Appendix
A Proof of Theorem 1
Proof
Now we have
$$\begin{aligned} F_{\lambda }=\frac{R_{\lambda }}{\widehat{\sigma }^{2}} \end{aligned}$$
where \(R_{\lambda }=\frac{\left( RRSS-URSS\right) }{n}\) and \(\widehat{\sigma }^{2}=\frac{URSS}{\left( nT-n\right) }\) using a set up which is similar to Orme and Yamagata (2006). Rearranging the terms, we have
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-1\right) =\frac{1}{\widehat{\sigma }^{2}}\sqrt{n} \left( R_{\lambda }-\widehat{\sigma }^{2}\right). \end{aligned}$$
Expanding the equations, we have
$$\begin{aligned} R_{\lambda }-\widehat{\sigma }^{2}&= \frac{\left( RRSS-URSS\right) }{n}-\frac{URSS}{\left( nT-n\right) } \\&= \frac{1}{n}\sum _{i=1}^{n}\sum _{t=1}^{T}\left[ -\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) ^{2}F_{t}^{2}+2\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) u_{it}F_{t}\right] \\&-\frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left[ u_{it}^{2}+\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) ^{2}F_{t}^{2}-2\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) u_{it}F_{t}\right] \\&= -\frac{1}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} ^{2}F_{t}^{2} \\&+\frac{2}{n\sqrt{T}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} u_{it}F_{t} \\&-\frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}-\frac{1}{nT(T-1)}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} ^{2}F_{t}^{2} \\&+\frac{2}{n\sqrt{T}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} u_{it}F_{t} \\&= I+II+III+IV+V. \end{aligned}$$
Consider \(I\).
$$\begin{aligned} I&= -\frac{1}{n}\sum _{i=1}^{n}\left[ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right] ^{2}\left[ \frac{1}{T}\sum _{t=1}^{T}F_{t}^{2}\right] \\&= -\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{1}{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T}\sum _{t=1}^{T}F_{t}^{2}\right] ^{-1}. \end{aligned}$$
For \({ II}\),
$$\begin{aligned} II&= \frac{2}{n\sqrt{T}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} u_{it}F_{t} \\&= \frac{2}{n}\sum _{i=1}^{n}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right] ^{-1}. \end{aligned}$$
Then
$$\begin{aligned} I+II=\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right] ^{-1}=O_{p}\left( 1\right). \end{aligned}$$
For \(III\),
$$\begin{aligned} III=-\frac{1}{n}\sum _{i=1}^{n}\frac{1}{T-1}\sum _{t=1}^{T}u_{it}^{2}=O_{p} \left( 1\right). \end{aligned}$$
For \(IV\) and \(V\), as already shown above,
$$\begin{aligned} IV=-\frac{1}{T-1}\frac{1}{n}\sum _{i=1}^{n}\left\{ \sqrt{T}\left( \widetilde{ \lambda }_{i}-\lambda _{i}\right) \right\} ^{2}\left( \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right) =O_{p}\left( \frac{1}{T}\right) \end{aligned}$$
and
$$\begin{aligned} V&= \frac{2}{n\sqrt{T}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} u_{it}F_{t} \\&= \frac{1}{T-1}\frac{2}{n}\sum _{i=1}^{n}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right] ^{-1} \\&= O_{p}\left( \frac{1}{T}\right) . \end{aligned}$$
After rearranging all the terms, one obtains
$$\begin{aligned} R_{\lambda }-\widehat{\sigma }^{2}&= \frac{1}{n}\sum _{i=1}^{n}\left[ \frac{1 }{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right] ^{-1} \\&-\frac{1}{n}\sum _{i=1}^{n}\frac{1}{T}\sum _{t=1}^{T}u_{it}^{2}+O_{p}\left( \frac{1}{T}\right) . \end{aligned}$$
It is easy to see that
$$\begin{aligned} \sqrt{n}\left( R_{\lambda }-\widehat{\sigma }^{2}\right) =\frac{1}{\sqrt{n}} \sum _{i=1}^{n}\left( \left( \frac{1}{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t} \right) ^{2}\phi _{F}^{-1}-\sigma ^{2}\right) +O_{p}\left( \frac{\sqrt{n}}{T} \right) . \end{aligned}$$
Now we obtain
$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \left( \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right) ^{2}\phi _{F}^{-1}-\sigma ^{2}\right) \overset{d}{\rightarrow }N\left( 0,2\sigma ^{4}\right) \end{aligned}$$
by Assumption 4.
Finally,
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-1\right) =\frac{1}{\widehat{\sigma }^{2}}\sqrt{n} \left( R_{\lambda }-\widehat{\sigma }^{2}\right) \overset{d}{\rightarrow } N(0,2) \end{aligned}$$
as \(\left( n,T\right) \rightarrow \infty \) if \(\frac{\sqrt{n}}{T}\rightarrow 0.\) Note that \(\frac{T}{n}\rightarrow c\) with \(0<c<\infty \) implies that \( \frac{\sqrt{n}}{T}=\frac{n}{T}\frac{1}{\sqrt{n}}\rightarrow 0.\)
B Proof of Theorem 2
Proof
First we consider
$$\begin{aligned} \frac{URSS}{\left( nT-n\right) }&= \frac{1}{n\left( T-1\right) } \sum _{i=1}^{n}\sum _{t=1}^{T}\left( y_{it}-\widehat{\lambda }_{i}\widehat{F} _{t}\right) ^{2} \\&= \frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left[ u_{it}-\left( \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) \right] ^{2} \\&= \frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}+\frac{ 1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{\lambda } _{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) ^{2} \\&-\frac{2}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it} \\&= I+II+III. \end{aligned}$$
Consider \(I\). One can easily verify that
$$\begin{aligned} I=\frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}=\sigma ^{2}+o_{p}\left( 1\right) \end{aligned}$$
as \(\left( n,T\right) \rightarrow \infty \). Note from (Bai (2003), p. 166), that \(\widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\) can be expanded as
$$\begin{aligned} \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}&= \frac{1}{\sqrt{n}} \lambda _{i}\left( \frac{\sum _{i=1}^{n}\lambda _{i}^{2}}{n}\right) ^{-1} \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt} \\&+\frac{1}{\sqrt{T}}F_{t}\left( \frac{\sum _{t=1}^{T}F_{t}^{2}}{T}\right) ^{-1}\frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}+O_{p}\left( \frac{1}{\delta _{nt}^{2}}\right) \\&= \phi _{\lambda }^{-1}\frac{1}{\sqrt{n}}\lambda _{i}\frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}+\phi _{F}^{-1}\frac{1}{\sqrt{T}}F_{t}\frac{1 }{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}+O_{p}\left( \frac{1}{\delta _{nt}^{2}} \right). \end{aligned}$$
For \(II\).
$$\begin{aligned} II&= \frac{1}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) ^{2}\!=\!\frac{1}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \phi _{\lambda }^{-1}\frac{ 1}{\sqrt{n}}\lambda _{i}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \right.\\&\left.+\,\phi _{F}^{-1}\frac{1}{\sqrt{T}}F_{t}\left( \frac{1}{ \sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\right) \right) ^{2}+o_{p}\left( 1\right) \\&= \phi _{\lambda }^{-2}\frac{1}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T}\frac{1}{n} \lambda _{i}^{2}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}\\&+\,\phi _{F}^{-2}\frac{1}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T} \frac{1}{T}F_{t}^{2}\left( \frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is} \right) ^{2} \\&+\,2\phi _{\lambda }^{-1}\phi _{F}^{-1}\frac{1}{nT}\sum _{i=1}^{n} \sum _{t=1}^{T}\frac{1}{\sqrt{nT}}\lambda _{i}F_{t}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) \!+\!o_{p}\left( 1\right)\\&= II_{a}+II_{b}+II_{c}+o_{p}\left( 1\right) . \end{aligned}$$
Consider \(II_{a}.\)
$$\begin{aligned} \phi _{\lambda }^{-2}\frac{1}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T}\frac{1}{n} \lambda _{i}^{2}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}&= \phi _{\lambda }^{-1}\frac{1}{nT}\sum _{t=1}^{T}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}+o_{p}\left( 1\right) \\&= \phi _{\lambda }^{-1}\frac{1}{n}\frac{1}{nT}\sum _{t=1}^{T}\sum _{i=1}^{n} \sum _{k=1}^{n}\lambda _{i}\lambda _{k}u_{it}u_{kt}+o_{p}\left( 1\right) \\&= \frac{1}{n}\phi _{\lambda }^{-1}\phi _{\lambda }\sigma ^{2}+o_{p}\left( 1\right) \\&= O_{p}\left( \frac{1}{n}\right) . \end{aligned}$$
Similarly,
$$\begin{aligned} II_{b}=O_{p}\left( \frac{1}{T}\right) . \end{aligned}$$
Consider \(II_{c}.\)
$$\begin{aligned} II_{c}&= 2\phi _{\lambda }^{-1}\phi _{F}^{-1}\frac{1}{nT}\left[ \frac{1}{ \sqrt{n}}\sum _{i=1}^{n}\left( \lambda _{i}\left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) \right) \right]\\&\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}\left( F_{t}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \right) \right] \\&= O_{p}\left( \frac{1}{nT}\right) . \end{aligned}$$
Hence
$$\begin{aligned} II=O_{p}\left( \frac{1}{\delta _{nt}^{2}}\right) +o_{p}\left( 1\right) . \end{aligned}$$
Consider \(III\).
$$\begin{aligned} III&= \frac{2}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it} \\&\!= \frac{2}{nT}\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \phi _{\lambda }^{-1}\frac{ 1}{\sqrt{n}}\lambda _{i}\frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\!+\!\phi _{F}^{-1}\frac{1}{\sqrt{T}}F_{t}\frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) u_{it}\!+\!o_{p}\left( 1\right) \\&= O_{p}\left( \frac{1}{n}\right) +O_{p}\left( \frac{1}{T}\right) =O_{p}\left( \frac{1}{\delta _{nt}^{2}}\right) +o_{p}\left( 1\right) . \end{aligned}$$
Now, we write
$$\begin{aligned}&\sqrt{n}\left( R_{\lambda }-\widehat{\sigma }^{2}\left( \sqrt{\frac{T}{n}} +1\right) \right) \\&\quad =\sqrt{n}\frac{\left( RRSS-URSS\right) }{n}-\sqrt{n} \frac{URSS}{\left( nT-n\right) }\left( \sqrt{\frac{T}{n}}+1\right) \\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left[ -\left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) ^{2}+2\left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it}\right] \\&\qquad -\frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left[ u_{it}^{2}+\left( \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) ^{2}-2\left( \widehat{\lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it}\right] \\&\quad =-\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) \right\} ^{2}+\frac{2}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it}\right\} \\&\qquad -\left( \sqrt{\frac{T}{n}}+1\right) \frac{1}{\sqrt{n}\left( T-1\right) } \sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2} \\&\qquad -\left( \sqrt{\frac{T}{n}}+1\right) \frac{1}{\sqrt{n}(T-1)} \sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda }_{i}\widehat{F} _{t}-\lambda _{i}F_{t}\right) \right\} ^{2} \\&\qquad +\left( \sqrt{\frac{T}{n}}+1\right) \frac{2}{\sqrt{n}\left( T-1\right) } \sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda }_{i}\widehat{F} _{t}-\lambda _{i}F_{t}\right) u_{it}\right\} \\&\quad =I+II+III+IV+V. \end{aligned}$$
Note that \(IV=O_{p}\left( \frac{\sqrt{T}}{\delta _{nt}^{2}}\right) \) and \( V=O_{p}\left( \frac{\sqrt{T}}{\delta _{nt}^{2}}\right) \) as shown above. Now we assume that
$$\begin{aligned} \frac{\sqrt{T}}{n}\rightarrow 0. \end{aligned}$$
Note that \(\frac{\sqrt{T}}{\delta _{nt}^{2}}\rightarrow 0\) when \(\frac{\sqrt{ T}}{n}\rightarrow 0.\) Also \(\frac{T}{n}\rightarrow c\) implies that \(\frac{ \sqrt{T}}{n}=\frac{T}{n}\frac{1}{\sqrt{T}}\rightarrow 0.\) Clearly,
$$\begin{aligned}&\sqrt{n}\left( R_{\lambda }-\widehat{\sigma }^{2}\left( \sqrt{\frac{T}{n}} +1\right) \right) =-\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) \right\} ^{2}\\&\quad +\frac{1}{ \sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda }_{i} \widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it}\right\} -\left( \sqrt{\frac{T }{n}}+1\right) \frac{1}{\sqrt{n}T}\sum _{i=1}^{n} \sum _{t=1}^{T}u_{it}^{2}+o_{p}\left( 1\right) . \end{aligned}$$
Consider the first term.
$$\begin{aligned}&-\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) \right\} ^{2} \\&\quad =-\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \phi _{\lambda }^{-1}\frac{1}{\sqrt{n}}\lambda _{i}\frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}+\phi _{F}^{-1}\frac{1}{\sqrt{T}}F_{t}\frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) ^{2}+o_{p}\left( 1\right) \\&\quad \!=\!-\phi _{\lambda }^{-2}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{t=1}^{T}\frac{ 1}{n}\lambda _{i}^{2}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}\!-\!\phi _{F}^{-2}\frac{1}{n}\sum _{i=1}^{n}\sum _{t=1}^{T} \frac{1}{T}F_{t}^{2}\left( \frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is} \right) ^{2} \\&\qquad -2\,\phi _{\lambda }^{-1}\phi _{F}^{-1}\frac{1}{n}\sum _{i=1}^{n} \sum _{t=1}^{T}\frac{1}{\sqrt{nT}}\lambda _{i}F_{t}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) +o_{p}\left( 1\right) \\&\quad =I+II+III. \end{aligned}$$
Consider \(I+II\).
$$\begin{aligned}&I+II =-\phi _{\lambda }^{-2}\frac{1}{\sqrt{n}}\left( \frac{1}{n} \sum _{i=1}^{n}\lambda _{i}^{2}\right) \sum _{t=1}^{T}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}\\&\qquad -\phi _{F}^{-2}\frac{1}{\sqrt{n}} \sum _{i=1}^{n}\left( \frac{1}{T}\sum _{t=1}^{T}F_{t}^{2}\right) \left( \frac{1 }{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\right) ^{2} \\&\quad =-\frac{1}{\sqrt{n}}\sum _{t=1}^{T}\phi _{\lambda }^{-1}\left( \frac{1}{ \sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}-\frac{1}{\sqrt{n}} \sum _{i=1}^{n}\phi _{F}^{-1}\left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) ^{2}+o_{p}\left( 1\right) . \end{aligned}$$
Consider \(III\).
$$\begin{aligned}&\phi _{\lambda }^{-1}\phi _{F}^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n} \sum _{t=1}^{T}\frac{1}{\sqrt{nT}}\lambda _{i}F_{t}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) \\&\quad =\phi _{\lambda }^{-1}\phi _{F}^{-1}\frac{1}{\sqrt{n}}\left( \frac{1}{ \sqrt{n}}\sum _{i=1}^{n}\lambda _{i}\left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) \right) \left( \frac{1}{\sqrt{T}} \sum _{t=1}^{T}F_{t}\left( \frac{1}{\sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) \right) \\&\quad =O_{p}\left( \frac{1}{\sqrt{n}}\right) . \end{aligned}$$
Consider the second term.
$$\begin{aligned} 2\frac{1}{n}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda } _{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it}\right\}&= 2\frac{1}{n}\sum _{t=1}^{T}\phi _{\lambda }^{-1}\left( \frac{1}{\sqrt{n}} \sum _{i=1}^{n}\lambda _{i}u_{it}\right) ^{2}\\&+2\frac{1}{n}\sum _{i=1}^{n}\phi _{F}^{-1}\left( \frac{1}{\sqrt{T}}\sum _{t=1}^{T}F_{t}u_{it}\right) ^{2}+o_{p}\left( 1\right) . \end{aligned}$$
Therefore
$$\begin{aligned}&-\frac{1}{n}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda } _{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) \right\} ^{2}+\frac{1}{n} \sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \left( \widehat{\lambda }_{i}\widehat{F} _{t}-\lambda _{i}F_{t}\right) u_{it}\right\} \\&\quad =\frac{1}{n}\sum _{t=1}^{T}\phi _{\lambda }^{-1}\left( \frac{1}{\sqrt{n}} \sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}+\frac{1}{n}\sum _{i=1}^{n}\phi _{F}^{-1}\left( \frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\right) ^{2}+o_{p}\left( 1\right) . \end{aligned}$$
We know that
$$\begin{aligned}&\frac{1}{\sqrt{n}}\sum _{t=1}^{T}\phi _{\lambda }^{-1}\left( \frac{1}{\sqrt{ n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}+\frac{1}{\sqrt{n}} \sum _{i=1}^{n}\phi _{F}^{-1}\left( \frac{1}{\sqrt{T}} \sum _{s=1}^{T}F_{s}u_{is}\right) ^{2} \\&\quad =\frac{1}{\sqrt{T}}\sum _{t=1}^{T}\phi _{\lambda }^{-1}\left( \frac{1}{ \sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}\left( \sqrt{\frac{T}{n} }\right) +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\phi _{F}^{-1}\left( \frac{1}{ \sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\right) ^{2}. \end{aligned}$$
Hence
$$\begin{aligned}&\frac{1}{\sqrt{n}}\sum _{t=1}^{T}\left[ \phi _{\lambda }^{-1}\left( \frac{1}{ \sqrt{n}}\sum _{k=1}^{n}\lambda _{k}u_{kt}\right) ^{2}-\sqrt{\frac{T}{n}} \sigma ^{2}\right]\\&\quad +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \phi _{F}^{-1}\left( \frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\right) ^{2}-\sigma ^{2}\right] \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \sigma ^{4}\right) \end{aligned}$$
by Assumption 4 and because \(\frac{1}{\sqrt{n}}\sum _{k=1}^{n} \lambda _{k}u_{kt}\) and \(\frac{1}{\sqrt{T}}\sum _{s=1}^{T}F_{s}u_{is}\) are asymptotically independent. Therefore
$$\begin{aligned} \sqrt{n}\left( R_{\lambda }-\widehat{\sigma }^{2}\left( \sqrt{\frac{T}{n}} +1\right) \right) \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \sigma ^{4}\right) \end{aligned}$$
as \(\left( n,T\right) \rightarrow \infty \) and \(\frac{T}{n}\rightarrow c.\) Finally
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-\left( \sqrt{\frac{T}{n}}+1\right) \right) \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \right) \end{aligned}$$
as required.\(\square \)
C Proof of Theorem 3
Proof
First we revisit Theorems 1 and 2 with
$$\begin{aligned} u_{it}=\sigma _{i}e_{it}. \end{aligned}$$
Now suppose \(F_{t}\) is known.
$$\begin{aligned} \sqrt{n}\left( R_{\lambda }-\widehat{\sigma }^{2}\right)&= \frac{\left( RRSS-URSS\right) }{\sqrt{n}}-\frac{URSS}{\sqrt{n}\left( T-1\right) } \\&= -\frac{1}{\sqrt{n}T}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} ^{2}F_{t}^{2} \\&+\,\frac{2}{\sqrt{n}\sqrt{T}}\sum _{i=1}^{n}\sum _{t=1}^{T}\left\{ \sqrt{T} \left( \widetilde{\lambda }_{i}\!-\!\lambda _{i}\right) \right\} u_{it}F_{t} -\frac{1}{\sqrt{n}\left( T\!-\!1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}\\&-\frac{1}{\sqrt{n}T(T-1)}\sum _{i=1}^{n} \sum _{t=1}^{T}\left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} ^{2}F_{t}^{2} \\&+\frac{2}{\sqrt{n}\sqrt{T}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T} \left\{ \sqrt{T}\left( \widetilde{\lambda }_{i}-\lambda _{i}\right) \right\} u_{it}F_{t} \\&= I+II+III+IV+V. \end{aligned}$$
Recall that
$$\begin{aligned} I+II&= \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}\left[ \frac{1}{T} \sum _{t=1}^{T}F_{t}^{2}\right] ^{-1} \\&= \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\phi _{F}^{-1}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}+o_{p}\left( 1\right) . \end{aligned}$$
We know Assumption 4 that
$$\begin{aligned} \frac{1}{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\overset{d}{\rightarrow }N\left( 0,\Phi _{i}\right) \end{aligned}$$
where
$$\begin{aligned} \Phi _{i}&= p\lim _{T\rightarrow \infty }\frac{1}{T}\sum _{s=1}^{T} \sum _{t=1}^{T}E\left[ F_{t}F_{s}u_{is}u_{it}\right] \\&= p\lim _{T\rightarrow \infty }\frac{1}{T}\sum _{s=1}^{T}\sum _{t=1}^{T}\sigma _{i}^{2}E\left[ F_{t}F_{s}e_{is}e_{it}\right] \\&= \sigma _{i}^{2}\phi _{F}. \end{aligned}$$
Consider III.
$$\begin{aligned} \frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}&= \frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2} \\&= \frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sigma _{i}^{2}\sum _{t=1}^{T}e_{it}^{2} \\&= \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sigma _{i}^{2}+o\left( 1\right) . \end{aligned}$$
It is easy to see that \(IV\) and \(V\) are \(O_{p}\left( \frac{\sqrt{n}}{T} \right) .\) Then
$$\begin{aligned}&\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\phi _{F}^{-1}\left[ \frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sigma _{i}^{2} \\&= \frac{1}{\sqrt{n}}\left( \sum _{i=1}^{n}\phi _{F}^{-1}\left[ \frac{1}{ \sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\sigma _{i}^{2}\right) . \end{aligned}$$
It follows that
$$\begin{aligned} \frac{\sum _{i=1}^{n}\left( \left[ \frac{\frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}}{\sqrt{\phi _{F}}}\right] ^{2}-\sigma _{i}^{2}\right) }{\sqrt{2\sum _{i=1}^{n}\sigma _{i}^{4}}}\overset{d}{ \rightarrow }N\left( 0,1\right) \end{aligned}$$
or
$$\begin{aligned} \frac{\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \phi _{F}^{-1}\left[ \frac{1}{ \sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\sigma _{i}^{2}\right) }{ \sqrt{2\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{4}}}\overset{d}{\rightarrow } N\left( 0,1\right) . \end{aligned}$$
Hence
$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \phi _{F}^{-1}\left[ \frac{1}{\sqrt{T} }\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\sigma _{i}^{2}\right) \overset{d}{ \rightarrow }N\left( 0,2\lim _{n\rightarrow \infty }\frac{1}{n} \sum _{i=1}^{n}\sigma _{i}^{4}\right) . \end{aligned}$$
Finally
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-1\right) \overset{d}{\rightarrow }N\left( 0,2 \frac{\lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{4}}{ \left( \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{2}\right) ^{2}}\right) \end{aligned}$$
if
$$\begin{aligned}&\frac{\sqrt{n}}{T}\rightarrow 0,\\&\lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{4}<\infty \end{aligned}$$
and
$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{2}<\infty . \end{aligned}$$
Next we assume \(F_{t}\) is unknown. Recall
$$\begin{aligned} \frac{URSS}{\left( nT-n\right) }&= \frac{1}{n\left( T-1\right) } \sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}+\frac{1}{n\left( T-1\right) } \sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{\lambda }_{i}\widehat{F} _{t}-\lambda _{i}F_{t}\right) ^{2} \\&-\frac{2}{n\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\left( \widehat{ \lambda }_{i}\widehat{F}_{t}-\lambda _{i}F_{t}\right) u_{it} \\&= I+II+III. \end{aligned}$$
We know that \(II+\)
\(III=O_{p}\left( \frac{1}{\delta _{nT}^{2}}\right) .\) Following similar steps as in the proof of Theorem 2 we obtain
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-\left( \sqrt{\frac{T}{n}}+1\right) \right) \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \frac{ \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{4}}{\left( \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\sigma _{i}^{2}\right) ^{2}}\right) . \end{aligned}$$
Next we allow
$$\begin{aligned} u_{it}=\sigma _{t}e_{it} \end{aligned}$$
and we examine
$$\begin{aligned}&\frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2} =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \lim _{n\rightarrow \infty }\frac{1 }{T}\sum _{t=1}^{T}\sigma _{t}^{2}\right) +o_{p}\left( 1\right) \end{aligned}$$
which lead to
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-1\right) \overset{d}{\rightarrow }N\left( 0,2\right) \end{aligned}$$
if \(F_{t}\) is known and
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-\left( \sqrt{\frac{T}{n}}+1\right) \right) \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \right) \end{aligned}$$
if \(F_{t}\) is unknown.
Finally we set \(u_{it}=\sigma _{it}e_{it}.\) Notice that
$$\begin{aligned} \frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}u_{it}^{2}&= \frac{1}{\sqrt{n}\left( T-1\right) }\sum _{i=1}^{n}\sum _{t=1}^{T}\sigma _{it}^{2}e_{it}^{2} \\&= \frac{1}{\sqrt{n}T}\sum _{i=1}^{n}\sum _{t=1}^{T}\sigma _{it}^{2}+o_{p}\left( 1\right) . \end{aligned}$$
Recall that
$$\begin{aligned} \frac{1}{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\overset{d}{\rightarrow }N\left( 0,\Phi _{i}\right) \end{aligned}$$
where
$$\begin{aligned} \Phi _{i}=\phi _{F}\omega _{i} \end{aligned}$$
with
$$\begin{aligned} \omega _{i}^{2}=\lim _{T\rightarrow \infty }\frac{1}{T}\sum _{t=1}^{T}\sigma _{it}^{2}. \end{aligned}$$
Recall
$$\begin{aligned} \left[ \frac{\frac{1}{\sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}}{\sqrt{\phi _{F}\omega _{i}^{2}}}\right] ^{2}\overset{d}{\rightarrow }\chi _{1}^{2} \end{aligned}$$
and
$$\begin{aligned} \frac{\sum _{i=1}^{n}\left( \left[ \frac{\frac{1}{\sqrt{T}} \sum _{t=1}^{T}u_{it}F_{t}}{\sqrt{\phi _{F}}}\right] ^{2}-\omega _{i}^{2}\right) }{\sqrt{2\sum _{i=1}^{n}\omega _{i}^{4}}}\overset{d}{ \rightarrow }N\left( 0,1\right) \end{aligned}$$
or
$$\begin{aligned} \frac{\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \phi _{F}^{-1}\left[ \frac{1}{ \sqrt{T}}\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\omega _{i}^{2}\right) }{ \sqrt{2\frac{1}{n}\sum _{i=1}^{n}\omega _{i}^{4}}}\overset{d}{\rightarrow } N\left( 0,1\right) . \end{aligned}$$
Hence
$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left( \phi _{F}^{-1}\left[ \frac{1}{\sqrt{T} }\sum _{t=1}^{T}u_{it}F_{t}\right] ^{2}-\omega _{i}^{2}\right) \overset{d}{ \rightarrow }N\left( 0,2\lim _{n\rightarrow \infty }\frac{1}{n} \sum _{i=1}^{n}\omega _{i}^{4}\right) . \end{aligned}$$
Finally
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-1\right) \overset{d}{\rightarrow }N\left( 0,2 \frac{\lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\omega _{i}^{4}}{ \left( \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\omega _{i}^{2}\right) ^{2}}\right) \end{aligned}$$
if \(F_{t}\) is known and
$$\begin{aligned} \sqrt{n}\left( F_{\lambda }-\left( \sqrt{\frac{T}{n}}+1\right) \right) \overset{d}{\rightarrow }N\left( 0,4\left( c+1\right) \frac{ \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\omega _{i}^{4}}{\left( \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\omega _{i}^{2}\right) ^{2}}\right) \end{aligned}$$
if \(F_{t}\) is unknown.\(\square \)
