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\begin{document}

\title{Supplement to: Adaptive Long Memory Testing under Heteroskedasticity}
\author{%
\begin{tabular}{c}
David Harris\thanks{%
This research is supported by the Australian Research Council Discovery
Grant DP1094010. Correspondence to: David Harris,\ E-mail: \texttt{%
david.harris3@monash.edu}}\ \ and Hsein Kew \\ 
Department of Econometrics and Business Statistics \\ 
Monash University%
\end{tabular}%
}
\maketitle

\bigskip 

This supplement provides additional results which are required for the
proofs given in Appendix A of the paper \textquotedblleft Adaptive Long
Memory Testing under Heteroskedasticity\textquotedblright . Throughout, let $%
C$ denote a generic positive constant.

\bigskip

\noindent \textbf{Lemma A}. Let $\bar{\sigma}_{t}^{2}=\sum_{i=1}^{T}w_{ti}%
\sigma _{i}^{2}$ and $\tilde{\sigma}_{t}^{2}=\sum_{i=1}^{T}w_{ti}e_{i}^{2}.$
Under Assumptions R, S, E and B,\newline
\ \newline
(a) $\hat{\psi}\overset{p}{\rightarrow }\psi _{0};$\newline
\ \newline
(b) $\sqrt{T}\left( \hat{\psi}-\psi _{0}\right) \rightsquigarrow N\left(
0,\nu ^{-2}V_{\psi \psi }^{-1}\right) ;$\newline
\ \newline
(c) let $t=\left[ Tr\right] ,$ for any fixed $r\in (0,1],$ $\frac{1}{Tb}%
\sum_{i=1}^{T}K_{ti}\rightarrow \int_{-\infty }^{\infty }K\left( z\right)
dz=1;$\newline
\ \newline
(d) $\max_{t,i}w_{ti}=O\left( 1/Tb\right) ;$\newline
\ \newline
(e) $\min_{1\leq t\leq T}\bar{\sigma}_{t}^{2}\geq C>0;$\newline
\ \newline
(f) $\max_{1\leq t\leq T}E\left \vert \tilde{\sigma}_{t}^{2}-\bar{\sigma}%
_{t}^{2}\right \vert ^{4}=O\left( 1/(Tb)^{2}\right) ;$\newline
\ \newline
(g) $\max_{t}\left \vert \tilde{\sigma}_{t}^{2}-\bar{\sigma}%
_{t}^{2}\right
\vert =O_{p}\left( T^{-1/4}b^{-1/2}\right) ;$\newline
\ \newline
(h) $\left( \min_{1\leq t\leq T}\tilde{\sigma}_{t}^{2}\right)
^{-1}=O_{p}\left( 1\right) ;$\newline
\ \newline
(i) $\max_{1\leq t\leq T}\left \vert \hat{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right \vert =O_{p}\left( 1/\sqrt{T}b\right) ;$\newline
\ \newline
(j) $\left( \min_{1\leq t\leq T}\hat{\sigma}_{t}^{2}\right)
^{-1}=O_{p}\left( 1\right) ;$\newline
\ \newline
(k) $\sum_{t=1}^{T}\left( \hat{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{2}=O_{p}\left( 1/\sqrt{T}\right) ;$\newline
\ \newline
(l) $T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}_{t}^{2}-\sigma
_{t}^{2}\right \vert =o\left( 1\right) .$

\bigskip

\pagebreak

\noindent \textbf{Lemma B}. Under Assumptions R, S, E and B,\newline
(a)%
\begin{equation*}
T^{-1/2}\sum_{t=1}^{T}\left( \hat{s}_{t}\left( 0,\psi _{0}\right)
-s_{t}\left( 0,\psi _{0}\right) \right) \overset{p}{\rightarrow }0,
\end{equation*}%
where $\hat{s}_{t}\left( 0,\psi _{0}\right) $ and $s_{t}\left( 0,\psi
_{0}\right) $ represent, respectively, the score vectors $\hat{s}_{t}\left(
\gamma \right) $ and $s_{t}\left( \gamma \right) $ but evaluated at $\theta
=0$ and $\psi =\psi _{0};$\newline
(b)%
\begin{equation*}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left( \hat{h}_{t}\left( 0,\psi
\right) -h_{t}\left( 0,\psi \right) \right) \overset{p}{\rightarrow }0,
\end{equation*}%
where $\hat{h}_{t}\left( 0,\psi \right) $ and $h_{t}\left( 0,\psi \right) $
represent, respectively, the matrix of second derivatives of $\hat{l}%
_{t}\left( \gamma \right) $ and $l_{t}\left( \gamma \right) $ but evaluated
at $\theta =0;$\newline
(c) Let $\bar{\psi}=\arg \max_{\psi \in \Psi ,\theta =0}L\left( \gamma
\right) .$ Then 
\begin{equation*}
\bar{\psi}\overset{p}{\rightarrow }\psi _{0};
\end{equation*}%
\newline
(d) 
\begin{equation*}
\tilde{\psi}\overset{p}{\rightarrow }\psi _{0}.
\end{equation*}

\bigskip

%TCIMACRO{\TeXButton{noindent}{\noindent}}%
%BeginExpansion
\noindent%
%EndExpansion
\textbf{Proof of Lemma A}

\noindent \textbf{\ Part (a)} The proof is given in Cavaliere, Nielsen and
Taylor (2015a, hereafter CNT) Lemma A.1.

\bigskip

\noindent \textbf{\ Part (b)} This result follows from Cavaliere, Nielsen
and Taylor (2015b) Lemma S.10, CNT Lemma A.2 and the standard Mean Value
Theorem.

\bigskip

\noindent \ \textbf{Parts (c) to (h)} follows directly from Xu and Phillips
(2008) Lemma A(c) to Lemma A(h).

\bigskip

\noindent \ \textbf{Part (i)} For $\hat{\psi},\psi _{0}\in \Psi $ and for
some $\psi ^{\ast }$ in between $\hat{\psi}$ and $\psi _{0},$%
\begin{eqnarray*}
\hat{e}_{i} &=&a\left( L;\hat{\psi}\right) y_{i} \\
&=&a\left( L;\psi _{0}\right) y_{i}+\left( a_{\psi }\left( L;\psi ^{\ast
}\right) y_{i}\right) ^{\prime }\left( \hat{\psi}-\psi _{0}\right) \\
&=&e_{i}+\left( a_{\psi }\left( L;\psi ^{\ast }\right) y_{i}\right) ^{\prime
}\left( \hat{\psi}-\psi _{0}\right)
\end{eqnarray*}%
and so%
\begin{equation*}
\hat{e}_{i}^{2}=e_{i}^{2}+2e_{i}\left( a_{\psi }\left( L;\psi ^{\ast
}\right) y_{i}\right) ^{\prime }\left( \hat{\psi}-\psi _{0}\right) +\left( 
\hat{\psi}-\psi _{0}\right) ^{\prime }\left( a_{\psi }\left( L;\psi ^{\ast
}\right) y_{i}\right) \left( a_{\psi }\left( L;\psi ^{\ast }\right)
y_{i}\right) ^{\prime }\left( \hat{\psi}-\psi _{0}\right) .
\end{equation*}

Denote $\hat{\psi}=\left( \hat{\psi}_{1,}\ldots ,\hat{\psi}_{p+q}\right)
^{\prime },\psi _{0}=\left( \psi _{0,1},\ldots ,\psi _{0,p+q}\right)
^{\prime }$ and $\psi ^{\ast }=\left( \psi _{1}^{\ast },\ldots \psi
_{p+q}^{\ast }\right) ^{\prime }.$ We have $a_{\psi }\left( L;\psi \right)
y_{i}=\left( z_{1,i}\left( \psi \right) ,\ldots ,z_{p+q,i}\left( \psi
\right) \right) ^{\prime }$ where $z_{k,i}\left( \psi \right)
=\sum_{j=1}^{i-1}\varphi _{k,j}\left( \psi \right) e_{i-j}$ and $\varphi
_{k,j}\left( \psi \right) $ decays exponentially in $j$ for all $\psi \in
\Psi $ under Assumption R and for $k=1,\ldots ,p+q.$

We write%
\begin{equation}
\hat{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}=\sum_{i=1}^{T}w_{ti}\left( \hat{e%
}_{i}^{2}-e_{i}^{2}\right) =A_{1,t}+A_{2,t},  \label{shatstilta}
\end{equation}%
where%
\begin{equation*}
A_{1,t}=2\sum_{i=1}^{T}w_{ti}e_{i}\left( a_{\psi }\left( L;\psi ^{\ast
}\right) y_{i}\right) ^{\prime }\left( \hat{\psi}-\psi _{0}\right)
=2\sum_{j=1}^{p+q}\left( \hat{\psi}_{j}-\psi _{0,j}\right)
\sum_{i=1}^{T}w_{ti}e_{i}z_{j,i}\left( \psi ^{\ast }\right)
\end{equation*}%
and%
\begin{eqnarray*}
A_{2,t} &=&\sum_{i=1}^{T}w_{ti}\left( \hat{\psi}-\psi _{0}\right) ^{\prime
}\left( a_{\psi }\left( L;\psi ^{\ast }\right) y_{i}\right) \left( a_{\psi
}\left( L;\psi ^{\ast }\right) y_{i}\right) ^{\prime }\left( \hat{\psi}-\psi
_{0}\right) \\
&=&\sum_{k=1}^{p+q}\sum_{l=1}^{p+q}\left( \hat{\psi}_{k}-\psi _{0,k}\right)
\left( \hat{\psi}_{l}-\psi _{0,l}\right) \sum_{i=1}^{T}w_{ti}z_{k,i}\left(
\psi ^{\ast }\right) z_{l,i}\left( \psi ^{\ast }\right) .
\end{eqnarray*}

Now the required results that%
\begin{equation*}
\max_{1\leq t\leq T}\left \vert \hat{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right \vert \leq \max_{1\leq t\leq T}\left \vert A_{1,t}\right
\vert +\max_{1\leq t\leq T}\left \vert A_{2,t}\right \vert =O_{p}\left( 
\frac{1}{\sqrt{T}b}\right)
\end{equation*}%
will follow from%
\begin{equation}
\max_{1\leq t\leq T}\left \vert A_{1,t}\right \vert =O_{p}\left( \frac{1}{%
\sqrt{T}b}\right)  \label{A1t}
\end{equation}%
and%
\begin{equation}
\max_{1\leq t\leq T}\left \vert A_{2,t}\right \vert =O_{p}\left( \frac{1}{Tb}%
\right) ,  \label{A2t}
\end{equation}%
as we will now show.

First note that%
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{i=1}^{T}\left\vert z_{k,i}\left( \psi
\right) z_{l,i}\left( \psi \right) \right\vert =O_{p}\left( 1\right) \ \ 
\text{\ for }k,l=1,\ldots ,p+q  \label{ZkZl}
\end{equation}%
and 
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{i=1}^{T}\left\vert e_{i}z_{j,i}\left( \psi
\right) \right\vert =O_{p}\left( 1\right) \ \ \ \text{for }j=1,\ldots ,p+q.
\label{eZl}
\end{equation}%
We will only show (\ref{ZkZl}) since (\ref{eZl}) follows as a special case.
To show (\ref{ZkZl})\ we will follow precisely the same lines as the proof
given in Cavaliere, Nielsen and Taylor (2015b) Lemma S.10 equation (S.16)
with $u_{1}=u_{2}=0$ in their $M_{T}\left( u_{1,}u_{2},\psi \right) $
notation. Following them, we write%
\begin{eqnarray}
T^{-1}\sum_{i=1}^{T}\left\vert z_{k,i}\left( \psi \right) z_{l,i}\left( \psi
\right) \right\vert &\leq
&T^{-1}\sum_{i=1}^{T}\sum_{m=1}^{i-1}\sum_{n=1}^{i-1}\left\vert \varphi
_{k,m}\left( \psi \right) \right\vert \left\vert \varphi _{l,n}\left( \psi
\right) \right\vert \left\vert e_{i-m}e_{i-n}\right\vert  \notag
\label{zkzl_T} \\
&=&\sum_{m=1}^{T-1}\sum_{n=1}^{T-1}\left\vert \varphi _{k,m}\left( \psi
\right) \right\vert \left\vert \varphi _{l,n}\left( \psi \right) \right\vert
T^{-1}\sum_{i=\max \left( m,n\right) +1}^{T}\left\vert
e_{i-m}e_{i-n}\right\vert .  \label{zkzl_T1}
\end{eqnarray}%
Since $T^{-1}\sum_{i=\max \left( m,n\right) +1}^{T}\left\vert
e_{i-m}e_{i-n}\right\vert =O_{p}\left( 1\right) $ uniformly in $m,n$ because 
$E\left\vert e_{i-m}e_{i-n}\right\vert \leq \sup_{r}\sigma \left( r\right)
^{2}<\infty $ it thus follows that (see the treatment of Cavaliere, Nielsen
and Taylor (2015b) equation (S.19))%
\begin{equation*}
\sup_{\psi \in \Psi }\left( \ref{zkzl_T1}\right) =O_{p}\left( \sup_{\psi \in
\Psi }\sum_{m=1}^{T-1}\sum_{n=1}^{T-1}\left\vert \varphi _{k,m}\left( \psi
\right) \right\vert \left\vert \varphi _{l,n}\left( \psi \right) \right\vert
\right) =O_{p}\left( 1\right) ,
\end{equation*}%
because $\sum_{j=1}^{\infty }\left\vert \varphi _{k,j}\left( \psi \right)
\right\vert <\infty $ uniformly in $\psi \in \Psi $ under Assumption R.

To show (\ref{A1t}), for $j=1,\ldots ,p+q$ and since $p$ and $q$ are known
fixed orders, we have%
\begin{eqnarray*}
\max_{t}\left \vert 2\left( \hat{\psi}_{j}-\psi _{0,j}\right)
\sum_{i=1}^{T}w_{ti}e_{i}z_{j,i}\left( \psi ^{\ast }\right) \right \vert
&\leq &\max_{t}2\left \vert \hat{\psi}_{j}-\psi _{0,j}\right \vert
\sum_{i=1}^{T}w_{ti}\left \vert e_{i}z_{j,i}\left( \psi ^{\ast }\right)
\right \vert \\
&\leq &2T\left \vert \hat{\psi}_{j}-\psi _{0,j}\right \vert \left(
\max_{t,i}w_{ti}\right) \sup_{\psi \in \Psi }T^{-1}\sum_{i=1}^{T}\left \vert
e_{i}z_{j,i}\left( \psi \right) \right \vert \\
&=&O_{p}\left( T^{1/2}\right) O\left( \frac{1}{Tb}\right) O_{p}\left(
1\right) =O_{p}\left( \frac{1}{\sqrt{T}b}\right) ,
\end{eqnarray*}%
by Lemma A(b,d) and (\ref{eZl}).

To show (\ref{A2t}), for $k,l=1,\ldots ,p+q$ and given that $p$ and $q$ are
known fixed orders, we have%
\begin{eqnarray*}
&&\max_{t}\left \vert \hat{\psi}_{k}-\psi _{0,k}\right \vert \left \vert 
\hat{\psi}_{l}-\psi _{0,l}\right \vert \sum_{i=1}^{T}w_{ti}\left \vert
z_{k,i}\left( \psi ^{\ast }\right) z_{l,i}\left( \psi ^{\ast }\right) \right
\vert \\
&\leq &T\left \vert \hat{\psi}_{k}-\psi _{0,k}\right \vert \left \vert \hat{%
\psi}_{l}-\psi _{0,l}\right \vert \left( \max_{t,i}w_{ti}\right) \sup_{\psi
\in \Psi }T^{-1}\sum_{i=1}^{T}\left \vert z_{k,i}\left( \psi \right)
z_{l,i}\left( \psi \right) \right \vert \\
&=&O_{p}\left( \frac{1}{Tb}\right) ,
\end{eqnarray*}%
by (\ref{ZkZl}). This concludes the proof for this part (i).

\bigskip

\noindent \ \textbf{Part (j)} This follows from Xu and Phillips (2008) Lemma
A(j) and our Lemma A(i) above.

\bigskip

\noindent \ \textbf{Part (k)} Recall equation (\ref{shatstilta}), we have%
\begin{equation*}
\sum_{t=1}^{T}\left( \hat{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{2}=\sum_{t=1}^{T}\left( A_{1,t}+A_{2,t}\right)
^{2}=\sum_{t=1}^{T}A_{1,t}^{2}+\sum_{t=1}^{T}A_{2,t}^{2}+2%
\sum_{t=1}^{T}A_{1,t}A_{2,t}
\end{equation*}%
and so the required result that $\sum_{t=1}^{T}\left( \hat{\sigma}_{t}^{2}-%
\tilde{\sigma}_{t}^{2}\right) ^{2}=O_{p}\left( 1/\sqrt{T}\right) \ $will
follow if we show that:%
\begin{eqnarray}
\left \vert \sum_{t=1}^{T}A_{1,t}^{2}\right \vert &=&O_{p}\left( \frac{1}{%
\sqrt{T}}\right) ;  \label{A1t_2} \\
\left \vert \sum_{t=1}^{T}A_{2,t}^{2}\right \vert &=&O_{p}\left( \frac{1}{T}%
\right) ;  \label{A2t_2} \\
\left \vert \sum_{t=1}^{T}A_{1,t}A_{2,t}\right \vert &=&O_{p}\left( \frac{1}{%
T^{3/4}}\right) .  \label{A1tA2t}
\end{eqnarray}

To show (\ref{A1t_2}), we write%
\begin{eqnarray*}
\sum_{t=1}^{T}A_{1,t}^{2} &=&4\sum_{t=1}^{T}\left( \sum_{k=1}^{p+q}\left( 
\hat{\psi}_{k}-\psi _{0,k}\right) \sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left(
\psi ^{\ast }\right) \right) ^{2} \\
&\leq &4\sum_{t=1}^{T}\left( \sum_{k=1}^{p+q}\left( \hat{\psi}_{k}-\psi
_{0,k}\right) ^{2}\right) \sum_{k=1}^{p+q}\left(
\sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi ^{\ast }\right) \right) ^{2} \\
&=&4\left( \sum_{k=1}^{p+q}T\left( \hat{\psi}_{k}-\psi _{0,k}\right)
^{2}\right) \sum_{k=1}^{p+q}T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi ^{\ast }\right) \right) ^{2}.
\end{eqnarray*}%
Since $p$ and $q$ are known fixed orders and $\left( \hat{\psi}_{k}-\psi
_{0,k}\right) =O_{p}\left( T^{-1/2}\right) $ by Lemma A(b), equation (\ref%
{A1t_2}) follows if, for $k=1,...,p+q,$ we show that%
\begin{equation}
\left \vert T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi ^{\ast }\right) \right)
^{2}-T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi
_{0}\right) \right) ^{2}\right \vert =O_{p}\left( \frac{1}{\sqrt{T}}\right)
\label{psi*psi0}
\end{equation}%
and 
\begin{equation}
T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi
_{0}\right) \right) ^{2}=O_{p}\left( \frac{1}{Tb}\right) .  \label{Epsi0}
\end{equation}

To show (\ref{psi*psi0}), for some $\breve{\psi}$ in between $\hat{\psi}$
and $\psi _{0},$%
\begin{eqnarray*}
&&\left\vert T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi ^{\ast }\right) \right)
^{2}-T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi
_{0}\right) \right) ^{2}\right\vert \\
&\leq &\sum_{l=1}^{p+q}\left\vert
T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{ti}w_{tj}e_{i}e_{j}\frac{%
\partial }{\partial \psi _{l}}\left( z_{k,i}\left( \breve{\psi}\right)
z_{k,j}\left( \breve{\psi}\right) \right) \right\vert \left\vert \psi
_{l}^{\ast }-\psi _{0,l}\right\vert \\
&\leq &\sum_{l=1}^{p+q}\sup_{\psi \in \Psi }\left\vert
T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{ti}w_{tj}e_{i}e_{j}\frac{%
\partial }{\partial \psi _{l}}\left( z_{k,i}\left( \psi \right)
z_{k,j}\left( \psi \right) \right) \right\vert \left\vert \psi _{l}^{\ast
}-\psi _{0,l}\right\vert .
\end{eqnarray*}%
Recall that $\psi ^{\ast }$ lies in between $\hat{\psi}$ and $\psi _{0}$
(see part (i)) and Lemma A(b) implies that $\left\vert \psi _{l}^{\ast
}-\psi _{0,l}\right\vert =O_{p}\left( T^{-1/2}\right) $ then (\ref{psi*psi0}%
) follows if, for $m,n=0,1,$ we show that%
\begin{equation}
\sup_{\psi \in \Psi }\left\vert
T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{ti}w_{tj}\left( e_{i}%
\frac{\partial ^{m}}{\partial \psi _{l}^{m}}z_{k,i}\left( \psi \right)
\right) \left( e_{j}\frac{\partial ^{n}}{\partial \psi _{l}^{n}}%
z_{k,j}\left( \psi \right) \right) \right\vert =O_{p}\left( 1\right) .
\label{sup_wrt_psi}
\end{equation}%
We show only when $m=n=0$ since the remaining cases may be shown in a
similar fashion given that the coefficients of the linear processes
resulting from the derivative wrt $\psi $ still decay exponentially under
Assumption R. To show (\ref{sup_wrt_psi}), it suffices to show that, for $%
k,l=1,\ldots ,p+q,$ 
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}z_{k,i}\left( \psi \right) z_{l,i}\left( \psi \right)
\right) ^{2}=O_{p}\left( 1\right) ,  \label{supwZkZl}
\end{equation}%
since (\ref{sup_wrt_psi}) is a special case. To show (\ref{supwZkZl}), we
will follow the same lines as those in the proof of Cavaliere, Nielsen and
Taylor (2015b) Lemma S.10 equation (S.16). Following them, we write%
\begin{eqnarray}
&&T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{T}w_{ti}z_{k,i}\left( \psi \right)
z_{l,i}\left( \psi \right) \right) ^{2}  \label{wZkZl_1} \\
&\leq
&\sum_{m=1}^{T-1}\sum_{n=1}^{T-1}\sum_{a=1}^{T-1}\sum_{b=1}^{T-1}\left\vert
\varphi _{k,m}\left( \psi \right) \right\vert \left\vert \varphi
_{l,n}\left( \psi \right) \right\vert \left\vert \varphi _{k,a}\left( \psi
\right) \right\vert \left\vert \varphi _{l,b}\left( \psi \right) \right\vert
\times  \label{wZkZl_2} \\
&&\ \ \ \ \ \ \ \ \ \ \ \ T^{-1}\sum_{t=1}^{T}\sum_{i=\max \left( m,n\right)
+1}^{T}\sum_{j=\max \left( a,b\right) +1}^{T}w_{ti}w_{tj}\left\vert
e_{i-m}e_{i-n}e_{j-a}e_{j-b}\right\vert .  \label{wZkZl_3}
\end{eqnarray}%
In (\ref{wZkZl_3}), $T^{-1}\sum_{t=1}^{T}\sum_{i=\max \left( m,n\right)
+1}^{T}\sum_{j=\max \left( a,b\right) +1}^{T}w_{ti}w_{tj}\left\vert
e_{i-m}e_{i-n}e_{j-a}e_{j-b}\right\vert =O_{p}\left( 1\right) $ uniformly in 
$m,n,a,b$ because%
\begin{eqnarray*}
&&E\left\vert T^{-1}\sum_{t=1}^{T}\sum_{i=\max \left( m,n\right)
+1}^{T}\sum_{j=\max \left( a,b\right)
+1}^{T}w_{ti}w_{tj}e_{i-m}e_{i-n}e_{j-a}e_{j-b}\right\vert \\
&\leq &T^{-1}\sum_{t=1}^{T}\sum_{i=\max \left( m,n\right)
+1}^{T}\sum_{j=\max \left( a,b\right) +1}^{T}w_{ti}w_{tj}E\left\vert
e_{i-m}e_{i-n}e_{j-a}e_{j-b}\right\vert \\
&\leq &T^{-1}\sum_{t=1}^{T}\sum_{i=\max \left( m,n\right)
+1}^{T}\sum_{j=\max \left( a,b\right) +1}^{T}w_{ti}w_{tj}\left( E\left(
e_{i-m}^{4}\right) E\left( e_{i-n}^{4}\right) E\left( e_{j-a}^{4}\right)
E\left( e_{j-b}^{4}\right) \right) ^{1/4} \\
&\leq &\tau _{0,0}\left( \sup_{r}\sigma \left( r\right) ^{2}\right)
^{2}T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{T}w_{ti}\sum_{j=1}^{T}w_{tj}=\tau
_{0,0}\left( \sup_{r}\sigma \left( r\right) ^{2}\right) ^{2}<\infty .
\end{eqnarray*}%
In (\ref{wZkZl_2}), since $\sum_{m=1}^{\infty }\left\vert \varphi
_{k,m}\left( \psi \right) \right\vert <\infty $ uniformly for all $\psi \in
\Psi $ under assumption R, it thus follows that (see the treatment of
Cavaliere, Nielsen and Taylor (2015b) equation (S.19)) 
\begin{equation*}
\sup_{\psi \in \Psi }\left( \ref{wZkZl_1}\right) =O_{p}\left( \left(
\sum_{m=1}^{T-1}\left\vert \varphi _{k,m}\left( \psi \right) \right\vert
\right) ^{2}\left( \sum_{n=1}^{T-1}\left\vert \varphi _{l,n}\left( \psi
\right) \right\vert \right) ^{2}\right) =O_{p}\left( 1\right) .
\end{equation*}

To show (\ref{Epsi0}), we write%
\begin{eqnarray*}
&&T^{-1}\sum_{t=1}^{T}E\left( \sum_{i=1}^{T}w_{ti}e_{i}z_{k,i}\left( \psi
_{0}\right) \right) ^{2} \\
&=&T^{-1}\sum_{t=1}^{T}E\left(
\sum_{i=2}^{T}w_{ti}e_{i}\sum_{m=1}^{i-1}\varphi _{k,m}\left( \psi
_{0}\right) e_{i-m}\right) ^{2} \\
&\leq
&T^{-1}\sum_{t=1}^{T}\sum_{i=2}^{T}\sum_{j=2}^{T}w_{ti}w_{tj}%
\sum_{m=1}^{i-1}\left \vert \varphi _{k,m}\left( \psi _{0}\right) \right
\vert \sum_{n=1}^{j-1}\left \vert \varphi _{k,n}\left( \psi _{0}\right)
\right \vert \left \vert E\left( e_{i-m}e_{j-n}e_{i}e_{j}\right) \right \vert
\\
&=&T^{-1}\sum_{t=1}^{T}\sum_{i=2}^{T}w_{ti}^{2}\sum_{m=1}^{i-1}\left \vert
\varphi _{k,m}\left( \psi _{0}\right) \right \vert \sum_{n=1}^{i-1}\left
\vert \varphi _{l,n}\left( \psi _{0}\right) \right \vert \left \vert E\left(
e_{i-m}e_{i-n}E\left( e_{i}^{2}|\tciFourier _{i-1}\right) \right) \right
\vert \\
&\leq &\left( \sup_{r}\sigma \left( r\right) ^{2}\right)
T^{-1}\sum_{t=1}^{T}\sum_{i=2}^{T}w_{ti}^{2}\sum_{m=1}^{i-1}\left \vert
\varphi _{k,m}\left( \psi _{0}\right) \right \vert \sum_{n=1}^{i-1}\left
\vert \varphi _{l,n}\left( \psi _{0}\right) \right \vert \left \vert E\left(
e_{i-m}e_{i-n}\right) \right \vert \\
&=&\left( \sup_{r}\sigma \left( r\right) ^{2}\right)
T^{-1}\sum_{t=1}^{T}\sum_{i=2}^{T}w_{ti}^{2}\sum_{m=1}^{i-1}\varphi
_{k,m}\left( \psi _{0}\right) ^{2}E\left( e_{i-m}^{2}\right) \\
&\leq &\left( \sup_{r}\sigma \left( r\right) ^{2}\right) ^{2}\left(
\max_{t,i}w_{ti}\right) \sum_{m=1}^{\infty }\varphi _{k,m}\left( \psi
_{0}\right) ^{2}T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{T}w_{ti}\right) \\
&\leq &\left( \sup_{r}\sigma \left( r\right) ^{2}\right) ^{2}\left(
\max_{t,i}w_{ti}\right) \sum_{m=1}^{\infty }\varphi _{k,m}\left( \psi
_{0}\right) ^{2}=O\left( \frac{1}{Tb}\right) .
\end{eqnarray*}

\bigskip

To show (\ref{A2t_2}), for $k,l=1,\ldots ,p+q$ and since $p$ and $q$ are
known fixed orders, we write%
\begin{eqnarray*}
&&\left( \hat{\psi}_{k}-\psi _{0,k}\right) ^{2}T\left( \hat{\psi}_{l}-\psi
_{0,l}\right) ^{2}T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}z_{k,i}\left( \psi ^{\ast }\right) z_{l,i}\left( \psi
^{\ast }\right) \right) ^{2} \\
&\leq &\left( \hat{\psi}_{k}-\psi _{0,k}\right) ^{2}T\left( \hat{\psi}%
_{l}-\psi _{0,l}\right) ^{2}\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left(
\sum_{i=1}^{T}w_{ti}z_{k,i}\left( \psi \right) z_{l,i}\left( \psi \right)
\right) ^{2} \\
&=&O_{p}\left( T^{-1}\right)
\end{eqnarray*}%
by Lemma A(b) and equation (\ref{supwZkZl}).

Finally (\ref{A1tA2t}) follows immediately from (\ref{A1t_2}), (\ref{A2t_2})
and Cauchy-Schwarz inequality.

\bigskip

\noindent \ \textbf{Part (l)} See Xu and Phillips (2008) Lemma A(l).

\bigskip

%TCIMACRO{\TeXButton{noindent}{\noindent}}%
%BeginExpansion
\noindent%
%EndExpansion
\textbf{Proof of Lemma B}

\noindent \ \textbf{Part (a)} We have%
\begin{equation}
T^{-1/2}\sum_{t=1}^{T}\left( \hat{s}_{\theta ,t}\left( 0,\psi _{0}\right)
-s_{\theta ,t}\left( 0,\psi _{0}\right) \right)
=T^{-1/2}\sum_{t=1}^{T}\left( \hat{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right)
\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t},  \label{shat_t}
\end{equation}%
and 
\begin{equation}
T^{-1/2}\sum_{t=1}^{T}\left( \hat{s}_{\psi ,t}\left( 0,\psi _{0}\right)
-s_{\psi ,t}\left( 0,\psi _{0}\right) \right) =T^{-1/2}\sum_{t=1}^{T}\left( 
\hat{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right) \sum_{j=1}^{t-1}b_{j}\left(
\psi _{0}\right) e_{t-j}e_{t},  \label{shat_c}
\end{equation}%
where $b_{j}\left( \psi _{0}\right) $ decays exponentially under Assumption
R. We will only show (\ref{shat_t}) $\overset{p}{\rightarrow }0$ since (\ref%
{shat_c}) $\overset{p}{\rightarrow }0$ follows immediately from Xu and
Phillips (2008). Let $\sigma =\left( \sigma _{1},\sigma _{2},\ldots ,\sigma
_{T}\right) ^{\prime }$ and define%
\begin{equation*}
B\left( \sigma \right) =T^{-1/2}\sum_{t=1}^{T}\sigma
_{t}^{-2}\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}.
\end{equation*}%
Following from Robinson (1987) and Xu and Phillips (2008), (\ref{shat_t}) $%
\overset{p}{\rightarrow }0$ follows if we show the following three results:%
\begin{equation}
B\left( \hat{\sigma}\right) -B\left( \tilde{\sigma}\right) \overset{p}{%
\rightarrow }0,  \label{B1}
\end{equation}%
\begin{equation}
B\left( \tilde{\sigma}\right) -B\left( \bar{\sigma}\right) \overset{p}{%
\rightarrow }0,  \label{B2}
\end{equation}%
\begin{equation}
B\left( \bar{\sigma}\right) -B\left( \sigma \right) \overset{p}{\rightarrow }%
0.  \label{B3}
\end{equation}%
Recall that 
\begin{equation*}
\tilde{\sigma}_{t}^{2}=\sum_{i=1}^{T}w_{ti}e_{i}^{2}\ \ \text{and}\ \ \bar{%
\sigma}_{t}^{2}=\sum_{i=1}^{T}w_{ti}\sigma _{i}^{2}.
\end{equation*}

To show (\ref{B1}), first note that 
\begin{equation}
T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right)
^{2}=O_{p}\left( 1\right)  \label{B1a}
\end{equation}%
because%
\begin{eqnarray*}
E\left( \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right) ^{2}
&=&\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1}{i}\frac{1}{j}E\left(
e_{t-i}e_{t-j}E\left( e_{t}^{2}|\tciFourier _{t-1}\right) \right) \\
&\leq &\sup_{r}\sigma \left( r\right) ^{2}\sum_{i=1}^{t-1}\frac{1}{i^{2}}%
E\left( e_{t-i}^{2}\right) \leq \left( \sup_{r}\sigma \left( r\right)
^{2}\right) ^{2}\frac{\pi ^{2}}{6}<\infty \ \ \text{uniformly in }t.
\end{eqnarray*}%
Then,%
\begin{eqnarray*}
\left \vert B\left( \hat{\sigma}\right) -B\left( \tilde{\sigma}\right)
\right \vert &=&\left \vert T^{-1/2}\sum_{t=1}^{T}\left( \hat{\sigma}%
_{t}^{-2}-\tilde{\sigma}_{t}^{-2}\right) \sum_{i=1}^{t-1}\frac{1}{i}%
e_{t-i}e_{t}\right \vert \\
&=&\left \vert T^{-1/2}\sum_{t=1}^{T}\frac{\left( \hat{\sigma}_{t}^{2}-%
\tilde{\sigma}_{t}^{2}\right) }{\hat{\sigma}_{t}^{2}\tilde{\sigma}_{t}^{2}}%
\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right \vert \\
&\leq &\left( \frac{1}{\min \hat{\sigma}_{t}^{2}}\right) \left( \frac{1}{%
\min \tilde{\sigma}_{t}^{2}}\right) \left( \sum_{t=1}^{T}\left( \hat{\sigma}%
_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) ^{2}\right) ^{1/2}\left(
T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right)
^{2}\right) ^{1/2} \\
&=&O_{p}\left( \frac{1}{T^{1/2}}\right) =o_{p}\left( 1\right) ,
\end{eqnarray*}%
where the second last equality uses Lemma A(h,j,k) and (\ref{B1a}).

To show (\ref{B2}), we have%
\begin{eqnarray}
&&T^{-1/2}\sum_{t=1}^{T}\left( \tilde{\sigma}_{t}^{-2}-\bar{\sigma}%
_{t}^{-2}\right) \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}  \notag \\
&=&T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) \bar{\sigma}_{t}^{-4}\sum_{i=1}^{t-1}\frac{1}{i}%
e_{t-i}e_{t}+T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) ^{2}\tilde{\sigma}_{t}^{-2}\bar{\sigma}_{t}^{-4}%
\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t},\ \ \ \ \ \   \label{B2a}
\end{eqnarray}%
where the last equality uses 
\begin{equation}
p^{-1}-q^{-1}=\left( q-p\right) q^{-2}+\left( q-p\right) ^{2}p^{-1}q^{-2}
\label{pq}
\end{equation}%
for any non-zero real numbers $p$ and $q$ (see Xu and Phillips (2008)
equation (21) page 276).

We now show that the first term in (\ref{B2a}) $\overset{p}{\rightarrow }0.$
To see this, let 
\begin{equation*}
x_{t-1}=\sum_{i=1}^{t-1}i^{-1}e_{t-i}.
\end{equation*}%
Then $E\left( \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) \bar{%
\sigma}_{t}^{-4}x_{t-1}e_{t}\right) =0$ follows the same arguments as in Xu
and Phillips (2008) equation (22). Now%
\begin{eqnarray}
&&E\left( T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) \bar{\sigma}_{t}^{-4}x_{t-1}e_{t}\right) ^{2}  \notag \\
&=&T^{-1}\sum_{t=1}^{T}E\left( \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) ^{2}\bar{\sigma}_{t}^{-8}x_{t-1}^{2}e_{t}^{2}\right)  \notag
\\
&&+T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}E\left( \bar{\sigma}_{t-s}^{2}-\tilde{%
\sigma}_{t-s}^{2}\right) \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) \bar{\sigma}_{t-s}^{-4}\bar{\sigma}%
_{t}^{-4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}.  \label{B2a_1}
\end{eqnarray}%
By Lemma A(e,f), the first term $\rightarrow 0$ since%
\begin{eqnarray*}
&&T^{-1}\sum_{t=1}^{T}E\left( \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) ^{2}\bar{\sigma}_{t}^{-8}x_{t-1}^{2}e_{t}^{2}\right) \\
&=&T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1}{i}\frac{1}{j}%
E\left( \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) ^{2}\bar{%
\sigma}_{t}^{-8}e_{t-i}e_{t-j}e_{t}^{2}\right) \\
&\leq &\left( \frac{1}{\min_{t}\bar{\sigma}_{t}^{2}}\right)
^{4}T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1}{i}\frac{1}{j%
}\left( E\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{4}E\left( e_{t-i}^{2}e_{t-j}^{2}e_{t}^{4}\right) \right) ^{1/2} \\
&\leq &\left( \frac{1}{\min_{t}\bar{\sigma}_{t}^{2}}\right) ^{4}\left(
\max_{t}E\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{4}\right) ^{1/2}T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1%
}{i}\frac{1}{j}\left( E\left( e_{t-i}^{4}e_{t-j}^{4}\right) E\left(
e_{t}^{8}\right) \right) ^{1/4} \\
&\leq &\left( \frac{1}{\min_{t}\bar{\sigma}_{t}^{2}}\right) ^{4}\left(
\max_{t}E\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{4}\right) ^{1/2}T^{-1}\sum_{t=1}^{T}\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1%
}{i}\frac{1}{j}\left( E\left( e_{t-i}^{8}\right) E\left( e_{t-j}^{8}\right)
\right) ^{1/8}E\left( e_{t}^{8}\right) ^{1/4} \\
&\leq &\left( \frac{1}{\min_{t}\bar{\sigma}_{t}^{2}}\right) ^{4}\left(
\max_{t}E\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right)
^{4}\right) ^{1/2}T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{t-1}\frac{1}{i}%
\right) ^{2} \\
&=&O\left( \frac{\left( \log T\right) ^{2}}{Tb}\right) .
\end{eqnarray*}

As for (\ref{B2a_1}), we have, by Lemma A(e),%
\begin{eqnarray}
&&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}E\left( \left( \bar{\sigma}_{t-s}^{2}-%
\tilde{\sigma}_{t-s}^{2}\right) \left( \bar{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right) \bar{\sigma}_{t-s}^{-4}\bar{\sigma}%
_{t}^{-4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&=&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{\left(
t-s\right) i}w_{tj}E\left( \left( e_{i}^{2}-\sigma _{i}^{2}\right) \left(
e_{j}^{2}-\sigma _{j}^{2}\right) \bar{\sigma}_{t-s}^{-4}\bar{\sigma}%
_{t}^{-4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&\leq
&CT^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{\left(
t-s\right) i}w_{tj}E\left( \left( e_{i}^{2}-\sigma _{i}^{2}\right) \left(
e_{j}^{2}-\sigma _{j}^{2}\right) x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag
\\
&=&CT^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{%
\left( t-s\right) i}w_{tj}E\left(
e_{i}^{2}e_{j}^{2}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \label{B2a_ww} \\
&&-CT^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{%
\left( t-s\right) i}w_{tj}\sigma _{j}^{2}E\left(
e_{i}^{2}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&&-CT^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{%
\left( t-s\right) i}w_{tj}\sigma _{i}^{2}E\left(
e_{j}^{2}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&&+CT^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}\sum_{j=1}^{T}w_{%
\left( t-s\right) i}w_{tj}\sigma _{i}^{2}\sigma _{j}^{2}E\left(
x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right) .  \notag
\end{eqnarray}

We will only show that (\ref{B2a_ww}) is zero since the remaining terms
follow immediately. First, when $i=j$ we write (\ref{B2a_ww}) as%
\begin{eqnarray}
&&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{T}w_{\left( t-s\right)
i}w_{ti}E\left( e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \label{B2a_ij}
\\
&=&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=1}^{t-1}w_{\left( t-s\right)
i}w_{ti}E\left( e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&&+T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}w_{\left( t-s\right) t}w_{tt}E\left(
e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right)  \notag \\
&&+T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=t+1}^{T}w_{\left( t-s\right)
i}w_{ti}E\left( e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right) .  \notag
\end{eqnarray}%
The first term is zero since 
\begin{equation*}
E\left( e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}|\tciFourier _{t-1}\right)
=e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}E\left( e_{t}|\tciFourier _{t-1}\right) =0,
\end{equation*}%
the second term is zero since $w_{tt}=0$ and finally the third term is zero
because%
\begin{eqnarray*}
&&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=t+1}^{T}w_{\left( t-s\right)
i}w_{ti}E\left( E\left( e_{i}^{4}x_{t-s-1}x_{t-1}e_{t-s}e_{t}|\tciFourier
_{t-1}\right) \right) \\
&=&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=t+1}^{T}w_{\left( t-s\right)
i}w_{ti}E\left( E\left( E\left(
x_{t-s-1}x_{t-1}e_{t-s}e_{t}e_{i}^{4}|\tciFourier _{i-1}\right) |\tciFourier
_{t-1}\right) \right) \\
&=&T^{-1}\sum_{t=2}^{T}\sum_{s=1}^{t-1}\sum_{i=t+1}^{T}w_{\left( t-s\right)
i}w_{ti}E\left( e_{i}^{4}\right) E\left( x_{t-s-1}x_{t-1}e_{t-s}E\left(
e_{t}|\tciFourier _{t-1}\right) \right) =0.
\end{eqnarray*}

Now when $i\neq j$ in (\ref{B2a_ww}), the arguments follow similarly from (%
\ref{B2a_ij}). For example if $i>j$ and both $i$ and $j$ greater than $t$
then 
\begin{eqnarray*}
E\left( e_{i}^{2}e_{j}^{2}x_{t-s-1}x_{t-1}e_{t-s}e_{t}\right) &=&E\left(
E\left( E\left( E\left(
e_{i}^{2}e_{j}^{2}x_{t-s-1}x_{t-1}e_{t-s}e_{t}|\tciFourier _{i-1}\right)
|\tciFourier _{j-1}\right) |\tciFourier _{t-1}\right) \right) \\
&=&0.
\end{eqnarray*}

For the second term in (\ref{B2a}),\ we have, by Lemma A(e,g,h) and (\ref%
{B1a}),%
\begin{eqnarray*}
&&\left \vert T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{2}-\tilde{\sigma%
}_{t}^{2}\right) ^{2}\tilde{\sigma}_{t}^{-2}\bar{\sigma}_{t}^{-4}%
\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right \vert \\
&\leq &\left( \frac{1}{\min \bar{\sigma}_{t}^{2}}\right) ^{2}\left( \frac{1}{%
\min \tilde{\sigma}_{t}^{2}}\right) \left \vert T^{-1/2}\sum_{t=1}^{T}\left( 
\bar{\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) ^{2}\sum_{i=1}^{t-1}\frac{%
1}{i}e_{t-i}e_{t}\right \vert \\
&\leq &\left( \frac{1}{\min \bar{\sigma}_{t}^{2}}\right) ^{2}\left( \frac{1}{%
\min \tilde{\sigma}_{t}^{2}}\right) \left( \sum_{t=1}^{T}\left( \bar{\sigma}%
_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) ^{4}\right) ^{1/2}\left(
T^{-1}\sum_{t=1}^{T}\left( \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right)
^{2}\right) ^{1/2} \\
&=&O_{p}\left( \frac{1}{T^{1/2}b}\right) =o_{p}\left( 1\right) .
\end{eqnarray*}

\bigskip

To show (\ref{B3}), we have%
\begin{equation*}
B\left( \bar{\sigma}\right) -B\left( \sigma \right)
=T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right)
\sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}.
\end{equation*}%
Note that 
\begin{equation*}
E\left( \left( \bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right) \tsum
\nolimits_{i=1}^{t-1}i^{-1}e_{t-i}e_{t}|\tciFourier _{t-1}\right) =\left( 
\bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right) \tsum
\nolimits_{i=1}^{t-1}i^{-1}e_{t-i}E\left( e_{t}|\tciFourier _{t-1}\right) =0
\end{equation*}%
so that $\left( \bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right)
\sum_{i=1}^{t-1}i^{-1}e_{t-i}e_{t}$ is a martingale difference sequence with
respect to $\tciFourier _{t-1}.$ Also note that 
\begin{equation*}
\sigma _{t}^{2}\left( \bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right) ^{2}\leq 
\bar{\sigma}_{t}^{-4}\sigma _{t}^{-2}\left \vert \bar{\sigma}_{t}^{2}+\sigma
_{t}^{2}\right \vert \cdot \left \vert \bar{\sigma}_{t}^{2}-\sigma
_{t}^{2}\right \vert \leq C\left \vert \bar{\sigma}_{t}^{2}-\sigma
_{t}^{2}\right \vert
\end{equation*}%
from Xu and Phillips (2008) equation (23), page 277. Then%
\begin{eqnarray*}
&&E\left( T^{-1/2}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{-2}-\sigma
_{t}^{-2}\right) \sum_{i=1}^{t-1}\frac{1}{i}e_{t-i}e_{t}\right) ^{2} \\
&=&T^{-1}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right)
^{2}\sum_{i=1}^{t-1}\sum_{j=1}^{t-1}\frac{1}{i}\frac{1}{j}E\left(
e_{t-i}e_{t-j}E\left( e_{t}^{2}|\tciFourier _{t-1}\right) \right) \\
&=&T^{-1}\sum_{t=1}^{T}\sigma _{t}^{2}\left( \bar{\sigma}_{t}^{-2}-\sigma
_{t}^{-2}\right) ^{2}\sum_{i=1}^{t-1}\frac{1}{i^{2}}E\left(
e_{t-i}^{2}\right) \\
&\leq &C\sup_{r}\sigma \left( r\right) ^{2}T^{-1}\sum_{t=1}^{T}\left \vert 
\bar{\sigma}_{t}^{2}-\sigma _{t}^{2}\right \vert \sum_{i=1}^{t-1}\frac{1}{%
i^{2}} \\
&\leq &C\sup_{r}\sigma \left( r\right) ^{2}\frac{\pi ^{2}}{6}%
T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}_{t}^{2}-\sigma _{t}^{2}\right
\vert \rightarrow 0,
\end{eqnarray*}%
by Lemma A(l).

\bigskip

\noindent\ \textbf{Part (b)} Recall $h_{t}\left( \gamma \right) $ given just
above Harris and Kew (2016) equation (A.3) and define its partitioned matrix
as%
\begin{eqnarray*}
h_{t}\left( \gamma \right) &=&-\frac{1}{\sigma _{t}^{2}}\left( c_{0}\left(
L;\gamma \right) e_{t}\cdot c_{2}\left( L;\gamma \right) e_{t}+\left(
c_{1}\left( L;\gamma \right) e_{t}\right) \left( c_{1}\left( L;\gamma
\right) e_{t}\right) ^{\prime }\right) \\
&=&\left( 
\begin{array}{cc}
h_{\theta \theta ,t}\left( \gamma \right) & h_{\theta \psi ,t}\left( \gamma
\right) \\ 
h_{\psi \theta ,t}\left( \gamma \right) & h_{\psi \psi ,t}\left( \gamma
\right)%
\end{array}%
\right) .
\end{eqnarray*}%
We will only deal with the term $h_{\theta \theta ,t}\left( \gamma \right) $
as the remaining terms are dealt with in the same fashion. Note that%
\begin{equation*}
h_{\theta \theta ,t}\left( \gamma \right) =-\frac{1}{\sigma _{t}^{2}}\left(
X_{0,t}\left( \gamma \right) X_{2,t-2}\left( \gamma \right) +X_{1,t-1}\left(
\gamma \right) ^{2}\right)
\end{equation*}%
where $X_{0,t}\left( \gamma \right) =c_{0}\left( L;\gamma \right)
e_{t},X_{2,t-2}\left( \gamma \right) $ is the $\left( 1,1\right) $-element
of the matrix $c_{2}\left( L;\gamma \right) e_{t}$ and $X_{1,t-1}\left(
\gamma \right) $ is the first element of the vector $c_{1}\left( L;\gamma
\right) e_{t}.$ Also $\hat{h}_{\theta \theta ,t}\left( \gamma \right) $ is
similarly defined by replacing $\sigma _{t}^{2}$ in $h_{\theta \theta
,t}\left( \gamma \right) $ with $\hat{\sigma}_{t}^{2}$.

For clarity we write $\left( \theta ,\psi \right) $ for $\gamma $ and we
will show that%
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left( \hat{h}_{\theta \theta
,t}\left( 0,\psi \right) -h_{\theta \theta ,t}\left( 0,\psi \right) \right) 
\overset{p}{\rightarrow }0.  \label{hhat_t}
\end{equation}%
To show (\ref{hhat_t}), we have%
\begin{equation*}
T^{-1}\sum_{t=1}^{T}\left( \hat{h}_{\theta \theta ,t}\left( 0,\psi \right)
-h_{\theta \theta ,t}\left( 0,\psi \right) \right)
=T^{-1}\sum_{t=1}^{T}\left( \hat{\sigma}_{t}^{-2}-\sigma _{t}^{-2}\right)
\left( X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left( 0,\psi \right)
+X_{1,t-1}\left( 0,\psi \right) ^{2}\right) .
\end{equation*}

First we will show that%
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left\vert X_{0,t}\left( 0,\psi
\right) X_{2,t-2}\left( 0,\psi \right) +X_{1,t-1}\left( 0,\psi \right)
^{2}\right\vert =O_{p}\left( 1\right) .  \label{supX123}
\end{equation}%
To show (\ref{supX123}) we write%
\begin{eqnarray*}
&&\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left\vert X_{0,t}\left( 0,\psi
\right) X_{2,t-2}\left( 0,\psi \right) +X_{1,t-1}\left( 0,\psi \right)
^{2}\right\vert \\
&\leq &\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left\vert X_{0,t}\left(
0,\psi \right) X_{2,t-2}\left( 0,\psi \right) \right\vert +\sup_{\psi \in
\Psi }T^{-1}\sum_{t=1}^{T}X_{1,t-1}\left( 0,\psi \right) ^{2} \\
&\leq &\left( \sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}X_{0,t}\left( 0,\psi
\right) ^{2}\right) ^{1/2}\left( \sup_{\psi \in \Psi
}T^{-1}\sum_{t=1}^{T}X_{2,t-2}\left( 0,\psi \right) ^{2}\right)
^{1/2}+\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}X_{1,t-1}\left( 0,\psi
\right) ^{2}.
\end{eqnarray*}%
\newline
Now from Cavaliere, Nielsen and Taylor (2015b) Lemma S.10, 
\begin{equation*}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}X_{0,t}\left( 0,\psi \right)
^{2}=O_{p}\left( 1\right)
\end{equation*}%
follows from setting $k=l=u_{1}=u_{2}=0$ in their Lemma S.10 equation
(S.16). Similarly for $X_{2,t-2}\left( 0,\psi \right) ^{2}$ with $k=l=2$ and 
$u_{1}=u_{2}=0$ and lastly for $X_{1,t-1}\left( 0,\psi \right) ^{2}$ with $%
k=l=1$ and $u_{1}=u_{2}=0.$

Second we define%
\begin{equation*}
A\left( \sigma ,\psi \right) =T^{-1}\sum_{t=1}^{T}\sigma _{t}^{-2}\left(
X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left( 0,\psi \right) +X_{1,t-1}\left(
0,\psi \right) ^{2}\right) .
\end{equation*}%
Now, following Xu and Phillips (2008), (\ref{hhat_t}) follows from showing
the following results:%
\begin{equation}
\sup_{\psi \in \Psi }\left \vert A\left( \hat{\sigma},\psi \right) -A\left( 
\tilde{\sigma},\psi \right) \right \vert \overset{p}{\rightarrow }0;
\label{A1}
\end{equation}%
\begin{equation}
\sup_{\psi \in \Psi }\left \vert A\left( \tilde{\sigma},\psi \right)
-A\left( \bar{\sigma},\psi \right) \right \vert \overset{p}{\rightarrow }0;
\label{A2}
\end{equation}%
\begin{equation}
\sup_{\psi \in \Psi }\left \vert A\left( \bar{\sigma},\psi \right) -A\left(
\sigma ,\psi \right) \right \vert \overset{p}{\rightarrow }0.  \label{A3}
\end{equation}

To show (\ref{A1}), by Lemma A(h,i,j) and equation (\ref{supX123}) above, we
write%
\begin{eqnarray*}
&&\sup_{\psi \in \Psi }\left \vert A\left( \hat{\sigma},\psi \right)
-A\left( \tilde{\sigma},\psi \right) \right \vert \\
&=&\sup_{\psi \in \Psi }\left \vert T^{-1}\sum_{t=1}^{T}\frac{\left( \hat{%
\sigma}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right) }{\tilde{\sigma}_{t}^{2}\hat{%
\sigma}_{t}^{2}}\left( X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left( 0,\psi
\right) +X_{1,t-1}\left( 0,\psi \right) ^{2}\right) \right \vert \\
&\leq &\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\frac{\left \vert \hat{\sigma%
}_{t}^{2}-\tilde{\sigma}_{t}^{2}\right \vert }{\tilde{\sigma}_{t}^{2}\hat{%
\sigma}_{t}^{2}}\left \vert X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left(
0,\psi \right) +X_{1,t-1}\left( 0,\psi \right) ^{2}\right \vert \\
&\leq &\frac{\max_{t}\left \vert \hat{\sigma}_{t}^{2}-\tilde{\sigma}%
_{t}^{2}\right \vert }{\left( \min_{t}\tilde{\sigma}_{t}^{2}\right) \left(
\min_{t}\hat{\sigma}_{t}^{2}\right) }\sup_{\psi \in \Psi
}T^{-1}\sum_{t=1}^{T}\left \vert X_{0,t}\left( 0,\psi \right)
X_{2,t-2}\left( 0,\psi \right) +X_{1,t-1}\left( 0,\psi \right) ^{2}\right
\vert \\
&=&O_{p}\left( \frac{1}{\sqrt{T}b}\right) .
\end{eqnarray*}

\bigskip

To show (\ref{A2}), using the same arguments as (\ref{A1}) but now using
Lemma A(e,g,h) and (\ref{supX123}), we have (\ref{A2}) $=O_{p}\left(
T^{-1/4}b^{-1/2}\right) .$

\bigskip

We now show (\ref{A3}). To simplify notation let $X_{t}\left( \psi \right)
=X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left( 0,\psi \right)
+X_{1,t-1}\left( 0,\psi \right) ^{2}.$ We write 
\begin{eqnarray*}
\sup_{\psi \in \Psi }\left \vert A\left( \bar{\sigma},\psi \right) -A\left(
\sigma ,\psi \right) \right \vert &=&\sup_{\psi \in \Psi }\left \vert
T^{-1}\sum_{t=1}^{T}\frac{\left( \bar{\sigma}_{t}^{2}-\sigma _{t}^{2}\right) 
}{\sigma _{t}^{2}\bar{\sigma}_{t}^{2}}X_{t}\left( \psi \right) \right \vert
\\
&\leq &\left( \frac{1}{\inf_{r}\sigma \left( r\right) ^{2}}\right) \left( 
\frac{1}{\min_{t}\bar{\sigma}_{t}^{2}}\right) \sup_{\psi \in \Psi
}T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}_{t}^{2}-\sigma _{t}^{2}\right
\vert \left \vert X_{t}\left( \psi \right) \right \vert .
\end{eqnarray*}%
Then, given Lemma A(e), (\ref{A3}) follows if we show 
\begin{equation}
\sup_{\psi \in \Psi }T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}%
_{t}^{2}-\sigma _{t}^{2}\right \vert \left \vert X_{t}\left( \psi \right)
\right \vert =o_{p}\left( 1\right) .  \label{sup_sbar}
\end{equation}%
To show (\ref{sup_sbar}), we first show pointwise convergence in $\psi $;
that is%
\begin{equation}
T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}_{t}^{2}-\sigma _{t}^{2}\right
\vert \left \vert X_{t}\left( \psi \right) \right \vert \overset{p}{%
\rightarrow }0.  \label{point_sbar}
\end{equation}%
To show this, we first note that 
\begin{eqnarray}
E\left \vert X_{t}\left( \psi \right) \right \vert &=&E\left \vert
X_{0,t}\left( 0,\psi \right) X_{2,t-2}\left( 0,\psi \right) +X_{1,t-1}\left(
0,\psi \right) ^{2}\right \vert  \notag \\
&\leq &\left( E\left( X_{0,t}\left( 0,\psi \right) \right) ^{2}\right)
^{1/2}\left( E\left( X_{2,t-2}\left( 0,\psi \right) \right) ^{2}\right)
^{1/2}+E\left( X_{1,t-1}\left( 0,\psi \right) ^{2}\right)  \notag \\
&<&C<\infty \ \ \text{uniformly in }t,  \label{EXpsi}
\end{eqnarray}%
because%
\begin{equation*}
E\left( X_{0,t}\left( 0,\psi \right) \right) ^{2}=E\left(
\sum_{j=0}^{t-1}\varphi _{j}\left( \psi \right) e_{t-j}\right)
^{2}=\sum_{j=0}^{t-1}\varphi _{j}\left( \psi \right) ^{2}E\left(
e_{t-j}^{2}\right) \leq \sup_{r}\sigma ^{2}\left( r\right)
\sum_{j=0}^{\infty }\varphi _{j}\left( \psi \right) ^{2}<C<\infty .
\end{equation*}%
Similarly for $\sup_{1\leq t\leq T}E\left( X_{2,t-2}\left( 0,\psi \right)
\right) ^{2}<\infty $ and $\sup_{1\leq t\leq T}E\left( X_{1,t-1}\left(
0,\psi \right) ^{2}\right) <\infty $ because the coefficients of these
linear processes are square summable. Now to show (\ref{point_sbar}), we have%
\begin{equation*}
E\left \vert T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}_{t}^{2}-\sigma
_{t}^{2}\right \vert \left \vert X_{t}\left( \psi \right) \right \vert
\right \vert \leq T^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}%
_{t}^{2}-\sigma _{t}^{2}\right \vert E\left \vert X_{t}\left( \psi \right)
\right \vert \leq CT^{-1}\sum_{t=1}^{T}\left \vert \bar{\sigma}%
_{t}^{2}-\sigma _{t}^{2}\right \vert =o\left( 1\right)
\end{equation*}%
by Lemma A(l) and (\ref{EXpsi}).

Finally, given (\ref{point_sbar}), (\ref{sup_sbar}) follows if we show that $%
\left\vert T^{-1}\sum_{t=1}^{T}\left( \bar{\sigma}_{t}^{2}-\sigma
_{t}^{2}\right) \sigma _{t}^{-2}\bar{\sigma}_{t}^{-2}X_{t}\left( \psi
\right) \right\vert $ is stochastically equicontinuous. By the same argument
that shows (\ref{supX123}) and Cavaliere, Nielsen and Taylor (2015b) Lemma
S.10, the stochastic equicontinuity holds since $\left( \bar{\sigma}%
_{t}^{2}-\sigma _{t}^{2}\right) \sigma _{t}^{-2}\bar{\sigma}_{t}^{-2}$ is a
non-stochastic sequence and does not depend on $\psi $ and hence does not
change the steps of their proof of Lemma S.10 equation (S.16).

\bigskip

\noindent \textbf{Part (c)} For clarity we write $\left( \theta ,\psi
\right) $ for $\gamma $ in what follows. Let 
\begin{equation*}
r\left( \psi \right) =\lim_{T\rightarrow \infty }T^{-1}\sum_{t=1}^{T}E\left( 
\frac{e_{t}\left( 0,\psi \right) }{\sigma _{t}}\right)
^{2}=\lim_{T\rightarrow \infty }T^{-1}\sum_{t=1}^{T}E\left( \frac{a\left(
L;\psi \right) a\left( L;\psi _{0}\right) ^{-1}e_{t}}{\sigma _{t}}\right)
^{2}
\end{equation*}%
and%
\begin{equation*}
r_{T}\left( \psi \right) =T^{-1}\sum_{t=1}^{T}\left( \frac{e_{t}\left(
0,\psi \right) }{\sigma _{t}}\right) ^{2}=T^{-1}\sum_{t=1}^{T}\left( \frac{%
a\left( L;\psi \right) a\left( L;\psi _{0}\right) ^{-1}e_{t}}{\sigma _{t}}%
\right) ^{2}.
\end{equation*}%
Following the proof of CNT Lemma A.1, the required result that $\bar{\psi}%
\overset{p}{\rightarrow }\psi _{0}$ holds if we show%
\begin{equation}
\sup_{\psi \in \Psi }\left \vert r_{T}\left( \psi \right) -r\left( \psi
\right) \right \vert \overset{p}{\rightarrow }0,  \label{con_a}
\end{equation}%
and%
\begin{equation}
\inf_{\psi \in \Psi \cap \left \{ \psi :\left \Vert \psi -\psi _{0}\right
\Vert \geq \epsilon \right \} }r\left( \psi \right) >r\left( \psi
_{0}\right) \ \ \ \text{for all }\epsilon >0,  \label{con_b}
\end{equation}%
where $\left \Vert x\right \Vert $ denote the usual Euclidean norm for any
vector $x.$

We first define%
\begin{equation*}
r^{\#}\left( \psi \right) =\lim_{T\rightarrow \infty
}T^{-1}\sum_{t=1}^{T}E\left( a\left( L;\psi \right) a\left( L;\psi
_{0}\right) ^{-1}\varepsilon _{t}\right) ^{2}
\end{equation*}%
and%
\begin{equation*}
r_{T}^{\#}\left( \psi \right) =T^{-1}\sum_{t=1}^{T}\left( a\left( L;\psi
\right) a\left( L;\psi _{0}\right) ^{-1}\varepsilon _{t}\right) ^{2}.
\end{equation*}%
To show (\ref{con_a}), we write%
\begin{eqnarray}
&&\sup_{\psi \in \Psi }\left \vert r_{T}\left( \psi \right) -r\left( \psi
\right) \right \vert  \notag \\
&=&\sup_{\psi \in \Psi }\left \vert r_{T}\left( \psi \right)
-r_{T}^{\#}\left( \psi \right) +r_{T}^{\#}\left( \psi \right) -r^{\#}\left(
\psi \right) +r^{\#}\left( \psi \right) -r\left( \psi \right) \right \vert 
\notag \\
&\leq &\sup_{\psi \in \Psi }\left \vert r_{T}\left( \psi \right)
-r_{T}^{\#}\left( \psi \right) \right \vert +\sup_{\psi \in \Psi }\left
\vert r_{T}^{\#}\left( \psi \right) -r^{\#}\left( \psi \right) \right \vert
+\sup_{\psi \in \Psi }\left \vert r^{\#}\left( \psi \right) -r\left( \psi
\right) \right \vert  \label{rT_r}
\end{eqnarray}

For the first term in (\ref{rT_r}) we have%
\begin{eqnarray}
&&\sup_{\psi \in \Psi }\left\vert r_{T}\left( \psi \right) -r_{T}^{\#}\left(
\psi \right) \right\vert  \notag \\
&=&\sup_{\psi \in \Psi }\left\vert T^{-1}\sum_{t=1}^{T}\left( \left(
\sum_{j=0}^{t-1}\varphi _{j}\left( \psi \right) \left( \frac{\sigma _{t-j}}{%
\sigma _{t}}\right) \varepsilon _{t-j}\right) ^{2}-\left(
\sum_{j=0}^{t-1}\varphi _{j}\left( \psi \right) \varepsilon _{t-j}\right)
^{2}\right) \right\vert .  \label{rT_rhash}
\end{eqnarray}%
Since $\varphi _{i}\left( \psi \right) $ decays exponentially in $i$
uniformly over all $\psi \in \Psi $ under Assumption R, (\ref{rT_rhash}) $%
\overset{p}{\rightarrow }0$ follows by the same arguments leading to (A.6)
in Harris and Kew (2016). The second term in (\ref{rT_r}) $\overset{p}{%
\rightarrow }0$ follows from CNT Lemma A.1 equation (A.1). The third term $%
\overset{p}{\rightarrow }0$ follows from the proof of (A.6) in Harris and
Kew (2016).

Finally, equation (\ref{con_b}) holds because the third term in (\ref{rT_r}) 
$\overset{p}{\rightarrow }0$ and CNT Lemma A.1 equation (A.2).

\bigskip

\noindent\ \textbf{Part (d)}. Define 
\begin{equation*}
\tilde{r}_{T}\left( \psi \right) =T^{-1}\sum_{t=1}^{T}\left( \frac{%
e_{t}\left( 0,\psi \right) }{\hat{\sigma}_{t}}\right) ^{2}.
\end{equation*}%
Recall $r_{T}\left( \psi \right) $ and $r\left( \psi \right) $ given in part
(c) above. The required result that $\tilde{\psi}\overset{p}{\rightarrow }%
\psi _{0}$ holds if we show%
\begin{equation*}
\sup_{\psi \in \Psi }\left\vert \tilde{r}_{T}\left( \psi \right) -r\left(
\psi \right) \right\vert \overset{p}{\rightarrow }0.
\end{equation*}%
To see this, we write%
\begin{equation*}
\sup_{\psi \in \Psi }\left\vert \tilde{r}_{T}\left( \psi \right)
-r_{T}\left( \psi \right) +r_{T}\left( \psi \right) -r\left( \psi \right)
\right\vert \leq \sup_{\psi \in \Psi }\left\vert \tilde{r}_{T}\left( \psi
\right) -r_{T}\left( \psi \right) \right\vert +\sup_{\psi \in \Psi
}\left\vert r_{T}\left( \psi \right) -r\left( \psi \right) \right\vert .
\end{equation*}%
The second term $\overset{p}{\rightarrow }0$ because of (\ref{con_a}). As
for the first term, 
\begin{equation*}
\sup_{\psi \in \Psi }\left\vert \tilde{r}_{T}\left( \psi \right)
-r_{T}\left( \psi \right) \right\vert =\sup_{\psi \in \Psi }\left\vert
T^{-1}\sum_{t=1}^{T}\left( \left( \frac{e_{t}\left( 0,\psi \right) }{\hat{%
\sigma}_{t}}\right) ^{2}-\left( \frac{e_{t}\left( 0,\psi \right) }{\sigma
_{t}}\right) ^{2}\right) \right\vert \overset{p}{\rightarrow }0,
\end{equation*}%
by the same arguments that give (\ref{hhat_t}) with $e_{t}\left( 0,\psi
\right) =X_{0,t}\left( \psi \right) .$

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{\Large References}

\begin{description}
\item {\small Cavaliere, G., M.\O . Nielsen and A.M.R. Taylor (2015a)
Bootstrap Score Tests for Fractional Integration in Heteroskedastic ARFIMA
Models, with an Application to Price Dynamics in Commodity Spot and Futures
Markets. \textit{Journal of Econometrics 187}, 557--579.}%
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\item {\small Cavaliere, G., M.\O . Nielsen and A.M.R. Taylor (2015b)
Quasi-Maximum Likelihood Estimation and Bootstrap Inference in Fractional
Time Series Models with Heteroskedasticity of Unknown Form. Working Papers
1324, Queen's University, Department of Economics.}%
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\item {\small Harris, D. and H. Kew (2016) Adaptive Long Memory Testing
under Heteroskedasticity. \textit{Econometric Thoery, }forthcoming.}%
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\item {\small Robinson, P.M. (1987). Asymptotically Efficient Estimation in
the Presence of Heteroskedasticity of Unknown Form. \textit{Econometrica}
55, 875-891.}%
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\item {\small Xu, K.-L. and P.C.B. Phillips (2008) Adaptive Estimation of
Autoregressive Models with Time-varying Variances. \textit{Journal of
Econometrics} 142, 265-280.}%
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\end{description}

\end{document}
