Proof of Theorem 1
Proof
Let \(S=\left\{ T^{-\delta }\left( T+1\right) \left( 1-{\hat{\rho }} _\mathrm{FE}\right) >3\right\} \) and \({\bar{S}}=\left\{ T^{-\delta }\left( T+1\right) \left( 1-{\hat{\rho }}_\mathrm{FE}\right) \le 3\right\} \). Consider (1). When \(\left| \rho \right| <1\), it suffices to show that
$$\begin{aligned}&\sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-\rho \right) -\sqrt{nT}\left( \hat{\rho }_\mathrm{FE}-\rho +\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \\&\quad = \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+\hat{\rho }_\mathrm{FE}}{T}\right) \overset{p}{\rightarrow }0. \end{aligned}$$
We have
$$\begin{aligned}&\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \right|>\epsilon \right) \\&\quad =\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \right|>\epsilon |S\right) \Pr \left( S\right) \\&\qquad +\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \right| >\epsilon |{\bar{S}}\right) \Pr \left( {\bar{S}}\right) . \end{aligned}$$
The first term is zero given that, if S is true, we have \({\hat{\rho }} _\mathrm{FEBC}={\hat{\rho }}_\mathrm{FE}+\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\) so that \(\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \right| >\epsilon |S\right) =0\). The second term is zero since
$$\begin{aligned}&T^{-\delta }\left( T+1\right) \left( 1-{\hat{\rho }}_\mathrm{FE}\right) -3\\&\quad =T^{-\delta }\left( T+1\right) \left( 1-\rho \right) -T^{-\delta } \sqrt{\frac{T}{n}}\frac{T+1}{T}\left[ \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FE} -\rho +\frac{1+\rho }{T}\right) \right] \\&\qquad -3+T^{-\delta }\frac{T+1}{T}\left( 1+\rho \right) \\&\quad =T^{-\delta }\left( T+1\right) \left( 1-\rho \right) +O_{p}\left( T^{-\delta }\sqrt{\frac{T}{n}}\frac{T+1}{T}\right) +O_{p}\left( T^{-\delta }\frac{T+1}{T}\right) \rightarrow \infty \end{aligned}$$
using\(\sqrt{nT}\left( {\hat{\rho }}_\mathrm{FE}-\rho +\frac{1+\rho }{T}\right) =O_{p}\left( 1\right) \) and \(T^{-\delta }\left( T+1\right) \left( 1-\rho \right) \rightarrow \infty \) as \(T\rightarrow \infty \). Hence \(\Pr \left( {\bar{S}}\right) \rightarrow 0\) as \(T\rightarrow \infty \). Therefore, \(\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }}_\mathrm{FEBC}-{\hat{\rho }}_\mathrm{FE}-\frac{1+{\hat{\rho }}_\mathrm{FE}}{T}\right) \right| >\epsilon \right) \) \(\rightarrow 0\) as \(\left( n,T\right) \rightarrow \infty \).
Consider (2). When \(\rho =1\), we have
$$\begin{aligned}&\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{n}T\left( {\hat{\rho }}_\mathrm{FEBC}-1\right) \right|>\epsilon \right) \\&\quad =\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{n}T\left( {\hat{\rho }}_\mathrm{FEBC}-1\right) \right|>\epsilon |S\right) \Pr \left( S\right) \\&\qquad +\lim _{\left( n,T\right) \rightarrow \infty }\Pr \left( \left| \sqrt{n}T\left( {\hat{\rho }}_\mathrm{FEBC}-1\right) \right| >\epsilon |{\bar{S}}\right) \Pr \left( {\bar{S}}\right) . \end{aligned}$$
Notice that
$$\begin{aligned} \Pr \left( S\right)&=\Pr \left( T^{-\delta }\left( T+1\right) \left( 1-{\hat{\rho }}_\mathrm{FE}\right)>3\right) \\&=\Pr \left( T^{-\delta }\left[ 3+\left( T+1\right) \left( 1-\hat{\rho }_\mathrm{FE}-\frac{3}{T+1}\right) \right]>3\right) \\&=\Pr \left( 3T^{-\delta }-\frac{1}{\sqrt{n}T^{\delta }}\frac{T+1}{T}\sqrt{n}T\left( {\hat{\rho }}_\mathrm{FE}-1+\frac{3}{T+1}\right) >3\right) \\&\rightarrow 0 \end{aligned}$$
as \(\left( n,T\right) \rightarrow \infty \) since \(\sqrt{n}T\left( \hat{\rho }_\mathrm{FE}-1+\frac{3}{T+1}\right) \) \(=O_{p}\left( 1\right) \), \(T^{-\delta }\rightarrow 0\) and \(\frac{1}{\sqrt{n}T^{\delta }}\rightarrow 0\). Hence the first term is zero. For the second term, if \({\bar{S}}\) is true, \(\hat{\rho }_\mathrm{FEBC}=1\) so that \(\Pr \left( \left| \sqrt{nT}\left( {\hat{\rho }} _\mathrm{FEBC}-1\right) \right| >\epsilon |{\bar{S}}\right) =0\). Thus, \(\Pr \left( \left| \sqrt{n}T\left( {\hat{\rho }}_\mathrm{FEBC}-1\right) \right| >\epsilon \right) \) \(\rightarrow 0\) as \(\left( n,T\right) \rightarrow \infty \). \(\square \)