,1}\left( x\right) \right\vert &\leq
&\max_{1\leq i\leq n}\left\vert \frac{n-1}{nh_{0}^{2}}E_{j}\left[
k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}%
\right) \right] \right\vert \times \sup_{x\in \mathcal{X}}\left\Vert \sigma
^{-1}\left( x\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( x\right) \right]
^{-1}\right\Vert \sup_{x\in \mathcal{X}}\left\Vert U_{1,n}\left( x\right)
\right\Vert \\
&=&O_{p}\left( 1\right) O_{p}\left( 1\right) O_{p}\left( n^{-1/2}h_{1}^{-d/2}%
\sqrt{\log n}\right) =O_{p}\left( n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}\right) ,
\end{eqnarray*}%
where we use the facts that $\sup_{x\in \mathcal{X}}\left\Vert U_{1,n}\left(
x\right) \right\Vert =O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n})$ by Masry
(1996a), $\max_{1\leq i\leq n}|h_{0}^{-2}$ $\times E_{j}[k_{0}^{\prime }((%
\bar{\varepsilon}_{i}-\varepsilon _{j})/h_{0})]-f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) |=O(h_{0}^{p+1})$ by standard bias calculation for
kernel estimates and $\max_{1\leq i\leq n}$ $\left\vert f^{\prime }\left( 
\bar{\varepsilon}_{i}\right) \right\vert \leq \sup_{\varepsilon }\left\vert
f^{\prime }\left( \varepsilon \right) \right\vert \leq C<\infty .$

Let $v_{j}\left( \bar{\varepsilon}_{i}\right) =k_{0}^{\prime }\left( (\bar{%
\varepsilon}_{i}-\varepsilon _{j})/h_{0}\right) -E[k_{0}^{\prime }\left( (%
\bar{\varepsilon}_{i}-\varepsilon _{j})/h_{0}\right) ].$ Then%
\begin{eqnarray*}
B_{31i,2}\left( x\right)  &=&\frac{1}{n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq
i}\sum_{l}v_{j}\left( \bar{\varepsilon}_{i}\right) \sigma ^{-1}\left(
X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}%
\mathbf{\tilde{Z}}_{l}K\left( \frac{X_{l}-X_{j}}{h_{1}}\right) u_{l} \\
&=&\frac{1}{n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq i}\sum_{l\neq
j,i}v_{j}\left( \bar{\varepsilon}_{i}\right) \sigma ^{-1}\left( X_{j}\right)
e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}\mathbf{\tilde{%
Z}}_{l}K\left( \frac{X_{l}-X_{j}}{h_{1}}\right) u_{l} \\
&&+\frac{1}{n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq i}v_{j}\left( \bar{%
\varepsilon}_{i}\right) \sigma ^{-1}\left( X_{j}\right) e_{1}^{^{\top }}%
\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}\mathbf{\tilde{Z}}_{j}K\left(
0\right) u_{j} \\
&&+\frac{1}{n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq i}v_{j}\left( \bar{%
\varepsilon}_{i}\right) \sigma ^{-1}\left( X_{j}\right) e_{1}^{^{\top }}%
\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}\mathbf{\tilde{Z}}_{i}K\left( 
\frac{X_{i}-X_{j}}{h_{1}}\right) u_{i} \\
&\equiv &B_{31i,2a}\left( x\right) +B_{31i,2b}\left( x\right)
+B_{31i,2c}\left( x\right) ,\text{ say.}
\end{eqnarray*}%
By construction, $B_{31i,2a}\left( x\right) $ is a second order degenerate $U
$-statistic (see, e.g., Lee (1990)) and we can bound it by straightforward
moment calculations. Let $\epsilon _{n}\equiv Cn^{-1/2}h_{1}^{-d/2}\sqrt{%
\log n}$ for some $C>0.$ By the Boole and Markov inequalities, 
\begin{equation*}
P\left( \max_{1\leq i\leq n}\left\vert B_{31i,2a}\left( x\right) \right\vert
\geq \epsilon _{n}\right) \leq \sum_{i=1}^{n}P\left( \left\vert
B_{31i,2a}\left( x\right) \right\vert \geq \epsilon _{n}\right) \leq
\sum_{i=1}^{n}\frac{E\left[ \left\vert B_{31i,2a}\left( x\right) \right\vert
^{4}\right] }{\epsilon _{n}^{4}}.
\end{equation*}%
Let $a_{lj}=e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}%
\mathbf{\tilde{Z}}_{l}K\left( (X_{l}-X_{j})/h_{1}\right) .$ Note that 
\begin{eqnarray*}
E\left[ \left\vert B_{31i,2}\left( x\right) \right\vert ^{4}\right]  &=&%
\frac{1}{\left( n^{2}h_{1}^{d}h_{0}^{2}\right) ^{4}}\sum_{j_{s}\neq
l_{s}\neq i\text{ for }s=1,2,3,4} \\
&&\times E\left\{
a_{l_{1}j_{1}}a_{l_{2}j_{2}}a_{l_{3}j_{3}}a_{l_{4}j_{4}}v_{j_{1}}\left( \bar{%
\varepsilon}_{i}\right) u_{l_{1}}v_{j_{2}}\left( \bar{\varepsilon}%
_{i}\right) u_{l_{2}}v_{j_{3}}\left( \bar{\varepsilon}_{i}\right)
u_{l_{3}}v_{j_{4}}\left( \bar{\varepsilon}_{i}\right) u_{l_{4}}\right\} ,
\end{eqnarray*}%
where the summations are only taken with respect to $j$ and $l$'s. Consider
the index set\ $S\equiv \{j_{s},l_{s},s=1,$ $2,$ $3,$ $4\}.$ If the number
of distinct elements in $S$ is larger than $4,$ then the expectation in the
last expression is zero by the IID condition in Assumption A1. We can
readily show that $E\left[ \left\vert B_{31i,2}\left( x\right) \right\vert
^{4}\right] =O(n^{-4}h_{1}^{-2d}h_{0}^{-6}).$ It follows that 
\begin{eqnarray*}
P\left( \max_{1\leq i\leq n}\left\vert B_{31i,2a}\left( x\right) \right\vert
\geq \epsilon _{n}\alpha _{n,0}\right)  &\leq &\frac{%
nO(n^{-4}h_{1}^{-2d}h_{0}^{-6})}{Cn^{-2}h_{1}^{-2d}\left( \log n\right)
^{2}\alpha _{n,0}^{4}}=\frac{O\left( n^{-1}h_{0}^{-6}\left( \log n\right)
^{-2}\right) }{\alpha _{n}^{4}} \\
&=&O\left( n^{-1}h^{-(2p+1)-d}\right) =O\left( 1\right) .
\end{eqnarray*}%
where recall $\alpha _{n,s}=h^{[(2p+d)/4-(s+1)]}(\log n)^{s+1}.$ Then $%
\max_{1\leq i\leq n}\left\vert B_{31i,2a}\left( x\right) \right\vert
=O_{p}(\alpha _{n,0}n^{-1/2}h_{1}^{-d/2}\sqrt{\log n})$ by the Markov
inequality. Analogously, we can show that $\max_{1\leq i\leq n}\left\vert
B_{31i,2c}\left( x\right) \right\vert $ $=o(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
).$ For $B_{31i,2b}\left( x\right) ,$ we continue to decompose it as follows%
\begin{eqnarray*}
B_{31i,2b}\left( x\right)  &=&\frac{K\left( 0\right) }{%
n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq i}\sigma ^{-1}\left( X_{j}\right)
e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}\mathbf{\tilde{%
Z}}_{j}\left\{ v_{j}\left( \bar{\varepsilon}_{i}\right) u_{j}-E_{j}\left[
v_{j}\left( \bar{\varepsilon}_{i}\right) u_{j}\right] \right\}  \\
&&+\frac{K\left( 0\right) }{n^{2}h_{1}^{d}h_{0}^{2}}\sum_{j\neq i}\sigma
^{-1}\left( X_{j}\right) e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) %
\right] ^{-1}\mathbf{\tilde{Z}}_{j}E_{j}\left[ v_{j}\left( \bar{\varepsilon}%
_{i}\right) u_{j}\right]  \\
&\equiv &B_{31i,2b1}\left( x\right) +B_{31i,2b2}\left( x\right) ,
\end{eqnarray*}%
where $E_{j}$ denotes expectation with respect to the variable indexed by $j.
$ We bound the second term first:%
\begin{eqnarray*}
\max_{1\leq i\leq n}\left\vert B_{31i,2b2}\left( x\right) \right\vert  &\leq
&\max_{1\leq i\leq n}\left\vert h_{0}^{-1}E_{j}\left[ v_{j}\left( \bar{%
\varepsilon}_{i}\right) u_{j}\right] \right\vert \frac{K\left( 0\right) }{%
n^{2}h_{1}^{d}h_{0}}\sum_{j=1}^{n}\sigma ^{-1}\left( X_{j}\right) \left\vert
e_{1}^{^{\top }}\left[ Mf_{X}\left( X_{j}\right) \right] ^{-1}\mathbf{\tilde{%
Z}}_{j}\right\vert  \\
&=&O_{p}\left( 1\right)
O(n^{-1}h_{1}^{-d}h_{0}^{-1})=O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1}).
\end{eqnarray*}%
By the Boole and Markov inequalities,%
\begin{eqnarray*}
P\left( \max_{1\leq i\leq n}\left\vert B_{31i,2b1}\left( x\right)
\right\vert \geq \epsilon _{n}\right)  &\leq &\sum_{i=1}^{n}\frac{E\left[
\left\vert B_{31i,2b1}\left( x\right) \right\vert ^{4}\right] }{\epsilon
_{n}^{4}}=\frac{nO(n^{-6}h_{1}^{-4d}h_{0}^{-6})}{Cn^{-2}h_{1}^{-2d}\left(
\log n\right) ^{2}} \\
&=&O(n^{-3}h_{1}^{-2d}h_{0}^{-6}\left( \log n\right) ^{-2})=o\left( 1\right)
,
\end{eqnarray*}%
implying that $\max_{1\leq i\leq n}\left\vert B_{31i,2b1}\left( x\right)
\right\vert =o_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}).$ Hence $\max_{1\leq
i\leq n}\left\vert B_{31i,2}\left( x\right) \right\vert
=O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1})$ $+o_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
).$ Consequently, we have shown that 
\begin{equation*}
\max_{1\leq i\leq n}\left\vert B_{31i}\left( x\right) \right\vert
=O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1})+\left( \alpha _{n,0}+o\left( 1\right)
\right) O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}).
\end{equation*}

By (\ref{a2}), the fact that $\sup_{x\in \mathcal{X}}\left\Vert
U_{1,n}\left( x\right) \right\Vert =O_{p}(n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}),
$ and the fact that $\max_{1\leq i\leq n}\frac{1}{nh_{0}^{2}}\sum_{j\neq i}$ 
$\left\vert k_{0}^{\prime }\left( (\overline{\varepsilon }_{i}-\varepsilon
_{j})/h_{0}\right) \right\vert =O(h_{0}^{-1}),$ we can readily show that $%
\max_{1\leq i\leq n}\left\vert B_{32i}\left( x\right) \right\vert
=O_{p}(\upsilon _{0n}n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}h_{0}^{-1}).$ For the
other terms, we have $\max_{1\leq i\leq n}\left\vert B_{33i}\left( x\right)
\right\vert $ $=O_{p}(h_{1}^{p+1}),$ and $\max_{1\leq i\leq n}\left\vert
B_{34i}\left( x\right) \right\vert =O_{p}(h_{1}^{p+1})O_{p}\left( \upsilon
_{0n}\right) $ $O_{p}\left( h_{0}^{-1}\right) =O_{p}(\upsilon
_{0n}h_{1}^{p+1}h_{0}^{-1}).$ Consequently, 
\begin{equation}
\underset{1\leq i\leq n}{\max }\left\vert B_{3i}\left( x\right) \right\vert
=O_{p}\left( n^{-1}h_{1}^{-d}h_{0}^{-1}+\upsilon _{1n}+\upsilon
_{0n}\upsilon _{1n}h_{0}^{-1}+\alpha _{n,0}n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
\right) .  \label{A3x}
\end{equation}

Now write%
\begin{eqnarray*}
B_{4i}\left( x\right) &=&\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right)
\varepsilon _{j}\frac{\tilde{\sigma}\left( X_{j}\right) -\sigma \left(
X_{j}\right) }{\tilde{\sigma}\left( X_{j}\right) } \\
&&+\frac{1}{nh_{0}^{2}}\sum_{j\neq i}k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{m\left( X_{j}\right) -%
\tilde{m}\left( X_{j}\right) }{\sigma \left( X_{j}\right) }\frac{\tilde{%
\sigma}\left( X_{j}\right) -\sigma \left( X_{j}\right) }{\tilde{\sigma}%
\left( X_{j}\right) } \\
&\equiv &B_{41i}\left( x\right) +B_{42i}\left( x\right) .
\end{eqnarray*}%
By (\ref{a2}) and Lemma \ref{LemC1}, it is easy to show that $\max_{1\leq
i\leq n}\left\vert B_{42i}\left( x\right) \right\vert =O_{p}\left( \upsilon
_{1n}^{2}h_{0}^{-1}\right) .$ Using analogous arguments as used in the
analysis of $B_{3i}\left( x\right) $ and Lemma \ref{LemC1}, we can show that 
$\max_{1\leq i\leq n}\left\vert B_{41i}\left( x\right) \right\vert $ $%
=O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1}+\upsilon _{0n}\upsilon
_{1n}h_{0}^{-1}+h_{1}^{p+1}).$ Consequently, 
\begin{equation}
\max_{1\leq i\leq n}\left\vert B_{4i}\left( x\right) \right\vert
=O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1}+\upsilon _{0n}\upsilon
_{1n}h_{0}^{-1}+h_{1}^{p+1}).  \label{A4x}
\end{equation}%
where we use the fact that $\upsilon
_{1n}^{2}h_{0}^{-1}=o_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1}+h_{1}^{p+1}).$ As
argued by Hansen (2008, pp.740-741), under Assumption A6 there exists an
integral function $k_{0}^{\ast }$ such that 
\begin{equation*}
\left\vert k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon
_{j}+wr_{1ij}}{h_{0}}\right) -k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \right\vert \leq wh_{0}^{-1}k_{0}^{\ast
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right)
\left\vert r_{1ij}\right\vert .
\end{equation*}%
It follows that 
\begin{eqnarray}
\max_{1\leq i\leq n}\left\vert B_{5i}\left( x\right) \right\vert &\leq &%
\frac{1}{nh_{0}^{3}}\sum_{j\neq i}k_{0}^{\ast }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) r_{1ij}^{2}=\frac{O_{p}\left( \upsilon
_{2n}^{2}\right) }{nh_{0}^{3}}\sum_{j\neq i}k_{0}^{\ast }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \left( \bar{\varepsilon}%
_{i}^{2}+\varepsilon _{j}^{2}\right)  \notag \\
&=&O_{p}\left( \upsilon _{2n}^{2}h_{0}^{-2}\right) .  \label{A5x}
\end{eqnarray}%
Combining (\ref{A1_5x}), (\ref{A1x}), (\ref{A2x}), (\ref{A3x}), (\ref{A4x}),
and (\ref{A5x}) and using the facts that $n^{-1}h_{1}^{-d}h_{0}^{-1}=o\left(
\upsilon _{1n}^{2}h_{0}^{-2}\right) $ and that $h_{1}^{p+1}=o(\upsilon
_{2n}) $ yield the desired result for $s=0.$

When $s>0,$ we can decompose $\tilde{f}_{i}^{\left( s\right) }\left( 
\overrightarrow{\varepsilon }_{i}\right) -\overline{f}^{\left( s\right)
}\left( \bar{\varepsilon}_{i}\right) $ as in (\ref{A1_5x}) with the
corresponding terms denoted as $B_{ri}^{\left( s\right) }\left( x\right) $
for $r=1,2,...,5.$ The probability orders of $B_{1i}^{\left( s\right)
}\left( x\right) $ and $B_{2i}^{\left( s\right) }\left( x\right) $ are the
same as those of $B_{1i}\left( x\right) $ and $B_{2i}\left( x\right) ,$
those of $B_{3i}^{\left( s\right) }\left( x\right) \ $and $B_{4i}^{\left(
s\right) }\left( x\right) $ become $O_{p}(n^{-1}h_{1}^{-d}h_{0}^{-1-s}+(%
\upsilon _{0n}h_{0}^{-1-s}+\alpha _{n,s})n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}%
+h_{1}^{p+1}),$ and the probability order of $B_{5i}^{\left( s\right)
}\left( x\right) $ is $O_{p}(\upsilon _{2n}^{2}h_{0}^{-2-s})$. Consequently, 
$\max_{1\leq i\leq n}|\tilde{f}_{i}^{\left( s\right) }\left( \overrightarrow{%
\varepsilon }_{i}\right) -\overline{f}^{\left( s\right) }\left( \bar{%
\varepsilon}_{i}\right) |=O_{p}(\upsilon _{2n}+(\upsilon
_{0n}h_{0}^{-1-s}+\alpha _{n,s})n^{-1/2}h_{1}^{-d/2}\sqrt{\log n}+\upsilon
_{2n}^{2}h_{0}^{-2-s}).$ ${\tiny \blacksquare }$\medskip\ \medskip

\noindent \textbf{Proof of Lemma A.2. }The proof is similar to but much
simpler than that of Lemma A.1 and thus omitted. ${\tiny \blacksquare }$%
\medskip 

\noindent \textbf{Proof of Lemma A.3. }The proof is analogous to that of
Lemma USSLN in Gozalo and Linton (2000) and thus we only sketch the proof
for the $r=1$ case. Let $\mathcal{C}_{n}=\left\{ q_{1n}\left( ^{.},\theta
\right) :\theta \in \Theta \right\} $. Under the permissibility and envelope
integrability of $\mathcal{C}_{n},$ the almost sure convergence of $%
\sup_{\theta \in \Theta }\left\vert h^{-d}\left[ P_{n}q_{n,1}\left( Z,\theta
\right) -Pq_{n,1}\left( Z_{i},\theta \right) \right] \right\vert $ is
equivalent to its convergence in probability. By the boundedness of $\Theta $
and measurability of the $q_{n,1}$, the class $\mathcal{C}_{n}$ is
permissible in the sense of Pollard (1984, p196). We now show the envelope
integrability of $\mathcal{C}_{n}.$ By Assumption A1 and the compactness of $%
K,$ $\left\vert \log \left( f\left( \varepsilon _{i}\left( \mathbf{\beta }%
\right) \right) \right) \right\vert \leq D(Y_{i})$ on the set $K_{ix}>0.$
Consequently, we can take the dominance function $\overline{q}_{n}=D\left(
Y\right) K\left( \left( x-X\right) /h\right) .$ Let $E\left[ D\left(
Y\right) |X\right] =\bar{D}\left( X\right) .$ Assumptions A1 and A3 ensure
that 
\begin{equation*}
P\overline{q}_{n}=E\left[ \bar{D}\left( X\right) K\left( \left( x-X\right)
/h\right) \right] =h^{d}\int \bar{D}\left( x-hu\right) f\left( x-hu\right)
K\left( u\right) du=O\left( h^{d}\right) .
\end{equation*}

The envelope integrability allows us to truncate the functions to a finite
range. Let $\alpha _{n}>1$ be a sequence of constants such that $\alpha
_{n}\rightarrow \infty $ as $n\rightarrow \infty .$ Define%
\begin{equation*}
\mathcal{C}_{\alpha _{n}}^{\ast }=\left\{ q_{\alpha _{n}}^{\ast }=\alpha
_{n}^{-1}q_{n,1}\mathbf{1}\left\{ \overline{q}_{n}\leq \alpha _{n}\right\}
:q_{n}\in \mathcal{C}_{n}\right\} .
\end{equation*}%
Let $b_{n}$ be a non-increasing sequence of positive numbers for which $%
nh^{d}b_{n}^{2}\gg \log n.$ By analysis similar to that of Gozalo and Linton
(2000) and Theorem II.37 of Pollard (1984, p.34), to show that $\underset{%
\mathcal{C}_{n}}{\sup }\left\vert P_{n}q_{n,1}-Pq_{n,1}\right\vert
=o_{p}\left( h^{d}b_{n}\right) ,$ it suffices to show%
\begin{equation}
\underset{\mathcal{C}_{\alpha _{n}}^{\ast }}{\sup }\left\vert P_{n}q_{\alpha
_{n}}^{\ast }-Pq_{\alpha _{n}}^{\ast }\right\vert =o_{p}\left(
h^{d}b_{n}\right) ,  \label{U3}
\end{equation}%
which holds provided 
\begin{equation}
\underset{\mathcal{C}_{\alpha _{n}}^{\ast }}{\sup }\left\{ P\left[ q_{\alpha
_{n}}^{\ast }\right] ^{2}\right\} ^{1/2}<h^{d/2}  \label{U4}
\end{equation}%
and%
\begin{equation}
\sup N_{1}\left( \epsilon ,G,\mathcal{C}_{\alpha _{n}}^{\ast }\right) \leq
C_{1}\epsilon ^{-C_{2}}\text{ for }0<\epsilon \leq 1,  \label{U5}
\end{equation}%
where $N_{1}\left( \epsilon ,G,\mathcal{C}_{\alpha _{n}}^{\ast }\right) $ is
the covering number of $\mathcal{C}_{\alpha _{n}}^{\ast },$ i.e., the
smallest value $J$ for which there exists functions $g_{1},...,g_{J}$ such
that $\min_{j\leq J}G\left\vert q-g_{j}\right\vert \leq \epsilon $ for each $%
q\in \mathcal{C}_{\alpha _{n}}^{\ast },$ the supremum is taken over all
probability measures $G,$ and $C_{1}$ and $C_{2}$ are positive constants
independent of $n.$

(\ref{U4}) holds by construction. For (\ref{U5}), we need to show that $%
\mathcal{C}_{\alpha _{n}}^{\ast }$ is a Euclidean class (Nolan and Pollard,
1987, p.789). Since the functions in $\mathcal{C}_{\alpha _{n}}^{\ast },$ $%
q_{\alpha _{n}}^{\ast }=\alpha _{n}^{-1}\log \left( f\left( \varepsilon
\left( \mathbf{\beta }\right) \right) \right) G_{b}\left( f\left(
\varepsilon \left( \mathbf{\beta }\right) \right) \right) K\left( \left(
x-X\right) /h\right) $ $\mathbf{1}\left\{ \overline{q}_{n}\leq \alpha
_{n}\right\} ,$ are composed from the classes of functions 
\begin{eqnarray*}
\mathcal{C}_{1} &=&\left\{ c_{1}\log f\left( \frac{y-P\left( \mathbf{\beta }%
_{1}\right) }{\sqrt{\exp \left( P\left( \mathbf{\beta }_{2}\right) \right) }}%
\right) :\left( \mathbf{\beta }_{1}^{^{\top }},\mathbf{\beta }_{2}^{^{\top
}}\right) ^{^{\top }}\in \mathcal{B},\text{ }c_{1}\leq 1\right\} , \\
\mathcal{C}_{2} &=&\left\{ c_{2}G_{b}\left( f\left( \frac{y-P\left( \mathbf{%
\beta }_{1}\right) }{\sqrt{\exp \left( P\left( \mathbf{\beta }_{2}\right)
\right) }}\right) \right) :\left( \mathbf{\beta }_{1}^{^{\top }},\mathbf{%
\beta }_{2}^{^{\top }}\right) ^{^{\top }}\in \mathcal{B},\text{ }c_{2}\leq
1\right\} , \\
\mathcal{C}_{3} &=&\left\{ K\left( x^{^{\top }}c_{3}+c_{4}\right) :c_{3}\in 
\mathbb{R}^{d},\text{ }c_{4}\in \mathbb{R}\right\} ,\text{ and }\mathcal{C}%
_{4}=\left\{ \mathbf{1}\left\{ c_{5}\overline{q}_{n}\leq 1\right\} :c_{5}\in 
\mathbb{R}\right\} ,
\end{eqnarray*}%
it suffices to show that the $\mathcal{C}_{j}^{\prime }s$ form Euclidean
classes by Nolan and Pollard (1987, pp. 796-797) and Pakes and Pollard
(1989, Lemmas 2.14 and 2.15).

First, for $j=1,2,$ $\{P(\mathbf{\beta }_{j})\}$ forms a polynomial class of
functions and is Euclidean by Lemma 2.12 of Pakes and Pollard (1989). By
Example 2.10 of Pakes and Pollard (1989) and the bounded variation
assumption on $f$, the class $\{f\left( \frac{y-m}{s}\right) :m\in \mathbb{R}
$, $s>0\}$ is Euclidean for the constant envelope $\sup_{\varepsilon
}\left\vert f\left( \varepsilon \right) \right\vert .$ It follows from Pakes
and Pollard (1989, Lemmas 2.15) that $\mathcal{C}_{1}$ is also Euclidean.
Similarly, $\mathcal{C}_{2}$ is Euclidean. By Nolan and Pollard (1987, Lemma
22) and the bounded variation of $K,$ $\mathcal{C}_{3}$ forms a Euclidean
class with constant envelope $\sup_{x}\left\vert K\left( x\right)
\right\vert .$ Finally, by Pollard (1984, Lemma II.25) and the Euclidean
property of $\mathcal{C}_{j},$ $j=1,2,3,$ $\mathcal{C}_{4}$ is Euclidean.
Consequently 
\begin{equation*}
\sup_{\theta }\left\vert \frac{1}{nh^{d}}\sum_{i=1}^{n}q_{1n}\left(
Z_{i},\theta \right) -Eq_{1n}\left( Z_{i},\theta \right) \right\vert
=o_{a.s.}\left( b_{n}\right) .
\end{equation*}%
Since Pollard's Theorem requires that $b_{n}\gg n^{-1/2}h^{-d/2}\sqrt{\log n}%
,$ we can take $b_{n}=n^{-1/2}h^{-d/2}\sqrt{\log n}$ to obtain the desired
result. ${\tiny \blacksquare }$\medskip

\noindent \textbf{Proof of Lemma A.4. }The proof is analogous to that of
Newey (1991, Corollary 3.2). We first show $\bar{P}_{n,1}\left( \theta
\right) $ is equicontinuous. Let $D_{n,i}\left( S\right) =\mathbf{1}\left\{
Y_{i}\notin S\right\} D\left( Y_{i}\right) K_{h}\left( x-X_{i}\right) $ for
a compact set $S$ on $\mathbb{R}$. By the H\"{o}lder inequality and the law
of iterated expectations,%
\begin{eqnarray}
ED_{n,i}\left( S\right)  &=&EE\left[ D_{n,i}\left( S\right) |X_{i}\right]  
\notag \\
&\leq &E\left[ \left\{ P\left( Y_{i}\notin S|X_{i}\right) \right\} ^{\left(
\gamma -1\right) /\gamma }\left\{ E\left[ D^{\gamma }\left( Y_{i}\right)
|X_{i}\right] \right\} ^{1/\gamma }K_{h}\left( x-X_{i}\right) \right]  
\notag \\
&=&E\left[ \left\{ P\left( Y_{i}\notin S|X_{i}\right) \right\} ^{\left(
\gamma -1\right) /\gamma }\left[ \bar{D}\left( X_{i}\right) \right]
^{1/\gamma }K_{h}\left( x-X_{i}\right) \right] .  \label{EC1}
\end{eqnarray}%
Note that 
\begin{equation}
E\left[ \left[ \bar{D}\left( X_{i}\right) \right] ^{1/\gamma }K_{h}\left(
x-X_{i}\right) \right] =\int \left[ \bar{D}\left( x-hv\right) \right]
^{1/\gamma }f\left( x-hv\right) K\left( v\right) dv\leq C\int K\left(
v\right) dv.  \label{EC2}
\end{equation}%
Consider $\epsilon ,\eta >0.$ By Assumption A2, we can choose $S$ large
enough such that $P\left( Y_{i}\notin S|X_{i}\right) $ is arbitrary small to
ensure $ED_{n,i}\left( S\right) <\epsilon \eta /4$. Also, $q_{n}\left(
z,\theta \right) $ is uniformly continuous on $(\mathcal{X}\times S)\times
\Theta $ for each compact set $\mathcal{X}\times S,$ implying that for any $%
\theta \in \Theta $ there exists $\mathcal{N\equiv N}\left( \theta \right) $
such that $\sup_{\left( z,\theta ^{\prime }\right) \in (\mathcal{X}\times
S)\times \mathcal{N}}|p_{1}\left( z,\theta ^{\prime }\right) -p_{1}(z,$ $%
\theta )|<\epsilon /2.$ Consequently%
\begin{equation}
\sup_{\theta ^{\prime }\in \mathcal{N}}\left\vert p_{1}\left( Z_{i},\theta
^{\prime }\right) -p_{1}\left( Z_{i},\theta \right) \right\vert <\epsilon
/2+2\cdot \mathbf{1}\left\{ Y_{i}\notin S\right\} D\left( Y_{i}\right)
K_{h}\left( x-X_{i}\right) .  \label{EC3}
\end{equation}%
Let $\triangle _{n}\left( \epsilon ,\eta \right) =\epsilon /2+2\bar{D}%
_{n}\left( S\right) ,$ where $\bar{D}_{n}\left( S\right)
=n^{-1}\sum_{i=1}^{n}D_{n,i}\left( S\right) .$ By (\ref{EC3}) and the
triangle inequality 
\begin{equation*}
\sup_{\theta ^{\prime }\in \mathcal{N}}\left\vert P_{n}p_{1}\left( Z,\theta
^{\prime }\right) -P_{n}p_{1}\left( Z,\theta \right) \right\vert <\triangle
_{n}\left( \epsilon ,\eta \right) .
\end{equation*}%
Also,%
\begin{equation*}
P\left( \triangle _{n}\left( \epsilon ,\eta \right) >\epsilon \right)
=P\left( \bar{D}_{n}\left( S\right) >\epsilon /4\right) \leq \frac{E\left[
D_{n,i}\left( S\right) \right] }{\epsilon /4}<\eta .
\end{equation*}%
Consequently 
\begin{eqnarray*}
\sup_{\theta ^{\prime }\in \mathcal{N}}\left\vert \bar{P}_{n,1}\left( \theta
^{\prime }\right) -\bar{P}_{n,1}\left( \theta \right) \right\vert 
&=&\sup_{\theta ^{\prime }\in \mathcal{N}}\left\vert E\left[
P_{n}p_{1}\left( Z,\theta ^{\prime }\right) -P_{n}p_{1}\left( Z,\theta
\right) \right] \right\vert  \\
&\leq &E\left[ \sup_{\theta ^{\prime }\in \mathcal{N}}\left\vert
P_{n}p_{1}\left( Z,\theta ^{\prime }\right) -P_{n}p_{1}\left( Z,\theta
\right) \right\vert \right] \leq E\left[ \triangle _{n}\left( \epsilon ,\eta
\right) \right] <\eta .
\end{eqnarray*}%
That is, $\left\{ \bar{P}_{n,1}\left( \theta \right) \right\} $ is
equicontinuous.

Notice that under our assumption on the compactness of $\mathcal{B}$ and the
support of $K$, $P_{i}\left( \mathbf{\beta }_{2}\right) $ is bounded. So the
proof for the equicontinuity of $\bar{P}_{n,2}\left( \theta \right) $ is
simpler than that of $\bar{P}_{n,1}\left( \theta \right) $ and thus omitted. 
${\tiny \blacksquare }$\medskip

\noindent \textbf{Proof of Lemma B.1. }We only prove the case $\left(
r,s\right) =\left( 1,1\right) \ $as the other cases are similar. For
notational simplicity, write $T_{1n\mathbf{j}}=T_{1n\mathbf{j}}\left(
1,1\right) .$ By the fact that $\varphi (P_{i}(\mathbf{\tilde{\beta}}%
_{2}))^{-1/2}-\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{-1/2}=O_{p}\left( \upsilon _{2n}\right) $ uniformly in $i$ on the set $%
\left\{ K_{ix}>0\right\} ,$ we can write 
\begin{eqnarray}
\tilde{q}_{1,i}(\mathbf{\tilde{\beta}})-q_{1,i}\left( \mathbf{\beta }%
^{0}\right)  &=&\frac{\tilde{f}_{i}^{\prime }\left( \overrightarrow{%
\varepsilon }_{i}\right) }{\tilde{f}_{i}\left( \overrightarrow{\varepsilon }%
_{i}\right) }\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{-1/2}-\frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2} 
\notag \\
&=&\left[ \frac{\tilde{f}_{i}^{\prime }\left( \overrightarrow{\varepsilon }%
_{i}\right) }{\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) }-%
\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{%
\varepsilon}_{i}\right) }\right] \varphi (P_{i}(\mathbf{\tilde{\beta}}%
_{2}))^{-1/2}+\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( 
\bar{\varepsilon}_{i}\right) }\left[ \varphi (P_{i}(\mathbf{\tilde{\beta}}%
_{2}))^{-1/2}-\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}%
\right]   \notag \\
&=&\left[ \frac{\tilde{f}_{i}^{\prime }\left( \overrightarrow{\varepsilon }%
_{i}\right) }{\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) }-%
\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{%
\varepsilon}_{i}\right) }\right] \left[ \varphi (P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) )^{-1/2}+O_{p}\left( \upsilon _{2n}\right) \right] +\frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }O_{p}\left( \upsilon _{2n}\right) .  \label{q11}
\end{eqnarray}%
Thus 
\begin{eqnarray*}
\left\vert T_{1n\mathbf{j}}\right\vert  &\leq &\frac{1}{nh^{d}}%
\sum_{i=1}^{n}\left\vert K_{ix,\mathbf{j}}\tilde{G}_{i}q_{1,i}\left( \mathbf{%
\beta }^{0}\right) \left[ \frac{\tilde{f}_{i}^{\prime }\left( 
\overrightarrow{\varepsilon }_{i}\right) }{\tilde{f}_{i}\left( 
\overrightarrow{\varepsilon }_{i}\right) }-\frac{f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\right]
\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}\right\vert  \\
&&+\frac{O_{p}\left( \upsilon _{2n}\right) }{nh^{d}}\sum_{i=1}^{n}\left\vert
K_{ix,\mathbf{j}}\tilde{G}_{i}q_{1,i}\left( \mathbf{\beta }^{0}\right) \left[
\frac{\tilde{f}_{i}^{\prime }\left( \overrightarrow{\varepsilon }_{i}\right) 
}{\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) }-\frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\right] \right\vert  \\
&&+\frac{O_{p}\left( \upsilon _{2n}\right) }{nh^{d}}\sum_{i=1}^{n}\left\vert
K_{ix,\mathbf{j}}\tilde{G}_{i}q_{1,i}\left( \mathbf{\beta }^{0}\right) \frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\right\vert .
\end{eqnarray*}%
Since the last two terms are of smaller order, it suffices to show the first
term (denoted as $|\bar{T}_{1n\mathbf{j}}|)$ is $O_{p}\left( h^{\epsilon
}\right) .$ By Lemma A.1, the definition of $\tilde{G}_{i},$ and Assumption
A7, 
\begin{eqnarray}
&&\left\vert \frac{\tilde{f}_{i}^{_{\prime }}\left( \overrightarrow{%
\varepsilon }_{i}\right) }{\tilde{f}_{i}\left( \overrightarrow{\varepsilon }%
_{i}\right) }-\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( 
\bar{\varepsilon}_{i}\right) }\right\vert \tilde{G}_{i}=\left\vert \frac{%
\tilde{f}_{i}^{_{\prime }}\left( \overrightarrow{\varepsilon }_{i}\right)
-f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{\tilde{f}_{i}\left( 
\overrightarrow{\varepsilon }_{i}\right) }+\frac{f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) \left[ f\left( \bar{\varepsilon}_{i}\right) -\tilde{f%
}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) \right] }{f\left( \bar{%
\varepsilon}_{i}\right) \tilde{f}_{i}\left( \overrightarrow{\varepsilon }%
_{i}\right) }\right\vert \tilde{G}_{i}  \notag \\
&\leq &O_{p}\left( b^{-1}\nu _{3n,1}\right) +\left( f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) /f\left( \bar{\varepsilon}_{i}\right) \right)
O_{p}\left( b^{-1}\nu _{3n,0}\right) =O_{p}\left( h^{\epsilon }\right)
\left\{ 1+\left\vert f^{\prime }\left( \bar{\varepsilon}_{i}\right) /f\left( 
\bar{\varepsilon}_{i}\right) \right\vert \right\} .  \label{q12}
\end{eqnarray}%
Therefore $\left\vert \bar{T}_{1n\mathbf{j}}\right\vert =\frac{O_{p}\left(
h^{\epsilon }\right) }{nh^{d}}\sum_{i=1}^{n}\left\vert K_{ix,\mathbf{j}%
}q_{1,i}\left( \mathbf{\beta }^{0}\right) \left( 1+\frac{f^{\prime }\left( 
\bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\right)
\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}\right\vert
=O_{p}\left( h^{\epsilon }\right) $ by Markov inequality and the fact that%
\begin{eqnarray*}
\frac{1}{h^{d}}E\left\vert K_{ix,\mathbf{j}}q_{1,i}\left( \mathbf{\beta }%
^{0}\right) \left( 1+\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{%
f\left( \bar{\varepsilon}_{i}\right) }\right) \varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1/2}\right\vert  &=&\frac{1}{h^{d}}E\left\vert
K_{ix,\mathbf{j}}\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{%
f\left( \bar{\varepsilon}_{i}\right) }\left( 1+\frac{f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\right)
\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1}\right\vert  \\
&=&\frac{1}{h^{d}}E\left\vert K_{ix,\mathbf{j}}\sigma ^{-2}\left(
X_{i}\right) \frac{f^{\prime }\left( \varepsilon _{i}\right) }{f\left(
\varepsilon _{i}\right) }\left( 1+\frac{f^{\prime }\left( \varepsilon
_{i}\right) }{f\left( \varepsilon _{i}\right) }\right) \right\vert \left\{
1+o\left( 1\right) \right\}  \\
&\leq &\frac{f_{X}\left( x\right) }{\sigma ^{2}\left( x\right) }\int
\left\vert K\left( u\right) u^{^{\mathbf{j}}}\right\vert du\left\{
I^{1/2}\left( f\right) +I\left( f\right) \right\} =O\left( 1\right) ,
\end{eqnarray*}%
where $I\left( f\right) \equiv E\left[ \psi ^{2}\left( \varepsilon
_{i}\right) \right] $ and we use the fact that $\varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )$ is the $p$-th order Taylor expansion of $\sigma
^{2}\left( X_{i}\right) $ around $x.$ This completes the proof of the lemma. 
${\tiny \blacksquare }$\medskip 

\noindent \textbf{Proof of Lemma B.2. }We only prove the case $\left(
r,s\right) =\left( 1,1\right) $ as the other cases are similar. For
notational simplicity, write $T_{2n\mathbf{j}}=T_{2n\mathbf{j}}\left(
1,1\right) .$ That is, we will show%
\begin{equation*}
T_{2n\mathbf{j}}=\frac{1}{nh^{d}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\tilde{G}%
_{i}\left\{ \tilde{q}_{1,i}(\mathbf{\tilde{\beta}})-q_{1}\left(
Y_{i};P_{i}\left( \mathbf{\beta }_{1}^{0}\right) ,P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) \right\} ^{2}=O_{p}\left( h^{\epsilon }\right) .
\end{equation*}%
By (\ref{q11}) and (\ref{q12}) in the proof of Lemma B.1, we can write 
\begin{eqnarray*}
&&\left\vert \tilde{q}_{1,i}(\mathbf{\tilde{\beta}})-q_{1}\left(
Y_{i};P_{i}\left( \mathbf{\beta }_{1}^{0}\right) ,P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) \right\vert ^{2}\tilde{G}_{i} \\
&=&\left[ O_{p}\left( h^{\epsilon }\right) \left( 1+\left\vert \frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\right\vert \right) ^{2}\left[ \varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1/2}+O_{p}(\upsilon _{2n})\right] ^{2}+\left( 
\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{%
\varepsilon}_{i}\right) }\right) ^{2}O_{p}(\upsilon _{2n}^{2})\right] \tilde{%
G}_{i} \\
&\leq &\left( 1+\left\vert \frac{f^{\prime }\left( \bar{\varepsilon}%
_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\right\vert \right) ^{2}%
\left[ \varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1}+1\right] 
\tilde{G}_{i}O_{p}\left( h^{\epsilon }\right) .
\end{eqnarray*}%
Thus 
\begin{equation*}
T_{2n\mathbf{j}}\leq \frac{O_{p}\left( h^{\epsilon }\right) }{nh^{d}}%
\sum_{i=1}^{n}\left\vert K_{ix,\mathbf{j}}\right\vert \left( 1+\left\vert 
\frac{f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{%
\varepsilon}_{i}\right) }\right\vert \right) ^{2}\left[ \varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1}+1\right] =O_{p}\left( h^{\epsilon
}\right) 
\end{equation*}%
by Markov inequality and the fact that 
\begin{eqnarray*}
&&\frac{1}{h^{d}}E\left\vert K_{ix,\mathbf{j}}\left( 1+\left\vert \frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\right\vert \right) ^{2}\left[ \varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1}+1\right] \right\vert  \\
&=&\frac{1}{h^{d}}E\left\vert K_{ix,\mathbf{j}}\left( 1+\left\vert \frac{%
f^{\prime }\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}%
_{i}\right) }\right\vert \right) ^{2}\left[ \sigma ^{-2}\left( X_{i}\right)
+1+o\left( 1\right) \right] \right\vert \left\{ 1+o\left( 1\right) \right\} 
\\
&\leq &f_{X}\left( x\right) \left[ \sigma ^{-2}\left( x\right) +1\right]
\int \left\vert K\left( u\right) u^{^{\mathbf{j}}}\right\vert du\left[
1+I\left( f\right) +2I^{1/2}\left( f\right) \right] \left\{ 1+o\left(
1\right) \right\} =O\left( 1\right) .
\end{eqnarray*}%
This completes the proof of the lemma. ${\tiny \blacksquare }$\medskip 

\noindent \textbf{Proof of Lemma B.3.} We only prove the case $\left(
r,s\right) =\left( 1,1\right) $ as the other cases are similar. For
notational simplicity, write $T_{3n\mathbf{j}}=T_{3n\mathbf{j}}\left(
1,1\right) .$ That is, we will show%
\begin{equation*}
T_{3n\mathbf{j}}=\frac{1}{nh^{d}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left( 1-%
\tilde{G}_{i}\right) q_{1,i}\left( \mathbf{\beta }^{0}\right)
^{2}=O_{p}\left( h^{\epsilon }\right) .
\end{equation*}%
We decompose $T_{3n\mathbf{j}}$ as follows 
\begin{eqnarray*}
T_{3n\mathbf{j}} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left[
1-G_{b}\left( f\left( \bar{\varepsilon}_{i}\right) \right) \right] \psi
^{2}\left( \bar{\varepsilon}_{i}\right)  \\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left[ G_{b}\left( f\left( 
\bar{\varepsilon}_{i}\right) \right) -G_{b}(\tilde{f}_{i}\left( 
\overrightarrow{\varepsilon }_{i}\right) )\right] \psi ^{2}\left( \bar{%
\varepsilon}_{i}\right)  \\
&\equiv &T_{3n\mathbf{j},1}+T_{3n\mathbf{j},2},\text{ say.}
\end{eqnarray*}%
By Lemma A.1, 
\begin{eqnarray}
\max_{1\leq i\leq n}|\tilde{G}_{i}-G_{i}| &=&\max_{1\leq i\leq n}|G_{b}(%
\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) )-G_{b}\left(
f\left( \bar{\varepsilon}_{i}\right) \right) |  \notag \\
&\leq &\frac{C}{b}\max_{1\leq i\leq n}|\tilde{f}_{i}\left( \overrightarrow{%
\varepsilon }_{i}\right) -f\left( \bar{\varepsilon}_{i}\right)
|=b^{-1}O_{p}\left( \upsilon _{3n,0}\right) =O_{p}\left( h^{\epsilon
}\right) .  \label{GG}
\end{eqnarray}%
With this, we can readily obtain $\left\vert T_{3n\mathbf{j},2}\right\vert
\leq O_{p}\left( h^{\epsilon }\right) \frac{1}{nh^{d}}\sum_{i=1}^{n}\left%
\vert K_{ix,\mathbf{j}}\right\vert \psi ^{2}\left( \bar{\varepsilon}%
_{i}\right) $ $=O_{p}\left( h^{\epsilon }\right) $ by Markov inequality. For 
$T_{3n\mathbf{j},1},$ we have 
\begin{eqnarray*}
E\left\vert T_{3n\mathbf{j},1}\right\vert  &\leq &E\left[ \frac{1}{h^{d}}%
\left\vert K_{ix,\mathbf{j}}\right\vert \left[ 1-G_{b}\left( f\left( \bar{%
\varepsilon}_{i}\right) \right) \right] \psi ^{2}\left( \bar{\varepsilon}%
_{i}\right) \right]  \\
&=&\frac{1}{h^{d}}E\left[ \left\vert K_{ix,\mathbf{j}}\right\vert \right]
E\left\{ \left[ 1-G_{b}\left( f\left( \varepsilon _{i}\right) \right) \right]
\psi ^{2}\left( \varepsilon _{i}\right) \right\} \left\{ 1+o\left( 1\right)
\right\} .
\end{eqnarray*}%
By the H\"{o}lder inequality, 
\begin{eqnarray*}
E\left\{ \left[ 1-G_{b}\left( f\left( \varepsilon _{i}\right) \right) \right]
\psi ^{2}\left( \varepsilon _{i}\right) \right\}  &\leq &E\left[ \psi
^{2}\left( \varepsilon _{i}\right) \text{ }\mathbf{1}\left\{ f\left(
\varepsilon \right) \leq 2b\right\} \right]  \\
&\leq &\left\{ E\left[ \psi ^{2\gamma }\left( \varepsilon _{i}\right) \right]
\right\} ^{1/\gamma }\left[ P\left( f\left( \varepsilon _{i}\right) \leq
2b\right) \right] ^{\left( \gamma -1\right) /\gamma } \\
&\leq &C\left[ P\left( f\left( \varepsilon _{i}\right) \leq 2b\right) \right]
^{\left( \gamma -1\right) /\gamma }=O\left( b^{\left( \gamma -1\right)
/(2\gamma )}\right) =O\left( h^{\epsilon }\right) ,
\end{eqnarray*}%
where the last line follows from Lemma 6 of Robinson (1988) and the Markov
inequality because by taking $\overline{B}=b^{-1/2},$ we have $P\left(
f\left( \varepsilon _{i}\right) \leq 2b\right) \leq 2b\int_{\left\vert
\varepsilon _{i}\right\vert \leq \overline{B}}dz+P\left( \left\vert
\varepsilon _{i}\right\vert >\overline{B}\right) $ $\leq
2b2b^{-1/2}+E\left\vert \varepsilon _{i}\right\vert b^{1/2}$ $=O\left(
b^{1/2}\right) =O\left( h^{\epsilon }\right) .$ This, in conjunction with
the fact that $\frac{1}{h^{d}}E\left[ \left\vert K_{ix,\mathbf{j}%
}\right\vert \right] =O\left( 1\right) ,$ implies that $T_{3n\mathbf{j}%
,1}=O_{p}\left( h^{\epsilon }\right) $ by Markov inequality. Consequently,
we have shown that $T_{3n\mathbf{j}}=O_{p}\left( h^{\epsilon }\right) $. $%
{\tiny \blacksquare }$\medskip 

\noindent \textbf{Proof of Lemma B.4. }Let $\tilde{f}_{i}=\tilde{f}%
_{i}\left( \overrightarrow{\varepsilon }_{i}\right) $ and\textbf{\ }$%
f_{i}=f\left( \bar{\varepsilon}_{i}\right) .$Note that $\tilde{f}_{i}^{-1}=$ 
$f_{i}^{-1}-(\tilde{f}_{i}-f_{i})/f_{i}^{2}+R_{2i},$ where $R_{2i}\equiv (%
\tilde{f}_{i}-f_{i})^{2}/\{(f_{i}^{2}\tilde{f}_{i}).$ First, we expand the
trimming function to the second order: 
\begin{equation}
G_{b}(\tilde{f}_{i})-G_{b}\left( f_{i}\right) =g_{b}\left( f_{i}\right)
\left( \tilde{f}_{i}-f_{i}\right) +\frac{1}{2}g_{b}^{\prime }\left(
f_{i}^{\ast }\right) \left( \tilde{f}_{i}-f_{i}\right) ^{2},  \label{Gb}
\end{equation}%
where $f_{i}^{\ast }$ is an intermediate value between $\tilde{f}_{i}$ and $%
f_{i}.$ Let $\rho _{i}\left( \mathbf{\beta }\right) \equiv \psi \left(
\varepsilon _{i}\left( \mathbf{\beta }\right) \right) \varepsilon _{i}\left( 
\mathbf{\beta }\right) +1,$ $\bar{\rho}_{i}\equiv \rho _{i}\left( \mathbf{%
\beta }^{0}\right) ,$ and $\rho _{i}\equiv \psi \left( \varepsilon
_{i}\right) \varepsilon _{i}+1.$ Let $\varphi _{i}\equiv \varphi ^{\prime
}(P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )/\varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) ).$ Then we have%
\begin{eqnarray*}
-\mathcal{S}_{1n\mathbf{j}} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\varphi _{i}\left\{ \bar{\rho}_{i}\left[ G_{b}(\tilde{f}_{i})-1%
\right] +\log \left( f\left( \bar{\varepsilon}_{i}\right) \varphi
(P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}\right) g_{b}\left(
f\left( \bar{\varepsilon}_{i}\right) \right) f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}\right\}  \\
&=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi
_{i}\left\{ \bar{\rho}_{i}\left[ G_{b}\left( f_{i}\right) -1\right] +\log
\left( f\left( \bar{\varepsilon}_{i}\right) \varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1/2}\right) g_{b}\left( f\left( \bar{\varepsilon}%
_{i}\right) \right) f^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{%
\varepsilon}_{i}\right\}  \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\bar{%
\rho}_{i}g_{b}\left( f_{i}\right) \left( \tilde{f}_{i}-f_{i}\right)  \\
&&+\frac{1}{4\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\bar{%
\rho}_{i}g_{b}^{\prime }\left( f_{i}^{\ast }\right) \left( \tilde{f}%
_{i}-f_{i}\right) ^{2} \\
&\equiv &\mathcal{S}_{1n\mathbf{j},1}+\mathcal{S}_{1n\mathbf{j},2}+\mathcal{S%
}_{1n\mathbf{j},3},\text{ say.}
\end{eqnarray*}%
Using a crude bound on the last term, we have $|\mathcal{S}_{1n\mathbf{j}%
,3}|=O_{p}\left( \upsilon _{3n,0}^{2}b^{-2}n^{1/2}h^{d/2}\right)
=o_{p}\left( 1\right) \ $by Lemma A.1, the fact that $\sup_{s}\left\vert
g_{b}^{\prime }\left( s\right) \right\vert =O(b^{-2}),$ and Assumption A7.

To show the first term is $o_{p}\left( 1\right) ,$ write 
\begin{equation*}
\mathcal{S}_{1n\mathbf{j},1}=\frac{-1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\xi _{1i}+\frac{1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}%
}\xi _{2i}\equiv -\mathcal{S}_{1n\mathbf{j},11}+\mathcal{S}_{1n\mathbf{j}%
,12},\text{ say,}
\end{equation*}%
where $\xi _{1i}=\frac{1}{2}\bar{\rho}_{i}\varphi _{i}$ and $\xi _{2i}=\frac{%
1}{2}\left\{ \bar{\rho}_{i}G_{b}\left( f_{i}\right) +\log (f\left( \bar{%
\varepsilon}_{i}\right) \varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{-1/2})g_{b}\left( f\left( \bar{\varepsilon}_{i}\right) \right) f^{\prime
}\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}\right\} \varphi
_{i}.$

Let $Q_{1n,i}\left( \mathbf{\beta }\right) \equiv \log \{f\left( \varepsilon
_{i}\left( \mathbf{\beta }\right) \right) \varphi (P_{i}\left( \mathbf{\beta 
}_{2}\right) )^{-1/2}\}K_{h}\left( x-X_{i}\right) ,$ $Q_{2n,i}\left( \mathbf{%
\beta }\right) \equiv \log \{f\left( \varepsilon _{i}\left( \mathbf{\beta }%
\right) \right) \varphi (P_{i}\left( \mathbf{\beta }_{2}\right) )^{-1/2}\}$ $%
\times G_{b}(f\left( \varepsilon _{i}\left( \mathbf{\beta }\right) \right)
K_{h}\left( x-X_{i}\right) ,$ and $\varsigma _{n}\left( \mathbf{\beta }%
\right) \equiv E\left[ Q_{2n,i}\left( \mathbf{\beta }\right) \right] -E\left[
Q_{1n,i}\left( \mathbf{\beta }\right) \right] .$ Then it is easy to show
that (i) $\varsigma _{n}\left( \mathbf{\beta }^{0}\right) \rightarrow 0,$
(ii) $\varsigma _{n}\left( \mathbf{\beta }\right) $ is differentiable in a
small $\epsilon _{0}$-neighborhood $N_{\epsilon _{0}}\left( \mathbf{\beta }%
^{0}\right) $ of $\mathbf{\beta }^{0}$ with $N_{\epsilon _{0}}\left( \mathbf{%
\beta }^{0}\right) \equiv \{\mathbf{\beta }:\left\Vert \mathbf{\beta }-%
\mathbf{\beta }^{0}\right\Vert \leq \epsilon _{0}\},$ (iii) $\varsigma
_{n}^{\prime }\left( \mathbf{\beta }\right) $ converges uniformly on $%
N_{\epsilon _{0}}\left( \mathbf{\beta }^{0}\right) .$ Then by Theorem 7.17
of Rudin (1976) and the fact that $h^{-\left\vert \mathbf{j}\right\vert
}\partial Q_{1n,i}\left( \mathbf{\beta }^{0}\right) /\partial \mathbf{\beta }%
_{2\mathbf{j}}=-h^{-d}\xi _{1i}K_{ix,\mathbf{j}}$ and $h^{-\left\vert 
\mathbf{j}\right\vert }\partial Q_{2n,i}\left( \mathbf{\beta }^{0}\right)
/\partial \mathbf{\beta }_{2\mathbf{j}}$ $=-h^{-d}\xi _{2i}K_{ix,\mathbf{j}%
}, $ we have%
\begin{eqnarray*}
E\left( \mathcal{S}_{1n\mathbf{j},12}\right) &=&-\sqrt{n}h^{d/2}h^{-\left%
\vert \mathbf{j}\right\vert }E\left[ \frac{\partial Q_{2n,i}\left( \mathbf{%
\beta }^{0}\right) }{\partial \mathbf{\beta }_{2\mathbf{j}}}\right] \\
&=&-\sqrt{n}h^{d/2}h^{-\left\vert \mathbf{j}\right\vert }E\left[ \frac{%
\partial Q_{1n,i}\left( \mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }%
_{2\mathbf{j}}}\right] \left\{ 1+o\left( 1\right) \right\} =E\left( \mathcal{%
S}_{1n\mathbf{j},11}\right) \left\{ 1+o\left( 1\right) \right\} .
\end{eqnarray*}%
Consequently, $E\left( \mathcal{S}_{1n\mathbf{j},1}\right) =o\left( 1\right)
E\left( \mathcal{S}_{1n\mathbf{j},11}\right) =o\left( 1\right) $ as $%
\mathcal{S}_{1n\mathbf{j},11}=n^{1/2}h^{-d/2}E\left[ K_{ix,\mathbf{j}}\xi
_{1i}\right] =O\left( n^{1/2}h^{d/2}h^{p+1}\right) $ $=O\left( 1\right) .$
By straightforward calculations and the IID assumption, we can readily show
that Var$\left( \mathcal{S}_{1n\mathbf{j},1}\right) =o\left( 1\right) .$
Therefore, $\mathcal{S}_{1n\mathbf{j},1}=o_{p}\left( 1\right) $ by the
Chebyshev inequality.\medskip\ 

Now, we show that $\mathcal{S}_{1n\mathbf{j},2}=o_{p}\left( 1\right) .$
Decompose $\mathcal{S}_{1n\mathbf{j},2}=\mathcal{S}_{1n\mathbf{j},21}+%
\mathcal{S}_{1n\mathbf{j},22},$ where $\mathcal{S}_{1n\mathbf{j},21}\equiv 
\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}$ $K_{ix,\mathbf{j}}\varphi
_{i}q_{2,i}\left( \mathbf{\beta }^{0}\right) g_{b}\left( f_{i}\right) \left( 
\tilde{f}_{i}\left( \overrightarrow{\varepsilon }_{i}\right) -\overline{f}%
\left( \bar{\varepsilon}_{i}\right) \right) ,$ and $\mathcal{S}_{1n\mathbf{j}%
,22}\equiv \frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi
_{i}q_{2,i}\left( \mathbf{\beta }^{0}\right) g_{b}\left( f_{i}\right) \left( 
\overline{f}\left( \bar{\varepsilon}_{i}\right) -f\left( \bar{\varepsilon}%
_{i}\right) \right) .$ It suffices to show that $\mathcal{S}_{1n\mathbf{j}%
,2s}=o_{p}\left( 1\right) ,$ $s=1,2.$ For $\mathcal{S}_{1n\mathbf{j},21},$
by a Taylor expansion and (\ref{r1ij1})-(\ref{r1ij2}), we have 
\begin{eqnarray*}
\mathcal{S}_{1n\mathbf{j},21} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) \frac{1}{nh_{0}%
}\sum_{j\neq i}\left[ k_{0}\left( \frac{\overrightarrow{\varepsilon }_{i}-%
\tilde{\varepsilon}_{j}}{h_{0}}\right) -k_{0}\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \right]  \\
&=&\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime
}\left( \frac{\varepsilon _{i}-\varepsilon _{j}}{h_{0}}\right) \left( 
\overrightarrow{\varepsilon }_{i}-\tilde{\varepsilon}_{j}-\bar{\varepsilon}%
_{i}+\varepsilon _{j}\right) +o_{p}\left( 1\right)  \\
&=&-\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{%
v_{1i}\left( x\right) }{\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{1/2}} \\
&&+\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{%
\tilde{m}\left( X_{j}\right) -m\left( X_{j}\right) }{\sigma \left(
X_{j}\right) } \\
&&+\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right)
\varepsilon _{j}\frac{\tilde{\sigma}\left( X_{j}\right) -\sigma \left(
X_{j}\right) }{\tilde{\sigma}\left( X_{j}\right) } \\
&&+\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,%
\mathbf{j}}\varphi _{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{%
u_{i}+\delta _{i}}{\varphi (P_{i}(\mathbf{\beta }_{2}^{0}))^{1/2}}\frac{%
\varphi (P_{i}(\mathbf{\beta }_{2}^{0}))^{1/2}-\varphi (P_{i}(\mathbf{\tilde{%
\beta}}_{2}))^{1/2}}{\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{1/2}} \\
&&+o_{p}\left( 1\right)  \\
&\equiv &-\mathcal{S}_{1n\mathbf{j},211}+\mathcal{S}_{1n\mathbf{j},212}+%
\mathcal{S}_{1n\mathbf{j},213}+\mathcal{S}_{1n\mathbf{j},214}+o_{p}\left(
1\right) .
\end{eqnarray*}%
For the first term, by Lemma A.2 and the fact that $v_{1i}\left( x\right)
=O_{p}\left( \upsilon _{2n}\right) ,$ $\varphi (P_{i}(\mathbf{\tilde{\beta}}%
_{2}))=\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )+O_{p}\left(
\upsilon _{2n}\right) $ uniformly on the set $\left\{ K_{ix}>0\right\} $, we
have%
\begin{eqnarray*}
\left\vert \mathcal{S}_{1n\mathbf{j},211}\right\vert  &=&\left\vert \frac{1}{%
2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\bar{\rho}%
_{i}g_{b}\left( f_{i}\right) \bar{f}_{i}^{\prime }\left( \varepsilon
_{i}\right) \frac{v_{1i}\left( x\right) }{\varphi (P_{i}(\mathbf{\tilde{\beta%
}}_{2}))^{1/2}}\right\vert  \\
&=&\left\vert \frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi
_{i}\bar{\rho}_{i}g_{b}\left( f_{i}\right) f^{\prime }\left( \varepsilon
_{i}\right) \frac{v_{1i}\left( x\right) }{\varphi (P_{i}\left( \mathbf{\beta 
}_{2}^{0}\right) )^{1/2}}\right\vert +o_{p}\left( 1\right)  \\
&\leq &\underset{\left\{ K_{ix}>0\right\} }{\max }\left\vert v_{1i}\left(
x\right) \right\vert \frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\left\vert K_{ix,%
\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{-1/2}\bar{\rho}_{i}g_{b}\left( f_{i}\right) f^{\prime }\left( \varepsilon
_{i}\right) \right\vert +o_{p}\left( 1\right) .
\end{eqnarray*}%
The first term in the last expression is $o_{p}\left( 1\right) $ if $%
n^{1/2}h^{-d/2}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}\bar{\rho}_{i}g_{b}\left(
f_{i}\right) f^{\prime }\left( \varepsilon _{i}\right) \right\vert =o\left(
\upsilon _{2n}^{-1}\right) $ by Markov inequality. Note that 
\begin{equation}
\bar{\varepsilon}_{i}-\varepsilon _{i}=\{\varepsilon _{i}[\sigma \left(
X_{i}\right) -\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{1/2}]+\delta _{i}\}/\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{1/2}=\varepsilon _{i}d_{i}+\overline{\delta }_{i}.  \label{epsilon1}
\end{equation}%
where $d_{i}\equiv \sigma \left( X_{i}\right) \varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1/2}-1=O_{p}\left( h^{p+1}\right) $ \ and $%
\overline{\delta }_{i}\equiv \delta _{i}\varphi (P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) )^{-1/2}=O_{p}\left( h^{p+1}\right) $ uniformly on the set $%
\left\{ K_{ix}>0\right\} $. Then by the triangle inequality, 
\begin{eqnarray*}
&&h^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}\bar{\rho}_{i}g_{b}\left(
f_{i}\right) f^{\prime }\left( \varepsilon _{i}\right) \right\vert  \\
&=&h^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}\left[ \bar{\rho}_{i}g_{b}\left(
f_{i}\right) f^{\prime }\left( \varepsilon _{i}\right) |X_{i}\right]
\right\vert  \\
&=&h^{-d}E\left\vert \frac{K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}}{d_{i}+1}\int_{b\leq f\left(
\varepsilon \right) \leq 2b}[\psi \left( \varepsilon \right) \varepsilon
+1]f^{\prime }\left( \varepsilon \right) g_{b}\left( f\left( \varepsilon
\right) \right) f\left( \frac{\varepsilon -\overline{\delta }_{i}}{d_{i}+1}%
\right) d\varepsilon \right\vert  \\
&\leq &h^{-d}E\left\vert \frac{K_{ix,\mathbf{j}}\varphi _{i}\varphi
(P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}}{d_{i}+1}\int_{b\leq
f\left( \varepsilon \right) \leq 2b}\rho \left( \varepsilon \right)
f^{\prime }\left( \varepsilon \right) g_{b}\left( f\left( \varepsilon
\right) \right) f\left( \varepsilon \right) d\varepsilon \right\vert  \\
&&+h^{-d}E\left\vert \frac{K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}}{d_{i}+1}\int_{b\leq f\left(
\varepsilon \right) \leq 2b}\rho \left( \varepsilon \right) f^{\prime
}\left( \varepsilon \right) g_{b}\left( f\left( \varepsilon \right) \right) %
\left[ f\left( \frac{\varepsilon -\overline{\delta }_{i}}{d_{i}+1}\right)
-f\left( \varepsilon \right) \right] d\varepsilon \right\vert  \\
&\equiv &S_{n1}+S_{n2},\text{ say.}
\end{eqnarray*}%
For $S_{n1},$ we have%
\begin{eqnarray*}
S_{n1} &=&h^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi
(P_{i}\left( \mathbf{\beta }_{2}^{0}\right) )^{-1/2}\left( d_{i}+1\right)
^{-1}\int_{b\leq f\left( \varepsilon \right) \leq 2b}\rho \left( \varepsilon
\right) f^{\prime }\left( \varepsilon \right) g_{b}\left( f\left(
\varepsilon \right) \right) f\left( \varepsilon \right) d\varepsilon
\right\vert  \\
&\leq &\underset{b\leq f\left( \varepsilon \right) \leq 2b}{\sup }[f\left(
\varepsilon \right) g_{b}\left( f\left( \varepsilon \right) \right)
]h^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )^{-1/2}\left( d_{i}+1\right) ^{-1}\int_{b\leq
f\left( \varepsilon \right) \leq 2b}\rho \left( \varepsilon \right)
f^{\prime }\left( \varepsilon \right) d\varepsilon \right\vert  \\
&\leq &Ch^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}\left( d_{i}+1\right)
^{-1}\right\vert \left\vert \int_{b\leq f\left( \varepsilon \right) \leq
2b}\rho \left( \varepsilon \right) f^{\prime }\left( \varepsilon \right)
d\varepsilon \right\vert  \\
&\leq &Ch^{-d}E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\varphi (P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) )^{-1/2}\left( d_{i}+1\right)
^{-1}\right\vert \left\{ \int_{b\leq f\left( \varepsilon \right) \leq
2b}\rho \left( \varepsilon \right) ^{2}f\left( \varepsilon \right)
d\varepsilon \int_{b\leq f\left( \varepsilon \right) \leq 2b}\psi \left(
\varepsilon \right) f\left( \varepsilon \right) d\varepsilon \right\} ^{1/2}
\\
&=&O\left( h^{\epsilon }\right) 
\end{eqnarray*}%
where the third inequality follows from the H\"{o}lder inequality and the
independence between $X_{i}$ and $\varepsilon _{i}$. By a Taylor expansion, $%
f\left( \frac{\varepsilon -\overline{\delta }_{i}}{1+d_{i}}\right) -f\left(
\varepsilon \right) $ $\simeq -f^{\prime }\left( \varepsilon \right) \left( 
\overline{\delta }_{i}+d_{i}\varepsilon \right) .$ With this, we can readily
show that $S_{n2}=O\left( h^{\epsilon }\right) .$ Consequently, $\left\vert 
\mathcal{S}_{1n\mathbf{j},211}\right\vert =O_{p}(\upsilon _{2n}\sqrt{nh^{d}}%
h^{\epsilon })=o_{p}\left( 1\right) .$

For $\mathcal{S}_{1n\mathbf{j},212},$ using (\ref{A2}) we can write 
\begin{eqnarray}
\mathcal{S}_{1n\mathbf{j},212} &=&\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}%
\sum_{i=1}^{n}\sum_{j\neq i}K_{ix,\mathbf{j}}\varphi _{i}\bar{\rho}%
_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{\tilde{m}\left( X_{j}\right)
-m\left( X_{j}\right) }{\sigma \left( X_{j}\right) }  \notag \\
&=&\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}\frac{K_{ix,%
\mathbf{j}}\varphi _{i}}{\sigma \left( X_{j}\right) }\bar{\rho}%
_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) e_{1}^{^{\top }}M_{n}^{-1}\left(
X_{j}\right) U_{1,n}\left( X_{j}\right)   \notag \\
&&-\frac{1}{2n^{3/2}h^{d/2}h_{0}^{2}}\sum_{i=1}^{n}\sum_{j\neq i}\frac{K_{ix,%
\mathbf{j}}\varphi _{i}}{\sigma \left( X_{j}\right) }\bar{\rho}%
_{i}g_{b}\left( f_{i}\right) k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) e_{1}^{^{\top }}M_{n}^{-1}\left(
X_{j}\right) B_{1,n}\left( X_{j}\right)   \notag \\
&\equiv &\mathcal{S}_{1n\mathbf{j},212a}+\mathcal{S}_{1n\mathbf{j},212b}.
\label{J12}
\end{eqnarray}%
Recall $\mathbf{\tilde{Z}}_{i}$ is defined analogously to $\mathbf{Z}_{i}$
with $h_{1}$ in place of $h.$ So $\mathcal{S}_{1n\mathbf{j},212a}$ can be
written as%
\begin{equation*}
\mathcal{S}_{1n\mathbf{j},212a}=\sum_{i=1}^{n}\sum_{j\neq i}\varsigma
_{2n}\left( \varepsilon _{i},\varepsilon _{j}\right)
+\sum_{i=1}^{n}\sum_{j\neq i}\sum_{l\neq i,l\neq j}\varsigma _{3n}\left(
\varepsilon _{i},\varepsilon _{j},\varepsilon _{l}\right) ,
\end{equation*}%
where $\varsigma _{2n}\left( \varepsilon _{i},\varepsilon _{j}\right) =\frac{%
1}{2n^{5/2}h^{d/2}h_{1}^{d}h_{0}^{2}}\frac{K_{ix,\mathbf{j}}\varphi _{i}}{%
\sigma \left( X_{j}\right) }\bar{\rho}_{i}g_{b}\left( f_{i}\right)
k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}%
\right) e_{1}^{^{\top }}M_{n}^{-1}\left( X_{j}\right) \mathbf{\tilde{Z}}%
_{i}K\left( \frac{X_{j}-X_{i}}{h_{1}}\right) u_{i}$ and $\varsigma
_{3n}(\varepsilon _{i},\varepsilon _{j},\varepsilon _{l})$ $=\frac{1}{%
2n^{5/2}h^{d/2}h_{1}^{d}h_{0}^{2}}\frac{K_{ix,\mathbf{j}}\varphi _{i}}{%
2\sigma \left( X_{j}\right) }\bar{\rho}_{i}g_{b}\left( f_{i}\right)
k_{0}^{\prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}%
\right) e_{1}^{^{\top }}M_{n}^{-1}\left( X_{j}\right) \mathbf{\tilde{Z}}%
_{i}K\left( \frac{X_{j}-X_{l}}{h_{1}}\right) u_{l}$. Let $\mathbb{X}\equiv
\{X_{1},...,X_{n}\}.$ Then $E[\mathcal{S}_{1n\mathbf{j},212a}|\mathbb{X}]=$ $%
\sum_{i=1}^{n}\sum_{j\neq i}E\left[ \varsigma _{2n}\left( z_{i},z_{j}\right)
|\mathbb{X}\right] =O_{p}\left( n^{-1/2}h^{d/2}b^{-1}\right) =o_{p}\left(
1\right) .$ For the variance of $\mathcal{S}_{1n\mathbf{j},212a},$ it is
easy to show that 
\begin{eqnarray*}
\text{Var}\left[ \sum_{i=1}^{n}\sum_{j\neq i}\varsigma _{2n}\left(
\varepsilon _{i},\varepsilon _{j}\right) |\mathbb{X}\right]  &=&O\left(
n^{2}\right) E\left[ \varsigma _{2n}\left( \varepsilon _{i},\varepsilon
_{j}\right) ^{2}+\varsigma _{2n}\left( \varepsilon _{i},\varepsilon
_{j}\right) \varsigma _{2n}\left( \varepsilon _{j},\varepsilon _{i}\right) |%
\mathbb{X}\right]  \\
&&+O\left( n^{3}\right) E\left[ \varsigma _{2n}\left( \varepsilon
_{i},\varepsilon _{j}\right) \varsigma _{2n}\left( \varepsilon
_{l},\varepsilon _{j}\right) +\varsigma _{2n}\left( \varepsilon
_{i},\varepsilon _{j}\right) \varsigma _{2n}\left( \varepsilon
_{i},\varepsilon _{i}\right) |\mathbb{X}\right]  \\
&=&O_{p}\left( n^{-3}h^{-d-4}b^{-2}\right) +O_{p}\left( n^{-2}b^{-2}\right)
=o_{p}\left( 1\right) .
\end{eqnarray*}%
Similarly, one can show that $E\left( \varsigma _{3n}\left(
z_{i},z_{j},z_{l}\right) |\mathbb{X}\right) =0$ and Var$\left[
\sum_{i=1}^{n}\sum_{j\neq i}\sum_{l\neq i,l\neq j}\varsigma _{3n}\left(
z_{i},z_{j},z_{l}\right) |\mathbb{X}\right] =o_{p}\left( 1\right) .$
Consequently, $\mathcal{S}_{1n\mathbf{j},212a}=o_{p}\left( 1\right) $ by the
conditional Chebyshev inequality. For $\mathcal{S}_{1n\mathbf{j},212b},$ we
have $\mathcal{S}_{1n\mathbf{j},212b}=O_{p}(n^{1/2}h^{d/2}h_{1}^{p+1})=o_{p}%
\left( 1\right) .$ Thus we have shown that $\mathcal{S}_{1n\mathbf{j}%
,212}=o_{p}\left( 1\right) .$ By analogous arguments, Lemma A.1, and (\ref%
{fact3}), we can show that $\mathcal{S}_{1n\mathbf{j},21s}=o_{p}\left(
1\right) $ for $s=3,4.$ It follows that $\mathcal{S}_{1n\mathbf{j}%
,21}=o_{p}\left( 1\right) .$ \medskip\ 

For $\mathcal{S}_{1n\mathbf{j},22},$ we make the following decomposition: 
\begin{equation*}
\mathcal{S}_{1n\mathbf{j},22}=\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\varphi _{i}q_{2,i}\left( \mathbf{\beta }^{0}\right) g_{b}\left(
f_{i}\right) \{\mathcal{V}\left( \bar{\varepsilon}_{i}\right) +\mathcal{B}%
\left( \bar{\varepsilon}_{i}\right) \}\equiv \mathcal{S}_{1n\mathbf{j},221}+%
\mathcal{S}_{1n\mathbf{j},222},
\end{equation*}%
where 
\begin{eqnarray}
\mathcal{V}\left( \bar{\varepsilon}_{i}\right) &=&\frac{1}{nh_{0}}%
\sum_{j\neq i}\left\{ k_{0}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon
_{j}}{h_{0}}\right) -E_{j}\left[ k_{0}\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \right] \right\} ,  \label{V1} \\
\mathcal{B}\left( \bar{\varepsilon}_{i}\right) &=&\frac{1}{nh_{0}}%
\sum_{j\neq i}E_{j}\left[ k_{0}\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \right] -f\left( \bar{\varepsilon}%
_{i}\right) ,  \label{B1}
\end{eqnarray}%
and $E_{j}$ indicates expectation with respect to the variable indexed by $%
j. $ Writing $\mathcal{S}_{1n\mathbf{j},221}$ as a second order degenerate
statistic we verify that $E\left[ \mathcal{S}_{1n\mathbf{j},221}\right]
^{2}=o\left( 1\right) $ and thus $\mathcal{S}_{1n\mathbf{j},221}=o_{p}\left(
1\right) .$ For $\mathcal{S}_{1n\mathbf{j},222},$ we verify that $\mathcal{S}%
_{1n\mathbf{j},222}=O_{p}(n^{1/2}h^{d/2}h_{0}^{p+1})=o_{p}\left( 1\right) .$
Consequently, $\mathcal{S}_{1n\mathbf{j},22}=o_{p}\left( 1\right) .$ This
concludes the proof of the lemma. ${\tiny \blacksquare }$\medskip

\noindent \textbf{Proof of Lemma B.5. }By a geometric expansion: $\tilde{f}%
_{i}=f^{-1}-(\tilde{f}_{i}-f)/f^{2}+(\tilde{f}_{i}-f)^{2}/(f^{2}\tilde{f}%
_{i})$ where $\tilde{f}_{i}=\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right)
,$ we have%
\begin{eqnarray*}
\mathcal{S}_{2n\mathbf{j}} &=&\frac{1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\left\{ \tilde{G}_{i}\left[ \tilde{q}_{2,i}\left( \mathbf{\beta }%
^{0}\right) -q_{2,i}\left( \mathbf{\beta }^{0}\right) \right] \right\}  \\
&=&-\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\frac{%
\tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) -f^{\prime
}\left( \bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }%
\bar{\varepsilon}_{i}\tilde{G}_{i} \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\frac{%
\tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \left[ \tilde{f}%
_{i}\left( \bar{\varepsilon}_{i}\right) -f\left( \bar{\varepsilon}%
_{i}\right) \right] }{f\left( \bar{\varepsilon}_{i}\right) }\bar{\varepsilon}%
_{i}\tilde{G}_{i} \\
&&-\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\frac{%
\tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \left[ \tilde{f}%
_{i}\left( \bar{\varepsilon}_{i}\right) -f\left( \bar{\varepsilon}%
_{i}\right) \right] ^{2}}{f^{2}\left( \bar{\varepsilon}_{i}\right) \tilde{f}%
_{i}\left( \bar{\varepsilon}_{i}\right) }\bar{\varepsilon}_{i}\tilde{G}_{i}
\\
&\equiv &-\mathcal{S}_{2n\mathbf{j},1}+\mathcal{S}_{2n\mathbf{j},2}-\mathcal{%
S}_{2n\mathbf{j},3}.
\end{eqnarray*}%
where recall $\varphi _{i}\equiv \varphi ^{\prime }(P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) )/\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
).$ It suffices to show that each of these three terms is $o_{p}\left(
1\right) .$ For $\mathcal{S}_{2n\mathbf{j},1},$ noticing that $G_{b}(\tilde{f%
}_{i})-G_{b}\left( f_{i}\right) $ $=g_{b}\left( f_{i}\right) (\tilde{f}%
_{i}-f_{i})+\frac{1}{2}g_{b}^{\prime }\left( f_{i}^{\ast }\right) (\tilde{f}%
_{i}-f_{i})^{2}$, we can apply Lemma A.2 and show that 
\begin{eqnarray*}
\mathcal{S}_{2n\mathbf{j},1} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\varphi _{i}\frac{\tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}%
_{i}\right) -\overline{f}^{\prime }\left( \bar{\varepsilon}_{i}\right) }{%
f\left( \bar{\varepsilon}_{i}\right) }\bar{\varepsilon}_{i}G_{b}\left(
f_{i}\right)  \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\frac{%
\overline{f}^{\prime }\left( \bar{\varepsilon}_{i}\right) -f^{\prime }\left( 
\bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\bar{%
\varepsilon}_{i}G_{b}\left( f_{i}\right) +o_{p}\left( 1\right)  \\
&\equiv &\mathcal{S}_{2n\mathbf{j},11}+\mathcal{S}_{2n\mathbf{j}%
,12}+o_{p}\left( 1\right) .
\end{eqnarray*}%
For the first term, we have 
\begin{eqnarray*}
\mathcal{S}_{2n\mathbf{j},11} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{%
K_{ix,\mathbf{j}}\varphi _{i}}{f\left( \bar{\varepsilon}_{i}\right) }\frac{1%
}{nh_{0}^{2}}\sum_{j\neq i}\left[ k_{0}^{\prime }\left( \frac{\bar{%
\varepsilon}_{i}-\tilde{\varepsilon}_{j}}{h_{0}}\right) -k_{0}^{\prime
}\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \right] 
\bar{\varepsilon}_{i}G_{b}\left( f_{i}\right)  \\
&=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\frac{1}{nh_{0}^{3}}\sum_{j\neq
i}k_{0}^{\prime \prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{%
h_{0}}\right) \left( \tilde{\varepsilon}_{j}-\varepsilon _{j}\right) \bar{%
\varepsilon}_{i}G_{b}\left( f_{i}\right) +o_{p}\left( 1\right)  \\
&=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\frac{1}{nh_{0}^{3}}\sum_{j\neq
i}k_{0}^{\prime \prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{%
h_{0}}\right) \frac{m\left( X_{j}\right) -\tilde{m}\left( X_{j}\right) }{%
\sigma \left( X_{j}\right) }\bar{\varepsilon}_{i}G_{b}\left( f_{i}\right)  \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\frac{1}{nh_{0}^{3}}\sum_{j\neq
i}k_{0}^{\prime \prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{%
h_{0}}\right) \varepsilon _{j}\frac{\sigma \left( X_{j}\right) -\tilde{\sigma%
}\left( X_{j}\right) }{\sigma \left( X_{j}\right) }\bar{\varepsilon}%
_{i}G_{b}\left( f_{i}\right) +o_{p}\left( 1\right)  \\
&\equiv &\mathcal{S}_{2n\mathbf{j},111}+\mathcal{S}_{2n\mathbf{j}%
,112}+o_{p}\left( 1\right) ,\text{ say.}
\end{eqnarray*}%
Write 
\begin{eqnarray*}
\mathcal{S}_{2n\mathbf{j},111} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}%
\frac{K_{ix,\mathbf{j}}\varphi _{i}}{f\left( \bar{\varepsilon}_{i}\right) }%
\frac{1}{nh_{0}^{3}}\sum_{j\neq i}k_{0}^{\prime \prime }\left( \frac{\bar{%
\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \frac{e_{1}^{^{\top
}}M_{n}^{-1}\left( X_{j}\right) U_{1,n}\left( X_{j}\right) }{\sigma \left(
X_{j}\right) }\bar{\varepsilon}_{i}G_{b}\left( f_{i}\right)  \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\frac{1}{nh_{0}^{3}}\sum_{j\neq
i}k_{0}^{\prime \prime }\left( \frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{%
h_{0}}\right) \frac{e_{1}^{^{\top }}M_{n}^{-1}\left( X_{j}\right)
B_{1,n}\left( X_{j}\right) }{\sigma \left( X_{j}\right) }\bar{\varepsilon}%
_{i}G_{b}\left( f_{i}\right)  \\
&\equiv &\mathcal{S}_{2n\mathbf{j},111a}+\mathcal{S}_{2n\mathbf{j},111b}.
\end{eqnarray*}%
Writing $\mathcal{S}_{2n\mathbf{j},111a}$ as a third order $U$-statistic, we
can show that $\mathcal{S}_{2n\mathbf{j},111a}=O_{p}(h^{d/2})=$ $o_{p}\left(
1\right) $ by conditional moment calculations and conditional Chebyshev
inequality. For $\mathcal{S}_{2n\mathbf{j},111b},$ we have $\mathcal{S}_{2n%
\mathbf{j},111b}=O_{p}(\sqrt{nh^{d}}h_{1}^{p+1})=o_{p}\left( 1\right) .$
Similarly, we can verify that $\mathcal{S}_{2n\mathbf{j},112}=o_{p}\left(
1\right) .$ Consequently $\mathcal{S}_{2n\mathbf{j},11}=o_{p}\left( 1\right)
.$ For $\mathcal{S}_{2n\mathbf{j},12},$ we have 
\begin{eqnarray*}
\mathcal{S}_{2n\mathbf{j},12} &=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{%
K_{ix,\mathbf{j}}\varphi _{i}}{f\left( \bar{\varepsilon}_{i}\right) }\frac{%
\overline{f}^{\prime }\left( \bar{\varepsilon}_{i}\right) -f^{\prime }\left( 
\bar{\varepsilon}_{i}\right) }{f\left( \bar{\varepsilon}_{i}\right) }\bar{%
\varepsilon}_{i}G_{b}\left( f_{i}\right)  \\
&=&\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\left\{ \frac{1}{nh_{0}^{2}}%
\sum_{j\neq i}\left\{ k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) -E_{j}\left[ k_{0}^{\prime }\left( 
\frac{\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \right] \right\}
\right\} \bar{\varepsilon}_{i}G_{b}\left( f_{i}\right)  \\
&&+\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{K_{ix,\mathbf{j}}\varphi _{i}%
}{f\left( \bar{\varepsilon}_{i}\right) }\left\{ \frac{1}{nh_{0}^{2}}%
\sum_{j\neq i}E_{j}\left[ k_{0}^{\prime }\left( \frac{\bar{\varepsilon}%
_{i}-\varepsilon _{j}}{h_{0}}\right) \right] -f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) \right\} \bar{\varepsilon}_{i}G_{b}\left(
f_{i}\right)  \\
&=&\mathcal{S}_{2n\mathbf{j},121}+\mathcal{S}_{2n\mathbf{j},122},
\end{eqnarray*}%
where $E_{j}$ indicates expectation with respect to the variable indexed by $%
j.$ Noting $\mathcal{S}_{2n\mathbf{j},121}$ is a second order statistic, it
is easy to verify that $E\left[ \mathcal{S}_{2n\mathbf{j},121}\right]
^{2}=O(h^{d})=o\left( 1\right) ,$ implying that $\mathcal{S}_{2n\mathbf{j}%
,121}=o_{p}\left( 1\right) .$ For $\mathcal{S}_{2n\mathbf{j},122},$ noticing
that 
\begin{equation*}
\frac{1}{nh_{0}^{2}}\sum_{j\neq i}E_{j}\left[ k_{0}^{\prime }\left( \frac{%
\bar{\varepsilon}_{i}-\varepsilon _{j}}{h_{0}}\right) \right] -f^{\prime
}\left( \bar{\varepsilon}_{i}\right) =h_{0}^{p+1}f^{\left( p+2\right)
}\left( \bar{\varepsilon}_{i}\right) \int k_{0}\left( u\right) u^{p+1}du,
\end{equation*}%
we can show that $\mathcal{S}_{2n\mathbf{j},122}=O_{p}(\sqrt{nh^{d}}%
h_{0}^{p+1})=o_{p}\left( 1\right) .$ Consequently, $\mathcal{S}_{2n\mathbf{j}%
,12}=o_{p}\left( 1\right) $ and $\mathcal{S}_{2n\mathbf{j},1}=o_{p}\left(
1\right) .$

For $\mathcal{S}_{2n\mathbf{j},2},$ we can easily show that 
\begin{equation*}
\mathcal{S}_{2n\mathbf{j},2}=\frac{1}{2\sqrt{nh^{d}}}\sum_{i=1}^{n}\frac{%
K_{ix,\mathbf{j}}\varphi _{i}f^{\prime }\left( \bar{\varepsilon}_{i}\right) 
}{f\left( \bar{\varepsilon}_{i}\right) }\left[ \tilde{f}_{i}\left( \bar{%
\varepsilon}_{i}\right) -f\left( \bar{\varepsilon}_{i}\right) \right] \bar{%
\varepsilon}_{i}G_{b}\left( f_{i}\right) +o_{p}\left( 1\right) .
\end{equation*}%
The rest of the proof is similar to that of $\mathcal{S}_{2n\mathbf{j},1}$
and thus omitted. For $\mathcal{S}_{2n\mathbf{j},3},$ by Lemma A.2, $%
\mathcal{S}_{2n\mathbf{j},3}=O_{p}(\sqrt{nh^{d}}b^{-2}\nu
_{3n}^{2})=o_{p}\left( 1\right) .$ This concludes the proof of the lemma. $%
{\tiny \blacksquare }$\medskip 

\noindent \textbf{Proof of Lemma B.6. }Write $\mathcal{S}_{3n\mathbf{j}}=%
\frac{1}{2}\{\mathcal{S}_{3n\mathbf{j,}1}-\mathcal{S}_{3n\mathbf{j,}2}+%
\mathcal{S}_{3n\mathbf{j,}3}-\mathcal{S}_{3n\mathbf{j,}4}\},$ where%
\begin{eqnarray*}
\mathcal{S}_{3n\mathbf{j,}1} &\equiv &\frac{1}{\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left\{ \log \left( \tilde{f}_{i}\left( 
\overrightarrow{\varepsilon }_{i}\right) \right) g_{b}\left( \tilde{f}%
_{i}\left( \overrightarrow{\varepsilon }_{i}\right) \right) \tilde{f}%
_{i}^{\prime }\left( \overrightarrow{\varepsilon }_{i}\right) 
\overrightarrow{\varepsilon }_{i}\varphi _{i}(\mathbf{\tilde{\beta}}%
_{2})\right.  \\
&&-\left. \log \left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right)
\right) g_{b}\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) \right) 
\tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}\varphi _{i}\right\} , \\
\mathcal{S}_{3n\mathbf{j,}2} &\equiv &\frac{1}{2\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left\{ \log \varphi \left( P_{i}(\mathbf{%
\tilde{\beta}}_{2})\right) g_{b}\left( \tilde{f}_{i}\left( \overrightarrow{%
\varepsilon }_{i}\right) \right) \tilde{f}_{i}^{\prime }\left( 
\overrightarrow{\varepsilon }_{i}\right) \overrightarrow{\varepsilon }%
_{i}\varphi _{i}(\mathbf{\tilde{\beta}}_{2})\right.  \\
&&-\left. \log \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right)
g_{b}\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) \right) \tilde{f%
}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}\varphi _{i}\right\} , \\
\mathcal{S}_{3n\mathbf{j,}3} &\equiv &\frac{1}{\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\left\{ \log \left( \tilde{f}_{i}\left( \bar{%
\varepsilon}_{i}\right) \right) g_{b}\left( \tilde{f}_{i}\left( \bar{%
\varepsilon}_{i}\right) \right) \tilde{f}_{i}^{\prime }\left( \bar{%
\varepsilon}_{i}\right) -\log \left( f\left( \bar{\varepsilon}_{i}\right)
\right) g_{b}\left( f\left( \bar{\varepsilon}_{i}\right) \right) f^{\prime
}\left( \bar{\varepsilon}_{i}\right) \right\} \bar{\varepsilon}_{i}\varphi
_{i}, \\
\mathcal{S}_{3n\mathbf{j,}4} &\equiv &\frac{1}{2\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\log \varphi \left( P_{i}\left( \mathbf{\beta 
}_{2}^{0}\right) \right) \left\{ g_{b}\left( \tilde{f}_{i}\left( \bar{%
\varepsilon}_{i}\right) \right) \tilde{f}_{i}^{\prime }\left( \bar{%
\varepsilon}_{i}\right) -g_{b}\left( f\left( \bar{\varepsilon}_{i}\right)
\right) f^{\prime }\left( \bar{\varepsilon}_{i}\right) \right\} \bar{%
\varepsilon}_{i}\varphi _{i}.
\end{eqnarray*}%
where $\varphi _{i}\left( \mathbf{\beta }_{2}\right) \equiv \varphi ^{\prime
}(P_{i}\left( \mathbf{\beta }_{2}\right) )/\varphi (P_{i}\left( \mathbf{%
\beta }_{2}\right) )$ and $\varphi _{i}=\varphi _{i}\left( \mathbf{\beta }%
_{2}^{0}\right) .$ We will only show that $\mathcal{S}_{3n\mathbf{j,}%
1}=o_{p}\left( 1\right) $ since the proofs of $\mathcal{S}_{3n\mathbf{j,}%
s}=o_{p}\left( 1\right) $ for $s=2,3,4$ are similar.

For $\mathcal{S}_{3n\mathbf{j,}1},$ noticing that $\tilde{v}_{2i}\left(
x\right) =(\varphi (P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
)^{1/2}-\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{1/2})/\varphi (P_{i}(%
\mathbf{\tilde{\beta}}_{2}))^{1/2}$ and $\tilde{v}_{1i}\left( x\right)
=v_{1i}\left( x\right) $ $/\varphi (P_{i}(\mathbf{\tilde{\beta}}_{2}))^{1/2}$
are both $O_{p}\left( \upsilon _{2n}\right) $ uniformly in $i$ on the set $%
\left\{ K_{ix}>0\right\} ,$ and $\overrightarrow{\varepsilon }_{i}-\bar{%
\varepsilon}_{i}=\bar{\varepsilon}_{i}\tilde{v}_{2i}\left( x\right) -\tilde{v%
}_{1i}\left( x\right) ,$ we can show that 
\begin{eqnarray*}
\mathcal{S}_{3n\mathbf{j,}1} &=&\frac{1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,%
\mathbf{j}}\varphi _{i}\left\{ \tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) g_{b}\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right)
\right) \tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{%
\varepsilon}_{i}\left\{ \bar{\varepsilon}_{i}\tilde{v}_{2i}\left( x\right) -%
\tilde{v}_{1i}\left( x\right) \right\} \right.  \\
&&+\frac{1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\log
\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) \right)
g_{b}^{\prime }\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right)
\right) \tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{%
\varepsilon}_{i}\left\{ \bar{\varepsilon}_{i}\tilde{v}_{2i}\left( x\right) -%
\tilde{v}_{1i}\left( x\right) \right\}  \\
&&+\frac{1}{\sqrt{nh^{d}}}\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\log
\left( \tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) \right) g_{b}\left( 
\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) \right) \tilde{f}%
_{i}^{\prime \prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}\left\{ \bar{\varepsilon}_{i}\tilde{v}_{2i}\left( x\right) -\tilde{v}%
_{1i}\left( x\right) \right\} +o_{p}\left( 1\right)  \\
&\equiv &\mathcal{S}_{3n\mathbf{j,}11}+\mathcal{S}_{3n\mathbf{j,}12}+%
\mathcal{S}_{3n\mathbf{j,}13}+o_{p}\left( 1\right) .
\end{eqnarray*}%
By Lemma A.1, \ we can show%
\begin{eqnarray}
\mathcal{S}_{3n\mathbf{j,}11} &\simeq &\frac{1}{\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\psi \left( \bar{\varepsilon}%
_{i}\right) g_{b}\left( f\left( \bar{\varepsilon}_{i}\right) \right)
f^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}\left\{ 
\bar{\varepsilon}_{i}\tilde{v}_{2i}\left( x\right) -\tilde{v}_{1i}\left(
x\right) \right\} ,  \label{J311} \\
\mathcal{S}_{3n\mathbf{j,}12} &\simeq &\frac{1}{\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\log \left( f\left( \bar{%
\varepsilon}_{i}\right) \right) g_{b}^{\prime }\left( f\left( \bar{%
\varepsilon}_{i}\right) \right) f^{\prime }\left( \bar{\varepsilon}%
_{i}\right) \bar{\varepsilon}_{i}\left\{ \bar{\varepsilon}_{i}\tilde{v}%
_{2i}\left( x\right) -\tilde{v}_{1i}\left( x\right) \right\} ,  \label{J312}
\\
\mathcal{S}_{3n\mathbf{j,}13} &\simeq &\frac{1}{\sqrt{nh^{d}}}%
\sum_{i=1}^{n}K_{ix,\mathbf{j}}\varphi _{i}\log \left( f\left( \bar{%
\varepsilon}_{i}\right) \right) g_{b}\left( f\left( \bar{\varepsilon}%
_{i}\right) \right) f^{\prime \prime }\left( \bar{\varepsilon}_{i}\right) 
\bar{\varepsilon}_{i}\left\{ \bar{\varepsilon}_{i}\tilde{v}_{2i}\left(
x\right) -\tilde{v}_{1i}\left( x\right) \right\} .  \label{J313}
\end{eqnarray}%
The rest of the proof relies on the repeated applications of the dominated
convergence arguments. For example, the right hand side of (\ref{J311}) is
smaller than 
\begin{eqnarray*}
&&\frac{1}{\sqrt{nh^{d}}}\underset{\left\{ K_{ix}>0\right\} }{\max }%
\left\vert \tilde{v}_{2i}\left( x\right) \right\vert
\sum_{i=1}^{n}\left\vert K_{ix,\mathbf{j}}\varphi _{i}\psi \left( \bar{%
\varepsilon}_{i}\right) g_{b}\left( f\left( \bar{\varepsilon}_{i}\right)
\right) f^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}^{2}\right\vert  \\
&&+\frac{1}{\sqrt{nh^{d}}}\underset{\left\{ K_{ix}>0\right\} }{\max }%
\left\vert \tilde{v}_{1i}\left( x\right) \right\vert
\sum_{i=1}^{n}\left\vert K_{ix,\mathbf{j}}\varphi _{i}\psi \left( \bar{%
\varepsilon}_{i}\right) g_{b}\left( f\left( \bar{\varepsilon}_{i}\right)
\right) f^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}\right\vert .
\end{eqnarray*}%
Noting that%
\begin{eqnarray*}
E\left\vert K_{ix,\mathbf{j}}\varphi _{i}\psi \left( \bar{\varepsilon}%
_{i}\right) g_{b}\left( f\left( \bar{\varepsilon}_{i}\right) \right)
f^{\prime }\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}%
_{i}^{r}\right\vert  &=&E\left[ \left\vert \frac{K_{ix,\mathbf{j}}\varphi
_{i}}{d_{i}+1}\int \psi \left( \varepsilon \right) g_{b}\left( f\left(
\varepsilon \right) \right) f^{\prime }\left( \varepsilon \right)
\varepsilon ^{r}\right\vert f\left( \frac{\varepsilon -\overline{\delta }_{i}%
}{d_{i}+1}\right) d\varepsilon \right]  \\
&\leq &\sup_{\varepsilon }\left[ g_{b}\left( f\left( \varepsilon \right)
\right) f\left( \varepsilon \right) \right] E\left\vert \frac{K_{ix,\mathbf{j%
}}\varphi _{i}}{d_{i}+1}\right\vert \int_{b\leq f\left( \varepsilon \right)
\leq 2b}\frac{f^{\prime }\left( \varepsilon \right) ^{2}}{f\left(
\varepsilon \right) }\left\vert \varepsilon ^{r}\right\vert d\varepsilon
+O\left( h^{\epsilon }\right)  \\
&\leq &Cf\left( \varepsilon \right) \int_{b\leq f\left( \varepsilon \right)
\leq 2b}\left\vert \frac{f^{\prime }\left( \varepsilon \right) ^{2}}{f\left(
\varepsilon \right) }\varepsilon ^{r}\right\vert d\varepsilon +O\left(
h^{\epsilon }\right)  \\
&\leq &C\int_{b\leq f\left( \varepsilon \right) \leq 2b}\left\vert \frac{%
f^{\prime }\left( \varepsilon \right) ^{2}}{f\left( \varepsilon \right) }%
\varepsilon ^{r}\right\vert d\varepsilon +O\left( h^{p+1}\right) =O\left(
b^{\left( \gamma -1\right) /(2\gamma )}+h^{\epsilon }\right) ,
\end{eqnarray*}%
where the last equality follows from similar argument to the proof of Lemma
B.3, we have $\mathcal{S}_{3n\mathbf{j,}11}=O_{p}(\upsilon _{2n}\sqrt{nh^{d}}%
(b^{\left( \gamma -1\right) /(2\gamma )}+h^{\epsilon }))=o_{p}\left(
1\right) .$ Similarly, we can show that $\mathcal{S}_{3n\mathbf{j,}%
1s}=o_{p}\left( 1\right) ,$ $s=2,3.$\textbf{\ }${\tiny \blacksquare }$%
\medskip 

\noindent \textbf{Proof of Lemma B.7. }Observe that 
\begin{eqnarray*}
\mathcal{R}_{1n} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}%
\right) \bar{H}^{-1}[\tilde{G}_{i}\tilde{s}_{i}\left( \mathbf{\beta }%
^{0}\right) \tilde{s}_{i}\left( \mathbf{\beta }^{0}\right) ^{\top
}-G_{i}s_{i}\left( \mathbf{\beta }^{0}\right) s_{i}\left( \mathbf{\beta }%
^{0}\right) ^{\top }]\otimes (\mathbf{\tilde{X}}_{i}\mathbf{\tilde{X}}%
_{i}^{\top })\bar{H}^{-1} \\
&=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}G_{i}[\tilde{s}_{i}\left( \mathbf{\beta }^{0}\right) \tilde{s}%
_{i}\left( \mathbf{\beta }^{0}\right) ^{\top }-s_{i}\left( \mathbf{\beta }%
^{0}\right) s_{i}\left( \mathbf{\beta }^{0}\right) ^{\top }]\otimes (\mathbf{%
\tilde{X}}_{i}\mathbf{\tilde{X}}_{i}^{\top })\bar{H}^{-1} \\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}\left( \tilde{G}_{i}-G_{i}\right) s_{i}\left( \mathbf{\beta }%
^{0}\right) s_{i}\left( \mathbf{\beta }^{0}\right) ^{\top }\otimes (\mathbf{%
\tilde{X}}_{i}\mathbf{\tilde{X}}_{i}^{\top })\bar{H}^{-1} \\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}\left( \tilde{G}_{i}-G_{i}\right) [\tilde{s}_{i}\left( \mathbf{\beta }%
^{0}\right) \tilde{s}_{i}\left( \mathbf{\beta }^{0}\right) ^{\top
}-s_{i}\left( \mathbf{\beta }^{0}\right) s_{i}\left( \mathbf{\beta }%
^{0}\right) ^{\top }]\otimes (\mathbf{\tilde{X}}_{i}\mathbf{\tilde{X}}%
_{i}^{\top })\bar{H}^{-1} \\
&\equiv &\mathcal{R}_{1n,1}+\mathcal{R}_{1n,2}+\mathcal{R}_{1n,3},\text{ say.%
}
\end{eqnarray*}%
It suffices to prove the lemma by showing that $\mathcal{R}%
_{1n,r}=o_{p}\left( 1\right) $ for $r=1,2,3.$ We only prove $\mathcal{R}%
_{1n,1}=o_{p}\left( 1\right) $ and $\mathcal{R}_{1n,2}=o_{p}\left( 1\right) $
as $\mathcal{R}_{1n,3}$ is a smaller order term and can be studied
analogously.

First, we show that $\mathcal{R}_{1n,1}=o_{p}\left( 1\right) .$ Note that 
\begin{equation*}
\tilde{s}_{i}\left( \mathbf{\beta }^{0}\right) \tilde{s}_{i}\left( \mathbf{%
\beta }^{0}\right) ^{^{\top }}=\left[ 
\begin{array}{cc}
\frac{\tilde{\psi}_{i}^{2}\left( \bar{\varepsilon}_{i}\right) }{\varphi
\left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) } & \frac{\varphi
^{\prime }\left( P_{i}\left( \mathbf{\beta }_{2}\right) \right) \tilde{\psi}%
_{i}\left( \bar{\varepsilon}_{i}\right) \left[ \tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] }{2\varphi \left(
P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{3/2}} \\ 
\frac{\varphi ^{\prime }\left( P_{i}\left( \mathbf{\beta }_{2}\right)
\right) \tilde{\psi}_{i}\left( \bar{\varepsilon}_{i}\right) \left[ \tilde{%
\psi}_{i}\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] 
}{2\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{3/2}}
& \frac{\left[ \varphi ^{\prime }\left( P_{i}\left( \mathbf{\beta }%
_{2}\right) \right) \right] ^{2}\left[ \tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] ^{2}}{4\varphi \left(
P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{2}}%
\end{array}%
\right] ,
\end{equation*}%
and $s_{i}\left( \mathbf{\beta }^{0}\right) s_{i}\left( \mathbf{\beta }%
^{0}\right) ^{\top }$ has a similar expression with $\psi \left( \bar{%
\varepsilon}_{i}\right) $ in the place of $\tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) .$ It follows that%
\begin{eqnarray*}
\mathcal{R}_{1n,1} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}%
\right) G_{i} \\
&&\times \left[ 
\begin{array}{cc}
\frac{\tilde{\psi}_{i}^{2}\left( \bar{\varepsilon}_{i}\right) -\psi
^{2}\left( \bar{\varepsilon}_{i}\right) }{\varphi \left( P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) \right) } & \frac{\varphi _{i}\left\{ \tilde{\psi}%
_{i}\left( \bar{\varepsilon}_{i}\right) \left[ \tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] -\psi \left( \bar{%
\varepsilon}_{i}\right) \left[ \psi \left( \bar{\varepsilon}_{i}\right) \bar{%
\varepsilon}_{i}+1\right] \right\} }{2\varphi \left( P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) \right) ^{1/2}} \\ 
\frac{\varphi _{i}\left\{ \tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) \left[ \tilde{\psi}_{i}\left( \bar{\varepsilon}_{i}\right) \bar{%
\varepsilon}_{i}+1\right] -\psi \left( \bar{\varepsilon}_{i}\right) \left[
\psi \left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right]
\right\} }{2\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
\right) ^{1/2}} & \frac{\varphi _{i}^{2}\left[ \tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] ^{2}-\left[ \psi
\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right] ^{2}}{4}%
\end{array}%
\right] \otimes (\mathbf{Z}_{i}\mathbf{Z}_{i}^{\top }) \\
&\equiv &\left[ 
\begin{array}{cc}
\mathcal{R}_{1n,1,11} & \mathcal{R}_{1n,1,12} \\ 
\mathcal{R}_{1n,1,21} & \mathcal{R}_{1n,1,22}%
\end{array}%
\right] ,\text{ say,}
\end{eqnarray*}%
where recall $\varphi _{i}=\varphi ^{\prime }\left( P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) \right) /\varphi \left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ,$ $\mathcal{R}_{1n,1,21}=\mathcal{R}%
_{1n,1,12}^{\top },$ and $\mathcal{R}_{1n,1,rs},$ $r,s=1,2,$ are all $%
N\times N$ matrices. We need to show that $\mathcal{R}_{1n,1,11},$ $\mathcal{%
R}_{1n,1,12}$ and $\mathcal{R}_{1n,1,22}$ are all $o_{p}\left( 1\right) .$
Noting that%
\begin{eqnarray*}
\tilde{\psi}_{i}^{2}\left( \bar{\varepsilon}_{i}\right) -\psi ^{2}\left( 
\bar{\varepsilon}_{i}\right)  &=&\frac{\tilde{f}_{i}^{\prime }\left( \bar{%
\varepsilon}_{i}\right) ^{2}f\left( \bar{\varepsilon}_{i}\right) ^{2}-\tilde{%
f}_{i}\left( \bar{\varepsilon}_{i}\right) ^{2}f^{\prime }\left( \bar{%
\varepsilon}_{i}\right) ^{2}}{\tilde{f}_{i}\left( \bar{\varepsilon}%
_{i}\right) ^{2}f\left( \bar{\varepsilon}_{i}\right) ^{2}} \\
&=&\frac{\left[ \tilde{f}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right)
^{2}-f_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) ^{2}\right] f\left( 
\bar{\varepsilon}_{i}\right) ^{2}+\left[ f\left( \bar{\varepsilon}%
_{i}\right) ^{2}-\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) ^{2}\right]
f_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) ^{2}}{\tilde{f}_{i}\left( 
\bar{\varepsilon}_{i}\right) ^{2}f\left( \bar{\varepsilon}_{i}\right) ^{2}},
\end{eqnarray*}%
we have%
\begin{eqnarray*}
\mathcal{R}_{1n,1,11} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}%
}{h}\right) G_{i}\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
\right) ^{-1}\left[ \tilde{\psi}_{i}^{2}\left( \bar{\varepsilon}_{i}\right)
-\psi ^{2}\left( \bar{\varepsilon}_{i}\right) \right] \mathbf{Z}_{i}\mathbf{Z%
}_{i}^{\top } \\
&=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right)
G_{i}\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{-1}%
\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) ^{-2}\left[ \tilde{f}%
_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) ^{2}-f_{i}^{\prime }\left( 
\bar{\varepsilon}_{i}\right) ^{2}\right] \mathbf{Z}_{i}\mathbf{Z}_{i}^{\top }
\\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right)
G_{i}\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{-1}%
\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right) ^{-2}\psi \left( \bar{%
\varepsilon}_{i}\right) ^{2}\left[ \tilde{f}_{i}\left( \bar{\varepsilon}%
_{i}\right) ^{2}-f_{i}\left( \bar{\varepsilon}_{i}\right) ^{2}\right] 
\mathbf{Z}_{i}\mathbf{Z}_{i}^{\top } \\
&\equiv &\mathcal{R}_{1n,1,11,a}+\mathcal{R}_{1n,1,11,b}\text{ say.}
\end{eqnarray*}%
Noting that $G_{i}\tilde{f}_{i}\left( \bar{\varepsilon}_{i}\right)
^{-2}=O\left( b^{-2}\right) ,$ by Lemma A.2, we have 
\begin{eqnarray*}
\left\Vert \mathcal{R}_{1n,1,11,a}\right\Vert  &\leq &O_{p}\left( \upsilon
_{3n,1}b^{-2}\right) \frac{1}{nh^{d}}\sum_{i=1}^{n}\left\Vert K\left( \frac{%
x-X_{i}}{h}\right) G_{i}\varphi \left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ^{-1}\mathbf{Z}_{i}\mathbf{Z}_{i}^{\top }\right\Vert 
\\
&=&O_{p}\left( \upsilon _{3n,1}b^{-2}\right) O_{p}\left( 1\right)
=O_{p}\left( \upsilon _{3n,1}b^{-2}\right) =o_{p}\left( 1\right) .
\end{eqnarray*}%
By the same token, $\left\vert \mathcal{R}_{1n,1,11,b}\right\vert
=o_{p}\left( 1\right) .$ Thus $\mathcal{R}_{1n,1,11}=o_{p}\left( 1\right) .$
Analogously, we can show $\mathcal{R}_{1n,1,12}=o_{p}\left( 1\right) $ and $%
\mathcal{R}_{1n,1,22}=o_{p}\left( 1\right) .$ Hence we have shown that $%
\mathcal{R}_{1n,1}=o_{p}\left( 1\right) .\medskip $

Now, we show that $\mathcal{R}_{1n,2}=o_{p}\left( 1\right) .$ By (\ref{GG})
and Markov inequality, we have%
\begin{eqnarray*}
\left\vert \mathcal{R}_{1n,2}\right\vert &\leq &O_{p}\left( h^{\epsilon
}\right) \frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right)
\left\Vert s_{i}\left( \mathbf{\beta }^{0}\right) s_{i}\left( \mathbf{\beta }%
^{0}\right) ^{\top }\otimes (\mathbf{Z}_{i}\mathbf{Z}_{i}^{\top })\right\Vert
\\
&=&O_{p}\left( h^{\epsilon }\right) O_{p}\left( 1\right) =O_{p}\left(
1\right) .
\end{eqnarray*}%
This completes the proof of the lemma. ${\tiny \blacksquare }$\medskip

\noindent \textbf{Proof of Lemma B.8. }Observe that%
\begin{eqnarray*}
\mathcal{R}_{2n} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}%
\right) \bar{H}^{-1}\left[ \tilde{G}_{i}\frac{\partial \tilde{s}_{i}\left( 
\mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }^{\top }}-G_{i}\frac{%
\partial s_{i}\left( \mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }%
^{\top }}\right] \otimes \mathbf{\tilde{X}}_{i}\bar{H}^{-1} \\
&=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}G_{i}\left\{ \left[ \frac{\partial \tilde{s}_{i}\left( \mathbf{\beta }%
^{0}\right) }{\partial \mathbf{\beta }^{\top }}-\frac{\partial s_{i}\left( 
\mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }^{\top }}\right]
\otimes \mathbf{\tilde{X}}_{i}\right\} \bar{H}^{-1} \\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}\left( \tilde{G}_{i}-G_{i}\right) \left\{ \frac{\partial s_{i}\left( 
\mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }^{\top }}\otimes 
\mathbf{\tilde{X}}_{i}\right\} \bar{H}^{-1} \\
&&+\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}\right) \bar{H}%
^{-1}\left( \tilde{G}_{i}-G_{i}\right) \left\{ \left[ \frac{\partial \tilde{s%
}_{i}\left( \mathbf{\beta }^{0}\right) }{\partial \mathbf{\beta }^{\top }}-%
\frac{\partial s_{i}\left( \mathbf{\beta }^{0}\right) }{\partial \mathbf{%
\beta }^{\top }}\right] \otimes \mathbf{\tilde{X}}_{i}\right\} \bar{H}^{-1}
\\
&\equiv &\mathcal{R}_{2n,1}+\mathcal{R}_{2n,2}+\mathcal{R}_{2n,3},\text{ say.%
}
\end{eqnarray*}%
We prove the lemma by showing that $\mathcal{R}_{2n,s}=o_{P}\left( 1\right) $
for $s=1,2,3.$ We will only show that $\mathcal{R}_{2n,1}=o_{P}\left(
1\right) $ as the other two cases can be proved analogously. Recall $%
c_{i\varphi }=\varphi ^{\prime }\left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ^{2}-\varphi ^{\prime \prime }\left( P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) \right) \varphi \left( P_{i}\left( \mathbf{%
\beta }_{2}^{0}\right) \right) $ and $\varphi _{i}\equiv \varphi ^{\prime
}\left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) /\varphi \left(
P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) .$ Noting that%
\begin{equation*}
\frac{\partial \tilde{s}_{i}\left( \mathbf{\beta }^{0}\right) }{\partial 
\mathbf{\beta }^{^{\top }}}=\left( 
\begin{array}{cc}
\frac{\tilde{\psi}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) }{%
\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) } & \frac{%
\varphi _{i}\left[ \tilde{\psi}_{i}^{\prime }\left( \bar{\varepsilon}%
_{i}\right) \bar{\varepsilon}_{i}+\tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) \right] }{2\varphi \left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ^{1/2}} \\ 
\frac{\varphi _{i}\left[ \tilde{\psi}_{i}^{\prime }\left( \bar{\varepsilon}%
_{i}\right) \bar{\varepsilon}_{i}+\tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) \right] }{2\varphi \left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ^{1/2}} & \frac{2c_{i\varphi }\left[ \tilde{\psi}%
_{i}\left( \bar{\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+1\right]
+\varphi ^{\prime }\left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right)
^{2}\bar{\varepsilon}_{i}\left[ \tilde{\psi}_{i}^{\prime }\left( \bar{%
\varepsilon}_{i}\right) \bar{\varepsilon}_{i}+\tilde{\psi}_{i}\left( \bar{%
\varepsilon}_{i}\right) \right] }{4\varphi \left( P_{i}\left( \mathbf{\beta }%
_{2}^{0}\right) \right) ^{2}}%
\end{array}%
\right) \otimes \mathbf{\tilde{X}}_{i}^{\top },
\end{equation*}%
and $\partial s_{i}\left( \mathbf{\beta }^{0}\right) /\partial \mathbf{\beta 
}^{\top }$ has similar expression with $\psi _{i}\left( \bar{\varepsilon}%
_{i}\right) $ in the place of $\tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) ,$ we have%
\begin{eqnarray*}
\mathcal{R}_{2n,1} &=&\frac{1}{nh^{d}}\sum_{i=1}^{n}K\left( \frac{x-X_{i}}{h}%
\right) G_{i} \\
&&\times \left( 
\begin{array}{cc}
\frac{\tilde{\psi}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) -\psi
^{\prime }\left( \bar{\varepsilon}_{i}\right) }{\varphi \left( P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) \right) } & \frac{\varphi _{i}\left\{ \left[ 
\tilde{\psi}_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) -\psi ^{\prime
}\left( \bar{\varepsilon}_{i}\right) \right] \bar{\varepsilon}_{i}+\left[ 
\tilde{\psi}_{i}\left( \bar{\varepsilon}_{i}\right) -\psi \left( \bar{%
\varepsilon}_{i}\right) \right] \right\} }{2\varphi \left( P_{i}\left( 
\mathbf{\beta }_{2}^{0}\right) \right) ^{1/2}} \\ 
\frac{\varphi _{i}\left\{ \left[ \tilde{\psi}_{i}^{\prime }\left( \bar{%
\varepsilon}_{i}\right) -\psi ^{\prime }\left( \bar{\varepsilon}_{i}\right) %
\right] \bar{\varepsilon}_{i}+\left[ \tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) -\psi \left( \bar{\varepsilon}_{i}\right) \right] \right\} }{%
2\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right) \right) ^{1/2}} & 
\frac{2c_{i\varphi }\left[ \tilde{\psi}_{i}\left( \bar{\varepsilon}%
_{i}\right) -\psi \left( \bar{\varepsilon}_{i}\right) \right] \bar{%
\varepsilon}_{i}}{4\varphi \left( P_{i}\left( \mathbf{\beta }_{2}^{0}\right)
\right) ^{2}}+\frac{\tilde{d}_{i}}{4}%
\end{array}%
\right) \otimes (\mathbf{Z}_{i}\mathbf{Z}_{i}^{\top }) \\
&\equiv &\left[ 
\begin{array}{cc}
\mathcal{R}_{2n,1,11} & \mathcal{R}_{2n,1,12} \\ 
\mathcal{R}_{2n,1,12}^{\top } & \mathcal{R}_{2n,1,22}%
\end{array}%
\right] ,\text{ say,}
\end{eqnarray*}%
where $\tilde{d}_{i}\equiv \varphi _{i}^{2}\bar{\varepsilon}_{i}[\tilde{\psi}%
_{i}^{\prime }\left( \bar{\varepsilon}_{i}\right) -\psi ^{\prime }\left( 
\bar{\varepsilon}_{i}\right) ]\bar{\varepsilon}_{i}+[\tilde{\psi}_{i}\left( 
\bar{\varepsilon}_{i}\right) -\psi \left( \bar{\varepsilon}_{i}\right) ]$.
As in the analysis of $\mathcal{R}_{1n,1},$ using Lemma A.2, we can readily
demonstrate that $\mathcal{R}_{2n,1,11}=o_{p}\left( 1\right) ,$ $\mathcal{R}%
_{2n,1,12}=o_{p}\left( 1\right) $ and $\mathcal{R}_{2n,1,22}=o_{p}\left(
1\right) .$ It follows that $\mathcal{R}_{2n,1}=o_{p}\left( 1\right) .$
Similarly, we can show that $\mathcal{R}_{2n,s}=o_{P}\left( 1\right) $ for $%
s=2,3.$ This completes the proof of the lemma. ${\tiny \blacksquare }$

\section{Derivative Matrices in the Proof of Proposition 2.1}

In this appendix, we give explicit expressions for the elements of some
derivative matrices of the log-likelihood function defined in the proof of
Proposition 2.1. The elements of the Hessian matrix are%
\begin{eqnarray*}
q_{11}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{\partial ^{2}\log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}%
\frac{1}{\varphi \left( \beta _{2}\right) }, \\
q_{12}\left( y;\beta _{1},\beta _{2}\right)  &=&\left\{ \frac{\partial
^{2}\log f\left( \varepsilon \left( \beta \right) \right) }{\partial
\varepsilon ^{2}}\varepsilon \left( \beta \right) +\frac{\partial \log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon }%
\right\} \frac{\varphi ^{\prime }\left( \beta _{2}\right) }{2\varphi \left(
\beta _{2}\right) ^{3/2}}, \\
q_{22}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{-\varphi ^{\prime
\prime }\left( \beta _{2}\right) }{2\varphi \left( \beta _{2}\right) }\left[ 
\frac{\partial \log f\left( \varepsilon \left( \beta \right) \right) }{%
\partial \varepsilon }\varepsilon \left( \beta \right) +1\right]  \\
&&+\frac{\varphi ^{\prime }\left( \beta _{2}\right) ^{2}}{4\varphi \left(
\beta _{2}\right) ^{2}}\left\{ \frac{\partial ^{2}\log f\left( \varepsilon
\left( \beta \right) \right) }{\partial \varepsilon ^{2}}\varepsilon \left(
\beta \right) ^{2}+3\frac{\partial \log f\left( \varepsilon \left( \beta
\right) \right) }{\partial \varepsilon }\varepsilon \left( \beta \right)
+2\right\} ,
\end{eqnarray*}%
and $q_{21}\left( y;\beta _{1},\beta _{2}\right) =q_{12}\left( y;\beta
_{1},\beta _{2}\right) $ by Young's theorem, where, e.g., $\frac{\partial
^{2}\log f\left( \varepsilon \right) }{\partial \varepsilon ^{2}}=\frac{%
f^{\prime \prime }(\varepsilon )f(\varepsilon )-f^{\prime }(\varepsilon )^{2}%
}{f^{2}(\varepsilon )}$ and $\frac{\partial ^{2}\log f\left( \varepsilon
\left( \beta \right) \right) }{\partial \varepsilon ^{2}}\equiv \left. \frac{%
\partial ^{2}\log f\left( \varepsilon \right) }{\partial \varepsilon ^{2}}%
\right\vert _{\varepsilon =\varepsilon \left( \beta \right) }.$ Note that
when we restrict our attention to the case $\varphi \left( u\right) =u$ or $%
\exp \left( u\right) ,$ the above formulae can be greatly simplified.

In addition, in the proof of Proposition 2.1, we also need that $%
q_{rst}\left( y;\beta _{1},\beta _{2}\right) \equiv \frac{\partial ^{3}}{%
\partial \beta _{r}\partial \beta _{s}\partial \beta _{t}}\log \left(
f\left( y;\beta _{1},\beta _{2}\right) \right) ,$ $r,s,t=1,2,$ should be
well behaved. Using the expressions%
\begin{equation*}
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta }=\left( 
\begin{array}{c}
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta _{1}} \\ 
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta _{2}}%
\end{array}%
\right) =\left( 
\begin{array}{c}
-\frac{1}{\varphi \left( \beta _{2}\right) ^{1/2}} \\ 
-\frac{\varphi ^{\prime }\left( \beta _{2}\right) }{2\varphi \left( \beta
_{2}\right) }\varepsilon \left( \beta \right) 
\end{array}%
\right) \text{ and }\frac{\partial ^{2}\log f\left( \varepsilon \right) }{%
\partial \varepsilon ^{2}}=\frac{f^{\prime \prime }(\varepsilon
)f(\varepsilon )-f^{\prime }(\varepsilon )^{2}}{f^{2}(\varepsilon )}
\end{equation*}%
and by straightforward calculations, we have 
\begin{eqnarray*}
q_{111}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{\partial ^{3}\log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{3}}%
\frac{1}{\varphi \left( \beta _{2}\right) }, \\
q_{112}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{\partial ^{3}\log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{3}}%
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta _{2}}\frac{1%
}{\varphi \left( \beta _{2}\right) }-\frac{\partial ^{2}\log f\left(
\varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}\frac{%
\varphi ^{\prime }\left( \beta _{2}\right) }{\varphi \left( \beta
_{2}\right) ^{2}}, \\
q_{121}\left( y;\beta _{1},\beta _{2}\right)  &=&\left\{ \frac{\partial
^{3}\log f\left( \varepsilon \left( \beta \right) \right) }{\partial
\varepsilon ^{3}}\frac{\partial \varepsilon \left( \beta \right) }{\partial
\beta _{1}}\varepsilon \left( \beta \right) +2\frac{\partial ^{2}\log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}%
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta _{1}}%
\right\} \frac{\varphi ^{\prime }\left( \beta _{2}\right) }{2\varphi \left(
\beta _{2}\right) ^{3/2}}=q_{112}\left( y;\beta _{1},\beta _{2}\right) , \\
q_{122}\left( y;\beta _{1},\beta _{2}\right)  &=&\left\{ \frac{\partial
^{3}\log f\left( \varepsilon \left( \beta \right) \right) }{\partial
\varepsilon ^{3}}\frac{\partial \varepsilon \left( \beta \right) }{\partial
\beta _{2}}\varepsilon \left( \beta \right) +2\frac{\partial ^{2}\log
f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}%
\frac{\partial \varepsilon \left( \beta \right) }{\partial \beta _{2}}%
\right\} \frac{\varphi ^{\prime }\left( \beta _{2}\right) }{2\varphi \left(
\beta _{2}\right) ^{3/2}}, \\
&&+\left\{ \frac{\partial ^{2}\log f\left( \varepsilon \left( \beta \right)
\right) }{\partial \varepsilon ^{2}}\varepsilon \left( \beta \right) +\frac{%
\partial \log f\left( \varepsilon \left( \beta \right) \right) }{\partial
\varepsilon }\right\} \frac{\varphi ^{\prime \prime }\left( \beta
_{2}\right) \varphi \left( \beta _{2}\right) ^{3/2}-\frac{3}{2}\varphi
^{\prime }\left( \beta _{2}\right) ^{2}\varphi \left( \beta _{2}\right)
^{1/2}}{2\varphi \left( \beta _{2}\right) ^{3}}, \\
q_{221}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{-\varphi ^{\prime
\prime }\left( \beta _{2}\right) }{2\varphi \left( \beta _{2}\right) }\left[ 
\frac{\partial ^{2}\log f\left( \varepsilon \left( \beta \right) \right) }{%
\partial \varepsilon ^{2}}\varepsilon \left( \beta \right) +\frac{\partial
\log f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon 
}\right] \frac{\partial \varepsilon \left( \beta \right) }{\partial \beta
_{1}}+\frac{\varphi ^{\prime }\left( \beta _{2}\right) ^{2}}{4\varphi \left(
\beta _{2}\right) ^{2}}\kappa \left( \beta \right) \frac{\partial
\varepsilon \left( \beta \right) }{\partial \beta _{1}} \\
&=&q_{122}\left( y;\beta _{1},\beta _{2}\right)  \\
q_{222}\left( y;\beta _{1},\beta _{2}\right)  &=&\frac{-\varphi ^{\prime
\prime }\left( \beta _{2}\right) }{2\varphi \left( \beta _{2}\right) }\left[ 
\frac{\partial ^{2}\log f\left( \varepsilon \left( \beta \right) \right) }{%
\partial \varepsilon ^{2}}\varepsilon \left( \beta \right) +\frac{\partial
\log f\left( \varepsilon \left( \beta \right) \right) }{\partial \varepsilon 
}\right] \frac{\partial \varepsilon \left( \beta \right) }{\partial \beta
_{2}} \\
&&-\frac{\varphi ^{\prime \prime \prime }\left( \beta _{2}\right) \varphi
\left( \beta _{2}\right) -\varphi ^{\prime \prime }\left( \beta _{2}\right)
\varphi ^{\prime }\left( \beta _{2}\right) }{2\varphi \left( \beta
_{2}\right) ^{2}}\left[ \frac{\partial \log f\left( \varepsilon \left( \beta
\right) \right) }{\partial \varepsilon }\varepsilon \left( \beta \right) +1%
\right]  \\
&&+\frac{\varphi ^{\prime }\left( \beta _{2}\right) ^{2}}{4\varphi \left(
\beta _{2}\right) ^{2}}\kappa \left( \beta \right) \frac{\partial
\varepsilon \left( \beta \right) }{\partial \beta _{2}} \\
&&+\frac{\varphi ^{\prime }\left( \beta _{2}\right) \varphi ^{\prime \prime
}\left( \beta _{2}\right) -\varphi ^{\prime }\left( \beta _{2}\right)
^{3}\varphi \left( \beta _{2}\right) }{2\varphi \left( \beta _{2}\right) ^{4}%
}\left\{ \frac{\partial ^{2}\log f\left( \varepsilon \left( \beta \right)
\right) }{\partial \varepsilon ^{2}}\varepsilon \left( \beta \right) ^{2}+3%
\frac{\partial \log f\left( \varepsilon \left( \beta \right) \right) }{%
\partial \varepsilon }\varepsilon \left( \beta \right) +2\right\} ,
\end{eqnarray*}%
$q_{211}=q_{121}=q_{112},$ and $q_{212}=q_{122}=q_{221}$ by Young's Theorem,
where $\kappa \left( \beta \right) \equiv \frac{\partial ^{3}\log f\left(
\varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{3}}%
\varepsilon \left( \beta \right) ^{2}$ $+2\frac{\partial ^{2}\log f\left(
\varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}%
\varepsilon \left( \beta \right) +3\frac{\partial ^{2}\log f\left(
\varepsilon \left( \beta \right) \right) }{\partial \varepsilon ^{2}}%
\varepsilon \left( \beta \right) +3\frac{\partial \log f\left( \varepsilon
\left( \beta \right) \right) }{\partial \varepsilon }.$ Note that under our
assumptions ($X_{i}$ has compact support, the parameter space is compact, $%
\sigma ^{2}\left( x\right) $ is bounded away from 0) the terms associated
with $\varphi \left( \cdot \right) $ or its derivatives are all well behaved
when $\varphi \left( \cdot \right) $ is evaluated in the neighborhood of $%
\beta _{2}^{0}\left( x\right) .$

\ \ \ 

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\end{document}