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\newtheorem{theorem}{Theorem}[section]
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\begin{document}

\title{Supplement to: Dynamic Panel Anderson-Hsiao Estimation with Roots
Near Unity\thanks{%
The author acknowledges support from the NSF under Grant No. SES 12-58258.}}
\author{Peter C. B. Phillips \\
%EndAName
\emph{Yale University, University of Auckland,}\\
\emph{Singapore Management University \& University of Southampton}}
\maketitle

This supplement provides detailed derivations and proofs of the results in
the paper \textquotedblleft Dynamic Panel Anderson-Hsiao Estimation with
Roots Near Unity\textquotedblright .\vspace{0.08in}

\begin{proof}[Proof of Theorem 1]
Part (i) follows by the Lindeberg L\'{e}vy CLT%
\begin{equation}
\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left( 
\begin{array}{c}
\sum_{t=2}^{T}\Delta u_{it}y_{it-2} \\ 
\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}%
\end{array}%
\right) \Rightarrow N\left( 0,V_{T}\right) ,  \label{aa1}
\end{equation}%
with 
\begin{equation}
V_{T}=\left( 
\begin{array}{cc}
\mathbb{E}\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) ^{2} & \mathbb{E}%
\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) \left(
\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}\right) \\ 
\mathbb{E}\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) \left(
\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}\right) & \mathbb{E}\left(
\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}\right) ^{2}%
\end{array}%
\right) .  \label{aa2}
\end{equation}%
To evaluate it is convenient to use partial summation 
\begin{equation}
\sum_{t=2}^{T}\Delta
u_{it}y_{it-2}=u_{iT}y_{iT-2}-u_{i1}y_{i0}-\sum_{t=3}^{T}u_{it-1}\Delta
y_{it-2}  \label{aa0}
\end{equation}%
To compute $V_{T},$ note that 
\begin{equation*}
\mathbb{E}\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) ^{2}=\mathbb{E}%
\left\{ \left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right)
-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} ^{2}
\end{equation*}%
\begin{eqnarray*}
&=&\mathbb{E}\left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) ^{2}-2\mathbb{E}%
\left\{ \left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) \left(
\sum_{t=3}^{T}u_{it-1}u_{it-2}\right) \right\} +\mathbb{E}\left(
\sum_{t=3}^{T}u_{it-1}u_{it-2}\right) ^{2} \\
&=&\mathbb{E}u_{iT}^{2}y_{iT-2}^{2}+\mathbb{E}u_{i1}^{2}y_{i0}^{2}+%
\sum_{t=3}^{T}\mathbb{E}\left( u_{it-1}^{2}u_{it-2}^{2}\right) \\
&=&\sigma ^{4}T_{2}+2\sigma ^{2}\mathbb{E}y_{i0}^{2}+\sigma
^{4}T_{2}=2\sigma ^{4}T_{2},
\end{eqnarray*}%
the final line following if the initial condition $y_{i0}=0,$ which will be
assumed in the calculations below. The large $n$ asymptotic results will
continue to hold for $y_{i0}=O_{p}\left( 1\right) $ even for finite $T$ with
some obvious minor adjustments to the variance matrix expressions involving
quantities of $O\left( 1\right) $ in $T.$ Next%
\begin{eqnarray*}
\mathbb{E}\left( \sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}\right) ^{2} &=&%
\mathbb{E}\left( \sum_{t=2}^{T}u_{it-1}y_{it-2}\right) ^{2}=\sigma
^{2}\sum_{t=2}^{T}\mathbb{E}y_{it-2}^{2}=\sigma ^{4}\sum_{t=2}^{T}\left(
t-2\right) \\
&=&\sigma ^{4}T_{2}T_{1}/2,
\end{eqnarray*}%
and, with $y_{i0}=0$ (or up to $O\left( 1\right) $ in $T$ if $y_{i0}\not=0)$ 
\begin{eqnarray*}
&&\mathbb{E}\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) \left(
\sum_{t=2}^{T}u_{it-1}y_{it-2}\right) =\mathbb{E}\left\{ \left( \left(
u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) -\sum_{t=3}^{T}u_{it-1}u_{it-2}\right)
\left( \sum_{t=2}^{T}u_{it-1}y_{it-2}\right) \right\} \\
&=&-\mathbb{E}\left( \sum_{t=3}^{T}u_{it-1}u_{it-2}\right) \left(
\sum_{t=2}^{T}u_{it-1}y_{it-2}\right) -\mathbb{E}\left( u_{i1}y_{i0}\right)
^{2} \\
&=&-\mathbb{E}\left\{ \sum_{t=3}^{T}\left( u_{it-1}u_{it-2}\right) \left(
u_{it-1}y_{it-2}\right) +\sum_{s,t=3;s\not=t}^{T}\left(
u_{it-1}u_{it-2}\right) \left( u_{is-1}y_{is-2}\right) \right\} \\
&=&-\sum_{t=3}^{T}\mathbb{E}u_{it-1}^{2}u_{it-2}^{2}=-\sigma
^{4}\sum_{t=3}^{T}1=-\sigma ^{4}T_{2}.
\end{eqnarray*}%
Then%
\begin{equation*}
V_{T}=\left( 
\begin{array}{cc}
2\sigma ^{4}T_{2} & -\sigma ^{4}T_{2} \\ 
-\sigma ^{4}T_{2} & \sigma ^{4}T_{2}T_{1}/2%
\end{array}%
\right) =\sigma ^{4}T_{2}\left( 
\begin{array}{cc}
2 & -1 \\ 
-1 & T_{1}/2%
\end{array}%
\right) ,
\end{equation*}%
as stated.

For Part (ii), simply write $\rho _{IV}-1=\frac{N_{nT}}{D_{nT}},$ and note
from (i) that $\left( N_{nT},D_{nT}\right) \underset{n\rightarrow \infty }{%
\Rightarrow }$ $\sigma ^{2}T_{2}^{1/2}\left( \xi _{N,T},\xi _{D,T}\right) $,
where $\left( \xi _{N,T},\xi _{D,T}\right) $ is bivariate $N\left( 0,\left[ 
\begin{array}{cc}
2 & -1 \\ 
-1 & T_{1}/2%
\end{array}%
\right] \right) .$ Next, decompose $\xi _{N,T}$ as $\xi _{N,T}=\xi _{N.D,T}+%
\frac{-1}{T_{1}/2}\xi _{D,T}$ where $\xi _{N.D,T}\equiv N\left( 0,2-\frac{%
\left( -1\right) ^{2}}{T_{1}/2}\right) =N\left( 0,2\left( 1-\frac{1}{T_{1}}%
\right) \right) $ is independent of $\xi _{D,T},$ so that%
\begin{equation*}
\left( 
\begin{array}{c}
\xi _{N.D,T} \\ 
\xi _{D,T}%
\end{array}%
\right) \equiv N\left( 0,\left[ 
\begin{array}{cc}
2\left( 1-\frac{1}{T_{1}}\right) & 0 \\ 
0 & T_{1}/2%
\end{array}%
\right] \right) .
\end{equation*}%
Combining these results, we have by joint weak convergence and continuous
mapping that as $n\rightarrow \infty $ with $T$ fixed,%
\begin{eqnarray}
\rho _{IV}-1 &=&\frac{N_{T}}{D_{T}}\underset{n\rightarrow \infty }{%
\Rightarrow }\frac{\xi _{N,T}}{\xi _{D,T}}=\frac{\xi _{N.D,T}-\frac{2}{T_{1}}%
\xi _{D,T}}{\xi _{D,T}}  \label{gg27} \\
&=&-\frac{2}{T_{1}}+\frac{\xi _{N.D,T}}{\xi _{D,T}}=-\frac{2}{T_{1}}+\frac{%
2\left( 1-\frac{1}{T_{1}}\right) ^{1/2}}{T_{1}^{1/2}}\frac{\zeta _{N}}{\zeta
_{D}}  \notag \\
&\equiv &-\frac{2}{T_{1}}+2\frac{\left( 1-\frac{1}{T_{1}}\right) ^{1/2}}{%
T_{1}^{1/2}}\mathbb{C}\text{,}  \label{gg7}
\end{eqnarray}%
where $\left( \zeta _{N},\zeta _{D}\right) \equiv N\left( 0,I_{2}\right) $
and $\mathbb{C}$ is a standard Cauchy variate. Thus%
\begin{equation}
\rho _{IV}-1\underset{n\rightarrow \infty }{\Rightarrow }-\frac{2}{T_{1}}+2%
\frac{\left( 1-\frac{1}{T_{1}}\right) ^{1/2}}{T_{1}^{1/2}}\mathbb{C}\text{,}
\label{gg17}
\end{equation}%
yielding the stated result.\vspace{0.08in}
\end{proof}

\begin{proof}[Proof of Theorem 2]
By definition we have%
\begin{eqnarray*}
\rho _{IV}-1 &=&\frac{\sum_{i=1}^{n}\sum_{t=2}^{T}\Delta u_{it}y_{it-2}}{%
\sum_{i=1}^{n}\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}} \\
&=&\frac{\sum_{i=1}^{n}\left\{ \left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right)
-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} }{\sum_{i=1}^{n}%
\sum_{t=2}^{T}u_{it-1}y_{it-2}},
\end{eqnarray*}%
and rescaling gives 
\begin{equation}
\sqrt{T}\left( \rho _{IV}-1\right) =\frac{\sum_{i=1}^{n}\frac{1}{\sqrt{T}}%
\left\{ \left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right)
-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} }{\sum_{i=1}^{n}\frac{1}{T}%
\sum_{t=2}^{T}u_{it-1}y_{it-2}}.  \label{gg12}
\end{equation}%
By partial summation 
\begin{equation*}
\sum_{t=1}^{T}u_{it}y_{it-1}=\sum_{t=1}^{T}u_{it}\left(
\sum_{s=1}^{t-1}u_{is}+y_{i0}\right) =\frac{1}{2}\left\{ \left(
\sum_{t=1}^{T}u_{it}\right) ^{2}-\sum_{t=1}^{T}u_{it}^{2}\right\}
+\sum_{t=1}^{T}u_{it}y_{i0}.
\end{equation*}%
Using the fact that $\mathbb{E}\left( u_{it}u_{is}u_{is-1}\right) =0$ for
all $\left( t,s\right) $, we have by standard functional limit theory for $%
r\in \left[ 0,1\right] $ 
\begin{equation*}
T^{-1/2}\sum_{t=1}^{\lfloor Tr\rfloor }\left[ 
\begin{array}{c}
u_{it} \\ 
u_{it}u_{it-1}%
\end{array}%
\right] \Rightarrow \left[ 
\begin{array}{c}
B_{i}\left( r\right) \\ 
G_{i}\left( r\right)%
\end{array}%
\right] \equiv BM\left( \left[ 
\begin{array}{cc}
\sigma ^{2} & 0 \\ 
0 & \sigma ^{4}%
\end{array}%
\right] \right) ,
\end{equation*}%
where $B_{i}$ and $G_{i}$ are independent Brownian motions for all $i.$
Then, since $y_{i0}=O_{p}\left( 1\right) $ and $T^{-1}%
\sum_{t=1}^{T}u_{it}=o_{p}\left( 1\right) ,$ we deduce the joint weak
convergence (Phillips, 1987a, 1989)%
\begin{equation}
\left[ 
\begin{array}{c}
T^{-1/2}\sum_{t=1}^{T}u_{it} \\ 
T^{-1}\sum_{t=1}^{T}u_{it}y_{it-1} \\ 
T^{-1}\sum_{t=1}^{T}u_{it}u_{it-1}%
\end{array}%
\right] \underset{T\rightarrow \infty }{\Rightarrow }\left[ 
\begin{array}{c}
B_{i}\left( 1\right) \\ 
\frac{1}{2}\left\{ B_{i}\left( 1\right) ^{2}-\sigma ^{2}\right\} \\ 
G_{i}\left( 1\right)%
\end{array}%
\right] =\left[ 
\begin{array}{c}
B_{i}\left( 1\right) \\ 
\int_{0}^{1}B_{i}dB_{i} \\ 
G_{i}\left( 1\right)%
\end{array}%
\right] .  \label{gg16}
\end{equation}%
Since $u_{it}$ is $iid$ over $t$ and $i,$ it follows that $u_{iT}\Rightarrow
u_{i\infty }$ as $T\rightarrow \infty ,$ where the limit variates $\left\{
u_{i\infty }\right\} $ are independent over $i$ and have the same
distribution as $u_{it}.$ Note that $u_{iT}$ is independent of $\left(
T^{-1/2}\sum_{t=1}^{T_{1}}u_{it},T^{-1}%
\sum_{t=1}^{T_{1}}u_{it}y_{it-1},T^{-1}\sum_{t=1}^{T_{1}}u_{it}u_{it-1}%
\right) $ and, hence, asymptotically independent of $\left(
T^{-1/2}\sum_{t=1}^{T}u_{it},T^{-1}\sum_{t=1}^{T}u_{it}y_{it-1},T^{-1}%
\sum_{t=1}^{T}u_{it}u_{it-1}\right) .$ It follows that $u_{i\infty }$ is
independent of the vector of limit variates (\ref{gg16}). We therefore have
the combined weak convergence%
\begin{equation}
\left[ 
\begin{array}{c}
T^{-1/2}\sum_{t=1}^{T}u_{it} \\ 
T^{-1}\sum_{t=1}^{T}u_{it}y_{it-1} \\ 
T^{-1/2}\sum_{t=1}^{T}u_{it}u_{it-1} \\ 
u_{iT}%
\end{array}%
\right] \underset{T\rightarrow \infty }{\Rightarrow }\left[ 
\begin{array}{c}
B_{i}\left( 1\right) \\ 
\frac{1}{2}\left\{ B_{i}\left( 1\right) ^{2}-\sigma ^{2}\right\} \\ 
G_{i}\left( 1\right) \\ 
u_{i\infty }%
\end{array}%
\right] =\left[ 
\begin{array}{c}
B_{i}\left( 1\right) \\ 
\int_{0}^{1}B_{i}dB_{i} \\ 
G_{i}\left( 1\right) \\ 
u_{i\infty }%
\end{array}%
\right] .  \label{gg19}
\end{equation}%
Setting $G_{i}=G_{i}\left( 1\right) ,$ the stated result 
\begin{equation}
\sqrt{T}\left( \rho _{IV}-1\right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{\sum_{i=1}^{n}\left\{ u_{i\infty }B_{i}\left( 1\right)
-G_{i}\right\} }{\sum_{i=1}^{n}\frac{1}{2}\left\{ B_{i}\left( 1\right)
^{2}-\sigma ^{2}\right\} }  \label{gg28}
\end{equation}%
follows from (\ref{gg12}) and (\ref{gg19}) by continuous mapping.

For part (ii) we consider sequential asymptotics in which $T\rightarrow
\infty $ is followed by $n\rightarrow \infty .$ Observe that $u_{i\infty
}B_{i}\left( 1\right) -G_{i}$ is $iid$ over $i$ with zero mean and variance 
\begin{equation*}
\mathbb{E}\left\{ u_{i\infty }B_{i}\left( 1\right) -G_{i}\left( 1\right)
\right\} ^{2}=\mathbb{E}\left( u_{i\infty }^{2}\right) \mathbb{E}\left(
B_{i}\left( 1\right) ^{2}\right) +\mathbb{E}\left( G_{i}\left( 1\right)
^{2}\right) =2\sigma ^{4},
\end{equation*}%
and is uncorrelated with $B_{i}\left( 1\right) ^{2}.$ Since $\left\{
B_{i}\left( 1\right) ^{2}-\sigma ^{2}\right\} $ is $iid$ with zero mean and
variance $2\sigma ^{4},$ application of the Lindeberg L\'{e}vy CLT as $%
n\rightarrow \infty $ gives%
\begin{equation}
\left[ 
\begin{array}{c}
n^{-1/2}\sum_{i=1}^{n}\left\{ u_{i\infty }B_{i}\left( 1\right) -G_{i}\right\}
\\ 
n^{-1/2}\sum_{i=1}^{n}\frac{1}{2}\left\{ B_{i}\left( 1\right) ^{2}-\sigma
^{2}\right\}%
\end{array}%
\right] \underset{n\rightarrow \infty }{\Rightarrow }\left[ 
\begin{array}{c}
\left( 2\sigma ^{4}\right) ^{1/2}\zeta _{N} \\ 
\left( \sigma ^{4}/2\right) ^{1/2}\zeta _{D}%
\end{array}%
\right] ,  \label{gg18}
\end{equation}%
where $\left( \zeta _{N},\zeta _{D}\right) \equiv N\left( 0,I_{2}\right) .$
Hence, 
\begin{equation}
\sqrt{T}\left( \rho _{IV}-1\right) \underset{\left( n,T\right) _{\func{seq}%
}\rightarrow \infty }{\Rightarrow }2\mathbb{C},  \label{gg15}
\end{equation}%
giving the required result. \vspace{0.08in}
\end{proof}

\begin{proof}[Proof of Theorem 3]
We proceed by examining a set of sufficient conditions for joint convergence
limit theory developed in Phillips and Moon (1999). In particular, we
consider conditions that suffice to ensure that sequential convergence as $%
\left( n,T\right) _{\mathrm{\func{seq}}}\rightarrow \infty $ (i.e., $%
T\rightarrow \infty $ followed by $n\rightarrow \infty )$ implies joint
convergence $\left( n,T\right) \rightarrow \infty $ where there is no
restriction on the diagonal path in which $n$ and $T$ pass to infinity.

We start by defining the vector of standardized components appearing in the
numerator and denominator of $\rho _{IV}$ 
\begin{equation}
X_{nT}=\left( n^{-1/2}\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{ \left(
u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) -\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\}
,n^{-1/2}\sum_{i=1}^{n}\left( \frac{1}{T}\sum_{t=2}^{T}u_{it-1}y_{it-2}%
\right) \right) ^{\prime }.  \label{gg21}
\end{equation}%
From (\ref{gg19}) and (\ref{gg18}) we have the sequential convergence 
\begin{eqnarray}
X_{nT}\underset{T\rightarrow \infty }{\Rightarrow }X_{n} &:&=\left(
n^{-1/2}\sum_{i=1}^{n}\left\{ u_{i\infty }B_{i}\left( 1\right) -G_{i}\left(
1\right) \right\} ,n^{-1/2}\sum_{i=1}^{n}\frac{1}{2}\left\{ B_{i}\left(
1\right) ^{2}-\sigma ^{2}\right\} \right) ^{\prime }  \notag \\
\underset{n\rightarrow \infty }{\Rightarrow }X &:&=\left( \left( 2\sigma
^{4}\right) ^{1/2}\zeta _{N},\left( \frac{\sigma ^{4}}{2}\right) ^{1/2}\zeta
_{D}\right) ,  \label{gg23}
\end{eqnarray}%
which in turn implies the sequential limit $\sqrt{T}\left( \rho
_{IV}-1\right) \underset{\left( n,T\right) _{\func{seq}}\rightarrow \infty }{%
\Rightarrow }2\mathbb{C}$ given in (\ref{gg15}). By Lemma 6(b) of Phillips
and Moon (1999), when $X_{nT}\underset{T\rightarrow \infty }{\Rightarrow }%
X_{n}\underset{n\rightarrow \infty }{\Rightarrow }X$ sequentially, joint
weak convergence $X_{nT}\Rightarrow X$ as $\left( n,T\right) \rightarrow
\infty $ holds if and only if%
\begin{equation}
\underset{n,T\rightarrow \infty }{\lim \sup }\left\vert \mathbb{E}f\left(
X_{nT}\right) -\mathbb{E}f\left( X_{n}\right) \right\vert =0  \label{gg20}
\end{equation}%
for all bounded, continuous real functions $f$ on $\mathbb{R}^{2}$.

Simple primitive conditions sufficient for (\ref{gg20}) to hold are
available in the case where the components of the random quantity $X_{nT}$
involve averages of $iid$ random variables as in the present case where we
have $X_{nT}=n^{-1/2}\sum_{i=1}^{n}Y_{iT}$ with the $Y_{iT}$ independent
over $i.$ Component-wise we have%
\begin{equation*}
X_{nT}=\left( X_{1nT},X_{2nT}\right) ^{\prime }:=\left(
n^{-1/2}\sum_{i=1}^{n}Y_{1iT},n^{-1/2}\sum_{i=1}^{n}Y_{2iT}\right) ,
\end{equation*}%
where $Y_{iT}=\left( Y_{1iT},Y_{2iT}\right) ^{\prime }$ with 
\begin{eqnarray*}
Y_{1iT} &=&\frac{1}{\sqrt{T}}\left\{ \left(
u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) -\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} 
\underset{T\rightarrow \infty }{\Rightarrow }Y_{1i}:=u_{i\infty }B_{i}\left(
1\right) -G_{i}\left( 1\right) , \\
Y_{2iT} &=&\frac{1}{T}\sum_{t=2}^{T}u_{it-1}y_{it-2}\underset{T\rightarrow
\infty }{\Rightarrow }Y_{2i}:=\frac{1}{2}\left\{ B_{i}\left( 1\right)
^{2}-\sigma ^{2}\right\} ,
\end{eqnarray*}%
for all $i.$ The working probability space can be expanded as needed to
ensure that the (limit) random quantities $Y_{i}:=\left(
Y_{1i},Y_{2i}\right) ^{\prime }$ are defined in the same space for all $i$
so that averages involving $\sum_{i=1}^{n}Y_{i}$ are meaningful. In this
framework we can use a result on joint convergence by Phillips and Moon
(1999) -- see lemma PM below -- to verify condition (\ref{gg20}). In what
follows we use the notation of lemma PM.

We proceed to verify these conditions for $Y_{iT}$ and $Y_{i}.$ First, $%
Y_{iT}$ is integrable since%
\begin{eqnarray*}
\mathbb{E}\left\vert u_{iT}y_{iT-2}\right\vert &\leq &\left( \mathbb{E}%
\left\vert u_{iT}\right\vert ^{2}\mathbb{E}\left\vert y_{iT-2}\right\vert
^{2}\right) ^{1/2}<\infty , \\
\mathbb{E}\left\vert \sum_{t=2}^{T}u_{it-1}u_{it-2}\right\vert &\leq &T%
\mathbb{E}\left\vert u_{it-1}u_{it-2}\right\vert \leq T\mathbb{E}\left(
u_{it}^{2}\right) <\infty , \\
\mathbb{E}\left\vert \sum_{t=2}^{T}u_{it-1}y_{it-2}\right\vert &\leq
&\sum_{t=2}^{T}\mathbb{E}\left\vert u_{it-1}y_{it-2}\right\vert \leq
\sum_{t=2}^{T}\left( \mathbb{E}u_{it-1}^{2}\mathbb{E}y_{it-2}^{2}\right)
^{1/2}<\infty .
\end{eqnarray*}%
To show (i) holds, observe that 
\begin{eqnarray}
\mathbb{E}\left\Vert Y_{iT}\right\Vert ^{2} &=&\mathbb{E}Y_{1iT}^{2}+\mathbb{%
E}Y_{2iT}^{2}  \notag \\
&=&\frac{1}{T}\mathbb{E}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} ^{2}+\frac{1}{T^{2}}%
\mathbb{E}\left( \sum_{t=2}^{T}u_{it-1}y_{it-2}\right) ^{2}  \notag \\
&=&2\sigma ^{4}\frac{T-2}{T}+\sigma ^{4}\frac{1}{T^{2}}\sum_{t=2}^{T}\left(
t-2\right) <\infty ,  \label{gg22}
\end{eqnarray}%
when $y_{i0}=0,$ with obviously valid extension to the case where $%
y_{i0}=O_{p}\left( 1\right) $ with finite second moments. Then 
\begin{equation*}
\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}\sum_{i=1}^{n}%
\mathbb{E}\left\Vert Y_{iT}\right\Vert =\underset{T\rightarrow \infty }{\lim
\sup }\mathbb{E}\left\Vert Y_{iT}\right\Vert \leq \underset{T\rightarrow
\infty }{\lim \sup }\left( \mathbb{E}\left\Vert Y_{iT}\right\Vert
^{2}\right) ^{1/2}<\infty ,
\end{equation*}%
as required. To show (ii) holds, simply observe that $\mathbb{E}Y_{iT}=%
\mathbb{E}Y_{i}=0.$ To show (iii) holds, note that 
\begin{equation*}
\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}\sum_{i=1}^{n}%
\mathbb{E}\left\Vert Y_{iT}\right\Vert \mathbf{1}\left\{ \left\Vert
Y_{iT}\right\Vert >n\epsilon \right\} =\underset{T\rightarrow \infty }{\lim
\sup }\mathbb{E}\left\Vert Y_{iT}\right\Vert \mathbf{1}\left\{ \left\Vert
Y_{iT}\right\Vert >n\epsilon \right\} =0,\text{ for all }\epsilon >0,
\end{equation*}%
since $\sup_{T}\mathbb{E}\left\Vert Y_{iT}\right\Vert ^{2}<\infty $ by
virtue of (\ref{gg22}). Finally, note that 
\begin{equation*}
\underset{n\rightarrow \infty }{\lim \sup }\frac{1}{n}\sum_{i=1}^{n}\mathbb{E%
}\left\Vert Y_{i}\right\Vert \mathbf{1}\left\{ \left\Vert Y_{i}\right\Vert
>n\epsilon \right\} =\underset{n\rightarrow \infty }{\lim \sup }\mathbb{E}%
\left\Vert Y_{i}\right\Vert \mathbf{1}\left\{ \left\Vert Y_{i}\right\Vert
>n\epsilon \right\} =0,
\end{equation*}%
since $\mathbb{E}\left\Vert Y_{i}\right\Vert ^{2}<\infty ,$ proving (iv).
Hence, condition (\ref{gg20}) holds and we\ have joint weak convergence%
\begin{equation*}
X_{nT}=n^{-1/2}\sum_{i=1}^{n}Y_{iT}\underset{n,T\rightarrow \infty }{%
\Rightarrow }X:=\left( \left( 2\sigma ^{4}\right) ^{1/2}\zeta _{N},\left( 
\frac{\sigma ^{4}}{2}\right) ^{1/2}\zeta _{D}\right) ,
\end{equation*}%
irrespective of the divergence rates of $n$ and $T$ to infinity. By
continuous mapping, the required result follows for the GMM estimator so
that $\sqrt{T}\left( \rho _{IV}-1\right) \underset{n,T\rightarrow \infty }{%
\Rightarrow }2\mathbb{C}$ jointly as $\left( n,T\right) \rightarrow \infty $
irrespective of the order and rates of divergence of the respective sample
sizes.
\end{proof}

\begin{description}
\item[Lemma PM (Phillips and Moon, 1999, theorem 1)] \textit{Suppose the }$%
m\times 1$\textit{\ random vectors }$Y_{iT}$\textit{\ are independent across 
}$i$\textit{\ for all }$T$\textit{\ and integrable. Assume that }$%
Y_{iT}\Rightarrow Y_{i}$\textit{\ as }$T\rightarrow \infty $\textit{\ for
all }$i.$\textit{\ Then, condition (\ref{gg20}) holds if the following hold:%
\newline
(i) }$\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}%
\sum_{i=1}^{n}E\left\Vert Y_{iT}\right\Vert <\infty ,$\textit{\newline
(ii) }$\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}%
\sum_{i=1}^{n}\left\Vert \mathbb{E}Y_{iT}-\mathbb{E}Y_{i}\right\Vert <\infty
,$\textit{\newline
(iii) }$\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}%
\sum_{i=1}^{n}E\left\Vert Y_{iT}\right\Vert 1\left\{ \left\Vert
Y_{iT}\right\Vert >n\epsilon \right\} =0,$\textit{\ for all }$\epsilon >0$%
\textit{\newline
(iv) }$\underset{n\rightarrow \infty }{\lim \sup }\frac{1}{n}%
\sum_{i=1}^{n}E\left\Vert Y_{i}\right\Vert 1\left\{ \left\Vert
Y_{i}\right\Vert >n\epsilon \right\} =0,$\textit{\ for all }$\epsilon >0%
\vspace{0.08in}$
\end{description}

\begin{proof}[Proof of Theorem 4]
In case (i) $T$ is fixed as well as $c<0,$ which implies that $\rho =1+\frac{%
c}{\sqrt{T}}$ is fixed. So large $n$ asymptotics follow as in the
(asymptotically) stationary case. By defintion we\ have $y_{it}=\alpha
_{i}\left( 1-\rho \right) +\rho y_{it-1}+u_{it}=-\frac{\alpha _{i}c}{\sqrt{T}%
}+\left( 1+\frac{c}{\sqrt{T}}\right) y_{it-1}+u_{it}$ and $\Delta
y_{it}=\rho \Delta y_{it-1}+\Delta u_{it}$ so that $\Delta y_{it}=\alpha
_{i}\left( 1-\rho \right) +\left( \rho -1\right) y_{it-1}+u_{it}=-\frac{%
\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-1}+u_{it}.$ Then, as usual, $%
\mathbb{E}\left( u_{it}y_{it-2}\right) =\mathbb{E}\left( \Delta
u_{it}y_{it-2}\right) =0$ and orthogonality holds. When $y_{i0}=0,$ back
substitution gives%
\begin{equation*}
y_{it}=\alpha _{i}\left( 1-\rho ^{t}\right) +\sum_{j=0}^{t-1}\rho
^{j}u_{it-j},
\end{equation*}%
and $\mathbb{E}\left( y_{it}\right) =\alpha _{i}\left( 1-\rho ^{t}\right) ,$ 
$\mathbb{V}\mathrm{ar}\left( y_{it}\right) =\sigma ^{2}\sum_{j=0}^{t-1}\rho
^{2j}=\sigma ^{2}\frac{1-\rho ^{2t}}{1-\rho ^{2}},$ and $\mathbb{E}\left(
y_{it}^{2}\right) =\sigma ^{2}\frac{1-\rho ^{2t}}{1-\rho ^{2}}+\alpha
_{i}^{2}\left( 1-\rho ^{t}\right) ^{2}.$ Instrument relevance is determined
by the magnitude of the moment 
\begin{eqnarray}
\mathbb{E}\left( \Delta y_{it-1}y_{it-2}\right)  &=&\mathbb{E}\left( \left\{
\alpha _{i}\left( 1-\rho \right) +\left( \rho -1\right)
y_{it-2}+u_{it-1}\right\} y_{it-2}\right)   \notag \\
&=&\alpha _{i}^{2}\left( 1-\rho \right) \left( 1-\rho ^{t-2}\right) +\left(
\rho -1\right) \left\{ \sigma ^{2}\frac{1-\rho ^{2\left( t-2\right) }}{%
1-\rho ^{2}}+\alpha _{i}^{2}\left( 1-\rho ^{t-2}\right) ^{2}\right\}   \notag
\\
&=&-\sigma ^{2}\frac{1-\rho ^{2\left( t-2\right) }}{1+\rho }-\alpha
_{i}^{2}\left( 1-\rho \right) \left( 1-\rho ^{t-2}\right) \rho ^{t-2}
\label{aa3}
\end{eqnarray}%
which is nonzero for $c<0$ and zero when $c=0,$ corresponding to the unit
root case $\left( \rho =1\right) $ considered earlier. Note that in the
fully stationary case where initial conditions are in the infinite past so
that $y_{i0}=\alpha _{i}+\sum_{j=0}^{\infty }\rho ^{j}u_{i,-j}$ and $%
y_{it}=\alpha _{i}+\sum_{j=0}^{\infty }\rho ^{j}u_{t-j}$ we have 
\begin{eqnarray*}
\mathbb{E}\left( \Delta y_{it-1}y_{it-2}\right)  &=&\alpha _{i}^{2}\left(
1-\rho \right) +\left( \rho -1\right) \mathbb{E}\left( y_{it}^{2}\right)
=\alpha _{i}^{2}\left( 1-\rho \right) -\left( 1-\rho \right) \left\{ \frac{%
\sigma ^{2}}{1-\rho ^{2}}+\alpha _{i}^{2}\right\}  \\
&=&-\frac{\sigma ^{2}}{1+\rho },
\end{eqnarray*}%
which corresponds with the leading term of (\ref{aa3}) when $t\rightarrow
\infty $ with $\left\vert \rho \right\vert <1$.

Now consider the numerator and denominator of the centred and scaled GMM
estimate%
\begin{equation}
\sqrt{n}\left( \rho _{IV}-\rho \right) =\frac{\frac{1}{\sqrt{n}}%
\sum_{i=1}^{n}\sum_{t=2}^{T}\Delta u_{it}y_{it-2}}{\frac{1}{n}%
\sum_{i=1}^{n}\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}}=:\frac{N_{nT}}{D_{nT}}.
\label{aa8}
\end{equation}%
First, noting that $\Delta y_{it-1}y_{it-2}$ is quadratic in $\alpha _{i},$
and using $T_{j}=T-j$ and $\sigma _{\alpha }^{2}=\lim_{n\rightarrow \infty }%
\frac{1}{n}\sum_{i=1}^{n}\alpha _{i}^{2},$ the denominator of (\ref{aa8})
takes the following form as $n\rightarrow \infty $%
\begin{eqnarray}
&&\frac{1}{n}\sum_{i=1}^{n}\sum_{t=2}^{T}\Delta y_{it-1}y_{it-2}\rightarrow
_{p}\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left(
\sum_{t=2}^{T}\Delta y_{it}y_{it-2}\right)  \notag \\
&=&\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\sum_{t=2}^{T}\left\{
-\sigma ^{2}\frac{1-\rho ^{2\left( t-2\right) }}{1+\rho }+\alpha
_{i}^{2}\left( 1-\rho \right) \left( 1-\rho ^{t-2}\right) \left[ 1-\left(
1-\rho ^{t-2}\right) \right] \right\}  \notag \\
&=&-\frac{\sigma ^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho ^{2T_{1}}}{1-\rho
^{2}}\right] +\sigma _{\alpha }^{2}\left( 1-\rho \right) \left[ T_{1}-\frac{%
1-\rho ^{T_{1}}}{1-\rho }\right] -\sigma _{\alpha }^{2}\left( 1-\rho \right) %
\left[ T_{1}-2\frac{1-\rho ^{T_{1}}}{1-\rho }+\frac{1-\rho ^{2T_{1}}}{1-\rho
^{2}}\right]  \notag \\
&=&-\frac{\sigma ^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho ^{2T_{1}}}{1-\rho
^{2}}\right] +\sigma _{\alpha }^{2}\left( 1-\rho \right) \left[ \frac{1-\rho
^{T_{1}}}{1-\rho }-\frac{1-\rho ^{2T_{1}}}{1-\rho ^{2}}\right]  \notag \\
&=&-\frac{\sigma ^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho ^{2T_{1}}}{1-\rho
^{2}}\right] +\sigma _{\alpha }^{2}\left[ 1-\rho ^{T_{1}}-\frac{1-\rho
^{2T_{1}}}{1+\rho }\right] ,  \label{aa10}
\end{eqnarray}%
which is again zero when $c=0$ ($\rho =1)$. Turning to the numerator, we
have $\mathbb{E}\left( \Delta u_{it}y_{it-2}\right) =0$ by orthogonality and
by a standard CLT\ argument for fixed $T$ as $n\rightarrow \infty $%
\begin{equation*}
\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left( \sum_{t=2}^{T}\Delta
u_{it}y_{it-2}\right) \Rightarrow N\left( 0,v_{T}\right)
\end{equation*}%
with%
\begin{equation*}
v_{T}=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left(
\sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) ^{2}
\end{equation*}%
We evaluate the above variance as follows. Using partial summation and $%
y_{i0}=0,$ we have%
\begin{equation}
\sum_{t=2}^{T}\Delta
u_{it}y_{it-2}=u_{iT}y_{iT-2}-u_{i1}y_{i0}-\sum_{t=3}^{T}u_{it-1}\Delta
y_{it-2}=u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\Delta y_{it-2},  \label{aa4}
\end{equation}%
with variance 
\begin{eqnarray*}
\mathbb{E}\left( u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\Delta y_{it-2}\right)
^{2} &=&\sigma ^{2}\mathbb{E}\left( y_{iT-2}\right) ^{2}+\sigma ^{2}\mathbb{E%
}\sum_{t=3}^{T}\left( \Delta y_{it-2}\right) ^{2} \\
&=&\sigma ^{4}\frac{1-\rho ^{2T_{2}}}{1-\rho ^{2}}+\alpha _{i}^{2}\sigma
^{2}\left( 1-\rho ^{T_{2}}\right) ^{2}+\sigma ^{2}\mathbb{E}%
\sum_{t=3}^{T}\left( \Delta y_{it-2}\right) ^{2}.
\end{eqnarray*}%
Using $\mathbb{E}\left( y_{it}\right) =\alpha _{i}\left( 1-\rho ^{t}\right)
, $ $\mathbb{V}\mathrm{ar}\left( y_{it}\right) =\sigma
^{2}\sum_{j=0}^{t-1}\rho ^{2j}=\sigma ^{2}\frac{1-\rho ^{2t}}{1-\rho ^{2}},$ 
$\mathbb{E}\left( y_{it}^{2}\right) =\sigma ^{2}\frac{1-\rho ^{2t}}{1-\rho
^{2}}+\alpha _{i}^{2}\left( 1-\rho ^{t}\right) ^{2},$ and $\Delta
y_{it}=\alpha _{i}\left( 1-\rho \right) +\left( \rho -1\right)
y_{it-1}+u_{it},$ the final term $\sum_{t=3}^{T}\mathbb{E}\left( \Delta
y_{it-2}\right) ^{2}$ above is 
\begin{eqnarray*}
&&\sum_{t=3}^{T}\left\{ \alpha _{i}^{2}\left( 1-\rho \right) ^{2}+\left(
1-\rho \right) ^{2}\mathbb{E}\left( y_{it-3}^{2}\right) +\sigma ^{2}-2\alpha
_{i}^{2}\left( 1-\rho \right) ^{2}\left( 1-\rho ^{t-3}\right) \right\} \\
&=&\sigma ^{2}T_{2}-T_{2}\alpha _{i}^{2}\left( 1-\rho \right) ^{2}+2\alpha
_{i}^{2}\left( 1-\rho \right) ^{2}\left( \frac{1-\rho ^{T_{2}}}{1-\rho }%
\right) +\left( 1-\rho \right) ^{2}\sum_{t=3}^{T}\mathbb{E}\left(
y_{it-3}^{2}\right) \\
&=&\sigma ^{2}T_{2}-T_{2}\alpha _{i}^{2}\left( 1-\rho \right) ^{2}+2\alpha
_{i}^{2}\left( 1-\rho \right) \left( 1-\rho ^{T_{2}}\right) +\left( 1-\rho
\right) ^{2}\frac{\sigma ^{2}}{1-\rho ^{2}}\left[ T_{2}-\frac{1-\rho
^{2T_{2}}}{1-\rho ^{2}}\right] \\
&&+\alpha _{i}^{2}\left( 1-\rho \right) ^{2}\left[ T_{2}-2\frac{1-\rho
^{T_{2}}}{1-\rho }+\frac{1-\rho ^{2T_{2}}}{1-\rho ^{2}}\right]
\end{eqnarray*}%
\begin{eqnarray*}
&=&\sigma ^{2}T_{2}\left[ 1+\frac{\left( 1-\rho \right) }{1+\rho }\right] -%
\frac{\sigma ^{2}\left( 1-\rho ^{2T_{2}}\right) }{\left( 1+\rho \right) ^{2}}%
+2\alpha _{i}^{2}\left( 1-\rho \right) \left( 1-\rho ^{T_{2}}\right) +\alpha
_{i}^{2}\left( 1-\rho \right) ^{2}\left[ -2\frac{1-\rho ^{T_{2}}}{1-\rho }+%
\frac{1-\rho ^{2T_{2}}}{1-\rho ^{2}}\right] \\
&=&\frac{2\sigma ^{2}T_{2}}{1+\rho }-\frac{\sigma ^{2}\left( 1-\rho
^{2T_{2}}\right) }{\left( 1+\rho \right) ^{2}}+\alpha _{i}^{2}\left( 1-\rho
\right) \frac{1-\rho ^{2T_{2}}}{1+\rho }.
\end{eqnarray*}%
Then%
\begin{eqnarray*}
&&\mathbb{E}\left( u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\Delta
y_{it-2}\right) ^{2}=\sigma ^{4}\frac{1-\rho ^{2T_{2}}}{1-\rho ^{2}}+\alpha
_{i}^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) ^{2}+\sigma ^{2}\mathbb{E}%
\sum_{t=3}^{T}\left( \Delta y_{it-2}\right) ^{2} \\
&=&\sigma ^{4}\frac{1-\rho ^{2T_{2}}}{1-\rho ^{2}}+\alpha _{i}^{2}\sigma
^{2}\left( 1-\rho ^{T_{2}}\right) ^{2}+\frac{2\sigma ^{4}T_{2}}{1+\rho }-%
\frac{\sigma ^{4}\left( 1-\rho ^{2T_{2}}\right) }{\left( 1+\rho \right) ^{2}}%
+\alpha _{i}^{2}\sigma ^{2}\left( 1-\rho \right) \frac{1-\rho ^{2T_{2}}}{%
1+\rho } \\
&=&\frac{2\sigma ^{4}T_{2}}{1+\rho }+\sigma ^{4}\frac{2\rho \left( 1-\rho
^{2T_{2}}\right) }{\left( 1-\rho ^{2}\right) \left( 1+\rho \right) }+\alpha
_{i}^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) ^{2}+\alpha _{i}^{2}\sigma
^{2}\left( 1-\rho \right) \frac{\left( 1-\rho ^{2T_{2}}\right) }{1+\rho } \\
&=&\frac{2\sigma ^{4}T_{2}}{1+\rho }+\sigma ^{4}\frac{2\rho \left( 1-\rho
^{2T_{2}}\right) }{\left( 1-\rho ^{2}\right) \left( 1+\rho \right) }+\alpha
_{i}^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) \left[ 1-\rho ^{T_{2}}+%
\frac{\left( 1-\rho \right) \left( 1+\rho ^{T_{2}}\right) }{1+\rho }\right] ,
\end{eqnarray*}%
and%
\begin{eqnarray}
&&\omega _{NT}=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}%
\left( \sum_{t=2}^{T}\Delta u_{it}y_{it-2}\right) ^{2}  \notag \\
&=&\frac{2\sigma ^{4}T_{2}}{1+\rho }+\sigma ^{4}\frac{2\rho \left( 1-\rho
^{2T_{2}}\right) }{\left( 1-\rho ^{2}\right) \left( 1+\rho \right) }+\sigma
_{\alpha }^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) \left[ 1-\rho
^{T_{2}}+\frac{\left( 1-\rho \right) \left( 1+\rho ^{T_{2}}\right) }{1+\rho }%
\right] .  \label{aa9}
\end{eqnarray}%
From (\ref{aa10}) we have%
\begin{equation}
\omega _{DT}=\left\{ -\frac{\sigma ^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho
^{2T_{1}}}{1-\rho ^{2}}\right] +\sigma _{\alpha }^{2}\left[ 1-\rho ^{T_{1}}-%
\frac{1-\rho ^{2T_{1}}}{1+\rho }\right] \right\} ^{2}=:v_{DT}^{2}
\label{aa11}
\end{equation}%
which leads to the asymptotic variance%
\begin{eqnarray}
\omega _{T}^{2} &=&\frac{\omega _{NT}}{\omega _{DT}}=\frac{\frac{2\sigma
^{4}T_{2}}{1+\rho }+\sigma ^{4}\frac{2\rho \left( 1-\rho ^{2T_{2}}\right) }{%
\left( 1-\rho ^{2}\right) \left( 1+\rho \right) }+\sigma _{\alpha
}^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) \left[ 1-\rho ^{T_{2}}+\frac{%
\left( 1-\rho \right) \left( 1+\rho ^{T_{2}}\right) }{1+\rho }\right] }{%
\left\{ -\frac{\sigma ^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho ^{2T_{1}}}{%
1-\rho ^{2}}\right] +\sigma _{\alpha }^{2}\left[ 1-\rho ^{T_{1}}-\frac{%
1-\rho ^{2T_{1}}}{1+\rho }\right] \right\} ^{2}}  \notag \\
&=&\frac{2\left( 1+\rho \right) }{T_{1}}+O\left( \frac{1}{T_{1}^{3/2}}%
\right) ,  \label{aa12}
\end{eqnarray}%
giving the stated result for (i). The error magnitude as $T\rightarrow
\infty $ in the asymptotic expansion (\ref{aa12}) is justified as follows.
Since $c<0$ is fixed we have $1-\rho ^{2}=-2\frac{c}{\sqrt{T}}-\frac{c^{2}}{T%
}$ and 
\begin{equation}
\frac{1-\rho ^{2T}}{1-\rho ^{2}}=\frac{1-\left[ 1+\frac{c}{\sqrt{T}}\right]
^{2T}}{-2\frac{c}{\sqrt{T}}-\frac{c^{2}}{T}}\thicksim \frac{1-e^{c\frac{2T}{%
\sqrt{T}}}}{-2\frac{c}{\sqrt{T}}-\frac{c^{2}}{T}}\thicksim \frac{\sqrt{T}}{%
-2c}.  \label{aa14}
\end{equation}%
Then, by direct calculation as $T\rightarrow \infty $%
\begin{eqnarray}
\frac{\omega _{NT}}{\omega _{DT}} &=&\frac{\frac{2\sigma ^{4}T_{2}}{1+\rho }%
+\sigma ^{4}\frac{2\rho }{\left( 1+\rho \right) }\left( \frac{\sqrt{T_{2}}}{%
-2c}\right) +\sigma _{\alpha }^{2}\sigma ^{2}\left( 1-e^{cT_{2}/\sqrt{T}%
}\right) \left[ 1-e^{cT_{2}/\sqrt{T}}-c\frac{1+e^{cT_{2}/\sqrt{T}}}{\sqrt{T}%
\left( 1+\rho \right) }\right] }{\left\{ -\frac{\sigma ^{2}}{1+\rho }\left[
T_{1}-\left( \frac{\sqrt{T}}{-2c}\right) \left( 1-e^{cT_{1}/\sqrt{T}}\right) %
\right] +\sigma _{\alpha }^{2}\left[ 1-e^{cT_{1}/\sqrt{T}}-\frac{%
1-e^{2cT_{1}/\sqrt{T}}}{1+\rho }\right] \right\} ^{2}}  \notag \\
&=&\frac{2\left( 1+\rho \right) }{T_{1}}+O\left( \frac{1}{T_{1}^{3/2}}%
\right) .  \label{aa13}
\end{eqnarray}%
The sequential limit theory (ii) follows directly from (i) and the
asymptotic expansion (\ref{aa12}) of $\omega _{T}^{2}.$

If $\rho =1+\frac{c}{T^{\gamma }}$ with $\gamma \in \left( 0,1\right) ,$ it
is clear that the above fixed $\left( T,c\right) $ limit theory as $%
n\rightarrow \infty $ continues to hold. Then, as $T\rightarrow \infty ,$ we
have in place of (\ref{aa14}) 
\begin{equation*}
\frac{1-\rho ^{2T}}{1-\rho ^{2}}=\frac{1-\left[ 1+\frac{c}{T^{\gamma }}%
\right] ^{2T}}{-2\frac{c}{T^{\gamma }}-\frac{c^{2}}{T^{\gamma }}}\thicksim 
\frac{1-e^{2cT^{1-\gamma }}}{-2\frac{c}{T^{\gamma }}-\frac{c^{2}}{T^{\gamma }%
}}\thicksim \frac{T^{\gamma }}{-2c}
\end{equation*}%
leading to 
\begin{equation*}
\frac{\omega _{NT}}{\omega _{DT}}=\frac{2\left( 1+\rho \right) }{T_{1}}%
+O\left( \frac{1}{T_{1}^{2-\gamma }}\right) \thicksim \frac{4}{T_{1}}%
+O\left( \frac{1}{T_{1}^{2-\gamma }}\right) .
\end{equation*}%
It follows that (ii) continues to hold with the same convergence rate $\sqrt{%
nT}$ and same limit variance $4$ for all $\gamma \in \left( 0,1\right) .$

When $\gamma =1,$ the sequential normal limit theory in (ii) still holds but
the variance of the limiting distribution changes. Observe that in this case 
\begin{equation*}
\frac{1-\rho ^{2T}}{1-\rho ^{2}}=\frac{1-\left[ 1+\frac{c}{T}\right] ^{2T}}{%
-2\frac{c}{T}-\frac{c^{2}}{T}}\thicksim \frac{1-e^{2c}}{-2\frac{c}{T}-\frac{%
c^{2}}{T}}\thicksim \frac{T\left( 1-e^{2c}\right) }{-2c}.
\end{equation*}%
Using (\ref{aa9}) we then have the following limit behavior as $T\rightarrow
\infty $%
\begin{equation*}
\frac{\omega _{NT}}{\omega _{DT}}\sim \frac{\sigma ^{4}T_{2}+\sigma ^{4}T_{2}%
\frac{\left( 1-e^{2c}\right) }{-2c}+O\left( 1\right) }{\left\{ -\frac{\sigma
^{2}T_{1}}{2}\left[ 1-\frac{\left( 1-e^{2c}\right) }{-2c}\right] +O\left(
1\right) \right\} ^{2}}=\frac{4}{T_{1}}\frac{1+\frac{\left( 1-e^{2c}\right) 
}{-2c}}{\left[ 1-\frac{\left( 1-e^{2c}\right) }{-2c}\right] ^{2}}\left\{
1+o\left( 1\right) \right\} ,
\end{equation*}%
so that%
\begin{equation*}
\omega _{T}^{2}=\frac{-8c}{T_{1}}\frac{\left( 1-2c-e^{2c}\right) }{\left(
1+2c-e^{2c}\right) ^{2}}\left\{ 1+o\left( 1\right) \right\} .
\end{equation*}%
Hence 
\begin{equation}
\sqrt{nT}\left( \rho _{IV}-\rho \right) \underset{\left( T,n\right) _{%
\mathrm{\func{seq}}}\rightarrow \infty }{\Rightarrow }N\left( 0,\left(
-8c\right) \frac{\left( 1-2c-e^{2c}\right) }{\left( 1+2c-e^{2c}\right) ^{2}}%
\right) ,  \label{aa45}
\end{equation}%
so the $\sqrt{nT}$ Gaussian limit theory holds but with a different variance
when $\rho =1+\frac{c}{T}.$ Observe that 
\begin{equation*}
\left( -8c\right) \frac{\left( 1-2c-e^{2c}\right) }{\left(
1+2c-e^{2c}\right) ^{2}}\sim \frac{8}{c^{2}}\rightarrow \infty \text{ \ as }%
c\rightarrow 0,
\end{equation*}%
indicating that the variance in (\ref{aa45}) diverges and the $\sqrt{nT}$
convergence rate fails as the unit root is approached via $c\rightarrow 0.$

Next, examine the case where $\rho =1+\frac{c}{T^{\gamma }}$ with $\gamma >1$
and $c<0,$ so that $\rho $ is in the immediate vicinity of unity, closer
than the LUR case but still satisfying $\rho <1$ for fixed $T.$ In that
case, we still have Gaussian limit theory as $n\rightarrow \infty $ because $%
\left\vert \rho \right\vert <1.$ To find the limit theory as $\left(
T,n\right) _{\mathrm{\func{seq}}}\rightarrow \infty $ we consider the
behavior of the numerator and denominator of $\omega _{T}.$ First, note that 
$\log \left[ 1+\frac{c}{T^{\gamma }}\right] ^{2T}=\frac{2c}{T^{\gamma -1}}-%
\frac{c^{2}}{T^{2\gamma -1}}+O\left( \frac{1}{T^{3\gamma -1}}\right) $ so
that $\left[ 1+\frac{c}{T^{\gamma }}\right] ^{2T}=1+\frac{2c}{T^{\gamma -1}}-%
\frac{c^{2}}{T^{2\gamma -1}}+\frac{1}{2}\left( \frac{2c}{T^{\gamma -1}}%
\right) ^{2}+O\left( \frac{1}{T^{2\gamma -2}}\right) $ giving 
\begin{eqnarray*}
&&\frac{1-\rho ^{2T}}{1-\rho ^{2}}=\frac{1-\left[ 1+\frac{c}{T^{\gamma }}%
\right] ^{2T}}{-2\frac{c}{T^{\gamma }}-\frac{c^{2}}{T^{2\gamma }}}=\frac{1-%
\left[ 1+\frac{2c}{T^{\gamma -1}}-\frac{c^{2}}{T^{2\gamma -1}}+\frac{1}{2}%
\left( \frac{2c}{T^{\gamma -1}}\right) ^{2}+O\left( \frac{1}{T^{3\left(
\gamma -1\right) }}\right) \right] }{-2\frac{c}{T^{\gamma }}-\frac{c^{2}}{%
T^{2\gamma }}} \\
&=&\frac{-2cT+\frac{c^{2}}{T^{\gamma -1}}-\frac{2c^{2}}{T^{\gamma -2}}%
+o\left( \frac{1}{T^{\gamma -2}}\right) }{-2c\left\{ 1+\frac{1}{2}\frac{c}{%
T^{\gamma }}+O\left( \frac{1}{T^{2\gamma }}\right) \right\} }=\left\{
T+cT^{2-\gamma }+O\left( T^{1-\gamma }\right) \right\} \left\{ 1+\frac{1}{2}%
\frac{c}{T^{\gamma }}+O\left( \frac{1}{T^{2\gamma }}\right) \right\} ^{-1} \\
&=&T+cT^{2-\gamma }+O\left( \frac{1}{T^{\gamma -1}}\right) .
\end{eqnarray*}%
Using this result and $\omega _{DT}=v_{DT}^{2}$ with $v_{DT}=-\frac{\sigma
^{2}}{1+\rho }\left[ T_{1}-\frac{1-\rho ^{2T_{1}}}{1-\rho ^{2}}\right]
+\sigma _{\alpha }^{2}\left[ 1-\rho ^{T_{1}}-\frac{1-\rho ^{2T_{1}}}{1+\rho }%
\right] ,$ we have 
\begin{eqnarray*}
v_{DT} &=&-\frac{\sigma ^{2}}{1+\rho }\left[ cT_{1}^{2-\gamma }\left\{
1+o\left( 1\right) \right\} \right] +\sigma _{\alpha }^{2}\left\{ -\left[ 
\frac{c}{T_{1}^{\gamma -1}}+\frac{T_{1}\left( T_{1}-1\right) }{2}\left( 
\frac{c}{T_{1}^{\gamma }}\right) ^{2}+O\left( \frac{1}{T_{1}^{3\left( \gamma
-1\right) }}\right) \right] \right.  \\
&&\left. -\frac{-\frac{2c}{T_{1}^{\gamma -1}}-\frac{T_{1}\left(
T_{1}-1\right) }{2}\left( \frac{c}{T_{1}^{\gamma }}\right) ^{2}+O\left( 
\frac{1}{T_{1}^{3\left( \gamma -1\right) }}\right) }{2+\frac{2c}{%
T_{1}^{\gamma -1}}\left\{ 1+o\left( 1\right) \right\} }\right\}  \\
&=&-\frac{c\sigma ^{2}}{1+\rho }T_{1}^{2-\gamma }\left\{ 1+o\left( 1\right)
\right\} +\sigma _{\alpha }^{2}\left[ -\frac{2c}{T_{1}^{\gamma -1}}-\frac{1}{%
2}\frac{c^{2}}{T_{1}^{2\left( \gamma -1\right) }}+O\left( \frac{1}{%
T_{1}^{3\left( \gamma -1\right) }}\right) \right.  \\
&&\left. +\left\{ \frac{c}{T_{1}^{\gamma -1}}+\frac{T_{1}\left(
T_{1}-1\right) }{4}\left( \frac{c}{T_{1}^{\gamma }}\right) ^{2}+O\left( 
\frac{1}{T_{1}^{3\left( \gamma -1\right) }}\right) \right\} \left\{ 1+\frac{c%
}{T_{1}^{\gamma -1}}\left\{ 1+o\left( 1\right) \right\} \right\} ^{-1}\right]
\\
&=&-\frac{c\sigma ^{2}}{1+\rho }T_{1}^{2-\gamma }\left\{ 1+o\left( 1\right)
\right\} +\sigma _{\alpha }^{2}\left[ -\frac{c}{T_{1}^{\gamma -1}}-\frac{1}{4%
}\frac{c^{2}}{T_{1}^{2\left( \gamma -1\right) }}+O\left( \frac{1}{%
T_{1}^{3\left( \gamma -1\right) }}\right) \right] \left\{ 1+o\left( 1\right)
\right\}  \\
&=&\left\{ -\frac{c\sigma ^{2}}{1+\rho }T_{1}^{2-\gamma }-c\sigma _{\alpha
}^{2}\frac{c}{T_{1}^{\gamma -1}}\right\} \left\{ 1+o\left( 1\right) \right\}
=-\frac{c\sigma ^{2}}{2}T_{1}^{2-\gamma }\left\{ 1+o\left( 1\right) \right\}
,
\end{eqnarray*}%
so that the denominator is $\omega _{DT}=\frac{c^{2}\sigma ^{4}}{4}%
T_{1}^{4-2\gamma }\left\{ 1+o\left( 1\right) \right\} .$ The numerator $%
\omega _{NT}$ is 
\begin{eqnarray*}
&&\frac{2\sigma ^{4}T_{2}}{1+\rho }+\sigma ^{4}\frac{2\rho \left( 1-\rho
^{2T_{2}}\right) }{\left( 1-\rho ^{2}\right) \left( 1+\rho \right) }+\sigma
_{\alpha }^{2}\sigma ^{2}\left( 1-\rho ^{T_{2}}\right) \left[ 1-\rho
^{T_{2}}+\frac{\left( 1-\rho \right) \left( 1+\rho ^{T_{2}}\right) }{1+\rho }%
\right]  \\
&=&\sigma ^{4}T_{2}\left\{ 1-\frac{c}{T_{1}^{\gamma -1}}\right\} \left\{
1+o\left( 1\right) \right\} +\sigma ^{4}\frac{\left( 2+\frac{2c}{T^{\gamma }}%
\right) }{\left( 2+\frac{c}{T^{\gamma }}\right) }\left\{ T+cT^{2-\gamma }-%
\frac{1}{2}\frac{c}{T^{\gamma -1}}+O\left( \frac{T}{T^{2\left( \gamma
-1\right) }}\right) \right\}  \\
&&+\sigma _{\alpha }^{2}\sigma ^{2}\left\{ -2cT_{2}-\frac{2T_{2}\left(
2T_{2}-1\right) }{2}\frac{c^{2}}{T_{2}^{\gamma }}+O\left( \frac{T^{3}}{%
T^{2\gamma }}\right) \right\}  \\
&&\times \left\{ -\left[ \frac{c}{T_{2}^{\gamma -1}}+\frac{T_{2}\left(
T_{2}-1\right) }{2}\left( \frac{c}{T_{2}^{\gamma }}\right) ^{2}\left\{
1+o\left( 1\right) \right\} \right] -\frac{\frac{2c}{T^{\gamma -1}}\left[ 1+%
\frac{2c}{T_{2}^{\gamma -1}}+\frac{2T_{2}\left( 2T_{2}-1\right) }{2}\left( 
\frac{c}{T_{2}^{\gamma }}\right) ^{2}\left\{ 1+o\left( 1\right) \right\} %
\right] }{2+\frac{2c}{T^{\gamma -1}}\left\{ 1+o\left( 1\right) \right\} }%
\right\}  \\
&=&\left\{ 2\sigma ^{4}T_{2}+2\sigma _{\alpha }^{2}\sigma ^{2}\left(
-c\right) T\left( \frac{-2c}{T_{2}^{\gamma -1}}\right) \right\} \left[
1+o\left( 1\right) \right] =2\sigma ^{4}T_{2}\left[ 1+o\left( 1\right) %
\right] .
\end{eqnarray*}%
Combining these results we obtain%
\begin{equation*}
\omega _{T}^{2}=\frac{\omega _{NT}}{\omega _{DT}}=\frac{2\sigma ^{4}T_{2}%
\left[ 1+o\left( 1\right) \right] }{\left\{ \frac{c^{2}\sigma ^{4}}{2}%
T_{1}^{4-2\gamma }\left\{ 1+o\left( 1\right) \right\} \right\} ^{2}\left\{
1+o\left( 1\right) \right\} }=\frac{8T_{2}}{c^{2}T_{1}^{2\left( 2-\gamma
\right) }}\left\{ 1+o\left( 1\right) \right\} .
\end{equation*}%
It now follows that for $\rho =1+\frac{c}{T^{\gamma }}$ with $c<0$ fixed and 
$\gamma >1$%
\begin{equation*}
\sqrt{nT^{3-2\gamma }}\left( \rho _{IV}-\rho \right) \underset{\left(
T,n\right) _{\mathrm{\func{seq}}}\rightarrow \infty }{\Rightarrow }N\left( 0,%
\frac{8}{c^{2}}\right) .
\end{equation*}%
Hence when $\rho $ is closer to unity than a local unit root, the $\sqrt{nT}$
rate of convergence is reduced to $\sqrt{n}T^{\frac{3-2\gamma }{2}}.$ When $%
\gamma =\frac{3}{2}$ the rate of convergence is simply $\sqrt{n}$ and for $%
\gamma >\frac{3}{2}$ the large $n$ Gaussian asymptotic distribution $N\left(
0,\omega _{T}^{2}\right) $ diverges as $T\rightarrow \infty $ because $%
\omega _{T}^{2}=\frac{8T_{2}}{c^{2}T_{1}^{2\left( 2-\gamma \right) }}\left\{
1+o\left( 1\right) \right\} $ diverges with $T.$ In this event, sequential $%
\left( T,n\right) _{\mathrm{\func{seq}}}\rightarrow \infty $ asymptotics
fail. In effect, the convergence rate is slower than $\sqrt{n}$ and the
non-Gaussian Cauchy limit theory cannot be captured in these $\left(
T,n\right) _{\mathrm{\func{seq}}}$ directional sequential asymptotics even
though $\rho =1+\frac{c}{T^{\gamma }}$ with $\gamma >1$ is in closer
proximity to a unit root than the usual local unit root case with $\gamma =1.
$ \vspace{0.08in}
\end{proof}

\begin{proof}[Proof of Theorem 5]
In the mildly integrated case where $\rho =1+\frac{c}{\sqrt{T}}$ we have $%
y_{it}=-\frac{\alpha _{i}c}{\sqrt{T}}+\left( 1+\frac{c}{\sqrt{T}}\right)
y_{it-1}+u_{it}$ and $\Delta y_{it}=\rho \Delta y_{it-1}+\Delta u_{it}$ so
that $\Delta y_{it}=-\frac{\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}}%
y_{it-1}+u_{it}=\alpha _{i}\left( 1-\rho \right) +\left( \rho -1\right)
y_{it-1}+u_{it}.$ By partial summation, as shown above, we have $%
\sum_{t=2}^{T}\Delta
u_{it}y_{it-2}=u_{iT}y_{iT-2}-u_{i1}y_{i0}-\sum_{t=3}^{T}u_{it-1}\Delta
y_{it-2}$, so that%
\begin{equation}
\rho _{IV}-\rho =\frac{\sum_{i=1}^{n}\left\{ \left(
u_{iT}y_{iT-2}-u_{i1}y_{i0}\right) -\sum_{t=3}^{T}u_{it-1}\left[ -\frac{%
\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-3}+u_{it-2}\right] \right\} 
}{\sum_{i=1}^{n}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{\sqrt{T}}+\frac{c%
}{\sqrt{T}}y_{it-2}+u_{it-1}\right\} y_{it-2}}.  \label{aa19}
\end{equation}%
Rescaling and using $y_{i0}=0$ gives 
\begin{equation}
\sqrt{T}\left( \rho _{IV}-\rho \right) =\frac{\sum_{i=1}^{n}\frac{1}{\sqrt{T}%
}\left\{ u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{%
\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-3}+u_{it-2}\right] \right\} }{%
\sum_{i=1}^{n}\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{\sqrt{T}}%
+\frac{c}{\sqrt{T}}y_{it-2}+u_{it-1}\right\} y_{it-2}}.  \label{aa16}
\end{equation}%
Since $\frac{1}{\sqrt{T}}\sum_{t=2}^{T}u_{it-1}u_{it-2}\Rightarrow
G_{i}\equiv N\left( 0,\sigma ^{4}\right) ,$ $\frac{1}{T}%
\sum_{t=2}^{T}u_{it-1}=o_{p}\left( 1\right) ,$ and $\frac{1}{T}%
\sum_{t=2}^{T}u_{it-1}y_{it-3}=o_{p}\left( 1\right) $ -- see (\ref{aa30})
below -- the numerator is%
\begin{eqnarray}
&&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{\sqrt{T}}+%
\frac{c}{\sqrt{T}}y_{it-3}+u_{it-2}\right] \right\}  \notag \\
&=&-\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\sum_{t=3}^{T}u_{it-1}u_{it-2}+o_{p}%
\left( 1\right) \Rightarrow -\sum_{i=1}^{n}G_{i}\left( 1\right) .
\label{aa17}
\end{eqnarray}%
Using Phillips and Magdalinos (2007, theorem 3.2) we find that 
\begin{equation}
T^{-3/2}\sum_{t=2}^{T}y_{it}^{2}\rightarrow _{p}\frac{\sigma ^{2}}{-2c}%
,T^{-3/4}\sum_{t=2}^{T}y_{it-1}u_{it}\Rightarrow N\left( 0,\frac{\sigma ^{4}%
}{-2c}\right) ,\text{ and }T^{-3/2}\sum_{t=2}^{T}y_{it}=o_{p}\left( 1\right)
.  \label{aa30}
\end{equation}%
The denominator of (\ref{aa16}) therefore satisfies 
\begin{equation}
\sum_{i=1}^{n}\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{\sqrt{T}}%
+\frac{c}{\sqrt{T}}y_{it-2}+u_{it-1}\right\} y_{it-2}\rightarrow
_{p}\sum_{i=1}^{n}\left\{ c\frac{\sigma ^{2}}{-2c}\right\} .  \label{aa18}
\end{equation}%
Hence, using (\ref{aa17}) and (\ref{aa18}) we have$\sqrt{T}\left( \rho
_{IV}-\rho \right) \underset{T\rightarrow \infty }{\Rightarrow }\frac{2}{%
n\sigma ^{2}}\sum_{i=1}^{n}G_{i}=\frac{2}{n}\sum_{i=1}^{n}\zeta _{i}$ where $%
\zeta _{i}\thicksim _{iid}N\left( 0,1\right) .$ Then 
\begin{equation}
\sqrt{nT}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{2}{\sqrt{n}}\sum_{i=1}^{n}\zeta _{i}\underset{%
n\rightarrow \infty }{\Rightarrow }N\left( 0,4\right) ,  \label{aa20}
\end{equation}%
which gives (i) and then leads directly to the sequential limit $\sqrt{nT}%
\left( \rho _{IV}-\rho \right) \underset{\left( n,T\right) _{\func{seq}%
}\rightarrow \infty }{\Rightarrow }N\left( 0,4\right) .$\vspace{0.08in}
\end{proof}

\begin{proof}[Proof of Theorem 6]
The proof follows the same lines as the proof of Theorem 3 above. As before,
we define the vector of standardized components appearing in the numerator
and denominator of $\sqrt{T}\left( \rho _{IV}-\rho \right) $ in (\ref{aa16}) 
\begin{equation*}
X_{nT}=\left( X_{1nT},X_{2nT}\right) ^{\prime }:=\left(
n^{-1/2}\sum_{i=1}^{n}Y_{1iT},n^{-1}\sum_{i=1}^{n}Y_{2iT}\right) ,
\end{equation*}%
where $Y_{iT}=\left( Y_{1iT},Y_{2iT}\right) ^{\prime }$ with 
\begin{eqnarray*}
Y_{1iT} &=&\frac{1}{\sqrt{T}}\left\{ u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}%
\left[ -\frac{\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-3}+u_{it-2}%
\right] \right\} \underset{T\rightarrow \infty }{\Rightarrow }%
Y_{1i}:=-G_{i}\left( 1\right) , \\
Y_{2iT} &=&\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{\sqrt{T}}+%
\frac{c}{\sqrt{T}}y_{it-2}+u_{it-1}\right\} y_{it-2}\underset{T\rightarrow
\infty }{\Rightarrow }Y_{2i}:=c\frac{\sigma ^{2}}{-2c}.
\end{eqnarray*}%
From (\ref{aa17}) and (\ref{aa18}) we have the sequential convergence 
\begin{eqnarray}
X_{nT}\underset{T\rightarrow \infty }{\Rightarrow }X_{n} &:&=\left(
-n^{-1/2}\sum_{i=1}^{n}G_{i}\left( 1\right) ,n^{-1}\sum_{i=1}^{n}\left\{ c%
\frac{\sigma ^{2}}{-2c}\right\} \right) ^{\prime }  \notag \\
\underset{n\rightarrow \infty }{\Rightarrow }X &:&=\left( \sigma ^{2}\zeta ,-%
\frac{\sigma ^{2}}{2}\right) ,\text{ where }\zeta =N\left( 0,1\right) ,
\label{aa21}
\end{eqnarray}%
which in turn implies the sequential limit $\sqrt{nT}\left( \rho _{IV}-\rho
\right) \underset{n\rightarrow \infty ,T\rightarrow \infty }{\Rightarrow }%
N\left( 0,4\right) $ given in (\ref{gg15}). Since $X_{nT}\underset{%
T\rightarrow \infty }{\Rightarrow }X_{n}\underset{n\rightarrow \infty }{%
\Rightarrow }X$ sequentially, joint weak convergence $X_{nT}\Rightarrow X$
as $\left( n,T\right) \rightarrow \infty $ holds in the same manner as
Theorem 3 with only minor definitional changes. First, $Y_{iT}$ is
integrable just as before. To show Lemma A(i) holds, observe that%
\begin{eqnarray*}
&&\mathbb{E}\left\Vert Y_{iT}\right\Vert ^{2}=\mathbb{E}Y_{1iT}^{2}+\mathbb{E%
}Y_{2iT}^{2} \\
&=&\frac{1}{T}\mathbb{E}\left\{ u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left( -%
\frac{\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-3}+u_{it-2}\right)
\right\} ^{2} \\
&&+\frac{1}{T^{2}}\mathbb{E}\left\{ \sum_{t=2}^{T}\left( -\frac{\alpha _{i}c%
}{\sqrt{T}}+\frac{c}{\sqrt{T}}y_{it-2}+u_{it-1}\right) y_{it-2}\right\} ^{2}
\\
&=&\frac{\sigma ^{2}}{T}\mathbb{E}y_{iT-2}^{2}+\frac{1}{T}\mathbb{E}\left\{
\sum_{t=3}^{T}u_{it-1}\left( -\frac{\alpha _{i}c}{\sqrt{T}}+\frac{c}{\sqrt{T}%
}y_{it-3}+u_{it-2}\right) \right\} ^{2} \\
&=&\frac{\sigma ^{2}}{T}\mathbb{E}y_{iT-2}^{2}+\frac{1}{T}\mathbb{E}\left(
\sum_{t=3}^{T}u_{it-1}u_{it-2}\right) ^{2}+\frac{c^{2}\alpha _{i}^{2}}{T^{2}}%
\mathbb{E}\left( \sum_{t=3}^{T}u_{it-1}\right) ^{2}+\frac{c^{2}}{T^{2}}%
\mathbb{E}\left( \sum_{t=3}^{T}u_{it-1}y_{it-3}\right) ^{2} \\
&&-\frac{2\alpha _{i}c}{T^{3/2}}\mathbb{E}\left(
\sum_{t=3}^{T}u_{it-1}\sum_{s=3}^{T}u_{is-2}\right) -\frac{2\alpha _{i}c^{2}%
}{T^{2}}\mathbb{E}\left( \sum_{t=3}^{T}u_{it-1}\sum_{s=3}^{T}y_{is-3}\right)
+\frac{2c}{T^{3/2}}\mathbb{E}\left(
\sum_{t=3}^{T}u_{it-1}u_{it-2}\sum_{s=3}^{T}y_{is-3}\right)
\end{eqnarray*}%
\begin{eqnarray*}
&=&\frac{\sigma ^{4}}{-2c}\frac{\sqrt{T_{2}}}{T}+\sigma ^{4}\frac{T_{2}}{T}%
+O\left( \frac{1}{T}\right) +\frac{c^{2}\sigma ^{2}}{T^{2}}\sum_{t=3}^{T}%
\mathbb{E}y_{it-3}^{2}+O\left( \frac{1}{\sqrt{T}}\right) +O\left( \frac{1}{T}%
\right) +O\left( \frac{1}{\sqrt{T}}\right) \\
&=&\sigma ^{4}\frac{T_{2}}{T}+o\left( 1\right) ,
\end{eqnarray*}%
since from (\ref{aa14}) $\mathbb{E}\left( y_{it}^{2}\right) =\sigma ^{2}%
\frac{1-\rho ^{2t}}{1-\rho ^{2}}+\alpha _{i}^{2}\left( 1-\rho ^{t}\right)
^{2}=\sigma ^{2}\frac{\sqrt{t}}{-2c}\left\{ 1+o\left( 1\right) \right\} .$
Then $\mathbb{E}\left\Vert Y_{iT}\right\Vert ^{2}<\infty $ and we deduce
that 
\begin{equation*}
\underset{n,T\rightarrow \infty }{\lim \sup }\frac{1}{n}\sum_{i=1}^{n}%
\mathbb{E}\left\Vert Y_{iT}\right\Vert =\underset{T\rightarrow \infty }{\lim
\sup }\mathbb{E}\left\Vert Y_{iT}\right\Vert \leq \underset{T\rightarrow
\infty }{\lim \sup }\left( \mathbb{E}\left\Vert Y_{iT}\right\Vert
^{2}\right) ^{1/2}<\infty ,
\end{equation*}%
as required. Condition (ii) holds, as we again have $\mathbb{E}Y_{iT}=%
\mathbb{E}Y_{i}=0;$ and condition (iii)\ and (iv) hold because $\sup_{T}%
\mathbb{E}\left\Vert Y_{iT}\right\Vert ^{2}<\infty $ and $\mathbb{E}%
\left\Vert Y_{i}\right\Vert ^{2}<\infty .$ We then have joint weak
convergence%
\begin{equation*}
X_{nT}=n^{-1/2}\sum_{i=1}^{n}Y_{iT}\underset{n,T\rightarrow \infty }{%
\Rightarrow }X:=\left( \sigma ^{2}\zeta ,\frac{\sigma ^{2}}{2}\right) ,
\end{equation*}%
irrespective of the divergence rates of $n$ and $T$ to infinity. By
continuous mapping, the required result follows for the IV estimator so that 
$\sqrt{T}\left( \rho _{IV}-\rho \right) \underset{n,T\rightarrow \infty }{%
\Rightarrow }N\left( 0,4\right) $ holds jointly as $\left( n,T\right)
\rightarrow \infty $ irrespective of the order and rates of divergence.%
\vspace{0.08in}
\end{proof}

\begin{proof}[Proof of Theorem 7]
We have $\rho =1+\frac{c}{T^{\gamma }}$\textit{\ }for some fixed\textit{\ }$%
c<0$ and let $T\rightarrow \infty .$ In this case, $y_{it}=-\frac{\alpha
_{i}c}{T^{\gamma }}+\left( 1+\frac{c}{T^{\gamma }}\right) y_{it-1}+u_{it}$
and $\Delta y_{it}=\rho \Delta y_{it-1}+\Delta u_{it}$ so that $\Delta
y_{it}=-\frac{\alpha _{i}c}{T^{\gamma }}+\frac{c}{T^{\gamma }}%
y_{it-1}+u_{it}.$ As before, we have%
\begin{equation}
\sqrt{T}\left( \rho _{IV}-\rho \right) =\frac{\frac{1}{\sqrt{T}}%
\sum_{i=1}^{n}\left\{ \left( u_{iT}y_{iT-2}-u_{i1}y_{i0}\right)
-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{T^{\gamma }}+\frac{c}{%
T^{\gamma }}y_{it-3}+u_{it-2}\right] \right\} }{\frac{1}{T}%
\sum_{i=1}^{n}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{T^{\gamma }}+\frac{c%
}{T^{\gamma }}y_{it-2}+u_{it-1}\right\} y_{it-2}}.  \label{aa23}
\end{equation}%
We use the following results from Phillips and Magdalinos (2007) and
Magdalinos and Phillips (2009), which hold for all $\gamma \in \left(
0,1\right) ,$ 
\begin{equation}
T^{-1-\gamma }\sum_{t=2}^{T}y_{it}^{2}\rightarrow _{p}\frac{\sigma ^{2}}{-2c}%
,T^{-\left( 1+\gamma \right) /2}\sum_{t=2}^{T}y_{it-1}u_{it}\Rightarrow
N\left( 0,\frac{\sigma ^{4}}{-2c}\right) ,\text{ and }T^{-1/2-\gamma
}\sum_{t=2}^{T}y_{it}=O_{p}\left( 1\right) .  \label{aa22}
\end{equation}%
Then, since $\frac{1}{\sqrt{T}}\sum_{t=2}^{T}u_{it-1}u_{it-2}\Rightarrow
G_{i}\equiv N\left( 0,\sigma ^{4}\right) ,$ $\frac{1}{T^{1/2+\gamma }}%
\sum_{t=2}^{T}u_{it-1}=o_{p}\left( 1\right) ,$ and $\frac{1}{T^{1/2+\gamma }}%
\sum_{t=2}^{T}u_{it-1}y_{it-3}=o_{p}\left( 1\right) $ when $\gamma \in
\left( 0,1\right) ,$ the numerator of (\ref{aa23}) is 
\begin{eqnarray*}
&&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{T^{\gamma }}%
+\frac{c}{T^{\gamma }}y_{it-3}+u_{it-2}\right] \right\} \\
&=&-\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\sum_{t=3}^{T}u_{it-1}u_{it-2}+o_{p}%
\left( 1\right) \Rightarrow -\sum_{i=1}^{n}G_{i}\left( 1\right) .
\end{eqnarray*}%
Using (\ref{aa22}), we find that the denominator of (\ref{aa23}) satisfies 
\begin{equation*}
\sum_{i=1}^{n}\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{%
T^{\gamma }}+\frac{c}{T^{\gamma }}y_{it-2}+u_{it-1}\right\}
y_{it-2}\rightarrow _{p}\sum_{i=1}^{n}\left\{ c\frac{\sigma ^{2}}{-2c}%
\right\} =-\frac{\sigma ^{2}n}{2}.
\end{equation*}%
Hence, as $T\rightarrow \infty $ 
\begin{equation*}
\sqrt{T}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{-\sum_{i=1}^{n}G_{i}}{-\frac{\sigma ^{2}}{2}n}=\frac{%
\sigma ^{2}\sum_{i=1}^{n}\zeta _{i}}{-\frac{\sigma ^{2}}{2}n},\text{ \ where 
}\zeta _{i}\thicksim _{iid}N\left( 0,1\right) .
\end{equation*}%
Then, as $T\rightarrow \infty $ is followed by $n\rightarrow ,$ we have 
\begin{equation*}
\sqrt{nT}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{2}{\sqrt{n}}\sum_{i=1}^{n}\zeta _{i}\underset{%
n\rightarrow \infty }{\Rightarrow }N\left( 0,4\right) ,\text{ for all }%
\gamma \in \left( 0,1\right) .
\end{equation*}

Next consider the case $\gamma =1.$ The numerator of (\ref{aa23}) is then 
\begin{eqnarray*}
&&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{T}+\frac{c}{%
T}y_{it-3}+u_{it-2}\right] \right\} \\
&=&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} +o_{p}\left( 1\right) 
\underset{T\rightarrow \infty }{\Rightarrow }u_{i\infty }K_{ci}\left(
1\right) -\sum_{i=1}^{n}G_{i}\left( 1\right) ,
\end{eqnarray*}%
since by standard functional limit theory for near integrated processes
(Phillips, 1987b) we have%
\begin{equation*}
\left( \frac{1}{T^{1/2}}y_{iT},\frac{1}{T}\sum_{t=2}^{T}y_{it-1}u_{it}%
\right) \underset{T\rightarrow \infty }{\Rightarrow }\left( K_{ci}\left(
r\right) ,\int_{0}^{1}K_{ci}dB_{i}\right) ,
\end{equation*}%
where $B_{i}\left( r\right) =:\sigma W_{i}\left( r\right) $ are $iid$
Brownian motions with common variance $\sigma ^{2},$ and $K_{ci}\left(
r\right) =\int_{0}^{r}e^{c\left( r-s\right) }dB_{i}\left( s\right) =:\sigma
J_{ci}\left( r\right) $ is a linear diffusion. The denominator of (\ref{aa23}%
) satisfies 
\begin{equation*}
\sum_{i=1}^{n}\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{T}+\frac{%
c}{T}y_{it-2}+u_{it-1}\right\} y_{it-2}\underset{T\rightarrow \infty }{%
\Rightarrow }\sum_{i=1}^{n}\left\{ c\int_{0}^{1}K_{ci}\left( r\right)
^{2}dr+\int_{0}^{1}K_{ci}dB_{i}\right\} .
\end{equation*}%
Hence%
\begin{equation}
\sqrt{T}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{\sum_{i=1}^{n}\left\{ u_{i\infty }K_{ci}\left( 1\right)
-\sum_{i=1}^{n}G_{i}\right\} }{\sum_{i=1}^{n}\left\{
c\int_{0}^{1}K_{ci}\left( r\right) ^{2}dr+\int_{0}^{1}K_{ci}\left( r\right)
dB_{i}\right\} }=\frac{\sum_{i=1}^{n}\left\{ \left( \sigma ^{-1}u_{i\infty
}\right) J_{ci}\left( 1\right) -\zeta _{i}\right\} }{\sum_{i=1}^{n}\left\{
c\int_{0}^{1}J_{ci}\left( r\right) ^{2}dr+\int_{0}^{1}J_{ci}dW_{i}\right\} },
\label{aa24}
\end{equation}%
where the $\zeta _{i}\thicksim _{iid}N\left( 0,1\right) $ and are
independent of the $W_{i}$ and $u_{i\infty }$ for all $i.$ This gives the
first part of (ii). Scaling the numerator and denominator of (\ref{aa24}),
noting that $\int_{0}^{1}J_{ci}\left( r\right) dW_{i}$ has zero mean and
finite variance, and using the independence of $\zeta _{i},u_{i\infty },$
and $W_{i},$ we obtain%
\begin{eqnarray*}
\sqrt{nT}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow } &&\frac{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left\{ \left( \sigma
^{-1}u_{i\infty }\right) J_{ci}\left( 1\right) -\zeta _{i}\right\} }{\frac{1%
}{n}\sum_{i=1}^{n}\left\{ c\int_{0}^{1}J_{ci}\left( r\right)
^{2}dr+\int_{0}^{1}J_{ci}\left( r\right) dW_{i}\right\} } \\
\underset{n\rightarrow \infty }{\Rightarrow }\frac{N\left( 0,\frac{%
1-2c-e^{2c}}{-2c}\right) }{c\mathbb{E}\left( \int_{0}^{1}J_{ci}\left(
r\right) ^{2}dr\right) } &=&N\left( 0,-8c\frac{1-2c-e^{2c}}{\left(
e^{2c}-1-2c\right) ^{2}}\right) ,
\end{eqnarray*}%
since, using results in Phillips (1987b), we have $\mathbb{E}\left(
\int_{0}^{1}J_{ci}\left( r\right) ^{2}dr\right) =\frac{e^{2c}-1-2c}{\left(
2c\right) ^{2}}$ and 
\begin{equation*}
\mathbb{E}\left\{ \left( \sigma ^{-1}u_{i\infty }\right) J_{ci}\left(
1\right) -\zeta _{i}\right\} ^{2}=\mathbb{E}\left( \sigma ^{-1}u_{i\infty
}\right) ^{2}\mathbb{E}J_{ci}\left( 1\right) ^{2}+\mathbb{E}\zeta _{i}^{2}=1+%
\frac{1-e^{2c}}{-2c}=\frac{1-2c-e^{2c}}{-2c}.
\end{equation*}%
Hence, when $\gamma =1,$ we have%
\begin{equation}
\sqrt{nT}\left( \rho _{IV}-\rho \right) \underset{\left( n,T\right)
\rightarrow \infty }{\Rightarrow }N\left( 0,\left( -8c\right) \frac{%
1-2c-e^{2c}}{\left( e^{2c}-1-2c\right) ^{2}}\right)  \label{aa36}
\end{equation}%
From Lemma 2 of Phillips (1987b)\ we have 
\begin{equation*}
\left( \left( -2c\right) \int_{0}^{1}J_{ci}\left( r\right) ^{2}dr,\left(
-2c\right) ^{1/2}\int_{0}^{1}J_{ci}\left( r\right) dW_{i}\right) \underset{%
c\rightarrow 0}{\Rightarrow }\left( 1,Z_{i}\right) ,\text{ \ }Z_{i}\thicksim
_{iid}N\left( 0,1,\right)
\end{equation*}%
and $\frac{1-2c-e^{2c}}{-2c}=2\left\{ 1+o\left( 1\right) \right\} $ as $%
c\rightarrow 0,$ so that 
\begin{equation}
\left( -8c\right) \frac{1-2c-e^{2c}}{\left( e^{2c}-1-2c\right) ^{2}}%
\thicksim \left( -8c\right) \frac{\left( -4c\right) }{\left\{ \frac{1}{2}%
\left( 2c\right) ^{2}\right\} ^{2}}=\frac{8}{c^{2}}\text{ \ for small }c\sim
0  \label{aa37}
\end{equation}%
which explodes as $c\rightarrow 0,$ consonant with the unit root case where
we only have $\sqrt{T}$ convergence. Observe that both (\ref{aa36}) and (\ref%
{aa37}) correspond to earlier results with the reverse order of sequential
convergence $\left( T,n\right) _{\func{seq}}\rightarrow \infty .$

Next suppose $\gamma >1$ so that $\rho =1+\frac{c}{T^{\gamma }}$ is closer
to unity than the LUR case with $\gamma =1.$ In this case, the numerator and
denominator of (\ref{aa23}) have the same limits as in the unit root case,
viz.,%
\begin{eqnarray*}
&&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}\left[ -\frac{\alpha _{i}c}{T^{\gamma }}%
+\frac{c}{T^{\gamma }}y_{it-3}+u_{it-2}\right] \right\} \\
&=&\sum_{i=1}^{n}\frac{1}{\sqrt{T}}\left\{
u_{iT}y_{iT-2}-\sum_{t=3}^{T}u_{it-1}u_{it-2}\right\} +o_{p}\left( 1\right) 
\underset{T\rightarrow \infty }{\Rightarrow }\sum_{i=1}^{n}\left\{
u_{i\infty }B_{i}\left( 1\right) -G_{i}\right\} ,
\end{eqnarray*}%
and%
\begin{equation*}
\sum_{i=1}^{n}\frac{1}{T}\sum_{t=2}^{T}\left\{ -\frac{\alpha _{i}c}{%
T^{\gamma }}+\frac{c}{T^{\gamma }}y_{it-2}+u_{it-1}\right\} y_{it-2}\underset%
{T\rightarrow \infty }{\Rightarrow }\sum_{i=1}^{n}\left\{
\int_{0}^{1}B_{i}dB_{i}\right\} .
\end{equation*}%
Then 
\begin{equation*}
\sqrt{T}\left( \rho _{IV}-\rho \right) \underset{T\rightarrow \infty }{%
\Rightarrow }\frac{\sum_{i=1}^{n}\left\{ u_{i\infty }B_{i}\left( 1\right)
-G_{i}\right\} }{\sum_{i=1}^{n}\int_{0}^{1}B_{i}dB_{i}}=\frac{\frac{1}{\sqrt{%
n}}\sum_{i=1}^{n}\left\{ \left( \sigma ^{-1}u_{i\infty }\right) W_{i}\left(
1\right) -\zeta _{i}\right\} }{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}%
\int_{0}^{1}W_{i}dW_{i}}\underset{n\rightarrow \infty }{\Rightarrow }2%
\mathbb{C},
\end{equation*}%
since $\left( \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\left\{ \left( \sigma
^{-1}u_{i\infty }\right) W_{i}\left( 1\right) -\zeta _{i}\right\} ,\frac{1}{%
\sqrt{n}}\sum_{i=1}^{n}\int_{0}^{1}W_{i}dW_{i}\right) \underset{n\rightarrow
\infty }{\Rightarrow }N\left( 0,\left[ 
\begin{array}{cc}
2 & 0 \\ 
0 & 1/2%
\end{array}%
\right] \right) .$ \newline
\end{proof}

\section{References}

\begin{description}
\item Magdalinos, T. and P. C. B. Phillips (2009). \textquotedblleft Limit
theory for cointegrated systems with moderately integrated and moderately
explosive regressors\textquotedblright , 25, 482-526.

\item Phillips, P. C. B. (1987a). \textquotedblleft Time Series Regression
with a Unit Root,\textquotedblright\ \textit{Econometrica,} 55, 277--302.

\item Phillips, P. C. B. (1987b). \textquotedblleft Towards a Unified
Asymptotic Theory for Autoregression,\textquotedblright\ \textit{Biometrika}
74, 535--547.

\item Phillips, P. C. B. (1989). \textquotedblleft Partially identified
econometric models,\textquotedblright\ \textit{Econometric Theory} 5,
181--240.

\item Phillips, P. C. B., T. Magdalinos (2007), "Limit Theory for Moderate
Deviations from a Unit Root," \textit{Journal of Econometrics} 136, 115-130.

\item Phillips, P.C.B. and H.R. Moon (1999) : Linear Regression Limit Theory
for Nonstationary Panel Data,\textquotedblright\ \textit{Econometrica, }67,
1057-1111.
\end{description}

\end{document}
