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\begin{document}

\title{
{\bf Supplementary Material}
 \\[20pt] 
Dynamic Linear Panel Regression Models \\ with Interactive Fixed Effects
}
\author{\setcounter{footnote}{2}
Hyungsik Roger Moon\footnote{
Department of Economics and USC Dornsife INET, University of Southern California,
Los Angeles, CA 90089-0253.
Email: {\tt moonr@usc.edu}.
Department of Economics, Yonsei University, Seoul, Korea.
}
\and Martin Weidner\footnote{
 Department of Economics,
 University College London,
 Gower Street,
 London WC1E~6BT, U.K.,
 and CeMMaP.
 Email: {\tt m.weidner@ucl.ac.uk}.
}}

\date{\today}

\maketitle


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Identification (Theorem~\ref{th:id})}
\label{app:identification}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{proof}[\bf Proof of Theorem~\ref{th:id}]
    Let  $Q(\beta,\lambda,f) \equiv  \mathbb{E}\left(\left\| Y \, - \, \beta \cdot X \, - \, \lambda \, f'
    \right\|^2_F \Big|  \lambda^0, f^0, w \right)$, where $\beta \in \mathbb{R}^K$,
    $\lambda \in \mathbb{R}^{N\times R}$ and
     $f \in  \mathbb{R}^{T\times R}$.
    We have
   \begin{align*}
        & Q(\beta,\lambda,f)
      \nonumber \\
         &= \mathbb{E} \left\{ \Tr \left[ \left( Y \, - \, \beta \cdot X \, - \, \lambda \, f' \right)'
                      \left( Y \, - \, \beta \cdot X \, - \, \lambda \, f' \right)
                      \right] \Big|  \lambda^0, f^0, w \right\}
      \nonumber \\
         &=  \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  + e \right)'
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  + e \right)
           \right] \Big|  \lambda^0, f^0, w \right\}
       \nonumber \\
          &=   \mathbb{E} \left[ \Tr \left( e' e \right) \Big|  \lambda^0, f^0, w \right]
       \nonumber \\
        & \qquad +
          \underbrace{ \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)'
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
          }_{ \equiv Q^*(\beta,\lambda,f) }
              .
    \end{align*}
    In the last step we used Assumption~\ref{ass:id}$(ii)$.
    Because $ \mathbb{E} \left[ \Tr \left( e' e \right) \Big|  \lambda^0, f^0, w \right]$
    is independent of $\beta,\lambda,f$, we find
   minimizing $Q(\beta,\lambda,f)$ is equivalent to
   minimizing $Q^*(\beta,\lambda,f)$. We decompose $Q^*(\beta,\lambda,f)$
   as follows
   \begin{align*}
       & Q^*(\beta,\lambda,f)
     \nonumber \\
       &=\mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)'
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
      \nonumber \\
        &=      \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)'
             M_{(\lambda,\lambda^0,w)}
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
        \nonumber \\
          & \quad +
              \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)'
             P_{(\lambda,\lambda^0,w)}
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
      \nonumber \\
        &=   \underbrace{   \mathbb{E} \left\{ \Tr \left[
          \left(  (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)'
             M_{(\lambda,\lambda^0,w)}
           \left(    (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
            }_{\equiv Q^{\rm high}(\beta^{\rm high},\lambda) }
        \nonumber \\
          & \quad +
             \underbrace{     \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)'
             P_{(\lambda,\lambda^0,w)}
           \left(  \lambda^0 f^{0 \prime} -  \lambda f' -  (\beta-\beta^0) \cdot X  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
             }_{\equiv Q^{\rm low}(\beta,\lambda,f) }    ,
   \end{align*}
   where $ (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high} =
   \sum_{m=K_1+1}^K (\beta_m-\beta^0_m) X_m$.
   A lower bound
   on $ Q^{\rm high}(\beta^{\rm high},\lambda)$ is given by
   \begin{align}
        & Q^{\rm high}(\beta^{\rm high},\lambda)
     \nonumber \\
        &\geq
        \min_{\widetilde \lambda \in \mathbb{R}^{N \times (R+R+{\rm rank}(w))}}
        \mathbb{E} \left\{ \Tr \left[
          \left(  (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)'
             M_{(\widetilde \lambda,\lambda,w)}
           \left(    (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
     \nonumber \\
        &=
        \sum_{r=R+R+{\rm rank}(w)}^{\min(N,T)}
       \mu_r\left\{  \mathbb{E}\left[ \left( (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)
             \left(  (\beta^{\rm high}-\beta^{0, {\rm high}}) \cdot X_{\rm high}  \right)'
               \Big|  \lambda^0, f^0, w \right] \right\}.
          \label{LowerBoundQhigh}
   \end{align}

   Because $Q^*(\beta,\lambda,f)$,
   $Q^{\rm high}(\beta^{\rm high},\lambda)$,
   and
   $Q^{\rm low}(\beta,\lambda,f)$,
   are expectations
   of traces of positive semi-definite matrices we have  $Q^*(\beta,\lambda,f) \geq 0$,
   $Q^{\rm high}(\beta^{\rm high},\lambda) \geq 0$,
   and $Q^{\rm low}(\beta,\lambda,f) \geq 0$
   for all $\beta$, $\lambda$, $f$.
  Let $\bar \beta$, $\bar \lambda$ and $\bar f$ be the parameter values
   that minimize $Q(\beta,\lambda,f)$, and thus also $Q^*(\beta,\lambda,f)$.
  Because
    $Q^*(\beta^0,\lambda^0,f^0)=0$
   we have
    $Q^*(\bar \beta,\bar \lambda,\bar f) = \min_{\beta,\lambda,f} Q^*(\beta,\lambda,f) = 0$.
   This implies
   $Q^{\rm high}(\bar \beta^{\rm high},\bar \lambda) = 0$
   and
   $Q^{\rm low}(\bar \beta,\bar \lambda,\bar f) = 0$.
   Assumption~\ref{ass:id}$(v)$,
   the lower bound \eqref{LowerBoundQhigh},
   and $Q^{\rm high}(\bar \beta^{\rm high},\bar \lambda) = 0$
   imply $\bar \beta^{\rm high} =  \beta^{0, {\rm high}}$.
   Using this, we find
   \begin{align}
       & Q^{\rm low}(\bar \beta,\bar \lambda,\bar f)
     \nonumber \\
       &= \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \bar \lambda \bar f' -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)'
           \left(  \lambda^0 f^{0 \prime} -  \bar \lambda \bar f'
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}   ,
     \nonumber \\
       &\geq \min_f \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -  \bar \lambda  f' -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)'
           \left(  \lambda^0 f^{0 \prime} -  \bar \lambda  f'
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\}
     \nonumber \\
      &=  \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)'
           M_{\bar \lambda}
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\} ,
        \label{QlowLowBound}
   \end{align}
   where
   $  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}
   =  \sum_{l=1}^{K_1} (\bar \beta_l-\beta^0_l) X_l$.
  Because
    $Q^{\rm low}(\bar \beta,\bar \lambda,\bar f) = 0$
    and the last expression in \eqref{QlowLowBound} is non-negative
    we must have
   \begin{align*}
        \mathbb{E} \left\{ \Tr \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)'
           M_{\bar \lambda}
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)
            \right] \Big|  \lambda^0, f^0, w \right\} &= 0.
   \end{align*}
   Using $M_{\bar \lambda} =M_{\bar \lambda} M_{\bar \lambda}$
   and  the cyclicality of the trace we
   obtain from the last equality:
   \begin{align*}
       \Tr \bigg\{
         M_{\bar \lambda}
          A M_{\bar \lambda}
                       \bigg\}  = 0,
   \end{align*}
   where
   $A =\mathbb{E} \left[
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)
\left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)' \Big|  \lambda^0, f^0, w \right]$.   
   The trace of a positive semi-definite matrix is only equal to zero
   if the matrix itself is equal to zero, so we find
   \begin{align*}
       M_{\bar \lambda}
          A M_{\bar \lambda}
           &= 0,
   \end{align*}
   This together with the fact that $A$ itself
   is positive semi definite implies
   (note that $A$ positive semi-definite implies $A=CC'$
   for some matrix $C$, and $M_{\bar \lambda}
          A M_{\bar \lambda}
           = 0$ then implies $M_{\bar \lambda} C = 0$,
           i.e., $C = P_{\bar \lambda} C$)
   \begin{align*}
      A &=  
          P_{\bar \lambda}
          A P_{\bar \lambda}  ,
   \end{align*}
   and therefore
   ${\rm rank}(A) \leq {\rm rank}( P_{\bar \lambda} )  \leq R$. 
   We have thus shown
    \begin{align*}
       {\rm rank} \left\{ \mathbb{E} \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)'
            \Big|  \lambda^0, f^0, w \right]
          \right\} \leq R.
   \end{align*}
    We furthermore find
    \begin{align*}
        R &\geq {\rm rank} \left\{ \mathbb{E} \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)'
            \Big|  \lambda^0, f^0, w \right] \right\}
       \nonumber \\
      &  \geq      {\rm rank} \left\{ M_{w} \mathbb{E} \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           P_{f^0}
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)' M_w
            \Big|  \lambda^0, f^0, w \right] \right\}
   \nonumber \\ & \quad
       +  {\rm rank} \left\{ P_{w} \mathbb{E} \left[
          \left(  \lambda^0 f^{0 \prime} -
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           M_{f^0}
           \left(  \lambda^0 f^{0 \prime}
              -  (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)' P_w
            \Big|  \lambda^0, f^0, w \right] \right\}
       \nonumber \\
      &  \geq      {\rm rank} \left[
      M_{w}   \lambda^0 f^{0 \prime} f^0 \lambda^{0 \prime} M_w
          \right]
   \nonumber \\ & \quad
       +  {\rm rank} \left\{   \mathbb{E} \left[
          \left(
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           M_{f^0}
           \left(
              (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)'
            \Big|  \lambda^0, f^0, w \right] \right\} .
   \end{align*}
    Assumption~\ref{ass:id}$(iv)$ guarantees
   ${\rm rank} \left( M_{w}   \lambda^0 f^{0 \prime} f^0 \lambda^{0 \prime} M_w \right)
    = {\rm rank} \left(  \lambda^0 f^{0 \prime} f^0 \lambda^{0 \prime} \right)
    = R$, that is, we must have
    \begin{align*}
         \mathbb{E} \left[
          \left(
           (\bar \beta^{\rm low}-\beta^{0, {\rm low}}) \cdot X_{\rm low}  \right)
           M_{f^0}
           \left(
              (\bar \beta^{\rm low}-\beta^{0,{\rm low}}) \cdot X_{\rm low}  \right)'
            \Big|  \lambda^0, f^0, w \right] =0.
    \end{align*}
    According to Assumption~\ref{ass:id}$(iii)$ this implies
    $\bar \beta^{\rm low} =\beta^{0,{\rm low}}$, i.e., we have
    $\bar \beta = \beta^0$.
    This also implies
    $Q^*(\bar \beta,\bar \lambda,\bar f) = \|  \lambda^0 f^{0 \prime} -  \bar \lambda \bar f'   \|_F^2
    =0$,
    and therefore $\bar \lambda \bar f' = \lambda^0 f^{0 \prime}$.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Examples of Error Distributions}
\label{app:error}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The following Lemma provides examples of error distributions that satisfy $\|e\|={\cal O}_p(\sqrt{\max(N,T)})$
as $N,T \rightarrow \infty$.
Example (i) is particularly relevant for us, because those
assumptions on $e_{it}$ are imposed in Assumption~\ref{ass:A5} in the main text, i.e., under 
those main text assumptions we indeed have $\|e\|={\cal O}_p(\sqrt{\max(N,T)})$.

\begin{lemma}
    \label{lemma:Enorm}
For each of the following distributional assumptions on the errors $e_{it}$, $i=1,\ldots,N$, $t=1,\ldots,T$,
we have $\|e\|={\cal O}_p(\sqrt{\max(N,T)})$.
   \begin{itemize}
      \item[(i)] The $e_{it}$ are independent across $i$ and $t$, conditional on ${\cal C}$, and
                 satisfy
                 $\mathbb{E}(e_{it} | {\cal C}) = 0$, and $\mathbb{E}(e_{it}^4 | {\cal C})$ is bounded uniformly by a non-random constant,
                 uniformly
                 over $i,t$ and $N,T$.
  Here ${\cal C}$ can be any conditioning sigma-field, including
  the empty one (corresponding to unconditional expectations).               
                                  
      \item[(ii)] The $e_{it}$ follow different ${\rm MA}(\infty)$ processes for each $i$, namely
                  \begin{align}
                     e_{it} &= \sum_{\tau=0}^\infty \, \psi_{i\tau} \, u_{i,t-\tau} \; , \qquad \text{for }
                     i=1\ldots N, \; t=1\ldots T \; ,  \label{errorMA}
                  \end{align}
                  where the $u_{it}$, $i=1\ldots N$, $t=-\infty \ldots T$ are independent random
                  variables with $\mathbb{E} u_{it} =0$ and $\mathbb{E} u_{it}^4$ uniformly bounded
                  across $i,t$ and $N,T$. The coefficients $\psi_{i\tau}$ satisfy
                  \begin{align}
                     \sum_{\tau=0}^\infty \, \tau \, \max_{i=1\ldots N} \, \psi_{i\tau}^2  \,
                            &< \, B \; , & \sum_{\tau=0}^\infty \, \max_{i=1\ldots N}  \left|
                      \psi_{i\tau} \right|  \, &< \, B \; ,
                      \label{MArestrictionsPSI}
                  \end{align}
                  for a finite constant $B$ which is independent of $N$ and $T$.
      \item[(iii)] The error matrix $e$ is generated as
         $e=\sigma^{1/2} \, u \, \Sigma^{1/2}$, where
                   $u$ is an $N \times T$ matrix with
                   independently distributed entries $u_{it}$ and $\mathbb{E} u_{it}=0$,
                   $\mathbb{E} u_{it}^2=1$, and $\mathbb{E} u_{it}^4$ is bounded uniformly across $i,t$
                   and $N,T$. Here $\sigma$ is the $N\times N$ cross-sectional covariance matrix,
                   and $\Sigma$ is the $T\times T$ time-serial covariance matrix, and they satisfy
                   \begin{align}
                      \max_{j=1\ldots N} \, \sum_{i=1}^N \, \left| \sigma_{ij} \right| \, &< \, B
                       \; , & \max_{\tau=1\ldots T} \, \sum_{t=1}^T \, \left| \Sigma_{t\tau}
                      \right| \, &< \, B \; ,
                   \end{align}
                   for some finite constant $B$ which is independent of $N$ and $T$.
                   In this example we have $\mathbb{E} e_{it} e_{j\tau} = \sigma_{ij} \Sigma_{t\tau}$.
\end{itemize}
\end{lemma}

 

\begin{proof}[\bf Proof of Lemma~\ref{lemma:Enorm}, Example (i)]
   Latala \cite*{Latala2006} showed that for a $N\times T$ matrix $e$ with independent entries,
   conditional on ${\cal C}$, we have
   \begin{align*}
      \mathbb{E}\left(\| e \| \, \big| {\cal C} \right)  \, \leq \, c \left\{ \max_i  \left[ \sum_t 
      \mathbb{E}\left( e_{it}^2  \, \big| {\cal C} \right) \right]^{1/2}
                                            +\max_j \left[ \sum_i \mathbb{E} \left( e_{it}^2  \, \big| {\cal C} \right) \right]^{1/2}
                                            + \left[ \sum_{i,t} \mathbb{E} \left( e_{it}^4 \, \big| {\cal C} \right) \right]^{1/4} 
                                            \right\} \; ,
   \end{align*}
   where $c$ is some universal constant.
   Because we assumed uniformly bounded $4$th conditional moments for $e_{it}$ we thus have
   $\|e\| = {\cal O}_P(\sqrt{T})+{\cal O}_P(\sqrt{N})+{\cal O}_P((TN)^{1/4}) = {\cal O}_p(\sqrt{\max(N,T)})$.
\end{proof}

\begin{proof}[\bf Example (ii)]

Let $\psi_j = (\psi_{1j}, \ldots , \psi_{Nj})$ be an $N \times 1$ vector for each $j \geq 0$.
Let $U_{-j}$ be an $N\times T$ sub-matrix of $(u_{it})$ consisting of $u_{it}$, $i=1\ldots N$, $t=1-j,\ldots,T-j$.
We can then write equation \eqref{errorMA} in matrix notation as
\begin{align*}
   e &= \sum_{j=0}^\infty \, \limfunc{diag}(\psi_j) \, U_{-j}
   \nonumber \\
     &= \sum_{j=0}^T \, \limfunc{diag}(\psi_j) \, U_{-j} + r_{NT} ,
\end{align*}
where we cut the sum at $T$, which results in the remainder
$r_{NT}= \sum_{j=T+1}^\infty \, \limfunc{diag}(\psi_j) \, U_{-j}$. When approximating
an ${\rm MA}(\infty)$ by a finite ${\rm MA}(T)$ process we have for the remainder
\begin{align*}
   \mathbb{E} \left(\| r_{NT} \|_{F}\right)^2 \,
     = \sum_{i=1}^N \, \sum_{t=1}^T \, \mathbb{E}  \left( r_{NT} \right)_{ij}^2 \,
      &\leq \, \sigma_u^2 \, \sum_{i=1}^N \, \sum_{t=1}^T \, \sum_{j=T+1}^\infty \, \psi_{ij}^2
      \nonumber \\
               &\leq  \sigma_u^2 \,  N \, T \, \sum_{j=T+1}^\infty \, \max_i\left( \psi_{ij}^2 \right)
      \nonumber \\
               &\leq  \sigma_u^2 \,  N \, \sum_{j=T+1}^\infty \, j \, \max_i\left( \psi_{ij}^2 \right) \;,
\end{align*}
where $\sigma_u^2$ is the variance of $u_{it}$.
Therefore, for $T \rightarrow \infty$ we have
\begin{align*}
   \mathbb{E} \left( \frac{ \left(\| r_{NT} \|_{F}\right)^2 } N \right) \, \longrightarrow \, 0 \; ,
\end{align*}
which implies $\left(\| r_{NT} \|_{F}\right)^2 = {\cal O}_p(N)$, and therefore
$\| r_{NT} \| \leq \| r_{NT} \|_{F} = {\cal O}_p(\sqrt{N})$.

Let $V$ be the $N\times 2T$ matrix consisting of $u_{it}$, $i=1\ldots N$, $t=1-T,\ldots,T$.
For $j=0\ldots T$ the matrices $U_{-j}$ are sub-matrices of $V$, and therefore $\| U_{-j} \| \leq \|V\|$.
From example (i) we know $\| V \| = {\cal O}_p(\sqrt{\max(N,2T)})$.
Furthermore, we know $\| \limfunc{diag}(\psi_j) \| \leq \max_i\left( \left| \psi_{ij} \right| \right)$.

Combining these results we find
\begin{align*}
    \left\| e  \right\|
       &\leq \, \sum_{j=0}^T \, \| \limfunc{diag}(\psi_j) \| \, \|U_{-j}\| + \|r_{NT}\|
     \nonumber \\
       &\leq \, \sum_{j=0}^T \, \max_i\left( \left| \psi_{ij} \right| \right) \|V\| + o_p(\sqrt{N})
     \nonumber \\
       &\leq \, \left[ \sum_{j=0}^\infty \, \max_i\left( \left| \psi_{ij} \right| \right) \right]
                      {\cal O}_p(\sqrt{\max(N,2T)})
                    + o_p(\sqrt{N})
     \nonumber \\
       &\leq \, {\cal O}_p(\sqrt{\max(N,T)}),
\end{align*}
as required for the proof.
\end{proof}

\begin{proof}[\bf Example (iii)]
  Because $\sigma$ and $\Sigma$ are positive definite, there exits a symmetric $N\times N$ matrix
  $\phi$ and a symmetric $T\times T$ matrix $\psi$ such that $\sigma=\phi^2$ and
  $\Sigma=\psi^2$. The error term can then be generated as $e= \phi u \psi$, where
  $u$ is an $N\times T$ matrix with iid entries $u_{it}$ such that
  $\mathbb{E}(u_{it})=0$ and $\mathbb{E}(u_{it}^4)<\infty$. Given this definition
  of $e$ we immediately have
  $\mathbb{E} e_{it} = 0$ and $\mathbb{E} e_{it} e_{j\tau} = \sigma_{ij} \Sigma_{t\tau}$.
  What is left to show is $\| e \|={\cal O}_p(\sqrt{\max(N,T)})$.
  From example (i) we know $\| u \|={\cal O}_p(\sqrt{\max(N,T)})$.
  Using the inequality
  $\| \sigma \| \leq \sqrt{ \| \sigma \|_1 \, \| \sigma \|_\infty} = \| \sigma \|_1$,
  where $\| \sigma \|_1 = \| \sigma \|_\infty$ because $\sigma$ is symmetric we find
  \begin{align*}
     \| \sigma \| \leq
     \| \sigma \|_1 \, \equiv \, \max_{j=1\ldots N} \, \sum_{i=1}^N \, \left| \sigma_{ij} \right| \, &< \, L \; , &
  \end{align*}
  and analogously $\| \Sigma \| < L$. Because $\| \sigma\| = \| \phi\|^2$ and $\|\Sigma\| = \| \psi \|^2$,
  we thus find $\|e \| \leq \| \phi \| \|u\| \|\psi \| \leq L {\cal O}_p(\sqrt{\max(N,T)})$,
  i.e., $\| e \|={\cal O}_p(\sqrt{\max(N,T)})$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Comments on Assumption \ref{ass:A4} on the regressors}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Consistency of the LS estimator $\widehat \beta$ requires the regressors not only satisfy the standard non-collinearity
condition in assumption \ref{ass:A4}(i), but also the additional conditions on high- and low-rank regressors
in assumption \ref{ass:A4}(ii). Bai \cite*{Bai2009} considers the special cases of only high-rank and only low-rank
regressors. As low-rank regressors he considers only cross-sectional invariant and time-invariant regressors,
and he shows that if only these two types of regressors are present, one can show consistency
under the assumption $\limfunc{plim}_{N,T\rightarrow \infty} W_{NT} > 0$ on the regressors
(instead of assumption \ref{ass:A4}), where $W_{NT}$ is the $K\times K$ matrix
defined by $W_{NT,k_1 k_2} = (NT)^{-1} \, {\rm Tr}(M_{f^0} \, X^{\prime}_{k_1} \, M_{\lambda^0} \, X_{k_2})$.
This matrix appears as the approximate Hessian in the profile objective expansion in theorem \ref{th:ass_expand},
i.e., the condition $\limfunc{plim}_{N,T\rightarrow \infty} W_{NT} > 0$ is very natural in the context
of the interactive fixed effect models, and one may wonder whether also for the general case one can replace
assumption \ref{ass:A4} with this weaker condition and still obtain consistency of the LS estimator.
Unfortunately, this is not the case, and below we present two simple counter
examples that show this.

\begin{itemize}
   \item[(i)] Let there only be one factor ($R=1$) $f^0_t$ with corresponding factor loadings $\lambda^0_i$.
              Let there only be one regressor ($K=1$) of the form $X_{it}=w_i v_t + \lambda^0_i f^0_t$.
              Assume the $N\times 1$ vector $w=(w_1,\ldots,w_N)'$, and the $T\times 1$ vector
              $v=(v_1,\ldots,v_N)'$ are such that the $N\times 2$ matrix $\Lambda=(\lambda^0,w)$ and
              and the $T\times 2$ matrix $F=(f^0,v)$ satisfy
              $\limfunc{plim}_{N,T \rightarrow \infty}\left(\Lambda^{\prime} \Lambda/N\right) > 0$, and
              $\limfunc{plim}_{N,T \rightarrow \infty} \left( F^{\prime} F / T \right) > 0$.
              In this case, we have
            $W_{NT}=(NT)^{-1} \, {\rm Tr}(M_{f^0} \, v w' \, M_{\lambda^0} \, w v')$,
              and therefore $\limfunc{plim}_{N,T\rightarrow \infty} W_{NT}
               =\limfunc{plim}_{N,T\rightarrow \infty} (NT)^{-1} \, {\rm Tr}(M_{f^0} \, v w' \, M_{\lambda^0} \, w v')
                          > 0$. However, $\beta$ is not identified
              because $\beta^0 X + \lambda^0 f^{0\prime} = (\beta^0+1) X - w v'$, i.e., it is not possible
              to distinguish $(\beta,\lambda,f)=(\beta^0,\lambda^0,f^0)$ and
              $(\beta,\lambda,f)=(\beta^0+1,-w,v)$. This implies that the LS estimator is not consistent
              (both $\beta^0$ and $\beta^0+1$ could be the true parameter, but the LS estimator cannot be consistent for both).
   \item[(ii)] Let there only be one factor ($R=1$) $f^0_t$ with corresponding factor loadings $\lambda^0_i$.
               Let the $N\times 1$ vectors $\lambda^0$, $w_1$ and $w_2$ be such that
               $\Lambda=(\lambda^0,w_1,w_2)$ satisfies
               $\limfunc{plim}_{N,T \rightarrow \infty}\left(\Lambda^{\prime} \Lambda/N\right) > 0$.
               Let the $T\times 1$ vectors $f^0$, $v_1$ and $v_2$ be such that
               $F=(f^0,v_1,v_2)$ satisfies
               $\limfunc{plim}_{N,T \rightarrow \infty} \left( F^{\prime} F / T \right) > 0$.
               Let there be four regressors ($K=4$) defined by
               $X_1=w_1 v_1'$, $X_2=w_2 v_2'$,
               $X_3=(w_1+\lambda^0)(v_2+f^0)'$, $X_4=(w_2+\lambda^0)(v_1+f^0)'$.
               In this case, one can easily check $\limfunc{plim}_{N,T\rightarrow \infty} W_{NT} > 0$.
               However, again $\beta_k$ is not identified, because
               $\sum_{k=1}^4 \beta^0_k X_k + \lambda^0 f^{0\prime} =
                \sum_{k=1}^4 (\beta^0_k+1) X_k - (\lambda^0+w_1+w_2) (f^{0\prime}+v_1+v_2)'$,
               i.e., we cannot distinguish between the true parameters
               and $(\beta,\lambda,f)=(\beta^0+1,\,-\lambda^0-w_1-w_2,\,f^{0\prime}+v_1+v_2)$.
               Again, as a consequence the LS estimator is not consistent in this case.
\end{itemize}

In example (ii), there are only low-rank regressors with ${\rm rank}(X_l)=1$.
One can easily check assumption \ref{ass:A4} is not satisfied for this example.
In example (i) the regressor is a low-rank regressor with ${\rm rank}(X)=2$.
In our present version of assumption \ref{ass:A4} we only consider low-rank regressors with ${\rm rank}(X)=1$,
but (as already noted in a footnote in the main paper) it is straightforward to extend the assumption and the
consistency proof to low-rank regressors with rank larger than one. Independent of whether we extend the assumption
or not, the regressor $X$ of example (i) fails to satisfy assumption \ref{ass:A4}.
This justifies our formulation of assumption \ref{ass:A4}, because it shows in general
the assumption cannot be replaced by the
weaker condition $\limfunc{plim}_{N,T\rightarrow \infty} W_{NT} > 0$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some Matrix Algebra (including Proof of Lemma~\ref{lemma:Optimization})}
\label{app:matrix}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The following statements are true for real matrices (throughout the whole paper and supplementary material
we never use complex numbers anywhere).
Let $A$ be an arbitrary $n\times m$ matrix.
In addition to the operator (or spectral) norm $\|A\|$ and to the Frobenius (or Hilbert-Schmidt) norm $\|A\|_{F}$,
it is also convenient to define the $1$-norm, the $\infty$-norm, and the $\max$-norm by
\begin{align*}
   \| A \|_1 \, &= \, \max_{j=1\ldots m} \, \sum_{i=1}^n \, \left| A_{ij} \right| \; , &
   \| A \|_\infty \, &= \, \max_{i=1\ldots n} \, \sum_{j=1}^m \, \left| A_{ij} \right| \; , &
   \| A \|_{\max} \, &= \, \max_{i=1\ldots n} \, \max_{j=1 \ldots m} \, \left| A_{ij} \right| \; . &
\end{align*}

\begin{lemma}[Some useful inequalities]
   \label{lemma:inequalities}
   Let A be an $n\times m$ matrix, $B$ be an $m\times p$ matrix, and $C$ and $D$ be $n\times n$ matrices.
   Then we have:
   \begin{align*}
       \text{(i)}& \qquad
             \left\| A\right\| \, \leq \, \left\| A\right\|_{F}
             \, \leq \, \left\| A\right\| \, \limfunc{rank}\left( A\right)^{1/2} \; ,
      \nonumber \\
       \text{(ii)}& \qquad
             \left\| AB \right\| \, \leq \, \left\| A\right\| \left\|B\right\|     \; ,
      \nonumber \\
       \text{(iii)}& \qquad
             \left\| AB \right\|_{F} \, \leq \, \left\| A\right\|_{F} \left\|B\right\|
                                          \, \leq \, \left\| A\right\|_{F} \left\|B\right\|_{F} \; ,
      \nonumber \\
       \text{(iv)}& \qquad
             |{\rm Tr}(AB)| \, \leq \, \left\| A\right\|_{F} \left\|B\right\|_{F} \; ,
                \qquad \text{for $n=p$,}
      \nonumber \\
       \text{(v)}& \qquad
                     \left| {\rm Tr}\left( C\right) \right|
                      \leq \left\| C\right\| \limfunc{rank}\left( C\right) \; ,
      \nonumber \\
       \text{(vi)}& \qquad
                     \left\| C\right\| \leq {\rm Tr}\left( C\right) \; ,
                \qquad \text{for $C$ symmetric and $C\geq0$,}
      \nonumber \\
       \text{(vii)}& \qquad
            \|A\|^2 \, \leq \, \|A\|_1 \, \|A\|_{\infty} \; ,
      \nonumber \\
       \text{(viii)}& \qquad
             \|A\|_{\max} \, \leq \, \|A\| \, \leq \, \sqrt{nm} \, \|A\|_{\max} \; ,
      \nonumber \\
       \text{(ix)}& \qquad
            \|A' C A \| \leq \|A' D A \| \; ,
                \qquad  \text{for $C$ symmetric and $C\leq D$.}
      \nonumber \\
      & \text{For $C$, $D$ symmetric, and $i=1,\ldots,n$ we have:}
      \nonumber \\
       \text{(x)} & \qquad   %\label{th:Weyl}
         {\mu}_i(C) + {\mu}_n(D) \, \leq \, {\mu}_i(C+D) \, \leq \,
        {\mu}_i(C) + {\mu}_1(D) \; ,
      \nonumber \\
       \text{(xi)} & \qquad
             {\mu}_i(C) \leq \, {\mu}_i(C+D) \; ,
                \qquad  \text{for $D\geq0$,}
      \nonumber \\
       \text{(xii)} & \qquad
             {\mu}_i(C) - \|D\| \, \leq \, {\mu}_i(C+D) \, \leq \,  {\mu}_i(C) + \|D\| \; .
   \end{align*}
\end{lemma}

\begin{proof}[\bf Proof]%[Proof of Lemma~\ref{lemma:inequalities}]
  Here we use notation $s_i(A)$ for the $i$th largest singular value of a matrix $A$.
\\
   (i) We have $\|A\|=s_1(A)$, and $\|A\|_F^2=\sum_{i=1}^{{\rm rank}(A)} (s_i(A))^2$.
       The inequalities follow directly from this representation.
   (ii) This inequality is true for all unitarily invariant norms, see, e.g., Bhatia \cite*{Bhatia97}.
   (iii) can be shown as follows
   \begin{align*}
       \left\| AB \right\|_{F}^2 &= {\rm Tr}(ABB'A')
               \nonumber \\
                        &= {\rm Tr}[\|B\|^2 \, AA'
                                    - A(\|B\|^2\mathbb{I}-BB')A']
               \nonumber \\
                        &\leq \|B\|^2 {\rm Tr}(AA') = \|B\|^2 \, \left\| A \right\|_{F}^2 \; ,
   \end{align*}
   where we used $A(\|B\|^2\mathbb{I}-BB')A'$ is positive definite.
   Relation (iv) is just the Cauchy Schwarz inequality.
   To show (v) we decompose $C=UDO'$ (singular value decomposition), where
   $U$ and $O$ are $n\times {\rm rank}(C)$ that satisfy $U'U=O'O=\mathbb{I}$
   and $D$ is a ${\rm rank}(C) \times {\rm rank}(C)$ diagonal matrix with entries
   $s_i(C)$. We then have $\|O\|=\|U\|=1$ and $\|D\|=\|C\|$ and therefore
   \begin{align*}
      |{\rm Tr}(C)| &= |{\rm Tr}(UDO')| = |{\rm Tr}(DO'U)|
               \nonumber \\
               &= \left|\sum_{i=1}^{{\rm rank}(C)} \, \eta_i' DO'U \eta_i \right|
               \nonumber \\
               &\leq \sum_{i=1}^{{\rm rank}(C)} \|D\| \|O'\| \|U\| = {\rm rank}(C) \|C\| \; .
   \end{align*}
   For (vi) let $e_1$ be a vector that satisfies $\|e_1\|=1$ and $\left\| C\right\| = e_1' C e_1$.
   Because $C$ is symmetric such an $e_1$ has to exist.
   Now choose $e_i$, $i=2,\ldots,n$, such that $e_i$, $i=1,\ldots,n$, becomes a orthonormal basis
   of the vector space of $n\times 1$ vectors. Because $C$ is positive semi definite we then have
   ${\rm Tr}\left( C\right) = \sum_{i} e_i' C e_i \geq e_1 C e_1 = \|C\|$, which is what we wanted to show.
   For (vii) we refer to Golub and van Loan \cite*{golubvanloan1996}, p.15.
   For (viii) let $e$ be the vector that satisfies $\|e\|=1$ and
   $\|A' C A\|=e' A' C A e$. Because $A' C A$ is symmetric such an $e$ has to exist.
   Because $C\leq D$ we then have $\|C\|= (e' A') C (A e) \leq (e' A') D (A e) \leq \|A'DA\|$.
   This is what we wanted to show.
   For inequality (ix) let $e_1$ be a vector that satisfied $\|e_1\|=1$ and $\left\| A'C A\right\| = e_1' A' C A e_1$.
   Then we have $\left\| A'C A\right\| = e_1' A' D A e_1 - e_1' A' (D-C) A e_1 \leq e_1' A' D A e_1 \leq \|A'DA\|$.
   Statement (x) is a special case of Weyl's inequality, see, e.g., Bhatia \cite*{Bhatia97}.
   The inequalities (xi) and (xii) follow directly from (ix) because ${\mu}_n(D)\geq 0$ for $D\geq 0$,
   and because $-\|D\| \leq {\mu}_i(D)\leq \|D\|$ for $i=1,\ldots,n$.
\end{proof}

\begin{definition}
   \label{def:angle}
   Let $A$ be an $n\times r_1$ matrix and $B$ be an $n \times r_2$ matrix
   with ${\rm rank}(A)=r_1$ and ${\rm rank}(B)=r_2$. The smallest principal angle
   $\theta_{A,B} \in [0,\pi/2]$ between the linear subspaces
   ${\rm span}(A)=\{A a | \, a \in \mathbb{R}^{r_1} \}$
   and ${\rm span}(B)=\{B b | \, b \in \mathbb{B}^{r_2} \}$ of $\mathbb{R}^n$ is defined by
   \begin{align*}
      \cos(\theta_{A,B}) &= \max_{0 \neq a \in \mathbb{R}^{r_1}} \max_{0\neq b \in \mathbb{R}^{r_2}}
                                  \frac{a' A' B b} {\|A a\| \|B b\|} \, .
   \end{align*}
\end{definition}

\begin{lemma}
   \label{lemma:angle}
   Let $A$ be an $n\times r_1$ matrix and $B$ be an $n \times r_2$ matrix
   with ${\rm rank}(A)=r_1$ and ${\rm rank}(B)=r_2$. Then we have the following alternative
   characterizations of the smallest principal angle between ${\rm span}(A)$ and ${\rm span}(B)$
   \begin{align*}
      \sin(\theta_{A,B}) &= \min_{0 \neq a \in \mathbb{R}^{r_1}} \, \frac{\| M_B \, A \, a \|} {\|A \, a\|}
         \nonumber \\
                        &= \min_{0 \neq b \in \mathbb{R}^{r_2}} \, \frac{\| M_A \, B \, b \|} {\|B \, b\|} \; .
   \end{align*}
\end{lemma}

\begin{proof}[\bf Proof]%[Proof of Lemma~\ref{lemma:angle}]
   Because $\| M_B \, A \, a \|^2 + \| P_B \, A \, a \|^2 = \|A\,a\|^2$ and
   $\sin(\theta_{A,B})^2 + \cos(\theta_{A,B})^2 = 1$, we find proving the theorem is equivalent to proving
   \begin{align*}
      \cos(\theta_{A,B}) &= \min_{0 \neq a \in \mathbb{R}^{r_1}} \, \frac{\| P_B \, A \, a \|} {\|A \, a\|}
                          = \min_{0 \neq b \in \mathbb{R}^{r_2}} \, \frac{\| P_A \, B \, b \|} {\|A \, b\|} \; .
   \end{align*}
   This last statement is theorem 8 in Galantai and Hegedus \cite*{GalantaiHegedus2006}, and the proof can be found there.
\end{proof}

\begin{proof}[\bf Proof of Lemma~\ref{lemma:Optimization}]
    Let
   \begin{align*}
      S_1(Z) &= \min_{f,\lambda} {\rm Tr}\left[ \left(Z-\lambda f'\right) \left(Z'-f \lambda'\right)\right] \; ,
 \nonumber \\
      S_2(Z) &= \min_f {\rm Tr}(Z \, M_f \, Z') \; ,
 \nonumber \\
      S_3(Z) &= \min_\lambda {\rm Tr}(Z' \, M_\lambda \, Z) \; ,
 \nonumber \\
      S_4(Z) &= \min_{\tilde \lambda,\tilde f} {\rm Tr}(M_{\widetilde \lambda} \, Z \, M_{\widetilde f} \, Z') \; ,
 \nonumber \\
      S_5(Z) &= \sum_{i=R+1}^{T} {\mu}_i(Z'Z) \; ,
 \nonumber \\
      S_6(Z) &= \sum_{i=R+1}^{N} {\mu}_i(ZZ') \; .
   \end{align*}
   The theorem claims
   \begin{align*}
      S_1(Z) \, &= \, S_2(Z) \, = \, S_3(Z) \, = \, S_4(Z) \, = \, S_5(Z) \, = \, S_6(Z) \; .
   \end{align*}
   We find:
   \begin{itemize}
      \item[(i)] The non-zero eigenvalues of $Z'Z$ and $ZZ'$ are identical, so in the sums in $S_5(Z)$ and
                 in $S_6(Z)$ we are summing over identical values, which shows $S_5(Z)=S_6(Z)$.
      \item[(ii)] Starting with $S_1(Z)$ and minimizing with respect to $f$ we obtain the first-order condition
                  \begin{align*}
                      \lambda^{\prime}\, Z &= \lambda^{\prime}\, \lambda \, f^{\prime} \; .
                  \end{align*}
                  Putting this into the objective function we can integrate out $f$, namely
                 \begin{align*}
{\rm Tr}\left[ \left( Z - \lambda f^{\prime }\right) ^{\prime
}\left( Z - \lambda f^{\prime }\right) \right] &= {\rm Tr}\left(
Z^{\prime} Z - Z^{\prime} \lambda f^{\prime}\right)
\notag \\
&= {\rm Tr}\left( Z^{\prime} Z - Z^{\prime} \lambda
(\lambda^{\prime}\lambda)^{-1} (\lambda^{\prime}\lambda) f^{\prime}\right)
\notag \\
&= {\rm Tr}\left( Z^{\prime} Z - Z^{\prime} \lambda
(\lambda^{\prime}\lambda)^{-1} (\lambda^{\prime}\lambda) \lambda^{\prime}\,
Z \right)  \notag \\
&= {\rm Tr}\left( Z^{\prime} \, M_\lambda \, Z \right) \; .
                 \end{align*}
                 This shows $S_1(Z)=S_3(Z)$.
                 Analogously, we can integrate out $\lambda$
                 to obtain $S_1(Z)=S_2(Z)$.
       \item[(iii)] Let $M_{\widehat \lambda}$ be the projector on the $N-R$ eigenspaces
                   corresponding to the $N-R$ smallest eigenvalues\footnote{%
If an eigenvalue has multiplicity $m$, we count it $m$ times when finding
the $N-R$ smallest eigenvalues. In this terminology we always have exactly $N$
eigenvalues of $ZZ^{\prime}$, but some may appear multiple
times.} of $ZZ^{\prime}$, let $P_{\widehat \lambda} = \mathbb{I}_N
- M_{\widehat \lambda}$, and let $\omega_{R}$ be the $R$'th largest eigenvalue
of $ZZ^{\prime}$. We then know the matrix $P_{\widehat
\lambda} [Z Z^{\prime}-\omega_R \mathbb{I}_N ] P_{\widehat
\lambda} -M_{\widehat \lambda} [Z Z^{\prime}-\omega_R \mathbb{I}_N
] M_{\widehat \lambda}$ is positive semi-definite. Thus, for an arbitrary $%
N\times R$ matrix $\lambda$ with corresponding projector $M_\lambda$ we have
\begin{align*}
0 &\leq {\rm Tr} \left\{ \left( P_{\widehat \lambda} [Z
Z^{\prime}-\omega_R \mathbb{I}_N ] P_{\widehat \lambda} -M_{\widehat \lambda}
[Z Z^{\prime}-\omega_R \mathbb{I}_N ] M_{\widehat \lambda} \right)
\left( M_{\lambda} - M_{\widehat \lambda} \right)^2 \right\}  \notag \\
&= {\rm Tr} \left\{ \left( P_{\widehat \lambda} [Z
Z^{\prime}-\omega_R \mathbb{I}_N ] P_{\widehat \lambda} + M_{\widehat
\lambda} [Z Z^{\prime}-\omega_R \mathbb{I}_N ] M_{\widehat
\lambda} \right) \left( M_{\lambda} - M_{\widehat \lambda} \right) \right\}
\notag \\
&= {\rm Tr} \left[ Z^{\prime} \, M_\lambda \, Z \right] -%
{\rm Tr} \left[ Z^{\prime} \, M_{\widehat \lambda} \, Z \right]
+ \omega_R \, \left[ \limfunc{rank}(M_\lambda) - \limfunc{rank}(M_{\widehat
\lambda}) \right] \; ,
\end{align*}
and because $\limfunc{rank}(M_{\widehat \lambda}) = N-R$ and $\limfunc{rank}%
(M_\lambda) \leq N-R$ we have
\begin{align*}
{\rm Tr} \left[ Z^{\prime} \, M_{\widehat \lambda} \, Z \right]
&\leq {\rm Tr} \left[ Z^{\prime} \, M_\lambda \, Z \right]
\; .
\end{align*}
This shows $M_{\widehat \lambda}$ is the optimal choice in the minimization problem
of $S_3(Z)$, i.e., the optimal $\lambda=\widehat \lambda$ is chosen such
that the span of the $N$-dimensional vectors $\widehat \lambda_r$ ($r=1\ldots R$)
equals to the span of the $R$ eigenvectors that correspond to the $R$
largest eigenvalues of $ZZ^{\prime}$. This shows $S_3(Z)=S_6(Z)$.
Analogously one can show $S_2(Z)=S_5(Z)$.
      \item[(iv)]
         In the minimization problem in $S_4(Z)$ we can choose
         $\widetilde \lambda$ such that the span of the $N$-dimensional vectors $\widetilde \lambda_r$ ($r=1\ldots R_1$) is
         equal to the span of the $R_1$ eigenvectors that correspond to the $R_1$
         largest eigenvalues of $ZZ^{\prime}$. In addition, we can choose
         $\widetilde f$ such that the span of the $T$-dimensional vectors $\widetilde f_r$ ($r=1\ldots R_2$) is
         equal to the span of the $R_2$ eigenvectors that correspond to the $(R_1+1)$-largest up to the
         $R$-largest eigenvalue of $Z^{\prime}Z$. With this choice of $\widetilde \lambda$ and $\widetilde f$
         we actually project out all the $R$ largest eigenvalues of $Z'Z$ and $ZZ'$. This shows that
         $S_4(Z) \leq S_5(Z)$. (This result is actually best understood by using the singular value decomposition
         of $Z$.)

         We can write $M_{\widetilde \lambda} \, Z \, M_{\widetilde f}=Z-\widetilde Z$, where
         \begin{align*}
            \widetilde Z &= P_{\widetilde \lambda} \, Z \, M_{\widetilde f} + Z \, P_{\widetilde f} \; .
         \end{align*}
         Because ${\rm rank}(Z)\leq {\rm rank}(P_{\widetilde \lambda} \, Z \, M_{\widetilde f})
                                 +{\rm rank}(Z \, P_{\widetilde f}) = R_1 + R_2 = R$, we can
         always write $\widetilde Z=\lambda f'$ for some appropriate $N\times R$ and $T\times R$ matrices
         $\lambda$ and $f$. This shows that
         \begin{align*}
            S_4(Z) &=    \min_{\bar \lambda,\bar f} {\rm Tr}(M_{\widetilde \lambda} \, Z \, M_{\widetilde f} \, Z')
               \nonumber \\
                   &\geq \min_{\{\widetilde Z \;:\; {\rm rank}(\widetilde Z)\leq R\}} {\rm Tr}((Z-\widetilde Z)(Z-\widetilde Z)')
               \nonumber \\
                   &= \min_{f,\lambda} {\rm Tr}\left[ \left(Z-\lambda f'\right) \left(Z'-f \lambda'\right)\right]
                    = S_1(Z) \; .
         \end{align*}
         Thus we have shown here $S_1(Z) \leq S_4(Z) \leq S_5(Z)$, and this holds with equality
         because $S_1(Z)=S_5(Z)$ was already shown above.
   \end{itemize}
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Supplement to the Consistency Proof (Appendix \ref{app:consistency})}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{lemma}
   \label{lemma:wv}
   Under assumptions \ref{ass:A1} and \ref{ass:A4} there exists a constant $B_0>0$ such that
   for the matrices $w$ and $v$ introduced in assumption \ref{ass:A4} we have
   \begin{align*}
      w' \, M_{\lambda^0} \, w  \, - \, B_0 \, w' \, w &\geq 0 \; , \qquad \text{wpa1,}
    \nonumber \\
      v' \, M_{f^0} \, v  \, - \, B_0 \, v' \, v &\geq 0  \; , \qquad \text{wpa1.}
   \end{align*}
\end{lemma}

\begin{proof}[\bf Proof]
   We can decompose $w=\widetilde w \, \bar w$, where
   $\widetilde w$ is an $N \times {\rm rank}(w)$ matrix and $\bar w$ is a ${\rm rank}(w) \times K_1$ matrix.
   Note $\widetilde w$ has full rank, and $M_w=M_{\widetilde w}$.

   By assumption \ref{ass:A1}(i) we know $\lambda^{0\prime}\lambda^0/N$ has a probability limit,
   i.e., there exists some $B_1>0$ such that $\lambda^{0\prime}\lambda^0/N < B_1 \mathbb{I}_R$ wpa1.
   Using this and assumption \ref{ass:A4} we find for any $R\times 1$ vector $a\neq0$ we have
   \begin{align*}
      \frac{\|M_{v} \, \lambda^0 \, a\|^2}
          {\|\lambda^0 \, a\|^2} \, =   \,
      \frac{a' \, \lambda^{0\prime} \, M_{v} \, \lambda^0 \, a }
          {a' \, \lambda^{0\prime} \, \lambda^0 \, a} &> \frac{B}{B_1} \;,  \qquad \text{wpa1.}
   \end{align*}
   Applying Lemma~\ref{lemma:angle} we find
   \begin{align*}
      \min_{0\neq b \in \mathbb{R}^{{\rm rank}(w)}} \, \frac{b' \, \widetilde w' \, M_{\lambda^0} \, \widetilde w \, b }
          {b' \, \widetilde w' \, \widetilde w \, b} \, =   \,
      \min_{0\neq a \in \mathbb{R}^R} \, \frac{a' \, \lambda^{0\prime} \, M_{w} \, \lambda^0 \, a }
          {a' \, \lambda^{0\prime} \, \lambda^0 \, a} &> \frac{B}{B_1} \; ,  \; \qquad \text{wpa1.}
   \end{align*}
   Therefore we find for every ${\rm rank}(w) \times 1$ vector $b$ that
$b' \left( \widetilde w' \, M_{\lambda^0} \,  \widetilde w - (B/B_1) \widetilde w' \widetilde w \,\right) b > 0$,
wpa1.
   Thus
$\widetilde w' \, M_{\lambda^0} \, \widetilde w - (B/B_1) \, \widetilde w' \, \widetilde w  >  0$, wpa1.
Multiplying from the left with $\bar w'$ and from the right with $\bar w$ we obtain
$w' \, M_{\lambda^0} \, w - (B/B_1) \, w' \, w  \geq  0$, wpa1.
This is what we wanted to show.
Analogously we can show the statement for $v$.
\end{proof}

As a consequence of the this lemma we obtain some properties of the low-rank regressors summarized in the following
lemma.
\begin{lemma}
   \label{lemma:lowrankprop}
   Let the assumptions \ref{ass:A1} and \ref{ass:A4} be satisfied and let
   $X_{{\rm low},\alpha}=\sum_{l=1}^{K_1} \alpha_l X_l$ be a linear combination of the low-rank regressors.
   Then there exists some constant $B>0$ such that
   \begin{align*}
         \min_{\{\alpha \in \mathbb{R}^{K_1}, \|\alpha\|=1\}}
         \frac{\left\|X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}'\right\|}{NT}
         &> B \; , \qquad \text{wpa1,}
      \nonumber \\
         \min_{\{\alpha \in \mathbb{R}^{K_1}, \|\alpha\|=1\}}
  \frac{\left\|M_{\lambda^0} \, X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}' \, M_{\lambda^0} \right\|}{NT}
         &> B \; , \qquad \text{wpa1.}
   \end{align*}
\end{lemma}

\begin{proof}[\bf Proof]
   Note
   $\left\|M_{\lambda^0} \, X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}' \, M_{\lambda^0} \right\|
     \leq \left\|X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}'\right\|$,
   because $\|M_{\lambda^0}\|=1$, i.e., if we can show the second inequality of the lemma we have also shown
   the first inequality.

   We can write $X_{{\rm low},\alpha} = w \, {\rm diag}(\alpha') \, v'$.
   Using Lemma~\ref{lemma:wv} and part (v), (vi) and (ix) of Lemma~\ref{lemma:inequalities} we find
   \begin{align*}
      \left\|M_{\lambda^0} \, X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}' \, M_{\lambda^0} \right\|
      &=  \left\|M_{\lambda^0} \, w \, {\rm diag}(\alpha')
          \, v' \, M_{f^0} \, v \, {\rm diag}(\alpha') \, w'  M_{\lambda^0} \right\|
      \nonumber \\  &
      \geq  B_0 \, \left\|M_{\lambda^0} \, w \, {\rm diag}(\alpha')
          \, v' \, \, v \, {\rm diag}(\alpha') \, w'  M_{\lambda^0} \right\|
      \nonumber \\  &
      \geq  \frac{B_0}{K_1} \,  {\rm Tr} \left[ M_{\lambda^0} \, w \, {\rm diag}(\alpha')
          \, v' \, \, v \, {\rm diag}(\alpha') \, w'  M_{\lambda^0} \right]
      \nonumber \\  &
      =  \frac{B_0}{K_1} \,  {\rm Tr} \left[ v \, {\rm diag}(\alpha') \, w'  M_{\lambda^0}
                w \, {\rm diag}(\alpha') \, v' \right]
      \nonumber \\  &
      \geq  \frac{B_0}{K_1} \, \left\| v \, {\rm diag}(\alpha') \, w'  M_{\lambda^0}
                w \, {\rm diag}(\alpha') \, v' \right\|
      \nonumber \\  &
      \geq  \frac{B_0^2}{K_1} \, \left\| v \, {\rm diag}(\alpha') \, w' w \, {\rm diag}(\alpha') \, v' \right\|
      \nonumber \\  &
      \geq  \frac{B_0^2}{K_1^2} \, {\rm Tr}\left[ v \, {\rm diag}(\alpha') \, w' w \, {\rm diag}(\alpha') \, v' \right]
      \nonumber \\  &
      = \frac{B_0^2}{K_1^2} {\rm Tr}\left[ X_{{\rm low},\alpha} X'_{{\rm low},\alpha} \right] \; .
   \end{align*}
   Thus we have
   $ \left\|M_{\lambda^0} \, X_{{\rm low},\alpha} \, M_{f^0} \, X_{{\rm low},\alpha}' \, M_{\lambda^0} \right\|
      /(NT) \geq (B_0/K_1)^2 \, \alpha' \, W^{\rm low}_{NT} \, \alpha$ ,
   where the $K_1 \times K_1$ matrix $W^{\rm low}_{NT}$ is defined by
     $W^{\rm low}_{NT,l_1 l_2} = (NT)^{-1} {\rm Tr}\left( X_{l_1} X'_{l_2} \right)$, i.e., it is a submatrix of $W_{NT}$.
   Because $W_{NT}$ and thus $W^{\rm low}_{NT}$ converges to a positive definite matrix the lemma is proven
   by the inequality above.
\end{proof}

Using the above lemmas we can now prove the lower bound on $\widetilde S^{(2)}_{NT}(\beta,f)$ that was used
in the consistency proof. Remember
\begin{align*}
  \widetilde S^{(2)}_{NT}(\beta,f)
    &=   \frac{1}{NT} \; {\rm Tr}\left[ \left(
       \lambda^0 \, f^{0\prime} + \sum_{k=1}^{K} (\beta^0_k-\beta_k) X_{k} \right)
       \, M_f \, \left( \lambda^0 \, f^{0\prime} + \sum_{k=1}^{K}
       (\beta^0_k-\beta_k) X_{k} \right)^{\prime}\, P_{(\lambda^0,w)} \right] \; .
\end{align*}
We want to show under the assumptions of theorem \ref{th:consistency} there exist
finite positive constants $a_0$, $a_1$, $a_2$, $a_3$ and $a_4$ such that
\begin{align*}
  \widetilde S^{(2)}_{NT}(\beta,f)
    &\geq  \, \frac{a_0  \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2 }
                { \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2
                  + a_1 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|
                  + a_2 }
        \nonumber \\ & \qquad \qquad
      -  a_3 \left\| \beta^{\rm high} - \beta^{0, {\rm high}} \right\|
  - a_4 \left\| \beta^{\rm high} - \beta^{0, {\rm high}} \right\| \, \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|
         \; , \qquad \text{wpa1.}
\end{align*}

\begin{proof}[\bf Proof of the lower bound on $\widetilde S^{(2)}_{NT}(\beta,f)$.]
Applying Lemma~\ref{lemma:Optimization} and part (xi) of Lemma~\ref{lemma:inequalities} we find
\begin{align*}
  \widetilde S^{(2)}_{NT}(\beta,f)
     &\geq \frac{1}{NT} \; {\mu}_{R+1} \left[
       \left( \lambda^0 \, f^{0\prime} + \sum_{k=1}^{K} (\beta^0_k-\beta_k) X_{k} \right)^{\prime}
        \, P_{(\lambda^0,w)} \,
        \left( \lambda^0 \, f^{0\prime} + \sum_{k=1}^{K} (\beta^0_k-\beta_k) X_{k} \right) \right]
     \nonumber \\
     &= \frac{1}{NT} \; {\mu}_{R+1} \Bigg[
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)^{\prime}
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)
     \nonumber \\
       &\qquad\qquad\qquad\qquad+ \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)^{\prime}
       P_{(\lambda^0,w)} \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m
     \nonumber \\
       &\qquad\qquad\qquad\qquad+ \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m' P_{(\lambda^0,w)}
         \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)
     \nonumber \\
       &\qquad\qquad\qquad\qquad+ \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m' P_{(\lambda^0,w)}
         \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m
     \Bigg]
     \nonumber \\
     &\geq \frac{1}{NT} \; {\mu}_{R+1} \Bigg[
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)^{\prime}
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)
     \nonumber \\
       &\qquad\qquad\qquad\qquad+ \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)^{\prime}
       P_{(\lambda^0,w)} \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m
     \nonumber \\
       &\qquad\qquad\qquad\qquad+ \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m' P_{(\lambda^0,w)}
         \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)
     \Bigg]
     \nonumber \\
     &\geq \frac{1}{NT} \; {\mu}_{R+1} \left[
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right)^{\prime}
       \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right) \right]
     \nonumber \\
          &\quad
          - a_3 \left\| \beta^{\rm high} - \beta^{0, {\rm high}} \right\|
          - a_4 \left\| \beta^{\rm high} - \beta^{0, {\rm high}} \right\|
                \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\| \; , \qquad \text{wpa1,}
\end{align*}
where $a_3>0$ and $a_4>0$ are appropriate constants.
For the last step we used part (xii) of Lemma~\ref{lemma:inequalities} and the fact that
\begin{align*}
   & \frac 1 {NT} \left\| \sum_{m=K_1}^{K} (\beta^0_m-\beta_m) X_m' P_{(\lambda^0,w)}
         \left( \lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) w_l \, v_l' \right) \right\|
   \nonumber \\
    & \qquad \leq K  \, \left\| \beta^{\rm high} - \beta^{0, {\rm high}} \right\|
          \max_m \left\| \frac {X_m}{\sqrt{NT}}  \right\|
         \left( \left\| \frac {\lambda^0 \, f^{0\prime}}{\sqrt{NT}}  \right\|
                + K \, \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\| \,
                      \max_l \left\| \frac {w_l v_l'}{\sqrt{NT}} \right\| \right) \; .
\end{align*}
Our assumptions guarantee the operator norms of $\lambda^0 \, f^{0\prime}/{\sqrt{NT}}$
and ${X_m}/{\sqrt{NT}}$ are bounded from above as $N,T \rightarrow \infty$, which results in
finite constants $a_3$ and $a_4$.

We write the above result as
$\widetilde S^{(2)}_{NT}(\beta,f)
 \geq {\mu}_{R+1}(A'A)/(NT) + \text{terms containing $\beta^{\rm high}$}$,
where we defined
$A=\lambda^0 \, f^{0\prime} + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) \, w_l \, v_l'$.
We also write $A = A_1 + A_2 + A_3$,
with $A_1 = M_w \, A \, P_{f^0} =  M_{w} \, \lambda^0 \, f^{0\prime}$,
$A_2 = P_w \, A \, M_{f^0} = \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) \, w_l \, v_l' \, M_{f^0}$,
$A_3 = P_w \, A \, P_{f^0} = P_w \, \lambda^0 \, f^{0\prime}
       + \sum_{l=1}^{K_1} (\beta^0_l-\beta_l) \, w_l \, v_l' \, P_f$.
We then find $A'A=A_1' A_1  + (A'_2 + A_3') (A_2 + A_3)$ and
\begin{align*}
   A' A \, &\geq \, A' A  - (a^{1/2} A'_3 + a^{-1/2} A_2') (a^{1/2} A_3 + a^{-1/2} A_2)
      \nonumber \\
           &=  \left[ A'_1 A_1 - (a-1) \, A'_3 A_3 \right] \, + \, (1-a^{-1}) A'_2 A_2 \; ,
\end{align*}
where $\geq$ for matrices refers to the difference being positive definite, and $a$ is a
positive number. We choose
$a = 1 + {\mu}_{R}(A'_1 A_1) / (2 \, \|A_3\|^2)$.
The reason for this choice becomes clear below.

Note $\left[A'_1 A_1 - (a-1) \, A'_3 A_3 \right]$ has at most rank $R$
(asymptotically it has exactly rank $R$).
The non-zero eigenvalues of $A'A$ are therefore given by the (at most) $R$ non-zero eigenvalues of
$\left[ A'_1 A_1 - (a-1) \, A'_3 A_3 \right]$
and the non-zero eigenvalues of $(1-a^{-1}) A'_2 A_2$,
the largest one of the latter being given given by the operator norm
$(1-a^{-1}) \|A_2 \|^2$. We therefore find
\begin{align*}
  \frac{1}{NT} \; {\mu}_{R+1} \left( A'A \right)
        &\geq \frac{1}{NT} \; {\mu}_{R+1}
           \left[ \left(  A'_1 A_1 - (a-1) \, A'_3 A_3 \right) \, + \,  (1-a^{-1}) A'_2 A_2 \right]
       \nonumber \\
        &\geq \frac{1}{NT} \,
          \min\left\{ (1-a^{-1}) \|A_2\|^2 \; , \; \; {\mu}_{R}\left[A'_1 A_1 - (a-1) \, A'_3 A_3 \right]  \right\} \; .
\end{align*}
Using Lemma~\ref{lemma:inequalities}(xii) and our particular choice of $a$ we find
\begin{align*}
   {\mu}_{R} \, \left[ A'_1 A_1 - (a-1) \, A'_3 A_3 \right]
      &\geq \, {\mu}_{R}(A'_1 A_1) -  \left\| (a-1) A'_3 A_3 \right\|
   \nonumber \\
      &= \, \frac{1}{2} \, {\mu}_{R}(A'_1 A_1) \; .
\end{align*}
Therefore
\begin{align*}
   \frac 1 {NT} \, {\mu}_{R+1}(A'A)
        &\geq \frac{1}{2 \,NT} \, {\mu}_{R}(A'_1 A_1) \,
          \min\left\{ 1 \; , \; \; \frac{2 \, \|A_2\|^2}
                           {2 \, \|A_3\|^2 + {\mu}_{R}(A'_1 A_1)} \right\}
   \nonumber \\
        &\geq \frac{1}{NT} \, \frac{\|A_2\|^2 \, {\mu}_{R}(A'_1 A_1)}
                           {2 \, \|A\|^2 + {\mu}_{R}(A'_1 A_1)} \; ,
\end{align*}
where we used $\|A\|\geq\|A_3\|$ and $\|A\|\geq\|A_2\|$.

Our assumptions guarantee there exist positive constants $c_0$, $c_1$, $c_2$, and $c_3$ such that
\begin{align*}
   \frac {\|A\|} {\sqrt{NT}}
        &\leq  \frac {\|\lambda^0 \, f^{0\prime}\|} {\sqrt{NT}}
              + \sum_{l=1}^{K_1} |\beta^0_l-\beta_l| \frac {\| w_l \, v_l' \|} {\sqrt{NT}}
        \leq  c_0 + c_1 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\| \, , \quad \text{wpa1} \; ,
  \nonumber \\
   \frac {{\mu}_{R}(A'_1 A_1)} {NT}
        &= \frac{{\mu}_{R}\left( f^0 \, \lambda^{0\prime} \, M_{w} \, \lambda^0 \, f^{0\prime} \right)}
                {NT}
           \geq c_2 \, , \quad \text{wpa1} \; ,
  \nonumber \\
   \frac{\|A_2\|^2}{NT} &= {\mu}_{1}
                     \left[ \sum_{l_1=1}^{K_1} (\beta^0_{l_1}-\beta_{l_1}) \, w_{l_1} \, v_{l_1}' \, M_{f^0} \,
                            \sum_{l_2=1}^{K_1} (\beta^0_{l_2}-\beta_{l_2}) \, v_{l_2} \, w_{l_2}' \right]
                 \nonumber \\
                      &\geq c_3 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2 \, , \quad \text{wpa1} \; ,
\end{align*}
were for the last inequality we used Lemma~\ref{lemma:lowrankprop}.

We thus have
\begin{align*}
  \frac{1}{NT} \; {\mu}_{R+1} \left( A'A \right)
        &\geq \frac{c_3 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2}
                   {1 + \frac{2}  {c_2} \left(c_0 + c_1 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|\right)^2
                            } \, , \quad \text{wpa1} \; .
\end{align*}
Defining $a_0=\frac{c_2 c_3}{2 c_1^2}$, $a_1=\frac{2 c_0}{c_1}$ and $a_2=\frac{c_2}{2 c_1^2}$ we thus obtain
\begin{align*}
  \frac{1}{NT} \; {\mu}_{R+1} \left( A'A \right)
        &\geq \frac{a_0  \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2 }
                { \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|^2
                  + a_1 \left\| \beta^{\rm low} - \beta^{0,{\rm low}} \right\|
                  + a_2 } \, , \quad \text{wpa1} \; ,
\end{align*}
i.e., we have shown the desired bound on $\widetilde S^{(2)}_{NT}(\beta,f)$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Regarding the Proof of Corollary \ref{cor:limit}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

As discussed in the main text, the proof of Corollary \ref{cor:limit}
is provided in Moon and Weidner~\cite*{MoonWeidner2015}.
All that is left to show here
is the matrix $W_{NT}=W_{NT}(\lambda^0,\, f^0,\, X_{k})$ does not become singular as $N,T \rightarrow \infty$
under our assumptions.

\begin{proof}[\bf Proof]
  Remember
  \begin{align*}
%      \tag{\ref{DefW}}
      W_{NT} &= \frac 1 {NT} {\rm Tr}(M_{f^0}
      \, X^{\prime}_{k_1} \, M_{\lambda^0} \, X_{k_2}) \; .
  \end{align*}
  The smallest eigenvalue of the symmetric matrix $W(\lambda^0,\, f^0,\, X_{k})$ is given by
  \begin{align*}
     {\mu}_K \left( W_{NT} \right)
          &= \min_{\{a \in \mathbb{R}^K, \; a \neq 0\}}
               \frac{a' \,  W_{NT} \, a}  {\|a\|^2}
       \nonumber \\
          &= \min_{\{a \in \mathbb{R}^K, \; a \neq 0\}}
            \frac 1 {NT \, \|a\|^2}
            {\rm Tr}\left[ M_{f^0} \, \left( \sum_{k_1=1}^K \,  a_{k_1} \, X^{\prime}_{k_1} \right)
               \, M_{\lambda^0} \, \left(\sum_{k_2=1}^K \, a_{k_2} \, X_{k_2} \right) \right]
       \nonumber \\
          &= \min_{\begin{minipage}{2.8cm}\begin{center}\scriptsize
                   $\{\alpha \in \mathbb{R}^{K_1}, \; \varphi \in \mathbb{R}^{K_2}$\\
                   $\alpha \neq 0, \; \varphi\neq 0\}$\end{center}\end{minipage}}
            \frac {
            {\rm Tr}\left[ M_{f^0} \, \left( X'_{{\rm low},\varphi} + X'_{{\rm high},\alpha} \right)
               \, M_{\lambda^0} \,  \left( X_{{\rm low},\varphi} + X_{{\rm high},\alpha} \right) \right] }
               {NT \, \left( \|\alpha\|^2 + \|\varphi\|^2 \right)} \; ,
  \end{align*}
  where we decomposed $a=(\varphi',\alpha')'$, with $\varphi$ and $\alpha$ being vectors of length $K_1$ and $K_2$,
  respectively, and we defined linear combinations
  of high- and low-rank regressors:
  \begin{align*}
     X_{{\rm low},\varphi} &=  \sum_{l=1}^{K_1} \, \varphi_{l} \, X_{l}  \; , &
     X_{{\rm high},\alpha} &=  \sum_{m=K_1+1}^{K} \, \alpha_{m} \, X_{m}  \; .
  \end{align*}
 Here, as in assumption \ref{ass:A4} the components of $\alpha$
  are denoted $\alpha_{K_1+1},\ldots,\alpha_{K}$ to simplify notation. 
  
  We have $M_{\lambda^0} =  M_{(\lambda^0,w)} + P_{(M_{\lambda^0} w)}$, where $w$ is the
  $N \times K_1$ matrix defined in assumption \ref{ass:A4}, i.e., $(\lambda^0,w)$
  is an $N \times (R+K_1)$ matrix, whereas $M_{\lambda^0} w$ is also an $N \times K_1$ matrix.
  Using this we obtain
  \begin{align}
      & {\mu}_K \left( W_{NT} \right)
      \nonumber \\
       & \quad = \min_{\begin{minipage}{2.8cm}\begin{center}\scriptsize
                   $\{\varphi \in \mathbb{R}^{K_1}, \; \alpha \in \mathbb{R}^{K_2}$\\
                   $\varphi \neq 0, \; \alpha\neq 0\}$\end{center}\end{minipage}}
            \frac 1 {NT \, \left( \|\varphi\|^2 + \|\alpha\|^2 \right)}
            \bigg\{
            {\rm Tr}\left[ M_{f^0} \, \left( X'_{{\rm low},\varphi} + X'_{{\rm high},\alpha} \right)
               \, M_{(\lambda^0,w)} \,  \left( X_{{\rm low},\varphi} + X_{{\rm high},\alpha} \right) \right]
         \nonumber \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad
           + {\rm Tr}\left[ M_{f^0} \, \left( X'_{{\rm low},\varphi} + X'_{{\rm high},\alpha} \right)
       \, P_{(M_{\lambda^0} w)} \,  \left( X_{{\rm low},\varphi} + X_{{\rm high},\alpha} \right) \right] \bigg\}
      \nonumber \\
       & \quad = \min_{\begin{minipage}{2.8cm}\begin{center}\scriptsize
                   $\{\varphi \in \mathbb{R}^{K_1}, \; \alpha \in \mathbb{R}^{K_2}$\\
                   $\varphi \neq 0, \; \alpha\neq 0\}$\end{center}\end{minipage}}
            \frac 1 {NT \, \left( \|\varphi\|^2 + \|\alpha\|^2 \right)}
            \bigg\{
            {\rm Tr}\left[ M_{f^0} \, X'_{{\rm high},\alpha} \, M_{(\lambda^0,w)} \,  X_{{\rm high},\alpha} \right]
         \nonumber \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
           + {\rm Tr}\left[ M_{f^0} \, \left( X'_{{\rm low},\varphi} + X'_{{\rm high},\alpha} \right)
       \, P_{(M_{\lambda^0} w)} \,  \left( X_{{\rm low},\varphi} + X_{{\rm high},\alpha} \right) \right] \bigg\} \, .
       \label{eq:boundEK1}
  \end{align}
  We note there exists finite positive constants $c_1$, $c_2$, and $c_3$ such that
  \begin{align}
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \, X'_{{\rm high},\alpha} \, M_{(\lambda^0,w)} \,  X_{{\rm high},\alpha} \right]
                &\geq  \, c_1 \| \alpha \|^2 \; , \quad \text{wpa1,}
   \nonumber \\
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \, \left( X'_{{\rm low},\varphi} + X'_{{\rm high},\alpha} \right)
    \, P_{(M_{\lambda^0} w)} \,  \left( X_{{\rm low},\varphi} + X_{{\rm high},\alpha} \right) \right] &\geq 0 \; ,
   \nonumber \\
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \,  X'_{{\rm low},\varphi}
    \, P_{(M_{\lambda^0} w)} \,  X_{{\rm low},\varphi}  \right] &\geq \, c_2 \,  \| \varphi \|^2 \; , \quad \text{wpa1,}
   \nonumber \\
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \, X'_{{\rm low},\varphi}
    \, P_{(M_{\lambda^0} w)} \,   X_{{\rm high},\alpha}  \right] &\geq - \frac {c_3} 2 \, \| \varphi \| \| \alpha \| \; ,
        \quad \text{wpa1,}
   \nonumber \\
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \,  X'_{{\rm high},\alpha}
    \, P_{(M_{\lambda^0} w)} \,   X_{{\rm high},\alpha} \right] &\geq 0 \; ,
    \label{inequ_highlow}
  \end{align}
  and we want to justify these inequalities now.
   The second and the last equation in \eqref{inequ_highlow} are true because,
  e.g., ${\rm Tr}\left[ M_{f^0} \,  X'_{{\rm high},\alpha} \, P_{(M_{\lambda^0} w)} \,   X_{{\rm high},\alpha} \right]
={\rm Tr}\left[ M_{f^0} \,  X'_{{\rm high},\alpha} \, P_{(M_{\lambda^0} w)} \,   X_{{\rm high},\alpha} \, M_{f^0} \right]$,
  and the trace of a symmetric positive semi-definite matrix is non-negative.
  The first inequality in \eqref{inequ_highlow} is true because ${\rm rank}(f^0)+{\rm rank}(\lambda^0,w)=2R+K_1$
  and using Lemma~\ref{lemma:Optimization} and assumption \ref{ass:A4} we have
  \begin{align*}
     \frac 1 {NT\|\alpha\|^2}
     {\rm Tr}\left[ M_{f^0} \, X'_{{\rm high},\alpha} \, M_{(\lambda^0,w)} \,  X_{{\rm high},\alpha} \right]
         \geq
     \frac 1 {NT\|\alpha\|^2}
     {\mu}_{2R+K_1+1}\left[  X_{{\rm high},\alpha} \, X'_{{\rm high},\alpha} \right]
         &> b \; , \quad \text{wpa1},
  \end{align*}
  i.e., we can set $c_1=b$.
  The third inequality in \eqref{inequ_highlow} is true because according Lemma~\ref{lemma:inequalities}(v) we have
  \begin{align*}
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \, X'_{{\rm low},\varphi}
    \, P_{(M_{\lambda^0} w)} \,   X_{{\rm high},\alpha}  \right]
       &\geq - \, \frac {K_1} {NT} \, \left\| X_{{\rm low},\varphi} \right\| \left\| X_{{\rm high},\alpha} \right\|
   \nonumber \\
       &\geq - \, \frac {K_1} {NT} \, \left\| X_{{\rm low},\varphi} \right\|_F \left\| X_{{\rm high},\alpha} \right\|_F
   \nonumber \\
       &\geq - \, K_1 \, K_1 \, K_2 \, \|\varphi\| \, \|\alpha\| \,
              \, \max_{k_1=1\ldots K_1} \left\| \frac{X_{k_1}} {\sqrt{NT}} \right\|_F
              \, \max_{k_2=K_1+1\ldots K} \left\| \frac{X_{k_2}} {\sqrt{NT}} \right\|_F
   \nonumber \\
       &\geq - \frac {c_3} 2 \, \|\varphi\| \, \|\alpha\| \; ,
  \end{align*}
  where we used that assumption \ref{ass:A4} implies $\left\| X_{k} / \sqrt{NT} \right\|_F < C$
  holds wpa1 for some constant $C$ as, and we set $c_3 = K_1 \, K_1 \, K_2 \, C^2$.
  Finally, we have to argue that the third inequality in \eqref{inequ_highlow} holds.
  Note $X'_{{\rm low},\varphi} \, P_{(M_{\lambda^0} w)} \,  X_{{\rm low},\varphi}
  = X'_{{\rm low},\varphi} \, M_{\lambda^0} \,  X_{{\rm low},\varphi}$, i.e., we need to show
  \begin{align*}
      \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \,  X'_{{\rm low},\varphi}
      \, M_{\lambda^0} \,  X_{{\rm low},\varphi}  \right] &\geq \, c_2 \,  \| \varphi \|^2 \; .
  \end{align*}
  Using part (vi) of Lemma~\ref{lemma:inequalities} we find
  \begin{align*}
     \frac 1 {NT} {\rm Tr}\left[ M_{f^0} \,  X'_{{\rm low},\varphi}
      \, M_{\lambda^0} \,  X_{{\rm low},\varphi}  \right]
   &= \frac 1 {NT}
   {\rm Tr}\left[ M_{\lambda^0} \,  X_{{\rm low},\varphi} \, M_{f^0} \,  X'_{{\rm low},\varphi} \, M_{\lambda^0} \right]
   \nonumber \\
   &\geq \frac 1 {NT}
         \left\| M_{\lambda^0} \,  X_{{\rm low},\varphi} \, M_{f^0} \,  X'_{{\rm low},\varphi} \, M_{\lambda^0} \right\| \; ,
  \end{align*}
  and according to Lemma~\ref{lemma:lowrankprop} this expression is bounded by some positive constant times
  $\| \varphi \|^2$ (in the lemma we have $\| \varphi \|=1$, but all expressions are homogeneous in $\|\varphi\|$).

  Using the inequalities \eqref{inequ_highlow} in equation \eqref{eq:boundEK1} we obtain
  \begin{align*}
     {\mu}_K \left( W_{NT} \right)
                  &\geq \min_{\begin{minipage}{2.8cm}\begin{center}\scriptsize
                   $\{\varphi \in \mathbb{R}^{K_1}, \; \alpha \in \mathbb{R}^{K_2}$\\
                   $\varphi \neq 0, \; \alpha\neq 0\}$\end{center}\end{minipage}}
            \frac 1 {\|\varphi\|^2 + \|\alpha\|^2}
               \left\{ c_1 \| \alpha \|^2 + \max\left[ 0, \, c_2 \| \varphi \|^2 - c_3 \|\varphi\| \|\alpha\| \right] \right\}
      \nonumber \\
      & \geq \min\left( \frac {c_2} 2 , \,  \frac{c_1 c_2^2} {c_2^2+c_3^2} \right)   \, , \quad \text{wpa1}.
  \end{align*}
  Thus, the smallest eigenvalue of $W_{NT}$ is bounded from below by a positive constant as $N,T \rightarrow \infty$,
  i.e., $W_{NT}$ is non-degenerate and invertible.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Examples for Assumption~\ref{ass:A5}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\begin{proof}[\bf Proof of Example 1.]

We want to show the conditions of Assumption~\ref{ass:A5} are satisfied.
Conditions (i)-(iii) are satisfied by the assumptions of the example.

For condition (iv), notice ${\rm Cov} \left( X_{it}, X_{is} | \mathcal{C} \right) = \mathbb{E} \left( U_{it} U_{is} \right)$. Because $|\beta^0| < 1$ and $\sup_{it} \mathbb{E}(e_{it}^2) < \infty$, it follows

\begin{eqnarray*}
    \frac{1}{NT} \sum_{i=1}^{N} \sum_{t,s=1}^T \left| {\rm Cov} \left( X_{it}, X_{is} | \mathcal{C} \right) \right| &=&
     \frac{1}{NT} \sum_{i=1}^{N} \sum_{t,s=1}^T \left| \mathbb{E} \left( U_{it} U_{is} \right) \right| \\
    &=&  \frac{1}{NT} \sum_{i=1}^{N} \sum_{t,s=1}^T \sum_{p,q = 0}^{\infty}  \left| (\beta^0)^{p+q} \mathbb{E} \left( e_{it-p} e_{is-q} \right) \right|
    < \infty .
\end{eqnarray*}

For condition (v), notice by the independence between the sigma field $%
\mathcal{C}$ and the error terms $\left\{ e_{it}\right\} $ that we have for
some finite constant $M,$
\begin{eqnarray*}
    && \frac{1}{NT^{2}} \sum_{i=1}^{N} \sum_{t,s,u,v=1}^{T} \left\vert {\rm Cov}\left(
e_{it}\widetilde{X}_{is},e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right) \right\vert  \\
   &=& \frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert {\rm Cov}\left(
e_{it}U_{is},e_{iu}U_{iv}\right) \right\vert  \\
   &=& \frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\sum_{p,q=0}^{\infty
}\left\vert \left( \beta ^{0}\right) ^{p+q}\mathbb{E}\left(
e_{it}e_{is-p}e_{iu}e_{iv-q}\right) -\left( \beta ^{0}\right) ^{p}\mathbb{E}%
\left( e_{it}e_{is-p}\right) \left( \beta ^{0}\right) ^{q}\mathbb{E}\left(
e_{iu}e_{iv-q}\right) \right\vert  \\
  &\leq & \frac{M}{T^{2}}\sum_{t,s,u,v=1}^{T}\sum_{p,q=0}^{\infty }\left\vert
\beta ^{0}\right\vert ^{p+q}\left[ \mathbb{I}\left\{ t=u\right\} \mathbb{I}%
\left\{ s-p=v-q\right\} +\mathbb{I}\left\{ t=v-q\right\} \mathbb{I}\left\{
s-p=u\right\} \right]  \\
  &=& \frac{M}{T^{2}}\sum_{t,u,s,v=1}^{T}\sum_{k=-\infty }^{s}\sum_{l=-\infty
}^{v}\left\vert \beta ^{0}\right\vert ^{s-k+v-l}\mathbb{I}\left\{
t=u\right\} \mathbb{I}\left\{ k=l\right\} +M\left( \frac{1}{T}\sum
_{\substack{ s,u=1 \\ s-u\geq 0}}^{T}\left\vert \beta ^{0}\right\vert
^{s-u}\right) \left( \frac{1}{T}\sum_{\substack{ v,t=1 \\ v-t\geq 0}}%
^{T}\left\vert \beta ^{0}\right\vert ^{v-t}\right)  \\
  &=& \frac{M}{T}\sum_{s,v=1}^{T}\sum_{k=-\infty }^{\min \left\{ s,v\right\}
}\left\vert \beta ^{0}\right\vert ^{s+v-2k}+M\left( \frac{1}{T}\sum
_{\substack{ s,u=1 \\ s-u\geq 0}}^{T}\left\vert \beta ^{0}\right\vert
^{s-u}\right) \left( \frac{1}{T}\sum_{\substack{ v,t=1 \\ v-t\geq 0}}%
^{T}\left\vert \beta ^{0}\right\vert ^{v-t}\right) .
\end{eqnarray*}%
Notice
\begin{eqnarray*}
&&\frac{1}{T}\sum_{s,v=1}^{T}\sum_{k=-\infty }^{\min \left\{ s,v\right\}
}\left\vert \beta ^{0}\right\vert ^{s+v-2k} \\
&=&\frac{2}{T}\sum_{s=2}^{T}\sum_{v=1}^{s}\sum_{k=-\infty }^{v}\left\vert
\beta ^{0}\right\vert ^{s-v+2(v-k)}+\frac{2}{T}\sum_{s=1}^{T}\sum_{k=-\infty
}^{s}\left\vert \beta ^{0}\right\vert ^{2(s-k)} \\
&=&\frac{2}{T}\sum_{s=2}^{T}\sum_{v=1}^{s}\left\vert \beta ^{0}\right\vert
^{s-v}\sum_{l=0}^{\infty }\left\vert \beta ^{0}\right\vert ^{2l}+\frac{2}{T}%
\sum_{s=1}^{T}\sum_{l=0}^{\infty }\left\vert \beta ^{0}\right\vert ^{2l} \\
&=&\frac{2}{1-\left\vert \beta ^{0}\right\vert ^{2}}\frac{1}{T}%
\sum_{s=2}^{T}\sum_{v=1}^{s}\left\vert \beta ^{0}\right\vert ^{s-v}+\frac{2}{%
1-\left\vert \beta ^{0}\right\vert ^{2}} \\
&=&\left( \frac{2}{1-\left\vert \beta ^{0}\right\vert ^{2}}\right)
\sum_{l=1}^{T-1}\left\vert \beta ^{0}\right\vert ^{l}\left( 1-\frac{l}{T}%
\right) +\frac{2}{1-\left\vert \beta ^{0}\right\vert ^{2}} \\
&=&O\left( 1\right) ,
\end{eqnarray*}%
and
\begin{equation*}
\frac{1}{T}\sum_{\substack{ s,u=1 \\ s-u\geq 0}}^{T}\left\vert \beta
^{0}\right\vert ^{s-u}=\frac{1}{T}\sum_{s=1}^{T}\sum_{u=1}^{s}\left\vert
\beta ^{0}\right\vert ^{s-u}=\sum_{l=0}^{T-1}\left\vert \beta
^{0}\right\vert ^{l}\left( 1-\frac{l}{T}\right) =O\left( 1\right) .
\end{equation*}%
Therefore, we have the desired result
\begin{equation*}
\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert {\rm Cov}\left( e_{it}%
\widetilde{X}_{is},e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right) \right\vert
={\cal O}_{p}\left( 1\right) .
\end{equation*}
\end{proof}

\textsc{Preliminaries for Proof of Example 2}

\begin{itemize}
\item Although we observe $X_{it}$ for $1\leq t\leq T,$ here we treat $%
Z_{it}=\left( e_{it},X_{it}\right) $ as having an infinite past and future.
Define
\begin{equation*}
\mathcal{G}_{\tau }^{t}\left( i\right) =
{\cal C} \vee \sigma \left( \left\{ X_{is}:\tau
\leq s\leq t\right\}  \right) \text{ and }\mathcal{H}_{\tau
}^{t}\left( i\right) =
{\cal C} \vee  \sigma \left( \left\{ Z_{it}:\tau \leq s\leq t\right\}\right) .
\end{equation*}%
Then, by definition, we have $\mathcal{G}_{\tau }^{t}\left( i\right) ,%
\mathcal{H}_{\tau }^{t}\left( i\right) \subset \mathcal{F}_{\tau }^{t}\left(
i\right) $ for all $\tau ,t,i.$ By Assumption (iv) of Example 2, the time
series of $\left\{ X_{it}:-\infty <t<\infty \right\} $ and $\left\{
Z_{it}:-\infty <t<\infty \right\} $ are conditional $\alpha$-mixing
conditioning on $\mathcal{C}$ uniformly in $i.$

\item Mixing inequality: The following inequality is a conditional version
of the $\alpha$-mixing inequality of Hall and Heyde~\cite*{HallHeyde1980}, p.~278. Suppose
 $X_{it}$ is a $\mathcal{F}_{t}$-measurable  random
variable with $\mathbb{E}\left( \left\vert X_{it}\right\vert ^{\max \left\{
p,q\right\} }|\mathcal{C}\right) <\infty ,$ where $p,q>1$ with $1/p+1/q<1.$
Denote $\left\Vert X_{it}\right\Vert _{\mathcal{C},p}=\left( \mathbb{E}%
\left( \left\vert X_{it}\right\vert ^{p}|\mathcal{C}\right) \right)
^{1/p}. $ Then, for each $i,$ we have
\begin{equation}
\left\vert {\rm Cov}\left( X_{it},X_{it+m}|\mathcal{C}\right) \right\vert \leq
8\left\Vert X_{it}\right\Vert _{\mathcal{C},p}\left\Vert
X_{it+m}\right\Vert _{\mathcal{C},q}\alpha _{m}^{1-\frac{1}{p}-\frac{1}{q}%
}\left( i\right) .  \label{eq:mixing inequality}
\end{equation}
\end{itemize}

\begin{proof}[\bf Proof of Example 2.]
Again, we want to show the conditions of Assumption~\ref{ass:A5} are satisfied.
Conditions (i)-(iii) are satisfied by the assumptions of the example.

 

 
For condition (iv), we apply the mixing inequality $\left( \ref{eq:mixing
inequality}\right) $ with $p=q>4$. Then, we have%
\begin{eqnarray*}
&&\frac{1}{NT}\sum_{i=1}^{N}\sum_{t,s=1}^{T}\left\vert {\rm Cov}\left(
X_{it},X_{is}|\mathcal{C}\right) \right\vert  \\
&\leq &\frac{2}{NT}\sum_{i=1}^{N}\sum_{t=1}^{T}\sum_{m=0}^{T-t}\left\vert
{\rm Cov}\left( X_{it},X_{it+m}|\mathcal{C}\right) \right\vert =\frac{2}{NT}%
\sum_{i=1}^{N}\sum_{m=0}^{T-1}\sum_{t=1}^{T-m}\left\vert {\rm Cov}\left(
X_{it},X_{it+m}|\mathcal{C}\right) \right\vert  \\
&=&\frac{16}{NT}\sum_{i=1}^{N}\sum_{m=0}^{T-1}\sum_{t=1}^{T-m}\left\Vert
X_{it}\right\Vert _{\mathcal{C},p}\left\Vert X_{it+m}\right\Vert _{\mathcal{C%
},p}\alpha _{m}\left( i\right) ^{\frac{p-2}{P}} \\
&\leq &16\left( \sup_{i,t}\left\Vert X_{it}\right\Vert _{\mathcal{C}%
,p}^{2}\right) \sum_{m=0}^{\infty }\alpha _{m}^{\frac{p-2}{P}} \\
&\leq &{\cal O}_{p}\left( 1\right),
\end{eqnarray*}%
where the last line holds because $\sup_{i,t}\left\Vert X_{it}\right\Vert _{\mathcal{C}%
,p}^{2}={\cal O}_{p}\left( 1\right) $ for some $p>4$ as assumed in the example (2), and $\sum_{m=0}^{\infty
}\alpha _{m}^{\frac{p-2}{P}}= \sum_{m=0}^{\infty} m^{-\zeta\frac{p-2}{P}} = {\cal O}\left( 1\right)$ because of $\zeta > 3\frac{4p}{4p-1}$ and $p>4$. 


For condition (v), we need to show
\begin{equation*}
\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert {\rm Cov}\left( e_{it}%
\widetilde{X}_{is},e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right) \right\vert
={\cal O}_{p}\left( 1\right) .
\end{equation*}%
Notice
\begin{eqnarray*}
&&\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert {\rm Cov}\left(
e_{it}\widetilde{X}_{is},e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right) \right\vert  \\
&=&\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert \mathbb{E}%
\left( e_{it}\widetilde{X}_{is}e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right) -\mathbb{E%
}\left( e_{it}\widetilde{X}_{is}|\mathcal{C}\right) \mathbb{E}\left( e_{iu}%
\widetilde{X}_{iv}|\mathcal{C}\right) \right\vert  \\
&\leq &\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t,s,u,v=1}^{T}\left\vert \mathbb{E%
}\left( e_{it}\widetilde{X}_{is}e_{iu}\widetilde{X}_{iv}|\mathcal{C}\right)
\right\vert +\frac{1}{N}\sum_{i=1}^{N}\left( \frac{1}{T}\sum_{t,s=1}^{T}%
\mathbb{E}\left( e_{it}\widetilde{X}_{is}|\mathcal{C}\right) \right) ^{2} \\
&=&I+II,\text{ say.}
\end{eqnarray*}%
First, for term $I,$ there are a finite number of different orderings among
the indices $t,s,u,v.$ We consider the case $t\leq s\leq u\leq v$ and
establish the desired result. The other cases can be shown analogously. Note
\begin{eqnarray*}
&&\frac{1}{NT^{2}}\sum_{i=1}^{N}\sum_{t=1}^{T}\sum_{k=0}^{T-t}%
\sum_{l=0}^{T-k}\sum_{m=0}^{T-l}\left\vert \mathbb{E}\left( e_{it}\widetilde{X}%
_{it+k}e_{it+k+l}\widetilde{X}_{it+k+l+m}|\mathcal{C}\right) \right\vert  \\
&\leq &\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{
0\leq l,m\leq k \\ 0\leq k+l+m\leq T-t}}\left\vert \mathbb{E}\left(
e_{it}\left( \widetilde{X}_{it+k}e_{it+k+l}\widetilde{X}_{it+k+l+m}\right) |\mathcal{%
C}\right) \right\vert  \\
&&+\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{ %
0\leq k,m\leq l \\ 0\leq k+l+m\leq T-t}}\bigg\vert \mathbb{E}\left[ \left(
e_{it}\widetilde{X}_{it+k}\right) \left( e_{it+k+l}\widetilde{X}_{it+k+l+m}\right) |%
\mathcal{C}\right] 
\\
&& \qquad \qquad \qquad \qquad  \qquad \qquad  \qquad \qquad 
 -\mathbb{E}\left( e_{it}\widetilde{X}_{it+k}|\mathcal{C}%
\right) \mathbb{E}\left( e_{it+k+l}\widetilde{X}_{it+k+l+m}|\mathcal{C}\right)
\bigg\vert  \\
&&+\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{ %
0\leq k,m\leq l \\ 0\leq k+l+m\leq T-t}}\mathbb{E}\left( e_{it}\widetilde{X}%
_{it+k}|\mathcal{C}\right) \mathbb{E}\left( e_{it+k+l}\widetilde{X}_{it+k+l+m}|%
\mathcal{C}\right)  \\
&&+\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{ %
0\leq p,l\leq m \\ 0\leq k+l+m\leq T-t}}\left\vert \mathbb{E}\left[ \left(
e_{it}\widetilde{X}_{it+k}e_{it+k+l}\right) \widetilde{X}_{it+k+l+m}|\mathcal{C}%
\right] \right\vert  \\
&=&I_{1}+I_{2}+I_{3}+I_{4},\text{ say.}
\end{eqnarray*}%
By applying the mixing inequality $\left( \ref{eq:mixing inequality}\right) $
to $\left\vert \mathbb{E}\left( e_{it}\left( \widetilde{X}_{it+k}e_{it+k+l}%
\widetilde{X}_{it+k+l+m}\right) |\mathcal{C}\right) \right\vert $ with $e_{it}$
and $\widetilde{X}_{it+k}e_{it+k+l}\widetilde{X}_{it+k+l+m},$ we have%
\begin{eqnarray*}
&&\left\vert \mathbb{E}\left( e_{it}\left( \widetilde{X}_{it+k}e_{it+k+l}\widetilde{X%
}_{it+k+l+m}\right) |\mathcal{C}\right) \right\vert  \\
&\leq &8\left\Vert e_{it}\right\Vert _{\mathcal{C},p}\left\Vert \widetilde{X}%
_{it+k}e_{it+k+l}\widetilde{X}_{it+k+l+m}\right\Vert _{\mathcal{C},q}\alpha
_{k}^{1-\frac{1}{p}-\frac{1}{q}}\left( i\right)  \\
&\leq &8\left\Vert e_{it}\right\Vert _{\mathcal{C},p}\left\Vert \widetilde{X}%
_{it+k}\right\Vert _{\mathcal{C},3q}\left\Vert e_{it+k+l}\right\Vert _{%
\mathcal{C},3q}\left\Vert \widetilde{X}_{it+k+l+m}\right\Vert _{\mathcal{C}%
,3q}\alpha _{k}^{1-\frac{1}{p}-\frac{1}{q}}\left( i\right) ,
\end{eqnarray*}%
where the last inequality follows by the generalized Holder's inequality.
Choose $p=3q>4.$ Then,
\begin{eqnarray*}
I_{1} &\leq &\frac{8}{N}\sum_{i=1}^{N}\frac{1}{T^{2}}\sum_{t=1}^{T}\sum
_{\substack{ 0\leq l,m\leq k \\ 0\leq k+l+m\leq T-t}}\left\Vert
e_{it}\right\Vert _{\mathcal{C},p}\left\Vert \widetilde{X}_{it+k}\right\Vert _{%
\mathcal{C},p}\left\Vert e_{it+k+l}\right\Vert _{\mathcal{C},p}\left\Vert
\widetilde{X}_{it+k+l+m}\right\Vert _{\mathcal{C},p}\alpha _{k}^{1-\frac{1}{4p}%
}\left( i\right)  \\
&\leq &8\left( \sup_{i,t}\left\Vert e_{it}\right\Vert _{\mathcal{C}%
,p}^{2}\right) \left( \sup_{i,t}\left\Vert \widetilde{X}_{it+k}\right\Vert _{%
\mathcal{C},p}^{2}\right) \frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{ %
0\leq l,m\leq k \\ 0\leq k+l+m\leq T-t}}\alpha _{k}^{1-\frac{1}{4p}} \\
&\leq &8\left( \sup_{i,t}\left\Vert e_{it}\right\Vert _{\mathcal{C}%
,p}^{2}\right) \left( \sup_{i,t}\left\Vert \widetilde{X}_{it+k}\right\Vert _{%
\mathcal{C},p}^{2}\right) \sum_{k=0}^{\infty }k^{2}\alpha _{k}^{1-\frac{1}{4p%
}} \\
&\leq &{\cal O}_{p}\left( 1\right),
\end{eqnarray*}%
where the last line holds because we assume in example (2) that 
$\left(\sup_{i,t}\left\Vert e_{it}\right\Vert _{\mathcal{C},p}^{2}\right) \left(
\sup_{i,t}\left\Vert \widetilde{X}_{it+k}\right\Vert _{\mathcal{C},p}^{2}\right)
={\cal O}_{p}\left( 1\right) $ for some $p>4,$, and $\sum_{m=0}^{\infty }m^{2}\alpha _{m}^{1-\frac{1%
}{4p}}= \sum_{m=0}^{\infty }m^{2-\zeta\frac{4p-1}{4p}}= O\left( 1\right)$ because of $\zeta > 3\frac{4p}{4p-1}$ and $p>4$.


By applying similar arguments, we can also show
\begin{equation*}
I_{2},I_{3},I_{4}={\cal O}_{p}\left( 1\right) .
\end{equation*}
\end{proof}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Supplement to the Proof of Theorem \ref{th:limdis}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\paragraph{\underline{Notation $\mathbb{E}_{\cal C}$ 
and ${\rm Var}_{\cal C}$ and ${\rm Cov}_{\cal C}$:}}
 In the remainder of this supplementary file we
write $\mathbb{E}_{\cal C}$, ${\rm Var}_{\cal C}$ and ${\rm Cov}_{\cal C}$
 for the expectation, variance and covariance operators conditional
on ${\cal C}$,~i.e., $\mathbb{E}_{\cal C}(A)=\mathbb{E}(A|{\cal C})$,
${\rm Var}_{\cal C}(A) = {\rm Var}(A|{\cal C})$
and ${\rm Cov}_{\cal C}(A,B) = {\rm Cov}(A,B |{\cal C})$.

\bigskip

What is left to show to complete the proof of Theorem \ref{th:limdis} is  
that Lemma~\ref{lemma:vanishing} and Lemma~\ref{lemma:denCLT} in the main text appendix hold.
Before showing this, we first present two further intermediate lemmas.

\begin{lemma}
   \label{lemma:normXweak}
   Under the assumptions of Theorem~\ref{th:limdis} we have for $k=1,\ldots ,K$,
   \begin{align*}
      \qquad && (a) && \| P_{\lambda^0} \widetilde X_k  \| &= o_p(\sqrt{NT})  \; ,
    \nonumber \\
     \qquad && (b) &&  \|  \widetilde X_k P_{f^0}\| &= o_p(\sqrt{NT}) \; ,
    \nonumber \\
      \qquad && (c) && \|P_{\lambda^0} e  X^{\prime}_k \| &= o_p(N^{3/2})          \, ,
    \nonumber \\
      \qquad && (d) &&  \|P_{\lambda^0} e P_{f^0} \| &= {\cal O}_p(1)  \; .
      && \qquad
   \end{align*}
\end{lemma}


\begin{proof}[\bf Proof of Lemma~\ref{lemma:normXweak}]
     \# Part (a): We have
       \begin{align*}
        \|P_{\lambda^0} \widetilde X_k \| &=
        \|\lambda^0 (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}  \widetilde X_k \|
        \nonumber \\
            &\leq
           \|\lambda^0 (\lambda^{0\prime}\lambda^0)^{-1}\| \| \lambda^{0\prime}  \widetilde X_k \|
        \nonumber \\
            &\leq
           \|\lambda^0 \| \| (\lambda^{0\prime}\lambda^0)^{-1}\| \| \lambda^{0\prime}  \widetilde X_k \|_F
           = {\cal O}_p(N^{-1/2})  \| \lambda^{0\prime}  \widetilde X_k \|_F \; ,
     \end{align*}
     where we used part (i) and (ii) of Lemma~\ref{lemma:inequalities}
     and Assumption~\ref{ass:A1}.
     We have
     \begin{align*}
        \mathbb{E}\left\{  \mathbb{E}_{\cal C}\left[  \| \lambda^{0\prime}  \widetilde X_k \|_F^2  \right] \right\}
          &=  \mathbb{E}\left\{  \sum_{r=1}^R \sum_{t=1}^T
          \mathbb{E}_{\cal C} \left[    \left(  \sum_{i=1}^N  \lambda^0_{ir}
          \widetilde X_{k,it} \right)^2
          \right] \right\}
        \\
             &=    \mathbb{E}\left\{  \sum_{r=1}^R \sum_{t=1}^T   \sum_{i=1}^N
               (\lambda^0_{ir})^2
          \mathbb{E}_{\cal C} \left(   \widetilde X_{k,it}^2  \right) \right\}
       \\
             &=     \sum_{r=1}^R \sum_{t=1}^T   \sum_{i=1}^N
               \mathbb{E}\left[ (\lambda^0_{ir})^2
           {\rm Var}_{\cal C} \left(  X_{k,it}   \right) \right]
      \\
           &=   {\cal O}_p( NT ) ,
     \end{align*}
     where we used $\widetilde X_{k,it}$ is mean zero and independent across $i$,
     conditional on ${\cal C}$, and our bounds on the moments
     of $\lambda^0_{ir}$ and $X_{k,it}$. We therefore have $ \| \lambda^{0\prime}  \widetilde X_k \|_F =
     {\cal O}_p(\sqrt{NT})$ and    the above inequality thus
     gives $\|P_{\lambda^0} \widetilde X_k \|  = {\cal O}_p(\sqrt{T} ) =  o_p(\sqrt{NT})$.

     \# The proof for part (b) is similar. As above we first obtain
     $ \|  \widetilde X_k P_{f^0}\| =   \|P_{f^0} \widetilde X_k'  \| \leq
       {\cal O}_p(T^{-1/2})  \| f^{0\prime}  \widetilde X_k' \|_F$.
    Next, we have
      \begin{align*}
          \mathbb{E}_{\cal C}\left[  \| f^{0\prime}  \widetilde X_k' \|_F^2 \right]
          &= \sum_{r=1}^R \sum_{i=1}^N
          \mathbb{E}_{\cal C} \left[    \left(  \sum_{t=1}^T  f^0_{tr}
          \widetilde X_{k,it} \right)^2  \right]
        \\
             &=
             \sum_{r=1}^R \sum_{i=1}^N   \sum_{t,s=1}^T f^0_{tr} f^0_{sr}
          \mathbb{E}_{\cal C} \left(
          \widetilde X_{k,it}    \widetilde X_{k,is}  \right)
        \\
            &\leq \left[ \sum_{r=1}^R
            \left(  \max_t | f^0_{tr} | \right)^2 \right]
             \sum_{i=1}^N   \sum_{t,s=1}^T
        \left|  {\rm Cov}_{\cal C} \left(
          X_{k,it} , X_{k,is} \right)  \right|
       \\
            &= {\cal O}_p(T^{2/(4+\epsilon)}) \, {\cal O}_p(NT) = o_p(N T^2) ,
     \end{align*}
     where we used that uniformly bounded $\mathbb{E} \| f^0_t \|^{4+\epsilon}$
     implies $\max_t | f^0_{tr} | = {\cal O}_p( T^{1/(4+\epsilon)} )$.
     We thus have $ \| f^{0\prime}  \widetilde X_k' \|_F^2 = o_p(T \sqrt{N})$
     and therefore $\|  \widetilde X_k P_{f^0}\|   = o_p(\sqrt{NT})$.

     \# Next, we show part (c).
      First, we  have
     \begin{align*}
        \mathbb{E} \left\{  \mathbb{E}_{\cal C} \left[ \left(  \|\lambda^{0\,\prime} e X'_k \|_F \right)^2  \right]
        \right\}
          &=   \mathbb{E} \left\{  \mathbb{E}_{\cal C}
      \left[ \sum_{r=1}^R \sum_{j=1}^N
           \left( \sum_{i=1}^N \sum_{t=1}^T \lambda^{0}_{ir} e_{it} X_{k,jt} \right)^2    \right] \right\}
        \nonumber \\
           &=   \mathbb{E} \left\{   \sum_{r=1}^R \sum_{i,j,l=1}^N
            \sum_{t,s=1}^T
           \lambda^{0}_{ir}    \lambda^{0}_{lr}
            \mathbb{E}_{\cal C}  \left( e_{it} e_{ls} X_{k,jt}   X_{k,js}      \right) \right\}
        \nonumber \\
          &=  \sum_{r=1}^R \sum_{i,j=1}^N    \sum_{t=1}^T
            \mathbb{E} \left[ (\lambda^{0}_{ir})^2
             \mathbb{E}_{\cal C} \left( e_{it}^2  X_{k,jt}^2     \right)
             \right]
         = {\cal O}(N^2 T) \; ,
     \end{align*}
     where we used that $ \mathbb{E}_{\cal C}  \left( e_{it} e_{ls} X_{k,jt}   X_{k,js}     \right)$
     is only non-zero if $i=l$ (because of cross-sectional independence conditional on ${\cal C}$)
     and $t=s$ (because regressors are pre-determined).
     We can thus conclude $\|\lambda^{0\,\prime} e X'_k \|_F = {\cal O}_p(N \sqrt{T})$.
    Using this we find
     \begin{align*}
        \|P_{\lambda^0} e X'_k \| &=
        \|\lambda^0 (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}  e X'_k \|
        \nonumber \\
            &\leq
           \|\lambda^0 (\lambda^{0\prime}\lambda^0)^{-1}\| \| \lambda^{0\prime}  e X'_k \|
        \nonumber \\
            &\leq
           \|\lambda^0 \|  \| (\lambda^{0\prime}\lambda^0)^{-1}\| \| \lambda^{0\prime}  e X'_k \|_F
           = {\cal O}_p(N^{-1/2})  {\cal O}_p(N \sqrt{T}) =  {\cal O}_p(\sqrt{NT}) \; .
     \end{align*}
     This is what we wanted to show.


      \# For part (d), we first find
      $\frac{1}{\sqrt{NT}}\left\Vert f^{0\prime}e\lambda^0 \right\Vert_{F}={\cal O}_{p}\left( 1\right)$, because
     \begin{eqnarray*}
        \mathbb{E} \left\{  \mathbb{E}_{\cal C} \left[ \left( \frac{\left\Vert f^{0\prime }e\lambda^0 \right\Vert _{F}}{\sqrt{NT}}%
            \right) ^{2}  \right] \right\}
            &=&\mathbb{E} \left\{  \frac{1}{NT}\mathbb{E}_{\cal C} \left[ \left(
        \sum_{i=1}^{N}\sum_{t=1}^{T}e_{it}f_{t}^{0\prime }\lambda^0_{i}\right) ^{2}  \right]  \right\}
           \nonumber \\
          &=& \mathbb{E} \left\{ \frac{1}{NT}\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{t=1}^{T}\sum_{s=1}^{T}\mathbb{E}_{\cal C}%
        \left( e_{it}e_{js}   \right) f_{t}^{0\prime }\lambda_{i}^0\lambda_{j}^{0\prime}f^0_{s} \right\}
            \nonumber \\
         &=&\frac{1}{NT}\sum_{i=1}^{N}\sum_{t=1}^{T}
         \mathbb{E}\left[ \mathbb{E}_{\cal C} \left( e_{it}^2  \right)
           f_{t}^{0\prime
       }\lambda^0_{i}\lambda_{i}^{0\prime }f^0_{t}  \right]
     \nonumber \\
        &=&{\cal O}\left( 1\right) ,
     \end{eqnarray*}
     where we used $e_{it}$ is independent across $i$ and over $t$,
     conditional on ${\cal C}$.
     Thus we obtain
     \begin{align*}
        \|P_{\lambda^0} e P_{f^0} \|
              &= \| \lambda^0 (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}
                     e f^0 (f^{0\prime}f^0)^{-1} \, f^{0\prime} \|
         \nonumber \\
              &\leq \| \lambda^0 \|  \left\| (\lambda^{0\prime}\lambda^0)^{-1} \right\|
                    \| \lambda^{0\prime} e f^0 \| \left\| (f^{0\prime}f^0)^{-1} \right\| \| f^{0\prime} \|
         \nonumber \\
              &\leq {\cal O}_p(N^{1/2})  {\cal O}_p(N^{-1})
                    \| \lambda^{0\prime} e f^0 \|_F {\cal O}_p(T^{-1}) {\cal O}_p(T^{1/2}) = {\cal O}_p(1) \;,
     \end{align*}
     where we used part (i) and (ii) of Lemma~\ref{lemma:inequalities}.
 \end{proof}

\begin{lemma}
  \label{lemma:eeterms}
Suppose $A$ and $B$ are  $T\times T$ and  $N\times N$ matrices that
are independent of $e$, conditional on ${\cal C}$, such that
$\mathbb{E}_{\cal C}\left( \left\Vert A\right\Vert _{F}^{2}   \right)={\cal O}_p\left(
NT\right) $ and $\mathbb{E}_{\cal C} \left( \left\Vert B\right\Vert _{F}^{2} \right)={\cal O}_p\left( NT\right)$,
and let Assumption~\ref{ass:A5} be satisfied.
Then there exists a finite non-random constant $c_0$ such that
\begin{align*}
  (a) &&  \mathbb{E}_{\cal C}\left( \left\{ \limfunc{Tr}\left[ \left( e^{\prime }e-\mathbb{E}_{\cal C} \left( e^{\prime
}e\right) \right) A\right]  \right\}^2  \right) &\leq c_0 \, N \,  \mathbb{E}_{\cal C} \left( \left\Vert
A\right\Vert _{F}^{2}   \right) \; ,
\nonumber \\
  (b) && \mathbb{E}_{\cal C}\left( \left\{ \limfunc{Tr}\left[ \left( ee^{\prime }-\mathbb{E}_{\cal C} \left( ee^{\prime
}\right) \right) B \right] \right\}^{2}  \right)  &\leq
c_0 \, T \,  \mathbb{E}_{\cal C} \left( \left\Vert B\right\Vert _{F}^{2}  \right) \; .
\end{align*}
\end{lemma}

\begin{proof}[\bf Proof]
\# Part (a): Denote $A_{ts}$ to be the $\left( t,s\right) ^{th}$ element of $A$.
We have
\begin{align*}
\limfunc{Tr}\left\{ \left( e^{\prime }e-\mathbb{E}_{\cal C}\left( e^{\prime }e\right)
\right) A\right\}
&= \sum_{t=1}^{T}\sum_{s=1}^{T}\left( e^{\prime }e-\mathbb{E}_{\cal C}\left( e^{\prime
}e\right) \right) _{ts}A_{ts} \nonumber \\
&= \sum_{t=1}^{T}\sum_{s=1}^{T}\left( \sum_{i=1}^{N}\left(
e_{it}e_{is}-\mathbb{E}_{\cal C}\left( e_{it}e_{is}\right) \right) \right) A_{ts}.
\end{align*}%
Therefore,
\begin{align*}
&\mathbb{E}_{\cal C}\left( \limfunc{Tr}\left\{ \left( e^{\prime }e-\mathbb{E}_{\cal C}\left( e^{\prime
}e\right) \right) A\right\} \right) ^{2} \nonumber \\
&\qquad  = \sum_{t=1}^{T}\sum_{s=1}^{T}\sum_{p=1}^{T}\sum_{q=1}^{T}\mathbb{E}_{\cal C}\left[ \left(
\sum_{i=1}^{N}\left( e_{it}e_{is}-\mathbb{E}_{\cal C}\left( e_{it}e_{is}\right) \right)
\right) \left( \sum_{j=1}^{N}\left( e_{jp}e_{jq}-\mathbb{E}_{\cal C}\left( e_{jp}e_{jq}\right)
\right) \right) \right]
\mathbbm{E}_{\cal C} \left( A_{ts}A_{pq}  \right).
\end{align*}%
Let $\Sigma_{it}=\mathbb{E}_{\cal C}(e_{it}^2)$. Then we find
\begin{align*}
&\mathbb{E}_{\cal C}\left\{ \left( \sum_{i=1}^{N}\left( e_{it}e_{is}-\mathbb{E}_{\cal C}\left(
e_{it}e_{is}\right) \right) \right) \left( \sum_{j=1}^{N}\left(
e_{jp}e_{jq}-\mathbb{E}_{\cal C}\left( e_{jp}e_{jq}\right) \right) \right) \right\}
  \nonumber \\ &\qquad\qquad\qquad\qquad
=\sum_{i=1}^{N}\sum_{j=1}^{N}\left\{ \mathbb{E}_{\cal C}\left(
e_{it}e_{is}e_{jp}e_{jq}\right) -\mathbb{E}_{\cal C}\left( e_{it}e_{is}\right) \mathbb{E}_{\cal C}\left(
e_{jp}e_{jq}\right) \right\}
  \nonumber \\ &\qquad\qquad\qquad\qquad
=\left\{
  \begin{array}{l@{\quad}l}
   \Sigma _{it}\Sigma _{is}  & \text{ if }\left( t=p\right) \neq \left(
s=q\right) \text{ and }\left( i=j\right)   \\
   \Sigma _{it}\Sigma _{is}  & \text{ if }\left( t=q\right) \neq \left(
s=p\right) \text{ and }\left( i=j\right)   \\
   \mathbb{E}_{\cal C}\left( e_{it}^{4}\right) -\Sigma _{it}^{2}   &   \text{ if }\left(
t=s=p=q\right) \text{ and }\left( i=j\right)  \\
    0 & \text{ otherwise.}
  \end{array} \right.
\end{align*}%
Therefore,
\begin{align*}
&\mathbb{E}_{\cal C}\left( \limfunc{Tr}\left\{ \left( e^{\prime }e-\mathbb{E}_{\cal C}\left( e^{\prime
}e\right) \right) A\right\} \right) ^{2}
\\
& \qquad
\leq \sum_{t=1}^{T}\sum_{s=1}^{T}\sum_{i=1}^{N}\Sigma _{it}\Sigma
_{is}\left( \mathbb{E}_{\cal C} \left( A_{ts}^{2}\right) +\mathbb{E}_{\cal C} \left( A_{ts}A_{st}\right) \right)
+\sum_{t=1}^{T}\sum_{i=1}^{N}\left( \mathbb{E}_{\cal C}\left( e_{it}^{4}\right) -\Sigma
_{it}^{2}\right) \mathbb{E}_{\cal C}  A_{tt}^{2}.
\end{align*}%
Define $\Sigma ^{i}={\rm diag}\left(  \Sigma _{i1},...,\Sigma _{iT} \right) .$ Then, we have%
\begin{eqnarray}
\sum_{t=1}^{T}\sum_{s=1}^{T}\sum_{i=1}^{N}\Sigma _{it}\Sigma _{is}\left(
\mathbb{E}_{\cal C}  A_{ts}^{2}\right) &=&\mathbb{E}_{\cal C} \left( \sum_{i=1}^{N}\limfunc{Tr}\left( A^{\prime
}\Sigma ^{i}A\Sigma ^{i}\right) \right) \nonumber \\
&\leq &\sum_{i=1}^{N}\mathbb{E}_{\cal C} \left\Vert A\Sigma ^{i}\right\Vert _{F}^{2}\leq
\sum_{i=1}^{N}\left\Vert \Sigma ^{i}\right\Vert ^{2}\mathbb{E}_{\cal C} \left\Vert A\right\Vert
_{F}^{2} \nonumber \\
&\leq &N\left( \sup_{it}\Sigma _{it}^{2}\right) \mathbb{E}_{\cal C} \left\Vert A\right\Vert
_{F}^{2}.
\end{eqnarray}%
Also,%
\begin{eqnarray}
\sum_{t=1}^{T}\sum_{s=1}^{T}\sum_{i=1}^{N}\Sigma _{it}\Sigma _{is}\mathbb{E}_{\cal C} \left(
A_{ts}A_{st}\right) &=&\mathbb{E}_{\cal C} \left[ \sum_{i=1}^{N}\limfunc{Tr}\left( \Sigma
^{i}AA\Sigma ^{i}\right) \right] \nonumber \\
&\leq &\sum_{i=1}^{N}\mathbb{E}_{\cal C} \left\Vert \Sigma ^{i}A\right\Vert _{F}\left\Vert
A\Sigma ^{i}\right\Vert _{F}\leq \sum_{i=1}^{N}\left\Vert \Sigma
^{i}\right\Vert ^{2}\mathbb{E}_{\cal C} \left\Vert A\right\Vert _{F}^{2} \nonumber \\
&\leq &N\left( \sup_{it}\Sigma _{it}^{2}\right) \mathbb{E}_{\cal C} \left\Vert A\right\Vert
_{F}^{2} \; .
\end{eqnarray}%
Finally,
\begin{eqnarray}
\sum_{t=1}^{T}\sum_{i=1}^{N}\left( \mathbb{E}_{\cal C}\left( e_{it}^{4}\right) -\Sigma
_{it}^{2}\right) \mathbb{E}_{\cal C}  A_{tt}^{2}
&\leq &N\left( \sup_{it}\mathbb{E}_{\cal C}\left( e_{it}^{4}\right) \right) \mathbb{E}_{\cal C} \left\Vert
A\right\Vert _{F}^{2},
\end{eqnarray}
and $\sup_{it}\mathbb{E}_{\cal C}\left( e_{it}^{4}\right)$ is assumed
bounded by Assumption~\ref{ass:A5}$(vi)$.

\# Part (b): The proof is analogous to the proof of part (a).
\end{proof}







\begin{proof}[\bf Proof of Lemma~\ref{lemma:vanishing}]
   \# For part (a) we have
     \begin{align*}
           \left|  \frac 1 {\sqrt{NT}} {\rm Tr} \left( P_{f^0} \, e^{\prime}\, P_{\lambda^0} \, \widetilde  X_k \right) \right|
             &=
             \left| \frac 1 {\sqrt{NT}} {\rm Tr} \left( P_{f^0} \, e^{\prime}\, P_{\lambda^0}  P_{\lambda^0} \widetilde  X_k P_{f^0} \right) \right|
            \nonumber \\
          & \leq  \frac R {\sqrt{NT}}
            \left\| P_{\lambda^0}  \, e \,  P_{f^0} \right\|
            \left\| P_{\lambda^0} \widetilde  X_k \right\|
            \left\| P_{f^0} \right\|
        \nonumber \\
          &=    \frac{1}{\sqrt{NT}} \; {\cal O}_p(1) \, o_p(\sqrt{NT}) \, {\cal O}_p(1)
        \nonumber \\
          &= o_p(1),
     \end{align*}
where the second-last equality follows by Lemma~\ref{lemma:normXweak} (a) and (d).


  \# To show statement (b) we define $\zeta_{k,ijt} = e_{it} \widetilde X_{k,jt}$. We then have
    \begin{align*}
          \frac 1 {\sqrt{NT}} {\rm Tr} \left(  P_{\lambda^0} \, e \,\widetilde X_k' \right)
             &=  \sum_{r,q=1}^R  \left[\left( \frac{\lambda^{0 \prime}  \lambda^{0}} N \right)^{-1} \right]_{rq}
              \underbrace{  \frac 1 {N \sqrt{NT}}
                \sum_{t=1}^T \sum_{i,j=1}^N  \lambda_{ir}^{0} \lambda_{jq}^{0} \zeta_{k,ijt}
                }_{ \equiv A_{k,rq} } .
    \end{align*}
    We only have $ \mathbb{E_{\cal C}}\left( \zeta_{k,ijt} \zeta_{k,lm s}  \right) \neq 0$
    if $t=s$ (because regressors are pre-determined)
    and $i=l$ and $j=m$ (because of cross-sectional independence). Therefore
    \begin{align*}
        \mathbb{E}\left\{  \mathbb{E}_{\cal C}\left( A_{k,rq}^2   \right) \right\}
         &=   \mathbb{E}\left\{ \frac 1 {N^3 T}    \sum_{t,s=1}^T \sum_{i,j,l,m=1}^N
         \lambda_{ir} \lambda_{jq} \lambda_{lr} \lambda_{mq}
         \, \mathbb{E}_{\cal C}\left( \zeta_{k,ijt} \zeta_{k,lm s}   \right) \right\}
       \\
          &=   \frac 1 {N^3 T}    \sum_{t=1}^T \sum_{i,j=1}^N
        \mathbb{E}\left[ \lambda_{ir}^2 \lambda_{jq}^2
         \, \mathbb{E}_{\cal C}\left( \zeta_{k,ijt}^2   \right) \right] = {\cal O}(1/N) = o_p(1).
    \end{align*}
    We thus have $A_{k,rq} = o_p(1)$ and
    therefore also $ \frac 1 {\sqrt{NT}} {\rm Tr} \left(  P_{\lambda^0} \, e \,\widetilde X_k' \right)
    = o_p(1)$.

 \# The proof for statement (c) is similar to the proof of statement (b).
 Define    $\xi_{k,its} = e_{it} \widetilde X_{k,is} - \mathbb{E}_{\cal C}\left( e_{it} \widetilde X_{k,is}  \right)$.
 We then have
 \begin{align*}
   \frac 1 {\sqrt{NT}} {\rm Tr} \left\{ P_{f^0} \, \left[ e^{\prime} \, \widetilde  X_k
                        - \mathbb{E}_{\cal C}\left( e^{\prime} \, \widetilde  X_k  \right) \right] \right\}
       &= \sum_{r,q=1}^R  \left[\left( \frac{f' f} T \right)^{-1} \right]_{rq}
              \underbrace{  \frac 1 {T \sqrt{NT}}
                \sum_{i=1}^N \sum_{t,s=1}^T  f_{tr} f_{sq} \xi_{k,its}
                }_{ \equiv B_{k,rq} } .
 \end{align*}
 Therefore
 \begin{align*}
      \mathbb{E}_{\cal C}\left( B_{k,rq}^2  \right)
         &=   \frac 1 {T^3 N}
          \sum_{i,j=1}^N \sum_{t,s,u,v=1}^T  f_{tr} f_{sq} f_{ur} f_{vq}  \mathbb{E}_{\cal C}\left(  \xi_{k,its} \xi_{k,juv}   \right)
    \\
        &\leq
         \left( \max_{t,\widetilde r} | f_{t \widetilde r} | \right)^4
         \frac 1 {T^3 N}
          \sum_{i,j=1}^N \sum_{t,s,u,v=1}^T
         \left|  {\rm Cov}_{\cal C}\left(  e_{it} \widetilde X_{k,is}  , e_{ju} \widetilde X_{k,jv}   \right) \right|
    \\
        &=
         \left( \max_{t,\widetilde r} | f_{t \widetilde r} | \right)^4
         \frac 1 {T^3 N}
          \sum_{i=1}^N \sum_{t,s,u,v=1}^T
         \left|  {\rm Cov}_{\cal C}\left(  e_{it} \widetilde X_{k,is}  , e_{iu} \widetilde X_{k,iv}    \right) \right|
  \\
       &= {\cal O}_p( T^{4/(4+\epsilon)} ) {\cal O}_p( 1/T )
   \\
      &= o_p(1),
 \end{align*}
 where we used uniformly bounded $\mathbb{E} \| f^0_t \|^{4+\epsilon}$
     implies $\max_t | f^0_{tr} | = {\cal O}_p( T^{1/(4+\epsilon)} )$.

  \#   Part (d) and (e):
  We have
   $\|\lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}\| = {\cal O}_p((NT)^{-1/2})$,
    $\|e\|={\cal O}_p(N^{1/2})$,
     $\|X_k\|={\cal O}_p(\sqrt{NT})$ and
     $\|P_{\lambda^0} e P_{f^0} \| = {\cal O}_p(1)$, which was shown in Lemma~\ref{lemma:normXweak}.
     Therefore:
     \begin{align*}
         &\frac 1 {\sqrt{NT}} {\rm Tr}\left(e P_{f^0} \, e' \, M_{\lambda^0} \, X_k \,
              f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right)
        \nonumber \\
         &\qquad \qquad =
           \frac 1 {\sqrt{NT}} {\rm Tr}\left(P_{\lambda^0} e P_{f^0} \, e' \, M_{\lambda^0} \, X_k \,
              f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right)
        \nonumber \\
         &\qquad \qquad \leq
           \frac R {\sqrt{NT}} \left\|P_{\lambda^0} e P_{f^0}\right\| \|e\| \|X_k\|
              \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\|
            = {\cal O}_p(N^{-1/2}) = o_p(1) \; .
     \end{align*}
     which shows statement (d). The proof for part (e) is analogous.

   \# To prove statement (f) we need to use in addition
        $ \| P_{\lambda^0} \, e  \, X'_k \| = o_p(N^{3/2})$, which was also
      shown in Lemma~\ref{lemma:normXweak}. We find
      \begin{align*}
         & \frac 1 {\sqrt{NT}} {\rm Tr}\left(e^{\prime}M_{\lambda^0} \, X_k \, M_{f^0} \, e^{\prime}
                \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \right)
       \nonumber \\
        & \qquad\qquad =
          \frac 1 {\sqrt{NT}} {\rm Tr}\left(e^{\prime}M_{\lambda^0} \, X_k \, e^{\prime} \, P_{\lambda^0}
                \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \right)
       \nonumber \\
        & \qquad\qquad \qquad -
          \frac 1 {\sqrt{NT}} {\rm Tr}\left(e^{\prime}M_{\lambda^0} \, X_k \, P_{f^0} \, e^{\prime}\, P_{\lambda^0}
                \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \right)
       \nonumber \\
        & \qquad\qquad \leq
         \frac R {\sqrt{NT}}  \| e \|  \| P_{\lambda^0} \, e  \, X'_k \|
                \, \| \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \|
       \nonumber \\
        & \qquad\qquad \qquad -
         \frac R {\sqrt{NT}} \| e   \| \| X_k \| \|   P_{\lambda^0}  \, e \, P_{f^0} \|
                \| \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \|
        \nonumber \\
          &\qquad\qquad =  o_p(1) \; .
      \end{align*}

\# Now we want to prove part (g) and (h) of the present lemma.
   For part (g) we have
   \begin{align*}
         & \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\}
          \nonumber \\
          &=     \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   \overline X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\}
          \\ & \qquad
            +
            \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   \widetilde X_k
            P_{f^0} \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\}
        \nonumber \\
        &= \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   \overline X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\}
          \\ & \qquad
          +  \frac 1 {\sqrt{NT}}
            \left\| e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right\|
            \left\| \widetilde X_k P_{f^0} \right\|
            \left\|  f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}  \right\|
        \nonumber \\
        &= \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   \overline X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\}
         + o_p(1) .
   \end{align*}
   Thus, what is left to prove is
   $ \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  \, M_{\lambda^0} \,   \overline X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\} = o_p(1)$.
   For this we define
   \begin{align*}
      B_k &=  M_{\lambda^0} \, \overline X_k \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \; .
   \end{align*}
   Using part (i) and (ii) of Lemma~\ref{lemma:inequalities} we find
   \begin{align*}
      \| B_k \|_F &\leq R^{1/2} \|B_k\|
        \nonumber \\
         &\leq R^{1/2} \| \overline X_k \|
           \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\|
        \nonumber \\
         &\leq R^{1/2} \| \overline X_k \|_F
           \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\| \; .
   \end{align*}
   and therefore
   \begin{align*}
      \mathbb{E}_{\cal C} \left( \| B_k \|_F^2   \right)
            &\leq R
                \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\|^2
                 \mathbb{E}_{\cal C} \left( \| \overline X_k \|_F^2  \right)
       \nonumber \\
             &= {\cal O}(1) \; ,
   \end{align*}
   where we used $ \mathbb{E}_{\cal C}\left( \| \overline X_k \|_F^2  \right) = {\cal O}(NT)$, which is true because we assumed uniformly
   bounded moments of $\overline X_{k,it}$.
   Applying Lemma~\ref{lemma:eeterms} we therefore find
   \begin{align*}
      \mathbb{E}_{\cal C} \left(
        \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  B_k \right\}  
      \right)^2
        &\leq c_0 \, \frac T {NT} \, \mathbb{E}_{\cal C} \left( \| B_k \|_F^2   \right)  = o(1) \; ,
   \end{align*}
   and thus
   \begin{align*}
      \frac 1 {\sqrt{NT}} {\rm Tr}\left\{ \left[ e e' - \mathbb{E}_{\cal C} \left( e e' \right) \right]  B_k \right\} &= o_p(1) \; ,
   \end{align*}
   which is what we wanted to show.
   The proof for part (h) is analogous.

   \#  Part (i): Conditional on ${\cal C}$ the expression
   $e_{it}^2  \mathfrak{X}_{it} \, \mathfrak{X}_{it}'
             -  \mathbb{E}_{\cal C} \left(  e_{it}^2 \, \mathfrak{X}_{it} \, \mathfrak{X}_{it}'  
                  \right)$
        is mean zero,
        and it is also uncorrelated across $i$.
        This together with the bounded moments that we assume implies
      \begin{align*}
          {\rm Var}_{\cal C}\left\{
          \frac 1 {NT} \, \sum_{i=1}^N \, \sum_{t=1}^T
             \left[  e_{it}^2  \, \mathfrak{X}_{it} \, \mathfrak{X}_{it}'
             -  \mathbb{E}_{\cal C} \left(  e_{it}^2 \, \mathfrak{X}_{it} \, \mathfrak{X}_{it}'  
                  \right)    \right]    \right\}  &= {\cal O}_p(1/N) = o_p(1) ,
      \end{align*}
      which shows the required result.

   \# Part (j):
     Define the $K \times K$ matrix $A =   \frac 1 {NT} \, \sum_{i=1}^N \, \sum_{t=1}^T
            \,    e_{it}^2 
            \left(  \mathfrak{X}_{it}   +  {\cal X}_{it}   \right)
            \left( \mathfrak{X}_{it}   -  {\cal X}_{it}  \right)'$.
      Then we have
      \begin{align*}
           \frac 1 {NT} \, \sum_{i=1}^N \, \sum_{t=1}^T
            \,   e_{it}^2  
            \left(  \mathfrak{X}_{it} \, \mathfrak{X}_{it}'   -  {\cal X}_{it} \, {\cal X}_{it}' \right)
             &=  \frac 1 2 \left( A + A' \right).
       \end{align*}
       Let $B_k$ be the $N \times T$ matrix with
        elements $B_{k,it} =  e_{it}^2  \left(  \mathfrak{X}_{k,it}   +  {\cal X}_{k,it}   \right)$.
       We have $\| B_k \| \leq \|B_k\|_F = {\cal O}_p( \sqrt{NT} )$, because the moments
       of  $B_{k,it}$ are uniformly bounded.
       The components of $A$ can be written as
       $A_{l k} = \frac 1 {NT} {\rm Tr}[ B_l  ( \mathfrak{X}_{k}   -  {\cal X}_{k})' ]$. We therefore have
       \begin{align*}
            | A_{l k} |
            \leq  \frac 1 {NT} {\rm rank}( \mathfrak{X}_{k}   -  {\cal X}_{k} )
                  \| B_l \| \left\|  \mathfrak{X}_{k}   -  {\cal X}_{k} \right\|   .
       \end{align*}
       We have $ \mathfrak{X}_{k}   -  {\cal X}_{k}  =
          \widetilde X_k \, P_{f^0} +  P_{\lambda^0} \, \widetilde X_k \, M_{f^0}$.
        Therefore   ${\rm rank}( \mathfrak{X}_{k}   -  {\cal X}_{k} ) \leq 2 R$ and
        \begin{align*}
             | A_{l k} |
           & \leq \frac{2 R} {NT}
                  \| B_l \|
                 \left(  \left\|   \widetilde X_k \, P_{f^0}
                  \right| + \left\| P_{\lambda^0} \, \widetilde X_k \, M_{f^0} \right\|   \right)
           \\
             &\leq  \frac{2 R} {NT}
                               \| B_l \|
                 \left(  \left\|   \widetilde X_k \, P_{f^0}
                  \right| + \left\| P_{\lambda^0} \, \widetilde X_k  \right\|   \right)
              = \frac{2 R} {NT}     {\cal O}_p( \sqrt{NT} )  o_p( \sqrt{NT} ) = o_p(1),
        \end{align*}
        where we used Lemma~\ref{lemma:normXweak}.
        This shows the desired result.
\end{proof}




\begin{proof}[\bf Proof of Lemma~\ref{lemma:denCLT}]
Let $c$ be a $K$-vector such that $\left\Vert c\right\Vert =1.$ The
required result follows by the Cramer-Wold device, if we show
\begin{align*}
\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\sum_{t=1}^{T}e_{it}\mathfrak{X}_{it}^{\prime }c \, \Rightarrow \, {\cal N}\left( 0,c^{\prime }\Omega c\right) \, .
\end{align*}
For this, define $\xi_{it} =e_{it}\mathfrak{X}_{it}^{\prime }c$.
Furthermore
define $\xi_m = \xi_{M,m} = \xi_{NT,it}$, with $M = NT$ and $m = T(i-1)+t \in \{1,\ldots,M\}$.
We then have the following:

\begin{itemize}
\item[(i)]
  Under Assumption~\ref{ass:A5}$(i)$, $(ii)$, $(iii)$ the sequence
 $\{ \xi_m, \, m=1,\ldots,M \}$  is a martingale difference sequence
 under the filtration 
 ${\cal F}_m = {\cal C} \vee \sigma(\{ \xi_{n}: n < m \})$.
 

\item[(ii)] $\mathbb{E}(\xi_{it}^4 )$ is uniformly bounded,
because by Assumption~\ref{ass:A5}$(vi)$
  $\mathbb{E}_{\cal C} e_{it}^8$ and $\mathbb{E}_{\cal C}\left( \| X_{it} \|^{8+\epsilon}   \right)$
  are uniformly bounded by a non-random constant
  (applying Cauchy-Schwarz and the law of iterated expectations).

\item[(iii)]
$\frac 1 {M}  \sum_{m=1}^M \xi_{m}^2 = c' \Omega c + o_p(1)$. \\
This is true, because firstly under our assumptions we have
$\mathbb{E}_{\cal C}
\left\{ \left[ \frac 1 {M}  \sum_{m=1}^M 
\left( \xi_{m}^2 - \mathbb{E}_{\cal C}( \xi_{m}^2) \right) \right]^2
\right\}
=
\mathbb{E}_{\cal C}
\left\{ \frac 1 {M^2}  \sum_{m=1}^M 
\left( \xi_{m}^2 - \mathbb{E}_{\cal C}( \xi_{m}^2) \right)^2
\right\} = {\cal O}_P(1/M) =  o_P(1)$, implying we have
$\frac 1 {M}  \sum_{m=1}^M \xi_{m}^2
=  \frac 1 {M}  \sum_{m=1}^M \mathbb{E}_{\cal C}( \xi_{m}^2  )
    + o_p(1)$.
We furthermore have
$\frac 1 {M}  \sum_{m=1}^M \mathbb{E}_{\cal C}( \xi_{m}^2  ) 
  = {\rm Var}_{\cal C}( M^{-1/2} \sum_{m=1}^M \xi_{m} )$,
and using the result in equation \eqref{VarEqOmega} of the main text
we find   
  ${\rm Var}_{\cal C}( M^{-1/2} \sum_{m=1}^M \xi_{m} )
  ={\rm Var}_{\cal C}( (NT)^{-1/2} \sum_{i=1}^N \sum_{t=1}^T \xi_{it} )
  = c' \Omega c + o_p(1)$.

\end{itemize}
These three properties of $\{ \xi_m, \, m=1,\ldots,M \}$ allow us to
apply Corollary~5.26 in White~\cite*{White2001}, which is based on
    Theorem~2.3 in Mcleish~\cite*{Mcleish1974},  to obtain
    $\frac 1 {  \sqrt{M}} \sum_{m=1}^M \xi_{m}
         \to_d {\cal N}(0,c' \Omega c)$.
    This concludes the proof, because
   $ \frac 1 {  \sqrt{M}} \sum_{m=1}^M \xi_{m}  =     \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\sum_{t=1}^{T}e_{it}\mathfrak{X}_{it}^{\prime }c$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Expansions of Projectors and Residuals}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The incidental parameter estimators  $\widehat f$ and $\widehat \lambda$
as well as the residuals $\widehat e$ enter into the asymptotic bias and variance
estimators for the LS estimator $\widehat \beta$.
To describe the properties of $\widehat f$, $\widehat \lambda$ and $\widehat e$,
it is convenient to have asymptotic  expansions of the projectors
$M_{\widehat \lambda}(\beta)$ and $M_{\widehat f}(\beta)$
that correspond to the minimizing parameters $\widehat \lambda(\beta)$
and $\widehat f(\beta)$ in equation \eqref{LNT123}. Note the minimizing $\widehat \lambda(\beta)$
and $\widehat f(\beta)$ can be defined for all values of $\beta$, not only for the
optimal value $\beta=\widehat \beta$.
The corresponding residuals are
$\widehat e(\beta) = Y \, - \, \beta \cdot X \, - \, \widehat \lambda(\beta) \, \widehat f'(\beta)$.

\begin{theorem}
   \label{theorem:expansions}
   Under Assumptions~\ref{ass:A1}, \ref{ass:A3}, and \ref{ass:A4}(i) we have the following expansions
  \begin{align*}
     M_{\widehat \lambda}(\beta) &= M_{\lambda^0} + M_{\widehat \lambda,e}^{(1)}
                                     + M_{\widehat \lambda,e}^{(2)}
                                     - \sum_{k=1}^K \left( \beta_k - \beta^0_k \right)  M_{\widehat \lambda,k}^{(1)}
                                     + M_{\widehat \lambda}^{({\rm rem})}(\beta) \; ,
    \nonumber \\
     M_{\widehat f}(\beta) &= M_{f^0} + M_{\widehat f,e}^{(1)}
                                     + M_{\widehat f,e}^{(2)}
                                     - \sum_{k=1}^K \left( \beta_k - \beta^0_k \right)  M_{\widehat f,k}^{(1)}
                                     + M_{\widehat f}^{({\rm rem})}(\beta) \; ,
    \nonumber \\
    \widehat e(\beta) &=  M_{\lambda^0} \, e \, M_{f^0}
            + \widehat e^{(1)}_e - \sum_{k=1}^K \left( \beta_k - \beta^0_k \right) \widehat e^{(1)}_k
            + \widehat e^{({\rm rem})}(\beta) \; ,
  \end{align*}
  where the spectral norms of the remainders satisfy for any series $\eta_{NT} \rightarrow 0$:
  \begin{align*}
     \sup_{\{\beta :\left\| \beta -\beta^{0} \right\| \leq \eta_{NT}\}}
        \frac{\left\| M_{\widehat \lambda}^{({\rm rem})}(\beta) \right\|}
        { \|\beta - \beta^0\|^2 + (NT)^{-1/2} \, \|e\| \, \|\beta - \beta^0\| \,  \,  + (NT)^{-3/2} \, \|e\|^3}
            &= {\cal O}_p\left(1\right) \, ,
     \nonumber \\
     \sup_{\{\beta :\left\| \beta -\beta^{0} \right\| \leq \eta_{NT}\}}
        \frac{\left\| M_{\widehat f}^{({\rm rem})}(\beta) \right\|}
        { \|\beta - \beta^0\|^2 + (NT)^{-1/2} \, \|e\| \, \|\beta - \beta^0\| \,  \,  + (NT)^{-3/2} \, \|e\|^3}
            &= {\cal O}_p\left(1\right) \, ,
     \nonumber \\
     \sup_{\{\beta :\left\| \beta -\beta^{0} \right\| \leq \eta_{NT}\}}
       \frac{ \left\| \widehat e^{({\rm rem})}(\beta) \right\| }
         { (NT)^{1/2} \|\beta - \beta^0\|^2 + \|e\| \, \|\beta - \beta^0\| + (NT)^{-1} \|e\|^3 }
           &= {\cal O}_p\left(1\right)  \; ,
  \end{align*}
  and we have ${\rm rank}(\widehat e^{({\rm rem})}(\beta)) \leq 7R$,
  and the expansion coefficients are given by
  \begin{align*}
     M^{(1)}_{\widehat \lambda,e} &= - \, M_{\lambda^0} \, e \,
      f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}
       \, - \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
              \, e' \, M_{\lambda^0}  \; ,
    \nonumber \\
   M^{(1)}_{\widehat \lambda,k} &= - \, M_{\lambda^0} \, X_k  \,
      f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}
       \, - \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
              \,  X'_k  \, M_{\lambda^0}  \; ,
    \nonumber \\
   M^{(2)}_{\widehat \lambda,e} &=
        M_{\lambda^0} \, e \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}
             \, e \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime}
       \nonumber \\ & \qquad
       +\lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
             \, e' \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
             \, e' \, M_{\lambda^0}
       \nonumber \\ & \qquad
          - M_{\lambda^0} \, e \, M_{f^0} \, e' \,
\lambda^0\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,(\lambda^{0\prime}\lambda^0)^{-1}\,\lambda^{0\prime}
       \nonumber \\ & \qquad
-\lambda^0\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,(\lambda^{0\prime}\lambda^0)^{-1}\,\lambda^{0\prime}
   \, e \, M_{f^0} \, e' \, M_{\lambda^0}
       \nonumber \\ & \qquad
          - M_{\lambda^0} \, e \,
      f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
         \, e' \, M_{\lambda^0}
       \nonumber \\ & \qquad
        + \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime}
            \, e' \, M_{\lambda^0} \, e \,
              f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \lambda^{0\prime} \, ,
  \end{align*}
  analogously
  \begin{align*}
   M^{(1)}_{\widehat f,e} &= \, - \, M_{f^0} \, e' \,
      \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} f^{0\prime}
       \, - \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
              \, e \, M_{f^0}   \; ,
    \nonumber \\
  M^{(1)}_{\widehat f,k} &=
       \, - \, M_{f^0} \,  X'_k  \,
      \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} f^{0\prime}
       \, - \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
              \,  X_k \,M_{f^0}  \; ,
    \nonumber \\
  M^{(2)}_{\widehat f,e} &=
       M_{f^0} \, e' \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} f^{0\prime}
             \, e' \, \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} f^{0\prime}
       \nonumber \\ & \qquad
       +f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
             \, e \, f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
             \, e \, M_{f^0}
       \nonumber \\ & \qquad
          - M_{f^0} \, e' \, M_{\lambda^0} \, e \,
f^0\,(f^{0\prime}f^0)^{-1}\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,f^{0\prime}
       \nonumber \\ & \qquad
-f^0\,(f^{0\prime}f^0)^{-1}\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,f^{0\prime}
   \, e' \, M_{\lambda^0} \, e \, M_{f^0}
       \nonumber \\ & \qquad
          - M_{f^0} \, e' \,
      \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
         \, e \, M_{f^0}
       \nonumber \\ & \qquad
        + f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
            \, e \, M_{f^0} \, e' \,
              \lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} f^{0\prime} \, ,
  \end{align*}
  and finally
  \begin{align*}
   \widehat e^{(1)}_k &= M_{\lambda^0} \, X_k \, M_{f^0} \; ,
             \nonumber \\
   \widehat e^{(1)}_e &= - M_{\lambda^0} \, e \, M_{f^0} \, e' \,
   \lambda^0\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,f^{0\prime}
             \nonumber \\ & \qquad
  - \lambda^0\,(\lambda^{0\prime}\lambda^0)^{-1}\,(f^{0\prime}f^0)^{-1}\,f^{0\prime}
   \, e' \, M_{\lambda^0} \, e \, M_{f^0}
             \nonumber \\ & \qquad
  - M_{\lambda^0} \, e \,
      f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
              \, e \, M_{f^0} \; .
  \end{align*}
\end{theorem}

\begin{proof}[\bf Proof]
   The general expansion of $M_{\widehat \lambda}(\beta)$ is given in
   Moon and Weidner \cite*{MoonWeidner2015}, and in the theorem we just make
   this expansion explicit up to a particular order.
   The result for $M_{\widehat f}(\beta)$
   is just obtained by symmetry ($N \leftrightarrow T$, $\lambda \leftrightarrow f$, $e \leftrightarrow e'$,
   $X_k \leftrightarrow X_k'$). For the residuals $\widehat e$ we have
   \begin{align*}
     \widehat e &= M_{\widehat \lambda} \, \left( Y - \sum_{k=1} \, \widehat \beta_k \, X_k \right)
   = M_{\widehat \lambda} \, \left[ e - 
     \left( \widehat \beta - \beta^0 \right) \cdot X
                                 +  \lambda^0 f^{0\prime} \right]  \; ,
   \end{align*}
   and plugging in the expansion of $M_{\widehat \lambda}$ gives the expansion of $\widehat e$.
   We have $\widehat e(\beta)
        = A_0  + \lambda^0 f^{0\prime} - \widehat \lambda(\beta) \widehat f'(\beta)$,
   where $A_0=e - \sum_k (\beta_k-\beta^0_k) X_k$.
   Therefore $\widehat e^{({\rm rem})}(\beta)=A_1+A_2+A_3$ with
   $A_1 = A_0 -  M_{\lambda^0} \, A_0 \, M_{f^0}$,
   $A_2 =\lambda^0 f^{0\prime} - \widehat \lambda(\beta) \widehat f'(\beta)$, and
   $A_3 =-\widehat e^{(1)}_e$.
   We find ${\rm rank}(A_1)\leq 2R$, ${\rm rank}(A_2)\leq 2R$, ${\rm rank}(A_3)\leq 3R$, and thus
   ${\rm rank}(\widehat e^{({\rm rem})}(\beta)) \leq 7R$, as stated in the theorem.
\end{proof}


Having expansions for $M_{\widehat \lambda}(\beta)$ and $M_{\widehat f}(\beta)$, we also have expansions
for $P_{\widehat \lambda}(\beta)=\mathbb{I}_N-M_{\widehat \lambda}(\beta)$ and $P_{\widehat f}(\beta)=\mathbb{I}_T-M_{\widehat f}(\beta)$.
The reason why we give expansions of the projectors and not expansions of $\widehat \lambda(\beta)$ and $\widehat f(\beta)$
directly is for the latter we would need to specify a normalization, whereas the projectors
are independent of any normalization choice. An expansion for $\widehat \lambda(\beta)$ can, for example, be defined by
$\widehat \lambda(\beta) = P_{\widehat \lambda}(\beta) \lambda^0$, in which case the normalization of $\widehat \lambda(\beta)$
is implicitly defined by the normalization of $\lambda^0$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Consistency Proof for Bias and Variance Estimators
  (Proof of Theorem~\ref{th:biascorrection})}
\label{app:ProofThBias}  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


It is convenient to introduce some alternative notation for
Definition~\ref{def:estimators}
in section~\ref{sec:BiasCorrection} of the main text.

\begin{paragraph}{\bf Definition}
\it
Let $\Gamma: \mathbb{R} \rightarrow \mathbb{R}$ be the truncation kernel defined by $\Gamma(x)=1$ for $|x|\leq 1$, and $\Gamma(x)=0$
otherwise. Let $M$ be a bandwidth parameter that depends on $N$ and $T$.
For an $N\times N$ matrix $A$ with elements $A_{ij}$ and a $T\times T$ matrix $B$ with elements
$B_{ts}$ we define
\begin{itemize}
   \item[(i)] the diagonal truncations
     $A^{\rm truncD} = {\rm diag}[ (A_{ii})_{i=1,\ldots,N} ]$
     and
     $B^{\rm truncD} =  {\rm diag}[ (B_{tt})_{t=1,\ldots,T} ]$.

   \item[(ii)] the right-sided  Kernel truncation of $B$,
          which is a $T \times T$ matrix $B^{\rm truncR}$
          with elements $B^{\rm truncR}_{ts} = \Gamma\left( \frac{s-t} M \right) B_{ts}$ for $t<s$,
          and $B^{\rm truncR}_{ts}=0$ otherwise.

\end{itemize}
\end{paragraph}

Here, we suppress the dependence of $B^{\rm truncR}$ on the bandwidth parameter $M$. Using this notation we can represent the 
estimators for the bias in Definition~\ref{def:estimators} as follows:
    \begin{align*}
       \widehat B_{1,k} &= \frac 1 N \, {\rm Tr}\left[ P_{\widehat f} \left( \widehat e' X_k \right)^{\rm truncR} \right]  \; ,
       \nonumber \\
       \widehat B_{2,k} &=
       \frac 1 T
         {\rm Tr}\left[ \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \, M_{\widehat \lambda} \, X_k \,
              \widehat f \, (\widehat f^{\prime} \widehat f)^{-1} \,
                             (\widehat \lambda^{\prime}\widehat \lambda)^{-1} \, \widehat \lambda^{\prime} \right] \; ,
       \nonumber \\
       \widehat B_{3,k} &=
       \frac 1 N
       {\rm Tr}\left[ \left( \widehat e' \, \widehat e \right)^{\rm truncD} \, M_{\widehat f} \, X^{\prime}_k \,
              \widehat \lambda \, (\widehat \lambda^{\prime} \widehat \lambda)^{-1} \,
                             (\widehat f^{\prime}\widehat f)^{-1} \, \widehat f^{\prime} \right]
                   \; .
    \end{align*}

Before proving Theorem~\ref{th:biascorrection} we establish some
preliminary results.


\begin{corollary}
   \label{lemma:sqrtNTcons}
   Under the Assumptions of Theorem~\ref{th:limdis}
    we have $\sqrt{NT} \left( \widehat \beta - \beta^0 \right) = {\cal O}_p(1)$.
\end{corollary}
This corollary directly follows from Theorem~\ref{th:limdis}.

\begin{corollary}
   \label{lemma:Pfhat}
   Under the Assumptions of Theorem~\ref{th:biascorrection} we have
  \begin{align*}
      \left\| P_{\widehat \lambda} - P_{\lambda^0} \right\| &= \left\| M_{\widehat \lambda} - M_{\lambda^0} \right\|
              = {\cal O}_p(N^{-1/2}) \; ,
      \nonumber \\
      \left\| P_{\widehat f} - P_{f^0} \right\| &= \left\| M_{\widehat f} - M_{f^0} \right\| = {\cal O}_p(T^{-1/2}) \; .
   \end{align*}
\end{corollary}

\begin{proof}[\bf Proof]
   Using $\|e\|={\cal O}_p(N^{1/2})$ and $\|X_k\|={\cal O}_p(N)$ we find
   the expansion terms in Theorem~\ref{theorem:expansions} satisfy
   \begin{align*}
      \left\|M_{\widehat \lambda,e}^{(1)}\right\| &= {\cal O}_p(N^{-1/2}) \; , &
      \left\|M_{\widehat \lambda,e}^{(2)}\right\| &= {\cal O}_p(N^{-1}) \; , &
      \left\|M_{\widehat \lambda,k}^{(1)}\right\| &= {\cal O}_p(1) \; .
   \end{align*}
   Together with corollary \ref{lemma:sqrtNTcons} the result for $\left\| M_{\widehat \lambda} - M_{\lambda^0} \right\|$
   immediately follows. In addition we have $P_{\widehat \lambda} - P_{\lambda^0}=-M_{\widehat \lambda}+M_{\lambda^0}$.
   The proof for $M_{\widehat f}$ and $P_{\widehat f}$ is analogous.
\end{proof}

\begin{lemma}
   \label{lemma:A1A2}
   Under the Assumptions of Theorem~\ref{th:biascorrection} we have
     \begin{align*}
         A_1 &\equiv \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T e_{it}^2
               \left( {\cal X}_{it} {\cal X}_{it}' - \widehat {\cal X}_{it} \widehat {\cal X}_{it}' \right)
                     = o_p(1)  \; ,
        \\
         A_2 &\equiv \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T \left( e_{it}^2 - \widehat e_{it}^2 \right)
                                               \widehat {\cal X}_{it} \widehat {\cal X}_{it}' = o_p(1) \; .
     \end{align*}
\end{lemma}

\begin{lemma}
   \label{lemma:lambdafINV}
   Let $\widehat f$ and $f^0$ be normalized as $\widehat f' \widehat f / T = \mathbb{I}_R$
   and $f^{0\prime} f^0 / T = \mathbb{I}_R$.
   Then, under the assumptions of Theorem~\ref{th:biascorrection}, there exists an $R\times R$ matrix $H=H_{NT}$
   such that
   \begin{align*}
      \left\| \widehat f - f^0 \, H \right\| &= {\cal O}_p\left(1\right) \; , &
      \left\| \widehat \lambda - \lambda^0 \, \left(H'\right)^{-1} \right\| &= {\cal O}_p\left(1\right) \; .
   \end{align*}
   Furthermore
   \begin{align*}
     \left\| \widehat \lambda \, (\widehat \lambda^{\prime}\widehat\lambda)^{-1} \, (\widehat f^{\prime}\widehat f)^{-1} \, \widehat f^{\prime}
        -\lambda^0 \, (\lambda^{0\prime}\lambda^0)^{-1} \, (f^{0\prime}f^0)^{-1} \, f^{0\prime} \right\|
    &= {\cal O}_p\left( N^{-3/2} \right) \; .
   \end{align*}
\end{lemma}
Here, the matrix $H$ depends on $N,T$, but we write $H$ instead of $H_{NT}$ to keep notation simple.

\begin{lemma}
   \label{lemma:exp}
   Under the Assumptions of Theorem~\ref{th:biascorrection} we have
   \begin{align*}
      {\rm(i)}&& N^{-1} \, \left\| \mathbb{E}_{\cal C}(e'   X_k    ) -
       \left( \widehat e' \, X_k \right)^{\rm truncR} \right\|
                         &= o_p(1) \; ,
     \nonumber \\
     {\rm(ii)}&& N^{-1} \, \left\| \mathbb{E}_{\cal C}(e'   e) -
         \left( \widehat e' \, \widehat e \right)^{\rm truncD} \right\|
                         &= o_p(1) \; ,
     \nonumber \\
     {\rm(iii)}&& T^{-1} \, \left\| \mathbb{E}_{\cal C}(e   e')
           - \left( \widehat e \, \widehat e' \right)^{\rm truncD}  \right\|
                         &= o_p(1) \; .
   \end{align*}
\end{lemma}

\begin{lemma}
   \label{lemma:normsTrunc}
   Under the Assumptions of Theorem~\ref{th:biascorrection} we have
   \begin{align*}
      {\rm(i)}&& N^{-1} \, \left\| \left( \widehat e' \, X_k \right)^{\rm truncR} \right\|
                         &= {\cal O}_p(M T^{1/8}) \; ,
     \nonumber \\
     {\rm(ii)}&& N^{-1} \, \left\| \left( \widehat e' \, \widehat e \right)^{\rm truncD} \right\|
                         &= {\cal O}_p(1) \; ,
     \nonumber \\
     {\rm(iii)}&& T^{-1} \, \left\| \left( \widehat e \, \widehat e' \right)^{\rm truncD}  \right\|
                         &= {\cal O}_p(1) \; .
   \end{align*}
\end{lemma}

The proof of the above lemmas is given section~\ref{sec:Intermed} below.
Using these lemmas we can now prove Theorem~\ref{th:biascorrection}.

\begin{proof}[\bf Proof of Theorem \ref{th:biascorrection}, Part I: show $\widehat W=W+o_p(1)$.]
   $\phantom{a}$
   \\
   Using $\left| {\rm Tr}\left( C\right) \right| \leq \left\| C\right\| \limfunc{rank}\left( C\right)$
   and corollary \ref{lemma:Pfhat}
   we find:
   \begin{align*}
      \big| \widehat W_{k_1 k_2} - & W_{NT,k_1 k_2} \big|
           \nonumber \\
         & = \bigg| (NT)^{-1} {\rm Tr}\left[ \left( M_{\widehat \lambda} - M_{\lambda^0} \right)
                                      \, X_{k_1} \, M_{\widehat f} \,  X_{k_2}' \right]
            + (NT)^{-1} {\rm Tr}\left[ M_{\lambda^0} \, X_{k_1} \,
                                       \left( M_{\widehat f} - M_{f^0} \right) \,  X_{k_2}' \right]
           \nonumber \\
         & \leq \frac{2R}{NT} \left\| M_{\widehat \lambda} - M_{\lambda^0} \right\|
                                      \| X_{k_1} \| \| X_{k_2} \|
                \frac{2R}{NT} \left\| M_{\widehat f} - M_{f^0} \right\|
                                      \| X_{k_1} \| \| X_{k_2} \|
           \nonumber \\
         & = \frac{2R}{NT} {\cal O}_p(N^{-1}) {\cal O}_p(NT)
             +\frac{2R}{NT} {\cal O}_p(T^{-1}) {\cal O}_p(NT)
           \nonumber \\
         &= o_p(1) \; .
   \end{align*}
   Thus we have $\widehat W=W_{NT}+o_p(1)=W+o_p(1)$.
\end{proof}

\begin{proof}[\bf Proof of Theorem \ref{th:biascorrection}, Part II: show $\widehat \Omega=\Omega+o_p(1)$.]
   $\phantom{a}$
   \\
    Let
    $\Omega_{NT} \equiv  \frac 1 {NT} \, \sum_{i=1}^N \, \sum_{t=1}^T
            \,   e_{it}^2   \, {\cal X}_{it} \, {\cal X}_{it}'$.
     We have
     $\Omega = \Omega_{NT} + o_P(1) 
     = \widehat \Omega + A_1 + A_2 + o_p(1) = \widehat \Omega + o_P(1)$, 
     where $A_1$ and $A_2$ are defined in Lemma \ref{lemma:A1A2},
     and the lemma states $A_1$ and $A_2$ are $o_p(1)$.
\end{proof}

\begin{proof}[\bf Proof of Theorem \ref{th:biascorrection}, Part III: show $\widehat B_1=B_1+o_p(1)$.]
  $\phantom{a}$
  \\
   Let $B_{1,k,NT}=N^{-1} \, {\rm Tr}\left[ P_{f^0} \, \mathbb{E}_{\cal C} \left( e' \, X_k   \right) \right]$.
   According to Assumption~\ref{ass:A6} we have $B_{1,k}=B_{1,k,NT}+o_p(1)$.
   What is left to show is $ B_{1,k,NT}=\widehat B_{1,k} + o_p(1)$.
   Using $\left| {\rm Tr}\left( C\right) \right| \leq \left\| C\right\| \limfunc{rank}\left( C\right)$ we find
   \begin{align*}
     \left|  B_{1,k,NT} - \widehat B_1 \right|
   &= \left| \mathbb{E}_{\cal C} \left[ \frac{1}{N} {\rm Tr}(P_{f^0} \, e^{\prime}\,   \, X_k)    \right]
         - \frac{1}{N} {\rm Tr}\left[ P_{\widehat f} \left( \widehat e' X_k \right)^{\rm truncR} \right] \right|
      \nonumber \\
   &\leq \left| \frac{1}{N} {\rm Tr}\left[
                \left( P_{f^0} - P_{\widehat f} \right) \left( \widehat e' X_k \right)^{\rm truncR} \right] \right|
      \nonumber \\ & \quad
    + \left| \frac{1}{N} {\rm Tr}\left\{
 P_{f^0} \left[ \mathbb{E}_{\cal C} \left(e^{\prime}\,  \, X_k  \right) - \left( \widehat e' X_k \right)^{\rm truncR} \right]
       \right\} \right|
      \nonumber \\
        &\leq \frac{2R}{N} \left\| P_{f^0} - P_{\widehat f} \right\| \,
                                   \left\| \left(\widehat e' X_k \right)^{\rm truncR} \right\|
      \nonumber \\ & \quad
             +\frac{R}{N}  \left\| P_{f^0} \right\| \,
             \left\| \mathbb{E}_{\cal C} \left(e^{\prime}\,  \, X_k   \right) - \left( \widehat e' X_k \right)^{\rm truncR}
             \right\| \; .
   \end{align*}
   We have $\left\| P_{f^0} \right\| = 1$.
   We now apply Lemmas \ref{lemma:exp}, \ref{lemma:Pfhat} and \ref{lemma:normsTrunc} to find
   \begin{align*}
      \left|  B_{1,k,NT} - \widehat B_1 \right|
        &=  N^{-1} \left( {\cal O}_p(N^{-1/2}) {\cal O}_p(M N T^{1/8}) + o_p(N) \right) = o_p(1) \; .
   \end{align*}
   This is what we wanted to show.
\end{proof}

\begin{proof}[\bf Proof of Theorem \ref{th:biascorrection}, final part: show $\widehat B_{2}=B_{2}+o_p(1)$ 
and $\widehat B_{3}=B_{3}+o_p(1)$.]
  $\phantom{a}$
  \\
   Define
   \begin{align*}
      B_{2,k,NT} &= \frac 1 T
         {\rm Tr}\left[ \mathbb{E}_{\cal C}\left( e e' \right) \, M_{\lambda^0} \, X_k \,
              f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right] \; .
   \end{align*}
   According to Assumption~\ref{ass:A6} we have $B_{2,k}=B_{2,k,NT}+o_p(1)$.
   What is left to show is $B_{2,k,NT}=\widehat B_{2,k} + o_p(1)$.
    We have
  \begin{align*}
      B_{2,k} - \widehat B_{2,k}
      =&
       \frac 1 T {\rm Tr}\left[ \mathbb{E}_{\cal C}\left( e   e' \right) \, M_{\lambda^0} \, X_k \,
              f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right]
       \nonumber \\ & \qquad
       -
       \frac 1 T
         {\rm Tr}\left[ \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \, M_{\widehat \lambda} \, X_k \,
              \widehat f \, (\widehat f^{\prime} \widehat f)^{-1} \,
                             (\widehat \lambda^{\prime}\widehat \lambda)^{-1} \, \widehat \lambda^{\prime} \right]
      \nonumber \\
      =& \frac 1 T
         {\rm Tr}\left[ \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \, M_{\widehat \lambda} \, X_k \,
           \left( f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
           - \widehat f \, (\widehat f^{\prime} \widehat f)^{-1} \,
                             (\widehat \lambda^{\prime}\widehat \lambda)^{-1} \, \widehat \lambda^{\prime} \right) \right]
      \nonumber \\
       & + \frac 1 T
         {\rm Tr}\left[ \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \,
                 \left( M_{\lambda^0} - M_{\widehat \lambda} \right) \, X_k \,
            f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
           \right]
      \nonumber \\
       & + \frac 1 T
         {\rm Tr}\left\{ \left[ \mathbb{E}_{\cal C}\left( e   e' \right)
                           - \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \right]\,
                  M_{\lambda^0}  \, X_k \,
            f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
           \right\} \; .
   \end{align*}
   Using $\left| {\rm Tr}\left( C\right) \right| \leq \left\| C\right\| \limfunc{rank}\left( C\right)$
   (which is true for every square matrix $C$) we find
\begin{align*}
   \left|   B_{2,k} - \widehat B_{2,k} \right|
   \leq &
       \frac R T
         \left\| \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \right\|
         \left\|  X_k \right\|
          \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime}
           - \widehat f \, (\widehat f^{\prime} \widehat f)^{-1} \,
                             (\widehat \lambda^{\prime}\widehat \lambda)^{-1} \, \widehat \lambda^{\prime} \right\|
      \nonumber \\
       & + \frac R T
        \left\| \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \right\|
                 \left\| M_{\lambda^0} - M_{\widehat \lambda} \right\| \left\| X_k \right\|
            \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\|
      \nonumber \\
       & + \frac R T
            \left\| \mathbb{E}_{\cal C}\left( e  e' \right)
                           - \left( \widehat e \, {\widehat e}^{\prime} \right)^{\rm truncD} \right\| \,
                 \left\| X_k \right\| \,
            \left\| f^0 \, (f^{0\prime}f^0)^{-1} \, (\lambda^{0\prime}\lambda^0)^{-1} \, \lambda^{0\prime} \right\|
       \, .
\end{align*}
Here we used $\left\| M_{f^0} \right\| = \left\| M_{\widehat f} \right\| = 1$.
Using $\left\| X_k \right\|={\cal O}_p(\sqrt{NT})$, and applying
Lemmas~\ref{lemma:Pfhat}, \ref{lemma:lambdafINV}, \ref{lemma:exp} and \ref{lemma:normsTrunc}, we now find
\begin{align*}
   \left|   B_{2,k} - \widehat B_{2,k} \right|
      &= T^{-1}
       \bigg[ {\cal O}_p(T) \, {\cal O}_p((NT)^{1/2}) \, {\cal O}_p(N^{-3/2})
   \nonumber \\ & \qquad \qquad \qquad
            +  {\cal O}_p(T) \, {\cal O}_p(N^{-1/2}) \, {\cal O}_p((NT)^{1/2}) \, {\cal O}_p( (NT)^{-1/2})
   \nonumber \\ & \qquad \qquad \qquad
          +  o_p(T) \,  {\cal O}_p((NT)^{1/2}) \, {\cal O}_p( (NT)^{-1/2}) \bigg] = o_p(1) \; .
\end{align*}
This is what we wanted to show. The proof of $\widehat B_3=B_3+o_p(1)$ is analogous.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Intermediate Lemma}
\label{sec:Intermed}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


Here we provide the proof of some intermediate lemmas that were stated and
used in section~\ref{app:ProofThBias}.
The following lemma gives a useful bound on the maximum of (correlated) random variables
\begin{lemma}
   \label{lemma:maxRV}
   Let $Z_i$, $i=1,2,\ldots,n$, be $n$ real valued random variables, and
   let $\gamma \geq 1$ and $B>0$ be finite constants (independent of $n$).
   Assume $\max_i \, \mathbb{E}_{\cal C} |Z_i|^{\gamma} \leq B$,
   i.e., the $\gamma$'th moment of the $Z_i$ are finite and uniformly bounded.
   For $n \rightarrow \infty$ we then have
   \begin{align}
      \max_i |Z_i| &= {\cal O}_p\left( n^{1/\gamma} \right) \; .
      \label{eq:Zbound}
   \end{align}
\end{lemma}

\begin{proof}[\bf Proof]
   Using Jensen's inequality one obtains
   $\mathbb{E}_{\cal C} \max_i |Z_i| \leq \left( \mathbb{E}_{\cal C} \max_i |Z_i|^\gamma \right)^{1/\gamma}
    \leq \left( \mathbb{E}_{\cal C} \sum_{i=1}^n |Z_i|^\gamma \right)^{1/\gamma}
    \leq \left( n \, \max_i \mathbb{E}_{\cal C} |Z_i|^\gamma \right)^{1/\gamma}
    \leq n^{1/\gamma} \, B^{1/\gamma}$.
   Markov's inequality then gives equation \eqref{eq:Zbound}.
\end{proof}

\begin{lemma}
   \label{lemma:barZ}
   Let
   \begin{align*}
      \bar Z^{(1)}_{k,t\tau}
            &= N^{-1/2} \sum_{i=1}^N \left[ e_{it} X_{k,i\tau} - \mathbb{E}_{\cal C} \left( e_{it} X_{k,i\tau} \right) \right] \; ,
      \nonumber \\
      \bar Z^{(2)}_{t}
            &= N^{-1/2} \sum_{i=1}^N \left[ e_{it}^2 - \mathbb{E}_{\cal C} \left( e_{it}^2 \right) \right] \; ,
      \nonumber \\
      \bar Z^{(3)}_{i}
            &= T^{-1/2} \sum_{t=1}^T \left[ e_{it}^2 - \mathbb{E}_{\cal C} \left( e_{it}^2 \right) \right] \; .
   \end{align*}
   Under assumption \ref{ass:A5} we have
   \begin{align*}
      \mathbb{E}_{\cal C} \left| \bar Z^{(1)}_{k,t\tau} \right|^4 &\leq B \; ,
      \nonumber \\
      \mathbb{E}_{\cal C} \left| \bar Z^{(2)}_{t\tau} \right|^4 &\leq B \; ,
      \nonumber \\
      \mathbb{E}_{\cal C} \left| \bar Z^{(3)}_{i} \right|^4 &\leq B \; ,
   \end{align*}
   for some $B>0$, i.e., the 
   conditional expectations $\bar Z^{(1)}_{k,t\tau}$, $\bar Z^{(2)}_{t\tau}$, and $\bar Z^{(3)}_{i}$
   are uniformly bounded over $t,\tau$, or $i$, respectively.
\end{lemma}

\begin{proof}[\bf Proof]
   \# We start with the proof for $\bar Z^{(1)}_{k,t\tau}$.
      Define
      $Z^{(1)}_{k,t\tau,i} = e_{it} X_{k,i\tau} - \mathbb{E}_{\cal C} \left( e_{it} X_{k,i\tau} \right)$.
      By assumption we have finite 8th moments for $e_{it}$ and $X_{k,i\tau}$ uniformly
      across $k,i,t,\tau$, and thus (using Cauchy Schwarz inequality) we have finite 4th moment
      of $Z^{(1)}_{k,t\tau,i}$ uniformly across $k,i,t,\tau$. For ease of notation we now
      fix $k,t,\tau$ and write $Z_i=Z^{(1)}_{k,t\tau,i}$.
      We have $\mathbb{E}_{\cal C}(Z_i)=0$ and
      $\mathbb{E}_{\cal C}(Z_{i} Z_{j} Z_{k} Z_{l})=0$ if $i \notin \{j,k,l\}$
      (and the same holds for permutations of $i,j,k,l$).
      Using this we compute
     \begin{align*}
        \mathbb{E}_{\cal C} \left( \sum_{i=1}^N \, Z_i \right)^4
         &= \sum_{i,j,k,l=1}^N \, \mathbb{E}_{\cal C}\left( Z_i Z_j Z_k Z_l\right)
        \nonumber \\
         &= 3 \, \sum_{i\neq j} \, \mathbb{E}_{\cal C}\left( Z_i^2 \, Z_j^2\right)
            + \sum_{i} \mathbb{E}_{\cal C}\left( Z_i^4 \right)
        \nonumber \\
         &= 3 \, \sum_{i,j=1}^N \, \mathbb{E}_{\cal C}\left( Z_i^2\right) \, \mathbb{E}_{\cal C}\left( Z_j^2\right)
    + \sum_{i=1}^N \, \left\{ \mathbb{E}_{\cal C}\left( Z_i^4 \right) - 3 \left[ \mathbb{E}_{\cal C}\left(Z_i^2\right) \right]^2 \right\}
       \; ,
     \end{align*}
     Because we argued $\mathbb{E}_{\cal C}\left( Z_i^4 \right)$ is bounded uniformly, the last equation shows
     $\bar Z^{(1)}_{k,t\tau} = N^{-1/2} \sum_{i=1}^N \, Z^{(1)}_{k,t\tau,i}$ is bounded uniformly
     across $k,t,\tau$. This is what we wanted to show.

   \# The proofs for $\bar Z^{(2)}_{t}$ and $\bar Z^{(3)}_{i}$ are analogous.
 \end{proof}

\begin{lemma}
   \label{Lemma:Trunc}
   For a $T\times T$ matrix $A$ we have
   \begin{align*}
      \left\| A^{\rm truncR} \right\| \, &\leq  \, M \left\| A^{\rm truncR} \right\|_{\max}
         \, \equiv \, M \, \max_t \, \max_{t<\tau\leq t+M} |A_{t\tau}| \, ,
   \end{align*}
\end{lemma}
Here, for the bounds on $\tau$ we could write
   $\max(1,t-M)$ instead of $t-M$, and $\min(T,t+M)$ instead of $t+M$, to guarantee $1\leq \tau \leq T$.
   Since this would complicate notation, we prefer the convention $A_{t\tau}=0$ for $t<1$ or $\tau<1$
   of $t>T$ or $\tau>T$.

\begin{proof}[\bf Proof]
   For the $1$-norm of $A^{\rm truncR}$ we find
   \begin{align*}
     \left\| A^{\rm truncR} \right\|_1  &= \max_{t=1\ldots T} \, \sum_{\tau=t+1}^{t+M} \, |A_{t\tau}|
       \nonumber \\
            &\leq M \,  \max_{t<\tau\leq t+M} \, |A_{t\tau}| = M \left\| A^{\rm truncR} \right\|_{\max}  \; ,
   \end{align*}
   and analogously we find the same bound for the $\infty$-norm $\left\| A^{\rm truncR} \right\|_\infty$.
   Applying part (vii) of Lemma~\ref{lemma:inequalities} we therefore also get this bound for the operator norm
   $\| A^{\rm truncR} \|$.
\end{proof}

\begin{proof}[\bf Proof of Lemma~\ref{lemma:A1A2}]
     \# We first show $A_1\equiv (NT)^{-1} \sum_{i=1}^N \sum_{t=1}^T e_{it}^2
               \left( {\cal X}_{it} {\cal X}_{it}' - \widehat {\cal X}_{it} \widehat {\cal X}_{it}' \right)
                     = o_p(1)$.
     Let $B_{1,it} =  {\cal X}_{it}- \widehat {\cal X}_{it}$,
     $B_{2,it} = e_{it}^2  {\cal X}_{it}$, and
     $B_{3,it} = e_{it}^2  \widehat {\cal X}_{it}$.
     Note $B_1$, $B_2$, and $B_3$ can either be viewed as $K$-vectors for each
     pair $(i,t)$, or equivalently as $N\times T$ matrices $B_{1,k}$, $B_{2,k}$, and $B_{3,k}$ for each $k=1,\ldots,K$.
     We have $A_1 = (NT)^{-1} \sum_i \sum_t \left( B_{1,it} B_{2,it}' +  B_{3,it} B_{1,it}' \right)$,
     or equivalently
     \begin{align*}
        A_{1,k_1 k_2} &= \frac 1 {NT} {\rm Tr}\left( B_{1,k_1} B_{3,k_2}' + B_{2,k_1} B_{1,k_2}' \right) \; .
     \end{align*}
     Using $\|M_{\widehat \lambda} - M_{\lambda^0}\| = {\cal O}_p(N^{-1/2})$,
     $\|M_{\widehat f} - M_{f^0}\| = {\cal O}_p(N^{-1/2})$,
     $\|X_k\| = {\cal O}_p(\sqrt{NT}) = {\cal O}_p(N)$,
     we find for $B_{1,k}  =
                     (M_{\lambda^0} - M_{\widehat \lambda}) X_k M_{f^0}
                    + M_{\widehat \lambda} X_k (M_{f^0} - M_{\widehat f}) $
    that $\|B_{1,k}\| = {\cal O}_p(N^{1/2})$.
     In addition we have ${\rm rank}(B_{1,k}) \leq 4R$.
      We also have
      \begin{align*}
        \| B_{2,k} \|^4 &\leq \| B_{2,k} \|_F^4
                   \nonumber \\
            &= \left( \sum_{i=1}^N \sum_{t=1}^T e_{it}^4  {\cal X}_{k,it}^2 \right)^2
                   \nonumber \\
            &\leq \left( \sum_{i=1}^N \sum_{t=1}^T e_{it}^8 \right)
                  \left( \sum_{i=1}^N \sum_{t=1}^T {\cal X}_{k,it}^4 \right)
             = {\cal O}_p(NT) \, {\cal O}_p(NT) \, ,
     \end{align*}
     which implies $\| B_{2,k} \|={\cal O}_p(\sqrt{NT})$, and analogously we find $\| B_{3,k} \|={\cal O}_p(\sqrt{NT})$.
     Therefore
     \begin{align*}
        | A_{1,k_1 k_2} | &\leq \frac {4R} {NT}
                      \left( \| B_{1,k_1} \| \| B_{3,k_2}\| + \|B_{2,k_1}\| \| B_{1,k_2}\| \right)
        \nonumber \\
           &= \frac {4R} {NT}
                      \left( {\cal O}_p(N^{1/2}) {\cal O}_p(\sqrt{NT}) + {\cal O}_p(\sqrt{NT}) {\cal O}_p(N^{1/2}) \right)
            = o_p(1) \; .
     \end{align*}
     This is what we wanted to show.

     \# Finally, we want to show $A_2 \equiv (NT)^{-1} \sum_{i=1}^N \sum_{t=1}^T \left( e_{it}^2 - \widehat e_{it}^2 \right)
                                               \widehat {\cal X}_{it} \widehat {\cal X}_{it}' = o_p(1) $.
     According to theorem \ref{theorem:expansions} we have $e - \widehat e = C_1 + C_2$,
     where we defined $C_1 = - \sum_{k=1}^K \left( \widehat \beta_k - \beta^0_k \right) \, X_k $,
     and $C_2=\sum_{k=1}^K \left( \widehat \beta_k - \beta^0_k \right)
               \left( P_{\lambda^0} \, X_k \, M_{f^0} +  X_k \, P_{f^0} \right)
             + P_{\lambda^0} \, e \, M_{f^0} + e \, P_{f^0}
            - \widehat e^{(1)}_e - \widehat e^{({\rm rem})}$,
     which satisfies $\|C_2\|={\cal O}_p(N^{1/2})$, and ${\rm rank}(C_2) \leq 11 R$
     (actually, one can easily prove $\leq 5R$, but this does not follow from theorem \ref{theorem:expansions}).
     Using this notation we have
     \begin{align*}
         A_2 &= \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T
                 (e_{it}+\widehat e_{it}) (C_{1,it} + C_{2,it})
                                               \widehat {\cal X}_{it} \widehat {\cal X}_{it}' \; ,
     \end{align*}
     which can also be written as
     \begin{align*}
        A_{2,k_1 k_2} &= - \,\sum_{k_3=1}^K \left( \widehat \beta_{k_3} - \beta^0_{k_3} \right)
                        \left( C_{5,k_1 k_2 k_3} + C_{6,k_1 k_2 k_3} \right)
                         + \frac 1 {NT} {\rm Tr}\left( C_2 \, C_{3,k_1 k_2}  \right)
                         + \frac 1 {NT} {\rm Tr}\left( C_2 \, C_{4,k_1 k_2}  \right) \; ,
     \end{align*}
     where we defined
     \begin{align*}
        C_{3,k_1 k_2,it} &= e_{it} \widehat {\cal X}_{k_1,it} \widehat {\cal X}_{k_2,it} \; ,
        \nonumber \\
        C_{4,k_1 k_2,it} &=\widehat e_{it} \widehat {\cal X}_{k_1,it} \widehat {\cal X}_{k_2,it} \; ,
        \nonumber \\
        C_{5,k_1 k_2 k_3}
       &=
 \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T \, e_{it} \widehat {\cal X}_{k_1,it} \widehat {\cal X}_{k_2,it} X_{k_3,it} \; ,
        \nonumber \\
    C_{6,k_1 k_2 k_3}
   &=\frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T \, \widehat e_{it} \widehat {\cal X}_{k_1,it} \widehat {\cal X}_{k_2,it} X_{k_3,it} \; .
     \end{align*}
     Again, because we have uniformly bounded $8$th moments for $e_{it}$ and $X_{k,it}$, we find
     \begin{align*}
        \| C_{3,k_1 k_2} \|^4 &\leq \| C_{3,k_1 k_2} \|_F^4
                  \nonumber \\
                &= \left( \sum_{i=1}^N \sum_{t=1}^T e_{it}^2 \widehat {\cal X}_{k_1,it}^2 \widehat {\cal X}_{k_2,it}^2
                   \right)^2
                  \nonumber \\
                &\leq  \left( \sum_{i=1}^N \sum_{t=1}^T e_{it}^4 \right)
             \left( \sum_{i=1}^N \sum_{t=1}^T \widehat {\cal X}_{k_1,it}^4 \widehat {\cal X}_{k_2,it}^4 \right)
                  \nonumber \\
                &= {\cal O}_p(N^2 T^2) \; ,
     \end{align*}
     i.e., $\| C_{3,k_1 k_2} \|={\cal O}_p(\sqrt{NT})$. Furthermore
     \begin{align*}
        \| C_{4,k_1 k_2} \|^2 &\leq \| C_{3,k_1 k_2} \|_F^2
                  \nonumber \\
                &= \sum_{i=1}^N \sum_{t=1}^T \widehat e_{it}^2 \widehat {\cal X}_{k_1,it}^2 \widehat {\cal X}_{k_2,it}^2
                  \nonumber \\
                &\leq \left( \sum_{i=1}^N \sum_{t=1}^T \widehat e_{it}^2 \right)
                       \max_{i=1\ldots N} \max_{t=1\ldots T}
                       \left( \widehat {\cal X}_{k_1,it}^2 \widehat {\cal X}_{k_2,it}^2 \right)
                  \nonumber \\
                &\leq \left( \sum_{i=1}^N \sum_{t=1}^T e_{it}^2 \right)
                       \max_{i=1\ldots N} \max_{t=1\ldots T}
                       \left( \widehat {\cal X}_{k_1,it}^2 \widehat {\cal X}_{k_2,it}^2 \right)
                  \nonumber \\
                &= {\cal O}_p(NT) {\cal O}_p((NT)^{(4/(8+\epsilon))}) = o_p((NT)^{(3/4)}) \; .
     \end{align*}
     Here we used the assumption that $X_k$ has uniformly bounded moments of order
     $8+\epsilon$ for some $\epsilon>0$. We also used
         $\sum_{i=1}^N \sum_{t=1}^T \widehat e_{it}^2 \leq \sum_{i=1}^N \sum_{t=1}^T e_{it}^2$.

     For $C_5$ we find
     \begin{align*}
        C_{5,k_1 k_2 k_3}^2
       &\leq
       \left( \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T \, e_{it}^2 \right)
       \left( \frac 1 {NT} \widehat {\cal X}_{k_1,it}^2 \widehat {\cal X}_{k_2,it}^2 X_{k_3,it}^2 \right)
       \nonumber \\
       &= {\cal O}_p(1) \; ,
     \end{align*}
     i.e., $C_{5,k_1 k_2 k_3} = {\cal O}_p(1)$, and analogously $C_{6,k_1 k_2 k_3} = {\cal O}_p(1)$, because
     $\sum_{i=1}^N \sum_{t=1}^T \widehat e_{it}^2 \leq \sum_{i=1}^N \sum_{t=1}^T e_{it}^2$.

     Using these results we obtain
     \begin{align*}
        | A_{2,k_1 k_2} | &\leq - \,\sum_{k_3=1}^K \left\| \widehat \beta_{k_3} - \beta^0_{k_3} \right\|
                        \left| C_{5,k_1 k_2 k_3} + C_{6,k_1 k_2 k_3} \right|
                         + \frac {11 R} {NT} \|C_2\| \|C_{3,k_1 k_2} \|
                         + \frac {11 R} {NT} \| C_2\| \| C_{4,k_1 k_2} \|
               \nonumber \\
                      &= {\cal O}_p((NT)^{-1/2}) {\cal O}_p(1)
                           + \frac {11 R} {NT} {\cal O}_p(N^{1/2}) {\cal O}_p(\sqrt{NT})
                           + \frac {11 R} {NT} {\cal O}_p(N^{1/2}) o_p((NT)^{3/4})
                       = o_p(1) \; .
     \end{align*}
     This is what we wanted to show.
\end{proof}

Remember, the truncation Kernel $\Gamma(.)$ is defined by $\Gamma(x)=1$ for $|x|\leq 1$ and $\Gamma(x)=0$ otherwise.
Without loss of generality we assume in the following the bandwidth parameter $M$ is a positive integer
(without this assumption, one needs to replace $M$ everywhere below by the largest integer contained in $M$,
but nothing else changes).

\begin{proof}[\bf Proof of Lemma \ref{lemma:lambdafINV}]
  By Lemma~\ref{lemma:Pfhat} we know asymptotically $P_{\widehat f}$ is close to $P_{f^0}$ and therefore
  ${\rm rank}(P_{\widehat f}P_{f^0})={\rm rank}(P_{f^0}P_{f^0})=R$ , i.e.,
  ${\rm rank}(P_{\widehat f}f^0)=R$ asymptotically. We can therefore write
  $\widehat f = P_{\widehat f} f^0 H$, where $H=H_{NT}$ is a non-singular $R\times R$ matrix.

  We now want to show $\|H\|={\cal O}_p(1)$ and $\|H^{-1}\|={\cal O}_p(1)$.
  Because of our normalization of $\widehat f$ and $f^0$ we have $H=(\widehat f' P_{\widehat f} f^0/T)^{-1}=(\widehat f' f^0/T)^{-1}$,
  and therefore $\|H^{-1}\|\leq \|\widehat f\| \|f^0\| /T ={\cal O}_p(1)$.
  We also have $\widehat f = f^0 H + (P_{\widehat f}-P_{f^0}) f^0 H$,
  and thus $H=f^{0\prime} \widehat f/T - f^{0\prime} (P_{\widehat f}-P_{f^0}) f^0 H /T$,
  i.e., $\|H\| \leq {\cal O}_p(1) + \|H\| {\cal O}_p\left(T^{-1/2}\right)$
  which shows $\|H\|={\cal O}_p(1)$.
  Note all the following results only require $\|H\|={\cal O}_p(1)$ and $\|H^{-1}\|={\cal O}_p(1)$, but apart from that
  are independent of the choice of normalization.

  The advantage of expressing $\widehat f$ in terms of $P_{\widehat f}$ as above is that
  the result $\left\| P_{\widehat f} - P_{f^0} \right\| = {\cal O}_p\left(T^{-1/2}\right)$
  of Lemma~\ref{lemma:Pfhat} immediately implies
  \begin{align*}
     \left\| \widehat f - f^0 \, H \right\| &= {\cal O}_p\left(1\right) \; .
  \end{align*}
  The FOC wrt $\lambda$ in the minimization of the first line in equation \eqref{LNT123} reads
  \begin{align}
     \widehat \lambda \, \widehat f' \widehat f &= \left(Y-\sum_{k=1}^{K} \widehat \beta_{k} X_{k} \right) \widehat f \; ,
     \label{hatf_close}
  \end{align}
  which yields
  \begin{align*}
    \widehat \lambda &= \left[ \lambda^0 f^{0\prime} - \sum_{k=1}^{K} \left( \widehat \beta_{k} - \beta^0_k \right) X_{k} \right]
                   \widehat f \left(\widehat f' \widehat f\right)^{-1}
      \nonumber \\
                &= \left[ \lambda^0 f^{0\prime} + \sum_{k=1}^{K} \left( \beta^0_k - \widehat \beta_{k} \right) X_{k} + e\right]
                   P_{\widehat f} f^0 \left(f^{0\prime} P_{\widehat f} f^0\right)^{-1} \, \left(H'\right)^{-1}
      \nonumber \\
                &= \lambda^0 \, \left(H'\right)^{-1}
            + \lambda^0 f^{0\prime} \left( P_{\widehat f}-P_{f^0} \right) f^0
                                    \left(f^{0\prime} P_{\widehat f} f^0\right)^{-1} \, \left(H'\right)^{-1}
      \nonumber \\ & \qquad \qquad \qquad
            + \lambda^0 f^{0\prime} f^0
                 \left[ \left(f^{0\prime} P_{\widehat f} f^0\right)^{-1} - \left(f^{0\prime} f^0\right)^{-1} \right]
                       \, \left(H'\right)^{-1}
      \nonumber \\ & \qquad \qquad \qquad
            + \left[ \sum_{k=1}^{K} \left( \beta^0_k - \widehat \beta_{k} \right) X_{k} + e\right]
                   P_{\widehat f} f^0 \left(f^{0\prime} P_{\widehat f} f^0\right)^{-1} \, \left(H'\right)^{-1} \; .
  \end{align*}
  We have
  $\left(f^{0\prime} P_{\widehat f} f^0\right/T)^{-1} - \left(f^{0\prime} f^0/T\right)^{-1}={\cal O}_p(T^{-1/2})$, because
  $\left\| P_{\widehat f} - P_{f^0} \right\| = {\cal O}_p\left(T^{-1/2}\right)$
  and $f^{0\prime} f^0/T$ by assumption is converging to a positive definite matrix (or given our particular choice of
  normalization is just the identity matrix $\mathbb{I}_R$).
  In addition, we have $\|e\|={\cal O}_p(\sqrt{T})$, $\|X_k\|={\cal O}_p(\sqrt{NT})$ and
  by corollary \ref{lemma:sqrtNTcons} also $\|\widehat \beta - \beta^0\|={\cal O}_p(1/\sqrt{NT})$. Therefore
  \begin{align}
     \left\| \widehat \lambda - \lambda^0 \, \left(H'\right)^{-1} \right\| &= {\cal O}_p\left(1\right) \; ,
     \label{hatlambda_close}
  \end{align}
  which is what we wanted to prove.

  Next, we want to show
  \begin{align}
     \label{lambdafsquare}
     \left\| \left( \frac{\widehat \lambda^{\prime} \, \widehat \lambda} N \right)^{-1}
   - \left( \frac{ \left(H\right)^{-1} \,\lambda^{0\prime} \, \lambda^0 \, \left(H'\right)^{-1}} N \right)^{-1} \right\|
       = {\cal O}_p\left(N^{-1/2}\right) \; ,
     \nonumber \\
     \left\| \left( \frac{\widehat f^{\prime} \, \widehat f} T \right)^{-1}
   -  \left( \frac{H' \,f^{0\prime} \, f^0 \,  H} T \right)^{-1} \right\|
       = {\cal O}_p\left(T^{-1/2}\right) \; .
  \end{align}
  Let $A=N^{-1} \, \widehat \lambda^{\prime} \, \widehat \lambda$ and
  $B=N^{-1} \, \left(H\right)^{-1} \, \lambda^{0\prime} \, \lambda^0 \, \left(H'\right)^{-1}$.
  Using \eqref{hatlambda_close} we find
  \begin{align*}
     \| A-B \| &= \frac 1 {2N} \left\|
                         \left[ \widehat \lambda^{\prime} + \left(H\right)^{-1} \, \lambda^{0\prime} \right]
                         \left[ \widehat \lambda - \lambda^0 \, \left(H'\right)^{-1} \right]
                        +\left[ \widehat \lambda^{\prime} - \left(H\right)^{-1} \, \lambda^{0\prime} \right]
                         \left[ \widehat \lambda + \lambda^0 \, \left(H'\right)^{-1} \right] \right\|
           \nonumber \\
               &= N^{-1} \, {\cal O}_p(N^{1/2}) \, {\cal O}_p(1) = {\cal O}_p \left( N^{-1/2} \right) \; .
  \end{align*}
  By assumption \ref{ass:A1} we know
  \begin{align*}
      \left\| \left( \frac{\lambda^{0\prime} \, \lambda^0} N \right)^{-1} \right\| &= {\cal O}_p(1) \; ,
  \end{align*}
  and thus also $\left\|B^{-1}\right\|= {\cal O}_p(1)$, and therefore $\left\|A^{-1}\right\|= {\cal O}_p(1)$
  (using $\| A-B \|=o_p(1)$ and applying Weyl's inequality to the smallest eigenvalue of $B$).
  Because $A^{-1} - B^{-1} = A^{-1} (B-A) B^{-1}$ we find
  \begin{align*}
     \left\| A^{-1} - B^{-1} \right\| &\leq  \left\|A^{-1}\right\| \, \left\|B^{-1}\right\| \, \left\|A-B\right\|
                        \nonumber \\
                                      &= {\cal O}_p\left( N^{-1/2} \right) \; .
  \end{align*}
  Thus, we have shown the first statement of \eqref{lambdafsquare}, and analogously one can show the second one.
  Combining \eqref{hatlambda_close}, \eqref{hatf_close} and \eqref{lambdafsquare} we obtain
  \begin{align*}
     & \left\| \frac{\widehat \lambda}{\sqrt{N}} \, \left( \frac{\widehat \lambda^{\prime}\widehat\lambda}{N} \right)^{-1}
            \, \left(\frac{\widehat f^{\prime}\widehat f}T\right)^{-1} \, \frac{\widehat f^{\prime}}{\sqrt{T}}
        - \frac{\lambda^0}{\sqrt{N}} \, \left(\frac{\lambda^{0\prime}\lambda^0}N\right)^{-1} \,
             \left(\frac{f^{0\prime}f^0}T\right)^{-1} \, \frac{f^{0\prime}} {\sqrt{T}} \right\|
    \nonumber \\  & =
       \left\| \frac{\widehat \lambda}{\sqrt{N}} \, \left( \frac{\widehat \lambda^{\prime}\widehat\lambda}{N} \right)^{-1}
            \, \left(\frac{\widehat f^{\prime}\widehat f}T\right)^{-1} \, \frac{\widehat f^{\prime}}{\sqrt{T}}
        - \frac{\lambda^0 \left(H'\right)^{-1}}{\sqrt{N}} \,
                \left(\frac{\left(H\right)^{-1}\lambda^{0\prime}\lambda^0 \left(H'\right)^{-1}}N\right)^{-1} \,
             \left(\frac{H' f^{0\prime}f^0 H}T\right)^{-1} \, \frac{H' f^{0\prime}} {\sqrt{T}} \right\|
    \nonumber \\  & \qquad \qquad =
       {\cal O}_p\left( N^{-1/2} \right) \; ,
  \end{align*}
  which is equivalent to the statement in the lemma.
  Note also
  $\widehat \lambda \, (\widehat \lambda^{\prime}\widehat\lambda)^{-1} \, (\widehat f^{\prime}\widehat f)^{-1} \, \widehat f^{\prime}$
  is independent of $H$, i.e., independent of the choice of normalization.
\end{proof}

\begin{proof}[\bf Proof of Lemma~\ref{lemma:exp}]
   \# Part A of the proof: We start by showing
   \begin{align}
       N^{-1} \, \left\|
           \mathbb{E}_{\cal C}\left[ e'  X_k - \left( e'  X_k \right)^{\rm truncR} \right] \right\|
                         &= o_p(1) \; .
       \label{exp_proof_part1}
   \end{align}
   Let $A=e'  X_k$ and $B=A-A^{\rm truncR}$.
   By definition of the left-sided truncation (using the truncation kernel $\Gamma(.)$ defined above)
   we have $B_{t\tau}=0$ for $t<\tau\leq t+M$ and $B_{t\tau}=A_{t\tau}$ otherwise.
   By assumption~\ref{ass:A5} we have $\mathbb{E}_{\cal C}(A_{t\tau})=0$ for $t \geq \tau$.
   For $t<\tau$ we have
   $\mathbb{E}_{\cal C}(A_{t\tau}) = \sum_{i=1}^N  \mathbb{E}_{\cal C}(e_{it}  X_{k,i\tau})$.
   We thus have $\mathbb{E}_{\cal C}(B_{t\tau})=0$ for $\tau\leq t+M$,
   and $\mathbb{E}_{\cal C} B_{t\tau} =  \sum_{i=1}^N  \mathbb{E}_{\cal C}(e_{it}  X_{k,i\tau}) $ for $\tau > t+M$.
   Therefore
   \begin{align*}
     \left\| \mathbb{E}_{\cal C}(B) \right\|_1  &= \max_{t=1\ldots T} \, \sum_{\tau=1}^T |\mathbb{E}_{\cal C}(B_{t\tau})|
                         \nonumber \\
                           &\leq    \max_{t=1\ldots T} \, \sum_{\tau=t+M+1}^T \,
                           \left|   \sum_{i=1}^N  \mathbb{E}_{\cal C}(e_{it}  X_{k,i\tau}) \right|
                     \leq     N  \max_{t=1\ldots T} \, \sum_{\tau=t+M+1}^T    c \, (\tau - t)^{- (1+ \epsilon)}
                            = o_p(N) \; ,
   \end{align*}
   where we used $M\rightarrow \infty$.
   Analogously we can show $\left\| \mathbb{E}_{\cal C}(B) \right\|_\infty  = o_p(N)$.
   Using part (vii) of Lemma~\ref{lemma:inequalities} we therefore also find
   $\left\| \mathbb{E}_{\cal C}(B) \right\| = o_p(N)$,
   which is equivalent to equation \eqref{exp_proof_part1} we wanted to show in this part of the proof.
   Analogously we can show
   \begin{align*}
       N^{-1} \, \left\|
           \mathbb{E}_{\cal C}\left[ e'  e - \left( e'  e \right)^{\rm truncD} \right] \right\|
                         &= o_p(1) \; ,
       \nonumber \\
       T^{-1} \, \left\|
           \mathbb{E}_{\cal C}\left[ e   e' - \left( e   e' \right)^{\rm truncD} \right] \right\|
                         &= o_p(1) \; .
   \end{align*}

   \# Part B of the proof: Next, we want to show
   \begin{align}
       N^{-1} \, \left\|
            \left[ e'  X_k \, - \,
                    \mathbb{E}_{\cal C}\left(e'  X_k\right) \right]^{\rm truncR} \right\|
                         &= o_p(1) \; .
       \label{exp_proof_part2}
   \end{align}
   Using Lemma~\ref{Lemma:Trunc} we have
   \begin{align*}
       N^{-1} \left\| \left[ e'  X_k \, - \,
                    \mathbb{E}_{\cal C}\left(e'  X_k\right) \right]^{\rm truncR} \right\|
          &\leq M \, \max_t \, \max_{t<\tau\leq t+M} \, N^{-1} \, \left| e_t'  X_{k,\tau}
                                                - \mathbb{E}_{\cal C} \left( e_t'  X_{k,\tau} \right) \right|
         \nonumber \\
          &\leq M \, \max_t \, \max_{t<\tau\leq t+M} \, N^{-1} \, \left| \sum_{i=1}^N \left[ e_{it} X_{k,i\tau}
                                       - \mathbb{E}_{\cal C} \left( e_{it} X_{k,i\tau} \right) \right] \right|
         \nonumber \\
          &\leq M \, N^{-1/2} \, \max_t \, \max_{t<\tau\leq t+M} \, \left| \bar Z^{(1)}_{k,t\tau} \right| .
   \end{align*}
   According to Lemma~\ref{lemma:barZ} we know $\mathbb{E}_{\cal C} \left| \bar Z^{(1)}_{k,t\tau} \right|^4$
   is bounded uniformly across $t$ and $\tau$. Applying Lemma~\ref{lemma:maxRV} we therefore find
   $\max_t \, \max_{t<\tau\leq t+M} \bar Z^{(1)}_{t\tau} = {\cal O}_p((MT)^{1/4})$. Thus we have
   \begin{align*}
      M \, N^{-1/2} \, \max_t \, \max_{t<\tau\leq t+M} \, \left| \bar Z^{(1)}_{t\tau} \right|
           &= {\cal O}_p\left(M \, N^{-1/2} \, (MT)^{1/4}\right) \, = \, o_p(1) \; .
   \end{align*}
   Here we used $M^5/T \rightarrow 0$.
   Analogously we can show
   \begin{align*}
       N^{-1} \, \left\|
            \left[ e'  e \, - \,
                    \mathbb{E}_{\cal C} \left(e'  e\right) \right]^{\rm truncD} \right\|
                         &= o_p(1) \; ,
       \nonumber \\
       T^{-1} \, \left\|
            \left[ e  e' \, - \,
                    \mathbb{E}_{\cal C}  \left(e  e'\right) \right]^{\rm truncD} \right\|
                         &= o_p(1) \; .
   \end{align*}

   \# Part C of the proof: Finally, we want to show
   \begin{align}
       N^{-1} \, \left\|
            \left[ e'  X_k \, - \,
                   \widehat e' \, X_k  \right]^{\rm truncR} \right\|
                         &= o_p(1) \; .
       \label{exp_proof_part3}
   \end{align}
   According to theorem \ref{theorem:expansions} we have
   $\widehat e =M_{\lambda^0}  e  M_{f^0}   + e_{\rm rem}$,
   where $e_{\rm rem} \equiv \widehat e^{(1)}_e
            - \sum_{k=1}^K \left( \widehat \beta_k - \beta^0_k \right) \widehat e^{(1)}_k
            + \widehat e^{({\rm rem})}$.
   We then have
   \begin{align*}
      & N^{-1} \, \left\|
            \left[ e'  X_k \, - \,
                   \widehat e' \, X_k  \right]^{\rm truncR} \right\|
      \\
       &\leq             N^{-1} \, \left\|
            \left[ e_{\rm rem}'  X_k    \right]^{\rm truncR} \right\|
          +   N^{-1} \, \left\|
            \left[ P_{f^0} e'  M_{\lambda^0}  X_k   \right]^{\rm truncR} \right\|
          +     N^{-1} \, \left\|
            \left[ e'  P_{\lambda^0} X_k   \right]^{\rm truncR} \right\| .
   \end{align*}
  Using
   corollary \ref{lemma:sqrtNTcons} we find the remainder term satisfies
   $\| e_{\rm rem} \| = {\cal O}_p(1)$.     Using Lemma~\ref{Lemma:Trunc} we find
   \begin{align*}
       N^{-1} \, \left\|
            \left[  e'_{\rm rem} \, X_k  \right]^{\rm truncR} \right\|
          &= \frac M N \, \max_{t,\tau} \, \widehat e'_{{\rm rem},t} \, X_{k,\tau}
      \nonumber \\
          &\leq \frac M N \, \max_{t,\tau} \, \|  e_{{\rm rem},t} \| \,  \| X_{k,\tau} \|
      \nonumber \\
          &\leq \frac M N \, \|   e_{\rm rem} \| \, \max_{\tau}  \| X_{k,\tau} \|
      \nonumber \\
          &\leq \frac M N  {\cal O}_p(1) {\cal O}_p(N^{1/2} T^{1/8}) = o_p(1) \; ,
   \end{align*}
   where we used the fact that the norm of each column $ e_{{\rm rem},t}$
   is smaller than the operator norm of the whole matrix $ e_{\rm rem}$.
   In addition we used Lemma~\ref{lemma:maxRV}
   and the fact that $N^{-1/2} \, \| X_{k,\tau} \| = \sqrt{ N^{-1} \sum_{i=1}^N X_{k,i\tau}^2 }$
   has finite 8'th moment to show $\max_{\tau}  \| X_{k,\tau} \| = {\cal O}_p(N^{1/2} T^{1/8})$.
      Using again Lemma~\ref{Lemma:Trunc} we find
   \begin{align*}
      N^{-1} \, \left\| \left[ P_{f^0} e' M_{\lambda^0} X_k  \right]^{\rm truncR} \right\|
        &\leq  N^{-1} \,  M \, \max_{t,\tau=1\ldots T}  \,
              \left| f^0_t \, (f^{0\prime} \, f^0)^{-1} \, f^{0\prime} \, e' M_{\lambda^0} X_{k,\tau} \right|
       \nonumber \\
        &\leq  N^{-1} \, M \, \|e\| \, \|f^0\| \, \left\|(f^{0\prime} \, f^0)^{-1} \right\| \,
                \max_t \, \| f^0_t \| \, \max_{\tau} \| X_{k,\tau} \|
       \nonumber \\
        &=  N^{-1} \, M \, {\cal O}_p(N^{1/2}) \, {\cal O}_p(T^{1/2}) \, {\cal O}_p(T^{-1})
               \, {\cal O}_p(N^{1/2} T^{1/8}) = o_p(1) \; ,
    \end{align*}
    and
    \begin{align*}
          \, \left\|
            \left[   e'  P_{\lambda^0} X_k   \right]^{\rm truncR} \right\|
            &\leq N^{-1/2} M \max_{t =1\ldots T}
             \left( N^{-1/2} \sum_i e_{it} \lambda^0_i \right)
             (N^{-1} \lambda^{0\prime} \, \lambda^0)^{-1}
            \max_{\tau =1\ldots T}    \left(  N^{-1} \sum_j \lambda^{0 \prime}_j  X_{jt}  \right)
         \\
           &= N^{-1/2} M {\cal O}_p(T^{1/8})   {\cal O}_p(1)  {\cal O}_p(T^{1/8})    = o_p(1).
    \end{align*}
   Thus, we proved equation \eqref{exp_proof_part3}.
    Analogously we obtain
   \begin{align*}
       N^{-1} \, \left\|
            \left[ e'  e \, - \,
                   \widehat e' \, \widehat e  \right]^{\rm truncD} \right\|
                         &= o_p(1) \; ,
      \nonumber \\
       T^{-1} \, \left\|
            \left[ e  e' \, - \,
                   \widehat e \, \widehat e'  \right]^{\rm truncD} \right\|
                         &= o_p(1) \; .
   \end{align*}

  \# Combining \eqref{exp_proof_part1}, \eqref{exp_proof_part2}, and \eqref{exp_proof_part3},
  we obtain
 $N^{-1} \, \left\| \mathbb{E}_{\cal C}(e'   X_k    ) -
       \left( \widehat e' \, X_k \right)^{\rm truncR} \right\|
                         = o_p(1)$. The proof of the other two statements of the lemma is analogous.
\end{proof}

\begin{proof}[\bf Proof of Lemma~\ref{lemma:normsTrunc}]
   Using theorem \ref{theorem:expansions} and \ref{lemma:sqrtNTcons} we find
   $\| \widehat e \| = {\cal O}_p(N^{1/2})$.
   Applying Lemma~\ref{Lemma:Trunc} we therefore find
   \begin{align*}
      N^{-1} \, \left\| \left( \widehat e' \, X_k \right)^{\rm truncR} \right\|
            &\leq \frac M N \, \max_{t,\tau} \, \left| \widehat e'_t \, X_{k,\tau} \right|
      \nonumber \\
          &\leq \frac M N \, \max_{t,\tau} \, \| \widehat e_{t} \| \,  \| X_{k,\tau} \|
      \nonumber \\
          &\leq \frac M N \, \| \widehat e \| \, \max_{\tau}  \| X_{k,\tau} \|
      \nonumber \\
          &\leq \frac M N  {\cal O}_p(N^{1/2}) {\cal O}_p(N^{1/2} T^{1/8}) = {\cal O}_p(M T^{1/8}) \; ,
   \end{align*}
   where we used the result $\max_{\tau}  \| X_{k,\tau} \| = {\cal O}_p(N^{1/2} T^{1/8})$ that was
   already obtained in the proof of the last theorem.

   The proof for the statement (ii) and (iii) is analogous.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proofs for Section \ref{sec:testing} (Testing)}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}[\bf Proof of Theorem \ref{th:gradient}]
   Using the expansion for $L_{NT}(\beta)$ in Lemma S.1
   in the supplementary material 
   of Moon and Weidner~\cite*{MoonWeidner2015} we find for the derivative
   (the sign convention $\epsilon_k=\beta^0_k - \beta_k$ results in the minus sign below)
   \begin{align*}
      \frac{\partial L_{NT}}{\partial \beta_k}
       &= \, - \, \frac{1} {NT} \, \sum_{g=2}^\infty \, g \,
      \sum_{\kappa_1=0}^K \, \sum_{\kappa_2=0}^K \, \ldots \sum_{\kappa_{g-1}=0}^K \,
     \epsilon_{\kappa_1} \, \epsilon_{\kappa_2} \, \ldots \, \epsilon_{\kappa_{g-1}}
      \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, X_{\kappa_1}, \ldots
     ,X_{\kappa_{g-1}}\right)
      \nonumber \\
      &=  \left[ 2 W_{NT} (\beta-\beta^0) \right]_k  - \, \frac 2 {\sqrt{NT}} C_{NT,k}
           + \frac 1 {NT} \nabla R_{1,NT,k} + \frac 1 {NT} \nabla R_{2,NT,k} \; ,
   \end{align*}
   where
   \begin{align*}
      W_{NT,k_1 k_2} &= \frac{1} {NT} \, L^{(2)}\left(\lambda^0,\, f^0,\, X_{k_1},\, X_{k_2} \right)  \; ,
     \nonumber \\
      C_{NT,k} &= \frac 1 {2\sqrt{NT}} \, \sum_{g=2}^{G_e} \, g \, (\epsilon_{0})^{g-1}
                         \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, X_{0}, \ldots ,X_{0}\right)
     \nonumber \\
               &=  \sum_{g=2}^{G_e} \, \frac g {2\sqrt{NT}} \,
                         \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, e, \ldots ,e\right) \; ,
   \end{align*}
   and
   \begin{align*}
      \nabla R_{1,NT,k}
               &= \, - \, \sum_{g=G_e+1}^\infty \, g \,
        (\epsilon_{0})^{g-1}
      \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, X_{0}, \ldots ,X_{0}\right)  \; ,
     \nonumber \\
               &= \, - \, \sum_{g=G_e+1}^\infty \, g \,
      \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, e, \ldots ,e\right)  \; ,
     \nonumber \\
      \nabla R_{2,NT,k}
          &= \, - \, \sum_{g=3}^\infty \, g \, \sum_{r=1}^{g-1} \, {g-1 \choose r}
      \sum_{k_1=1}^K \, \ldots \sum_{{k_r}=1}^K \,
      \epsilon_{k_1} \, \ldots \,\epsilon_{k_r} \, (\epsilon_0)^{g-r-1} \,
    \nonumber \\
        & \qquad \qquad \qquad \qquad \qquad \qquad
      \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, X_{k_1}, \ldots
     ,X_{k_r},X_0,\ldots,X_0\right)  \; .
    \nonumber \\
          &= \, - \, \sum_{g=3}^\infty \, g \, \sum_{r=1}^{g-1} \, {g-1 \choose r}
      \sum_{k_1=1}^K \, \ldots \sum_{{k_r}=1}^K \,
      \,(\beta^0_{k_1}-\beta_{k_1}) \, \ldots \, (\beta^0_{k_r}-\beta_{k_r})
    \nonumber \\
        & \qquad \qquad \qquad \qquad \qquad \qquad
      \, L^{(g)}\left(\lambda^0,\, f^0,\, X_{k},\, X_{k_1}, \ldots
     ,X_{k_r},e,\ldots,e\right)  \; .
   \end{align*}
   The above expressions for $W_{NT}$ and $C_{NT}$ are equivalent to their definitions given in theorem
   \ref{th:ass_expand}. Using the bound on $L^{(g)}$ and ${{n \choose k}} \leq 4^n$
   we find 
   \begin{align*}
      |\nabla R_{1,NT,k}|
               &\leq  c_0 \, NT \,  \frac{\|X_k\|}{\sqrt{NT}} \, \sum_{g=G_e+1}^\infty \, g^2 \,
                                \left(\frac{c_1 \|e\|}{\sqrt{NT}} \right)^{g-1}
    \nonumber \\
               &\leq 2 \, c_0 \, (1+G_e)^2 \, NT \,  \frac{\|X_k\|}{\sqrt{NT}} \,
                         \left(\frac{c_1 \|e\|}{\sqrt{NT}} \right)^{G_e} \,
                          \left[1-\left(\frac{c_1 \|e\|}{\sqrt{NT}} \right)\right]^{-3}
               = o_p(\sqrt{NT}) \; ,
    \nonumber \\
      |\nabla R_{2,NT,k}|
          &\leq c_0 \, NT \, \frac{\|X_{k}\|}{\sqrt{NT}} \,
                \sum_{g=3}^\infty \, g^2 \, \sum_{r=1}^{g-1} \, {g-1 \choose r}
                        \, c_1^{g-1} \,
                   \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_k|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                             \right)
              \nonumber \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \times
                    \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_k|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                            + \frac{\|e\|}{\sqrt{NT}} \right)^{g-2}
        \nonumber \\
          &\leq c_0 \, NT \, \frac{\|X_{k}\|}{\sqrt{NT}} \,
                \sum_{g=3}^\infty \, g^3
                        \, (4c_1)^{g-1} \,
                   \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_k|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                             \right)
                    \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_{\widetilde k}|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                            + \frac{\|e\|}{\sqrt{NT}} \right)^{g-2}
        \nonumber \\
          &\leq c_2 \, NT \, \frac{\|X_{k}\|}{\sqrt{NT}} \,
                    \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_k|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                             \right)
                     \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_{\widetilde k}|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                            + \frac{\|e\|}{\sqrt{NT}} \right) \; ,
  \end{align*}
  where $c_0=8 R d_{\max}(\lambda^0,f^0)/2$ and $c_1=16 d_{\max}(\lambda^0,f^0)/d_{\min}^2(\lambda^0,f^0)$
  both converge to a constants as $N,T \rightarrow \infty$, and the very last inequality is only true
  if $4 c_1 \left( \sum_{\widetilde k=1}^K |\beta_{\widetilde k}-\beta^0_{\widetilde k}|  \frac{\|X_{\widetilde k}\|}{\sqrt{NT}}
                            + \frac{\|e\|}{\sqrt{NT}} \right)<1$, and $c_2>0$ is an appropriate positive constant.
  To show $\nabla R_{1,NT,k}=o_p(NT)$ we used Assumption~\ref{ass:A3}$^*$.
  From the above inequalities we find for $\eta_{NT} \rightarrow \infty$
  \begin{align*}
      \sup_{\{\beta :\left\| \beta -\beta^{0} \right\| \leq \eta_{NT}\}}
                  \frac{ \left\| \nabla R_{1,NT}(\beta) \right\| }
       { \sqrt{NT} } = o_{p}\left( 1 \right) ,
     \nonumber \\
      \sup_{\{\beta :\left\| \beta -\beta^{0} \right\| \leq \eta_{NT}\}}
                  \frac{ \left\| \nabla R_{2,NT}(\beta) \right\| }
       { NT \, \left\| \beta -\beta^{0} \right\|  } = o_{p}\left( 1 \right) .
  \end{align*}
  Thus $R_{NT}(\beta)=R_{1,NT}(\beta)+R_{2,NT}(\beta)$ satisfies the bound
  in the theorem.
\end{proof}
\begin{proof}[\bf Proof of Theorem \ref{th:testing}]
   Using Theorem~\ref{th:limdis} it is straightforward
   to show $WD_{NT}^*$ has limiting distribution $\chi^2_r$.

    For the LR test we have to show the estimator
   $\widehat c = (NT)^{-1} {\rm Tr} (\widehat e(\widehat \beta) \, \widehat e'(\widehat \beta))$
   is consistent for  $c=\mathbb{E}_{\cal C} e_{it}^2$.
   As already noted in the main text we have $\widehat c = L_{NT}\left( \widehat \beta \right)$,
   and using our expansion and $\sqrt{NT}$-consistency of $\widehat \beta$ we immediately obtain
   \begin{align*}
      \widehat c &= \frac 1 {NT} \,  {\rm Tr} (M_{\lambda^0} e M_{f^0} e') + o_p(1) \; .
   \end{align*}
   Alternatively, one could use the expansion of $\widehat e$ in Theorem~\ref{theorem:expansions} to show this.
   From the above result we find
   \begin{align*}
      \left| \widehat c - \frac 1 {NT} {\rm Tr} (ee') \right|
          &= \frac 1 {NT} \left|
                   {\rm Tr} (P_{\lambda^0} e M_{f^0} e')
                   + {\rm Tr} (e P_{f^0} e') \right| + o_p(1)
       \nonumber \\
             &\leq \frac {2R} {NT} \, \|e\|^2 + o_p(1) = o_p(1)  \; .
   \end{align*}
   By the weak law of large numbers we thus have
   \begin{align*}
      \widehat c = \frac 1 {NT} \sum_{i=1}^N \sum_{t=1}^T e_{it}^2 + o_p(1) = c + o_p(1) \; ,
   \end{align*}
   i.e., $\widehat c$ is indeed consistent for $c$.
   Having this one immediately obtains the result
   for the limiting distribution of $LR_{NT}^*$.


    For the LM test we first want to show equation \eqref{EquivGrads} holds.
      Using the expansion of $\widehat e$ in Theorem~\ref{theorem:expansions} one obtains
      \begin{align*}
         \sqrt{NT} (\widetilde \nabla {\cal L}_{NT})_k \, &= \, - \, \frac 2 {\sqrt{NT}} \,  {\rm Tr}\left(X'_k \widetilde e\right)
             \nonumber \\
                     &= \left[ 2 \, \sqrt{NT}
                \, W_{NT} \, \left( \widetilde \beta - \beta^0\right)  \right]_k + \frac 2 {NT} C^{(1)}(\lambda^0,f^0,X_k,e)
                                  + \frac 2 {NT} C^{(2)}(\lambda^0,f^0,X_k,e)
                       \nonumber \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
                                     \qquad \qquad \qquad \qquad
                         \, - \, \frac 2 {\sqrt{NT}} \,  {\rm Tr}\left(X'_k \widetilde e^{(\rm rem)}\right)
                \nonumber \\
                     &= \left[ 2 \,\sqrt{NT} \, W_{NT} \, \left( \widetilde \beta - \beta^0\right) + \frac 2 {NT} C_{NT}
                        \right]_k + o_p(1)
                \nonumber \\
                     &= \sqrt{NT}  \left[ \nabla L_{NT}(\widetilde \beta) \right]_k  + o_p(1) \, ,
      \end{align*}
      which is what we wanted to show.
      Here we used $|{\rm Tr}\left(X'_k \widetilde e^{(\rm rem)}\right)| \leq 7 R \|X_k\| \|\widetilde e^{(\rm rem)}\|
                                               = {\cal O}_p(N^{3/2})$.
      Note that $\|X_k\|={\cal O}_p(N)$, and Theorem~\ref{theorem:expansions}, and $\sqrt{NT}$-consistency of
       $\widetilde \beta$, together imply $\|\widetilde e^{(\rm rem)}\| = {\cal O}_p(\sqrt{N})$.
       We also used the expression for $\nabla L_{NT}(\widetilde \beta)$ given in Theorem~\ref{th:gradient},
       and the bound on $\nabla R_{NT}(\beta)$ given there.

    We now use
    equation \eqref{limNablaL} and $\widetilde W=W+o_p(1)$, $\widetilde \Omega=\Omega+o_p(1)$, and $\widetilde B=B+o_p(1)$
   to obtain
   \begin{align*}
      LM^*_{NT} \; \;  \limfunc{\longrightarrow}_d \; \;
                  (C - B)'
                   W^{-1} H' (H W^{-1} \Omega W^{-1} H')^{-1} H W^{-1}
                      (C - B) \; .
   \end{align*}
   Under $H_0$ we thus find $LM^*_{NT} \; \limfunc{\rightarrow}_d \; \chi^2_r$.
\end{proof}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Additional Monte Carlo Results}
%\label{sec:AddMC}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We consider an ${\rm AR}(1)$ model with $R$ factors
\begin{align*}
   Y_{it} \, &= \, \rho^0 \, Y_{i,t-1} \, + \, \sum_{r=1}^R \lambda^0_{ir} \, f^0_{tr} \, + \, e_{it} \; .
\end{align*}
We draw the $e_{it}$ independently and identically distributed from a t-distribution with
five degrees of freedom.
The   $\lambda^0_{ir}$ are  independently distributed as ${\cal N}(1,1)$, and we generate the factors
from an ${\rm AR}(1)$ specification, namely $f^0_{tr}=\rho_f \, f^0_{t-1,r} + u_{tr}$, for each $r=1,\ldots,R$,
where $u_{tr} \sim {\rm iid} {\cal N}(0,(1-\rho_f^2)\sigma_f^2)$.   For all simulations we generate 1,000 initial time periods for $f^0_t$ and $Y_{it}$ that are not used for estimation.
  This guarantees the simulated data used for estimation are distributed according to the stationary distribution
  of the model.
  
For $R=1$ this is exactly the simulation design used in the main text Monte Carlo section, but DGPs with $R>1$
were not considered in the main text.
Table~\ref{tab:extra1} reports results for which $R=1$ is used both in the DGP and for the LS estimation.
Table~\ref{tab:extra2} reports results for which $R=1$ is used in the DGP, but $R=2$ is used for the LS estimation.
Table~\ref{tab:extra3} reports results for which $R=2$ is used both in the DGP and for the LS estimation.
The results in Table~\ref{tab:extra1} and~\ref{tab:extra2} are identical to those reported in the main text
Table~\ref{tab:T1} and~\ref{tab:T2}, except we also report results for the CCE estimator.
The results in Table~\ref{tab:extra3} are not contained in the main text.

The CCE estimator is obtained by using $\widehat f^{\rm proxy}_t =  N^{-1} \sum_i ( Y_{it}, \,   Y_{i,t-1} )'$ as a proxy for the
factors and then estimating the parameters $\rho$, $\lambda_{i1}$, $\lambda_{i2}$, $i=1,\ldots,N$,
via OLS in the linear regression model
$ Y_{it}  =  \rho   Y_{i,t-1}  +   \lambda_{i1}  \widehat f^{\rm proxy}_{t1}
   +   \lambda_{i2}  \widehat f^{\rm proxy}_{t2}  +  e_{it} $.
   
The performance of the CCE estimator in Table~\ref{tab:extra1} and~\ref{tab:extra2} are identical
(up to random MC noise), because the number of factors need not be specified for the CCE estimator, and the DGPs
in Table~\ref{tab:extra1} and~\ref{tab:extra2} are identical. These tables show for $R=1$ in the DGP,
the CCE estimator performs very well. From Chudik and Pesaran~\cite*{ChudikPesaran2015} we expect the
CCE estimator to have a bias of order $1/T$ in a dynamic model, which is confirmed in the simulations: 
the bias of the CCE estimator shrinks roughly in inverse proportion to $T$, as $T$ becomes larger.
The $1/T$ bias of the CCE estimator could be corrected for, and we would expect the bias-corrected CCE estimator
to perform similarly to the bias-corrected LS estimator.

However, if there are $R=2$ factors in the true DGP, then it turns out the proxies $\widehat f^{\rm proxy}_t$
do not pick those up correctly.
Table~\ref{tab:extra3} shows for some parameter values and sample sizes (e.g., $\rho^0=0.3$ and $T=10$,
or $\rho^0 = 0.9$ and $T=40$) the CCE estimator is almost unbiased, but for other values, including $T=80$,
the CCE estimator is heavily biased if $R=2$. In particular,   the bias of the CCE estimator does not seem to converge
to zero as $T$ becomes large in this case. By contrast, the correctly specified  LS estimators (i.e., correctly 
using $R=2$ factors in the estimation)
performs very well according to Table~\ref{tab:extra3}. However, an incorrectly specified
LS estimator, which would underestimate the number of factors (e.g., using $R=1$ factors in estimation 
instead of the correct number $R=2$)
would probably perform similarly to the CCE estimator, because not all factors would be corrected for.
Overestimating the number of factors (i.e., using $R=3$ factors in estimation 
instead of the correct number $R=2$) should, however, not pose a problem for the LS estimator, according to Moon and Weidner~\cite*{MoonWeidner2015}.


\theendnotes

\begin{thebibliography}{}

\bibitem[\protect\astroncite{Bai}{2009}]{Bai2009}
Bai, J. (2009).
\newblock Panel data models with interactive fixed effects.
\newblock {\em Econometrica}, 77(4):1229--1279.

\bibitem[\protect\astroncite{Bhatia}{1997}]{Bhatia97}
Bhatia, R. (1997).
\newblock {\em Matrix {A}nalysis}.
\newblock Springer-Verlag, New York.

\bibitem[\protect\astroncite{Chudik and Pesaran}{2015}]{ChudikPesaran2015}
Chudik, A. and Pesaran, M.~H. (2015).
\newblock Common correlated effects estimation of heterogeneous dynamic panel
  data models with weakly exogenous regressors.
\newblock {\em Journal of Econometrics}.

\bibitem[\protect\astroncite{Galantai and Hegedus}{2006}]{GalantaiHegedus2006}
Galantai, A. and Hegedus, C.~J. (2006).
\newblock {Jordan's principal angles in complex vector spaces}.
\newblock {\em Num. Lin. Alg. Appl.}, 13:589--598.

\bibitem[\protect\astroncite{Golub and Van~Loan}{1996}]{golubvanloan1996}
Golub, G.~H. and Van~Loan, C.~F. (1996).
\newblock {\em Matrix Computations (Johns Hopkins Studies in Mathematical
  Sciences), Third Edition}.
\newblock The Johns Hopkins University Press.

\bibitem[\protect\astroncite{Hall and Heyde}{1980}]{HallHeyde1980}
Hall, P. and Heyde, C. (1980).
\newblock Martingale limit theory and applications.
\newblock {\em Academic Press, New York}.

\bibitem[\protect\astroncite{Latala}{2005}]{Latala2006}
Latala, R. (2005).
\newblock Some estimates of norms of random matrices.
\newblock {\em Proc. Amer. Math. Soc.}, 133:1273--1282.

\bibitem[\protect\astroncite{McLeish}{1974}]{Mcleish1974}
McLeish, D. (1974).
\newblock Dependent central limit theorems and invariance principles.
\newblock {\em the Annals of Probability}, pages 620--628.

\bibitem[\protect\astroncite{Moon and Weidner}{2015}]{MoonWeidner2015}
Moon, H.~R. and Weidner, M. (2015).
\newblock Linear regression for panel with unknown number of factors as
  interactive fixed effects.
\newblock {\em Econometrica}, 83(4):1543--1579.

\bibitem[\protect\astroncite{White}{2001}]{White2001}
White, H. (2001).
\newblock {\em Asymptotic theory for econometricians}.
\newblock Academic press New York.

\end{thebibliography}


\newpage
\section*{Tables}

%%%%%%%%%
% DATA FOR TABLES:
%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% TABLE DATA:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\bOLSEbiasAA}{0.1232}
\newcommand{\bOLSEstdeAA}{0.1444}
\newcommand{\bOLSErmseAA}{0.1898}
\newcommand{\bQMLEbiasAA}{-0.1419}
\newcommand{\bQMLEstdeAA}{0.1480}
\newcommand{\bQMLErmseAA}{0.2050}
\newcommand{\bBCQMbiasAA}{-0.0713}
\newcommand{\bBCQMstdeAA}{0.0982}
\newcommand{\bBCQMrmseAA}{0.1213}
\newcommand{\bCCEPbiasAA}{-0.1755}
\newcommand{\bCCEPstdeAA}{0.1681}
\newcommand{\bCCEPrmseAA}{0.2430}

\newcommand{\bOLSEbiasAB}{0.0200}
\newcommand{\bOLSEstdeAB}{0.0723}
\newcommand{\bOLSErmseAB}{0.0750}
\newcommand{\bQMLEbiasAB}{-0.3686}
\newcommand{\bQMLEstdeAB}{0.1718}
\newcommand{\bQMLErmseAB}{0.4067}
\newcommand{\bBCQMbiasAB}{-0.2330}
\newcommand{\bBCQMstdeAB}{0.1301}
\newcommand{\bBCQMrmseAB}{0.2669}
\newcommand{\bCCEPbiasAB}{-0.3298}
\newcommand{\bCCEPstdeAB}{0.2203}
\newcommand{\bCCEPrmseAB}{0.3966}

\newcommand{\bOLSEbiasAC}{0.1339}
\newcommand{\bOLSEstdeAC}{0.1148}
\newcommand{\bOLSErmseAC}{0.1764}
\newcommand{\bQMLEbiasAC}{-0.0542}
\newcommand{\bQMLEstdeAC}{0.0596}
\newcommand{\bQMLErmseAC}{0.0806}
\newcommand{\bBCQMbiasAC}{-0.0201}
\newcommand{\bBCQMstdeAC}{0.0423}
\newcommand{\bBCQMrmseAC}{0.0469}
\newcommand{\bCCEPbiasAC}{-0.0819}
\newcommand{\bCCEPstdeAC}{0.0593}
\newcommand{\bCCEPrmseAC}{0.1011}

\newcommand{\bOLSEbiasAD}{0.0218}
\newcommand{\bOLSEstdeAD}{0.0513}
\newcommand{\bOLSErmseAD}{0.0557}
\newcommand{\bQMLEbiasAD}{-0.1019}
\newcommand{\bQMLEstdeAD}{0.1094}
\newcommand{\bQMLErmseAD}{0.1495}
\newcommand{\bBCQMbiasAD}{-0.0623}
\newcommand{\bBCQMstdeAD}{0.0747}
\newcommand{\bBCQMrmseAD}{0.0973}
\newcommand{\bCCEPbiasAD}{-0.1436}
\newcommand{\bCCEPstdeAD}{0.0972}
\newcommand{\bCCEPrmseAD}{0.1734}

\newcommand{\bOLSEbiasAE}{0.1441}
\newcommand{\bOLSEstdeAE}{0.0879}
\newcommand{\bOLSErmseAE}{0.1687}
\newcommand{\bQMLEbiasAE}{-0.0264}
\newcommand{\bQMLEstdeAE}{0.0284}
\newcommand{\bQMLErmseAE}{0.0388}
\newcommand{\bBCQMbiasAE}{-0.0070}
\newcommand{\bBCQMstdeAE}{0.0240}
\newcommand{\bBCQMrmseAE}{0.0250}
\newcommand{\bCCEPbiasAE}{-0.0405}
\newcommand{\bCCEPstdeAE}{0.0277}
\newcommand{\bCCEPrmseAE}{0.0491}

\newcommand{\bOLSEbiasAF}{0.0254}
\newcommand{\bOLSEstdeAF}{0.0353}
\newcommand{\bOLSErmseAF}{0.0434}
\newcommand{\bQMLEbiasAF}{-0.0173}
\newcommand{\bQMLEstdeAF}{0.0299}
\newcommand{\bQMLErmseAF}{0.0345}
\newcommand{\bBCQMbiasAF}{-0.0085}
\newcommand{\bBCQMstdeAF}{0.0219}
\newcommand{\bBCQMrmseAF}{0.0235}
\newcommand{\bCCEPbiasAF}{-0.0617}
\newcommand{\bCCEPstdeAF}{0.0406}
\newcommand{\bCCEPrmseAF}{0.0739}

\newcommand{\bOLSEbiasAG}{0.1517}
\newcommand{\bOLSEstdeAG}{0.0657}
\newcommand{\bOLSErmseAG}{0.1654}
\newcommand{\bQMLEbiasAG}{-0.0130}
\newcommand{\bQMLEstdeAG}{0.0170}
\newcommand{\bQMLErmseAG}{0.0214}
\newcommand{\bBCQMbiasAG}{-0.0021}
\newcommand{\bBCQMstdeAG}{0.0160}
\newcommand{\bBCQMrmseAG}{0.0161}
\newcommand{\bCCEPbiasAG}{-0.0200}
\newcommand{\bCCEPstdeAG}{0.0166}
\newcommand{\bCCEPrmseAG}{0.0260}

\newcommand{\bOLSEbiasAH}{0.0294}
\newcommand{\bOLSEstdeAH}{0.0250}
\newcommand{\bOLSErmseAH}{0.0386}
\newcommand{\bQMLEbiasAH}{-0.0057}
\newcommand{\bQMLEstdeAH}{0.0105}
\newcommand{\bQMLErmseAH}{0.0119}
\newcommand{\bBCQMbiasAH}{-0.0019}
\newcommand{\bBCQMstdeAH}{0.0089}
\newcommand{\bBCQMrmseAH}{0.0091}
\newcommand{\bCCEPbiasAH}{-0.0281}
\newcommand{\bCCEPstdeAH}{0.0162}
\newcommand{\bCCEPrmseAH}{0.0324}

\newcommand{\bOLSEbiasAI}{0.1552}
\newcommand{\bOLSEstdeAI}{0.0487}
\newcommand{\bOLSErmseAI}{0.1627}
\newcommand{\bQMLEbiasAI}{-0.0066}
\newcommand{\bQMLEstdeAI}{0.0112}
\newcommand{\bQMLErmseAI}{0.0130}
\newcommand{\bBCQMbiasAI}{-0.0007}
\newcommand{\bBCQMstdeAI}{0.0109}
\newcommand{\bBCQMrmseAI}{0.0109}
\newcommand{\bCCEPbiasAI}{-0.0100}
\newcommand{\bCCEPstdeAI}{0.0111}
\newcommand{\bCCEPrmseAI}{0.0149}

\newcommand{\bOLSEbiasAJ}{0.0326}
\newcommand{\bOLSEstdeAJ}{0.0179}
\newcommand{\bOLSErmseAJ}{0.0372}
\newcommand{\bQMLEbiasAJ}{-0.0026}
\newcommand{\bQMLEstdeAJ}{0.0056}
\newcommand{\bQMLErmseAJ}{0.0062}
\newcommand{\bBCQMbiasAJ}{-0.0006}
\newcommand{\bBCQMstdeAJ}{0.0053}
\newcommand{\bBCQMrmseAJ}{0.0053}
\newcommand{\bCCEPbiasAJ}{-0.0136}
\newcommand{\bCCEPstdeAJ}{0.0073}
\newcommand{\bCCEPrmseAJ}{0.0154}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% TABLE DATA:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\bbOLSEbiasAA}{0.1239}
\newcommand{\bbOLSEstdeAA}{0.1454}
\newcommand{\bbOLSErmseAA}{0.1910}
\newcommand{\bbQMLEbiasAA}{-0.5467}
\newcommand{\bbQMLEstdeAA}{0.1528}
\newcommand{\bbQMLErmseAA}{0.5676}
\newcommand{\bbBCQMbiasAA}{-0.3721}
\newcommand{\bbBCQMstdeAA}{0.1299}
\newcommand{\bbBCQMrmseAA}{0.3942}
\newcommand{\bbCCEPbiasAA}{-0.1767}
\newcommand{\bbCCEPstdeAA}{0.1678}
\newcommand{\bbCCEPrmseAA}{0.2437}

\newcommand{\bbOLSEbiasAB}{0.0218}
\newcommand{\bbOLSEstdeAB}{0.0731}
\newcommand{\bbOLSErmseAB}{0.0763}
\newcommand{\bbQMLEbiasAB}{-0.9716}
\newcommand{\bbQMLEstdeAB}{0.1216}
\newcommand{\bbQMLErmseAB}{0.9792}
\newcommand{\bbBCQMbiasAB}{-0.7490}
\newcommand{\bbBCQMstdeAB}{0.1341}
\newcommand{\bbBCQMrmseAB}{0.7609}
\newcommand{\bbCCEPbiasAB}{-0.3289}
\newcommand{\bbCCEPstdeAB}{0.2203}
\newcommand{\bbCCEPrmseAB}{0.3958}

\newcommand{\bbOLSEbiasAC}{0.1343}
\newcommand{\bbOLSEstdeAC}{0.1145}
\newcommand{\bbOLSErmseAC}{0.1765}
\newcommand{\bbQMLEbiasAC}{-0.1874}
\newcommand{\bbQMLEstdeAC}{0.1159}
\newcommand{\bbQMLErmseAC}{0.2203}
\newcommand{\bbBCQMbiasAC}{-0.1001}
\newcommand{\bbBCQMstdeAC}{0.0758}
\newcommand{\bbBCQMrmseAC}{0.1256}
\newcommand{\bbCCEPbiasAC}{-0.0816}
\newcommand{\bbCCEPstdeAC}{0.0592}
\newcommand{\bbCCEPrmseAC}{0.1008}

\newcommand{\bbOLSEbiasAD}{0.0210}
\newcommand{\bbOLSEstdeAD}{0.0518}
\newcommand{\bbOLSErmseAD}{0.0559}
\newcommand{\bbQMLEbiasAD}{-0.4923}
\newcommand{\bbQMLEstdeAD}{0.1159}
\newcommand{\bbQMLErmseAD}{0.5058}
\newcommand{\bbBCQMbiasAD}{-0.3271}
\newcommand{\bbBCQMstdeAD}{0.0970}
\newcommand{\bbBCQMrmseAD}{0.3412}
\newcommand{\bbCCEPbiasAD}{-0.1414}
\newcommand{\bbCCEPstdeAD}{0.0971}
\newcommand{\bbCCEPrmseAD}{0.1715}

\newcommand{\bbOLSEbiasAE}{0.1451}
\newcommand{\bbOLSEstdeAE}{0.0879}
\newcommand{\bbOLSErmseAE}{0.1696}
\newcommand{\bbQMLEbiasAE}{-0.0448}
\newcommand{\bbQMLEstdeAE}{0.0469}
\newcommand{\bbQMLErmseAE}{0.0648}
\newcommand{\bbBCQMbiasAE}{-0.0168}
\newcommand{\bbBCQMstdeAE}{0.0320}
\newcommand{\bbBCQMrmseAE}{0.0362}
\newcommand{\bbCCEPbiasAE}{-0.0407}
\newcommand{\bbCCEPstdeAE}{0.0277}
\newcommand{\bbCCEPrmseAE}{0.0492}

\newcommand{\bbOLSEbiasAF}{0.0255}
\newcommand{\bbOLSEstdeAF}{0.0354}
\newcommand{\bbOLSErmseAF}{0.0436}
\newcommand{\bbQMLEbiasAF}{-0.1822}
\newcommand{\bbQMLEstdeAF}{0.0820}
\newcommand{\bbQMLErmseAF}{0.1999}
\newcommand{\bbBCQMbiasAF}{-0.1085}
\newcommand{\bbBCQMstdeAF}{0.0528}
\newcommand{\bbBCQMrmseAF}{0.1207}
\newcommand{\bbCCEPbiasAF}{-0.0618}
\newcommand{\bbCCEPstdeAF}{0.0404}
\newcommand{\bbCCEPrmseAF}{0.0739}

\newcommand{\bbOLSEbiasAG}{0.1511}
\newcommand{\bbOLSEstdeAG}{0.0663}
\newcommand{\bbOLSErmseAG}{0.1650}
\newcommand{\bbQMLEbiasAG}{-0.0161}
\newcommand{\bbQMLEstdeAG}{0.0209}
\newcommand{\bbQMLErmseAG}{0.0264}
\newcommand{\bbBCQMbiasAG}{-0.0038}
\newcommand{\bbBCQMstdeAG}{0.0177}
\newcommand{\bbBCQMrmseAG}{0.0181}
\newcommand{\bbCCEPbiasAG}{-0.0199}
\newcommand{\bbCCEPstdeAG}{0.0167}
\newcommand{\bbCCEPrmseAG}{0.0260}

\newcommand{\bbOLSEbiasAH}{0.0300}
\newcommand{\bbOLSEstdeAH}{0.0250}
\newcommand{\bbOLSErmseAH}{0.0390}
\newcommand{\bbQMLEbiasAH}{-0.0227}
\newcommand{\bbQMLEstdeAH}{0.0342}
\newcommand{\bbQMLErmseAH}{0.0410}
\newcommand{\bbBCQMbiasAH}{-0.0128}
\newcommand{\bbBCQMstdeAH}{0.0225}
\newcommand{\bbBCQMrmseAH}{0.0258}
\newcommand{\bbCCEPbiasAH}{-0.0282}
\newcommand{\bbCCEPstdeAH}{0.0164}
\newcommand{\bbCCEPrmseAH}{0.0326}

\newcommand{\bbOLSEbiasAI}{0.1550}
\newcommand{\bbOLSEstdeAI}{0.0488}
\newcommand{\bbOLSErmseAI}{0.1625}
\newcommand{\bbQMLEbiasAI}{-0.0072}
\newcommand{\bbQMLEstdeAI}{0.0123}
\newcommand{\bbQMLErmseAI}{0.0143}
\newcommand{\bbBCQMbiasAI}{-0.0011}
\newcommand{\bbBCQMstdeAI}{0.0115}
\newcommand{\bbBCQMrmseAI}{0.0116}
\newcommand{\bbCCEPbiasAI}{-0.0100}
\newcommand{\bbCCEPstdeAI}{0.0111}
\newcommand{\bbCCEPrmseAI}{0.0149}

\newcommand{\bbOLSEbiasAJ}{0.0325}
\newcommand{\bbOLSEstdeAJ}{0.0182}
\newcommand{\bbOLSErmseAJ}{0.0372}
\newcommand{\bbQMLEbiasAJ}{-0.0030}
\newcommand{\bbQMLEstdeAJ}{0.0064}
\newcommand{\bbQMLErmseAJ}{0.0071}
\newcommand{\bbBCQMbiasAJ}{-0.0010}
\newcommand{\bbBCQMstdeAJ}{0.0057}
\newcommand{\bbBCQMrmseAJ}{0.0058}
\newcommand{\bbCCEPbiasAJ}{-0.0136}
\newcommand{\bbCCEPstdeAJ}{0.0074}
\newcommand{\bbCCEPrmseAJ}{0.0155}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% TABLE EXTRA DATA:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\bbbbOLSEbiasAA}{0.1861}
\newcommand{\bbbbOLSEstdeAA}{0.1562}
\newcommand{\bbbbOLSErmseAA}{0.2429}
\newcommand{\bbbbQMLEbiasAA}{-0.4968}
\newcommand{\bbbbQMLEstdeAA}{0.1910}
\newcommand{\bbbbQMLErmseAA}{0.5322}
\newcommand{\bbbbBCQMbiasAA}{-0.3323}
\newcommand{\bbbbBCQMstdeAA}{0.1580}
\newcommand{\bbbbBCQMrmseAA}{0.3680}
\newcommand{\bbbbCCEPbiasAA}{-0.1002}
\newcommand{\bbbbCCEPstdeAA}{0.2063}
\newcommand{\bbbbCCEPrmseAA}{0.2294}
 
\newcommand{\bbbbOLSEbiasAB}{0.0309}
\newcommand{\bbbbOLSEstdeAB}{0.0801}
\newcommand{\bbbbOLSErmseAB}{0.0859}
\newcommand{\bbbbQMLEbiasAB}{-0.9305}
\newcommand{\bbbbQMLEstdeAB}{0.1644}
\newcommand{\bbbbQMLErmseAB}{0.9449}
\newcommand{\bbbbBCQMbiasAB}{-0.7057}
\newcommand{\bbbbBCQMstdeAB}{0.1754}
\newcommand{\bbbbBCQMrmseAB}{0.7272}
\newcommand{\bbbbCCEPbiasAB}{-0.2750}
\newcommand{\bbbbCCEPstdeAB}{0.2302}
\newcommand{\bbbbCCEPrmseAB}{0.3586}
 
\newcommand{\bbbbOLSEbiasAC}{0.1989}
\newcommand{\bbbbOLSEstdeAC}{0.1185}
\newcommand{\bbbbOLSErmseAC}{0.2315}
\newcommand{\bbbbQMLEbiasAC}{-0.1569}
\newcommand{\bbbbQMLEstdeAC}{0.1018}
\newcommand{\bbbbQMLErmseAC}{0.1870}
\newcommand{\bbbbBCQMbiasAC}{-0.0758}
\newcommand{\bbbbBCQMstdeAC}{0.0700}
\newcommand{\bbbbBCQMrmseAC}{0.1031}
\newcommand{\bbbbCCEPbiasAC}{0.0036}
\newcommand{\bbbbCCEPstdeAC}{0.1074}
\newcommand{\bbbbCCEPrmseAC}{0.1074}
 
\newcommand{\bbbbOLSEbiasAD}{0.0326}
\newcommand{\bbbbOLSEstdeAD}{0.0543}
\newcommand{\bbbbOLSErmseAD}{0.0633}
\newcommand{\bbbbQMLEbiasAD}{-0.4209}
\newcommand{\bbbbQMLEstdeAD}{0.1607}
\newcommand{\bbbbQMLErmseAD}{0.4505}
\newcommand{\bbbbBCQMbiasAD}{-0.2732}
\newcommand{\bbbbBCQMstdeAD}{0.1235}
\newcommand{\bbbbBCQMrmseAD}{0.2998}
\newcommand{\bbbbCCEPbiasAD}{-0.1040}
\newcommand{\bbbbCCEPstdeAD}{0.1070}
\newcommand{\bbbbCCEPrmseAD}{0.1492}
 
\newcommand{\bbbbOLSEbiasAE}{0.2096}
\newcommand{\bbbbOLSEstdeAE}{0.0884}
\newcommand{\bbbbOLSErmseAE}{0.2274}
\newcommand{\bbbbQMLEbiasAE}{-0.0592}
\newcommand{\bbbbQMLEstdeAE}{0.0377}
\newcommand{\bbbbQMLErmseAE}{0.0702}
\newcommand{\bbbbBCQMbiasAE}{-0.0185}
\newcommand{\bbbbBCQMstdeAE}{0.0287}
\newcommand{\bbbbBCQMrmseAE}{0.0341}
\newcommand{\bbbbCCEPbiasAE}{0.0520}
\newcommand{\bbbbCCEPstdeAE}{0.0711}
\newcommand{\bbbbCCEPrmseAE}{0.0881}
 
\newcommand{\bbbbOLSEbiasAF}{0.0366}
\newcommand{\bbbbOLSEstdeAF}{0.0356}
\newcommand{\bbbbOLSErmseAF}{0.0511}
\newcommand{\bbbbQMLEbiasAF}{-0.0741}
\newcommand{\bbbbQMLEstdeAF}{0.0859}
\newcommand{\bbbbQMLErmseAF}{0.1134}
\newcommand{\bbbbBCQMbiasAF}{-0.0406}
\newcommand{\bbbbBCQMstdeAF}{0.0552}
\newcommand{\bbbbBCQMrmseAF}{0.0686}
\newcommand{\bbbbCCEPbiasAF}{-0.0310}
\newcommand{\bbbbCCEPstdeAF}{0.0512}
\newcommand{\bbbbCCEPrmseAF}{0.0599}
 
\newcommand{\bbbbOLSEbiasAG}{0.2174}
\newcommand{\bbbbOLSEstdeAG}{0.0649}
\newcommand{\bbbbOLSErmseAG}{0.2269}
\newcommand{\bbbbQMLEbiasAG}{-0.0275}
\newcommand{\bbbbQMLEstdeAG}{0.0192}
\newcommand{\bbbbQMLErmseAG}{0.0335}
\newcommand{\bbbbBCQMbiasAG}{-0.0054}
\newcommand{\bbbbBCQMstdeAG}{0.0170}
\newcommand{\bbbbBCQMrmseAG}{0.0179}
\newcommand{\bbbbCCEPbiasAG}{0.0759}
\newcommand{\bbbbCCEPstdeAG}{0.0500}
\newcommand{\bbbbCCEPrmseAG}{0.0908}
 
\newcommand{\bbbbOLSEbiasAH}{0.0404}
\newcommand{\bbbbOLSEstdeAH}{0.0239}
\newcommand{\bbbbOLSErmseAH}{0.0469}
\newcommand{\bbbbQMLEbiasAH}{-0.0134}
\newcommand{\bbbbQMLEstdeAH}{0.0166}
\newcommand{\bbbbQMLErmseAH}{0.0214}
\newcommand{\bbbbBCQMbiasAH}{-0.0047}
\newcommand{\bbbbBCQMstdeAH}{0.0122}
\newcommand{\bbbbBCQMrmseAH}{0.0131}
\newcommand{\bbbbCCEPbiasAH}{-0.0012}
\newcommand{\bbbbCCEPstdeAH}{0.0281}
\newcommand{\bbbbCCEPrmseAH}{0.0281}
 
\newcommand{\bbbbOLSEbiasAI}{0.2232}
\newcommand{\bbbbOLSEstdeAI}{0.0472}
\newcommand{\bbbbOLSErmseAI}{0.2281}
\newcommand{\bbbbQMLEbiasAI}{-0.0134}
\newcommand{\bbbbQMLEstdeAI}{0.0118}
\newcommand{\bbbbQMLErmseAI}{0.0179}
\newcommand{\bbbbBCQMbiasAI}{-0.0016}
\newcommand{\bbbbBCQMstdeAI}{0.0113}
\newcommand{\bbbbBCQMrmseAI}{0.0114}
\newcommand{\bbbbCCEPbiasAI}{0.0873}
\newcommand{\bbbbCCEPstdeAI}{0.0364}
\newcommand{\bbbbCCEPrmseAI}{0.0946}
 
\newcommand{\bbbbOLSEbiasAJ}{0.0433}
\newcommand{\bbbbOLSEstdeAJ}{0.0164}
\newcommand{\bbbbOLSErmseAJ}{0.0463}
\newcommand{\bbbbQMLEbiasAJ}{-0.0052}
\newcommand{\bbbbQMLEstdeAJ}{0.0066}
\newcommand{\bbbbQMLErmseAJ}{0.0084}
\newcommand{\bbbbBCQMbiasAJ}{-0.0012}
\newcommand{\bbbbBCQMstdeAJ}{0.0058}
\newcommand{\bbbbBCQMrmseAJ}{0.0059}
\newcommand{\bbbbCCEPbiasAJ}{0.0125}
\newcommand{\bbbbCCEPstdeAJ}{0.0176}
\newcommand{\bbbbCCEPrmseAJ}{0.0216}




\begin{table}[H]
   \begin{center}
   \caption{\label{tab:extra1} %\footnotesize
                      Same as Table~\ref{tab:T1} in main paper, but also reporting pooled CCE estimator of
                      Pesaran~(2006). }
   \vspace{0.5cm}
   \begin{tabular}{l@{\;\,}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l}
     \hline 
                  &   &  \multicolumn{4}{c}{$\rho^0=0.3$}    & \multicolumn{4}{c}{$\rho^0=0.9$}    \\[0.1cm]
                  &   & OLS & FLS & BC-FLS & CCE &  OLS & FLS & BC-FLS  & CCE \\
      \hline 
    $T=5$  &  bias &   \bOLSEbiasAA &  \bQMLEbiasAA & \bBCQMbiasAA & \bCCEPbiasAA &  \bOLSEbiasAB &  \bQMLEbiasAB & \bBCQMbiasAB & \bCCEPbiasAB \\
     $(M=2)$             &  std  &   \bOLSEstdeAA &  \bQMLEstdeAA & \bBCQMstdeAA & \bCCEPstdeAA &  \bOLSEstdeAB &  \bQMLEstdeAB & \bBCQMstdeAB & \bCCEPstdeAB \\
                  &  rmse &   \bOLSErmseAA &  \bQMLErmseAA & \bBCQMrmseAA & \bCCEPrmseAA &  \bOLSErmseAB &  \bQMLErmseAB & \bBCQMrmseAB & \bCCEPrmseAB
                   \\[8pt]
   $T=10$  &  bias &   \bOLSEbiasAC &  \bQMLEbiasAC & \bBCQMbiasAC & \bCCEPbiasAC &  \bOLSEbiasAD &  \bQMLEbiasAD & \bBCQMbiasAD & \bCCEPbiasAD \\
   $(M=3)$               &  std  &   \bOLSEstdeAC &  \bQMLEstdeAC & \bBCQMstdeAC & \bCCEPstdeAC &  \bOLSEstdeAD &  \bQMLEstdeAD & \bBCQMstdeAD & \bCCEPstdeAD \\
                  &  rmse &   \bOLSErmseAC &  \bQMLErmseAC & \bBCQMrmseAC & \bCCEPrmseAC &  \bOLSErmseAD &  \bQMLErmseAD & \bBCQMrmseAD & \bCCEPrmseAD 
                   \\[8pt]
   $T=20$  &  bias &   \bOLSEbiasAE &  \bQMLEbiasAE & \bBCQMbiasAE & \bCCEPbiasAE &  \bOLSEbiasAF &  \bQMLEbiasAF & \bBCQMbiasAF & \bCCEPbiasAF \\
    $(M=4)$              &  std  &   \bOLSEstdeAE &  \bQMLEstdeAE & \bBCQMstdeAE & \bCCEPstdeAE &  \bOLSEstdeAF &  \bQMLEstdeAF & \bBCQMstdeAF & \bCCEPstdeAF \\
                  &  rmse &   \bOLSErmseAE &  \bQMLErmseAE & \bBCQMrmseAE & \bCCEPrmseAE &  \bOLSErmseAF &  \bQMLErmseAF & \bBCQMrmseAF & \bCCEPrmseAF 
                   \\[8pt]
        $T=40$  &  bias &   \bOLSEbiasAG &  \bQMLEbiasAG & \bBCQMbiasAG & \bCCEPbiasAG &  \bOLSEbiasAH &  \bQMLEbiasAH & \bBCQMbiasAH & \bCCEPbiasAH \\
    $(M=5)$              &  std  &   \bOLSEstdeAG &  \bQMLEstdeAG & \bBCQMstdeAG & \bCCEPstdeAG &  \bOLSEstdeAH &  \bQMLEstdeAH & \bBCQMstdeAH & \bCCEPstdeAH \\
                  &  rmse &   \bOLSErmseAG &  \bQMLErmseAG & \bBCQMrmseAG & \bCCEPrmseAG &  \bOLSErmseAH &  \bQMLErmseAH & \bBCQMrmseAH & \bCCEPrmseAH 
             \\[8pt]
                $T=80$ &  bias &   \bOLSEbiasAI &  \bQMLEbiasAI & \bBCQMbiasAI & \bCCEPbiasAI &  \bOLSEbiasAJ &  \bQMLEbiasAJ & \bBCQMbiasAJ & \bCCEPbiasAJ \\
    $(M=6)$              &  std  &   \bOLSEstdeAI &  \bQMLEstdeAI & \bBCQMstdeAI & \bCCEPstdeAI &  \bOLSEstdeAJ &  \bQMLEstdeAJ & \bBCQMstdeAJ & \bCCEPstdeAJ \\
                  &  rmse &   \bOLSErmseAI &  \bQMLErmseAI & \bBCQMrmseAI & \bCCEPrmseAI &  \bOLSErmseAJ &  \bQMLErmseAJ & \bBCQMrmseAJ & \bCCEPrmseAJ \\
                  \hline
   \end{tabular}
   \end{center}
\end{table}

\begin{table}[H]
   \begin{center}
   \caption{\label{tab:extra2} %\footnotesize
                      Same as Table~\ref{tab:T2} in main paper, but also reporting pooled CCE estimator of
                      Pesaran~(2006). }
   \vspace{0.5cm}
   \begin{tabular}{l@{\;\,}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l}
     \hline 
                  &   &  \multicolumn{4}{c}{$\rho^0=0.3$}    & \multicolumn{4}{c}{$\rho^0=0.9$}    \\[0.1cm]
                  &   & OLS & FLS & BC-FLS & CCE &  OLS & FLS & BC-FLS  & CCE \\
      \hline 
    $T=5$  &  bias &   \bbOLSEbiasAA &  \bbQMLEbiasAA & \bbBCQMbiasAA & \bbCCEPbiasAA &  \bbOLSEbiasAB &  \bbQMLEbiasAB & \bbBCQMbiasAB & \bbCCEPbiasAB \\
     $(M=2)$             &  std  &   \bbOLSEstdeAA &  \bbQMLEstdeAA & \bbBCQMstdeAA & \bbCCEPstdeAA &  \bbOLSEstdeAB &  \bbQMLEstdeAB & \bbBCQMstdeAB & \bbCCEPstdeAB \\
                  &  rmse &   \bbOLSErmseAA &  \bbQMLErmseAA & \bbBCQMrmseAA & \bbCCEPrmseAA &  \bbOLSErmseAB &  \bbQMLErmseAB & \bbBCQMrmseAB & \bbCCEPrmseAB 
                   \\[8pt]
                      $T=10$  &  bias &   \bbOLSEbiasAC &  \bbQMLEbiasAC & \bbBCQMbiasAC & \bbCCEPbiasAC &  \bbOLSEbiasAD &  \bbQMLEbiasAD & \bbBCQMbiasAD & \bbCCEPbiasAD \\
   $(M=3)$               &  std  &   \bbOLSEstdeAC &  \bbQMLEstdeAC & \bbBCQMstdeAC & \bbCCEPstdeAC &  \bbOLSEstdeAD &  \bbQMLEstdeAD & \bbBCQMstdeAD & \bbCCEPstdeAD \\
                  &  rmse &   \bbOLSErmseAC &  \bbQMLErmseAC & \bbBCQMrmseAC & \bbCCEPrmseAC &  \bbOLSErmseAD &  \bbQMLErmseAD & \bbBCQMrmseAD & \bbCCEPrmseAD 
                   \\[8pt]
                      $T=20$  &  bias &   \bbOLSEbiasAE &  \bbQMLEbiasAE & \bbBCQMbiasAE & \bbCCEPbiasAE &  \bbOLSEbiasAF &  \bbQMLEbiasAF & \bbBCQMbiasAF & \bbCCEPbiasAF \\
    $(M=4)$              &  std  &   \bbOLSEstdeAE &  \bbQMLEstdeAE & \bbBCQMstdeAE & \bbCCEPstdeAE &  \bbOLSEstdeAF &  \bbQMLEstdeAF & \bbBCQMstdeAF & \bbCCEPstdeAF \\
                  &  rmse &   \bbOLSErmseAE &  \bbQMLErmseAE & \bbBCQMrmseAE & \bbCCEPrmseAE &  \bbOLSErmseAF &  \bbQMLErmseAF & \bbBCQMrmseAF & \bbCCEPrmseAF 
                   \\[8pt]
                      $T=40$  &  bias &   \bbOLSEbiasAG &  \bbQMLEbiasAG & \bbBCQMbiasAG & \bbCCEPbiasAG &  \bbOLSEbiasAH &  \bbQMLEbiasAH & \bbBCQMbiasAH & \bbCCEPbiasAH \\
    $(M=5)$              &  std  &   \bbOLSEstdeAG &  \bbQMLEstdeAG & \bbBCQMstdeAG & \bbCCEPstdeAG &  \bbOLSEstdeAH &  \bbQMLEstdeAH & \bbBCQMstdeAH & \bbCCEPstdeAH \\
                  &  rmse &   \bbOLSErmseAG &  \bbQMLErmseAG & \bbBCQMrmseAG & \bbCCEPrmseAG &  \bbOLSErmseAH &  \bbQMLErmseAH & \bbBCQMrmseAH & \bbCCEPrmseAH
                   \\[8pt]
                      $T=80$ &  bias &   \bbOLSEbiasAI &  \bbQMLEbiasAI & \bbBCQMbiasAI & \bbCCEPbiasAI &  \bbOLSEbiasAJ &  \bbQMLEbiasAJ & \bbBCQMbiasAJ & \bbCCEPbiasAJ \\
    $(M=6)$              &  std  &   \bbOLSEstdeAI &  \bbQMLEstdeAI & \bbBCQMstdeAI & \bbCCEPstdeAI &  \bbOLSEstdeAJ &  \bbQMLEstdeAJ & \bbBCQMstdeAJ & \bbCCEPstdeAJ \\
                  &  rmse &   \bbOLSErmseAI &  \bbQMLErmseAI & \bbBCQMrmseAI & \bbCCEPrmseAI &  \bbOLSErmseAJ &  \bbQMLErmseAJ & \bbBCQMrmseAJ & \bbCCEPrmseAJ \\
                  \hline
   \end{tabular}
   \end{center}
\end{table}

\begin{table}[H]
   \begin{center}
   \caption{\label{tab:extra3} %\footnotesize
                      Analogous to Table~\ref{tab:T2} in main paper, but with $R=2$ correctly specified, and
                      also reporting pooled CCE estimator of
                      Pesaran~(2006). }
      \vspace{0.5cm}
    \begin{tabular}{l@{\;\,}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l@{\quad}l@{\;\;}l@{\;\;}l@{\;\;}l}
     \hline 
                  &   &  \multicolumn{4}{c}{$\rho^0=0.3$}    & \multicolumn{4}{c}{$\rho^0=0.9$}    \\[0.1cm]
                  &   & OLS & FLS & BC-FLS & CCE &  OLS & FLS & BC-FLS  & CCE \\
      \hline 
    $T=5$  &  bias &   \bbbbOLSEbiasAA &  \bbbbQMLEbiasAA & \bbbbBCQMbiasAA & \bbbbCCEPbiasAA &  \bbbbOLSEbiasAB &  \bbbbQMLEbiasAB & \bbbbBCQMbiasAB & \bbbbCCEPbiasAB \\
     $(M=2)$             &  std  &   \bbbbOLSEstdeAA &  \bbbbQMLEstdeAA & \bbbbBCQMstdeAA & \bbbbCCEPstdeAA &  \bbbbOLSEstdeAB &  \bbbbQMLEstdeAB & \bbbbBCQMstdeAB & \bbbbCCEPstdeAB \\
                  &  rmse &   \bbbbOLSErmseAA &  \bbbbQMLErmseAA & \bbbbBCQMrmseAA & \bbbbCCEPrmseAA &  \bbbbOLSErmseAB &  \bbbbQMLErmseAB & \bbbbBCQMrmseAB & \bbbbCCEPrmseAB 
                   \\[8pt]
                      $T=10$  &  bias &   \bbbbOLSEbiasAC &  \bbbbQMLEbiasAC & \bbbbBCQMbiasAC & \bbbbCCEPbiasAC &  \bbbbOLSEbiasAD &  \bbbbQMLEbiasAD & \bbbbBCQMbiasAD & \bbbbCCEPbiasAD \\
   $(M=3)$               &  std  &   \bbbbOLSEstdeAC &  \bbbbQMLEstdeAC & \bbbbBCQMstdeAC & \bbbbCCEPstdeAC &  \bbbbOLSEstdeAD &  \bbbbQMLEstdeAD & \bbbbBCQMstdeAD & \bbbbCCEPstdeAD \\
                  &  rmse &   \bbbbOLSErmseAC &  \bbbbQMLErmseAC & \bbbbBCQMrmseAC & \bbbbCCEPrmseAC &  \bbbbOLSErmseAD &  \bbbbQMLErmseAD & \bbbbBCQMrmseAD & \bbbbCCEPrmseAD 
                   \\[8pt]
                      $T=20$  &  bias &   \bbbbOLSEbiasAE &  \bbbbQMLEbiasAE & \bbbbBCQMbiasAE & \bbbbCCEPbiasAE &  \bbbbOLSEbiasAF &  \bbbbQMLEbiasAF & \bbbbBCQMbiasAF & \bbbbCCEPbiasAF \\
    $(M=4)$              &  std  &   \bbbbOLSEstdeAE &  \bbbbQMLEstdeAE & \bbbbBCQMstdeAE & \bbbbCCEPstdeAE &  \bbbbOLSEstdeAF &  \bbbbQMLEstdeAF & \bbbbBCQMstdeAF & \bbbbCCEPstdeAF \\
                  &  rmse &   \bbbbOLSErmseAE &  \bbbbQMLErmseAE & \bbbbBCQMrmseAE & \bbbbCCEPrmseAE &  \bbbbOLSErmseAF &  \bbbbQMLErmseAF & \bbbbBCQMrmseAF & \bbbbCCEPrmseAF 
                   \\[8pt]
                      $T=40$  &  bias &   \bbbbOLSEbiasAG &  \bbbbQMLEbiasAG & \bbbbBCQMbiasAG & \bbbbCCEPbiasAG &  \bbbbOLSEbiasAH &  \bbbbQMLEbiasAH & \bbbbBCQMbiasAH & \bbbbCCEPbiasAH \\
    $(M=5)$              &  std  &   \bbbbOLSEstdeAG &  \bbbbQMLEstdeAG & \bbbbBCQMstdeAG & \bbbbCCEPstdeAG &  \bbbbOLSEstdeAH &  \bbbbQMLEstdeAH & \bbbbBCQMstdeAH & \bbbbCCEPstdeAH \\
                  &  rmse &   \bbbbOLSErmseAG &  \bbbbQMLErmseAG & \bbbbBCQMrmseAG & \bbbbCCEPrmseAG &  \bbbbOLSErmseAH &  \bbbbQMLErmseAH & \bbbbBCQMrmseAH & \bbbbCCEPrmseAH
                   \\[8pt]
                      $T=80$ &  bias &   \bbbbOLSEbiasAI &  \bbbbQMLEbiasAI & \bbbbBCQMbiasAI & \bbbbCCEPbiasAI &  \bbbbOLSEbiasAJ &  \bbbbQMLEbiasAJ & \bbbbBCQMbiasAJ & \bbbbCCEPbiasAJ \\
    $(M=6)$              &  std  &   \bbbbOLSEstdeAI &  \bbbbQMLEstdeAI & \bbbbBCQMstdeAI & \bbbbCCEPstdeAI &  \bbbbOLSEstdeAJ &  \bbbbQMLEstdeAJ & \bbbbBCQMstdeAJ & \bbbbCCEPstdeAJ \\
                  &  rmse &   \bbbbOLSErmseAI &  \bbbbQMLErmseAI & \bbbbBCQMrmseAI & \bbbbCCEPrmseAI &  \bbbbOLSErmseAJ &  \bbbbQMLErmseAJ & \bbbbBCQMrmseAJ & \bbbbCCEPrmseAJ \\
                  \hline
   \end{tabular}
      \end{center}
\end{table}


\end{document}

