Estimation of Panel Data Models with Nonlinear Factor Structure

abstractPanel data models with unobserved heterogeneity in the form of interactive effects standardly assume that the time effects -- or “common factors” -- enter linearly. This assumption is unnatural in the sense that it pertains to the unobserved component of the model, and there is rarely any reason to believe that this component takes on a particular functional form. This is in stark contrast to the relationship between the observables, which can often be credibly argued to be linear. Linearity in the factors has persevered mainly because it is convenient, and that it is better than standard fixed effects. The present paper relaxes this assumption. It does so by combining the common correlated effects (CCE) approach to standard interactive effects with the method of sieves. The new estimator -- abbreviated “SCCE” -- retains many of the advantages of CCE, including its computational simplicity, and good small-sample and asymptotic properties, but is applicable under a much broader class of factor structures that includes the linear one as a special case. This makes it well-suited for a wide range of empirical applications.

step[Initial estimation]\ \normalfont Compute $\widehat{\*F}$. This is our “mock” proxy of the factor component.

step[Sieve approximation of the factor space]\ \normalfont As pointed out earlier, $\widehat{\*F}$ is likely inconsistent for the space spanned by $\mathcal{G}_i(\cdot)$. To address this issue, in the second step we employ the method of sieves. The main idea is to approximate $\mathcal{G}_i(\cdot)$ using a set of user-specified basis functions with $\widehat{\*F}$ as their argument. Let us therefore denote by $\*p=[p_1(\cdot), \dots, p_K(\cdot)]' \in \mathbb{R}^{K}$ a vector of $K$ such basis functions. The sought sieve basis matrix is given by $\widehat{\*P} = \left[\*p(\widehat{\*f}_1), \dots, \*p(\widehat{\*f}_T)\right]' \in \mathbb{R}^{T \times K}$.

step[Estimation of $\+\beta$] \normalfont Given $\widehat{\*P}$, our proposed estimator of $\boldsymbol{\beta}$, denoted by $\widehat{\boldsymbol{\beta}}_{SCCE}$, is given by \begin{align} \widehat{\boldsymbol{\beta}}_{SCCE} \equiv \left(\sum_{i=1}^N \mathbf{X}_{i}'\mathbf{M}_{\widehat{P}} \mathbf{X}_{i}\right)^{-1}\sum_{i=1}^N \mathbf{X}_{i}' \mathbf{M}_{\widehat{P}} \mathbf{y}_{i}. \end{align}

assumption[Idiosyncratic errors] \leavevmode \begin{enumerate}[label=(\alph*)] • For each $i$, $\{(\varepsilon_{i,t}, \*v_{i,t}): t \geq 1\}$ is a strictly stationary and $\alpha$-mixing process with mixing coefficients $\alpha_{i}(j)$ satisfying $\sum_{j=1}^\infty j^2 \alpha_i(j)^{\frac{\eta}{4+\eta}} \leq C$ for some $\eta > 0$. The process is also independent across $i$ with $\mathbb{E}(\varepsilon_{i,t}) = 0$, $\mathbb{E}(\*v_{i,t}) = \*0_{d \times 1}$, $\max_{t=1,\dots,T} \max_{i=1,\dots,N} \mathbb{E}(\|\*v_{i,t}\|^{4+\eta}) \leq C$ and $\max_{t=1,\dots,T} \max_{i=1,\dots,N}\mathbb{E}(|\varepsilon_{i,t}|^{4 + \eta}) \leq C$. • Let $\sigma^2_{\varepsilon, i} \equiv \lim_{T \to \infty} \frac{1}{T} \mathbb{E}(\+\varepsilon_i'\+\varepsilon_i) \in \mathbb{R}$ and $\+\Sigma_{v,i} \equiv \lim_{T \to \infty} \frac{1}{T} \mathbb{E}(\*V_i' \*V_i) \in \mathbb{R}^{d \times d}$. The following applies to these variances: $\inf_{i=1, \dots, N} \sigma_{\varepsilon, i}^2 > 0$, $\sup_{i=1, \dots, N} \sigma_{\varepsilon, i}^2 \leq C$, $\sigma^2_{\varepsilon} = \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^N \sigma_{\varepsilon, i}^2 \leq C$, $\inf_{i=1, \dots, N}\lambda_{\min}(\+\Sigma_{v,i}) > 0$, $\sup_{i=1, \dots, N} \|\+\Sigma_{v,i} \| \leq C$ and $\+\Sigma_v \equiv \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^N \+\Sigma_{v,i} \in \mathbb{R}^{d \times d}$ is positive definite. Moreover, the covariance matrices $\+\Omega_{\varepsilon, i} \equiv \mathbb{E}(\+\varepsilon_i \+\varepsilon_i')\in \mathbb{R}^{T \times T}$ and $\+\Omega_{v, i} \equiv \mathbb{E}(\*V_{i}\*V'_{i})\in \mathbb{R}^{T \times T}$ have eigenvalues bounded away from zero and infinity. \end{enumerate}

assumption[Common factors] $\{\*f_t: t \geq 1\}$ is a strictly stationary and $\alpha$-mixing process with mixing coefficient $\alpha_0(j)$ satisfying $\sum_{j=1}^{\infty} j^2 \alpha_0(j)^{\frac{\eta}{4+ \eta}} \leq C$ with $\eta > 0$. Also, $\sup_{t=1, \dots, T}\mathbb{E}(\|\*f_{t}\|^{4+\eta}) \leq C$ and $\+\Sigma_F \equiv \lim_{T \to \infty} \frac{1}{T} \mathbb{E}(\*F'\*F)\in \mathbb{R}^{m \times m}$ is positive definite.

assumption[Independence] $\*f_t$, $\varepsilon_{i, s}$ and $\*v_{j, l}$ are mutually independent for all $t$, $i$, $j$, $s$ and $l$.

definition[H\"older-smooth functions] Let $\Lambda^\lambda(\mathcal{R}_f, \omega)$ denote a weighted H\"older space of functions $g: \mathcal{R}_f \to \mathbb{R}$ such that $g(\cdot)[1 + \|\cdot\|^2]^{-\frac{\omega}{2}}$ is in $\Lambda^\lambda(\mathcal{R}_f)$. A weighted H\"older ball with radius $r$ is given by $\Lambda_r^\lambda(\mathcal{R}_f, \omega) \equiv \{g \in \Lambda^\lambda(\mathcal{R}_f, \omega): \|g(\cdot)[1 + \|\cdot\|^2]^{-\frac{\omega}{2}}\|_{\Lambda^\lambda} \leq r $\}. A function $g(\cdot)$ is said to be $H(\lambda, \omega)$-smooth on $\mathcal{R}_f$ if it belongs to the weighted H\"older ball $\Lambda_r^\lambda(\mathcal{R}_f, \omega)$ for some $\lambda>0$, $r \in (0,\infty)$ and $\omega \geq0$.

assumption[Function class] Define the set $\mathcal{S}\equiv \{g_i(\cdot),G_{j,i}(\cdot): i=1, \dots, N, j=1, \dots, d\}$, where the mappings $g_i(\cdot)$ and $G_{j,i}(\cdot)$ are defined as before. \begin{enumerate}[label=(\alph*)] • For any $\widetilde{g}_i(\cdot)\in \mathcal{S}$, $\widetilde{g}_i(\cdot) \in \Lambda_r^{\lambda_i}(\mathcal{R}_f, \omega_i)$ for some $\lambda_i > 0$, $\omega_i \geq 0$. • $\sup_{t=1,\hdots,T}\int_{\mathbb{R}^m} [1+\|\*f_t\|^{2}]^{\overline{\omega}} d\mu(\*f_t) < \infty$ for some $\overline{\omega} > \sup_{i=1,\dots, N}\left(\omega_i + \lambda_i\right)$, where $d\mu(\*f_t) = w(\*f_t)d\*f_t$ and $w(\*f_t)$ is the probability density function of $\*f_t$. • For any $\widetilde{g}_i(\cdot) \in \Lambda_r^{\lambda_i}(\mathcal{R}_f, \omega_i)$, there is a function $\Pi_{\infty K} \widetilde{g}_i(\cdot)\equiv \+\alpha_{\widetilde{g}_i}' \*p(\cdot)$ in the sieve space $\mathfrak{G}_K \equiv \{h(\cdot) = \*a'\*p(\cdot)\}$ such that the approximation error satisfies $\| \widetilde{g}_i(\cdot) - \Pi_{\infty K}\widetilde{g}_i(\cdot) \|_{\infty, \overline{\omega}} = O\left(K^{-\frac{\lambda_i}{m}}\right)$. • $\mathbb{E}\left[\widetilde{g}_i(\*f_t)\right] = 0$. \end{enumerate}

assumption[Rank condition] $\overline{\mathcal{G}}(\cdot): \mathbb{R}^m \to \overline{\mathcal{G}}\left(\mathbb{R}^m\right) \subseteq \mathbb{R}^{d+1}$ is $H(\lambda, \omega)$-smooth and injective. Further, $\nabla \overline{\mathcal{G}}\left(\*f_t\right) \in \mathbb{R}^{(d+1) \times m}$ exists and is of full column-rank $m$ w.p.1, such that for all $\*f_t \in \mathbb{R}^{m}$, $\operatorname{rank} \left(\nabla \overline{\mathcal{G}}\left(\*f_t\right)\right) = m \leq d+1$.

assumption[Basis functions] Define the set of sieve matrices $\mathcal{P} \equiv \{\widetilde{\*P}, \widehat{\*P}\}$, where $\widetilde{\mathbf{P}} = \left[\widetilde{\*p}\left(\*f_1\right), \dots, \widetilde{\*p}\left(\*f_T\right)\right]' = \left[\*p\left(\overline{\mathcal{G}}\left(\*f_1\right)\right), \dots, \*p\left(\overline{\mathcal{G}}\left(\*f_T\right)\right) \right]'$ and $\widehat{\*P}$ is as before. \begin{enumerate}[label=(\alph*)] • For any $\*P \in \mathcal{P}$, uniformly in $K$ and $T$, there exist constants $0 < \underline{C} < \overline{C} < \infty$, such that $\underline{C} \leq \lambda_{\min}[{\mathbb{E}(\frac{1}{T}\*P' \*P})]\leq \lambda_{\max}[\mathbb{E}(\frac{1}{T}\*P' \*P)] \leq \overline{C}$. • $\sup_{k=1, \dots, K} \mathbb{E}[|p_k(\cdot)|^{8 + \eta}]\leq C$, where $\eta>0$ is as in Assumption (ref) and $p_k(\cdot)$ is the $k$-th sieve basis function of $\*P \in \mathcal{P}$. • For large enough $K$, there exists a sequence of constants $\zeta_0(K)$ and $\zeta_1(K)$ satisfying $\sup_{\*f}\|\*p(\*f)\|\leq \zeta_0(K)$, $\sup_{\*f}\|\nabla\*p(\*f)\|\leq \zeta_1(K)$ with $\zeta_0(K)\geq 1, \, \zeta_1(K) \geq 1$. Further, $\sqrt{NT}K^{-\frac{\lambda_i}{m}}\to 0$ with $\lambda_i > 0$, $\frac{\sqrt{T}K^{\frac{1}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2} \to 0$, $\sqrt{\frac{K}{T}}\zeta_{0}(K)^{\frac{3}{2}}\to 0$, as $N,T\to \infty$. \end{enumerate}

assumption[Gram matrices of sieve coefficients] Define $\+\Sigma_{\alpha_g} \equiv \frac{1}{N}\sum_{i=1}^N \+\alpha_{g_i}\+\alpha_{g_i}' \in \mathbb{R}^{K\times K}$ and $\+\Sigma_{\alpha_G} \equiv \frac{1}{N}\sum_{i=1}^N \+\alpha_{G_i}\+\alpha_{G_i}' \in \mathbb{R}^{K \times K}$. Then, there exist constants $0 < \underline{C} < \overline{C} < \infty$, such that for all $j\in \{\alpha_g, \alpha_G\}$, $\liminf_{N\to \infty}\lambda_{\min}\left(\+\Sigma_{j}\right) \geq \underline{C}$ and $\limsup_{N\to \infty} \lambda_{\max}\left(\+\Sigma_{j}\right) \leq \overline{C}$.

theoremUnder Assumptions (ref)--(ref), as $N,\,T\to\infty$, \begin{eqnarray} \sqrt{NT}(\widehat{\+\beta}_{SCCE} - \+\beta) \overset{d}{\to} \mathcal{N}(\*0_{d\times 1},\+\Sigma_v^{-1}\+\Theta \+\Sigma_v^{-1}), \end{eqnarray} where \begin{align} \+\Theta \equiv \lim_{N,T\to\infty}\frac{1}{NT}\sum_{i=1}^{N}\mathbb{E}\left(\mathbf{V}_{i}'\boldsymbol{\Omega}_{\varepsilon,i}\mathbf{V}_{i}\right) \end{align} with $\boldsymbol{\Omega}_{\varepsilon,i} \equiv \mathbb{E}\left(\+\varepsilon_i\+\varepsilon_i'\right)$.

landscape\begin{table}[h] \begin{threeparttable} \caption{Estimation results.} \begin{tabular}{lc c c c c c c c c} \toprule & \multicolumn{4}{c}{Narrow model} & \multicolumn{4}{c}{Full model} \\ \cmidrule(lr){2-5} \cmidrule(lr){6-9} Regressor & SCCE & CCEP & CCEMG & & SCCE & CCEP & CCEMG & \\ \midrule $\ln(\sigma_{i,t})$ & -0.34 & -0.61 & -0.59 & & -0.33 & -0.52 & -0.71 & \\ & (-0.67; -0.03) & (-0.97; -0.25) & (-1.01; -0.15) & & (-0.70; 0.00) & (-0.88; -0.17) & (-1.17; -0.24) & \\ $\ln\left(\frac{H_{i,t}}{L_{i,t}}\right)$ & 0.56 & 0.40 & 0.36 & & 0.61 & 0.48 & 0.45 & \\ & (0.50; 0.61) & (0.33; 0.46) & (0.33; 0.38) & & (0.55; 0.66) & (0.42; 0.55) & (0.43; 0.48) &\\ $z_{1,i,t}$ & & & & & -0.03 & -0.19 & -1.64 &\\ & & & & & (-0.47; 0.51) & (-0.44; 0.06) & (-2.74; -0.63) &\\ $z_{2,i,t}$ & & & & & 3.15 & 1.23 & -3.29 & \\ & & & & & (0.40; 6.02) & (-0.37; 2.78) & (-6.66; 0.10) & \\ $z_{3,i,t}$ & & & & & -1.03 & 0.25 & 2.28 & \\ & & & & & (-2.57; 0.42) & (-0.43; 1.01) & (0.19; 4.28) & \\ $z_{4,i,t}$ & & & & & -0.20 & 0.20 & 1.44 & \\ & & & & & (-0.74; 0.35) & (-0.29; 0.64) & (-0.23; 3.04) & \\ $z_{5,i,t}$ & & & & & -0.17 & -0.08 & -1.20 & \\ & & & & & (-0.46; 0.08) & (-0.19; 0.06) & (-2.28; -0.22) & \\ $z_{6,i,t}$ & & & & & 0.07 & -0.09 & 0.20 & \\ & & & & & (-0.23; 0.38) & (-0.26; 0.08) & (-0.50; 0.98) & \\ \bottomrule \end{tabular} \begin{tablenotes} • \scriptsize Notes: The table reports point estimates of the coefficients of the model in (ref) with the associated 95% bootstrap confidence intervals appearing within parentheses. Results are reported for three estimators, SCCE, CCEP and CCEMG. Results with 95% confidence intervals based on HAC standard errors for our SCCE estimator can be found in appendix, Table (ref). The regressors of interest are $\ln\left(\sigma_{i,t}\right)$, which is a measure of the input skill intensity, and $\ln\left(\frac{H_{i,t}}{L_{i,t}}\right)$, which is the ratio of high-skilled to low-skilled labor. The six control variables are: real capital equipment per worker ($z_{1,i,t}$), the sectoral share of office, computing and accounting equipment ($z_{2,i,t}$), the difference between high-tech and computer capital share ($z_{3,i,t}$), R&D intensity ($z_{4,i,t}$), narrow outsourcing ($z_{5,i,t}$), and the difference between broad and narrow outsourcing ($z_{6,i,t}$). The dependent variable is the log of the relative wage of low-skilled to high-skilled workers. \end{tablenotes} \end{threeparttable} \end{table}

appendix\setcounter{lemma}{0} \setcounter{theorem}{0} \setcounter{table}{0} \setcounter{figure}{0} \section{Auxiliary Results and Proof of Theorem (ref)} \@startsection{subsection}{2} \z@{-.5\linespacing\@plus-.7\linespacing}{.5\linespacing} {\normalfont}{Rate Results} Before we start our asymptotic analysis, it is useful to establish several auxiliary rate results. These results will be invoked repeatedly throughout this appendix. We begin by noting that the basis functions satisfy the following standard growth rates (see, for example, newey97): \begin{align} \zeta_0(K)=\begin{cases} O\left(\sqrt{K}\right) & for splines\\ O\left(K\right) & for power series \end{cases}, \qquad\zeta_{1}(K)=\begin{cases} O\left(K^{\frac{3}{2}}\right) & for splines\\ O\left(K^{3}\right) & for power series \end{cases}. \end{align} We also note that under the conditions of Assumption (ref)(ref), \begin{align} \sqrt{\frac{T}{N}}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}=o(1)&\iff\frac{1}{\sqrt{N}}=o\left(\frac{1}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}}\right)\notag \\ &\iff\frac{1}{N}=o\left(\frac{1}{T\sqrt{K}\zeta_{0}(K)^{3}\zeta_{1}(K)^{4}}\right),\\ \sqrt{\frac{K}{T}}\zeta_{0}(K)^{\frac{3}{2}}=o(1)&\iff\frac{1}{\sqrt{T}}=o\left(\frac{1}{\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}}\right)\notag \\ &\iff\frac{1}{T}=o\left(\frac{1}{K\zeta_{0}(K)^{3}}\right),\\ \sqrt{NT}K^{-\frac{\lambda_{i}}{m}}=o(1)&\iff K^{-\frac{\lambda_{i}}{m}}=o\left(\frac{1}{\sqrt{NT}}\right). \end{align} By using these results and the rates in (ref), we can show the following: \begin{align} \frac{\sqrt{KT}\zeta_{1}(K)^{2}}{N\zeta_{0}(K)}&=o\left(\frac{\sqrt{KT}\zeta_{1}(K)^{2}}{T\sqrt{K}\zeta_{0}(K)^{4}\zeta_{1}(K)^{4}}\right)=o\left(\frac{1}{\sqrt{T}\zeta_{0}(K)^{4}\zeta_{1}(K)^{2}}\right)=o(1),\\ \frac{K^{\frac{1}{4}}\zeta_{1}(K)}{\sqrt{N\zeta_{0}(K)}}&=o\left(\frac{K^{\frac{1}{4}}\zeta_{1}(K)}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{2}\zeta_{1}(K)^{2}}\right)=o\left(\frac{1}{\sqrt{T}\zeta_{0}(K)^{2}\zeta_{1}(K)}\right)=o(1),\\ \frac{K}{NT}\zeta_{1}(K)^{2}&=o\left(\frac{K}{NK\zeta_{0}(K)^{3}\zeta_{1}(K)^{2}}\right)=o\left(\frac{1}{N\zeta_{0}(K)^{3}\zeta_{1}(K)^{2}}\right)=o(1),\\ \frac{K^{\frac{1}{4}}}{\sqrt{T}}\zeta_{0}(K)^{\frac{3}{2}}&=o\left(\frac{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}{\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}}\right)=o\left(\frac{1}{K^{\frac{1}{4}}}\right)=o(1),\\ \frac{K^{\frac{5}{2}}}{N^{\frac{3}{2}}}\zeta_{1}(K)^{3}&=o\left(\frac{K^{\frac{5}{2}}\zeta_{1}(K)^{3}}{T^{\frac{3}{2}}K^{\frac{3}{4}}\zeta_{0}(K)^{\frac{9}{2}}\zeta_{1}(K)^{6}}\right)=o\left(\frac{K^{\frac{7}{4}}}{K^{\frac{3}{2}}\zeta_{0}(K)^{\frac{9}{2}}\zeta_{0}(K)^{\frac{9}{2}}\zeta_{1}(K)^{3}}\right)\notag \\ &=o\left(\frac{K^{\frac{1}{4}}}{\zeta_{0}(K)^{9}\zeta_{1}(K)^{3}}\right)=o(1),\\ \frac{K^{\frac{4m-4\lambda_{i}}{4m}}}{N}\zeta_{1}(K)^{2}&=o\left(\frac{KK^{-\frac{\lambda_{i}}{m}}\zeta_{1}(K)^{2}}{T\sqrt{K}\zeta_{0}(K)^{3}\zeta_{1}(K)^{4}}\right)=o\left(\frac{\sqrt{K}K^{-\frac{\lambda_{i}}{m}}}{K\zeta_{0}(K)^{3}\zeta_{0}(K)^{3}\zeta_{1}(K)^{2}}\right)\notag \\ &=o\left(\frac{K^{-\frac{\lambda_{i}}{m}}}{\sqrt{K}\zeta_{0}(K)^{6}\zeta_{1}(K)^{2}}\right)=o\left(\frac{1}{\sqrt{NTK}\zeta_{0}(K)^{6}\zeta_{1}(K)^{2}}\right)\notag \\ &=o(1),\\ \frac{K^{\frac{13m-8\lambda_{i}}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)&=o\left(\frac{K^{\frac{13}{8}}K^{-\frac{\lambda_{i}}{m}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}}\right)=o\left(\frac{K^{\frac{11}{8}}K^{-\frac{\lambda_{i}}{m}}}{\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)}\right)\notag \\ &=o\left(\frac{K^{\frac{7}{8}}K^{-\frac{\lambda_{i}}{m}}}{\zeta_{0}(K)^{\frac{9}{4}}\zeta_{1}(K)}\right)=o(1),\\ \frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)&=o\left(\frac{K^{\frac{3}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}}\right)=o\left(\frac{\sqrt{K}}{\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)}\right)\notag \\ &=o\left(\frac{1}{\zeta_{0}(K)^{\frac{5}{2}}}\right)=o(1),\\ K^{\frac{m-4\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}&=o\left(\frac{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}{\sqrt{NT}}\right)=o\left(\frac{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}{\sqrt{NK}\zeta_{0}(K)^{\frac{3}{2}}}\right)\notag \\ &=o\left(\frac{1}{\sqrt{N}K^{\frac{1}{4}}}\right)=o(1),\\ \sqrt{\frac{T}{N}}\zeta_{1}(K)^{2}&=o\left(\frac{\sqrt{T}\zeta_{1}(K)^{2}}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}}\right)=o\left(\frac{1}{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}\right)=o(1),\\ \frac{\sqrt{T}K}{N^{\frac{3}{2}}}\zeta_{1}(K)^{4}&=o\left(\frac{\sqrt{T}K\zeta_{1}(K)^{4}}{T^{\frac{3}{2}}K^{\frac{3}{4}}\zeta_{0}(K)^{\frac{9}{2}}\zeta_{1}(K)^{6}}\right)=o\left(\frac{K^{\frac{1}{4}}}{\sqrt{T}\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{0}(K)^{\frac{9}{2}}\zeta_{1}(K)^{2}}\right)\notag\\ &=o\left(\frac{1}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{11}{2}}\zeta_{1}(K)^{2}}\right)=o(1),\\ \frac{\sqrt{T}K^{\frac{m-2\lambda_{i}}{m}}}{\sqrt{N}}\zeta_{1}(K)^{2}&=\left(\frac{K^{\frac{m-2\lambda_{i}}{m}}}{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}\right)=o\left(\frac{K^{\frac{3}{4}}}{NT\zeta_{0}(K)^{\frac{3}{2}}}\right)\notag \\ &=o\left(\frac{K^{\frac{3}{4}}}{NK\zeta_{0}(K)^{3}\zeta_{0}(K)^{\frac{3}{2}}}\right)=o\left(\frac{1}{NK^{\frac{1}{4}}\zeta_{0}(K)^{\frac{9}{2}}}\right)=o(1),\\ \sqrt{NT}K^{\frac{m-8\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}&=o\left(\frac{\sqrt{NT}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}{NT}\right)=o\left(\frac{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}{\sqrt{NT}}\right)\notag \\ &=o\left(\frac{1}{\sqrt{N}K^{\frac{1}{4}}}\right)=o(1),\\ \sqrt{\frac{T}{N}}K\zeta_{1}(K)^{2}&=o\left(\frac{K}{K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}}\right)=o\left(\frac{K^{\frac{3}{4}}}{\zeta_{0}(K)^{\frac{3}{2}}}\right)=o(1),\\ TK^{-\frac{4\lambda_{i}}{m}}&=o\left(\frac{T}{N^{4}T^{4}}\right)=o\left(\frac{1}{N^{4}T^{3}}\right)=o(1),\\ \frac{K^{\frac{3}{2}}}{T^{\frac{3}{2}}}\zeta_{0}(K)&=o\left(\frac{K^{\frac{3}{2}}\zeta_{0}(K)}{K^{\frac{3}{2}}\zeta_{0}(K)^{\frac{9}{2}}}\right)=o\left(\frac{1}{\zeta_{0}(K)^{\frac{7}{2}}}\right)=o(1), \\ \frac{K^{\frac{17}{8}}}{N}\zeta_0(K)^{\frac{3}{4}}\zeta_1(K)^2 &= o\left(\frac{K^{\frac{17}{8}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}}{T\sqrt{K}\zeta_{0}(K)^{3}\zeta_{1}(K)^{4}}\right)=o\left(\frac{K^{\frac{13}{8}}}{K\zeta_{0}(K)^{3}\zeta_{0}(K)^{\frac{9}{4}}\zeta_{1}(K)^{2}}\right)\notag \\ &=o\left(\frac{K^{\frac{5}{8}}}{\zeta_{0}(K)^{\frac{21}{4}}\zeta_{1}(K)^{2}}\right) = o(1), \\ \sqrt{T}K^{\frac{1}{2}-\frac{\lambda_{i}}{m}}\zeta_{1}(K)&=o\left(\frac{\sqrt{TK}}{\sqrt{NT}}\zeta_{1}(K)\right)=o\left(\frac{\sqrt{K}\zeta_{1}(K)}{\sqrt{T}K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}}\right)\notag \\ &=o\left(\frac{K^{\frac{1}{4}}}{\sqrt{K}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)}\right)=o\left(\frac{1}{K^{\frac{1}{4}}\zeta_{0}(K)^{3}\zeta_{1}(K)}\right)\notag \\ &=o(1). \end{align} \@startsection{subsection}{2} \z@{-.5\linespacing\@plus-.7\linespacing}{.5\linespacing} {\normalfont}{A Useful Lemma} \begin{lemma} Let $\*f_t \in \mathbb{R}^m$, $\widehat{\*f}_t \in \mathbb{R}^{d+1}$, $\mathcal{G}_i(\cdot): \mathbb{R}^m \to \mathcal{G}_i\left(\mathbb{R}^m\right) \subseteq\mathbb{R}^{d+1}$, and $\overline{\mathcal{G}}(\cdot): \mathbb{R}^m \to \overline{\mathcal{G}}\left(\mathbb{R}^m\right) \subseteq\mathbb{R}^{d+1}$. Under our Assumptions (ref)--(ref), as $N,T \to \infty$, \begin{enumerate}[label=(\roman*)] • $\left\|\mathcal{G}_i\left(\*f_t\right) - \+\alpha_{\mathcal{G}_i}' \*p\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| = o_p(1)$, • $\left\|\*p (\widehat{\*f}_t) - \*p\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| = O_p\left(\frac{\zeta_1(K)}{\sqrt{N}}\right)=o_p(1)$. \end{enumerate} \end{lemma} Proof of Lemma (ref). Consider (ref). Let us define a mapping $h_i(\cdot): \mathbb{R}^{d+1} \supseteq \mathcal{R}_{\overline{\mathcal{G}}} \to \mathbb{R}^{d+1}$ and $h_i(\*u) \equiv \mathcal{G}_i\left(\overline{\mathcal{G}}^{-1}\left(\*u\right)\right)$ for some vector $\*u \in \mathcal{R}_{\overline{\mathcal{G}}}$, with $\*u \equiv \overline{\mathcal{G}}\left(\*f_t\right)$ and $\mathcal{R}_{\overline{\mathcal{G}}} \equiv \overline{\mathcal{G}}\left(\mathbb{R}^m\right)$. The mappings $\overline{\mathcal{G}}(\cdot)$ and $\mathcal{G}_i(\cdot)$ are defined as before. By Assumption (ref), $\overline{\mathcal{G}}(\cdot)$ is injective, $H(\lambda, \omega)$-smooth, and has full column rank. Hence, by inverse function theorem, for each $\*u \in \mathcal{R}_{\overline{\mathcal{G}}}$, there exists a neighborhood in which the inverse $\overline{\mathcal{G}}^{-1}(\cdot): \mathbb{R}^{d+1}\supseteq \mathcal{R}_{\overline{\mathcal{G}}} \to \mathbb{R}^m$ is well defined and $H(\lambda, \omega)$-smooth. Consequently, $h_i\left(\overline{\mathcal{G}}\left(\*f_t\right)\right) = \mathcal{G}_i\left(\overline{\mathcal{G}}^{-1}\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right) = \mathcal{G}_i\left(\*f_t\right)$. Since a composition of smooth functions is smooth, $h_i(\cdot)$ itself is H\"older-smooth on $\mathcal{R}_{\overline{\mathcal{G}}}$. This means that by Assumption (ref)(ref) there exists a function $\Pi_{\infty K}h_i \equiv \+\alpha_{h_i}'\*p(\cdot)$, such that \begin{align} \left\|h_i(\*u) - \Pi_{\infty K}h_i(\*u) \right\|_{\infty, \widetilde{\omega}} = \sup_{\*u} \left\|h_i(\*u) - \+\alpha_{h_i}'\*p(\*u)\right\| \left[1 + \left\|\*u\right\|^2\right]^{-\frac{\tilde{\omega}}{2}} = O\left(K^{-\frac{\lambda_i}{m}}\right). \end{align} Before we proceed, let us first state the following theorem (also known as the "Changing Variables Theorem"), which is necessary in order to prove Lemma (ref)(ref) (see for example, evans2015measure, Theorem 3.9): \begin{theorem} Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be Lipschitz continuous, $n \leq m$. Then, for each $\mathcal{L}^n$-summable function $g: \mathbb{R}^n \to \mathbb{R}$, \begin{align} \int_{\mathbb{R}^n} g(x) J(f(x))dx = \int_{\mathbb{R}^m} \left[\sum_{x \in f^{-1}\{y\}} g(x)\right] d \mathcal{H}^n(y), \end{align} where a function $g$ is called $\mathcal{L}^n$-summable if $\left|g\right|$ has a finite integral. Here, $J(f(x))$ is a Jacobian of $f(x)$ defined as $J(f(x)) \equiv \sqrt{\det \nabla f(x)'\nabla f(x)}$, and $\mathcal{H}^n(y)$ is the Hausdorff measure.\footnote{The Hausdorff measure generalizes length/area/volume to possibly lower (or even non-integer) dimensions, which allows a measurement in lower-dimensional sets within a space $\mathbb{R}^n$. In contrast, Lebesgue measure can only measure $n$-dimensional volume in this case. For a formal definition of the Hausdorff measure, see evans2015measure.} \end{theorem} We would like to point out two things regarding Theorem (ref). First, note that if the function $f$ is injective, then every $y$ in $f\left(\mathbb{R}^n\right)$ can be mapped to at most one point in $\mathbb{R}^n$, such that \begin{align} \sum_{x \in f^{-1}\{y\}}g(x) = \begin{cases} g\left(f^{-1}\left(y\right)\right) &\text{if} \ y \in f\left(\mathbb{R}^n\right) \\ 0 \ &\text{otherwise}\end{cases}. \end{align} Hence, equation (ref) reduces to \begin{align} \int_{\mathbb{R}^n} g(x) J(f(x))dx = \int_{f\left(\mathbb{R}^n\right)} g\left(f^{-1}\left(y\right)\right) d\mathcal{H}^n(y). \end{align} Second, by Assumption (ref), $\overline{\mathcal{G}}(\cdot)$ is H\"older-smooth, and $\nabla \overline{\mathcal{G}}(f_t)$ exists and has full column rank $m$. Together, these conditions imply that $\overline{\mathcal{G}}(\cdot)$ is locally Lipschitz on $\mathbb R^m$, which is sufficient for the application of Theorem (ref). Now consider \begin{align} \mathbb{E}\left(\left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2\right) = \int_{\mathcal{R}_{\overline{\mathcal{G}}}} \left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2 w_u(\*u) d \mathcal{H}^m(\*u), \end{align} where $h_i(\*u) = \left[h_{i,1}(\*u), \dots, h_{i,d+1}(\*u)\right]' \in \mathbb{R}^{d+1}$, $\+\alpha_{h_i} \in \mathbb{R}^{K \times (d+1)}$ is a sieve coefficient matrix, $\*p(\cdot) \in \mathbb{R}^{K}$ is a vector of basis functions, $w_u(\*u)$ is a probability density function of the vector $\*u$, and $\mathcal{H}^m(\*u)$ is a Hausdorff measure since $m \leq d+1$. By multiplying and dividing (ref) by $\left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}}$, we obtain \begin{align} &\mathbb{E}\left(\left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2\right) \notag\\ & = \int_{\mathcal{R}_{\overline{\mathcal{G}}}} \left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2 w_u(\*u) d \mathcal{H}^m(\*u)\notag \\ &= \int_{\mathcal{R}_{\overline{\mathcal{G}}}} \left[ \left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|\left(1 + \left\|\*u\right\|^2\right)^{-\frac{\tilde{\omega}}{2}}\right]^2 \left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}}w_u(\*u) d \mathcal{H}^m(\*u) \notag\\ & \leq \left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2_{\infty, \widetilde{\omega}} \int_{\mathcal{R}_{\overline{\mathcal{G}}}}\left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}} w_u(\*u) d\mathcal{H}^m(\*u) \notag\\ & = O\left(K^{-\frac{2\lambda_i}{m}}\right)\int_{\mathcal{R}_{\overline{\mathcal{G}}}}\left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}} w_u(\*u) d\mathcal{H}^m(\*u), \end{align} where the last equality holds by Assumption (ref)(ref) for a weight $\widetilde{\omega} \geq 0$. By Assumption $\ref{A5}$, $\overline{\mathcal{G}}(\cdot) \in \Lambda_r^{\lambda}(\mathcal{R}_f, \omega)$ and hence, for some generic $\*f_t$, \begin{align} \sup_{\*f_t} \left\|\overline{\mathcal{G}}(\*f_t) \left(1+ \left\|\*f_t\right\|^2\right)^{-{\frac{\omega}{2}}}\right\|_{\Lambda^\lambda} &\leq C, \notag\\ \Longleftrightarrow \left\|\overline{\mathcal{G}}(\*f_t)\right\| &\leq C \left(1+ \left\|\*f_t\right\|^2\right)^{{\frac{\omega}{2}}} \notag\\ \Longleftrightarrow \left\|\overline{\mathcal{G}}(\*f_t)\right\|^2 &\leq C \left(1+ \left\|\*f_t\right\|^2\right)^{\omega} \notag\\ \Longleftrightarrow \left(1 + \left\|\overline{\mathcal{G}}(\*f_t)\right\|^2\right)^{\widetilde{\omega}} &\leq C' \left(1+ \left\|\*f_t\right\|^2\right)^{\omega \widetilde{\omega}} = C' \left(1+ \left\|\*f_t\right\|^2\right)^{\overline{\omega}}, \end{align} for some constants $C, C' \in (0, \infty)$ and a weight $\overline{\omega} \geq 0$. This means that $\overline{\mathcal{G}}(\cdot)$ is bounded by the polynomial growth of $\*f_t$. Thus, \begin{align} \left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}} \equiv \left(1 + \left\|\overline{\mathcal{G}}\left(\*f_t\right)\right\|^2\right)^{\widetilde{\omega}} \leq C' \left(1+ \left\|\*f_t\right\|^2\right)^{\overline{\omega}}. \end{align} As discussed before, the inverse $\overline{\mathcal{G}}^{-1}(\cdot)$ exists and is well defined under Assumption (ref), such that \begin{align} \left(1+\left\|\overline{\mathcal{G}}^{-1}\left(\*u\right)\right\|^2\right)^{\overline{\omega}} &= \left(1+\left\|\overline{\mathcal{G}}^{-1}\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\|^2\right)^{\overline{\omega}} \notag \\ &=\left(1+\left\|\left(\overline{\mathcal{G}}^{-1}\circ\overline{\mathcal{G}}\right)\left(\*f_t\right)\right\|^2\right)^{\overline{\omega}} = \left(1+\left\|\*f_t\right\|^2\right)^{\overline{\omega}}. \end{align} We can make use of the results in (ref) and (ref) to arrive at the following expression for (ref): \begin{align} \mathbb{E}\left(\left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2\right) &= O\left(K^{-\frac{2\lambda_i}{m}}\right)\int_{\mathcal{R}_{\overline{\mathcal{G}}}}\left(1 + \left\|\*u\right\|^2\right)^{\widetilde{\omega}} w_u(\*u) d\mathcal{H}^m(\*u) \notag \\ &\leq O\left(K^{-\frac{2\lambda_i}{m}}\right)\int_{\mathcal{R}_{\overline{\mathcal{G}}}}\left(1 + \left\|\overline{\mathcal{G}}^{-1}\left(\*u\right)\right\|^2\right)^{\overline{\omega}} w_u(\*u) d\mathcal{H}^m(\*u) \notag \\ &= O\left(K^{-\frac{2\lambda_i}{m}}\right) \int_{\mathbb{R}^m} \left(1 + \left\|\*f_t\right\|^2\right)^{\overline{\omega}}w_u(\overline{\mathcal{G}}(\*f_t)) J\left( \overline{\mathcal{G}}(\*f_t)\right) d\*f_t \\ &= O\left(K^{-\frac{2\lambda_i}{m}}\right) \int_{\mathbb{R}^m} \left(1 + \left\|\*f_t\right\|^2\right)^{\overline{\omega}} w_f(\*f_t)d\*f_t, \end{align} where we use Theorem (ref) for (ref) and define $w_f(\*f_t) \equiv w_u(\overline{\mathcal{G}}(\*f_t))J\left( \overline{\mathcal{G}}(\*f_t)\right) $, with $J\left(\overline{\mathcal{G}}(\*f_t)\right) \equiv \sqrt{\det \left[\nabla \overline{\mathcal{G}}(\*f_t)'\nabla \overline{\mathcal{G}}(\*f_t)\right]}$ being the Jacobian of $\overline{\mathcal{G}}(\*f_t)$. By Assumption (ref)(ref), $\sup_{t=1, \dots, T}\int_{\mathcal{R}_f}\left(1+\left\|\*f_t\right\|^2\right)^{\overline{\omega}}w_f(\*f_t)d\*f_t < \infty$, which allows us to conclude that \begin{align} \mathbb{E}\left(\left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\|^2\right)= O\left(K^{-\frac{2\lambda_i}{m}}\right) O(1) = O\left(K^{-\frac{2\lambda_i}{m}}\right) . \end{align} Hence, \begin{align} \left\|h_i (\*u) - \+\alpha_{h_i}'\*p(\*u)\right\| &= \left\|h_i\left(\overline{\mathcal{G}}\left(\*f_t\right)\right) - \+\alpha_{h_i}'\*p\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| \notag \\ &= \left\|\mathcal{G}_i\left(\*f_t\right) - \+\alpha_{\mathcal{G}_i}'\*p\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| \notag\\ &=O_p\left(K^{-\frac{\lambda_i}{m}}\right)=o_p(1), \end{align} where the last equality follows from the fact that $K^{-\frac{\lambda_i}{m}}=o(1)$. This established part (ref) of the lemma. We move on to part (ref). By the mean-value theorem, there exists a vector $\*f_t^* \in \mathbb{R}^{d+1}$ which lies element-wise between $\overline{\mathcal{G}}(\*f_t)$ and $\widehat{\*f}_t$, such that \begin{align} \left\|\*p(\widehat{\*f}_t) - \*p\left(\overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| &= \left\|\nabla \*p\left(\*f_t^*\right)\left(\widehat{\*f}_t - \overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| \notag\\ & \leq \left\|\nabla \*p\left(\*f_t^*\right) \right\| \left\|\left(\widehat{\*f}_t - \overline{\mathcal{G}}\left(\*f_t\right)\right)\right\| \notag\\ & \leq \zeta_1\left(K\right) \left\|\overline{\*u}_t\right\|\notag \\ &= O_p\left(\frac{\zeta_1\left(K\right)}{\sqrt{N}}\right) \notag \\ &=O_p\left(o(1)\right) = o_p(1), \end{align} where we have used that $\overline{\*u}_t = O_p\left(\frac{1}{\sqrt{N}}\right)$ and the result in (ref). This establishes (ref), and hence the proof of Lemma (ref) is complete. $\blacksquare$ \@startsection{subsection}{2} \z@{-.5\linespacing\@plus-.7\linespacing}{.5\linespacing} {\normalfont}{Proof of Theorem (ref)} Recall from Assumption (ref) that $\widetilde{\mathbf{P}} = \left[\widetilde{\*p}\left(\*f_1\right), \dots, \widetilde{\*p}\left(\*f_T\right)\right]' = \left[\*p\left(\overline{\mathcal{G}}\left(\*f_1\right)\right), \dots, \*p\left(\overline{\mathcal{G}}\left(\*f_T\right)\right) \right]'$, where the mapping $\overline{\mathcal{G}}(\cdot)$ is defined in Assumption (ref). Further, we define $\+\alpha_{g_i}\in \mathbb{R}^{K \times 1}$ and $\+\alpha_{G_i} \in \mathbb{R}^{K \times d}$ as the sieve coefficient vector and matrix, when approximating $\*g_i(\*F)$ and $\*G_i(\*F)$, respectively. We continue by normalizing $\widehat{\mathbf{P}}$. As a convenient way to partly deal with the growing dimension of $\widetilde{\mathbf{P}}$, similarly to newey97, we replace $\widehat{\mathbf{P}}$ by $\widehat{\mathbf{P}}\left(\mathbb{E}\left[\frac{1}{T}\widetilde{\mathbf{P}}' \widetilde{\mathbf{P}}\right]\right)^{-\frac{1}{2}}$, where $\*A^{\frac{1}{2}}$ is such that $\left(\*A^{\frac{1}{2}}\right)'\*A^{\frac{1}{2}}=\*A$ for any positive semidefinite matrix $\*A$. This is possible because, by Assumption (ref)(ref), $\lambda_{\min}\left[\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right] > 0$. Moreover, this normalization preserves projection operators such that $\*M_{\widehat{P}} = \*M_{\widehat{P}\left(\mathbb{E}[\frac{1}{T} \widetilde{P}'\widetilde{P}]\right)^{-\frac{1}{2}}}$. Because of this normalization, we can assume without loss of generality that $\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right) = \*I_K$. We begin by re-writing equation (ref) for $\*y_{i}$ in the following way: \begin{align} \*y_i &= \*X_i\+\beta + \*g_i\left(\*F\right) + \+\varepsilon_i \notag \\ &= \*X_i \+\beta + \widehat{\*P}\+\alpha_{g_i} + \+\varepsilon_i - \left(\widehat{\*P}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right), \end{align} where \begin{align} \widehat{\*P}\+\alpha_{g_i} - \*g_i\left(\*F\right) &= \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \+\alpha_{g_i} + \left(\widetilde{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right). \end{align} Insertion into $\widehat{\+\beta}_{SCCE}$ yields \begin{align} \sqrt{NT}\left(\widehat{\+\beta}_{SCCE} - \+\beta\right) = \left(\frac{1}{NT}\sum_{i=1}^N \*X_i'\*M_{\widehat{P}}\*X_i \right)^{-1} \frac{1}{\sqrt{NT}}\sum_{i=1}^N\*X_i'\*M_{\widehat{P}}\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right]. \end{align} A key first step in establishing the asymptotic distribution of $\sqrt{NT}\left(\widehat{\+\beta}_{SCCE} - \+\beta\right)$ is to derive a cleaned up asymptotic representation that is free of all errors coming from the functional approximation. We start by considering $\*M_{\widehat P} \*X_i$. Equation (ref) for $\*X_i$ can be manipulated in the same way as (ref) for $\*y_i$ to obtain \begin{align} \*X_i = \*G_i\left(\*F\right) + \*V_i = \widehat{\mathbf{P}} \+\alpha_{G_i} + \*V_i - \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right), \end{align} where analogously to $\widehat{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)$ we can write \begin{align} \widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right) &= \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \+\alpha_{G_i} + \left(\widetilde{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right). \end{align} It follows that \begin{align} \*M_{\widehat{P}}\*X_i = \*M_{\widehat{P}}\left[\*V_i - \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right]. \end{align} We now make use of this last expression for $\*M_{\widehat{P}}\*X_i$ in order to evaluate the numerator of (ref). The following expansion will be used: \begin{align} &\frac{1}{\sqrt{NT}}\sum_{i=1}^N\*X_i'\*M_{\widehat{P}}\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right) \right] \notag \\ &=\frac{1}{\sqrt{NT}}\sum_{i=1}^N \left[\*V_i - \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right]'\*M_{\widehat{P}}\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right) \right] \notag \\ &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left[\*V_i - \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right]'\*M_{\widetilde{P}}\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right) \right] \notag \\ &- \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left[\*V_i - \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right]'\left(\*M_{\widetilde{P}} - \*M_{\widehat{P}}\right)\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right) \right] \notag \\ &= \sum_{j=1}^8 \*D_j, \end{align} where \begin{align} \*D_1 &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i' \*M_{\widetilde{P}} \+\varepsilon_i, \notag \\ \*D_2 &= - \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i' \*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right),\notag \\ \*D_3 &= - \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)'\*M_{\widetilde{P}}\+\varepsilon_i, \notag\\ \*D_4 &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right),\notag \\ \*D_5 &= - \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\*M_{\widetilde{P}} - \*M_{\widehat{P}}\right)\+\varepsilon_i, \notag\\ \*D_6 &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i' \left(\*M_{\widetilde{P}} - \*M_{\widehat{P}}\right)\left(\widehat{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right),\notag \\ \*D_7 &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left(\widehat{\mathbf{P}}\+\alpha_{G_i} - \*G_i\left(\*F\right)\right)'\left(\*M_{\widetilde{P}} - \*M_{\widehat{P}}\right)\+\varepsilon_i,\notag \\ \*D_8 &= - \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left(\widehat{\mathbf{P}} \+\alpha_{G_i} - \*G_i\left(\*F\right)\right)'\left(\*M_{\widetilde{P}} - \*M_{\widehat{P}}\right)\left(\widehat{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right). \end{align} We now proceed to evaluate each of the terms appearing above, starting with $\*D_5$. Analogous to (A.12) in westerlundpetrova2019, we have \begin{align} \*M_{\widetilde P} - \*M_{\widehat P} &= \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)' + \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \notag \\ & \quad + \widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)' + \widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}', \end{align} from which it follows that \begin{align} -\*D_5 & \equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\*M_{\widetilde P} - \*M_{\widehat{P}}\right)\+\varepsilon_i \notag \\ &= \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)' \+\varepsilon_i + \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\+\varepsilon_i \notag \\ &\quad + \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\+\varepsilon_i + \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\+\varepsilon_i \notag \\ &\equiv \*d_{5,1} + \*d_{5,2} + \*d_{5,3} + \*d_{5,4} , \end{align} with an implicit definition of $\*d_{5,1}$, $\*d_{5,2}$, $\*d_{5,3}$ and $\*d_{5,4}$. In what remains of this appendix, implicit definitions like these will be used repeatedly to save space. Before we begin evaluating each of the four terms in (ref), let us state the following lemma, Lemma (ref), which will be used throughout this proof. The proof of Lemma (ref) is provided after the proof of Theorem (ref). \begin{lemma} Under the conditions of Theorem (ref), \begin{enumerate}[label=(\roman*)] • $ \left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\| = O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right)$, • $\lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=1+o_p(1)$, $\lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=1+o_p(1)$, • $\lambda_{\min}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)=1+o_p(1)$, $\lambda_{\max}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)=1+o_p(1)$. \end{enumerate} \end{lemma} We start by taking the second term of $\*D_5$, $\*d_{5,2} \equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\+\varepsilon_i$, in detail. The evaluation of the other terms is similar. Consider some matrices $\*A$ and $\*B$. Then, $\mathrm{tr}(\*B\*A\*B')\leq \lambda_{\max}(\*A)\mathrm{tr}(\*B\*B')$ whenever $\*A$ is positive semidefinite (see lut, result (12) on page 44). Under Assumptions (ref) and (ref) this allows us to obtain \begin{align} & \mathbb{E}\left(\left\|\*d_{5,2}\right\|^2\right) \notag\\ &=\mathbb{E}\left(\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i' \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \+\varepsilon_i \right\|^2\right) \notag\\ & \leq \mathbb{E}\left[\mathrm{tr}\left(\frac{1}{NT}\sum_{i=1}^N\sum_{j=1}^N\+\varepsilon_i'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\mathbb{E}\left(\*V_i\*V_j'\right) \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \+\varepsilon_j\right)\right] \notag \\ &\leq \sum_{i=1}^N \lambda_{\max}\left(\+\Omega_{v,i}\right)\mathbb{E}\left[\mathrm{tr}\left(\frac{1}{NT}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \mathbb{E}\left(\+\varepsilon_i\+\varepsilon_i'\right)\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widetilde{\mathbf{P}} - \widehat{\mathbf{P}}\right)'\right)\right] \notag \\ &\leq \sum_{i=1}^N\lambda_{\max}\left(\+\Omega_{v,i}\right)\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right)\mathbb{E}\left[\frac{1}{NT}\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right] \notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right)\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^N\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right] \notag \\ &= O(1)\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^N\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right] \notag \\ &= O\left(\frac{1}{T}\right)\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right], \end{align} where the last equality holds by Assumption (ref)(ref), since the covariance matrices $\+\Omega_{\varepsilon,i}$ and $\+\Omega_{v,i}$ have eigenvalues bounded away from zero and infinity. We are left to analyze $\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right]$ in (ref). For that, add and subtract $\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\widetilde{\mathbf{P}}'$ from $\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'$ and obtain \begin{align} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' & = \frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}} \right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \notag\\ & = \frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\widetilde{\mathbf{P}}' + \frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+} -\*I_K\right] \widetilde{\mathbf{P}}'. \end{align} Hence, \begin{align} \mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \right\|^2\right] &\leq 2\mathbb{E}\left[\left\|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \widetilde{\mathbf{P}}' \right\|^2\right] \notag \\ &\quad + 2\mathbb{E}\left[\left \|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right] \widetilde{\mathbf{P}}' \right\|^2\right], \end{align} since $(a+b)^2 \leq 2(a^2+b^2)$. We begin by evaluating $\mathbb{E}\left[\left\|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \widetilde{\mathbf{P}}' \right\|^2\right]$. First, note how \begin{align} &\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^4\right] \notag\\ & = \mathbb{E}\left[\left(\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^2\right)^2\right] \notag \\ &=\mathbb{E}\left[\left(\frac{1}{T}\sum_{t=1}^T \left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^2\right)^2\right] \notag \\ &=\mathbb{E}\left[\frac{1}{T^2}\sum_{t=1}^T\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4+\frac{2}{T^2}\sum_{t <s}\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^2\left\|\*p(\widehat{\*f}_s)-\widetilde{\*p}(\*f_s)\right\|^2\right] \notag \\ &=\mathbb{E}\left[\frac{1}{T^2}\sum_{t=1}^T\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4\right]+\mathbb{E}\left[\frac{2}{T^2}\sum_{t< s}\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^2\left\|\*p(\widehat{\*f}_s)-\widetilde{\*p}(\*f_s)\right\|^2\right] \notag \\ &= \frac{1}{T^2}\sum_{t=1}^T \mathbb{E}\left(\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4\right) + \frac{2}{T^2}\sum_{t< s} \mathbb{E}\left[\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^2\left\|\*p(\widehat{\*f}_s)-\widetilde{\*p}(\*f_s)\right\|^2\right], \end{align} where we have used $\left(\sum_{i}a_i\right)^2 = \sum_ia_i^2+2\sum_{i< j}a_ia_j$. Using this fact again and the Cauchy-Schwarz inequality, the first term in (ref) can be expressed as \begin{align} &\frac{1}{T^{2}}\sum_{t=1}^{T}\mathbb{E}\left[\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\right] \notag\\ & =\frac{1}{T^{2}}\sum_{t=1}^{T}\mathbb{E}\left[\left(\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\right)^2\right]\notag \\ &=\frac{1}{T^{2}}\sum_{t=1}^{T}\mathbb{E}\left[\left(\sum_{k=1}^{K}\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{2}\right)^{2}\right] \notag \\ &=\frac{1}{T^{2}}\sum_{t=1}^{T}\mathbb{E}\left[\sum_{k=1}^{K}\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}+2\sum_{k< l}\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{2}\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{2}\right]\notag \\ &=\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{k=1}^{K}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]+\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{k< l}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{2}\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{2}\right]\notag \\ &\leq \frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{k=1}^{K}\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]+\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{k< l}\sup_{k,l}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{2}\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{2}\right]\notag \\ &\leq\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{k=1}^{K}\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]\notag \\ &\quad +\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{k< l}\sup_{k,l}\sqrt{\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]}\sqrt{\mathbb{E}\left[\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{4}\right]}\notag \\ &\leq\frac{K}{T^{2}}\sum_{t=1}^{T}\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]\notag \\ &\quad +\frac{2K^{2}}{T^{2}}\sum_{t=1}^{T}\sqrt{\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]}\sqrt{\sup_{l}\mathbb{E}\left[\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{4}\right]}\notag \\ &=\frac{K}{T}O\left(\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]\right)\notag \\ &\quad +\frac{2K^{2}}{T}O\left(\sqrt{\sup_{k}\mathbb{E}\left[\left|p_{k}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{4}\right]}\sqrt{\sup_{l}\mathbb{E}\left[\left|p_{l}(\widehat{\mathbf{f}}_{t})-\widetilde{p}_{l}(\mathbf{f}_{t})\right|^{4}\right]}\right)\notag \\ &=\frac{K}{T}O\left(\frac{\zeta_{1}(K)^{4}}{N^{2}}\right)+\frac{K^{2}}{T}O\left(\frac{\zeta_{1}(K)^{4}}{N^{2}}\right)\notag \\ &=O\left(\frac{K^{2}}{TN^{2}}\zeta_{1}(K)^{4}\right), \end{align} which holds because under Assumption (ref)(ref), we have $\sup_k \mathbb{E}\left(\left|p_k(\cdot)\right|^{8+\eta}\right)\leq C$ and $\sup_k \mathbb{E}\left(\left|\widetilde{p}_k(\cdot)\right|^{8+\eta}\right)\leq C$ for some $\eta > 0$. This implies that the $(8+\eta)$-th moments of $\left|p_k(\cdot)-\widetilde{p}(\cdot)\right|$ are uniformly bounded as well because by Minkowski inequality, $\sup_k\mathbb{E}\left(\left|p_k(\cdot)-\widetilde{p}(\cdot)\right|^{8+\eta}\right)^{\frac{1}{8+\eta}}\leq \sup_k\left[\mathbb{E}\left(\left|p_k(\cdot)\right|^{8+\eta}\right)\right]^{\frac{1}{8+\eta}}+\sup_k\left[\mathbb{E}\left(\left|\widetilde{p}_k(\cdot)\right|^{8+\eta}\right)\right]^{\frac{1}{8+\eta}}$. Consequently, the random variable $\left\{\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|\right\}$ is uniformly integrable (see for example Theorem 6.1 in dasgupta). This is important because uniform integrability combined with convergence in probability implies convergence in moments. This in turn implies that we can use the bound in Lemma (ref)(ref) to bound $\mathbb{E}\left(\left|p_k(\widehat{\*f}_t)-\widetilde{p}_k(\*f_t)\right|^4\right)$. The argument goes as follows: Since by Lemma (ref)(ref) $\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|=O_p\left(\frac{\zeta_1(K)}{\sqrt{N}}\right)$, we have $\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4=O_p\left(\frac{\zeta_1(K)^4}{N^2}\right)$. Further, it is easy to see that for any $k\in \left\{1, \dots, K\right\}$, \begin{align} \left|p_k(\widehat{\*f}_t) -\widetilde{p}_k(\*f_t)\right|^4 \leq \sum_{k=1}^K\left|p_k(\widehat{\*f}_t) -\widetilde{p}_k(\*f_t)\right|^4 = \left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4, \end{align} implying component-wise convergence in probability of the vector $\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)$, such that $\left|p_k(\widehat{\*f}_t) -\widetilde{p}_k(\*f_t)\right|^4 =O_p\left(\frac{\zeta_1(K)^4}{N^2}\right)$. We can therefore conclude that $\mathbb{E}\left(\left|p_k(\widehat{\*f}_t)-\widetilde{p}_k(\*f_t)\right|^4\right)=O\left(\frac{\zeta_1(K)^4}{N^2}\right)$ for all $k=1, \dots, K$. For the second term in (ref), we use the Cauchy-Schwarz inequality and the result we obtained in (ref). We arrive at \begin{align} &\frac{2}{T^2}\sum_{t\leq s} \mathbb{E}\left[\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^2\left\|\*p(\widehat{\*f}_s)-\widetilde{\*p}(\*f_s)\right\|^2\right] \notag \\ &\leq \frac{2}{T^2}\sum_{t\leq s}\sqrt{\mathbb{E}\left[\left\|\*p(\widehat{\*f}_t)-\widetilde{\*p}(\*f_t)\right\|^4\right]}\sqrt{\mathbb{E}\left[\left\|\*p(\widehat{\*f}_s)-\widetilde{\*p}(\*f_s)\right\|^4\right]} \notag \\ &= \frac{2}{T^2}\frac{T(T-1)}{2}O\left(\frac{K^2}{N^2}\zeta_1(K)^4\right) \notag \\ &= O\left(\frac{K^2}{N^2}\zeta_1(K)^4\right). \end{align} Consequently, \begin{align} \mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^4\right] &= O\left(\frac{K^{2}}{TN^{2}}\zeta_{1}(K)^{4}\right) + O\left(\frac{K^2}{N^2}\zeta_1(K)^4\right) \notag \\ &=O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right). \end{align} We now move on to $\mathbb{E}\left[\|\frac{1}{T}(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}})\widetilde{\mathbf{P}}'\|^2\right]$. To bound this term, we can use (ref) and the Cauchy-Schwarz inequality to obtain \begin{align} \mathbb{E}\left[\left\|\frac{1}{T}(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}})\widetilde{\mathbf{P}}'\right\|^2\right] & \le\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{2}\right]\notag \\ &\leq\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]}\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{4}\right]} \notag \\ &= \sqrt{O\left(\frac{K^2}{N^2}\zeta_1(K)^4\right)}\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{4}\right]} \notag \\ &=O\left(\frac{K}{N}\zeta_1(K)^2\right)\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{4}\right]}, \end{align} where by using Jensen's inequality and the normalization condition $\mathbb{E}(\widetilde{\*p}(\*f_t)\widetilde{\*p}(\*f_t)')=\*I_K$, we can show that \begin{align} \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{4}\right] &=\mathbb{E}\left[\left(\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{2}\right)^{2}\right]\notag\\ &=\mathbb{E}\left[\left(\frac{1}{T}\sum_{t=1}^{T}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\right)^{2}\right]\notag \\ &=\mathbb{E}\left[\frac{1}{T^{2}}\sum_{t=1}^{T}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}+\frac{2}{T^{2}}\sum_{t<s}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\right]\notag \\ &=\mathbb{E}\left[\frac{1}{T^{2}}\sum_{t=1}^{T}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\right]+\mathbb{E}\left[\frac{2}{T^{2}}\sum_{t<s}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\right]\notag \\ &\leq\sup_{\mathbf{f}_{t}}\mathbb{E}\left[\frac{1}{T^{2}}\sum_{t=1}^{T}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\right]+\sup_{\mathbf{f}_{t}}\mathbb{E}\left[\frac{2}{T^{2}}\sum_{t<s}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\right]\notag \\ &\leq\frac{1}{T^{2}}\sum_{t=1}^{T}\sup_{\mathbf{f}_{t}}\mathbb{E}\left[\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\right]+\frac{2}{T^{2}}\sum_{t<s}\sup_{\mathbf{f}_{t}}\mathbb{E}\left[\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\right]\notag \\ &\leq\frac{1}{T^2}\sum_{t=1}^T \mathrm{tr}\left[\mathbb{E}\left(\widetilde{\mathbf{p}}(\mathbf{f}_{t})\widetilde{\mathbf{p}}(\mathbf{f}_{t})'\right)\right]\zeta_0(K)^2 \notag\\ & \quad+\frac{2}{T^{2}}\sum_{t<s}\sqrt{\sup_{\*f_t}\mathbb{E}\left(\left\|\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\|^4\right)}\sqrt{\sup_{\*f_s}\mathbb{E}\left(\left\|\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\|^4\right)}\notag \\ &\leq\frac{K\zeta_{0}(K)^{2}}{T}+\frac{2}{T^{2}}\frac{T(T-1)}{2}K\zeta_{0}(K)^{2}\notag \\ &=\frac{K\zeta_{0}(K)^{2}}{T}+\frac{T-1}{T}K\zeta_{0}(K)^{2}\notag \\ &=\frac{K\zeta_{0}(K)^{2}}{T}+K\zeta_{0}(K)^{2}-\frac{K\zeta_{0}(K)^{2}}{T}\notag \\ &=K\zeta_{0}(K)^{2}, \end{align} and hence, \begin{align} \mathbb{E}\left[\left\|\frac{1}{T}(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}})\widetilde{\mathbf{P}}'\right\|^2\right] &\leq O\left(\frac{K}{N}\zeta_1(K)^2\right)\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{4}\right]} \notag \\ &= O\left(\frac{K}{N}\zeta_1(K)^2\right)\sqrt{O\left(K\zeta_{0}(K)^{2}\right)}\notag \\ &=O\left(\frac{K^{\frac{3}{2}}}{N}\zeta_1(K)^2\zeta_0(K)\right). \end{align} We now analyze the second term in (ref) in detail. First, applying H\"older's inequality, we get \begin{align} &\mathbb{E}\left\{ \left\Vert \frac{1}{T}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\widetilde{\mathbf{P}}'\right\Vert ^{2}\right\} \notag \\ &\leq\mathbb{E}\left\{ \left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{2}\right\} \notag \\ &\leq\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right]^{\frac{1}{3}}\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{6}\right]^{\frac{1}{3}}\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{6}\right]^{\frac{1}{3}}. \end{align} The first term in (ref) can be written as \begin{align} &\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right] \notag\\ &=\mathbb{E}\left[\left(\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\right)^{3}\right]\notag \\ &=\mathbb{E}\left[\left(\frac{1}{T}\sum_{t=1}^{T}\left\|\mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\|^{2}\right)^{3}\right]\notag \\ &=\mathbb{E}\left[\frac{1}{T^{3}}\sum_{t=1}^{T}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{6}+\frac{3}{T^{3}}\sum_{t\neq s}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{s})-\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert^2\right. \notag \\ &\quad \left.+\frac{6}{T^{3}}\sum_{t<s<r}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert^2 \left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{s})-\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert^2 \left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{r})-\widetilde{\mathbf{p}}(\mathbf{f}_{r})\right\Vert^2\right] \notag \\ &=\mathbb{E}\left[\frac{1}{T^{3}}\sum_{t=1}^{T}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{6}\right]\notag \\ &\quad+\mathbb{E}\left[\frac{3}{T^{3}}\sum_{t\neq s}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{s})-\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert^2 \right]\notag \\ &\quad+\mathbb{E}\left[\frac{6}{T^{3}}\sum_{t<s<r}\left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{t})-\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert^2 \left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{s})-\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert^2 \left\Vert \mathbf{p}(\widehat{\mathbf{f}}_{r})-\widetilde{\mathbf{p}}(\mathbf{f}_{r})\right\Vert^2 \right], \end{align} where we have used $\left(\sum_i a_i\right)^3=\sum_i a_i^3+3\sum_{i\neq l}a_i^2a_l + 6\sum_{i<l<m}a_ia_la_m$. While the first term in (ref) can be bounded using the same arguments as in (ref), the second and third terms can be bounded by arguments analogous to those used to evaluate (ref). It follows that \begin{align} \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right]&=O\left(\frac{K^{3}}{N^{3}T^{2}}\zeta_{1}(K)^{6}\right) + O\left(\frac{K^{3}}{N^{3}T}\zeta_{1}(K)^{6}\right) + O\left(\frac{K^{3}}{N^{3}}\zeta_{1}(K)^{6}\right)\notag \\ &= O\left(\frac{K^{3}}{N^{3}}\zeta_{1}(K)^{6}\right). \end{align} Terms of the form in (ref) and (ref) will reappear with different exponents throughout the proof. The key point is that their asymptotic behavior follows the same pattern; after pre-multiplying by $\frac{1}{T^\delta}$ with $\delta>0$, only the terms involving the maximal number of distinct indices survive. These sums scale with $O(T^\delta)$ while all other components grow more slowly, typically $O(T^{\delta-\eta})$ with $0<\eta\leq \delta$. After dividing by $T^\delta$ those lower-order terms vanish. From now on we will make use of this observation and Assumption (ref)(ref) to bound higher-order moments, instead of computing them explicitly. We have evaluated the first term on the right-hand side of (ref). For the second term in the same equation, we use \begin{align} \widehat{\mathbf{P}}'\widehat{\mathbf{P}} = \widetilde{\mathbf{P}}'\widetilde{\mathbf{P}} + \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\widetilde{\mathbf{P}} + \widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) + \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right). \end{align}This allows us to write \begin{align} \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\*I_K\right\| &\le \left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\| +2\left\|\frac{1}{T}(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}})'\widetilde{\mathbf{P}}\right\| +\left\|\frac{1}{\sqrt{T}}(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}})\right\|^2. \end{align}By Lemma (ref)(ref), and the known orders of the variances of $\left\|\frac{1}{T}(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}})'\widetilde{\mathbf{P}}\right\|$ from (ref) and $\left\|\frac{1}{\sqrt{T}}(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}})\right\|^2$ from (ref), (ref) can be bounded as \begin{align} \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\*I_K\right\| &= O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right) + O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_1(K)\sqrt{\zeta_0(K)}\right) + O_p\left(\frac{K}{N}\zeta_1(K)^2\right) \notag \\ &= O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right) + O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_1(K)\sqrt{\zeta_0(K)}\right), \end{align} since $\frac{\frac{K}{N}\zeta_1(K)^2}{\sqrt{\frac{K}{T}}\zeta_0(K)} = \frac{\sqrt{KT}}{N}\frac{\zeta_1(K)^2}{\zeta_0(K)}=o(1)$ by (ref) and $\frac{\frac{K}{N}\zeta_1(K)^2}{\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_1(K)\sqrt{\zeta_0(K)}} = \frac{K^{\frac{1}{4}}\zeta_1(K)}{\sqrt{N}\sqrt{\zeta_0(K)}}=o(1)$ by (ref). Hence, the third term above is dominated by the first two. The two terms that remain will appear again and again. It is therefore convenient to introduce \begin{align} \delta_{NTK} \equiv \max\left\{\sqrt{\frac{K}{T}}\zeta_0(K), \, \frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_1(K)\sqrt{\zeta_0(K)}\right\}. \end{align} By (ref) and (ref) we have \begin{align} \delta_{NTK}=o(1). \end{align} This result will be used repeatedly in the sequel. We continue by noting that \begin{align} \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\*I_K\right\| =O_p\left(\delta_{NTK}\right). \end{align} By Lemma (ref)(ref), asymptotically $\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}$ will have full rank $K$, as $N \to \infty$, and therefore the Moore-Penrose inverse of this matrix reduces to the standard inverse. This implies that \begin{align} \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K = \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1} - \*I_K = -\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right). \end{align} If $\lambda$ is an eigenvalue of $\*A$, then $\lambda^2$ is an eigenvalue of $\*A^2$ and $\lambda^{-1}$ is an eigenvalue of $\*A^{-1}$. Moreover, because $\lambda_{\max}(\*A) \geq \lambda_{\min}(\*A)$, we have $\lambda_{\max}(\*A)^{-1} \leq \lambda_{\min}(\*A)^{-1}$. Together with $\mathrm{tr}\,(\*B\*A\*B') \leq \lambda_{\max}(\*A) \mathrm{tr}\,(\*B\*B')$, these results yield \begin{align} \left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right\|^2 &= \mathrm{tr}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right] \notag\\ &= \mathrm{tr}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-2}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right] \notag \\ &\leq \lambda_{\max}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1}\right]^2\mathrm{tr}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right] \notag \\ &\leq \lambda_{\max}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-2}\mathrm{tr}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right] \notag \\ & \leq \lambda_{\min}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-2}\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right\|^2. \end{align} Hence, by Lemma (ref)(ref) and the result in (ref) \begin{align} \left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right\|& \leq \lambda_{\min}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1}\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}} - \*I_K\right)\right\|\notag \\ &= O_p(1) O_p\left(\delta_{NTK}\right) \notag \\ &=O_p\left(\delta_{NTK}\right). \end{align} For convenience, let us define $\left\|\*E_K\right\|\equiv \left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right\|=\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{-1} - \*I_K \right\|=O_p(\delta_{NTK})$. To bound the second term in (ref), we use that $\left\|\*E_K\right\|^6$ is non-negative. It follows that \begin{align} \mathbb{E}\left[\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\right] =\int_0^\infty \mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\geq x\right)dx, \end{align} where $\mathbb{P}(A)$ is the probability of the event $A$. Performing change of variables, such that $x=t^6$ and consequently $dx=6t^5 dt$ yields \begin{align} \mathbb{E}\left[\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\right]& = \int_0^\infty \mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\geq x\right)dx = 6\int_0^\infty t^5\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\geq t^6\right)dt\notag \\ &=6\int_{0}^{\infty}t^{5}\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)dt\notag \\ &=6\int_{0}^{C_1\delta_{NTK}}t^{5}\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)dt+6\int_{C_1\delta_{NTK}}^{\infty}t^{5}\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)dt\notag\\ &\equiv I_{1}+I_{2}, \end{align} with implicit definitions of $I_1$ and $I_2$. We can bound $I_1$ as \begin{align} I_{1}&\equiv6\int_{0}^{C_{1}\delta_{NTK}}t^{5}\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)dt \leq6\int_{0}^{C_{1}\delta_{NTK}}t^{5}\cdot1dt =\left[t^{5}\right]_{0}^{C_{1}\delta_{NTK}}+C_{2}\notag \\ &=C_{1}^{5}\delta_{NTK}^{5}+C_{2}, \end{align} since $\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)\leq 1$. Therefore, $I_{1}=O\left(\delta_{NTK}^{5}\right)$. Let us now consider $I_2$; \begin{align} I_{2}&\equiv6\int_{C_{2}\delta_{NTK}}^{\infty}t^{5}\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)dt\le6\int_{C_{2}\delta_{NTK}}^{\infty}t^{5}C_{3}\exp(-C_{4}t^{2})dt\notag \\ &=6C_{3}\int_{C_{2}\delta_{NTK}}^{\infty}t^{4}\exp(-C_{4}t^{2})tdt, \end{align} where we have used the fact that bounded random variables belong to the class of sub-gaussian variables, such that one can bound $\mathbb{P}\left(\left\Vert \mathbf{E}_{K}\right\Vert \geq t\right)\leq C_{3}\exp(-C_{4}t^{2})$ (see, for example, rudel08). Let $u=t^2$ implying $du=2tdt$ and in turn $tdt=\frac{1}{2}du$. Integration by substitution and by parts now yields \begin{align} I_{2}&\leq6C_{3}\int_{C_{2}\delta_{NTK}}^{\infty}t^{4}\exp(-C_{4}t^{2})tdt\notag \\ &=3C_{3}\lim_{r\to\infty}\int_{C_{2}^{2}\delta_{NTK}^{2}}^{r}u^{2}\exp(-C_{4}u)du\notag \\ &=-\frac{3C_3}{C_{4}}\lim_{r\to\infty}\left[u^{2}\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}+\frac{3C_3}{C_{4}}\lim_{r\to\infty}\int_{C_{2}^{2}\delta_{NTK}^{2}}^{r}2u\exp(-C_{4}u)du\notag \\ &=-\frac{3C_3}{C_{4}}\lim_{r\to\infty}\left[u^{2}\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}-\frac{6C_3}{C_{4}^2}\lim_{r\to\infty}\left[u\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}\notag \\ &\quad +\frac{6C_3}{C_{4}^2}\lim_{r\to\infty}\int\exp(-C_{4}u)du\notag \\ &=-\frac{3C_3}{C_{4}}\lim_{r\to\infty}\left[u^{2}\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}-\frac{6C_3}{C_{4}^2}\lim_{r\to\infty}\left[u\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}\notag \\ &\quad -\frac{6C_3}{C_{4}^{3}}\lim_{r\to\infty}\left[\exp(-C_{4}u)\right]_{C_{2}^{2}\delta_{NTK}^{2}}^{r}\notag \\ &=\lim_{r\to\infty}\left[-\left(\frac{3C_3}{C_4}u^2+\frac{6C_3}{C_4^2}u+\frac{6C_3}{C_4^2}\right)\exp\left(-C_4 u\right)\right]^r_{C_2^2\delta_{NTK}^2}\notag \\ &=\left(\frac{3C_3C_2^2}{C_4}\delta_{NTK}^4 + \frac{6C_3C_2^2}{C_4^2}\delta_{NTK}^2+\frac{6C_3}{C_4^2}\right)\exp\left(-C_4C_2^2\delta_{NTK}^2\right), \end{align} where $C_1,\dots, C_4$ are nonnegative constants. Since $\delta_{NTK}\to0$ and by the continuous mapping theorem, $\exp(-\delta_{NTK}^{2})\to\exp(0)=1$, we can show that $I_2=O(1)$. Thus, \begin{align} \mathbb{E}\left[\left\Vert \mathbf{E}_{K}\right\Vert ^{6}\right] \equiv \mathbb{E}\left[\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right\|^6\right]=I_1 + I_2 = O\left(\delta_{NTK}^5\right) + O(1) = O(1), \end{align} since $\delta_{NTK}\to 0$. We are left to bound the third term in (ref). Here, we can rely on the same steps used to derive (ref) to arrive at \begin{align} \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{6}\right] &= \mathbb{E}\left[\left(\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{2}\right)^3\right] = \mathbb{E}\left[\left(\frac{1}{T}\sum_{t=1}^{T}\left\Vert \widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\right)^{3}\right]\notag \\ &\leq K\zeta_0(K)^4. \end{align} By combining the results, we obtain the following bound for (ref): \begin{align} &\mathbb{E}\left\{ \left\Vert \frac{1}{T}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\widetilde{\mathbf{P}}'\right\Vert ^{2}\right\} \notag \\ &\leq\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right]^{\frac{1}{3}}\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{6}\right]^{\frac{1}{3}}\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{6}\right]^{\frac{1}{3}} \notag \\ &= \left[O\left(\frac{K^3}{N^3}\zeta_1(K)^6\right)\right]^{\frac{1}{3}}\left[O\left(1\right)\right]^{\frac{1}{3}}\left[O\left(K\zeta_0(K)^4\right)\right]^{\frac{1}{3}} \notag \\ &=O\left(\frac{K^{\frac{4}{3}}}{N}\zeta_1(K)^2\zeta_0(K)^{\frac{4}{3}}\right), \end{align} and consequently, \begin{align} \mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}' \right\|^2\right] &\leq 2\mathbb{E}\left[\left\|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \widetilde{\mathbf{P}}' \right\|^2\right]\notag \\ &\quad + 2\mathbb{E}\left[\left \|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \*I_K \right] \widetilde{\mathbf{P}}' \right\|^2\right] \notag\\ &=O\left(\frac{K^{\frac{3}{2}}}{N}\zeta_1(K)^2\zeta_0(K)\right) + O\left(\frac{K^{\frac{4}{3}}}{N}\zeta_1(K)^2\zeta_0(K)^{\frac{4}{3}}\right). \end{align} Therefore, \begin{align} \mathbb{E}\left(\left\|\*d_{5,2}\right\|^2\right) &\equiv \mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\+\varepsilon_i\right\|^2\right] \notag \\ &\leq O\left(\frac{1}{T}\right)\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+ \widetilde{\mathbf{P}}'\right\|^2\right]\notag \\ &= O\left(\frac{K^{\frac{3}{2}}}{NT}\zeta_1(K)^2\zeta_0(K)\right) + O\left(\frac{K^{\frac{4}{3}}}{NT}\zeta_1(K)^2\zeta_0(K)^{\frac{4}{3}}\right). \end{align} This allows us to conclude that \begin{align} \left\|\*d_{5,2}\right\| = O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{NT}}\sqrt{\zeta_0(K)}\zeta_1(K)\right) + O_p\left(\frac{K^{\frac{2}{3}}}{\sqrt{NT}}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right). \end{align} The above arguments can be applied also to the other terms in (ref). The third term, $\*d_{5,3} \equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\+\varepsilon_i$, is of the same order as $\left\|\*d_{5,2}\right\|$; \begin{align} \left\|\*d_{5,3}\right\| = O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{NT}}\sqrt{\zeta_0(K)}\zeta_1(K)\right) + O_p\left(\frac{K^{\frac{2}{3}}}{\sqrt{NT}}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right). \end{align} For $\*d_{5,4}\|$, we can use the same steps as for analyzing $\mathbb{E}\left(\left\|\*d_{5,2}\right\|^2\right)$ to obtain \begin{align} &\mathbb{E}\left(\left\|\*d_{5,4}\right\|^2\right) \notag\\ & = \mathbb{E} \left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\+\varepsilon_i\right\|^2\right] \notag \\ &\leq \mathbb{E}\left[\mathrm{tr}\left(\frac{1}{NT}\sum_{i=1}^N\sum_{j=1}^N\+\varepsilon_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\mathbb{E}\left(\*V_i \*V_j'\right)\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\+\varepsilon_j\right)\right] \notag \\ &= \mathbb{E}\left[\mathrm{tr}\left(\frac{1}{NT}\sum_{i=1}^N\+\varepsilon_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\+\Omega_{v,i}\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\+\varepsilon_i\right)\right] \notag \\ &\leq \frac{1}{NT} \sum_{i=1}^N \lambda_{\max}\left(\+\Omega_{v,i}\right)\mathbb{E}\left[\mathrm{tr}\left(\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\mathbb{E}\left(\+\varepsilon_i\+\varepsilon_i'\right)\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right)\right] \notag \\ &\leq \frac{1}{NT} \sum_{i=1}^N \lambda_{\max}\left(\+\Omega_{v,i}\right)\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right)\notag\\ &\quad \times \mathbb{E}\left[\mathrm{tr}\left(\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right)\right] \notag \\ &= \frac{1}{NT} \sum_{i=1}^N \lambda_{\max}\left(\+\Omega_{v,i}\right)\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right)\mathbb{E}\left(\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right\|^2\right) \notag \\ & = \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right) \frac{1}{NT}\sum_{i=1}^N\mathbb{E}\left(\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right\|^2\right) \notag \\ &= \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{\varepsilon,i}\right) \frac{1}{T}\mathbb{E}\left(\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right\|^2\right) \notag \\ &= O\left(\frac{1}{T}\right)\mathbb{E}\left(\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right\|^2\right). \end{align} Further use of \begin{align} \left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\|&=\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}+\mathbf{I}_{K}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\|\notag\\ &=\left\|\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]-\left[\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\right\|\notag\\ &\leq\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|+\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|, \end{align} and the Cauchy-Schwarz inequality yields \begin{align} &\mathbb{E}\left(\left\Vert \widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\right\Vert ^{2}\right) \notag \\ &=\mathbb{E}\left\{ \mathrm{tr}\left[\frac{1}{T}\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\right]\right\}\notag \\ &=\mathbb{E}\left\{ \mathrm{tr}\left[\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\right]\right\}\notag \\ &=\mathbb{E}\left\{ \left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\right\|^{2}\right\}\notag \\ &\leq\mathbb{E}\left[\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\|^{2}\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\|^{2}\right]\notag\\ &\leq\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\Vert ^{4}\right]}\sqrt{\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\Vert ^{4}\right]}\notag\\ &\leq \sqrt{\mathbb{E}\left[\left\Vert \frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\Vert ^{4}\right]}\sqrt{\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{4}\right]+\mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|^{4}\right]}\notag \\ &\leq \sqrt{\mathbb{E} \left[\left\|\frac{1}{\sqrt{T}} \widetilde{\mathbf{P}}\right\|^4 \left\|\frac{1}{\sqrt{T}} \widetilde{\mathbf{P}}\right\|^4\right]}\sqrt{\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{4}\right]+\mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|^{4}\right]} \notag\\ &\leq \sqrt{\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^8\right]}\sqrt{\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{4}\right]+\mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|^{4}\right]}. \end{align} From our previous derivations in (ref) and (ref), we have \begin{align} \mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^8\right]=K\zeta_0(K)^6, \end{align} and \begin{align} \mathbb{E}\left[\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+-\*I_K\right\|^4\right] = O(1). \end{align} Thus, we are left to bound $\mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+ - \*I_K\right\|^4\right]$. Here, we can employ exactly the same strategy we used to bound (ref). By Lemma (ref)(ref), asymptotically $\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}$ will have full rank $K$, as $N\to \infty$ and therefore, the Moore-Penrose inverse of this matrix reduces to the standard inverse also in this case. Thus, similarly to (ref)--(ref), we obtain \begin{align} \left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+ - \*I_K\right\| &\leq \lambda_{\min} \left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{-1}\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\| \notag\\ &=O_p(1) O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right)\notag\\ &=O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right), \end{align} which holds by Lemma (ref)(ref) and (ref)(ref). Moreover, we can use exactly the same arguments as in (ref)--(ref) to show that \begin{align} \mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+ - \*I_K\right\|^4\right] = O\left(\frac{K^2}{T^2}\zeta_0(K)^4\right)+O(1)=O(1), \end{align} since $\frac{K^2}{T^2}\zeta_0(K)^4=o\left(\frac{K^2\zeta_0(K)^4}{K^2\zeta_0(K)^6}\right)=o\left(\frac{1}{\zeta_0(K)^2}\right)=o(1)$ under Assumption (ref)(ref). Therefore, the bound for the expression in (ref) is given by \begin{align} &\mathbb{E}\left(\left\Vert \widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\right\Vert ^{2}\right)\notag\\ &\leq \sqrt{\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^8\right]}\sqrt{\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{4}\right]+\mathbb{E}\left[\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|^{4}\right]}\notag \\ &=\sqrt{O\left(K\zeta_0(K)^6\right)}\sqrt{O(1)}\notag\\ &=O\left(\sqrt{K}\zeta_0(K)^3\right), \end{align} yielding \begin{align} \mathbb{E}\left(\left\|\*d_{5,4}\right\|^2\right) &\leq O\left(\frac{1}{T}\right)\mathbb{E}\left(\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\right\|^2\right)\notag\\ &=O\left(\frac{\sqrt{K}}{T}\zeta_0(K)^3\right), \end{align} and thus, \begin{align} \left\|\*d_{5,4}\right\| = O_p\left(\frac{K^{\frac{1}{4}}}{\sqrt{T}}\zeta_0(K)^{\frac{3}{2}}\right). \end{align} The first term in (ref), $\*d_{5,1}\equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}' \widehat{\mathbf{P}}\right)^+\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\+\varepsilon_i$, can be analyzed very similarly to (ref). In particular, \begin{align} \mathbb{E}\left(\left\|\mathbf{d}_{5,1}\right\|^{2}\right) &=\mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\boldsymbol{\varepsilon}_{i}\right\|^{2}\right] \notag\\ &\leq O\left(\frac{1}{T}\right)\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\|^{2}\right], \end{align} where \begin{align} &\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\Vert ^{2}\right]\notag\\ &\leq2\mathbb{E}\left[\left\|\frac{1}{T}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\|^{2}\right]+2\mathbb{E}\left[\left\|\frac{1}{T}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\|^{2}\right]\notag\\ &\leq2\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{4}\right]+2\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\left\Vert \left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\right]\notag\\ &\leq2\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]+2\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right]\right\} ^{\frac{1}{3}}\left\{ \mathbb{E}\left[\left\Vert \left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\right\Vert ^{6}\right]\right\} ^{\frac{1}{3}}\notag\\ & \quad \times \left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{6}\right]\right\} ^{\frac{1}{3}}\notag\\ &=O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right)+O\left(\frac{K^{3}}{N^{3}}\zeta_{1}(K)^{6}\right)^{\frac{2}{3}}O(1)\notag\\ &=O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right)+O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right)\notag\\ &=O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right). \end{align} Here we have used the results in (ref), (ref) and (ref). Thus, \begin{align} \left\|\*d_{5,1}\right\| = O_p\left(\frac{K}{NT}\zeta_1(K)^2\right). \end{align} By adding the results, \begin{align} \left\|\*D_5\right\| &= \left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_{i}' (\*M_{\widetilde{P}} - \*M_{\widehat P}) \+\varepsilon_{i} \right\| \notag \\ &\leq \left\|\*d_{5,1}\right\| + \left\|\*d_{5,2}\right\| + \left\|\*d_{5,3}\right\| + \left\|\*d_{5,4}\right\| \notag \\ &=O_p\left(\frac{K}{NT}\zeta_1(K)^2\right) + O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{NT}}\sqrt{\zeta_0(K)}\zeta_1(K)\right) \notag \\ &\quad + O_p\left(\frac{K^{\frac{2}{3}}}{\sqrt{NT}}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right) + O_p\left(\frac{K^{\frac{1}{4}}}{\sqrt{T}}\zeta_0(K)^{\frac{3}{2}}\right)\notag \\ &= o_p(1), \end{align} where we have used (ref), (ref) and (ref) to show the last equality. We now move to $\mathbf{D}_{6},$ which we again expand using (ref); \begin{align} \mathbf{D}_{6} & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}})\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \equiv \*d_{6,1}+ \*d_{6,2} + \*d_{6,3} + \*d_{6,4}. \end{align} As before, we proceed by evaluating each of the four terms on the right-hand side of (ref). In light of the decomposition in (ref), it is important to note that each of these four terms is actually comprised of two terms, as each component in $\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)$ must be evaluated separately. We start by considering the second term in (ref): \begin{align} \*d_{6,2} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right) \notag \\ &\equiv \*d^{(1)}_{6,2} + \*d^{(2)}_{6,2}. \end{align} Both terms on the right can be treated the same way as $\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{i}.$ The treatment of the second term is actually almost identical. The reason is that $\mathbf{V}_{i}$ is mean zero and independent across $i$. It is also independent of $\mathbf{F}$, and thus of any function thereof, including $\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})$. Hence, using the same arguments as for $\*d_{5,2}$, we can consider the following variance expression for $\*d_{6,2}^{(2)}$: \begin{align} & \mathbb{E}\left(\|\*d_{6,2}^{(2)}\|^2 \right)\\ & = \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert^2\right] \notag \\ & \leq \mathbb{E}\left[\mathrm{tr}\left(\frac{1}{NT}\sum_{i=1}^N\sum_{j=1}^N\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbb{E}\left(\*V_i\mathbf{V}_{j}'\right) \right.\right. \notag\\ & \quad \times \left. \left. \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widetilde{\mathbf{P}}\+\alpha_{g_j}-\*g_j(\mathbf{F})\right)\right)\right] \notag \\ &\leq \frac{1}{NT}\sum_{i=1}^N\lambda_{\max}\left(\+\Omega_{v,i}\right) \notag\\ & \quad \times \mathbb{E}\left[\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right)\right] \notag \\ &=\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\|^{2}\right]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})\notag\\ &\quad\times\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{2}\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^{2}\frac{1}{N}\sum_{i=1}^{N}\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\|^{2}\right]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}+\mathbf{I}_{K}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{8}\right]+\mathbb{E}\left(\left\Vert \mathbf{I}_{K}\right\Vert ^{8}\right)\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &=O(1)\left[O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right]^{\frac{1}{4}}\left[O(1)+O\left(K^{8}\right)\right]^{\frac{1}{4}}\left[K\zeta_{0}(K)^{6}\right]^{\frac{1}{4}}\notag\\ &\quad\times\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &=O\left(\frac{K^{\frac{13}{4}}}{N}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}. \end{align} Most of the terms in (ref) are either known from previous derivations or can be extrapolated using those. The only new term is \begin{align} &\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\notag\\ &=\mathbb{E}\left[\left(\frac{1}{T}\sum_{t=1}^{T}\left\|g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\|^{2}\right)^{4}\right]\notag\\ &=\frac{1}{T^{4}}\sum_{t=1}^{T}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{8}\right]\notag\\ &\quad+\frac{4}{T^{4}}\sum_{t\neq s}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{6}\left\Vert g_{i}(\mathbf{f}_{s})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\right]\notag\\ &\quad+\frac{6}{T^{4}}\sum_{t<s}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\left\Vert g_{i}(\mathbf{f}_{s})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{4}\right]\notag\\ &\quad+\frac{12}{T^{4}}\sum_{t\neq s\neq r}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{4}\left\Vert g_{i}(\mathbf{f}_{s})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\left\Vert g_{i}(\mathbf{f}_{r})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{r})\right\Vert ^{2}\right]\notag\\ &\quad+\frac{24}{T^{4}}\sum_{t<s<r<q}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\left\Vert g_{i}(\mathbf{f}_{s})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{s})\right\Vert ^{2}\left\Vert g_{i}(\mathbf{f}_{r})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{r})\right\Vert ^{2} \right.\notag \\ &\quad \left. \times \left\Vert g_{i}(\mathbf{f}_{q})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{q})\right\Vert ^{2}\right]. \end{align} By Assumption (ref)(ref) and (ref), for $g_{i}(\cdot)\in \Lambda_r^{\lambda_i}(\mathcal{R}_f, \omega_i)$ there exists $\Pi_{\infty K}g_{i}(\cdot)=\+\alpha'_{g_i}\*p(\cdot)\in \mathfrak{G}_K$, such that for any $\overline{\omega} > \sup_{i=1,\dots, N}(\omega_i + \lambda_i)$ we can write \begin{align} \left\|g_i(\cdot)-\Pi_{\infty K}g_i(\cdot)\right\|_{\infty, \overline{\omega}} = \sup_{\*f \in \mathcal{R}_f} \left|\left[g_i(\*f)-\+\alpha'_{g_i}\*p(\*f)\right]\left[1+\left\|\*f\right\|^2\right]^{-\frac{\overline{\omega}}{2}}\right| \leq CK^{-\frac{\lambda_i}{m}}, \end{align} where we have used the definition of the weighted sup-norm from (ref). Then, by Assumption (ref), \begin{align} &\frac{1}{T^{4}}\sum_{t=1}^{T}\mathbb{E}\left[\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{8}\right] = \frac{1}{T^{4}}\sum_{t=1}^{T}\mathbb{E}\left[\left(\left\Vert g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right\Vert ^{2}\right)^{4}\right]\notag\\ &=\frac{1}{T^{3}}\int_{\mathcal{R}_f}\left[\left|g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right|^{2}\right]^{4}w_{f}(\mathbf{f}_{t})d\mathbf{f}_{t}\notag\\ &=\frac{1}{T^{3}}\int_{\mathcal{R}_f}\left\{ \left[\left(g_{i}(\mathbf{f}_{t})-\boldsymbol{\alpha}'_{g_{i}}\widetilde{\mathbf{p}}(\mathbf{f}_{t})\right)\left(1+\left\Vert \mathbf{f}_{t}\right\Vert ^{2}\right)^{-\frac{\overline{\omega}}{2}}\right]^{2}\left(1+\left\Vert \mathbf{f}_{t}\right\Vert ^{2}\right)^{\overline{\omega}}\right\} ^{4}w_{f}(\mathbf{f}_{t})d\mathbf{f}_{t}\notag\\ &\leq\frac{1}{T^{3}}\left(\left\|g_{i}(\cdot)-\Pi_{\infty K}g_{i}(\cdot)\right\|_{\infty,\overline{\omega}}\right)^{8}\int_{\mathcal{R}_f}\left(1+\left\Vert \mathbf{f}_{t}\right\Vert ^{2}\right)^{4\overline{\omega}}w_{f}(\mathbf{f}_{t})d\mathbf{f}_{t}\notag\\ &=O\left(\frac{K^{-\frac{8\lambda_{i}}{m}}}{T^{3}}\right)O(1)\notag\\ &=O\left(\frac{K^{-\frac{8\lambda_{i}}{m}}}{T^{3}}\right), \end{align} with $w_f(\*f_t)$ denoting the density of $\*f_t$. The steps above can be applied to every component in (ref). Using exactly the same arguments as for (ref), only the sum with the maximum number of cross-terms will survive. We can thereforee conclude that \begin{align} \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]&=O\left(K^{-\frac{8\lambda_{i}}{m}}\right), \end{align} and consequently, \begin{align} \mathbb{E}\left(\left\|\*d_{6,2}^{(2)}\right\|^2\right)&\leq O\left(\frac{K^{\frac{13}{4}}}{N}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)\frac{1}{N}\sum_{i=1}^{N}\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}} \\ &=O\left(\frac{K^{\frac{13}{4}}}{N}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)\left[O\left(K^{-\frac{8\lambda_{i}}{m}}\right)\right]^{\frac{1}{4}}\\ &=O\left(\frac{K^{\frac{13m-8\lambda_i}{4m}}}{N}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right), \end{align} which implies \begin{align} \left\|\*d_{6,2}^{(2)}\right\|=O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right). \end{align} The analysis of the first term in (ref), $\*d_{6,2}^{(1)}$, is similar. We can follow the same steps and assumptions as in (ref) to obtain \begin{align} & \mathbb{E} \left(\|\*d_{6,2}^{(1)}\|^2\right) \notag\\ & \equiv \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\Vert^2\right] \nonumber \\ &= \mathbb{E}\Bigg[\mathrm{tr}\Bigg(\frac{1}{NT}\sum_{i=1}^N \+\alpha_{g_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\widetilde{\mathbf{P}} \left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbb{E}\left(\*V_i\mathbf{V}_{i}'\right)\notag\\ & \quad \times \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\Bigg)\Bigg] \notag \\ &\leq \frac{1}{NT}\sum_{i=1}^N\lambda_{\max}\left(\+\Omega_{v,i}\right)\notag\\ & \quad \times \mathbb{E}\left[\mathrm{tr}\left(\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\+\alpha_{g_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\widetilde{\mathbf{P}} \left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right)\right] \notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right) \notag \\ & \quad \times \frac{1}{T}\mathbb{E}\left[\mathrm{tr}\left(\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\frac{1}{N}\sum_{i=1}^N\+\alpha_{g_i}\+\alpha_{g_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\widetilde{\mathbf{P}} \left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right)\right] \notag \\ &= \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right) \notag \\ &\quad \times \frac{1}{T}\mathbb{E} \left[\mathrm{tr}\left(\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\Sigma_{\alpha_g}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\widetilde{\mathbf{P}} \left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right)\right] \notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right) \lambda_{\max}\left(\+\Sigma_{\alpha_g}\right) \frac{1}{T}\mathbb{E}\left(\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\right) \notag \\ & =O\left(\frac{1}{T}\right)\mathbb{E}\left(\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\right), \end{align} where $\+\Omega_{v,i}$ and $\+\Sigma_{\alpha_g}$ are defined as in Assumption (ref)(ref) and (ref), respectively. For the term in expectations in (ref), we use the H\"older and Cauchy-Schwarz inequalities repeatedly to arrive at \begin{align} &\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\right]\notag\\ &\leq\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\|^{2}\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\right]\notag\\ &\leq\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{4}\right]}\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]}\notag\\ &\leq\sqrt{\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{4}\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\|^{4}\right]}\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]}\notag\\ &\leq\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]}\notag\\ &\leq\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{8}\right]+\mathbb{E}\left(\left\|\mathbf{I}_{K}\right\|^{8}\right)\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\sqrt{\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{4}\right]}\notag\\ &\leq\left\{ O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right\} ^{\frac{1}{4}}\left\{ O(1)+O\left(K^{8}\right)\right\} ^{\frac{1}{4}}\notag\\ &\quad\times\sqrt{\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\right\|^{4}\left\|\frac{\sqrt{T}}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{4}\right]}\notag\\ &\leq O\left(\frac{K}{N}\zeta_{1}(K)^{2}\right)O\left(K^{2}\right)\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \frac{\sqrt{T}}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &=O\left(\frac{K^{3}}{N}\zeta_{1}(K)^{2}\right)\left[O\left(K\zeta_{0}(K)^{6}\right)\right]^{\frac{1}{4}}\left[T^{4}O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right]^{\frac{1}{4}}\notag\\ &=O\left(\frac{K^{3}}{N}\zeta_{1}(K)^{2}\right)O\left(K^{\frac{1}{4}}\zeta_{0}(K)^{\frac{3}{2}}\right)O\left(\frac{TK}{N}\zeta_{1}(K)^{2}\right)\notag\\ &=O\left(\frac{TK^{\frac{17}{4}}}{N^{2}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{4}\right), \end{align} leading to \begin{align} \mathbb{E}\left(\left\|\*d_{6,2}^{(1)}\right\|^2\right)&\leq O\left(\frac{1}{T}\right)\mathbb{E}\left(\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\right)\notag\\ &=O\left(\frac{K^{\frac{17}{4}}}{N^{2}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{4}\right). \end{align} Consequently, \begin{align} \left\|\*d_{6,2}^{(1)}\right\|=O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right). \end{align} Insertion into (ref) yields the following bound for $\*d_{6,2}$: \begin{align} \left\|\*d_{6,2}\right\| &\equiv \left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert \nonumber \\ & \leq\left\|\*d_{6,2}^{(1)}\right\| +\left\|\*d_{6,2}^{(2)}\right\|\nonumber \\ & =O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right). \end{align} The order of $\*d_{6,3}$ is the same as that of $\*d_{6,2}$; \begin{align} \left\Vert \*d_{6,3}\right\Vert & \equiv\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\|\nonumber \\ & =O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right). \end{align} To bound $\*d_{6,1}$, let us again expand using (ref); \begin{align} \*d_{6,1} & \equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right) \nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\nonumber \\ &\quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right) \notag \\ &\equiv \*d_{6,1}^{(1)} + \*d_{6,1}^{(2)}. \end{align} We start with $\*d_{6,1}^{(1)}$. By using exactly the same steps as for $\*d_{6,2}^{(1)}$ in (ref), we can show that \begin{align} \mathbb{E}\left(\|\*d_{6,1}^{(1)}\|^2\right) &=\mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\|^2\right] \notag \\ &\leq \sup_{i=1, \dots, N} \lambda_{\max}\left(\+\Omega_{v,i}\right) \lambda_{\max}\left(\+\Sigma_{\alpha_g}\right) \notag \\ &\quad \times \frac{1}{T}\mathbb{E}\left(\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\right) \notag \\ &\leq O\left(\frac{1}{T}\right) \mathbb{E}\left(\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\right) \notag \\ &= O\left(\frac{1}{T}\right)\mathbb{E}\left(\left\|\frac{1}{T}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\right)\notag \\ &\leq O\left(\frac{1}{T}\right)\mathbb{E}\Bigg(\left\|\frac{\sqrt{T}}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\left[\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}+\mathbf{I}_{K}\right\|^{2}\right]\notag\\ & \quad \times\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^2\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\Bigg)\notag \\ &\leq O\left(\frac{1}{T}\right)\left\{ \mathbb{E}\left[\left\Vert \frac{\sqrt{T}}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{8}\right]+\mathbb{E}\left[\left\Vert \mathbf{I}_{K}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag \\ &\quad \times \left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{2}}\notag \\ &=O\left(\frac{1}{T}\right)\left[T^{4}O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right]^{\frac{1}{4}}\left[O(1)+O\left(K^{8}\right)\right]^{\frac{1}{4}}\left[O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right]^{\frac{1}{2}}\notag \\ &=O\left(\frac{1}{T}\right)O\left(\frac{TK}{N}\zeta_{1}(K)^{2}\right)O\left(K^{2}\right)O\left(\frac{K^2}{N^2}\zeta_{1}(K)^4\right)\notag \\ &=O\left(\frac{K^5}{N^3}\zeta_{1}(K)^6\right). \end{align} It follows that \begin{align} \left\|\*d_{6,1}^{(1)}\right\| &\leq \left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\| \notag \\ &= O\left(\frac{K^{\frac{5}{2}}}{N^{\frac{3}{2}}}\zeta_{1}(K)^3\right). \end{align} $\*d_{6,1}^{(2)}$ can be analyzed in exactly the same fashion as $\*d_{6,2}^{(2)}$, resulting in \begin{align} &\mathbb{E}\left(\left\Vert \mathbf{d}_{6,1}^{(2)}\right\Vert ^{2}\right)\notag\\ &=\mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\|^{2}\right]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\frac{1}{T}\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\|^{2}\right]\notag\\ &=O(1)\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\frac{1}{\sqrt{N}}\sum_{i=1}^{N}\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{2}\right]\notag\\ &\leq\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{2}\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{2}\right]\notag\\ &\leq\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right]\right\} ^{\frac{2}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}} \notag\\ & \quad \times \left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &=\sqrt{O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)}\left\{ O\left(K^{8}\right)\right\} ^{\frac{1}{4}}\left\{ O\left(K^{-\frac{8\lambda_{i}}{m}}\right)\right\} ^{\frac{1}{4}}\notag\\ &=O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right)O\left(K^{2}\right)O\left(K^{-\frac{2\lambda_{i}}{m}}\right)\notag\\ &=O\left(\frac{K^{\frac{4m-4\lambda_{i}}{2m}}}{N^{2}}\zeta_{1}(K)^{4}\right), \end{align} with $\lambda_{\max}\left(\+\Omega_{v,i}\right)$ defined as in Assumption (ref)(ref). Consequently, \begin{align} \left\|\*d_{6,1}^{(2)}\right\| = O_p\left(\frac{K^{\frac{4m-4\lambda_i}{4m}}}{N}\zeta_1(K)^2\right). \end{align} Since the bounds for all terms in (ref) are now known, we can conclude that \begin{align} \left\Vert \*d_{6,1}\right\Vert & \equiv\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert \notag \\ & \leq \left\|\*d_{6,1}^{(1)}\right\| + \left\|\*d_{6,1}^{(2)}\right\| \notag \\ &= O\left(\frac{K^{\frac{5}{2}}}{N^{\frac{3}{2}}}\zeta_{1}(K)^3\right) + O_p\left(\frac{K^{\frac{4m-4\lambda_i}{4m}}}{N}\zeta_1(K)^2\right). \end{align} $\*d_{6,4}$ can be expanded in the following fashion: \begin{align} \left\|\*d_{6,4}\right\| &\equiv \left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\| \notag \\ &\leq \left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}})\+\alpha_{g_i}\right\| \notag \\ &\quad +\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'(\widetilde{\mathbf{P}}\+\alpha_{g_i} -\*g_i(\*F))\right\| \notag \\ & \equiv \left\|\*d_{6,4}^{(1)}\right\| + \left\|\*d_{6,4}^{(2)}\right\|. \end{align} Repeated use of the same steps as before gives \begin{align} &\mathbb{E}\left(\left\|\*d_{6,4}^{(1)}\right\|^2\right)\notag\\ & = \mathbb{E} \left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}})\+\alpha_{g_i}\right\|^2\right] \notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\lambda_{\max}\left(\+\Sigma_{\alpha_g}\right)\notag\\ & \quad \times \mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\frac{1}{\sqrt{T}}(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}})\right\|^2\right] \notag \\ & \leq O(1)\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^{2}\left\|\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\right\|^{2}\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\right\|^{2}\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\|^{2}\right]\notag \\ &\leq\left\{ \mathbb{E}\left(\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{8}\right)\right\} ^{\frac{2}{4}}\left\{ \mathbb{E}\left[\left\Vert \left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\left\{ \left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert ^{8}\right\} ^{\frac{1}{4}}\notag \\ &\leq\sqrt{O\left(K\zeta_{0}(K)^{6}\right)}\left\{ \mathbb{E}\left[\left\|\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|^{8}\right]+\mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag \\ &\quad\times\left[O\left(\frac{K^{4}}{N^{4}}\zeta_{1}(K)^{8}\right)\right]^{\frac{1}{4}}\notag \\ &=O\left(\sqrt{K}\zeta_{0}(K)^{3}\right)O(1)O\left(\frac{K}{N}\zeta_{1}(K)^{2}\right)\notag \\ &=O\left(\frac{K^{\frac{3}{2}}}{N}\zeta_{0}(K)^{3}\zeta_{1}(K)^{2}\right), \end{align} and \begin{align} &\mathbb{E}\left(\left\|\*d_{6,4}^{(2)}\right\|^2\right) \notag\\ & = \mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N \*V_i'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'(\widetilde{\mathbf{P}}\+\alpha_{g_i} -\*g_i(\*F))\right\|^2\right] \notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\frac{1}{N}\sum_{i=1}^N\mathbb{E}\left[\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^+ - \left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^+\right]\widetilde{\mathbf{P}}'\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}} \+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\|^2\right] \notag \\ &\leq O(1)\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\|^{2}\right]\notag\\ &\leq\mathbb{E}\left[\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\|^{2}\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{2}\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{2}\right]\notag \\ &\leq\left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}\right\Vert ^{8}\right]\right\} ^{\frac{2}{4}}\left\{ \mathbb{E}\left[\left\Vert \left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ & \quad\times \left\{ \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\boldsymbol{\alpha}_{g_{i}}-\mathbf{g}_{i}(\mathbf{F})\right)\right\Vert ^{8}\right]\right\} ^{\frac{1}{4}}\notag\\ &=\sqrt{O\left(K\zeta_{0}(K)^{6}\right)}O(1)\left[O\left(K^{-\frac{8\lambda_{i}}{m}}\right)\right]^{\frac{1}{4}}\notag\\ &=O\left(\sqrt{K}\zeta_{0}(K)^{3}\right)O\left(K^{-\frac{2\lambda_{i}}{m}}\right)\notag\\ &=O\left(K^{\frac{m-4\lambda_{i}}{2m}}\zeta_{0}(K)^{3}\right). \end{align} Consequently, \begin{align} \left\|\*d_{6,4}^{(1)}\right\| = O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)\right), \end{align} and \begin{align} \left\|\*d_{6,4}^{(2)}\right\| = O_p\left(K^{\frac{m-4\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right), \end{align} which establishes the following bound for $\*d_{6,4}$: \begin{align} \left\|\*d_{6,4}\right\| &\leq \left\|\*d_{6,4}^{(1)}\right\| + \left\|\*d_{6,4}^{(2)}\right\| \notag \\ &= O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)\right) + O_p\left(K^{\frac{m-4\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right). \end{align} Therefore, \begin{align} \left\|\*D_6\right\| &\equiv \left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}})\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\| \notag \\ &\leq \left\|\*d_{6,1}\right\| + \left\|\*d_{6,2}\right\| + \left\|\*d_{6,3}\right\| + \left\|\*d_{6,4}\right\| \notag \\ &= O\left(\frac{K^{\frac{5}{2}}}{N^{\frac{3}{2}}}\zeta_{1}(K)^3\right) + O_p\left(\frac{K^{\frac{4m-4\lambda_i}{4m}}}{N}\zeta_1(K)^2\right) \notag \\ &\quad +O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right)\notag \\ &\quad +O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)\right) + O_p\left(K^{\frac{m-4\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right)\notag \\ &=o_p(1), \end{align} where we have made use of (ref), (ref), (ref), (ref), (ref) and (ref) to establish the last equality. $\mathbf{D}_{7}$ has the same basic structure as $\mathbf{D}_{6}$; \begin{align} \mathbf{D}_{7} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\boldsymbol{\varepsilon}_{i}\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\boldsymbol{\varepsilon}_{i}\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{i}\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\boldsymbol{\varepsilon}_{i}\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{i}\nonumber \\ & \equiv \*d_{7,1} + \*d_{7,2} + \*d_{7,3} + \*d_{7,4}, \end{align} where it is straightforward to show using the same arguments as before that \begin{align} \left\|\*d_{7,1}\right\| & =O_{p}\left(\left\Vert \*d_{6,1}\right\Vert \right)=O\left(\frac{K^{\frac{5}{2}}}{N^{\frac{3}{2}}}\zeta_{1}(K)^3\right) + O_p\left(\frac{K^{\frac{4m-4\lambda_i}{4m}}}{N}\zeta_1(K)^2\right),\nonumber \\ \left\Vert \*d_{7,2}\right\Vert & =O_{p}\left(\left\Vert \*d_{6,2}\right\Vert \right) =O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right),\nonumber \\ \left\Vert \*d_{7,3}\right\Vert & =O_{p}\left(\left\Vert \*d_{6,3}\right\Vert \right)= O_p\left(\frac{K^{\frac{17}{8}}}N\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{13m-8\lambda_i}{8m}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{4}}\zeta_{1}(K)\right),\nonumber \\ \left\Vert \*d_{7,4}\right\Vert & =O_{p}\left(\left\Vert \*d_{6,4}\right\Vert \right)= O_p\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)\right) + O_p\left(K^{\frac{m-4\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right), \end{align} which under our conditions implies \begin{align} \left\|\mathbf{D}_{7}\right\| & \leq\left\Vert \*d_{7,1}\right\Vert +\left\Vert \*d_{7,2}\right\Vert +\left\Vert \*d_{7,3}\right\Vert +\left\Vert \*d_{7,4}\right\Vert = o_p(1). \end{align} We now continue with $\mathbf{D}_{8}.$ Application of (ref) yields \begin{align} -\mathbf{D}_{8} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ &\quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ &\quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ &\quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \equiv \*d_{8,1}+\*d_{8,2}+ \*d_{8,3}+\*d_{8,4}. \end{align} Consider the second term, $\*d_{8,2}$. The evaluation of this term is similar to that of $\*d_{6,2}$, except that we cannot rely on the same law of large numbers argument to take care of the cross-sectional sum. We instead use the following approach: \begin{align} & \left\Vert \*d_{8,2}\right\Vert \notag\\ & \equiv\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert \nonumber \\ & \leq\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left\Vert \left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert \nonumber \\ & \leq\sqrt{NT}\left\Vert \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\right\Vert \frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right\Vert \left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert \nonumber \\ & \leq\sqrt{NT}\left\Vert \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\right\Vert \sqrt{\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right\Vert ^{2}} \notag \\ &\quad \times\sqrt{\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2}}\nonumber \\ & =\sqrt{NT}\left[O_{p}\left(\frac{K^{\frac{3}{4}}}{\sqrt{N}}\sqrt{\zeta_0(K)}\zeta_1(K)\right)+ O_p\left(\frac{K^{\frac{2}{3}}}{\sqrt{N}}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right)\right]\notag \\ &\quad \times \left[O_p\left(\frac{\zeta_{1}(K)^{2}}{N}\right) + O_p\left(K^{-\frac{2\lambda_i}{m}}\right)\right] \notag \\ &= \sqrt{NT}o_p(1)\left[O_p\left(\frac{\zeta_{1}(K)^{2}}{N}\right) + O_p\left(K^{-\frac{2\lambda_i}{m}}\right)\right]\notag \\ &= o_p\left(\frac{\sqrt{T}\zeta_1(K)^2}{\sqrt{N}}\right)+o_p\left(\sqrt{NT}K^{-\frac{2\lambda_i}{m}}\right), \end{align} where we have used the results in (ref) and (ref), as well as the Cauchy-Schwarz inequality followed by \begin{align} &\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2} \notag \\ & \leq\frac{2}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\Vert ^{2}+\frac{2}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2}\nonumber \\ &= O_p\left(\frac{\zeta_1(K)^2}{N}\right) + O_p\left(K^{-\frac{2\lambda_i}{m}}\right). \end{align} This last result holds because by Assumption (ref) and Lemma (ref)(ref) we have \begin{align} \frac{1}{N}\sum_{i=1}^N\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\|^2 &= \mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\frac{1}{N}\sum_{i=1}^N\+\alpha_{g_i}\+\alpha_{g_i}'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\right] \notag \\ &= \mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\+\Sigma_{\alpha_g}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\right] \notag \\ &\leq \lambda_{\max}\left(\+\Sigma_{\alpha_g}\right) \mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\right] \notag \\ &=\lambda_{\max}\left(\+\Sigma_{\alpha_g}\right)\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^2 \notag \\ &= O_p(1) O_p\left(\frac{\zeta_1(K)^2}{N}\right) \notag \\ &= O_p\left(\frac{\zeta_1(K)^2}{N}\right). \end{align} Exactly the same arguments can be used to show that $\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right\Vert ^{2}$ is of the same order as $\frac{1}{N}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2}$. $\*d_{8,3}$ is of the same order as $\*d_{8,2}$; that is, \begin{align} \left\|\*d_{8,3}\right\|=o_p\left(\frac{\sqrt{T}\zeta_1(K)^2}{\sqrt{N}}\right)+o_p\left(\sqrt{NT}K^{-\frac{2\lambda_i}{m}}\right). \end{align} As for $\*d_{8,1}$, by the usual arguments together with $\left\|\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)' \right\| = O_p\left(\frac{K}{N}\zeta_1(K)^2\right)$ from (ref), we can show that \begin{align} \left\Vert \*d_{8,1}\right\Vert & =\sqrt{NT}O_{p}\left(\frac{K}{N}\zeta_1(K)^2\right) \left[O_{p}\left(\frac{\zeta_{1}(K)^{2}}{N}\right) + O_p\left(K^{-\frac{2\lambda_i}{m}}\right)\right]\nonumber \\ & = O_{p}\left(\sqrt{NT}\frac{K}{N^{2}}\zeta_{1}(K)^{4}\right)+O_{p}\left(\sqrt{NT}\frac{K^{\frac{m-2\lambda_{i}}{m}}}{N}\zeta_{1}(K)^{2}\right)\notag \\ &=O_{p}\left(\frac{\sqrt{T}K}{N^{\frac{3}{2}}}\zeta_{1}(K)^{4}\right)+O_{P}\left(\frac{\sqrt{T}K^{\frac{m-2\lambda_{i}}{m}}}{\sqrt{N}}\zeta_{1}(K)^{2}\right). \end{align} Similarly, by using the known order of $\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\right\| = O_p\left(K^{\frac{1}{4}}\zeta_0(K)^{\frac{3}{2}}\right)$ from (ref), we obtain \begin{align} \left\|\*d_{8,4}\right\| & =\sqrt{NT}O_p\left(K^{\frac{1}{4}}\zeta_0(K)^{\frac{3}{2}}\right)\left[O_{p}\left(\frac{\zeta_{1}(K)^{2}}{N}\right) + O_p\left(K^{-\frac{2\lambda_i}{m}}\right)\right] \notag \\ &= O_{p}\left(\sqrt{NT}\frac{K^{\frac{1}{4}}}{N}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)+O_{P}\left(\sqrt{NT}K^{\frac{m-8\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right)\notag\\ &=O_{p}\left(\frac{\sqrt{T}K^{\frac{1}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)+O_{P}\left(\sqrt{NT}K^{\frac{m-8\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right). \end{align} This allows us to arrive at the order of $\mathbf{D}_{8}$; \begin{align} \left\|\mathbf{D}_{8}\right\| & \leq\left\|\*d_{8,1}\right\|+\left\Vert \*d_{8,2}\right\Vert +\left\Vert \*d_{8,3}\right\Vert +\left\Vert \*d_{8,4}\right\Vert \nonumber \\ & = O_{p}\left(\frac{\sqrt{T}K}{N^{\frac{3}{2}}}\zeta_{1}(K)^{4}\right)+O_{P}\left(\frac{\sqrt{T}K^{\frac{m-2\lambda_{i}}{m}}}{\sqrt{N}}\zeta_{1}(K)^{2}\right) \notag \\ &\quad + o_p\left(\frac{\sqrt{T}\zeta_1(K)^2}{\sqrt{N}}\right)+o_p\left(\sqrt{NT}K^{-\frac{2\lambda_i}{m}}\right) \notag \\ & \quad +O_{p}\left(\frac{\sqrt{T}K^{\frac{1}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}\right)+O_{P}\left(\sqrt{NT}K^{\frac{m-8\lambda_{i}}{4m}}\zeta_{0}(K)^{\frac{3}{2}}\right) \notag \\ &=o_p(1), \end{align} where the last equality follows from (ref), (ref), (ref), (ref), and conditions $\sqrt{NT}K^{-\frac{\lambda_i}{m}}=o(1)$ and $\frac{\sqrt{T}K^{\frac{1}{4}}}{\sqrt{N}}\zeta_{0}(K)^{\frac{3}{2}}\zeta_{1}(K)^{2}=o(1)$ in Assumption (ref)(ref). Let us now consider $\mathbf{D}_{2}$, which we can expand in the following fashion using (ref): \begin{align} -\mathbf{D}_{2} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}+\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \equiv \*d_{2,1}+\*d_{2,2}. \end{align} We start with the second term on the right, $\*d_{2,2}$. Note that $\mathbf{M}_{\widetilde{P}}$ is an idempotent matrix whose eigenvalues are either zero or one (see abadir2005, Exercise 8.56). Therefore, $\lambda_{\max}(\mathbf{M}_{\widetilde{P}})=1$. Then, using Assumptions (ref)(ref), (ref), and the definitions of the variances in Assumption (ref)(ref), \begin{align} &\mathbb{E}\left[\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\|^{2}\right] \notag \\ & =\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{j}'\right)\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_j}-\*g_j(\mathbf{F})\right)\right)\right]\nonumber \\ & =\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T}\mathbb{E}\left[\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\boldsymbol{\Omega}_{v,i}\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right)\right]\nonumber \\ & \leq\frac{1}{N}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\frac{1}{T}\mathbb{E}\left[\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right)\right]\nonumber \\ & \leq\frac{1}{N}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\frac{1}{T}\mathbb{E}\left[\lambda_{\max}(\mathbf{M}_{\widetilde{P}})\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right)\right]\nonumber \\ & =\frac{1}{N}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\frac{1}{T}\mathbb{E}\left[\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right)\right]\nonumber \\ & \leq\frac{1}{N}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2}\right]\nonumber \\ & \leq\sup_{i=1,\dots,N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\frac{1}{N}\sum_{i=1}^{N}\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\Vert ^{2}\right]\nonumber \\ & =O(1)O\left(K^{-\frac{2\lambda_i}{m}}\right)\nonumber \\ & =O\left(K^{-\frac{2\lambda_i}{m}}\right), \end{align} implying \begin{equation} \left\Vert \*d_{2,2}\right\Vert =O_{p}\left(K^{-\frac{\lambda_i}{m}}\right). \end{equation} For $\*d_{2,1}$, we use \begin{align} \mathbb{E} \left(\|\*d_{2,1}\|^2\right) & = \mathbb{E}\left[\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\Vert ^{2}\right] \notag \\ & =\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathrm{tr}\left(\+\alpha_{g_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{M}_{\widetilde{P}}\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{j}'\right)\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_j}\right)\right]\nonumber \\ & =\frac{1}{N}\sum_{i=1}^{N}\frac{1}{T}\mathbb{E}\left[\mathrm{tr}\left(\+\alpha_{g_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{M}_{\widetilde{P}}\boldsymbol{\Omega}_{v,i}\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right)\right]\nonumber \\ & \leq \sup_{i=1, \dots, N} \lambda_{\max} \left(\+\Omega_{v,i}\right)\mathbb{E} \left[\mathrm{tr}\left(\frac{1}{T} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \frac{1}{N}\sum_{i=1}^N\+\alpha_{g_i}\+\alpha_{g_i}'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\right)\right] \notag \\ &= \sup_{i=1, \dots, N} \lambda_{\max}\left(\+\Omega_{v,i}\right)\mathbb{E} \left[\mathrm{tr}\left(\frac{1}{T} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right) \+\Sigma_{\alpha_g}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\right)\right]\notag \\ &\leq \sup_{i=1, \dots, N}\lambda_{\max}\left(\+\Omega_{v,i}\right)\lambda_{\max}\left(\+\Sigma_{\alpha_g}\right) \mathbb{E} \left[\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^2\right]\notag \\ &=O(1)O(1)O\left(\frac{\zeta_1(K)^2}{N}\right)=O\left(\frac{\zeta_1(K)^2}{N}\right). \end{align} Hence, \begin{equation} \left\|\*d_{2,1}\right\|=O_{p}\left(\frac{\zeta_{1}(K)}{\sqrt{N}}\right). \end{equation} Insertion into (ref) leads to the following: \begin{align} \left\|\mathbf{D}_{2}\right\| & \leq\left\|\*d_{2,1}\right\|+\left\Vert \*d_{2,2}\right\Vert \nonumber \\ & =O_{p}\left(\frac{\zeta_{1}(K)}{\sqrt{N}}\right)+O_{p}\left(K^{-\frac{\lambda_i}{m}}\right)\nonumber \\ & = o_p(1), \end{align} which holds by Assumption (ref)(ref) and the result in (ref). $\mathbf{D}_{3}$ has the same basic structure as $\mathbf{D}_{2}$. Note in particular how \begin{align} -\mathbf{D}_{3} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\boldsymbol{\varepsilon}_{i}\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\+\alpha_{G_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{M}_{\widetilde{P}}\boldsymbol{\varepsilon}_{i}+\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\boldsymbol{\varepsilon}_{i}\nonumber \\ & \equiv \*d_{3,1} + \*d_{3,2}, \end{align} where under our assumptions it is straightforward to show that \begin{align} \left\|\*d_{3,1}\right\| & =O_{p}\left(\left\Vert \*d_{2,1}\right\Vert \right)=O_{p}\left(\frac{\zeta_{1}(K)}{\sqrt{N}}\right),\\ \left\Vert \*d_{3,2}\right\Vert & =O_{p}\left(\left\Vert \*d_{2,2}\right\Vert \right)=O_{p}\left(K^{-\frac{\lambda_i}{m}}\right), \end{align} and thus, \begin{align} \left\|\mathbf{D}_{3}\right\| & \leq\left\Vert \*d_{3,1}\right\Vert +\left\Vert \*d_{3,2}\right\Vert \nonumber \\ & =O_{p}\left(\frac{\zeta_{1}(K)}{\sqrt{N}}\right)+O_{p}\left(K^{-\frac{\lambda_i}{m}}\right)\nonumber \\ & = o_p(1), \end{align} where the last equality holds similarly to the one in $\left\|\*D_2\right\|$. We continue with $\mathbf{D}_{4}$; \begin{align} \mathbf{D}_{4} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left[\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{G_i}\right]'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\notag \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left[\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{G_i}\right]'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i} \notag \\ & \quad +\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\nonumber \\ & \equiv \*d_{4,1}+\*d_{4,2}+\*d_{4,3}+\*d_{4,4}. \end{align} For the first term on the right, $\*d_{4,1}$, we can use the Cauchy-Schwarz inequality to obtain \begin{align} \left\|\*d_{4,1}\right\|&=\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^N\+\alpha_{G_i}' \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\| \notag \\ & \leq \frac{1}{\sqrt{NT}} \sum_{i=1}^N \left\|\+\alpha_{G_i}' \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\| \notag \\ & \leq \sqrt{NT} \left\|\frac{1}{T} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\| \frac{1}{NK}\sum_{i=1}^N \left\|\+\alpha_{G_i}\right\|\left\|\+\alpha_{g_i}\right\| \notag \\ & \leq \sqrt{NT} \left\|\frac{1}{T} \left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\| \sqrt{\frac{1}{NK}\sum_{i=1}^N\left\|\+\alpha_{G_i}\right\|^2}\sqrt{\frac{1}{NK}\sum_{i=1}^N\left\|\+\alpha_{g_i}\right\|^2} \notag \\ & = \sqrt{NT} K O_p\left(\frac{\zeta_1(K)^2}{N}\right)O_p(1) \notag \\ &= O_p\left(\sqrt{\frac{T}{N}} K \zeta_1(K)^2\right), \end{align} which holds under Assumption (ref) and the fact that the trace of a square matrix is the sum of its eigenvalues (abadir2005, Exercise 7.27). Note how \begin{align} \frac{1}{NK}\sum_{i=1}^N\left\|\+\alpha_{g_i}\right\|^2 &= \frac{1}{NK}\sum_{i=1}^N \mathrm{tr}\left(\+\alpha_{g_i}\+\alpha_{g_i}'\right) = \mathrm{tr}\left(\frac{1}{NK}\sum_{i=1}^N \+\alpha_{g_i}\+\alpha_{g_i}'\right) \notag \\ &= \frac{1}{K}\mathrm{tr}\left(\+\Sigma_{\alpha_g}\right) = \frac{1}{K}\sum_{k=1}^K \lambda_k\left(\+\Sigma_{\alpha_g}\right) \leq \frac{1}{K} K \lambda_{\max}\left(\+\Sigma_{\alpha_g}\right)= O(1), \end{align} and we can similarly show that $\frac{1}{NK}\sum_{i=1}^N \left\|\+\alpha_{G_i}\right\|^2 = O(1)$. Further, \begin{align} & \left\|\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\| \notag\\ &= \sqrt{\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]} \notag \\ &\leq \sqrt{\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]}\notag \\ &\leq \lambda_{\max}\left(\*M_{\widetilde{P}}\right)\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]\notag \\ &=\lambda_{\max}\left(\*M_{\widetilde{P}}\right)\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^2 \notag \\ &= O_p(1) O_p\left(\frac{\zeta_1(K)^2}{N}\right) = O_p\left(\frac{\zeta_1(K)^2}{N}\right), \end{align} with which we can bound $\*d_{4,1}$. We can apply exactly the same steps to \begin{align} \*d_{4,2}\equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left[\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{G_i}\right]'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right). \end{align} The only difference here is that instead of a second $O_{p}\left(\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\right\Vert \right)$, we need to bound $O_{p}\left(\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\|\right)$ instead. From (ref) we know that this term is of order $O_{p}\left(K^{-\frac{\lambda_i}{m}}\right)$. Therefore, using the same argumentation as before, \begin{align} \left\|\*d_{4,2}\right\| & \equiv\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\+\alpha_{G_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\|\nonumber \\ & \leq \frac{1}{\sqrt{NT}}\sum_{i=1}^N \left\|\+\alpha_{G_i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\| \notag \\ &\leq \sqrt{NTK}\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\right\| \frac{1}{N\sqrt{K}}\sum_{i=1}^N \left\|\+\alpha_{G_i}\right\| \left\|\frac{1}{\sqrt{T}} \left(\widetilde{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\| \notag \\ &\leq \sqrt{NTK}\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\right\|\sqrt{\frac{1}{NK}\sum_{i=1}^N \left\|\+\alpha_{G_i}\right\|^2}\notag\\ & \quad \times\sqrt{\frac{1}{N}\sum_{i=1}^N \left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\|^2}, \end{align} where \begin{align} \left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\right\| &= \sqrt{\mathrm{tr}\left[\frac{1}{T}\*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\right]} \notag \\ &= \sqrt{\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)' \*M_{\widetilde{P}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]} \notag \\ &\leq \sqrt{\lambda_{\max}\left(\*M_{\widetilde{P}}\right)\mathrm{tr}\left[\frac{1}{T}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right]} \notag \\ &= \sqrt{\lambda_{\max}\left(\*M_{\widetilde{P}}\right)\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)\right\|^2} \notag \\ &= O_p(1) O_p\left(\frac{\zeta_1(K)}{\sqrt{N}}\right) = O_p\left(\frac{\zeta_1(K)}{\sqrt{N}}\right), \end{align} and therefore, \begin{align} \left\|\*d_{4,2}\right\| &\leq \sqrt{NTK}\left\|\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}} - \widetilde{\mathbf{P}}\right)'\*M_{\widetilde{P}}\right\|\sqrt{\frac{1}{NK}\sum_{i=1}^N \left\|\+\alpha_{G_i}\right\|^2}\sqrt{\frac{1}{N}\sum_{i=1}^N \left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\|^2} \notag \\ &= \sqrt{NTK}O_p\left(\frac{\zeta_1(K)}{\sqrt{N}}\right)O_p(1)O_p\left(K^{-\frac{\lambda_i}{m}}\right) \notag \\ &= O_p\left(\sqrt{T}K^{\frac{1}{2}-\frac{\lambda_i}{m}}\zeta_1(K)\right). \end{align} The third term is obviously of the same order as the second and hence, \begin{equation} \left\|\*d_{4,3}\right\|\equiv\left\|\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\+\alpha_{g_i}\right\|=O_p\left(\sqrt{T}K^{\frac{1}{2}-\frac{\lambda_i}{m}}\zeta_1(K)\right). \end{equation} It remains to consider $\*d_{4,4}$. Note how \begin{align} \left\|\*d_{4,4}\right\|^2 &= \left\|\frac{\sqrt{T}}{\sqrt{N}T}\sum_{i=1}^{N}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right\|^2 \notag \\ &=\mathrm{tr}\Bigg[\frac{T}{NT^2}\sum_{i=1}^N\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\*M_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\notag\\ & \quad \times\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\mathbf{M}_{\widetilde{P}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\Bigg] \notag \\ &\leq \frac{T}{N} \sum_{i=1}^N \mathrm{tr}\left[\frac{1}{T}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\left(\widetilde{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)'\*M_{\widetilde{P}}\right] \notag \\ &\quad \times \mathrm{tr}\left[\frac{1}{T}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\left(\widetilde{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)'\*M_{\widetilde{P}}\right] \notag \\ &\leq \frac{T}{N}\sum_{i=1}^N \lambda_{\max}\left(\*M_{\widetilde{P}}\right)^2\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\|^2\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right\|^2 \notag \\ &\leq T \frac{1}{N}\sum_{i=1}^N\sqrt{\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right)\right\|^4} \sqrt{\left\|\frac{1}{\sqrt{T}}\left(\widetilde{\mathbf{P}}\+\alpha_{G_i} - \*G_i\left(\*F\right)\right)\right\|^4} \notag \\ &= T O_p\left(K^{-\frac{4\lambda_i}{m}}\right) \notag \\ &= O_p\left(T K^{-\frac{4\lambda_i}{m}}\right). \end{align} By adding the above terms, we can show that \begin{align} \left\|\mathbf{D}_{4}\right\| & \leq\left\Vert \*d_{4,1}\right\Vert +\left\Vert \*d_{4,2}\right\Vert +\left\Vert \*d_{4,3}\right\Vert +\left\Vert \*d_{4,4}\right\Vert \nonumber \\ & =O_p\left(\sqrt{\frac{T}{N}}K\zeta_1(K)^2\right) + O_p\left(\sqrt{T}K^{\frac{1}{2}-\frac{\lambda_i}{m}}\zeta_1(K)\right) + O_p\left(T K^{-\frac{4\lambda_i}{m}}\right) \notag \\ &=o_p(1), \end{align} where we have used (ref), (ref) and (ref) to establish the last equality. The only term that remains now is $\mathbf{D}_{1}$, which we write as \begin{align} \mathbf{D}_{1} & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\boldsymbol{\varepsilon}_{i}\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\left[\mathbf{I}_{T}-\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\right]\boldsymbol{\varepsilon}_{i}\nonumber \\ & =\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}-\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{i}\nonumber \\ & \equiv \*d_{1,1}+\*d_{1,2}, \end{align} where we have used the definition of the projection matrix $\mathbf{M}_{\widetilde{P}}\equiv\mathbf{I}_{T}-\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'$. Consider the second term on the right-hand side, $\*d_{1,2}$. Here, we proceed as follows: \begin{align} \mathbb{E}\left(\|\*d_{1,2}\|^2\right) &= \mathbb{E}\left(\left\Vert \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{i}\right\Vert ^{2}\right)\nonumber \\ & =\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathrm{tr}\left(\boldsymbol{\varepsilon}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\mathbf{V}_{j}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\boldsymbol{\varepsilon}_{j}\right)\right]\nonumber \\ & =\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\mathrm{tr}\left[\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{i}'\right)\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbb{E}\left(\boldsymbol{\varepsilon}_{i}\boldsymbol{\varepsilon}_{i}'\right)\right]\right]\nonumber \\ & \leq\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\mathbb{E}\left(\boldsymbol{\varepsilon}_{i}'\boldsymbol{\varepsilon}_{i}\right)\right)\right]\nonumber \\ & \leq\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\mathbb{E}\left(\boldsymbol{\varepsilon}_{i}'\boldsymbol{\varepsilon}_{i}\right)\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right]\nonumber \\ & \leq\mathbb{E}\left[\frac{1}{NT}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\lambda_{\max}\left(\boldsymbol{\Omega_{\varepsilon,i}}\right)\mathrm{tr}\left(\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right]\nonumber \\ & =\frac{1}{NT}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\lambda_{\max}\left(\boldsymbol{\Omega}_{\varepsilon,i}\right)\mathrm{tr}\left(\mathbf{I}_{K}\right)\nonumber \\ & =\frac{1}{N}\sum_{i=1}^{N}\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\lambda_{\max}\left(\boldsymbol{\Omega}_{\varepsilon,i}\right)\frac{K}{T}\nonumber \\ & =O(1)\frac{K}{T}\nonumber \\ & =O\left(\frac{K}{T}\right). \end{align} This allows us to conclude that \begin{equation} \left\|\*d_{1,2}\right\|=O_{p}\left(\sqrt{\frac{K}{T}}\right), \end{equation} which in turn implies the following: \begin{align} \mathbf{D}_{1} & \equiv \*d_{1,1}+\*d_{1,2}\nonumber \\ & \equiv\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}+O_{p}\left(\sqrt{\frac{K}{T}}\right), \end{align} where the reminder is negligible provided that $\frac{K}{T}=o(1)$, which holds because $\sqrt{\frac{K}{T}}\zeta_0(K)^{\frac{3}{2}}=o(1)$ by Assumption (ref)(ref). We have shown that $\*D_2 -\*D_8$ are negligible. Therefore, (ref) reduces to \begin{align} \frac{1}{\sqrt{NT}}\sum_{i=1}^N\*X_i'\*M_{\widehat{P}}\left[\+\varepsilon_i - \left(\widehat{\mathbf{P}}\+\alpha_{g_i} - \*g_i\left(\*F\right)\right) \+\gamma_i\right] \equiv \sum_{j=1}^8 \*D_j \equiv \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i} + o_p(1). \end{align} This is the sought cleaned-up asymptotic representation of the numerator of $\sqrt{NT}(\widehat{\+\beta}_{SCCE} - \+\beta)$. Let us now consider the denominator, which we expand in the usual fashion: \begin{align} \frac{1}{NT}\sum_{i=1}^{N} & \mathbf{X}_{i}'\mathbf{M}_{\widehat{P}}\mathbf{X}_{i}=\frac{1}{NT}\sum_{i=1}^{N}\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]'\mathbf{M}_{\widehat{P}}\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]\nonumber \\ & =\frac{1}{NT}\sum_{i=1}^{N}\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]'\mathbf{M}_{\widetilde{P}}\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]\nonumber \\ & \quad -\frac{1}{NT}\sum_{i=1}^{N}\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\left[\mathbf{V}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{G_i}-\*G_i(\mathbf{F})\right)\right]. \end{align} Most of the terms in (ref) are either known from before or similar to those considered earlier, and are negligible under our assumptions. The only new terms are \begin{align} \*b_1 &\equiv \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\mathbf{V}_{i}, \\ \*b_2 &\equiv \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\mathbf{V}_{i}. \end{align} Consider the second term, $\*b_2$. From (ref), \begin{align} \*b_2 &\equiv \frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\mathbf{V}_{i}\nonumber \\ & =\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i} +\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\nonumber \\ & \quad +\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i} +\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\mathbf{V}_{i} \notag \\ &\equiv \*b_{2,1} + \*b_{2,2} + \*b_{2,3} + \*b_{2,4}. \end{align} We start with the first term, $\*b_{2,1} \equiv \frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i}$, where \begin{align} \mathbb{E}\left(\left\|\*b_{2,1}\right\|^2\right) &=\mathbb{E}\left[\left\|\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i}\right\|^{2}\right]\notag\\ &=\mathbb{E}\Bigg[\frac{1}{N^{2}T^{2}}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathrm{tr}\Big(\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{j}'\right) \notag\\ & \quad \times \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{j}\Big)\Bigg]\notag\\ &\leq\frac{1}{N^{2}T^2}\sum_{i=1}^{N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})^{2}\mathbb{E}\left[\left\|\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\|^{2}\right]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})^{2}\frac{1}{N^{2}T^2}\sum_{i=1}^{N}\mathbb{E}\left[\left\Vert \frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\frac{1}{\sqrt{T}}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\right\Vert ^{2}\right]\notag\\ &=O(1)\frac{1}{NT^2}O\left(\frac{K^{2}}{N^{2}}\zeta_{1}(K)^{4}\right)\notag\\ &=O\left(\frac{K^{2}}{N^{3}T^2}\zeta_{1}(K)^{4}\right), \end{align} where we have used the result in (ref) and Assumption (ref). Hence, \begin{align} \left\|\mathbf{b}_{2,1}\right\|&=\left\Vert \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i}\right\Vert \notag \\ &=O_{p}\left(\frac{K}{N^{\frac{3}{2}}T}\zeta_{1}(K)^{2}\right). \end{align} The second term on the right-hand site of (ref), $\*b_{2,2}\equiv \frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}$, can be evaluated in a similar way. Hence, \begin{align} \mathbb{E}\left(\left\|\mathbf{b}_{2,2}\right\|^{2}\right)&=\mathbb{E}\left[\left\|\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|^{2}\right]\notag\\ &=\mathbb{E}\left[\frac{1}{N^{2}T^{2}}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathrm{tr}\left[\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{j}'\right)\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{j}\right]\right]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})^{2}\frac{1}{N^{2}T^{2}}\sum_{i=1}^{N}\mathbb{E}\left[\left\Vert \left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\right\Vert ^{2}\right]\notag\\ &=O(1)\frac{1}{NT^{2}}\left[O\left(\frac{K^{\frac{3}{2}}}{N}\zeta_1(K)^2\zeta_0(K)\right) + O\left(\frac{K^{\frac{4}{3}}}{N}\zeta_1(K)^2\zeta_0(K)^{\frac{4}{3}}\right)\right]\notag\\ &=O\left(\frac{K^{\frac{3}{2}}}{N^{2}T^{2}}\zeta_{0}(K)\zeta_{1}(K)^{2}\right) + O\left(\frac{K^{\frac{4}{3}}}{N^2T^2}\zeta_1(K)^2\zeta_0(K)^{\frac{4}{3}}\right), \end{align} where we have used the result from (ref). Hence, \begin{align} \left\Vert \mathbf{b}_{2,2}\right\Vert &=\left\Vert \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert \notag \\ &=O_{p}\left(\frac{K^{\frac{3}{4}}}{NT}\sqrt{\zeta_{0}(K)}\zeta_{1}(K)\right) + O_p\left(\frac{K^{\frac{2}{3}}}{NT}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right). \end{align} The third term on the right-hand site of (ref) is of the same order as the second, implying \begin{align} \left\|\*b_{2,3}\right\| &\equiv \left\|\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}\left(\widehat{\mathbf{P}}-\widetilde{\mathbf{P}}\right)'\mathbf{V}_{i}\right\| \notag \\ & =O_{p}\left(\frac{K^{\frac{3}{4}}}{NT}\sqrt{\zeta_{0}(K)}\zeta_{1}(K)\right). \end{align} We are left to analyze the last term. Following the same steps as before yields \begin{align} &\mathbb{E}\left(\left\|\mathbf{b}_{2,4}\right\|^{2}\right)=\mathbb{E}\left[\left\|\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|^{2}\right]\notag\\ &=\frac{1}{N^{2}T^{2}}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathbb{E}\Bigg[\mathrm{tr}\Big(\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{j}'\right) \notag\\ & \quad\times \widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\mathbf{V}_{j}\Big)\Bigg]\notag\\ &\leq\sup_{i=1,\dots,N}\lambda_{\max}(\boldsymbol{\Omega}_{v,i})^{2}\frac{1}{N^{2}T^{2}}\sum_{i=1}^{N}\mathbb{E}\left[\left\|\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\right\|^{2}\right]\notag\\ &=O(1)\frac{1}{NT^{2}}O\left(\sqrt{K}\zeta_{0}(K)^{3}\right)\notag\\ &=O\left(\frac{\sqrt{K}}{NT^{2}}\zeta_{0}(K)^{3}\right), \end{align} where we have used (ref). Hence, \begin{align} \left\Vert \mathbf{b}_{2,4}\right\Vert &=\left\Vert \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)^{+}-\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\right]\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert \notag\\ &=O_{p}\left(\frac{K^{\frac{1}{4}}}{\sqrt{N}T}\zeta_{0}(K)^{\frac{3}{2}}\right). \end{align} Collecting the results, \begin{align} \left\|\*b_{2}\right\| &\equiv \left\|\frac{1}{NT}\sum_{i=1}^N\mathbf{V}_{i}'\left(\mathbf{M}_{\widetilde{P}}-\mathbf{M}_{\widehat{P}}\right)\mathbf{V}_{i}\right\| \notag \\ & \leq \left\|\*b_{2,1}\right\| + \left\|\*b_{2,2}\right\| + \left\|\*b_{2,3}\right\| + \left\|\*b_{2,4}\right\| \notag \\ & =O_{p}\left(\frac{K}{N^{\frac{3}{2}}T}\zeta_{1}(K)^{2}\right) + O_{p}\left(\frac{K^{\frac{3}{4}}}{NT}\sqrt{\zeta_{0}(K)}\zeta_{1}(K)\right)\notag \\ &\quad + O_p\left(\frac{K^{\frac{2}{3}}}{NT}\zeta_0(K)^{\frac{2}{3}}\zeta_1(K)\right) + O_{p}\left(\frac{K^{\frac{1}{4}}}{\sqrt{N}T}\zeta_{0}(K)^{\frac{3}{2}}\right) \notag \\ &=o_p(1), \end{align} where the last equality holds by (ref), (ref) and (ref). The above results can be used to determine the bound of the first term, $\*b_1$. Note how \begin{align} \left\|\*b_1\right\| & \equiv \left\|\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\mathbf{V}_{i}\right\| \notag \\ &= \left\Vert \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert \nonumber \\ & =\left\|\frac{1}{NT^{2}}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|\nonumber \\ & \leq\frac{1}{NT}\sum_{i=1}^{N}\left\|\frac{1}{T}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|+\frac{1}{NT}\sum_{i=1}^{N}\left\|\frac{1}{T}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left[\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right]\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|\nonumber \\ & \leq\frac{1}{NT}\sum_{i=1}^{N}\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|^{2}+\frac{1}{NT}\sum_{i=1}^{N}\left\|\frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\|^{2}\left\|\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\|. \end{align} The order of $\left\Vert \left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert $ is known by (ref). Hence, we are only left to investigate the first term in (ref). Using Assumption (ref) and (ref), as well as the normalization condition $\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=\*I_K$, we can show \begin{align} \mathbb{E}\left(\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert ^{2}\right) & =\mathbb{E}\left[\mathrm{tr}\left(\frac{1}{T}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right)\right]\nonumber \\ & =\mathbb{E}\left\{\mathrm{tr}\left[\frac{1}{T}\widetilde{\mathbf{P}}'\mathbb{E}\left(\mathbf{V}_{i}\mathbf{V}_{i}'\right)\widetilde{\mathbf{P}}\right]\right\}\nonumber \\ & =\mathbb{E}\left[\mathrm{tr}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\boldsymbol{\Omega}_{v,i}\widetilde{\mathbf{P}}\right)\right]\nonumber \\ & \leq\lambda_{\max}\left(\boldsymbol{\Omega}_{v,i}\right)\mathrm{tr}\left[\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right]\nonumber \\ &= O(1)O(K) \notag \\ &=O(K). \end{align} Consequently, \begin{align} \left\|\*b_1\right\| &\equiv \left\Vert \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert \notag \\ & \leq\frac{1}{NT}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert ^{2}+\frac{1}{NT}\sum_{i=1}^{N}\left\Vert \frac{1}{\sqrt{T}}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\right\Vert ^{2}\left\Vert \left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}-\mathbf{I}_{K}\right\Vert \nonumber \\ & =\frac{1}{NT}\sum_{i=1}^{N}O_{p}(K)+\frac{1}{NT}\sum_{i=1}^{N}O_{p}(K)O_{p}\left(\sqrt{\frac{K}{T}}\zeta_{0}(K)\right)\nonumber \\ & =O_{p}\left(\frac{K}{T}\right)+O_{p}\left(\frac{K\sqrt{K}}{T\sqrt{T}}\zeta_{0}(K)\right)\nonumber \\ & =O_p\left(\frac{K}{T}\right) + O_p \left(\frac{K^{\frac{3}{2}}}{T^{\frac{3}{2}}}\zeta_0(K)\right) \notag \\ &=o_p(1), \end{align} which holds by Assumption (ref)(ref) and the result in (ref). By using this and applying a law of large numbers to $\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{V}_{i}$, \begin{align} \frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{M}_{\widetilde{P}}\mathbf{V}_{i} & =\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{V}_{i}-\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\widetilde{\mathbf{P}}\left(\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)^{+}\widetilde{\mathbf{P}}'\mathbf{V}_{i}\nonumber \\ & =\frac{1}{NT}\sum_{i=1}^{N}\mathbf{V}_{i}'\mathbf{V}_{i}+o_{p}(1)\nonumber \\ & =\boldsymbol{\Sigma}_{v}+o_{p}(1), \end{align} where $\boldsymbol{\Sigma}_{v}$ is as defined in Assumption (ref)(ref). Insertion into (ref) yields \begin{equation} \frac{1}{NT}\sum_{i=1}^{N}\mathbf{X}_{i}'\mathbf{M}_{\widehat{P}}\mathbf{X}_{i}=\boldsymbol{\Sigma}_{v}+o_{p}(1), \end{equation} which, together with (ref) and (ref), leads to the following cleaned-up asymptotic representation for $\sqrt{NT}\left(\widehat{\boldsymbol{\beta}}_{SCCE}-\boldsymbol{\beta}\right)$: \begin{align} \sqrt{NT}\left(\widehat{\boldsymbol{\beta}}_{SCCE}-\boldsymbol{\beta}\right) & =\left(\frac{1}{NT}\sum_{i=1}^{N}\mathbf{X}_{i}'\mathbf{M}_{\widehat{P}}\mathbf{X}_{i}\right)^{-1}\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{X}_{i}'\mathbf{M}_{\widehat{P}}\left[\boldsymbol{\varepsilon}_{i}-\left(\widehat{\mathbf{P}}\+\alpha_{g_i}-\*g_i(\mathbf{F})\right)\right]\nonumber \\ & =\boldsymbol{\Sigma}_{v}^{-1}\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}+o_{p}(1). \end{align} In order to arrive at the sought asymptotic distribution, we apply a central limit law for independent processes to $\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}$. This gives \begin{equation} \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}\overset{d}{\to}N(\mathbf{0}_{d\times1},\boldsymbol{\Theta}) \end{equation} as $N,T\to\infty$, where \begin{align} \boldsymbol{\Theta} & =\lim_{N,T\to\infty}\mathbb{E}\left[\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}\left(\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}\right)'\right]\nonumber \\ & =\lim_{N,T\to\infty}\frac{1}{NT}\sum_{i=1}^{N}\sum_{j=1}^{N}\mathbb{E}\left[\mathbf{V}_{i}'\mathbb{E}\left(\boldsymbol{\varepsilon}_{i}\boldsymbol{\varepsilon}_{j}\right)\mathbf{V}_{j}\right]\nonumber \\ & =\lim_{N,T\to\infty}\frac{1}{NT}\sum_{i=1}^{N}\mathbb{E}\left[\mathbf{V}_{i}'\mathbb{E}\left(\boldsymbol{\varepsilon}_{i}\boldsymbol{\varepsilon}_{i}'\right)\mathbf{V}_{i}\right]\nonumber \\ & =\lim_{N,T\to\infty}\frac{1}{NT}\sum_{i=1}^{N}\mathbb{E}\left(\mathbf{V}_{i}'\boldsymbol{\Omega}_{\varepsilon,i}\mathbf{V}_{i}\right). \end{align} We can therefore show that under Assumptions (ref)--(ref), \begin{equation} \sqrt{NT}\left(\widehat{\boldsymbol{\beta}}_{SCCE}-\boldsymbol{\beta}\right)=\boldsymbol{\Sigma}_{v}^{-1}\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}\mathbf{V}_{i}'\boldsymbol{\varepsilon}_{i}+o_{p}(1)\overset{d}{\to}N(\mathbf{0}_{d\times1},\boldsymbol{\Sigma}_{v}^{-1}\boldsymbol{\Omega}\boldsymbol{\Sigma}_{v}^{-1}), \end{equation} as $N,T\to\infty.$ $\blacksquare$ \@startsection{subsection}{2} \z@{-.5\linespacing\@plus-.7\linespacing}{.5\linespacing} {\normalfont}{Proof of Lemma (ref)} Consider (ref). In view of $\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}' \widetilde{\mathbf{P}}\right) = \*I_K$, $\mathbb{E}\left(\widetilde{\*p}(\*f_t)\widetilde{\*p}(\*f_t)'\right)= \*I_K$, for all $t=1, \dots, T$. Hence, we can consider the following expression \begin{align} &\mathbb{E} \left[\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\|^2\right]=\mathbb{E}\left\{\mathrm{tr}\left[\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right)'\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right)\right]\right\} \notag \\ &=\mathbb{E}\left[\frac{1}{T^2}\sum_{t=1}^T\sum_{s=1}^T\mathrm{tr}\left(\widetilde{\*p}(\*f_t)\widetilde{\*p}(\*f_t)'\widetilde{\*p}(\*f_s)\widetilde{\*p}(\*f_s)'\right)\right] -2\mathrm{tr}\left[\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right]+\mathbb{E}\left[\mathrm{tr}(\*I_K)\right] \notag \\ &=\mathbb{E}\left[\frac{1}{T^2}\sum_{t=1}^T\sum_{s=1}^T\mathrm{tr}\left(\widetilde{\*p}(\*f_t)\widetilde{\*p}(\*f_t)'\widetilde{\*p}(\*f_s)\widetilde{\*p}(\*f_s)'\right)\right] -2\mathrm{tr}\left(\*I_K\right)+\mathrm{tr}(\*I_K) \notag \\ &=\mathbb{E}\left[\frac{1}{T^2}\sum_{t=1}^T\sum_{s=1}^T\mathrm{tr}\left(\widetilde{\*p}(\*f_t)'\widetilde{\*p}(\*f_s)\widetilde{\*p}(\*f_s)'\widetilde{\*p}(\*f_t)\right)\right]-K \notag \\ &=\frac{1}{T^2}\sum_{t=1}^T\mathbb{E}\left(\left\|\widetilde{\*p}(\*f_t)\right\|^4\right)+\frac{1}{T^2}\sum_{t=1}^T\sum_{\substack{s=1 \\ s\neq t}}^T\mathbb{E}\left(\widetilde{\*p}(\*f_t)'\widetilde{\*p}(\*f_s)\widetilde{\*p}(\*f_s)'\widetilde{\*p}(\*f_t)\right)-K \notag \\ &\equiv A_1 + A_2. \end{align} Starting with $A_1$, let us derive \begin{align} A_1 &\equiv \frac{1}{T^2}\sum_{t=1}^T\mathbb{E}\left(\left\|\widetilde{\*p}(\*f_t)\right\|^4\right) \leq \frac{1}{T^2}\sum_{t=1}^T\mathbb{E}\left(\left\|\widetilde{\*p}(\*f_t)\right\|^2\right)\zeta_0(K)^2 \notag \\ &\leq \frac{1}{T^2} \sum_{t=1}^T \mathrm{tr}\left[\mathbb{E}\left(\widetilde{\*p}(\*f_t)\widetilde{\*p}(\*f_t)'\right)\right]\zeta_0(K)^2 =\frac{1}{T^2} \sum_{t=1}^T \mathrm{tr}\left[\*I_K\right]\zeta_0(K)^2 \notag \\ &=O\left(\frac{K \zeta_0(K)^2}{T}\right). \end{align} Now, let us continue with $A_2$. By Davydov's inequality (see for example bosq, pp. 20-21) for $\alpha$-mixing processes, we have \begin{align} A_2 &\equiv \frac{1}{T^2}\sum_{t=1}^T\sum_{\substack{s=1 \\ s\neq t}}^T\mathbb{E}\left(\widetilde{\*p}(\*f_t)'\widetilde{\*p}(\*f_s)\widetilde{\*p}(\*f_s)'\widetilde{\*p}(\*f_t)\right)-K \notag \\ &= \frac{1}{T^2}\sum_{t=1}^T\sum_{\substack{s=1 \\ s\neq t}}^T\sum_{j=1}^K\sum_{k=1}^K\mathbb{E}\left(\widetilde{p}_j(\*f_t)\widetilde{p}_j(\*f_s)\widetilde{p}_k(\*f_t)\widetilde{p}_k(\*f_s)\right)-K \notag \\ &=\frac{1}{T^2}\sum_{t=1}^T\sum_{\substack{s=1 \\ s\neq t}}^T\sum_{j=1}^K\sum_{k=1}^K \text{Cov}\left(\widetilde{p}_j(\*f_t)\widetilde{p}_k(\*f_t), \widetilde{p}_j(\*f_s)\widetilde{p}_k(\*f_s)\right) \notag \\ &\quad + \frac{1}{T^2}\sum_{t=1}^T\sum_{\substack{s=1 \\ s\neq t}}^T\sum_{j=1}^K\sum_{k=1}^K \mathbb{E}\left(\widetilde{p}_j(\*f_t)\widetilde{p}_k(\*f_t)\right)\mathbb{E}\left(\widetilde{p}_j(\*f_s)\widetilde{p}_k(\*f_s)\right)-K\notag \\ &=\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{\substack{s=1\\ s\neq t } }^{T}\sum_{j=1}^{K}\sum_{k=1}^{K}\text{Cov}\left(\widetilde{p}_{j}(\mathbf{f}_{t})\widetilde{p}_{k}(\mathbf{f}_{t}),\widetilde{p}_{j}(\mathbf{f}_{s})\widetilde{p}_{k}(\mathbf{f}_{s})\right)+\frac{T(T-1)}{T^{2}}\sum_{j=1}^{K}1-K\notag \\ &\leq\frac{1}{T^{2}}\frac{8+2\eta}{8+\eta}2^{\frac{\eta}{8+\eta}}\sum_{t=1}^{T}\sum_{\substack{s=1\\ s\neq t } }^{T}\sum_{j=1}^{K}\sum_{k=1}^{K}\alpha\left(\left|t-s\right|\right)^{\frac{\eta}{8+\eta}}\mathbb{E}\left(\left|\widetilde{p}_{j}(\mathbf{f}_{t})\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{\frac{8+\eta}{2}}\right)^{\frac{2}{8+\eta}}\notag \\ &\quad \times \mathbb{E}\left(\left|\widetilde{p}_{j}(\mathbf{f}_{s})\widetilde{p}_{k}(\mathbf{f}_{s})\right|^{\frac{8+\eta}{2}}\right)^{\frac{2}{8+\eta}}-\frac{K}{T}\notag \\ &\leq\frac{c_{\eta}}{T^{2}}\sum_{t=1}^{T}\sum_{s\neq t}^{T-1}\sum_{j=1}^{K}\sum_{k=1}^{K}\alpha\left(\left|t-s\right|\right)^{\frac{\eta}{8+\eta}}\left\{ \mathbb{E}\left(\left|\widetilde{p}_{j}(\mathbf{f}_{t})\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{\frac{8+\eta}{2}}\right)\right\} ^{\frac{4}{8+\eta}}-\frac{K}{T}\notag \\ &\leq\frac{c_{\eta}}{T^{2}}\sum_{t=1}^{T}\sum_{s\neq t}^{T-1}\sum_{j=1}^{K}\sum_{k=1}^{K}\alpha\left(\left|t-s\right|\right)^{\frac{\eta}{8+\eta}}\left\{ \mathbb{E}\left(\left|\widetilde{p}_{j}(\mathbf{f}_{t})\right|^{8+\eta}\right)\mathbb{E}\left(\left|\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{8+\eta}\right)\right\} ^{\frac{2}{8+\eta}}-\frac{K}{T}\notag \\ &\leq\frac{c_{\eta}}{T}\Bigg[\sum_{l=1}^{T-1}\alpha(l)^{\frac{\eta}{8+\eta}}\Bigg]\sum_{j=1}^{K}\sum_{k=1}^{K}\left\{ \mathbb{E}\left(\left|\widetilde{p}_{j}(\mathbf{f}_{t})\right|^{8+\eta}\right)\mathbb{E}\left(\left|\widetilde{p}_{k}(\mathbf{f}_{t})\right|^{8+\eta}\right)\right\} ^{\frac{2}{8+\eta}}-\frac{K}{T}\notag \\ &=O\left(\frac{K^{2}}{T}\right), \end{align} which holds under Assumptions (ref) and (ref)(ref). Here, we define $c_{\eta}\equiv \frac{8+2\eta}{8+\eta}2^{\frac{\eta}{8+\eta}}$ and $l\equiv \left|t-s\right|$. Taking the results for $A_1$ and $A_2$ together, we can bound $\mathbb{E}\left[\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\|^2\right]$ as \begin{align} \mathbb{E}\left[\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\|^2\right] &\equiv A_1 + A_2 =O\left(\frac{K\zeta_0(K)^2}{T}\right) + O\left(\frac{K^2}{T}\right) = O\left(\frac{K\zeta_0(K)^2}{T}\right), \end{align} since typically, $\zeta_0(K)=O(\sqrt{K})$ or $\zeta_0(K)=O(K)$. Therefore, \begin{align} \left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\| O_p\left(\sqrt{\frac{K}{T}}\zeta_0(K)\right) = o_p(1), \end{align} which holds under the Assumption (ref)(ref) condition that $\sqrt{\frac{K}{T}}\zeta_0(K)^{\frac{3}{2}}=o(1)$. Now consider (ref). By Lemma (ref)(ref), we have $\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=\*I_K$. Moreover, for any matrix $\*A\in \mathbb{R}^{K\times K}$, every eigenvalue $\lambda_j(A)$ satisfies $\left|\lambda_j(A)\right|\leq \left\|\*A\right\|$, for all $j=1, \dots, K$ (see, for example, horn). Using this we can show that \begin{align} \left|\lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)-\lambda_{\max}\left[\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right]\right| &= \left|\lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)-\lambda_{\max}\left(\*I_K\right)\right| \notag \\ &\leq \left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\| = o_p(1), \end{align} which holds by (ref). This implies that \begin{align} \lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)-\lambda_{\max}\left(\*I_K\right) = \lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)-1 = o_p(1), \end{align} and \begin{align} \lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right) = 1+o_p(1) = O_p(1). \end{align} Since exactly the same argumentation applies to $\lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)$, \begin{align} \lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right) = 1+o_p(1) = O_p(1), \end{align}which establishes (ref). To prove (ref), we can first consider the following expression \begin{align} \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\| &= \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\*I_K-\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right)\right\| \notag \\ &\leq \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\*I_K\right\|+\left\|\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}-\*I_K\right\| \notag \\ &= O_p\left(\delta_{NTK}\right)+ o_p(1) = O_p\left(\delta_{NTK}\right) =o_p(1), \end{align} where the last equality holds under Lemma (ref)(ref), as well as the results in (ref) and (ref). Consequently, by the same arguments used to prove (ref), \begin{align} \left|\lambda_{\min}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)-\lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)\right| \leq \left\|\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}-\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right\|= o_p(1). \end{align} Under Lemma (ref)(ref), and the normalization $\mathbb{E}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=\*I_K$, \begin{align} \lambda_{\max}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=\lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)=1+o_p(1), \end{align} and hence, \begin{align} \lambda_{\min}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)=\lambda_{\min}\left(\frac{1}{T}\widetilde{\mathbf{P}}'\widetilde{\mathbf{P}}\right)+o_p(1)=1+o_p(1)=O_p(1), \end{align} and also, \begin{align} \lambda_{\max}\left(\frac{1}{T}\widehat{\mathbf{P}}'\widehat{\mathbf{P}}\right)=1+o_p(1)=O_p(1), \end{align} which holds under exactly the same arguments as for the proof of (ref). This establishes (ref). The proof of Lemma (ref) is therefore complete. $\blacksquare$ \section{Additional Results: Empirical Application} In this section, we provide additional illustrations for our empirical application. Figure (ref) shows the development of wage inequality in U.S. manufacturing over the sample period. Table (ref) presents results from a Dickey-Fuller test for the estimated factors, supporting our hypothesis that they are unit-root non stationary. Figure (ref) plots $\widehat{\*f}_t = [\overline{\ln{\sigma_t}}, \overline{\ln \left(\frac{H_t}{L_t}\right)}, \overline{\*z}_t]'$ over time, after taking first differences, indicating mean reversion. Table (ref) presents SCCE estimates with 95% confidence intervals based on HAC standard errors, and Figure (ref) shows the estimated effects of our main regressors, $\ln(\sigma_{i,t})$ and $\ln\left(\frac{H_{i,t}}{L_{i,t}}\right)$, across different number of knots, $J$, using HAC standard errors. \setcounter{figure}{0} \begin{figure}[H] \caption{Average wage inequality in U.S. manufacturing sector for the time period 1958 -- 2005.} \begin{minipage}{\textwidth} \scriptsize \textit{Notes:} We calculate the average wage inequality as $\overline{\ln\left(\frac{w_t^{(L)}}{w_t^{(H)}}\right)}=\frac{1}{N}\sum_{i=1}^N \ln\left(\frac{w_{i,t}^{(L)}}{w_{i,t}^{(H)}}\right)$, for manufacturing sectors $i=1, \dots, N$. Here, $\frac{w_{i,t}^{(L)}}{w_{i,t}^{(H)}}$ is the relative wage of low-skilled workers to high-skilled workers in a U.S. manufacturing sector $i$ at time $t$. This illustration is based on a balanced version of the data from voig. \end{minipage} \end{figure} \begin{table} \begin{threeparttable} \caption{Results from Augmented Dickey-Fuller test.} \begin{tabular*}{0.7\textwidth}{@{\extracolsep{\fill}}lcc} \toprule Factor & ADF & $p$-value \\ \midrule $\overline{\ln(\sigma_{t})}$ & -1.02 & 0.75 \rule{0pt}{3ex}\\ $\overline{\ln\left(\frac{H_{t}}{L_{t}}\right)}$ & -0.57 & 0.88 \rule{0pt}{3ex}\\ $\overline{z}_{1,t}$ &0.98 & 0.99 \rule{0pt}{3ex}\\ $\overline{z}_{2,t}$ & 0.90 & 0.99 \rule{0pt}{3ex}\\ $\overline{z}_{3,t}$ & 0.76 & 0.99 \rule{0pt}{3ex}\\ $\overline{z}_{4,t}$ & -1.49 & 0.54 \rule{0pt}{3ex}\\ $\overline{z}_{5,t}$ & 2.39 & 0.99 \rule{0pt}{3ex}\\ $\overline{z}_{6,t}$ & 2.45 & 0.99 \rule{0pt}{3ex}\\ \bottomrule \end{tabular*} \begin{tablenotes} • \scriptsize \textit{Notes}: The lag order in the ADF test is selected using BIC. $\overline{z}_{1,t}, \dots, \overline{z}_{6,t}$ denote auxiliary regressors used as controls and are further specified in Table (ref). \end{tablenotes} \end{threeparttable} \end{table} \begin{figure}[p] \caption{Plotting $\widehat{\*f}_t = [\overline{\ln{\sigma_t}}, \overline{\ln \left(\frac{H_t}{L_t}\right)}, \overline{\*z}_t]'$ over time, after taking first differences.} \begin{subfigure}{0.49\textwidth} \caption{$\overline{\ln{\sigma_t}}$ and $\overline{\ln \left(\frac{H_t}{L_t}\right)}$.} \end{subfigure} \begin{subfigure}{0.49\textwidth} \caption{The elements of $\overline{\mathbf{z}}_t$.} \end{subfigure} \scriptsize \begin{minipage}[t]{\textwidth} \textit{Notes:} The figure plots the estimated factors over the 1958-2005 period. Figure B2.(A) plots the cross-sectional averages of the two key regressors, $\ln\left(\sigma_{i,t}\right)$ and $\ln\left(\frac{H_{i,t}}{L_{i,t}}\right)$, while Figure B2.(B) plots the controls included in $\mathbf{z}_{i,t}$. \end{minipage} \end{figure} \begin{table}[p] \begin{threeparttable} \caption{Estimation results for SCCE with HAC standard errors.} \begin{tabular*}{0.7\textwidth}{@{\extracolsep{\fill}}lcc} \toprule & Narrow model & Full model \\ \midrule $\ln(\sigma_{i,t})$ & -0.34 & -0.33 \\ & (-0.62; -0.06) & (-0.62; -0.05) \\ $\ln\!\left(\frac{H_{i,t}}{L_{i,t}}\right)$ & 0.56 & 0.61 \\ & (0.52; 0.59) & (0.58; 0.64) \\ $z_{1,i,t}$ & & -0.03 \\ & & (-0.37; 0.31) \\ $z_{2,i,t}$ & & 3.15 \\ & & (1.20; 5.11) \\ $z_{3,i,t}$ & & -1.03 \\ & & (-2.09; 0.04) \\ $z_{4,i,t}$ & & -0.20 \\ & & (-0.55; 0.16) \\ $z_{5,i,t}$ & & -0.17 \\ & & (-0.33; -0.00) \\ $z_{6,i,t}$ & & 0.07 \\ & & (-0.14; 0.28) \\ \bottomrule \end{tabular*} \begin{tablenotes} • \scriptsize \textit{Notes}: The table reports point estimates of the coefficients of the model in (ref) with the associated 95% confidence intervals appearing within parentheses, based on HAC standard errors. Results are reported for our SCCE estimator for the narrow and full model. The regressors of interest are $\ln\left(\sigma_{i,t}\right)$, which is a measure of the input skill intensity, and $\ln\left(\frac{H_{i,t}}{L_{i,t}}\right)$, which is the ratio of high-skilled to low-skilled labor. The six control variables are; real capital equipment per worker ($z_{1,i,t}$), the sectoral share of office, computing and accounting equipment ($z_{2,i,t}$), the difference between high-tech and computer capital share ($z_{3,i,t}$), R&D intensity ($z_{4,i,t}$), narrow outsourcing ($z_{5,i,t}$), and the difference between broad and narrow outsourcing ($z_{6,i,t}$). The dependent variable is the log of the relative wage of low-skilled to high-skilled workers. \end{tablenotes} \end{threeparttable} \end{table} \begin{figure}[p] \caption{Estimated effects of $\ln \left(\sigma_{i,t}\right)$ and $\ln \left(\frac{H_{i,t}}{L_{i,t}}\right)$ for different number of knots, $J$, using HAC standard errors.} \begin{subfigure}{0.49\textwidth} \caption{The effect of $\ln (\sigma_{i,t})$ in the narrow model.} \end{subfigure} \begin{subfigure}{0.49\textwidth} \caption{The effect of $\ln \left(\sigma_{i,t}\right)$ in the full model.} \end{subfigure} \begin{subfigure}{0.49\textwidth} \caption{The effect of $\ln \left(\frac{H_{i,t}}{L_{i,t}}\right)$ in the narrow model.} \end{subfigure} \begin{subfigure}{0.49\textwidth} \caption{The effect of $\ln \left(\frac{H_{i,t}}{L_{i,t}}\right)$ in the full model.} \end{subfigure} \scriptsize \begin{minipage}{\textwidth} \textit{Notes:} The figure plots point estimates obtained by reestimating (ref) by SCCE for $J\in \{2, 4, 6, 8, 10\}$. The grey area represents 95% confidence intervals based on HAC standard errors. \end{minipage} \end{figure} \section{Additional Results: Monte Carlo Simulations} This section presents additional results from a small-scale Monte Carlo simulation exercise, with a focus on various extensions that show the flexibility of our SCCE estimator. The specifications are basically Experiment \textbf{E1} in Section (ref) with minor deviations. Hence, we refer the reader there for more details. The following experiments are considered: \begin{enumerate} • Here, we report the numerical results of our estimator for the data-generating process as in Experiment \textbf{E1} in Section (ref). • Here, we report the numerical results of our estimator, assuming interactive effects as in Experiment E2 in section (ref). • Here, we allow the factors to be unit root non-stationary. More specifically, we generate $\*f_t = \*f_{t-1}+\*w_t$, with $\*w_t \sim \mathcal{N}(\*0_{m \times 1}, 0.05 \cdot \*I_m)$ and $\*f_0=\*0_{m \times 1}$. • Here, we consider an infeasible estimator, where we use the true factors as input for the sieve basis. • Here, we consider a DGP exactly as in E1 but use Hermite polynomials instead of splines as a sieve basis. For a definition of Hermite polynomials see, for example, chen2007. • Here, relax the independence assumption on the error terms in (ref) and (ref), and allow for weak serial and cross-sectional correlations as well as cross-sectional heteroscedasticity. We follow kad25, and generate $\varepsilon_{i,t}=\pi \varepsilon_{i,t-1} + \theta_{i,t}+\sum_{l=1}^L \pi (\theta_{i-l,t} +\theta_{i+l,t})$, where $\varepsilon_{1,0}= \dots = \varepsilon_{N,0}=0$, $L=5$, and $\theta_{i,1} \sim \mathcal{N}(0, \sigma_i^2)$ with $\sigma_i \sim \mathcal{U}(0.5,1)$. Further, $\pi \in \{0.2, 0.5\}$, where $\pi = 0.2$ implies "low", and $\pi = 0.5 $ implies "high" dependence. $\*v_{i,t}$ is generated in the same way. • Here, we consider different number of knots for the spline basis in Experiment \textbf{E1}. More specifically, we will set $J = C \lfloor T^{\frac{1}{4}} \rfloor$, with $C\in \{2,3,4\}$. Further, we also consider different rates for $T$ i.e., $J=\{T^{\frac{1}{3}}, T^{\frac{1}{5}}, T^{\frac{1}{10}}\}$. \end{enumerate} We ran $1{,}000$ simulations with $N,T \in \{20,50,100,200,300\}$. Given the computational load from high-dimensional, nonlinear estimation, the Monte Carlo experiments were executed on LUNARC's COSMOS cluster at Lund University. COSMOS comprises 182 compute nodes, each with two AMD 7413 processors (Milan), resulting in 48 cores per node. All simulations were implemented in Python 3.9.18. \begin{landscape} \begin{table}[p] \begin{threeparttable} \caption{Simulation results for Experiments E1.A, E1.B, E1.C, E1.D, and E1.E.} \begin{tabular}{rrrrrrrrrrrr} \toprule & & \multicolumn{2}{c}{E1.A} & \multicolumn{2}{c}{E1.B} & \multicolumn{2}{c}{E1.C} & \multicolumn{2}{c}{E1.D} & \multicolumn{2}{c}{E1.E} \\ \cmidrule(lr){3-4}\cmidrule(lr){5-6}\cmidrule(lr){7-8}\cmidrule(lr){9-10}\cmidrule(lr){11-12} N & T & Bias & RMSE & Bias & RMSE & Bias & RMSE & Bias & RMSE & Bias & RMSE \\ \midrule 20 & 20 & 0.0014 & 0.1420 & 0.0016 & 0.1183 & 0.0017 & 0.1166 & 0.0001 & 0.0660 & 0.0016 & 0.0861 \\ 20 & 50 & 0.0011 & 0.0478 & 0.0008 & 0.0420 & 0.0009 & 0.0406 & 0.0003 & 0.0332 & 0.0013 & 0.0449 \\ 20 & 100 & 0.0016 & 0.0335 & 0.0022 & 0.0273 & 0.0001 & 0.0260 & 0.0000 & 0.0235 & 0.0017 & 0.0333 \\ 20 & 200 & 0.0003 & 0.0258 & 0.0002 & 0.0220 & 0.0005 & 0.0188 & 0.0003 & 0.0177 & 0.0001 & 0.0269 \\ 20 & 300 & 0.0008 & 0.0226 & 0.0005 & 0.0201 & 0.0009 & 0.0169 & 0.0003 & 0.0140 & 0.0006 & 0.0228 \\ 50 & 20 & 0.0042 & 0.0859 & 0.0011 & 0.0704 & 0.0002 & 0.0714 & 0.0019 & 0.0423 & 0.0001 & 0.0534 \\ 50 & 50 & 0.0015 & 0.0317 & 0.0002 & 0.0241 & 0.0008 & 0.0247 & 0.0008 & 0.0224 & 0.0009 & 0.0309 \\ 50 & 100 & 0.0009 & 0.0215 & 0.0004 & 0.0159 & 0.0011 & 0.0162 & 0.0001 & 0.0149 & 0.0008 & 0.0226 \\ 50 & 200 & 0.0003 & 0.0148 & 0.0002 & 0.0109 & 0.0003 & 0.0121 & 0.0000 & 0.0104 & 0.0001 & 0.0156 \\ 50 & 300 & 0.0003 & 0.0131 & 0.0003 & 0.0089 & 0.0001 & 0.0100 & 0.0002 & 0.0086 & 0.0003 & 0.0138 \\ 100 & 20 & 0.0054 & 0.0584 & 0.0004 & 0.0484 & 0.0005 & 0.0504 & 0.0007 & 0.0310 & 0.0028 & 0.0383 \\ 100 & 50 & 0.0008 & 0.0210 & 0.0000 & 0.0175 & 0.0011 & 0.0169 & 0.0001 & 0.0157 & 0.0009 & 0.0212 \\ 100 & 100 & 0.0005 & 0.0143 & 0.0001 & 0.0112 & 0.0001 & 0.0120 & 0.0007 & 0.0107 & 0.0002 & 0.0147 \\ 100 & 200 & 0.0003 & 0.0102 & 0.0002 & 0.0075 & 0.0003 & 0.0080 & 0.0002 & 0.0076 & 0.0003 & 0.0108 \\ 100 & 300 & 0.0004 & 0.0089 & 0.0002 & 0.0062 & 0.0001 & 0.0066 & 0.0002 & 0.0057 & 0.0005 & 0.0093 \\ 200 & 20 & 0.0041 & 0.0568 & 0.0008 & 0.0356 & 0.0006 & 0.0350 & 0.0001 & 0.0206 & 0.0017 & 0.0272 \\ 200 & 50 & 0.0006 & 0.0149 & 0.0004 & 0.0125 & 0.0006 & 0.0127 & 0.0002 & 0.0108 & 0.0007 & 0.0145 \\ 200 & 100 & 0.0004 & 0.0100 & 0.0003 & 0.0084 & 0.0002 & 0.0081 & 0.0002 & 0.0077 & 0.0002 & 0.0102 \\ 200 & 200 & 0.0001 & 0.0069 & 0.0001 & 0.0054 & 0.0002 & 0.0056 & 0.0001 & 0.0050 & 0.0003 & 0.0074 \\ 200 & 300 & 0.0000 & 0.0062 & 0.0001 & 0.0043 & 0.0002 & 0.0046 & 0.0002 & 0.0042 & 0.0001 & 0.0067 \\ 300 & 20 & 0.0000 & 0.0326 & 0.0024 & 0.0288 & 0.0003 & 0.0291 & 0.0008 & 0.0170 & 0.0000 & 0.0214 \\ 300 & 50 & 0.0006 & 0.0120 & 0.0004 & 0.0097 & 0.0001 & 0.0100 & 0.0007 & 0.0088 & 0.0004 & 0.0118 \\ 300 & 100 & 0.0002 & 0.0084 & 0.0003 & 0.0067 & 0.0003 & 0.0066 & 0.0003 & 0.0062 & 0.0003 & 0.0087 \\ 300 & 200 & 0.0001 & 0.0058 & 0.0002 & 0.0042 & 0.0000 & 0.0046 & 0.0001 & 0.0043 & 0.0002 & 0.0062 \\ 300 & 300 & 0.0000 & 0.0051 & 0.0002 & 0.0035 & 0.0000 & 0.0037 & 0.0001 & 0.0036 & 0.0000 & 0.0054 \\ \bottomrule \end{tabular} \begin{tablenotes} • \scriptsize \textit{Notes}: The absolute bias and RMSE are computed as $\frac{1}{S} \sum_{s=1}^S |\widehat{\beta}_{1,s} - \beta_1 |$ and $\sqrt{\frac{1}{S}\sum_{s=1}^S (\widehat{\beta}_{1,s} - \beta_1)^2}$, respectively, where $S =1,000$ is the number of simulations, and $\widehat{\beta}_{1,s}$ is the estimate of $\beta_1$ from replication $s$. While we focus here on the first coefficient, $\beta_1$, the results for the second coefficient, $\beta_2$, were similar and can be made are available upon request. \end{tablenotes} \end{threeparttable} \end{table} \end{landscape} \begin{table}[p] \begin{threeparttable} \caption{Simulation results for Experiment E1.F.} \begin{tabular*}{0.7\textwidth}{@{\extracolsep{\fill}}cccccc} \toprule &&\multicolumn{2}{c}{$\pi =0.2$} & \multicolumn{2}{c}{$\pi =0.5$}\\ \cmidrule(lr){3-4} \cmidrule(lr){5-6} N & T & Bias & RMSE & Bias & RMSE \\ \toprule 20 & 20 & 0.0119 & 0.1911 & 0.0057 & 0.2733 \\ 20 & 50 & 0.0003 & 0.0675 & 0.0022 & 0.0971 \\ 20 & 100 & 0.0010 & 0.0481 & 0.0018 & 0.0678 \\ 20 & 200 & 0.0005 & 0.0341 & 0.0015 & 0.0465 \\ 20 & 300 & 0.0008 & 0.0300 & 0.0001 & 0.0402 \\ 50 & 20 & 0.0066 & 0.1251 & 0.0049 & 0.1804 \\ 50 & 50 & 0.0006 & 0.0433 & 0.0016 & 0.0655 \\ 50 & 100 & 0.0003 & 0.0282 & 0.0014 & 0.0421 \\ 50 & 200 & 0.0000 & 0.0206 & 0.0003 & 0.0298 \\ 50 & 300 & 0.0002 & 0.0174 & 0.0006 & 0.0253 \\ 100 & 20 & 0.0018 & 0.1010 & 0.0006 & 0.1303 \\ 100 & 50 & 0.0009 & 0.0296 & 0.0005 & 0.0456 \\ 100 & 100 & 0.0003 & 0.0187 & 0.0001 & 0.0293 \\ 100 & 200 & 0.0004 & 0.0142 & 0.0003 & 0.0201 \\ 100 & 300 & 0.0005 & 0.0120 & 0.0008 & 0.0170 \\ 200 & 20 & 0.0003 & 0.0856 & 0.0025 & 0.0893 \\ 200 & 50 & 0.0004 & 0.0214 & 0.0002 & 0.0329 \\ 200 & 100 & 0.0001 & 0.0135 & 0.0002 & 0.0208 \\ 200 & 200 & 0.0004 & 0.0091 & 0.0000 & 0.0137 \\ 200 & 300 & 0.0003 & 0.0082 & 0.0006 & 0.0120 \\ 300 & 20 & 0.0029 & 0.0496 & 0.0047 & 0.0706 \\ 300 & 50 & 0.0000 & 0.0172 & 0.0001 & 0.0256 \\ 300 & 100 & 0.0003 & 0.0112 & 0.0008 & 0.0169 \\ 300 & 200 & 0.0003 & 0.0083 & 0.0004 & 0.0115 \\ 300 & 300 & 0.0002 & 0.0067 & 0.0005 & 0.0098 \\ \bottomrule \end{tabular*} \begin{tablenotes} • \scriptsize \textit{Notes}: See Table 1 for an explanation. Here, $\pi$ refers to serial and cross-sectional correlation of the errors in the equations (ref) and (ref). \end{tablenotes} \end{threeparttable} \end{table} \begin{landscape} \begin{table}[p] \begin{threeparttable} \caption{Simulation results for Experiment E1.G.} \begin{tabular}{rrrrrrrrrrrrrr} \toprule &&\multicolumn{2}{c}{$J = 2\lfloor T^{\frac{1}{4}}\rfloor$} & \multicolumn{2}{c}{$J = 3\lfloor T^{\frac{1}{4}}\rfloor$} & \multicolumn{2}{c}{$J = 4\lfloor T^{\frac{1}{4}}\rfloor$} & \multicolumn{2}{c}{$J=\lfloor T^{\frac{1}{3}}\rfloor$} & \multicolumn{2}{c}{$J=\lfloor T^{\frac{1}{5}}\rfloor$} & \multicolumn{2}{c}{$J=\lfloor T^{\frac{1}{10}}\rfloor$}\\ \cmidrule(lr){3-4} \cmidrule(lr){5-6} \cmidrule(lr){7-8} \cmidrule(lr){9-10} \cmidrule(lr){11-12} \cmidrule(lr){13-14} N & T & Bias & RMSE & Bias & RMSE & Bias & RMSE & Bias & RMSE & Bias & RMSE & Bias & RMSE \\ \midrule 20 & 20 & 0.2027 & 6.5197 & 0.1158 & 3.3448 & 0.7988 & 28.9344 & 0.0014 & 0.1420 & 0.0004 & 0.1029 & 0.0004 & 0.1029 \\ 20 & 50 & 0.0018 & 0.0513 & 0.0022 & 0.0573 & 0.0015 & 0.0648 & 0.0017 & 0.0499 & 0.0011 & 0.0478 & 0.0014 & 0.0459 \\ 20 & 100 & 0.0010 & 0.0353 & 0.0011 & 0.0367 & 0.0008 & 0.0377 & 0.0015 & 0.0340 & 0.0018 & 0.0335 & 0.0014 & 0.0328 \\ 20 & 200 & 0.0005 & 0.0261 & 0.0003 & 0.0264 & 0.0003 & 0.0266 & 0.0005 & 0.0258 & 0.0004 & 0.0267 & 0.0002 & 0.0264 \\ 20 & 300 & 0.0008 & 0.0230 & 0.0007 & 0.0232 & 0.0006 & 0.0234 & 0.0009 & 0.0227 & 0.0007 & 0.0221 & 0.0003 & 0.0224 \\ 50 & 20 & 0.2721 & 8.9585 & 0.0924 & 4.8175 & 0.1307 & 2.4790 & 0.0042 & 0.0859 & 0.0001 & 0.0644 & 0.0001 & 0.0644 \\ 50 & 50 & 0.0019 & 0.0335 & 0.0019 & 0.0375 & 0.0028 & 0.0427 & 0.0023 & 0.0324 & 0.0015 & 0.0317 & 0.0006 & 0.0312 \\ 50 & 100 & 0.0012 & 0.0221 & 0.0011 & 0.0230 & 0.0011 & 0.0241 & 0.0009 & 0.0217 & 0.0010 & 0.0219 & 0.0009 & 0.0221 \\ 50 & 200 & 0.0003 & 0.0152 & 0.0003 & 0.0154 & 0.0003 & 0.0156 & 0.0003 & 0.0150 & 0.0002 & 0.0153 & 0.0003 & 0.0155 \\ 50 & 300 & 0.0002 & 0.0132 & 0.0001 & 0.0133 & 0.0002 & 0.0134 & 0.0004 & 0.0131 & 0.0002 & 0.0129 & 0.0004 & 0.0135 \\ 100 & 20 & 0.0034 & 3.6462 & 0.0082 & 1.2543 & 0.0746 & 3.6349 & 0.0054 & 0.0584 & 0.0031 & 0.0461 & 0.0031 & 0.0461 \\ 100 & 50 & 0.0009 & 0.0230 & 0.0017 & 0.0247 & 0.0015 & 0.0269 & 0.0007 & 0.0217 & 0.0008 & 0.0210 & 0.0010 & 0.0212 \\ 100 & 100 & 0.0004 & 0.0146 & 0.0004 & 0.0153 & 0.0003 & 0.0157 & 0.0006 & 0.0142 & 0.0004 & 0.0144 & 0.0003 & 0.0146 \\ 100 & 200 & 0.0003 & 0.0102 & 0.0003 & 0.0102 & 0.0003 & 0.0103 & 0.0002 & 0.0101 & 0.0002 & 0.0107 & 0.0003 & 0.0106 \\ 100 & 300 & 0.0004 & 0.0089 & 0.0004 & 0.0089 & 0.0004 & 0.0090 & 0.0004 & 0.0088 & 0.0004 & 0.0087 & 0.0004 & 0.0092 \\ 200 & 20 & 0.0274 & 1.5439 & 0.0200 & 1.2613 & 0.0042 & 2.0946 & 0.0041 & 0.0568 & 0.0020 & 0.0309 & 0.0020 & 0.0309 \\ 200 & 50 & 0.0004 & 0.0158 & 0.0009 & 0.0175 & 0.0014 & 0.0205 & 0.0004 & 0.0152 & 0.0006 & 0.0149 & 0.0006 & 0.0145 \\ 200 & 100 & 0.0004 & 0.0109 & 0.0004 & 0.0104 & 0.0004 & 0.0107 & 0.0004 & 0.0101 & 0.0004 & 0.0099 & 0.0002 & 0.0099 \\ 200 & 200 & 0.0001 & 0.0070 & 0.0000 & 0.0071 & 0.0000 & 0.0072 & 0.0001 & 0.0070 & 0.0001 & 0.0070 & 0.0002 & 0.0071 \\ 200 & 300 & 0.0000 & 0.0062 & 0.0001 & 0.0063 & 0.0001 & 0.0064 & 0.0000 & 0.0062 & 0.0000 & 0.0062 & 0.0001 & 0.0064 \\ 300 & 20 & 0.0271 & 0.4488 & 0.1172 & 5.2088 & 0.1167 & 2.3462 & 0.0000 & 0.0326 & 0.0002 & 0.0249 & 0.0002 & 0.0249 \\ 300 & 50 & 0.0004 & 0.0126 & 0.0003 & 0.0139 & 0.0005 & 0.0155 & 0.0006 & 0.0122 & 0.0006 & 0.0120 & 0.0006 & 0.0117 \\ 300 & 100 & 0.0001 & 0.0086 & 0.0000 & 0.0089 & 0.0000 & 0.0092 & 0.0001 & 0.0085 & 0.0002 & 0.0085 & 0.0002 & 0.0085 \\ 300 & 200 & 0.0001 & 0.0058 & 0.0002 & 0.0060 & 0.0001 & 0.0060 & 0.0001 & 0.0058 & 0.0001 & 0.0060 & 0.0001 & 0.0061 \\ 300 & 300 & 0.0000 & 0.0051 & 0.0000 & 0.0051 & 0.0000 & 0.0051 & 0.0000 & 0.0051 & 0.0000 & 0.0051 & 0.0000 & 0.0053 \\ \bottomrule \end{tabular} \begin{tablenotes} • \scriptsize \textit{Notes}: See Table 1 for an explanation. Here, $J$ is the number of knots and $T$ is the time dimension. \end{tablenotes} \end{threeparttable} \end{table} \end{landscape}