Asymptotic equivalence of Principal Components and Quasi Maximum Likelihood estimators in Large Approximate Factor Models

abstractThis paper investigates the properties of Quasi Maximum Likelihood estimation of an approximate factor model for an $n$-dimensional vector of stationary time series. We prove that the factor loadings estimated by Quasi Maximum Likelihood are asymptotically equivalent, as $n\to\infty$, to those estimated via Principal Components. Both estimators are, in turn, also asymptotically equivalent, as $n\to\infty$, to the unfeasible Ordinary Least Squares estimator we would have if the factors were observed. We also show that the usual sandwich form of the asymptotic covariance matrix of the Quasi Maximum Likelihood estimator is asymptotically equivalent to the simpler asymptotic covariance matrix of the unfeasible Ordinary Least Squares. All these results hold in the general case in which the idiosyncratic components are cross-sectionally heteroskedastic, as well as serially and cross-sectionally weakly correlated. The intuition behind these results is that as $n\to\infty$ the factors can be considered as observed, thus showing that factor models enjoy a blessing of dimensionality. Keywords: Approximate Dynamic Factor Model; Principal Component Analysis; Quasi Maximum Likelihood.

ass[common component] $\,$ \begin{compactenum}[(a)] • $\lim_{n\to\infty}\Vert n^{-1}\bm\Lambda^\prime\bm\Lambda-\bm\Sigma_{\Lambda}\Vert=0$, where $\bm\Sigma_{\Lambda}$ is $r\times r$ positive definite, and, for all $i\in\mathbb N$, $\Vert\bm\lambda_i\Vert\le M_\Lambda$ for some finite positive real $M_\Lambda$ independent of $i$. • For all $t\in\mathbb Z$, $\mathbb{E}[\mathbf F_{t}]=\mathbf 0_r$ and $\bm\Gamma^F=\mathbb{E}[\mathbf F_t\mathbf F_t^\prime]$ is $r\times r$ positive definite and $\Vert\bm\Gamma^F\Vert\le M_F$ for some finite positive real $M_F$ independent of $t$. • \begin{inparaenum} • For all $t\in\mathbb Z$, $\mathbb{E}[\Vert \mathbf F_t\Vert^4]\le K_F$ for some finite positive real $K_F$ independent of $t$; \\ • for all $i,j=1,\ldots, r$, all $s=1,\ldots, T$, and all $T\in\mathbb N$, \[ \mathbb{E}\left[\left\vert\frac 1{\sqrt{T}}\sum_{t=1}^T\left\{F_{is}F_{jt}-\mathbb{E}[F_{is}F_{jt}]\right\} \right\vert^2\right] \le C_F \] for some finite positive real $C_F$ independent of $i$, $j$, $s$, and $T$. \end{inparaenum} • There exists an integer $N$ such that for all $n> N$, $r$ is a finite positive integer, independent of $n$. \end{compactenum}

ass[idiosyncratic component] $\,$ \begin{compactenum}[(a)] • For all $i\in\mathbb N$ and all $t\in\mathbb Z$, $\mathbb{E}[\xi_{it}]= 0$ and $\sigma_i^2=\mathbb{E}[\xi_{it}^2]$ is such that $C_\xi\le \sigma_i^2\le C_\xi^\prime$ for some finite positive reals $C_\xi$ and $C_\xi^\prime$ independent of $i$ and $t$. • For all $i,j\in\mathbb N$, all $t\in\mathbb Z$, and all $k\in\mathbb Z$, $\vert \mathbb{E}[\xi_{it}\xi_{j,t-k}]\vert\le \rho^{\vert k\vert} M_{ij}$, where $\rho$ and $M_{ij}$ are finite positive reals independent of $t$ such that $0\le \rho <1$, $\sum_{j=1,j\ne i}^n M_{ij}\le M_\xi$, and $\sum_{i=1, i\ne j}^n M_{ij}\le M_\xi$ for some finite positive real $M_{\xi}$ independent of $i$, $j$, and $n$. • \begin{inparaenum} • For all $i=1,\ldots, n$, all $t=1,\ldots, T$, and all $n,T\in\mathbb N$, $\mathbb{E}[\xi_{it}^4]\le Q_\xi$ for some finite positive real $Q_\xi$ independent of $i$ and $t$;\\ • for all $j=1,\ldots, n$, all $s=1,\ldots, T$, and all $n,T\in\mathbb N$, \[ \mathbb{E}\left[\left\vert\frac 1{\sqrt{nT}} \sum_{i=1}^n\sum_{t=1}^T\left\{\xi_{is}\xi_{jt}-\mathbb{E}[\xi_{is}\xi_{jt}]\right\} \right\vert^2\right] \le K_\xi \] for some finite positive real $K_\xi$ independent of $j$, $s$, $n$, and $T$. \end{inparaenum} \end{compactenum}

ass[independence] The processes $\{\xi_{it},\, i\in\mathbb N,\, t\in\mathbb Z\}$ and $\{F_{jt},\, j=1,\ldots, r,\, t\in\mathbb Z\}$ are mutually independent.

ass[distinct eigenvalues] For all $n\in\mathbb N$ and all $i=1,\ldots, r-1$, $\mu_i^\chi > \mu_{i+1}^\chi$.

ass[Identification] $\,$ \begin{inparaenum}[(a)] • For all $n\in\mathbb N$, $n^{-1}{\bm\Lambda^\prime\bm\Lambda}$ diagonal. • For all $T\in\mathbb N$, $T^{-1}{\bm F^\prime\bm F}=\mathbf I_r$. \end{inparaenum}

ass[Global identification] For all $j=1,\ldots, r$, one of the two following conditions holds: \begin{inparaenum}[(a)] • $\lambda_{1j}> 0$; or • $F_{j1}> 0$. \end{inparaenum}

ass[Central limit theorem] For all $i\in\mathbb N$, as $T\to\infty$, \[ \frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t \xi_{it} \to_d \mathcal N\left(\mathbf 0_r, \bm\Phi_i \right), \] where $\bm\Phi_i=\lim_{T\to\infty}T^{-1}\sum_{t,s=1}^T \mathbb{E}_{}[\mathbf F_t\mathbf F_s^\prime\xi_{it}\xi_{is}]$.

rem{ As discussed above, all assumptions for PC estimation by Bai03 are implied or equivalent to ours. Regarding QML estimation and the assumptions in baili16, we do not need their Assumption D, which requires the QML estimators of the idiosyncratic variances, $\sigma_i^2$, $i=1,\ldots, n$, to be finite and strictly positive, and we do not require the moment condition E.3, which is needed only for estimation of $\sigma_i^2$ and it is not necessary to prove our results. Assumptions E.4, E.5, E.6, F.2, and F.3 in baili16 are needed only for studying the behavior of the estimated factors so are not needed here. All other assumptions in baili16 are equivalent or nested into ours, with the crucial exception of their identifying condition on the loadings which, as noted above, differs from ours.}

rem{ Allowing for data with non-zero mean simply amounts to adding a constant term, $\alpha_i\ne 0$, to (ref), so that $x_{it}=\alpha_i+\bm\lambda_i^\prime\mathbf F_t+\xi_{it}$. All estimators described in the following retain the same properties even in this case. Indeed, estimation of such model by PCs simply requires to center data first, i.e., to work with $x_{it}-\bar {x}_i$ where $\bar x_i=T^{-1}\sum_{t=1}^T x_{it}$ is clearly a consistent estimator of $\alpha_i$. As for QML estimation, it is straightforward to see that $\bar x_i$ is precisely the QML estimator of $\alpha_i$ and thus it is enough to work with the likelihood of the centered data baili12. }

inparaenum• based on the $n\times n$ matrix $\bm X^\prime\bm X$ solve first for loadings and then get the factors by projecting $\bm X$ onto the estimated loadings; • based on the $T\times T$ matrix $\bm X\bm X^\prime$ solve first for the factors and then get the loadings by projecting $\bm X$ onto the estimated factors.

theorem[] Under Assumptions (ref) through (ref), as $n,T\to\infty$, \begin{compactenum}[(a)] • $\min(n,\sqrt T)\Vert\widehat{\bm\lambda}_i-{\bm\lambda}_i\Vert=O_{\mathrm P}(1)$, for any given $i=1,\ldots,n$, and, if $n^{-1}\sqrt T\to0$, $ \sqrt T(\widehat{\bm\lambda}_i-{\bm\lambda}_i) \to_d\mathcal N\left(\mathbf 0_r, \bm\Phi_i\right), $ where $\bm\Phi_i$ is defined in Assumption (ref); • $\min(n,\sqrt T)\Vert n^{-1/2}({\widehat{\bm\Lambda}-\bm\Lambda})\Vert=O_{\mathrm P}(1)$. \end{compactenum}

cor[] Under Assumptions (ref) through (ref), as $n,T\to\infty$, \begin{inparaenum}[(a)] • $ \min(n,\sqrt{nT})\Vert\widehat{\bm\lambda}_i - {\bm\lambda}_i^{\text{\tiny \upshape OLS}}\Vert = O_{\mathrm P}\left(1\right) $, for any given $i=1,\ldots,n$; • $\min(n,\sqrt {nT})\Vert n^{-1/2}({\widehat{\bm\Lambda}-{\bm\Lambda}^{\text{\tiny \upshape OLS}}})\Vert=O_{\mathrm P}(1)$. \end{inparaenum}

theorem[] Assume $n^{-1}{\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda}$ to be diagonal for all $n\in\mathbb N$ and $T^{-1}{\bm F^\prime\bm F}=\mathbf I_r$ for all $T\in\mathbb N$. Then, under Assumptions (ref), (ref), (ref), and (ref), if $n^{-1}\sqrt T\to0$, as $n,T\to\infty$, for any given $i=1,\ldots,n$, $ \sqrt T(\widehat{\bm\lambda}_i^{\text{\tiny \upshape QML,E}} -\bm\lambda_i)\to_d\mathcal N\left(\mathbf 0_r, \bm\Phi_i\right), $ where $\bm\Phi_i$ is defined in Assumption (ref).

rem{ Notice that the fact that we consider a mis-specified log-likelihood with serially and cross-sectionally uncorrelated idiosyncratic components, does not affect consistency of the estimated loadings, but only their asymptotic covariance. And in particular, notice that what matters for the asymptotic covariance is just the idiosyncratic serial correlation. Indeed, if the idiosyncratic components were cross-sectionally uncorrelated but serially correlated the asymptotic covariance in Theorem (ref) would be the same, while if they were serially uncorrelated the asymptotic covariance would be $ \sigma_i^2 \mathbf I_r$ (due to the identification $\bm\Gamma^F=\mathbf I_r$), regardless of the presence or not of cross-sectional correlation. This result, which is a special case of Theorem (ref), does not require any constraint between the rates of divergence of $n$ and $T$ and it is proved by baili12. }

rem{ The case of QML estimation when $\bm\Omega^F$ depends explicitly on additional parameters capturing the autocorrelations in the factors is considered in DGRqml. However, in that case QML estimation requires the use of the EM algorithm jointly with the Kalman smoother and the results of this paper do not directly apply unless we first prove convergence of the EM to the same QML estimator considered here. }

theoremUnder Assumptions (ref) through (ref) and assuming also that $\bm\Gamma^\xi$ is positive definite, as $n,T\to\infty$, \begin{inparaenum}[(a)] • $n\Vert n^{-1/2}({\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}}-\widehat{\bm\Lambda}})\Vert= O_{\mathrm {P}}(1)$; • $n \Vert\widehat{\bm\lambda}_i^{\text{\tiny \upshape QML}}-\widehat{\bm\lambda}_i \Vert = O_{\mathrm {P}}(1)$, for any given $i=1,\ldots,n$. \end{inparaenum}

theoremUnder Assumptions (ref) through (ref) and assuming also that $\bm\Gamma^\xi$ is positive definite, as $n,T\to\infty$: \begin{compactenum}[(a)] • $\min(n,\sqrt T)\Vert\widehat{\bm\lambda}_i^{\text{\tiny \upshape QML}}-{\bm\lambda}_i\Vert = O_{\mathrm {P}}(1)$, for any given $i=1,\ldots,n$,, and, if $n^{-1}\sqrt T\to 0$ $ \sqrt T(\widehat{\bm\lambda}_i^{\text{\tiny \upshape QML}}-\bm\lambda_i)\to_d\mathcal N\left(\mathbf 0_r, \bm\Phi_i\right), $ where $\bm\Phi_i$ is defined in Assumption (ref); • $\min(n,\sqrt T)\Vert n^{-1/2}({\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}}-\bm\Lambda})\Vert=O_{\mathrm P}(1)$. \end{compactenum}

corUnder Assumptions (ref) through (ref), as $n,T\to\infty$, \begin{inparaenum}[(a)] • $n\Vert \widehat{\bm\lambda}_i - \widehat{\bm\lambda}_i^{\text{\tiny \upshape QML,E}}\Vert=O_{\mathrm P}(1)$; • $n\Vert \widehat{\bm\lambda}_i - \widehat{\bm\lambda}_i^{\text{\tiny \upshape QML,E}_0}\Vert=O_{\mathrm P}(1)$. \end{inparaenum}

rem{ The appealing feature of the log-likelihood (ref), is that it does not depend on the factors, and, under the identifying constraint $T^{-1}{\bm F'\bm F}=\mathbf I_r$ for all $T\in\mathbb N$, it does not depend on the second moments of the factors either. Still, if we follow the classical approach to consider the vector of factors $\bm{\mathcal F}$ as an $rT$-dimensional sequence of deterministic constants lawleymaxwell71, then a log-likelihood alternative to (ref) can be considered, namely: \begin{equation} \ell_{\tiny E}(\bm{\mathcal X};\bm\varphi,\bm{\mathcal F})= -\frac T2\log\det(\bm\Gamma^\xi)-\frac 12\sum_{t=1}^T (\mathbf x_t-\bm\Lambda\,\mathbf F_t)^\prime (\underline{\bm\Gamma}^\xi)^{-1} (\mathbf x_t-\underline{\bm\Lambda}\,\underline{\mathbf F}_t). \end{equation} It is straightforward to see that, for given factors, the loadings maximizing (ref) are given by their OLS estimator, while, for given loadings, the factors maximizing (ref) are given by their GLS estimator. Hence, full maximization of (ref) requires knowing, or estimating, the factors too. Although seemingly simpler this approach presents at least two major drawbacks with respect to the approach followed in the proof of Theorem (ref) (see also the comments in baili12, and AR56). First, by iterating, between OLS and GLS we can think of finding a solution. This approach is similar to the one recently considered by pz23, but the convergence properties of such algorithm are not proved nor discussed, so it is unclear how this estimator is related to the maximizer of the full likelihood (ref). Moreover, this strategy would require a positive definite estimator of $\bm\Gamma^\xi$ in order to compute the GLS estimator of the factors, a hard task when $n$ is large. This, in general, requires again mis-specifying or regularizing the log-likelihood (ref), e.g., by replacing $\bm\Gamma^\xi$ with the diagonal $\bm\Sigma^\xi$. The asymptotic properties of the estimator of the loadings will then explicitly depend on the properties of an estimator of the idiosyncratic covariance, or at least of its diagonal elements. As shown in Theorem (ref), this is not the case for the QML estimator maximizing the log-likelihood (ref). Second, if the factors are treated as random variables, as, e.g., in the popular Factor Augmented VAR models BBE05, then they cannot be considered as constant parameters, which means that (ref) is not the full log-likelihood of the data $\bm{\mathcal X}$ but it is just the conditional log-likelihood of $\bm{\mathcal X}$ given the factors, so, in principle, not all information is used to estimate the loadings. Indeed, if the factors are random variables then the log-likelihood (ref) is decomposed as \begin{equation} \ell_{\text{\tiny E}}(\bm{\mathcal X};\underline{\bm\varphi})=\ell_{\text{\tiny E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}) +\ell_{\text{\tiny E}}(\bm{\mathcal F};\underline{\bm\varphi})-\ell_{\text{\tiny E}}(\bm{\mathcal F}|\bm{\mathcal X};\underline{\bm\varphi}), \end{equation} where $\ell_{\text{\tiny E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})$ coincides with (ref) but it does not coincide with $\ell_{\text{\tiny E}}(\bm{\mathcal X};\underline{\bm\varphi})$ anymore. This last point has both a theoretical and an applied implication. From the theory point of view, to directly show from (ref) that the QML estimator of the loadings maximizing $\ell_{\text{\tiny E}}(\bm{\mathcal X};\underline{\bm\varphi})$ is asymptotically equivalent to the unfeasible OLS maximizing $\ell_{\text{\tiny E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})$ would require showing that $\ell_{\text{\tiny E}}(\bm{\mathcal F};\underline{\bm\varphi})$ and $\ell_{\text{\tiny E}}(\bm{\mathcal F}|\bm{\mathcal X};\underline{\bm\varphi})$ are asymptotically negligible. This is the argument sketched by BT11, but to make it formal is not an easy task. Consider the simplest case in which the factors are treated as serially uncorrelated, then, while $\ell_{\text{\tiny E}}(\bm{\mathcal F};\underline{\bm\varphi})$ does not depend on any parameter and can be discarded, the expression of $\ell_{\text{\tiny E}}(\bm{\mathcal F}|\bm{\mathcal X};\underline{\bm\varphi})$ will depend on the conditional moments (mean and covariance) of the factors given $\bm{\mathcal X}$. These in turn have simple expressions only if we are willing to assume Gaussianity, in which case the conditional mean is just a linear projection. Otherwise computation of those moments is not straightforward. The proof of Theorem (ref) relies instead only on $\ell_{\text{\tiny E}}(\bm{\mathcal X};\underline{\bm\varphi})$, so, as noticed above, it does not require knowing the factors or their conditional moments. Finally, from a practical point of view, we could use the right-hand-side of (ref) to compute the QML estimator by means of the Expectation Maximization (EM) algorithm, which simplifies our task since it allows us to discard $\ell_{\text{\tiny E}}(\bm{\mathcal F}|\bm{\mathcal X};\underline{\bm\varphi})$ wu83. However, once again we would still have to compute the conditional moments of the factors when in the E-step we need compute the conditional expectation of $\ell_{\text{\tiny E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}) +\ell_{\text{\tiny E}}(\bm{\mathcal F};\underline{\bm\varphi})$ given $\bm{\mathcal X}$. This introduces a correction in the estimation of the loadings and the resulting estimator is not given by a simple OLS anymore. }

theoremUnder Assumptions (ref) through (ref), as $n,T\to\infty$, for any given $i=1,\ldots,n$, \begin{compactenum}[(a)] • \[ \frac 1{\sqrt T}\left\Vert \left.\frac{\partial \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}={\bm\varphi}} -\left.\frac{\partial \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}={\bm\varphi}} \right\Vert=O_{\mathrm P}\left(\max\left(\frac 1{\sqrt n},\frac {\sqrt T}{n}\right)\right); \] • \[ \frac 1T\left\Vert \left.\frac{\partial^2 \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial \underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}={\bm\varphi}} -\left.\frac{\partial^2 \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial \underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}={\bm\varphi}} \right\Vert=O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}\right)\right); \] • \[ \frac 1{\sqrt T}\left\Vert \left.\frac{\partial \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}=\widehat{\bm\varphi}^{\text{\tiny \upshape QML,E}}} -\left.\frac{\partial \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}=\widehat{\bm\varphi}^{\text{\tiny \upshape QML,E}}} \right\Vert=O_{\mathrm P}\left(\max\left(\frac 1{\sqrt n},\frac {\sqrt T}{n}\right)\right); \] • \[ \frac 1T\left\Vert \left.\frac{\partial^2 \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial \underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}=\widehat{\bm\varphi}^{\text{\tiny \upshape QML,E}}} -\left.\frac{\partial^2 \ell_{\text{\tiny \upshape E}}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial \underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}=\widehat{\bm\varphi}^{\text{\tiny \upshape QML,E}}} \right\Vert=O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}\right)\right). \] \end{compactenum}

rem{ The sandwich form of the asymptotic covariances (ref) and (ref) comes from the fact that in the log-likelihood we treat the idiosyncratic components as serially uncorrelated, while, in fact, they might be autocorrelated. Indeed, if we assumed $\mathbb{E}[\xi_{it}\xi_{is}]=0$ for all $s,t=1,\ldots, T$ with $s\ne t$ and all $i=1,\ldots, T$, then $\bm \Phi_i = \sigma_i^2\mathbf I_r$ (see also Remark (ref)). It follows that: $\bm{\mathcal I}_i(\bm{\mathcal X}|\bm{\mathcal F};\bm\varphi)={\sigma_i^{-2}}\mathbf I_r=-\bm{\mathcal H}_{i}(\bm{\mathcal X}|\bm{\mathcal F};\bm\varphi)$. And, by virtue of Theorem (ref), $\bm{\mathcal I}_i(\bm{\mathcal X};\bm\varphi)$ must coincide asymptotically with $-\bm{\mathcal H}_{i}(\bm{\mathcal X};\bm\varphi)$, thus, from (ref), we have $\text{A$\!\mathbb V$ar}_0(\sqrt T \widehat{\bm\lambda}_i^{\text{\tiny QML,E}})=\{\bm{\mathcal I}_i(\bm{\mathcal X};\bm\varphi)\}^{-1}$. }

rem{ If we explicitly model the dynamics of $\mathbf F_t$ then the expression of $\widehat{\mathbf F}_t^{\text{\tiny LP}}$ in (ref) is replaced by the Kalman smoother DGRfilter. This, in fact, can be shown to be asymptotically equivalent, as $n\to\infty$, to the GLS estimator in (ref) baili16,PR22. }

rem{ In the case of deterministic factors, we could also write the mis-specified log-likelihood (ref) of an exact factor model as: \begin{equation} \ell_{\tiny E}(\bm{\mathcal X};\bm\varphi)= -\frac 1 2\sum_{i=1}^n \log\det (\bm F\,\bm \lambda_i\bm \lambda_i^\prime\bm F^\prime+\underline{\sigma}_i^2\mathbf I_T)-\frac 12\sum_{i=1}^n \bm x_i^\prime(\underline{\bm F}\,\underline{\bm \lambda}_i\underline{\bm \lambda}_i^\prime\underline{\bm F}^\prime+\underline{\sigma}_i^2\mathbf I_T)^{-1}\bm x_i, \end{equation} where $\bm x_i=(x_{i1}\cdots x_{iT})'$, thereby exchanging the role of $n$ and $T$. Then, we can conjecture that the QML estimator of the factors maximizing (ref) will be asymptotically equivalent, this time as $T\to\infty$, to their PC estimator. This is approach is also considered in fortin2023latent. However, since, as noted above, the PC estimator of the factors is asymptotically equivalent to the OLS, it is not the most efficient estimator because it does not account for possible cross-sectional heteroskedasticity of idiosyncratic components. }

inparaenum• $\bm\ell_i$ has entries ${\ell}_{ij}{\sim}iid\,\mathcal{N}(1,1)$, $i=1,\ldots, N$, $j=1,\ldots, r$; • ${\bm A}=0.9 \check{\bm A} \Vert{\bm A}\Vert^{-1}$, where $\check{\bm A}$ is $r\times r$ with entries $\check{a}_{jj}{\sim} iid\,\mathcal U[0.5,0.8]$ for all $j$, and $\check{ a}_{jk}{\sim} iid\, \mathcal U[0,0.3]$ if $j\ne k$; • $u_{jt}{\sim}(0,1)$, $j=1,\ldots, r$, $t=1,\ldots, t$, with either a Gaussian or an Asymmetric Laplace distribution and with $\mathbb{C}\mathrm{ov}(u_{it},u_{jt})=0$ if $i\ne j$, and $\mathbb{C}\mathrm{ov}(u_{it},u_{jt-k})=0$ for all $i,j$ and if $k\ne 0$; • $e_{it}{\sim}(0,\sigma_{ei}^2)$, with either a Gaussian or an Asymmetric Laplace distribution and with $\sigma_{ei}^2\sim \mathcal U[0.5, 1.5]$ for all $i$, $\mathbb{C}\mathrm{ov}(e_{it},e_{jt})=\tau^{\vert i-j\vert }\mathbb I(\vert i-j\vert \le 10)$ with $\tau\in\{0,0.5\}$ if $i\ne j$, and $\mathbb{C}\mathrm{ov}(e_{it},e_{jt-k})=0$ for all $i,j$ and if $k\ne 0$; • $\delta_i{\sim}iid\,\mathcal{U}(0,\delta)$, and $\delta\in\{0,0.5\}$; • $\phi_i=\{{\theta_i (\sum_{t=1}^T \chi_{it}^2)/(\sum_{t=1}^T \xi_{it}^2)}\}^{1/2}$, and $\theta_i{\sim}iid\,\mathcal{U}(0.25,0.5)$.

appendix\numberwithin{equation}{section} \numberwithin{prop}{section} \section{Proof of main results} \subsection{Proof of Theorem (ref)} From (ref) and (ref), and since by (ref) we have $\widehat{\bm\Lambda}^\prime\widehat{\bm\Lambda}=\widehat{\mathbf M}^x$ and $\widehat{\mathbf V}^x=\widehat{\bm\Lambda}(\widehat{\mathbf M}^x)^{-1/2}$ which is well defined because of Lemma (ref)(iv), we get \begin{equation} \frac{\bm X^\prime\bm X}{nT}\widehat{\bm\Lambda}=\widehat{\bm\Lambda}\frac{\widehat{\mathbf M}^x}{n}. \end{equation} Then, substituting $\bm X^\prime\bm X=(\bm\Lambda\bm F^\prime+\bm\Xi^\prime)^\prime(\bm F\bm\Lambda^\prime+\bm\Xi)$ into (ref) \begin{align} \frac{\bm\Lambda\bm F^\prime\bm F\bm\Lambda^\prime\widehat{\bm\Lambda}}{nT} +\frac{\bm\Lambda\bm F^\prime\bm \Xi\widehat{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm F\bm\Lambda^\prime\widehat{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm \Xi\widehat{\bm\Lambda}}{nT}=\widehat{\bm\Lambda}\frac{\widehat{\mathbf M}^x}{n}. \end{align} Define \begin{equation} \widehat{\mathbf H}= \left(\frac{\bm F^\prime\bm F}{T}\right) \left(\frac{\bm\Lambda^\prime\widehat{\bm\Lambda}}{n}\right) \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}= \left(\frac{\bm\Lambda^\prime\widehat{\bm\Lambda}}{n}\right) \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}, \end{equation} by Assumption (ref)(b). Notice that, as $n,T\to\infty$, $\widehat{\mathbf H}$ is well defined because $\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}$ is well defined because of Lemma (ref)(iv). From (ref) and (ref) \begin{align} \widehat{\bm\Lambda}&-\bm\Lambda\widehat{\mathbf H}= \left(\frac{\bm\Lambda\bm F^\prime\bm \Xi\widehat{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm F\bm\Lambda^\prime\widehat{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm \Xi\widehat{\bm\Lambda}}{nT}\right)\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\nonumber\\ =&\, \left(\frac{\bm\Lambda\bm F^\prime\bm \Xi{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm F\bm\Lambda^\prime{\bm\Lambda}}{nT} +\frac{\bm\Xi^\prime\bm \Xi{\bm\Lambda}}{nT}\right)\bm{\mathcal H} \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\!\!\!\!\!+ \left(\frac{\bm\Lambda\bm F^\prime\bm \Xi}{nT} +\frac{\bm\Xi^\prime\bm F\bm\Lambda^\prime}{nT} +\frac{\bm\Xi^\prime\bm \Xi}{nT}\right)(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}) \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}. \end{align} Taking the $i$th row of (ref) \begin{align} \widehat{\bm\lambda}_i^\prime-{\bm\lambda}_i^\prime\widehat{\mathbf H}=&\, \left( \underbrace{\frac 1{nT}{\bm\lambda}_i^\prime\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime}_{(1.a)} +\underbrace{\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime\sum_{j=1}^n\bm\lambda_j\bm\lambda_j^\prime}_{(1.b)} +\underbrace{\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} \bm\lambda_j^\prime}_{(1.c)} \right)\bm{\mathcal H} \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\nonumber\\ &+ \left( \underbrace{\frac 1{nT}{\bm\lambda}_i^\prime\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}(\widehat{\bm\lambda}_j^\prime-\bm\lambda_j^\prime\bm{\mathcal H})}_{(1.d)} +\underbrace{\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime\sum_{j=1}^n\bm\lambda_j(\widehat{\bm\lambda}_j^\prime-\bm\lambda_j^\prime\bm{\mathcal H})}_{(1.e)}\right.\nonumber\\ &\;\;\left.+\underbrace{\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} (\widehat{\bm\lambda}_j^\prime-\bm\lambda_j^\prime\bm{\mathcal H})}_{(1.f)}\right) \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}. \end{align} From Proposition (ref) we see that, under Assumptions (ref) through (ref), the terms $\text{\upshape (1.a)}$, $\text{\upshape (1.c)}$, $\text{\upshape (1.d)}$, $\text{\upshape (1.e)}$, and $\text{\upshape (1.f)}$ are all $o_{\mathrm P}\left(\frac 1{\sqrt T}\right)$. In particular, from (ref), for any $i=1,\ldots, n$, we get \begin{align} \widehat{\bm\lambda}_i-\widehat{\mathbf H}^\prime{\bm\lambda}_i = \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1} \bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right) \left(\frac 1T\sum_{t=1}^T \mathbf F_t\xi_{it}\right) + O_{\mathrm {P}}\left(\max\left(\frac 1 n, \frac 1{\sqrt{nT}}\right)\right). \end{align} Consistency follows immediately since the first term in (ref) is $O_{\mathrm P}\left(\frac 1{\sqrt T}\right)$ because of Assumption (ref) (see also (ref) in the proof of Theorem (ref)) and since $\left\Vert \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1} \bm{\mathcal H}^\prime \right\Vert = O_{\mathrm {P}}(1)$ because of Lemma (ref)(iv) and (ref). Then, from (ref), by Proposition (ref)(b): \begin{align} \sqrt T(\widehat{\bm\lambda}_i-\widehat{\mathbf H}^\prime{\bm\lambda}_i) &= \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1} \bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right) \left(\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}\right) + o_{\mathrm {P}}(1)\nonumber\\ &= \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\left(\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}\right) \left(\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}\right) + o_{\mathrm {P}}(1). \end{align} Now, from Proposition (ref) we have that, as $n,T\to\infty$, $\min(\sqrt{n},\sqrt T)\Vert \widehat{\mathbf H}-\bm J\Vert = o_{\mathrm P}(1)$, and, by using the definition of $\widehat{\mathbf H}$ in (ref), it follows that (ref) is equivalent to (note that $\Vert\bm\lambda_i\Vert=O(1)$ by Assumption (ref)(a)) \begin{align} \sqrt T(\widehat{\bm\lambda}_i-\bm J\bm\lambda_i)&=\sqrt T(\widehat{\bm\lambda}_i-\widehat{\mathbf H}^\prime\bm\lambda_i)+\sqrt T(\widehat{\mathbf H}^\prime-\bm J)\bm\lambda_i+o_{\mathrm P}(1)\nonumber\\ &= \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1} \left(\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}\right) \left(\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}\right)+ o_{\mathrm {P}}(1) \nonumber\\ &=\widehat{\mathbf H}^\prime \left(\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}\right)+o_{\mathrm P}(1)\nonumber\\ &= \bm J\left(\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}\right)+o_{\mathrm P}(1)\nonumber\\ &\to_d\mathcal N\left(\mathbf 0_r, \bm\Phi_i\right), \end{align} where we used Slutsky's theorem and Assumption (ref). Notice that, $\bm J$ plays no role in the covariance since it is diagonal and $\bm J^2=\mathbf I_r$. Because of Assumption (ref) the sign indeterminacy on the left-hand-side can be easily fixed so that $\bm J=\mathbf I_r$. By substituting $\mathbf I_r$ in place of $\widehat{\mathbf H}$ in (ref) and using Assumption (ref) it follows also that: \[ \left\Vert \widehat{\bm\lambda}_i-{\bm\lambda}_i \right\Vert = O_{\mathrm {P}}\left(\max\left(\frac 1 n, \frac 1{\sqrt{T}}\right)\right). \] This completes the proof of part (a). For part (b), from (ref) \begin{align} \left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H}}{\sqrt n}\right\Vert\le&\, \left(\left\Vert\frac{\bm F^\prime\bm \Xi}{\sqrt nT}\right\Vert\,\left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^2 +\left\Vert \frac{\bm\Xi^\prime\bm F}{\sqrt {n}T}\right\Vert\,\left\Vert \frac{\bm\Lambda^\prime{\bm\Lambda}}{n}\right\Vert +\left\Vert\frac{\bm\Xi^\prime\bm \Xi\bm\Lambda}{n^{3/2}T}\right\Vert\right)\left\Vert\bm{\mathcal H} \right\Vert \,\left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\nonumber\\ &+ \left(\left\Vert\frac{\bm F^\prime\bm \Xi}{\sqrt nT}\right\Vert\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert +\left\Vert\frac{\bm\Xi^\prime\bm F}{\sqrt nT}\right\Vert\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert +\left\Vert\frac{\bm\Xi^\prime\bm \Xi}{nT}\right\Vert\right)\left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert,\nonumber \end{align} and the proof of part (b) follows from Proposition (ref) and Lemma (ref)(i), (ref), (ref)(i), (ref)(ii), (ref)(iii), (ref)(iv), (ref). This proves part (b) and completes the proof. $\Box$ \subsection{Proof of Corollary (ref)} From (ref) and the second last line of (ref) in the proof of Theorem (ref) and by imposing the identification constraint of orthonormal factors in Assumption (ref)(b) and $\bm J=\mathbf I_r$ by Assumption (ref), we have \begin{align} \widehat{\bm\lambda}_i-\bm\lambda_i &= \frac 1{ T}\sum_{t=1}^T \mathbf F_t\xi_{it}+ O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}\right)\right), \end{align} where the rate of the last term comes from (ref) in the proof of Theorem (ref)(a). By definition of OLS, and again imposing Assumption (ref)(b), we have: \begin{align} {\bm\lambda}_i^{\text{\tiny OLS}}-\bm\lambda_i = \frac 1{ T}\sum_{t=1}^T \mathbf F_t\xi_{it}. \end{align} By comparing (ref) and (ref) we complete the proof of part (a). Part (b) follows also directly from Theorem (ref)(b). This completes the proof. $\Box$ \subsection{Proof of Theorem (ref)} In principle, we could try to replicate the proofs by baili16 under our identifying constraints and using only our assumptions. However, there is a much simpler and intuitive way to proceed. Consider the log-likelihood (ref). The parameters to be estimated are given by ${\bm\varphi}=(\mathrm{vec}({\bm\Lambda})^\prime, \mathrm{vech}({\bm\Gamma}^\xi)^\prime)^\prime$. Let also $\widehat{\bm\varphi}^{\text{\tiny \upshape QML}}=(\mathrm{vec}(\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}})^\prime, \mathrm{vech}(\widehat{\bm\Gamma}^{\xi,\text{\tiny \upshape QML}})^\prime)^\prime$ denote the maximizer of (ref) and $\underline{\bm\varphi}=(\mathrm{vec}(\underline{\bm\Lambda})^\prime, \mathrm{vech}(\underline{\bm\Gamma}^\xi)^\prime)^\prime$ denote a generic value of the parameters. Whenever we consider $\underline{\bm\varphi}$ it is intended that its elements satisfy Assumptions (ref) through (ref). Then, the elements of $\widehat{\bm\varphi}^{\text{\tiny \upshape QML}}$ are such that: \begin{equation} \widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}=\widehat{\bm\Gamma}^x, \end{equation} where $\widehat{\bm\Gamma}^x=T^{-1} {\bm X^\prime\bm X}$. To see that (ref) defines the global maximum of the log-likelihood we proceed in two steps. First, notice that the first order conditions derived from the log-likelihood (ref) are satisfied when (ref) holds: \begin{align} &\left.\frac{\partial \ell(\bm{\mathcal X};\underline{\bm\varphi})}{\partial\underline{\bm\Lambda}}\right\vert_{\underline{\bm\Lambda}=\widehat{\bm\Lambda}^{\text{\tiny QML}}}\nonumber\\ &= T \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Gamma}^x \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}} -T\left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}}\nonumber\\ &= T \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}} -T\left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}}=\mathbf 0_{r\times n},\nonumber\\ &\left.\frac{\partial \ell(\bm{\mathcal X};\underline{\bm\varphi})}{\partial\underline{\bm\Gamma}^\xi}\right\vert_{\underline{\bm\Gamma}^\xi=\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}}\nonumber\\ &= \frac T2 \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Gamma}^x \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} - \frac T2 \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1}\nonumber\\ &= \frac T2 \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} - \frac T2 \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1}=\mathbf 0_{n\times n}.\nonumber \end{align} Notice also that the conditions given in baili12, which are derived from the first order conditions above, are also satisfied. Namely, it holds that: \begin{align} &\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}\left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1}\left\{\widehat{\bm\Gamma}^x-\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}-\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right\}=\mathbf 0_{r\times n},\nonumber\\ &\left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1}= \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm \Gamma}^x \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1},\nonumber\\ &\widehat{\bm\Lambda}^{\text{\tiny QML}\prime} \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}}= \widehat{\bm\Lambda}^{\text{\tiny QML}\prime} \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm \Gamma}^x \left(\widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)^{-1} \widehat{\bm\Lambda}^{\text{\tiny QML}}.\nonumber \end{align} Second, given the log-likelihood (ref), for any $\underline{\bm\varphi}$, we have: \begin{align} \ell(\bm{\mathcal X};\widehat{\bm\varphi}^{\text{\tiny QML}})&-\ell(\bm{\mathcal X};\underline{\bm\varphi})\nonumber\\ &=-\frac{T}2\log\frac{\det\left( \widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)} {\det\left(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right)}-\frac{nT}{2} +\frac T2 \text{tr}\left\{ \widehat{\bm\Gamma}^x \left(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right)^{-1} \right\}\\ &=-\frac{T}2\log\frac{\det\left( \widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)} {\det\left(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right)}-\frac{nT}{2} +\frac T2 \text{tr}\left\{ \left( \widehat{\bm\Lambda}^{\text{\tiny QML}}\widehat{\bm\Lambda}^{\text{\tiny QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}} \right) \left(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right)^{-1} \right\},\nonumber \end{align} because of (ref). Now, denote as $\zeta_j$, $j=1,\ldots, n$, the $n$ roots of \[ \det\left\{\left(\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}}\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny \upshape QML}}\right)-\zeta\left( \underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right) \right\}=0, \] which are all real since $\left(\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}}\widehat{\bm\Lambda}^{\text{\tiny \upshape QML}\prime}+\widehat{\bm\Gamma}^{\xi,\text{\tiny \upshape QML}}\right)\left( \underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Gamma}^\xi \right)^{-1} $ is a symmetric matrix. Then, (ref) reads: \begin{equation} \ell(\bm{\mathcal X};\widehat{\bm\varphi}^{\text{\tiny QML}})-\ell(\bm{\mathcal X};\underline{\bm\varphi}) = \frac T2\sum_{j=1}^n \left\{-\log\zeta_j-1+\zeta_j\right\}\ge 0, \end{equation} since $x\le e^{x-1}$ and so $-\log x-1+x\ge 0$. Therefore, from (ref) we see that (ref) defines indeed the global maximum of the log-likleihood (ref). Now, consider the Singular Value Decomposition (SVD) of the true loadings: \begin{equation} \frac{{\bm\Lambda}}{\sqrt n}={\mathbf V}{\mathbf D}{\mathbf U}, \end{equation} where ${\mathbf V}$ is $n\times r$ and such that ${\mathbf V}^\prime{\mathbf V}=\mathbf I_r$ for all $n\in\mathbb N$, ${\mathbf D}$ is $r\times r$ diagonal with strictly positive entries, and ${\mathbf U}$ is $r\times r$ such that ${\mathbf U}{\mathbf U}^\prime={\mathbf U}^{\prime}{\mathbf U}=\mathbf I_r$. We know that under Assumption (ref) and (ref), $\bm\Lambda$ is globally identified. Let us show that ${\mathbf V}$, ${\mathbf D}$, and ${\mathbf U}$ in (ref) are also globally identified. First, notice that, given Assumption (ref)(a) which requires $n^{-1}{{\bm\Lambda}^\prime{\bm\Lambda}}$ to be diagonal, in order to estimate ${\bm\Lambda}$ we need to estimate $nr-{r(r-1)}/{2}$ parameters. Then, by looking at the right-hand-side of (ref) we see that to estimate ${\mathbf V}$ we need to estimate $nr-\frac{r(r+1)}{2}$ parameters and to estimate ${\mathbf D}$ we need to estimate $r$ parameters, thus ${\mathbf V}{\mathbf D}$ depends on $nr-{r(r+1)}/{2}+r=nr-{r(r-1)}/{2}$ parameters as ${\bm\Lambda}$. However, in principle the right-hand-side of (ref) depends also on ${\mathbf U}$ which in turn requires estimating ${r(r+1)}/2$ parameters more. But if we impose Assumption (ref)(a) also to the right-hand-side of (ref) we have that $n^{-1}{{\bm\Lambda}^\prime{\bm\Lambda}}={\mathbf U}^\prime{\mathbf D}^2{\mathbf U}$ has to be diagonal, and since ${\mathbf D}$ is diagonal, without loss of generality we can set ${\mathbf U}=\mathbf J$, a diagonal $r\times r$ matrix with entries $\pm 1$. Furthermore, from Proposition (ref)(a) we also see that we must have $n^{-1}{{\bm\Lambda}^\prime{\bm\Lambda}}=n^{-1}{\mathbf M^\chi}=\mathbf D^2$. Hence, \begin{equation} \mathbf D = \left(\frac{\mathbf M^\chi}n\right)^{1/2}, \end{equation} and by Lemma (ref)(iv) the entries of $\mathbf D$, denoted as $d_j$, $j=1,\ldots,r$, are such that \begin{equation} \sqrt{\underline C_j}\!\le \lim\inf_{n\to\infty} d_j \le\lim\sup_{n\to\infty} d_j \le\! \sqrt{\overline C_j}, \end{equation} where $\underline C_j$ and $\overline C_j$ are finite positive reals. Last, from (ref) and Proposition (ref)(b), we must have $n^{-1/2}{{\bm\Lambda}}={\mathbf V}{\mathbf D}\mathbf J = \mathbf V^\chi \left(n^{-1}{\mathbf M^\chi}\right)^{1/2}$, and by (ref) it follows that \begin{equation} {\mathbf V} = \mathbf V^\chi \mathbf J. \end{equation} This shows that under Assumption (ref)(a), the parameters in ${\mathbf D}$ are globally identified while the parameters in $\mathbf V$ are uniquely identified up to a right-multiplication by $\mathbf J$, which can be pinned down by means of Assumption (ref), thus achieving global identification of $\mathbf V$ as well. Given this discussion, hereafter, we can directly set $\mathbf U=\mathbf I_r$. It is clear that the problem of QML estimation of the loadings can be rewritten as a problem of QML estimation of their SVD in (ref), namely of $\mathbf V$ and $\mathbf D$. To this end, we introduce also the SVDs of the QML estimator of the loadings and of a generic value of the loadings: \begin{align} &\frac{\widehat{\bm\Lambda}^{\text{\tiny QML}}}{\sqrt n}=\widehat{\mathbf V}^{\text{\tiny QML}}\widehat{\mathbf D}^{\text{\tiny QML}}\widehat{\mathbf U}^{\text{\tiny QML}},\qquad\frac{\underline{\bm\Lambda}}{\sqrt n}=\underline{\mathbf V}\,\underline{\mathbf D}\,\underline{\mathbf U}, \end{align} where $\widehat{\mathbf V}^{\text{\tiny \upshape QML}}$ and $\underline{\mathbf V}$ have the same properties as ${\mathbf V}$, $\widehat{\mathbf D}^{\text{\tiny \upshape QML}}$ and $\underline{\mathbf D}$ have the same properties as ${\mathbf D}$, and $\widehat{\mathbf U}^{\text{\tiny \upshape QML}}$ and $\underline{\mathbf U}$ have the same properties as ${\mathbf U}$. Because we set $\mathbf U=\mathbf I_r$, it follows that we can set $\widehat{\mathbf U}^{\text{\tiny \upshape QML}} = \mathbf I_r$, and we are left with the task of finding $\widehat{\mathbf V}^{\text{\tiny \upshape QML}}$ and $\widehat{\mathbf D}^{\text{\tiny \upshape QML}}$. From (ref) and since we must have $\widehat{\mathbf V}^{\text{\tiny \upshape QML}\prime}\widehat{\mathbf V}^{\text{\tiny \upshape QML}}=\mathbf I_r$, it follows that \begin{equation} \frac{\widehat{\bm\Gamma}^x}{n}-\widehat{\mathbf V}^{\text{\tiny QML}}\left(\widehat{\mathbf D}^{\text{\tiny QML}}\right)^2 \widehat{\mathbf V}^{\text{\tiny QML}\prime}-\frac{\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}}{n}=\mathbf 0_{n\times n},\nonumber \end{equation} which is equivalent to \begin{equation} \frac{\widehat{\mathbf V}^{\text{\tiny QML}\prime}\widehat{\bm\Gamma}^x\widehat{\mathbf V}^{\text{\tiny QML}}}{n}-\left(\widehat{\mathbf D}^{\text{\tiny QML}}\right)^2 -\frac{\widehat{\mathbf V}^{\text{\tiny QML}\prime}\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\widehat{\mathbf V}^{\text{\tiny QML}}}{n}=\mathbf 0_{r\times r}. \end{equation} Now, let $\widehat{\bm v}_j^{\text{\tiny \upshape QML}}$, $j=1,\ldots, r$, be the $j$th column of $\widehat{\mathbf V}^{\text{\tiny \upshape QML}}$ and let $\widehat{d}_j^{\,\text{\tiny \upshape QML}}$, $j=1,\ldots, r$, be the $j$th diagonal entry of $\widehat{\mathbf D}^{\text{\tiny \upshape QML}}$. Then, from (ref), for all $j=1,\ldots, r$, we have \begin{equation} \frac{\widehat{\bm v}_j^{\text{\tiny QML}\prime}\widehat{\bm\Gamma}^x\widehat{\bm v}_j^{\text{\tiny QML}}}{n}-\left(\widehat{ d}_j^{\,\text{\tiny QML}}\right)^2 -\frac{\widehat{\bm v}_j^{\text{\tiny QML}\prime}\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\widehat{\bm v}_j^{\text{\tiny QML}}}{n}=0. \end{equation} Then, trivially, the QML estimators are such that: \begin{equation} \left(\widehat{\bm v}_j^{\text{\tiny QML}}, \widehat{d}_j^{\,\text{\tiny QML}},\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}\right)=\arg\!\!\!\!\min_{\underline{\bm v}_j,\underline{d}_j,\underline{\bm\Gamma}^\xi} \left(\frac{\underline{\bm v}_j^{\prime}\widehat{\bm\Gamma}^x\underline{\bm v}_j}{n}-\underline{ d}_j^2 - \frac{\underline{\bm v}_j^{\prime}\underline{\bm\Gamma}^{\xi}\underline{\bm v}_j}n \right)^2=\arg\!\!\!\!\min_{\underline{\bm v}_j,\underline{d}_j,\underline{\bm\Gamma}^\xi} \mathcal L_{0j}\left(\underline{\bm v}_j, \underline d_j,\underline{\bm\Gamma}^{\xi} \right), \;\text{say,} \end{equation} where $\underline{\bm v}_j$, $j=1,\ldots, r$, is the $j$th column of $\underline{\mathbf V}$ and $\underline{d}_j$, $j=1,\ldots, r$, is the $j$th diagonal entry of $\underline{\mathbf D}$. Define also the estimators $\widetilde{\bm v}_j$ and $\widetilde{d}_j$ such that: \begin{equation} \left(\widetilde{\bm v}_j, \widetilde{d}_j\right)=\arg\!\min_{\underline{\bm v}_j,\underline{d}_j} \left(\frac{\underline{\bm v}_j^{\prime}\widehat{\bm\Gamma}^x\underline{\bm v}_j}{n}-\underline{ d}_j^2 \right)^2=\arg\!\min_{\underline{\bm v}_j,\underline{d}_j}\mathcal L_{1j}\left(\underline{\bm v}_j, \underline d_j\right), \;\text{say.} \end{equation} Consistently with (ref) and (ref), these define an estimator of $\bm\Lambda$ by means of its SVD: \begin{equation} \frac{\widetilde{\bm\Lambda}}{\sqrt n} = \widetilde{\mathbf V}\widetilde{\mathbf D}\widetilde{\mathbf U}, \end{equation} where $\widetilde{\mathbf V}$ has columns $\widetilde{\bm v}_j$, $j=1,\ldots, r$, and it has the same properties as ${\mathbf V}$ (because $\underline{\mathbf V}$ does), and $\widetilde{\mathbf D}$ has entries $\widetilde{ d}_j$, $j=1,\ldots, r$, and it has the same properties as ${\mathbf D}$ (because $\underline{\mathbf D}$ does). We set $\widetilde{\mathbf U}=\mathbf I_r$ consistently with the fact that $\mathbf U=\mathbf I_r$. Then, it is easily seen that \begin{equation} \mathcal L_{0j}\left(\underline{\bm v}_j, \underline d_j,\underline{\bm\Gamma}^{\xi} \right)= \mathcal L_{1j}\left(\underline{\bm v}_j, \underline d_j\right)+\left(\frac{\underline{\bm v}_j^{\prime}\underline{\bm\Gamma}^{\xi}\underline{\bm v}_j}n\right)^2-2\left(\frac{\underline{\bm v}_j^{\prime}\underline{\bm\Gamma}^{\xi}\underline{\bm v}_j}n\right)\sqrt{\mathcal L_{1j}\left(\underline{\bm v}_j, \underline d_j\right)}. \end{equation} Now, we know that for any generic value of the idiosyncratic covariance matrix satisfying Assumption (ref)(b), it holds that \begin{equation} \frac{\underline{\bm v}_j^{\prime}\underline{\bm\Gamma}^{\xi}\underline{\bm v}_j}n\le \max_{\bm w\, :\, \bm w^\prime\bm w=1}\frac{\bm w'\underline{\bm\Gamma}^{\xi}\bm w}{n}=\frac{\underline {\mu}^\xi_1}{n}\le \frac{M_{2\xi}}n, \end{equation} since $\underline{\bm v}_j^{\prime}\underline{\bm v}_j=1$ and because of Lemma (ref)(v) which implies also that $M_{2\xi}$ is independent of $n$. Moreover, for any generic $\underline{\bm \Lambda}$ satisfying Assumption (ref)(a): \begin{equation} \sqrt{\mathcal L_{1j}\left(\underline{\bm v}_j, \underline d_j\right)} = \left\vert \frac{\underline{\bm v}_j^{\prime}\widehat{\bm\Gamma}^x\underline{\bm v}_j}{n}-\underline{ d}_j^2 \right\vert \le \frac{\underline{\bm v}_j^{\prime}\widehat{\bm\Gamma}^x\underline{\bm v}_j}{n}+ \underline{ d}_j^2 \le \frac{\widehat{\mu}_1^x}{n} + \underline{ d}_j^2 = O_{\mathrm P}(1), \end{equation} since $\underline d_j= O(1)$ because $\Vert n^{-1/2}{\underline{\bm \Lambda}}\Vert= O(1)$ by Lemma (ref)(i), and $\widehat{\mu}_1^x = O_{\mathrm P}(n)$ because of Lemma (ref)(iii). Therefore, from (ref), (ref), and (ref), we have \begin{equation} \left\vert\mathcal L_{0j}\left(\underline{\bm v}_j, \underline d_j,\underline{\bm\Gamma}^{\xi} \right)-\mathcal L_{1j}\left(\underline{\bm v}_j, \underline d_j \right)\right\vert = O_{\mathrm P}\left(\frac 1n\right), \end{equation} which holds for any $\underline{\bm\Gamma}^{\xi}$ satisfying Assumption (ref)(b). By continuity of these loss functions, from (ref) it follows that their minima satisfy: \begin{align} \left\Vert \widehat{\bm v}_j^{\text{\tiny QML}}-\widetilde{\bm v}_j \right\Vert = O_{\mathrm P}\left(\frac 1n\right),\qquad \left\vert \widehat{d}_j^{\,\text{\tiny QML}}-\widetilde{d}_j \right\vert = O_{\mathrm P}\left(\frac 1n\right), \end{align} for all $j=1,\ldots, r$. Let us now find $\widetilde{\bm v}_j $ and $\widetilde{d}_j $. From (ref), it is clear that the solutions must be such that: \begin{equation} \frac{\widetilde{\bm v}_j^{\prime}\widehat{\bm\Gamma}^x\widetilde{\bm v}_j}{n}=\widetilde{ d}_j^{\,2}, \end{equation} which means that $\widetilde{ d}_j^{\,2}$ must be an eigenvalue of $n^{-1}{\widehat{\bm\Gamma}^x}$ and $\widetilde{\bm v}_j$ is the corresponding normalized eigenvector. Obviously, the solution in (ref) defines a global minimum of the loss $\mathcal L_1(\underline{\bm v}_j,\underline d_j)$ since $\mathcal L_1(\widetilde{\bm v}_j,\widetilde d_j)=0$ while $\mathcal L_1(\underline{\bm v}_j,\underline d_j)> 0$ for any other value $\underline{\bm v}_j\ne \widetilde{\bm v}_j$ and $\underline d_j\ne \widetilde d_j$. Now, let us show that indeed it must be that $\widetilde{ d}_j^{\,2}=n^{-1}{\widehat{\mu}_j^x}$, $j=1,\ldots, r$, i.e., they have to be the $r$ largest eigenvalues of $n^{-1}{\widehat{\bm\Gamma}^x}$. First, by Lemma (ref)(i) and Weyl's inequality, for all $k=1,\ldots, r$, as $n,T\to\infty$, \begin{equation} \left\vert\frac{\widehat{\mu}_k^x}n-\frac{{\mu}_k^x}n\right\vert\le \left\Vert\frac{\widehat{\bm\Gamma}^x}n-\frac{{\bm\Gamma}^x}n\right\Vert= O_{\mathrm P}\left(\frac 1{\sqrt T}\right). \end{equation} Therefore, if for any given $j=1,\ldots,r $ we were to choose $\widetilde{ d}_j^{\,2}=n^{-1}{\widehat{\mu}_k^x}$ for, say, $k=r+1$, then, from (ref), we would have \begin{align} &\lim\inf_{n\to\infty} \widetilde{ d}_j^{\,2} =\lim\inf_{n\to\infty} \frac{\widehat{\mu}_{r+1}^x}n =\lim\inf_{n\to\infty} \frac{{\mu}_{r+1}^x}n+O_{\mathrm P}\left(\frac 1{\sqrt T}\right) \ge \lim\inf_{n\to\infty} \frac{{\mu}_{n}^{\xi}}n+O_{\mathrm P}\left(\frac 1{\sqrt T}\right)= O_{\mathrm P}\left(\frac 1{\sqrt T}\right),\nonumber\\ &\lim\sup_{n\to\infty} \widetilde{ d}_j^{\,2} =\lim\sup_{n\to\infty} \frac{\widehat{\mu}_{r+1}^x}n =\lim\sup_{n\to\infty} \frac{{\mu}_{r+1}^x}n+O_{\mathrm P}\left(\frac 1{\sqrt T}\right) \le \lim_{n\to\infty} \frac{M_\xi}n+O_{\mathrm P}\left(\frac 1{\sqrt T}\right)= O_{\mathrm P}\left(\frac 1{\sqrt T}\right),\nonumber \end{align} since we assumed $\bm\Gamma^{\xi}$ to be positive definite, so $\mu^\xi_n>0$ for all $n\in\mathbb N$, and by Lemma (ref)(vi). Therefore, as $n,T\to\infty$, this choice for $\widetilde{ d}_j$ cannot be a consistent estimator of $d_j$ (nor an approximation of $\widehat{d}_j^{\,\text{\tiny QML}}$), since while $\widetilde{ d}_j\to0$, as $n,T\to\infty$, it must be that $d_j>0$ for all $n\in\mathbb N$, as required in (ref). So from (ref) and the above reasoning it follows that, for all $j=1,\ldots, r$, \begin{equation} \widetilde{ d}_j^{\,2}= \frac{\widehat{\mu}_j^x}{n},\qquad \widetilde{\bm v}_j = \widehat{\bm v}_j^x. \end{equation} where $ \widehat{\bm v}_j^x$ is the eigenvector of $n^{-1}{\widehat{\bm\Gamma}^x}$ corresponding to its $j$th largest eigenvalue. Then, from (ref), first we have \begin{align} &\left\Vert \widehat{\mathbf V}^{\text{\tiny QML}}-\widehat{\mathbf V}^x \right\Vert=\left\Vert \widehat{\mathbf V}^{\text{\tiny QML}}-\widetilde{\mathbf V} \right\Vert = O_{\mathrm P}\left(\frac 1n\right),\\ &\left\Vert \widehat{\mathbf D}^{\,\text{\tiny QML}}-\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2} \right\Vert= \left\Vert \widehat{\mathbf D}^{\,\text{\tiny QML}}-\widetilde{\mathbf D} \right\Vert = O_{\mathrm P}\left(\frac 1n\right), \end{align} because of (ref), and, second, we have \begin{equation} \frac{\widetilde{\bm\Lambda}}{\sqrt n} = \widehat{\mathbf V}^x\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2}=\frac{\widehat{\bm\Lambda}}{\sqrt n} , \end{equation} which shows that the estimator $\widetilde{\bm\Lambda}$ is the PC estimator defined in (ref). Therefore, from (ref), (ref), and (ref), and by using the SVD of the QML estimator in (ref), it follows that: \begin{align} \left\Vert \frac{\widehat{\bm\Lambda}^{\text{\tiny QML}}}{\sqrt n}-\frac{\widehat{\bm\Lambda}}{\sqrt n} \right\Vert =&\, \left\Vert \widehat{\mathbf V}^{\text{\tiny QML}}\widehat{\mathbf D}^{\text{\tiny QML}} - \widehat{\mathbf V}^x\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2} \right\Vert\nonumber\\ \le&\, \left\Vert \widehat{\mathbf V}^{\text{\tiny QML}}- \widehat{\mathbf V}^x \right\Vert\, \left\Vert \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2} \right\Vert + \left\Vert \widehat{\mathbf D}^{\text{\tiny QML}}-\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2} \right\Vert\, \left\Vert \widehat{\mathbf V}^x \right\Vert\nonumber\\ &+\left\Vert \widehat{\mathbf V}^{\text{\tiny QML}}- \widehat{\mathbf V}^x \right\Vert\, \left\Vert \widehat{\mathbf D}^{\text{\tiny QML}}-\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{1/2} \right\Vert = O_{\mathrm P}\left(\frac 1n\right),\nonumber \end{align} because $ \Vert\widehat{\mathbf V}^x \Vert=1$ (it is a matrix of normalized eigenvectors) and $\Vert (n^{-1}{\widehat{\mathbf M}^x})^{1/2}\Vert=O_{\mathrm P}(1)$ by Lemma (ref)(iii). This proves part (a). Finally, consider the $r$ dimensional $i$th rows, $i=1,\ldots, n$, of ${\mathbf V}^\chi$, $\widehat{\mathbf V}^x$, $\underline {\mathbf V}$, and $\widehat{\mathbf V}^{\text{\tiny \upshape QML}}$, denoted as ${\mathbf v}_i^{\chi\prime}$, $\widehat{\mathbf v}_i^{x\prime}$, $\underline{\mathbf v}_i^\prime$ , and $\widehat{\mathbf v}_i^{\text{\tiny \upshape QML}\prime}$, respectively. From Lemma (ref)(ii) and (ref)(iii), we know that $\sqrt n\Vert {\mathbf v}_i^{\chi\prime}\Vert = O(1)$ and $\sqrt n\Vert\widehat{\mathbf v}_i^{x\prime}\Vert=O_{\mathrm P}(1)$, which means that we must have $\sqrt n\Vert \underline {\mathbf v}_i^\prime\Vert= O(1)$ since any generic parameter we consider is assumed to satisfy the same assumptions as the corresponding true parameters (${\mathbf v}_i^{\chi\prime}$ in this case). Therefore, since the search of the elements of $\widehat{\mathbf V}^{\text{\tiny \upshape QML}}$ is made over all $\underline {\mathbf V}$ satisfying the same assumptions as ${\mathbf V}^\chi$, it follows that we must have also $\sqrt n\Vert \widehat{\mathbf v}_i^{\text{\tiny \upshape QML}\prime}\Vert=O_{\mathrm P}(1)$. To see that this is the case, notice also that because of Assumption (ref)(a) the QML estimator must be such that $n^{-1}{\widehat{\bm \Lambda}^{\text{\tiny \upshape QML}\prime}\widehat{\bm \Lambda}^{\text{\tiny \upshape QML}}}$ is diagonal and positive definite. Thus, we must have $\Vert n^{-1/2}{\widehat{\bm \Lambda}^{\text{\tiny \upshape QML}}}\Vert = O_{\mathrm P}(1)$ (see also Lemma (ref)(i)). From the SVD in (ref) when $\widehat{\mathbf U}^{\text{\tiny \upshape QML}}=\mathbf I_r$ and since it must be that $\widehat{\mathbf V}^{\text{\tiny \upshape QML}\prime}\widehat{\mathbf V}^{\text{\tiny \upshape QML}}=\mathbf I_r$ for all $n\in\mathbb N$, then it follows that $\Vert \widehat{\mathbf V}^{\text{\tiny \upshape QML}}\Vert =O_{\mathrm P}(1)$ and $\Vert \widehat{\mathbf D}^{\text{\tiny \upshape QML}}\Vert =O_{\mathrm P}(\sqrt n)$. Moreover, since by Assumption (ref)(a) the $i$th row of $\widehat{\bm \Lambda}^{\text{\tiny \upshape QML}}$ must be such that $\Vert\widehat{\bm \lambda}_i^{\text{\tiny \upshape QML}}\Vert=O_{\mathrm P}(1)$ and $\widehat{\bm \lambda}_i^{\text{\tiny \upshape QML}\prime}= \widehat{\mathbf v}_i^{\text{\tiny \upshape QML}\prime} \widehat{\mathbf D}^{\text{\tiny \upshape QML}}$, then it must be that $\sqrt n\Vert \widehat{\mathbf v}_i^{\text{\tiny \upshape QML}\prime}\Vert=O_{\mathrm P}(1)$. From this reasoning and (ref) it follows that: \begin{equation} \sqrt n \left\Vert \widehat{\mathbf v}_i^{\text{\tiny QML}}-\widehat{\mathbf v}_i^x \right\Vert = O_{\mathrm P}\left(\frac 1n\right). \end{equation} Now, by taking the $i$th row of the PC estimator in (ref) and the SVD of the QML estimator in (ref), for any $i=1,\ldots, n$, we get \begin{align} \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML}\prime}-\widehat{\bm\lambda}_i^{\prime} \right\Vert =&\, \left\Vert \widehat{\mathbf v}_i^{\text{\tiny QML}\prime} \sqrt n \widehat{ \mathbf D}^{\text{\tiny QML}}- \widehat{\mathbf v}_i^{x\prime}(\widehat{\mathbf M}^x)^{1/2} \right\Vert\nonumber\\ \le&\, \left\Vert \widehat{\mathbf v}_i^{\text{\tiny QML}} - \widehat{\mathbf v}_i^x \right\Vert \, \left\Vert \left({\widehat{\mathbf M}^x}\right)^{1/2} \right\Vert + \left\Vert \sqrt n \widehat{ \mathbf D}^{\text{\tiny QML}}-\left({\widehat{\mathbf M}^x}\right)^{1/2} \right\Vert\, \left\Vert \widehat{\mathbf v}_i^x \right\Vert\nonumber\\ &+\left\Vert \widehat{\mathbf v}_i^{\text{\tiny QML}} - \widehat{\mathbf v}_i^x \right\Vert \, \left\Vert \sqrt n \widehat{ \mathbf D}^{\text{\tiny QML}}-\left({\widehat{\mathbf M}^x}\right)^{1/2} \right\Vert= O_{\mathrm P}\left(\frac 1n\right),\nonumber \end{align} because of (ref), (ref), and since $\Vert \widehat{\mathbf v}_i^x \Vert=O_{\mathrm P}(n^{-1/2})$ by Lemma (ref)(ii) and (ref)(iii), and $\Vert ({\widehat{\mathbf M}^x})^{1/2}\Vert=O_{\mathrm P}(\sqrt n)$ by Lemma (ref)(iii). This proves part (b) and completes the proof. $\Box$\\ \subsection{Proof of Theorem (ref)} From Theorem (ref)(b) and Corollary (ref) it follows that: \begin{align} \Vert\widehat{\bm\lambda}_i^{\text{\tiny QML}}-{\bm\lambda}_i\Vert&=\Vert\widehat{\bm\lambda}_i-{\bm\lambda}_i\Vert+O_{\mathrm {P}}\left(\frac 1{n}\right)\nonumber\\ &=\Vert\widehat{\bm\lambda}_i-{\bm\lambda}_i^{\text{\tiny OLS}}\Vert+\Vert{\bm\lambda}_i^{\text{\tiny OLS}}-{\bm\lambda}_i\Vert+O_{\mathrm {P}}\left(\frac 1{n}\right)\nonumber\\ &=\Vert{\bm\lambda}_i^{\text{\tiny OLS}}-{\bm\lambda}_i\Vert+O_{\mathrm {P}}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}\right)\right)\nonumber\\ &=O_{\mathrm {P}}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1{\sqrt {T}}\right)\right), \end{align} since the unfeasible OLS estimator is $\sqrt T$-consistent. Moreover, if $\sqrt T/n\to 0$ as $n,T\to\infty$, by imposing the identification constraint of orthonormal factors in Assumption (ref)(b), we have \begin{align} \sqrt T(\widehat{\bm\lambda}_i^{\text{\tiny QML}}-\bm\lambda_i)&= \sqrt T(\widehat{\bm\lambda}_i^{\text{\tiny OLS}}-\bm\lambda_i)+ o_{\mathrm P}(1) =\frac 1{\sqrt T}\sum_{t=1}^T \mathbf F_t\xi_{it}+ o_{\mathrm P}(1)\to_d\mathcal N\left(\mathbf 0_r, \bm\Phi_i\right).\nonumber \end{align} by Slutsky's theorem and Assumption (ref). This completes the proof of part (a). Part (b) follows similarly from Theorem (ref)(a). This completes the proof. $\Box$ \subsection{Proof of Corollary (ref)} The proof of part (a) follows by using the log-likelihood (ref) of an exact factor model in place of the log-likelihood (ref), then, by noticing that $\bm{\Sigma}^{\xi}$ is positive definite by Assumption (ref)(a), and fnally by replacing in the proof of Theorem (ref), $\widehat{\bm\Gamma}^{\xi,\text{\tiny QML}}$ and $\underline{\bm\Gamma}^{\xi}$ with $\widehat{\bm\Sigma}^{\xi,\text{\tiny QML}}$ and $\underline{\bm\Sigma}^{\xi}$ respectively. The proof of part (b) is the same but when using $\sigma^2\mathbf I_n$, with $\sigma^2>0$, in place of $\bm\Sigma^\xi$. $\Box$ \subsection{Proof of Theorem (ref)} First of all, denote the log-likelihoods for one observation: \begin{align} &\ell_{t}(\mathbf x_t;\underline{\bm\varphi})=-\frac 12\log \det(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Sigma}^\xi)-\frac 12 \mathbf x_t^\prime (\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime+\underline{\bm\Sigma}^\xi)^{-1}\mathbf x_t,\\ &\ell_{t}(\mathbf x_t|\mathbf F_t;\underline{\bm\varphi})=-\frac 12\log \det(\underline{\bm\Sigma}^\xi)-\frac 12 (\mathbf x_t-\underline{\bm\Lambda}\mathbf F_t)^\prime (\underline{\bm\Sigma}^\xi)^{-1}(\mathbf x_t-\underline{\bm\Lambda}\mathbf F_t), \end{align} Let us consider part (a). For any fixed value of the parameters, say $\widetilde{\bm\varphi}$, let \begin{align} \bm S(\bm{\mathcal X};\widetilde{\bm\varphi}) &= \sum_{t=1}^T\left.\frac{\partial \ell_t(\mathbf x_t;\underline{\bm\varphi})}{\partial \underline{\bm\Lambda}^\prime}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left(\begin{array}{c} \bm s_1^\prime(\bm{\mathcal X};\widetilde{\bm\varphi})\\ \vdots\\ \bm s_n^\prime(\bm{\mathcal X};\widetilde{\bm\varphi})\\ \end{array} \right),\nonumber\\ \bm S(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi}) &= \sum_{t=1}^T\left.\frac{\partial \ell_t(\mathbf x_t|\mathbf F_t;\underline{\bm\varphi})}{\partial \underline{\bm\Lambda}^\prime}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left(\begin{array}{c} \bm s_1^\prime(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})\\ \vdots\\ \bm s_n^\prime(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})\\ \end{array} \right),\nonumber \end{align} which are $n\times r$ matrices of first derivatives, and where, for any given $i=1,\ldots, n$, \begin{equation} \bm s_i(\bm{\mathcal X};\widetilde{\bm\varphi})=\sum_{t=1}^T\left.\frac{\partial \ell_t(\mathbf x_t;\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}}, \quad \bm s_i(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})=\sum_{t=1}^T\left.\frac{\partial \ell_t(\mathbf x_t|\mathbf F_t;\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}}, \end{equation} which are $r$-dimensional column vectors. Then, recalling that $\widehat{\bm\Gamma}^x = \frac 1T\sum_{t=1}^T\mathbf x_t\mathbf x_t^\prime$, by computing the first derivatives of the log-likelihood (ref), we have \begin{align} \bm S(\bm{\mathcal X};\underline{\bm\varphi})=&\, -T (\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}+ (\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1} \widehat{\bm\Gamma}^x (\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\nonumber\\ =&\, -T (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1} + T (\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1} \widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1}\nonumber\\ =&\, -T (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1} + T (\underline{\bm\Sigma}^\xi)^{-1} \widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1}\nonumber\\ &-T(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda} \{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1} \underline{\bm\Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1} \widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1}, \end{align} where we used the Woodbury identities \begin{align} &(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda} = (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1},\\ &(\underline{\bm\Lambda}\,\underline{\bm\Lambda}^\prime + \underline{\bm\Sigma}^\xi)^{-1}= (\underline{\bm\Sigma}^\xi)^{-1}-(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda} \{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\}^{-1} \underline{\bm\Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}. \end{align} In what follows, we make use of the following results. Let $\widehat{\bm\Gamma}^\xi=\frac 1T\sum_{t=1}^T \bm\xi_t\bm\xi_t^\prime$ and let $\widehat{\bm\Sigma}^\xi=\text{diag}(\widehat{\bm\Gamma}^\xi)$, the diagonal matrix having as diagonal entries the diagonal entries of $\widehat{\bm\Gamma}^\xi$. Then, by using twice Lemma (ref)(ii) and by Lemma (ref)(v), we have \begin{align} \frac 1n\left\Vert \bm\Sigma^\xi -\widehat{\bm\Gamma}^\xi\right\Vert &\le \frac 1n\left\Vert \bm\Sigma^\xi - \widehat{\bm\Sigma}^\xi\right\Vert +\frac 1n\left\Vert\widehat{\bm\Gamma}^\xi-\widehat{\bm\Sigma}^\xi \right\Vert \le O_{\mathrm P}\left(\frac 1{\sqrt T}\right) + \frac 1n\left\Vert\widehat{\bm\Gamma}^\xi \right\Vert\nonumber\\ &\le O_{\mathrm P}\left(\frac 1{\sqrt T}\right) + \frac 1n \left\Vert\bm\Gamma^\xi\right\Vert + O_{\mathrm P}\left(\frac 1{\sqrt T}\right) = O_{\mathrm P}\left(\frac 1{\sqrt T}\right) + O\left(\frac 1n\right). \end{align} which implies: \begin{equation} \left\Vert (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi - \mathbf I_n\right\Vert= O_{\mathrm P}\left(\max\left(\frac 1{n},\frac 1{\sqrt T}\right)\right). \end{equation} Moreover, \begin{equation} \left\Vert\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}-\mathbf I_r\right\Vert= O\left(\frac 1n\right), \end{equation} because of Lemma (ref), \begin{equation} \left\Vert \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\right\Vert \le \frac 1{n\underline C_r \Vert({\bm\Sigma}^\xi)^{-1}\Vert}=\frac 1{n\underline C_r \min_{i=1,\ldots,n} \sigma_i^2}=\frac 1{n\underline C_r C_\xi}= O\left(\frac 1n\right), \end{equation} because of Lemma (ref)(iv)-(ref)(vi), Assumption (ref)(a), and MK04, \begin{align} \left\Vert\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}-\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\right\Vert= O\left(\frac 1{n^2}\right),\end{align} because of (ref) and (ref), \begin{align} &\left\Vert \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime\right\Vert = O\left(\frac 1{\sqrt n}\right),\\ &\left\Vert\bm\Lambda\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}-\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\right\Vert= O\left(\frac 1{n^{3/2}}\right), \end{align} because of (ref), (ref) and since $\Vert {\bm\Lambda}\Vert = O(\sqrt n)$ by Lemma (ref)(i), and, last, \begin{align} &\left\Vert {\bm\Lambda}{\bm\Lambda}^\prime (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}-\bm\Lambda\right\Vert= O\left(\frac 1{\sqrt n}\right), \end{align} because of (ref) and since $\Vert {\bm\Lambda}\Vert = O(\sqrt n)$ by Lemma (ref)(i). Now, denote $\widehat{\bm\Gamma}^{\xi F}=\frac 1T\sum_{t=1}^T \bm\xi_t\mathbf F_t^\prime$ and $\widehat{\bm\Gamma}^{F\xi}=\widehat{\bm\Gamma}^{\xi F\prime}$, so that we can write \begin{equation} \widehat{\bm\Gamma}^x = \bm\Lambda\left(\frac 1T\sum_{t=1}^T\mathbf F_t\mathbf F_t^\prime\right)\bm\Lambda^\prime + \widehat{\bm\Gamma}^\xi+ \bm\Lambda\widehat{\bm\Gamma}^{F\xi}+\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime = \bm\Lambda\bm\Lambda^\prime + \widehat{\bm\Gamma}^\xi+ \bm\Lambda\widehat{\bm\Gamma}^{F\xi}+\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime, \end{equation} because of Assumption (ref)(b). Let us consider (ref) when computed in the true value of the parameters. By means of (ref)-(ref) and (ref) we have \begin{align} \bm S(\bm{\mathcal X};{\bm\varphi})=&\, T(\bm\Sigma^\xi)^{-1} \left\{ \bm\Lambda\bm\Lambda^\prime + \widehat{\bm\Gamma}^\xi + \bm\Lambda\widehat{\bm\Gamma}^{F\xi} +\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime \right\} (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &- T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \left\{ \bm\Lambda\bm\Lambda^\prime + \widehat{\bm\Gamma}^\xi+ \bm\Lambda\widehat{\bm\Gamma}^{F\xi}+\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime \right\} (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O\left(\frac T{\sqrt n}\right)\nonumber\\ =&\, T(\bm\Sigma^\xi)^{-1} \bm\Lambda\bm\Lambda^\prime (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+T(\bm\Sigma^\xi)^{-1} \widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &+T(\bm\Sigma^\xi)^{-1} \bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+T(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O\left(\frac T{\sqrt n}\right)\nonumber\\ =&\,T(\bm\Sigma^\xi)^{-1} \bm\Lambda+T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+T(\bm\Sigma^\xi)^{-1} \bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &+T(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}-T(\bm\Sigma^\xi)^{-1}\bm\Lambda-T(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}-T(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}-T(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &+ O\left(\frac T{\sqrt n}\right)+ O_{\mathrm P}\left(\frac {\sqrt T}{\sqrt n}\right)+ O\left(\frac T{n^{3/2}}\right)\nonumber\\ =&\,T(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}-T(\bm\Sigma^\xi)^{-1} \bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\nonumber\\ &-T(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O\left(\frac T{\sqrt n}\right)+ O_{\mathrm P}\left(\frac {\sqrt T}{\sqrt n}\right)+ O\left(\frac T{n^{3/2}}\right). \end{align} Then, notice that $[(\bm\Sigma^\xi)^{-1}]_{i\cdot}\bm\xi_t= [(\bm\Sigma^\xi)^{-1}]_{ii} \xi_{it}= \frac{ \xi_{it}}{\sigma_i^{2}}$, and $[(\bm\Sigma^\xi)^{-1}]_{i\cdot}\bm\Lambda = \frac{\bm\lambda_i^\prime}{\sigma_i^{2} }$, $i=1,\ldots, n$. The following holds \begin{align} & \frac 1{\sigma_i^2}{\bm\lambda}_i^\prime \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}= O\left(\frac 1n\right),\\ &T\left[(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1} \right]_{i\cdot} = \frac T{\sigma_i^2}{\bm\lambda}_i^\prime \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O_{\mathrm P}\left(\frac{\sqrt T}{n} \right)+ O\left(\frac{ T}{n^2} \right),\\ &T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1} \right]_{i\cdot}=\frac T{\sigma_i^2}{\bm\lambda}_i^\prime \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}\nonumber\\ &+ O_{\mathrm P}\left(\frac{\sqrt T}{n} \right)+ O\left(\frac{ T}{n^2} \right),\\ &T\bm\lambda_i^\prime\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}=T\bm\lambda_i^\prime\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O\left(\frac T{n^{2}}\right),\\ &T{\bm\lambda}_i^\prime{\bm\Lambda}^\prime (\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}=T\bm\lambda_i^\prime + O\left(\frac T{n}\right), \end{align} where we used Assumption (ref)(a) and then (ref) follows from (ref), (ref) and (ref) follow from (ref) and (ref), (ref) follows from (ref), and (ref) follows from and (ref). Therefore, by using (ref)-(ref) in (ref) we have that the $i$th row of $\bm S(\bm{\mathcal X};\bm{\varphi})$ is such that \begin{align} \bm s_i^\prime(\bm{\mathcal X};\bm{\varphi})=&\,\frac 1{\sigma_i^2}\sum_{t=1}^T {\xi}_{it}\mathbf F_t^\prime-\frac 1{\sigma_i^2}\bm\lambda_i^\prime \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}{\bm\Lambda}^\prime(\bm\Sigma^\xi)^{-1}\sum_{t=1}^T {\bm\xi}_{t}\mathbf F_t^\prime\nonumber\\ &-\frac T{\sigma_i^2}{\bm\lambda}_i^\prime\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1}+ O\left(\frac T{ n}\right)+ O_{\mathrm P}\left(\frac {\sqrt T}{n}\right)+ O\left(\frac T{ n^2}\right). \end{align} Hence, $\Vert\bm s_i^\prime(\bm{\mathcal X};\bm{\varphi})\Vert=O_{\mathrm P}(\sqrt T)$, because of Assumption (ref)(a), (ref) and since \begin{align} \mathbb{E}\left[\left\Vert \sum_{t=1}^T \frac{\mathbf F_t\bm\xi_t^\prime \bm\Lambda}{ nT}\right\Vert^2\right] = O\left(\frac 1{nT}\right) \;\text{ and }\; \mathbb{E}\left[\left\Vert \sum_{t=1}^T \frac{\mathbf F_t\bm\xi_t^\prime }{\sqrt nT}\right\Vert^2\right] = O\left(\frac 1{\sqrt T}\right) , \end{align} because of (ref) in the proof of Proposition (ref)(a) and by Lemma (ref). Moreover, from (ref) and by noticing that $(\bm\Sigma^\xi)^{-1}{\bm\Lambda}$ has the same properties as ${\bm\Lambda}$ since $\Vert (\bm\Sigma^\xi)^{-1}\Vert = O(1)$ by Assumption (ref)(a), we get \begin{align} \frac 1{\sqrt T}\left\Vert \bm s_i^\prime(\bm{\mathcal X};\bm{\varphi})-\frac 1{\sigma_i^2}\sum_{t=1}^T {\xi}_{it}\mathbf F_t^\prime \right\Vert\le &\, \frac 1{C_\xi}\left\Vert {\bm\lambda}_i^\prime \left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1} \right\Vert\, \left\Vert\frac 1{\sqrt T} {\bm\Lambda}^\prime(\bm\Sigma^\xi)^{-1}\sum_{t=1}^T {\bm\xi}_{t}\mathbf F_t^\prime \right\Vert\nonumber\\ &+ \frac{\sqrt T}{C_\xi} \left\Vert {\bm\lambda}_i^\prime\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}^{-1} \right\Vert+ O\left(\frac {\sqrt T}{n}\right)+ O_{\mathrm P}\left(\frac {1}{n}\right)+ O\left(\frac {\sqrt T}{n^2}\right)\nonumber\\ =&\, O\left(\frac 1n\right) O_{\mathrm P}\left(\sqrt {n}\right)+ O\left(\frac {\sqrt T}{n}\right)+O_{\mathrm P}\left(\frac 1{n}\right)+ O\left(\frac {\sqrt T}{n^2}\right)\nonumber\\ =&\, O_{\mathrm P}\left(\max\left( \frac 1{\sqrt n},\frac{\sqrt T} {n}\right)\right), \end{align} . Finally, by computing the first derivatives of the log-likelihood (ref), we have \begin{align} \bm S(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})&= (\underline{\bm\Sigma}^\xi)^{-1}\sum_{t=1}^T (\mathbf x_t-\underline{\bm\Lambda}\mathbf F_t)\mathbf F_t^\prime. \end{align} And (ref) when computed in the true value of the parameters is \begin{align} \bm S(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})&= ({\bm\Sigma}^\xi)^{-1}\sum_{t=1}^T \bm\xi_t\mathbf F_t^\prime. \end{align} From (ref), the $i$th row of $\bm S(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})$ is then: \begin{equation} \bm s_i^\prime(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})=\frac 1{\sigma_i^2}\sum_{t=1}^T {\xi}_{it}\mathbf F_t^\prime. \end{equation} Hence, $\Vert \bm s_i^\prime(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})\Vert = O_{\mathrm P}(\sqrt T)$ because of (ref), and, by using (ref) in (ref), for any given $i=1,\ldots, n$ we have: \[ \frac 1{\sqrt T}\left\Vert \bm s_i^\prime(\bm{\mathcal X};\bm{\varphi})-\bm s_i^\prime(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi}) \right\Vert = O_{\mathrm P}\left(\max\left( \frac 1{\sqrt n},\frac{\sqrt T} {n}\right)\right), \] which proves part (a). Turning to part (b). For any specific value of the parameters, say $\widetilde{\bm\varphi}$, let \begin{align} \bm H(\bm{\mathcal X};\widetilde{\bm\varphi}) &= \sum_{t=1}^T\left.\frac{\partial^2 \ell_t(\mathbf x_t;\underline{\bm\varphi})}{\partial \text{vec}(\underline{\bm\Lambda})^\prime\partial \text{vec}(\underline{\bm\Lambda})}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left. \frac{\partial \text{vec}(\bm S(\bm{\mathcal X};\underline{\bm\varphi}))^\prime}{\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left(\begin{array}{cccc} \bm h_{11}(\bm{\mathcal X};\widetilde{\bm\varphi})&\ldots& \bm h_{1n}(\bm{\mathcal X};\widetilde{\bm\varphi})\\ \vdots&\ddots&\vdots\\ \bm h_{n1}(\bm{\mathcal X};\widetilde{\bm\varphi})&\ldots& \bm h_{nn}(\bm{\mathcal X};\widetilde{\bm\varphi})\\ \end{array} \right),\nonumber\\ \bm H(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi}) &= \sum_{t=1}^T\left.\frac{\partial^2 \ell_t(\mathbf x_t|\mathbf F_t;\underline{\bm\varphi})}{\partial \text{vec}(\underline{\bm\Lambda})^\prime\partial \text{vec}(\underline{\bm\Lambda})}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left. \frac{\partial \text{vec}(\bm S(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}))^\prime}{\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left(\begin{array}{cccc} \bm h_{11}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})&\ldots& \bm h_{1n}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})\\ \vdots&\ddots&\vdots\\ \bm h_{n1}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})&\ldots& \bm h_{nn}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})\\ \end{array} \right),\nonumber \end{align} which are $nr\times nr$ matrices obtained from the matricization of the 4th order tensor of second derivatives, and where, for any given $ i=1,\ldots, n$, \begin{align} \bm h_{ii}(\bm{\mathcal X};\widetilde{\bm\varphi})&= \sum_{t=1}^T\left.\frac{\partial^2 \ell_t(\mathbf x_t;\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial\underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left. \frac{\partial \bm s_i^\prime(\bm{\mathcal X};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime} \right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}},\\ \bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi})&= \sum_{t=1}^T\left.\frac{\partial^2 \ell_t(\mathbf x_t|\mathbf F_t;\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime\partial \underline{\bm\lambda}_i}\right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}} = \left. \frac{\partial \bm s_i^\prime(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})}{\partial \underline{\bm\lambda}_i^\prime} \right\vert_{\underline{\bm\varphi}={\widetilde{\bm\varphi}}},\nonumber \end{align} which are $r\times r$ matrices. We now use the following relation: \begin{align} \text{vec}\left(\mathrm d \bm S^\prime(\bm{\mathcal X};\underline{\bm\varphi})\right) &=\left( \frac{\partial \text{vec}(\bm S(\bm{\mathcal X};\underline{\bm\varphi}))^\prime}{\partial \text{vec}(\underline{\bm\Lambda})^\prime}\right)^\prime \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)= \bm H(\bm{\mathcal X};\underline{\bm\varphi}) \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime), \end{align} Then, by denoting $\underline{\bm P}=\left\{\mathbf I_r+\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\right\}$, from (ref) we have \begin{align} \text{vec}\left(\mathrm d \bm S^\prime(\bm{\mathcal X};\underline{\bm\varphi})\right)=&\, -T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\otimes \mathbf I_r \right\} \text{vec}\left(\mathrm d \underline{\bm P}^{-1}\right)\nonumber\\ &-T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\otimes \underline{\bm P}^{-1} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)\nonumber\\ &+T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\otimes \mathbf I_r \right\} \text{vec}\left(\mathrm d \underline{\bm P}^{-1}\right)\nonumber\\ &+T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\otimes \underline{\bm P}^{-1} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)\nonumber\\ &-T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda} \otimes \mathbf I_r \right\} \text{vec}\left(\mathrm d \underline{\bm P}^{-1}\right)\nonumber\\ &-T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)\nonumber\\ &-T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \right\} \bm C_{n,r} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)\nonumber\\ &-T \left\{ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda} \right\} \text{vec}\left(\mathrm d \underline{\bm P}^{-1}\right)\nonumber\\ &-T \left\{(\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\underline{\bm P}^{-1} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime). \end{align} Moreover, \begin{align} \text{vec}\left(\mathrm d \underline{\bm P}^{-1}\right) &= -\left( \underline{\bm P}^{-1}\otimes \underline{\bm P}^{-1} \right) \left\{ \left[ \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\otimes \mathbf I_r \right] + \left[ \mathbf I_r \otimes \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)\nonumber\\ &= - \left\{ \left[ \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\otimes \underline{\bm P}^{-1} \right] +\left[ \underline{\bm P}^{-1}\otimes\underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r} \right\} \text{vec}(\mathrm d \underline{\bm \Lambda}^\prime), \end{align} where $\bm C_{n,r}$ is the $nr\times nr$ commutation matrix such that $\text{vec}(\mathrm d \underline{\bm \Lambda})=\bm C_{n,r}\text{vec}(\mathrm d \underline{\bm \Lambda}^\prime)$. Therefore, from (ref), (ref), and (ref) we get \begin{align} \bm H(\bm{\mathcal X};\underline{\bm\varphi}) =&\, T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it A.1}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it A.2}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it A.3}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \right]\quad \text{\it A.4}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right] \bm C_{n,r}\quad \text{\it A.5}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right] \bm C_{n,r}\quad \text{\it A.6}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right] \bm C_{n,r}\quad \text{\it A.7}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right] \bm C_{n,r}\quad \text{\it A.8}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it B.1}\nonumber\\ &+T\left[ (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it B.2}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\quad \text{\it B.3}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm \Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it B.4}\nonumber\\ &-T\left[ (\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \right].\quad \text{\it B.5} \end{align} Let us consider (ref) when computed in the true value of the parameters. By means of (ref) we have: { \begin{align} \bm H(\bm{\mathcal X};\bm\varphi)=&\, T\left[(\bm\Sigma^\xi)^{-1}\bm\Lambda \bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.2.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.2.2} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.2.3} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.2.4} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.3.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.3.2} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.3.3} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it A.3.4} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it A.4.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it A.4.2} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it A.4.3} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1} \bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it A.4.4} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.5} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.6.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.6.2} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.6.3} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.6.4} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.2} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.3} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime (\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.4} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.8.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.8.2} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F \xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.8.3} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it A.8.4} \nonumber\\ &-T\left[(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.2.1} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.2.2} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.2.3} \nonumber\\ &+T\left[ (\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.2.4} \nonumber\\ &[\ldots]\nonumber \end{align} \begin{align} &[\ldots]\nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.3.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.3.2} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.3.3} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \right]\quad \text{\it B.3.4} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it B.4.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it B.4.2} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it B.4.3} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1} \right]\bm C_{n,r}\quad \text{\it B.4.4} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it B.5.1} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it B.5.2} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right]\quad \text{\it B.5.3} \nonumber\\ &-T\left[ (\bm\Sigma^\xi)^{-1} \otimes \bm P^{-1} \bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\widehat{\bm\Gamma}^{\xi F}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\bm P^{-1} \right].\quad \text{\it B.5.4} \end{align} } By using (ref)-(ref) 30 terms of (ref) cancel out asymptotically, namely: \begin{enumerate} • $\Vert \text{\it A.1}-\text{\it A.2.2}\Vert = O_{\mathrm P}(\sqrt T n^{-1})+ O_{\mathrm P}(T n^{-2})$; • $\Vert \text{\it A.2.1}-\text{\it A.3.1}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.2.3}-\text{\it A.3.3}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.5}-\text{\it A.6.2}\Vert = O_{\mathrm P}(\sqrt T n^{-1})+ O_{\mathrm P}(T n^{-2})$; • $\Vert \text{\it A.6.1}-\text{\it A.7.1}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.6.3}-\text{\it A.7.3}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.2.4}-\text{\it B.2.4}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it B.2.1}-\text{\it B.3.1}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.3.2}-\text{\it B.3.2}\Vert = O_{\mathrm P}(\sqrt T n^{-1})+ O_{\mathrm P}(T n^{-2})$; • $\Vert \text{\it B.2.3}-\text{\it B.3.3}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.3.4}-\text{\it B.3.4}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.8.1}-\text{\it B.4.1}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it A.8.2}-\text{\it B.4.2}\Vert = O_{\mathrm P}(T n^{-1})+O_{\mathrm P}(\sqrt T n^{-1})+ O_{\mathrm P}(T n^{-2})$; • $\Vert \text{\it A.8.4}-\text{\it B.4.4}\Vert = O_{\mathrm P}(T n^{-1})$; • $\Vert \text{\it B.2.2}-\text{\it B.5.2}\Vert = O_{\mathrm P}(T n^{-1})+O_{\mathrm P}(\sqrt T n^{-1})+ O_{\mathrm P}(T n^{-2})$. \end{enumerate} and by using again (ref) we are left with the following 13 terms (ordered differently than in the previous expression): \begin{align} \bm H(\bm{\mathcal X};\bm\varphi)=&\, -T \left[ (\bm\Sigma^{\xi})^{-1} \otimes \mathbf I_r \right]\quad \text{\it B.5.1}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \otimes \mathbf I_r \right]\quad \text{\it A.4.1}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \right]\quad \text{\it A.4.2}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \otimes \widehat{\bm\Gamma}^{F\xi} (\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \right]\quad \text{\it A.4.3}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \widehat{\bm\Gamma}^{\xi F} \right]\quad \text{\it A.4.4}\nonumber\\ &-T\left[ (\bm\Sigma^{\xi})^{-1} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \right]\quad \text{\it B.1}\nonumber\\ &-T\left[ (\bm\Sigma^{\xi})^{-1} \otimes \widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \right]\quad \text{\it B.5.2}\nonumber\\ &-T\left[ (\bm\Sigma^{\xi})^{-1} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\widehat{\bm\Gamma}^{\xi F} \right]\quad \text{\it B.5.3}\nonumber\\ &-T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \otimes \widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^{\xi})^{-1} \right]\bm C_{n,r}\quad \text{\it B.4.3}\nonumber\\ &-T\left[ (\bm\Sigma^{\xi})^{-1}\widehat{\bm\Gamma}^{\xi F} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\bm\Lambda^\prime (\bm\Sigma^{\xi})^{-1} \right]\bm C_{n,r}\quad \text{\it A.6.4}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.1}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\widehat{\bm\Gamma}^{\xi F} \otimes \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \right]\bm C_{n,r}\quad \text{\it A.7.2}\nonumber\\ &+T\left[ (\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \otimes \widehat{\bm\Gamma}^{F\xi}(\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1} \right]\bm C_{n,r}\quad \text{\it A.8.3}\nonumber\\ &+O\left(\frac {T}n\right)+O_{\mathrm P}\left(\frac {\sqrt T}n\right)+O\left(\frac {T}{n^2}\right). \end{align} Therefore, by using again arguments as those in (ref)-(ref) in (ref) we have that the $i$th $r\times r$ sub-matrix of $\bm H(\bm{\mathcal X};\bm{\varphi})$ is such that \begin{align} \bm h_{ii}&(\bm{\mathcal X};\bm\varphi)= -\frac T{\sigma_i^2} \mathbf I_r\nonumber\\ & + \frac T{\sigma_i^4}{\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}{\bm\lambda}_i+\nonumber\\ & + \frac T{\sigma_i^4}{\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}{\bm\lambda}_i \left\{ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}+ \frac 1T\sum_{t=1}^T\mathbf F_t{\bm\xi}_t^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right.\nonumber\\ &\;\;+\left.\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\frac 1T\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \right\}\nonumber\\ &- \frac T{\sigma_i^2} \left\{ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}+ \frac 1T\sum_{t=1}^T\mathbf F_t{\bm\xi}_t^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}+ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\frac 1T\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \right\}\nonumber\\ &- \frac T{\sigma_i^4}\left\{ \frac 1T \sum_{t=1}^T\mathbf F_t{\xi}_{it} \otimes {\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \right\} - \frac T{\sigma_i^4}\left\{ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}{\bm\lambda}_i \otimes \frac 1T \sum_{t=1}^T{\xi}_{it}\mathbf F_t^\prime \right\}\nonumber\%&- \frac T{\sigma_i^4}\left\{ &+ \frac T{\sigma_i^4} \Bigg\{ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} {\bm\lambda}_i \otimes {\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \Bigg\}\nonumber\\ &+ \frac T{\sigma_i^4}\Bigg\{ \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}{\bm\lambda}_i \otimes {\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\frac 1T\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \Bigg\}\nonumber\\ &+ \frac T{\sigma_i^4}\Bigg\{ \frac 1T\sum_{t=1}^T\mathbf F_t{\bm\xi}_t^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}{\bm\lambda}_i \otimes {\bm\lambda}_i^\prime\left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1} \Bigg\}\nonumber\\ &+O\left(\frac{T}{n^2}\right)+O_{\mathrm P}\left(\frac{\sqrt T}{n^2}\right)+O\left(\frac{T}{n^4}\right). \end{align} Hence, $\Vert \bm h_{ii}(\bm{\mathcal X};\bm\varphi)\Vert=O_{\mathrm P}(T)$, because of Assumption (ref)(a), (ref), and (ref). Moreover, from (ref) and by noticing that $(\bm\Sigma^\xi)^{-1}{\bm\Lambda}$ has the same properties as ${\bm\Lambda}$ since $\Vert (\bm\Sigma^\xi)^{-1}\Vert = O(1)$ by Assumption (ref)(a), we get \begin{align} \frac 1T&\left\Vert\bm h_{ii}(\bm{\mathcal X};\bm\varphi)- \left(-\frac T{\sigma_i^2} \mathbf I_r\right)\right\Vert\le \frac{M_\Lambda^2}{C_\xi^2} \left\Vert \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right\Vert\nonumber\\ & + \frac{M_\Lambda^2}{C_\xi^2} \left\Vert \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right\Vert^2 \Bigg\{ 1 + 2\left\Vert\frac 1T\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \right\Vert \Bigg\}\nonumber\\ & \frac 1{C_\xi} \left\Vert \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right\Vert \Bigg\{ 1 + 2\left\Vert\frac 1T\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \right\Vert \Bigg\}\nonumber\\ &+ \frac{2M_\Lambda}{C_\xi^2} \left\Vert \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right\Vert\,\left\Vert\frac 1T\sum_{t=1}^T\mathbf F_t{\xi}_{it}\right\Vert+ \frac{M_\Lambda^2}{C_\xi^2} \left\Vert \left\{\bm\Lambda^\prime(\bm\Sigma^{\xi})^{-1}\bm\Lambda \right\}^{-1}\right\Vert^2\Bigg\{ 1 + 2\left\Vert\frac 1T\bm\Lambda^\prime(\bm\Sigma^\xi)^{-1}\sum_{t=1}^T{\bm\xi}_t\mathbf F_t^\prime \right\Vert \Bigg\}\nonumber\\ &+O\left(\frac{1}{n^2}\right)+O_{\mathrm P}\left(\frac{1}{n^2\sqrt T}\right)+O\left(\frac{1}{n^4}\right)\nonumber\\ =&\,O\left(\frac{1}{n}\right) +O\left(\frac{1}{n^2}\right) \Bigg\{1+O_{\mathrm P}\left(\frac{\sqrt n}{\sqrt T}\right)\Bigg\}+O\left(\frac{1}{n}\right) \Bigg\{1+O_{\mathrm P}\left(\frac{\sqrt n}{\sqrt T}\right)\Bigg\}+O\left(\frac{1}{n}\right)O_{\mathrm P}\left(\frac{1}{\sqrt T}\right)\nonumber\\ &+O\left(\frac{1}{n^2}\right) \Bigg\{1+O_{\mathrm P}\left(\frac{\sqrt n}{\sqrt T}\right)\Bigg\}+O\left(\frac{1}{n^2}\right)+O_{\mathrm P}\left(\frac{1}{n^2\sqrt T}\right)+O\left(\frac{1}{n^4}\right)= O_{\mathrm P}\left(\max\left(\frac{1}{n},\frac 1{\sqrt{nT}}\right)\right). \end{align} Finally, from (ref) we have \begin{align} \mathrm d \bm S^\prime(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})&=\mathrm d\left(\sum_{t=1}^T\mathbf F_t(\mathbf x_t-\underline{\bm\Lambda}\mathbf F_t)^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right) = -\sum_{t=1}^T\mathbf F_t\mathbf F_t^\prime\left(\mathrm d\underline{\bm\Lambda}^\prime\right)(\underline{\bm\Sigma}^\xi)^{-1},\nonumber \end{align} so \begin{equation} \text{vec}(\mathrm d \bm S^\prime(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi})) = - (\underline{\bm\Sigma}^\xi)^{-1}\otimes \left(\sum_{t=1}^T\mathbf F_t\mathbf F_t^\prime\right)\text{vec}(\mathrm d \underline{\bm \Lambda}^\prime),\nonumber \end{equation} and \begin{equation} \bm H(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}) = - (\underline{\bm\Sigma}^\xi)^{-1}\otimes \left(\sum_{t=1}^T\mathbf F_t\mathbf F_t^\prime\right)=-T (\underline{\bm\Sigma}^\xi)^{-1}\otimes \mathbf I_r, \end{equation} because of Assumption (ref)(b). And (ref) computed in the true value of the parameters is \begin{equation} \bm H(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi}) = -T ({\bm\Sigma}^\xi)^{-1}\otimes \mathbf I_r. \end{equation} From (ref) the $i$th $r\times r$ sub-matrix of $\bm H(\bm{\mathcal X}|\bm{\mathcal F};\bm{\varphi})$ is then: \begin{equation} \bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi}) = -\frac T{\sigma_i^2}\mathbf I_r. \end{equation} Hence, $\Vert\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi}) \Vert=O_{\mathrm P}(T)$, and, by using (ref) into (ref), for any given $i=1,\ldots, n$, we have: \begin{equation}\nonumber \frac 1T\left\Vert\bm h_{ii}(\bm{\mathcal X};\bm\varphi)- \bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})\right\Vert=O_{\mathrm P}\left(\max\left(\frac{1}{n},\frac 1{\sqrt{nT}}\right)\right), \end{equation} which proves part (b). For part (c) we provide two different equivalent proofs. First, \begin{align} \left\Vert \bm s_{i}(\bm{\mathcal X};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}})-\bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}}) \right\Vert \le&\, \left\Vert \bm s_{i}(\bm{\mathcal X};{\bm{\varphi}})-\bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};{\bm{\varphi}}) \right\Vert\nonumber\\ &+\left\Vert \bm h_{ii}(\bm{\mathcal X};{\bm{\varphi}})-\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm{\varphi}}) \right\Vert\, \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i \right\Vert+O_{\mathrm P}\left(\left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i \right\Vert^2\right)\nonumber\\ =&\,O_{\mathrm P}\left(\max\left(\frac {T}n,\frac {\sqrt T}{\sqrt{n}}\right)\right)+O_{\mathrm P}\left(\max\left(\frac {T}n,\frac {\sqrt T}{\sqrt{n}}\right)\right)O_{\mathrm P}\left(\max\left(\frac {1}n,\frac {1}{\sqrt{T}}\right)\right).\nonumber \end{align} because of parts (a) and (b), and (ref), which follows from Theorem (ref), which, in turn, holds even if we use the simpler log-likelihood (ref) in place of the log-likelihood (ref). As a consequence, \[ \frac 1{\sqrt T}\left\Vert \bm s_{i}(\bm{\mathcal X};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}})-\bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}}) \right\Vert= O_{\mathrm P}\left(\max\left(\frac {\sqrt T}n,\frac 1{\sqrt{n}}\right)\right). \] Alternatively, let $\widehat{\bm\varphi}^{\text{\tiny OLS}}=(\mathrm{vec}(\bm\Lambda^{\text{\tiny OLS}})^\prime, \mathrm{vech}(\underline{\bm\Gamma}^\xi)^\prime)^\prime$ which is the vector of parameters made of the OLS estimator of the loadings and any generic value of the idiosyncratic covariance matrix satisfying Assumption (ref). Then, notice that $\bm s_{i}(\bm{\mathcal X};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}})=\mathbf 0_r$ by definition of QML estimator, and $\bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny OLS}})=\mathbf 0_r$ because the OLS estimator is the QML estimator when maximizing the conditional log-likelihood, and the OLS estimator of the loadings does not depend on the estimator of the idiosyncratic covariance. Recall also that \[ \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i^{\text{\tiny OLS}} \right\Vert\le \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-\widehat{\bm\lambda}_i \right\Vert+\left\Vert \widehat{\bm\lambda}_i-{\bm\lambda}_i^{\text{\tiny OLS}} \right\Vert = O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt{nT}}\right)\right), \] because of Theorem (ref) and (ref) which hold even if we use the log-likelihood (ref) in place of the log-likelihood (ref). Moreover, from (ref) it is straightforward to see that $\Vert\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm{\varphi}})\Vert=O_{\mathrm P}(T)$, for any $\underline{\bm{\varphi}}\in\mathcal O_n$. Therefore, \begin{align} \left\Vert \bm s_{i}(\bm{\mathcal X};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}})-\bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}}) \right\Vert =&\, \left\Vert \bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}}) \right\Vert\nonumber\\ \le&\, \left\Vert \bm s_{i}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny OLS}}) \right\Vert+ \left\Vert \bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny OLS}}) \right\Vert\, \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i^{\text{\tiny OLS}} \right\Vert\nonumber\\ &+ O_{\mathrm P}\left(\left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i^{\text{\tiny OLS}} \right\Vert^2\right)\nonumber\\ \le &\, O_{\mathrm P}\left(T\right) O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt{nT}}\right)\right)=O_{\mathrm P}\left(\max\left(\frac {T}n,\frac {\sqrt T}{\sqrt{n}}\right)\right).\nonumber \end{align} This proves part (c). For part (d), for any specific value of the parameters, say $\widetilde{\bm\varphi}$, define the $n^2r^2\times nr$ matrices of third derivatives \[ \bm T(\bm{\mathcal X};\widetilde{\bm\varphi}) =\left.\frac{\partial \text{vec}(\bm H(\bm{\mathcal X};\underline{\bm\varphi}))^\prime}{\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right|_{\underline{\bm\varphi}=\widetilde{\bm\varphi}},\quad \bm T(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi}) =\left.\frac{\partial \text{vec}(\bm H(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}))^\prime}{\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right|_{\underline{\bm\varphi}=\widetilde{\bm\varphi}}, \] with components \[ \bm t_{iii}(\bm{\mathcal X};\widetilde{\bm\varphi}) = \left.\frac{\partial \text{vec}(\bm h_{ii}(\bm{\mathcal X};\underline{\bm\varphi}))^\prime}{\partial \underline{\bm\lambda}_i^\prime} \right|_{\underline{\bm\varphi}=\widetilde{\bm\varphi}},\quad \bm t_{iii}(\bm{\mathcal X}|\bm{\mathcal F};\widetilde{\bm\varphi}) = \left.\frac{\partial \text{vec}(\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};\underline{\bm\varphi}))^\prime}{\partial \underline{\bm\lambda}_i^\prime} \right|_{\underline{\bm\varphi}=\widetilde{\bm\varphi}},\quad i=1,\ldots, n, \] which are $r^2\times r$ matrices. Then, from (ref) the $i$th $r\times r$ sub-matrix of $\bm H(\bm{\mathcal X};\underline{\bm\varphi})$ is given by: \begin{align} \bm h_{ii}(\bm{\mathcal X};\underline{\bm\varphi})=&\, \frac T{\underline{\sigma}_i^4} \left[ \left\{\underline{\bm\lambda}_i^\prime \underline{\bm P}^{-1}\underline{\bm\lambda}_i\right\} \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^4} \left[ \left\{[\widehat{\bm\Gamma}^x]_{i\cdot}(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\underline{\bm P}^{-1}\underline{\bm\lambda}_i \right\} \underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ \left\{\underline{\bm \lambda}_i\underline{\bm P}^{-1} \underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda}\,\underline{\bm P}^{-1} \underline{\bm \lambda}_i \right\} \underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ \left\{\underline{\bm\lambda}_i^\prime \underline{\bm P}^{-1}\underline{\bm\lambda}_i \right\} \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\underline{\bm\Sigma}^\xi)^{-1} \underline{\bm\Lambda}\,\underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ \underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes \underline{\bm\lambda}_i^\prime \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^4} \left[ \underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes [\widehat{\bm\Gamma}^x]_{i\cdot}(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ \underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes \underline{\bm\lambda}_i^\prime \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^\xi(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes \underline{\bm\lambda}_i^\prime \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^2} \left[ \underline{\bm P}^{-1} \right] \nonumber\\ &+\frac T{\underline{\sigma}_i^4} \left[ [\widehat{\bm\Gamma}^x]_{ii} \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^4} \left[ \left\{\underline{\bm\lambda}_i^\prime\underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}[\widehat{\bm\Gamma}^x]_{\cdot i} \right\} \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^4} \left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}[\widehat{\bm\Gamma}^x]_{\cdot i} \otimes \underline{\bm\lambda}_i^\prime\underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^2} \left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\underline{\bm P}^{-1} \right], \end{align} where $[\widehat{\bm\Gamma}^x]_{i\cdot}$ and $[\widehat{\bm\Gamma}^x]_{\cdot i}$ are the $i$th row and column of $\widehat{\bm\Gamma}^x$, respectively, and $[\widehat{\bm\Gamma}^x]_{ii}$ is the $i$th term on its diagonal. Since, \begin{align} &\left\Vert \underline{\bm\lambda}_i\right\Vert \le M_\Lambda,\quad \left\Vert \underline{\bm\Lambda}\right\Vert = O(\sqrt n),\nonumber\\ &\left\Vert (\underline{\bm\Sigma}^\xi)^{-1}\right\Vert \le \frac 1{C_\xi},\quad \left\vert \frac 1{\sigma_i^2}\right\vert\le \frac 1{C_\xi},\nonumber\\ &\left\Vert \underline{\bm P}^{-1}\right\Vert =\left\Vert \left\{\mathbf I_r+\underline{\bm \Lambda}^\prime (\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm \Lambda} \right\}^{-1}\right\Vert = O\left(\frac 1n\right),\nonumber\\ &\left\Vert \widehat{\bm\Gamma}^x\right\Vert \le \left\Vert \widehat{\bm\Gamma}^x-\bm\Gamma^x\right\Vert+\left\Vert {\bm\Gamma}^x\right\Vert = O_{\mathrm P}\left(\frac n{\sqrt T}\right) +O(n),\nonumber\\ & \left\Vert [\widehat{\bm\Gamma}^x]_{i\cdot }\right\Vert = \left\Vert [\widehat{\bm\Gamma}^x]_{i\cdot }-[{\bm\Gamma}^x]_{i\cdot }\right\Vert+\left\Vert [{\bm\Gamma}^x]_{i\cdot }\right\Vert= O_{\mathrm P}\left(\frac{\sqrt n}{\sqrt T}\right) +O(\sqrt n), \end{align} because of Assumption (ref)(a), Lemma (ref)(i), Assumption (ref)(a), Lemma (ref)(i), and Lemma (ref)(vi), respectively From (ref) and (ref) it is clear that the leading term in $\bm h_{ii}(\bm{\mathcal X};\underline{\bm\varphi})$ is the last one, which is $O_{\mathrm P}(T)$ while all others are $o_{\mathrm P}(T)$, specifically, \begin{align} \frac 1T \bm h_{ii}(\bm{\mathcal X};\underline{\bm\varphi})&= -\frac 1{\underline{\sigma}_i^2} \left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\underline{\bm P}^{-1} \right]+O_{\mathrm P}\left(\frac 1n\right)+O\left(\frac 1{n^2}\right)\nonumber\\ &=\frac 1T \bm h^*_{ii}(\bm{\mathcal X};\underline{\bm\varphi})+O_{\mathrm P}\left(\frac 1n\right)+O\left(\frac 1{n^2}\right), \;\text{say.} \end{align} From (ref) \begin{align} \text{vec}\left(\mathrm d \bm h_{ii}^{*\prime} (\bm{\mathcal X};\underline{\bm\varphi})\right) =&\, -\frac T{\underline{\sigma}_i^2}\left\{ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda} \otimes \mathbf I_r \right\}\text{vec}\left(\mathrm d\underline{\bm P}^{-1}\right)\nonumber\\ &-\frac T{\underline{\sigma}_i^2}\left\{ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right\} \text{vec}\left(\mathrm d\underline{\bm\Lambda}^\prime\right)\nonumber\\ &-\frac T{\underline{\sigma}_i^2}\left\{ \underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1} \right\}\bm C_{n,r}\text{vec}\left(\mathrm d\underline{\bm\Lambda}^\prime\right)\nonumber\\ &-\frac T{\underline{\sigma}_i^2}\left\{\mathbf I_r \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda} \right\}\text{vec}\left(\mathrm d\underline{\bm P}^{-1}\right). \end{align} And, since \begin{equation} \text{vec}\left(\mathrm d \bm h_{ii}^{*\prime} (\bm{\mathcal X};\underline{\bm\varphi})\right) =\left( \frac{\partial \text{vec}(\bm h_{ii}^* (\bm{\mathcal X};\underline{\bm\varphi}))^\prime} {\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right)^\prime \text{vec}\left(\mathrm d\underline{\bm\Lambda}^\prime\right),\nonumber \end{equation} from (ref) and (ref), \begin{align} \left( \frac{\partial \text{vec}(\bm h_{ii}^* (\bm{\mathcal X};\underline{\bm\varphi}))^\prime} {\partial \text{vec}(\underline{\bm\Lambda})^\prime} \right)^\prime=&\, \frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r}\nonumber\\ &-\frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1} \right]\nonumber\\ &-\frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r}\nonumber\\ &+\frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1} \right]\nonumber\\ &+\frac T{\underline{\sigma}_i^2}\left[ \underline{\bm P}^{-1} \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1} \right]\bm C_{n,r}, \end{align} which is an $r^2\times nr$ matrix. From (ref) we have \begin{align} \frac{\partial \text{vec}(\bm h^*_{ii}(\bm{\mathcal X};\underline{\bm\varphi}))^\prime}{\partial \underline{\bm\lambda}_i^\prime}=&\, \frac {2T}{\underline{\sigma}_i^4}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes \underline{\bm P}^{-1} \right]\nonumber\\ &+\frac {2T}{\underline{\sigma}_i^4}\left[ \underline{\bm P}^{-1}\underline{\bm\lambda}_i \otimes \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}\widehat{\bm\Gamma}^x(\underline{\bm\Sigma}^\xi)^{-1}\underline{\bm\Lambda}\,\underline{\bm P}^{-1} \right]\nonumber\\ &-\frac {2T}{\underline{\sigma}_i^4}\left[ \underline{\bm P}^{-1}\underline{\bm\Lambda}^\prime(\underline{\bm\Sigma}^\xi)^{-1}[\widehat{\bm\Gamma}^x]_{\cdot i} \otimes \underline{\bm P}^{-1} \right]. \end{align} By using (ref) into (ref) and because of (ref), we have \begin{equation} \left\Vert \text{vec}\left(\bm t_{iii}(\bm{\mathcal X};\bm\varphi)\right)\right\Vert = \left\Vert \bm t_{iii}(\bm{\mathcal X};\bm\varphi)\right\Vert = O_{\mathrm P}\left(\frac T{ n}\right). \end{equation} Moreover, from (ref) it immediately follows that $\text{vec}\left(\bm T(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})\right) = \mathbf 0_{n^3r^3}$. and, thus, $\text{vec}\left(\bm t_{iii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi})\right) = \mathbf 0_{r^3} $, or, equivalently, \begin{equation} \bm t_{iii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm\varphi}) = \mathbf 0_{r\times r\times r}, \end{equation} i.e., a 3rd order tensor of zeros. Finally, \begin{align} \left\Vert \bm h_{ii}(\bm{\mathcal X};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}})-\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};\widehat{\bm{\varphi}}^{\text{\tiny QML,E}}) \right\Vert \le&\, \left\Vert \bm h_{ii}(\bm{\mathcal X};{\bm{\varphi}})-\bm h_{ii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm{\varphi}}) \right\Vert\nonumber\\ &+\left\Vert \bm t_{iii}(\bm{\mathcal X};{\bm{\varphi}})-\bm t_{iii}(\bm{\mathcal X}|\bm{\mathcal F};{\bm{\varphi}}) \right\Vert\, \left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i \right\Vert+O_{\mathrm P}\left(\left\Vert \widehat{\bm\lambda}_i^{\text{\tiny QML,E}}-{\bm\lambda}_i \right\Vert^2\right)\nonumber\\ =&\,O_{\mathrm P}\left(\max\left(\frac {T}n,\frac {\sqrt T}{\sqrt{n}}\right)\right)+O_{\mathrm P}\left(\frac {T}{n}\right)O_{\mathrm P}\left(\max\left(\frac {1}n,\frac {1}{\sqrt{T}}\right)\right),\nonumber \end{align} which follows from part (b), (ref), (ref), and (ref) in the proof of Corollary (ref), which, in turn, holds even if we use the log-likelihood (ref) in place of thel log-likelihood (ref). This completes the proof. $\Box$ \setcounter{equation}{0} \section{Auxiliary propositions} \begin{prop} Under Assumptions (ref) and (ref), \begin{compactenum}[(a)] • $\frac{\bm\Lambda^\prime\bm\Lambda}{n} = \frac{\mathbf M^\chi}{n}$, for all $n\in\mathbb N$; • $\bm\Lambda=\mathbf V^\chi(\mathbf M^\chi)^{1/2}$, for all $n\in\mathbb N$. \end{compactenum} \end{prop} \textbf{Proof of Proposition (ref).} For part (a), first notice that Assumptions (ref)(c) and (ref)(b), imply $\bm\Gamma^F=\mathbf I_r$ which, in turn, implies $\bm\Gamma^\chi=\bm\Lambda\bm\Lambda^\prime$. Therefore, since the non-zero eigenvalues of $\frac{\bm\Gamma^\chi}n$ are the same as the $r$ eigenvalues of $\frac{\bm\Lambda^\prime\bm\Lambda}{n}$, which is diagonal by Assumption (ref)(a). Then, we must have, for all $n\in\mathbb N$, $\frac{\bm\Lambda^\prime\bm\Lambda}{n} = \frac{\mathbf M^\chi}{n}$. Equivalently, for a given $T$, from Assumption (ref)(b), we have $\widehat{\bm\Gamma}^\chi=\frac 1T \sum_{t=1}^T\bm\chi_t\bm\chi_t^\prime=\bm\Lambda\bm\Lambda^\prime$ which implies that $\frac{\bm\Lambda^\prime\bm\Lambda}{n} = \frac{\widehat{\mathbf M}^\chi}{n}$. However, since $\widehat{\bm\Gamma}^\chi = {\bm\Gamma}^\chi$ it must be that $\frac{\widehat{\mathbf M}^\chi}{n}=\frac{{\mathbf M}^\chi}{n}$. This proves part (a). For part (b), since $\bm\Gamma^\chi=\mathbf V^\chi\mathbf M^\chi\mathbf V^{\chi\prime}$, it must be that \begin{equation} \bm\Lambda\mathbf K_*= \mathbf V^\chi(\mathbf M^\chi)^{1/2}, \end{equation} for some $r\times r$ invertible $\mathbf K_*$. Now, since $\text{rk}(\frac{\bm\Lambda} {\sqrt n})=r$ for all $n>N$ (see the proof of Proposition (ref)): \begin{equation} \mathbf K_*= (\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2}=(\mathbf M^\chi)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2}, \end{equation} which is also obtained by linear projection, and \begin{equation} \mathbf K^{-1}_*= (\mathbf M^\chi)^{-1/2}{\mathbf V^{\chi\prime}\bm\Lambda}. \end{equation} Notice that since $\mathbf K_*$ is a special case of the matrix $\mathbf K$ defined in (ref) in the proof of Proposition (ref), then $\mathbf K_*$ is finite and positive definite because of Lemma (ref)(i) and (ref)(ii), respectively, hence, $\mathbf K^{-1}_*$ in (ref) is well defined. Moreover, from (ref), because of Assumption (ref)(b) and part (a): \begin{align} \mathbf K_*\mathbf K_*^\prime&=(\mathbf M^\chi)^{-1}\bm\Lambda^\prime\mathbf V^\chi\mathbf M^\chi\mathbf V^{\chi\prime}\bm\Lambda(\mathbf M^\chi)^{-1}\nonumber\\ &=(\mathbf M^\chi)^{-1}\bm\Lambda^\prime\bm\Gamma^\chi\bm\Lambda(\mathbf M^\chi)^{-1}\nonumber\\ &=(\mathbf M^\chi)^{-1}\bm\Lambda^\prime\bm\Lambda\bm\Lambda^\prime\bm\Lambda(\mathbf M^\chi)^{-1}\nonumber\\ &=\mathbf I_r. \end{align} So because of (ref), we have that $\mathbf K_*$ is an orthogonal matrix, i.e., $\mathbf K_*=\mathbf K_*^{-1}$. Finally, by (ref) we also have \begin{equation} \mathbf V^\chi = \bm\Lambda\mathbf K_*(\mathbf M^\chi)^{-1/2} \end{equation} and by substituting (ref) into (ref), because of (ref), \[ \mathbf K_*^{-1} = (\mathbf M^\chi)^{-1}\mathbf K_*^\prime \bm\Lambda^\prime \bm\Lambda=(\mathbf M^\chi)^{-1}\mathbf K_*^{-1}\bm\Lambda^\prime \bm\Lambda, \] which is equivalent to: \begin{equation} \mathbf K_*^{-1}\bm\Lambda^\prime \bm\Lambda \mathbf K_*= \mathbf M^\chi, \end{equation} and by part (a), we must have $\mathbf K_*=\mathbf I_r$. Alternatively, by multiplying on the right both sides of (ref) by their transposed: \begin{equation} \mathbf K_*^\prime\bm\Lambda^\prime \bm\Lambda \mathbf K_*= \mathbf M^\chi, \end{equation} since eigenvectors are normalized, and again by part (a), we must have $\mathbf K_*=\mathbf I_r$. So from (ref) or (ref) we prove part (b). This completes the proof. $\Box$ \begin{prop} Under Assumptions (ref) through (ref), as $n,T\to\infty$ $\min(n,\sqrt T)\left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert= O_{\mathrm P}(1)$, where $\bm{\mathcal H}=(\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi (\mathbf M^\chi)^{1/2}\mathbf J$ and $\mathbf J$ is an $r\times r$ diagonal matrix with entries $\pm 1$. \end{prop} \begin{proof} Notice that $\text{rk}\left(\frac{\bm\Lambda}{\sqrt n}\right)=r$ for all $n$, since $\text{rk}(\bm\Gamma^F)=r$ by Assumption (ref)(b) and $\text{rk}\left(\frac{\bm\Gamma^\chi}n \right)=r$ by Lemma (ref)(iv). Indeed, $\text{rk}\left(\frac{\bm\Gamma^\chi}n \right)\le \min (\text{rk}(\bm\Gamma^F),\text{rk}\left(\frac{\bm\Lambda}{\sqrt n}\right))$. This holds for all $n>N$ and since eigenvalues are an increasing sequence in $n$. Therefore, $(\frac{\bm\Lambda^\prime\bm\Lambda}n)^{-1}$ is well defined for all $n$ and $\bm\Lambda$ admits a left inverse. Second, since \begin{equation} \frac{\bm\Gamma^\chi}n = \mathbf V^\chi\frac{\mathbf M^\chi}n\mathbf V^{\chi\prime}= \frac{\bm\Lambda}{\sqrt n}\bm\Gamma^F\frac{\bm\Lambda^\prime}{\sqrt n}. \end{equation} the columns of $\frac{\mathbf V^\chi(\mathbf M^\chi)^{1/2}}{\sqrt n}$ and the columns of $\frac{\bm\Lambda(\bm\Gamma^F)^{1/2}}{\sqrt n}$ must span the same space. So there exists an $r\times r$ invertible matrix $\mathbf K$ such that \begin{equation} {\bm\Lambda(\bm\Gamma^F)^{1/2}}\mathbf K= {\mathbf V^\chi(\mathbf M^\chi)^{1/2}} \end{equation} Therefore, from (ref) \begin{equation} \mathbf K=(\bm\Gamma^F)^{-1/2} (\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2} \end{equation} which is also obtained by linear projection, and also \begin{equation} \mathbf K^{-1}= (\mathbf M^\chi)^{-1/2}{\mathbf V^{\chi\prime}\bm\Lambda}(\bm\Gamma^F)^{1/2} \end{equation} which are both finite and positive definite because of Lemma (ref). Now, from (ref) and (ref) \begin{align} {\mathbf V}^\chi ( {\mathbf M}^\chi)^{1/2}= \bm\Lambda(\bm\Gamma^F )^{1/2}\mathbf K = \bm\Lambda(\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2} \end{align} Let \begin{equation} \bm{\mathcal H}=(\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2}\mathbf J, \end{equation} which is finite and positive definite because of Lemma (ref). Now, because of Lemmas (ref)(iii), (ref)(iv), (ref)(i), using (ref), (ref), and (ref), \begin{align} \left\Vert\frac { \widehat{\bm\Lambda} - \bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert=&\, \left\Vert\widehat{\mathbf V}^x\left(\frac{\widehat{\mathbf M}^x}n\right)^{1/2} - {\mathbf V}^\chi \left(\frac{ {\mathbf M}^\chi}n\right)^{1/2} \mathbf J\right\Vert=\ \left\Vert\widehat{\mathbf V}^x\left(\frac{\widehat{\mathbf M}^x}n\right)^{1/2} - {\mathbf V}^\chi\mathbf J \left(\frac{ {\mathbf M}^\chi}n\right)^{1/2} \right\Vert\nonumber\\ \le&\, \left\Vert \widehat{\mathbf V}^x-{\mathbf V}^\chi\mathbf J\right\Vert\,\left\Vert\frac{\mathbf M^\chi}{n} \right\Vert+ \left\Vert\frac 1 {\sqrt n}\left\{\left(\widehat{\mathbf M}^x\right)^{1/2}-\left(\mathbf M^\chi\right)^{1/2}\right\}\right\Vert \,\left\Vert \mathbf V^\chi\right\Vert\nonumber\\ &+ \left\Vert \widehat{\mathbf V}^x-{\mathbf V}^\chi\mathbf J\right\Vert\, \left\Vert\frac 1 {\sqrt n}\left\{\left(\widehat{\mathbf M}^x\right)^{1/2}-\left(\mathbf M^\chi\right)^{1/2}\right\}\right\Vert\nonumber\\ &= O_{\mathrm P}\left(\max\left(\frac 1 n,\frac 1{\sqrt T}\right)\right)+ O_{\mathrm P}\left(\max\left(\frac 1 {n^2},\frac 1{T}\right)\right), \nonumber \end{align} since $\Vert \mathbf V^\chi\Vert=1$ because eigenvectors are normalized. This completes the proof. \end{proof} \begin{prop} Under Assumptions (ref) through (ref) the terms in (ref) are such that, as $n,T\to\infty$, \begin{compactenum}[(a)] • $\sqrt {nT}\left\Vert \text{\upshape (1.a)}\right\Vert = O_{\mathrm P}(1)$; • $\sqrt T \left\Vert \text{\upshape (1.b)}\right\Vert = O_{\mathrm P}(1)$; • $\min(n,\sqrt{nT})\left\Vert \text{\upshape (1.c)}\right\Vert = O_{\mathrm P}(1)$; • $\min(\sqrt{nT},T)\left\Vert \text{\upshape (1.d)}\right\Vert = O_{\mathrm P}(1)$; • $\min(\sqrt{nT},T)\left\Vert \text{\upshape (1.e)}\right\Vert = O_{\mathrm P}(1)$; • $\min(n,\sqrt{nT},T)\left\Vert \text{\upshape (1.f)}\right\Vert = O_{\mathrm P}(1)$; \end{compactenum} uniformly in $i$. \end{prop} \begin{proof} For part (a), for any $i=1,\ldots,n$, by Assumption (ref)(a), \begin{equation} \left\Vert\frac 1{nT}{\bm\lambda}_i^\prime\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime\right\Vert \le \left\Vert \bm\lambda_i\right\Vert\,\left\Vert\frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime \right\Vert\le M_\Lambda\left\Vert\frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime \right\Vert. \end{equation} Then, by Assumptions (ref)(a) and (ref) \begin{align} \mathbb{E}\left[\left\Vert\frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime \right\Vert^2\right] &\le \mathbb{E}\left[\left\Vert\frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime \right\Vert_F^2\right] =\frac 1{n^2T^2} \sum_{k=1}^r\sum_{h=1}^r \mathbb{E}\left[\left(\sum_{t=1}^T F_{kt}\sum_{j=1}^n \xi_{jt}[\bm\Lambda_{jh}]\right)^2\right] \nonumber\\ &\le \frac{r^2}{n^2T^2} \max_{h,k=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}\left[F_{kt} \left(\sum_{j=1}^n \xi_{jt}[\bm\Lambda_{jh}]\right) F_{ks}\left(\sum_{\ell=1}^n \xi_{\ell s}[\bm\Lambda_{\ell h}]\right) \right]\nonumber\\ &= \frac{r^2}{n^2T^2} \max_{h,k=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[F_{kt} F_{ks}]\,\sum_{j=1}^n\sum_{\ell=1}^n \mathbb{E}\left[\xi_{jt} \xi_{\ell s} \right] [\bm\Lambda_{jh}][\bm\Lambda_{\ell h}]\nonumber\\ &\le \left\{\frac{r^2}{T}\max_{k=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}\left[F_{kt}F_{ks}\right]\right\} \,\left\{\max_{t,s=1\ldots,T} \frac{M_{\Lambda}^2}{n^2T}\sum_{j=1}^n\sum_{\ell=1}^n\left\vert \mathbb{E}[\xi_{jt}\xi_{\ell s}]\right\vert\right\}. \end{align} Now, by Cauchy-Schwarz inequality, for any $k=1,\ldots, r$, \begin{equation} \left\vert \frac{1}{T}\sum_{t=1}^T\sum_{s=1}^TF_{kt}F_{ks}\right\vert\le \left(\frac 1 T\sum_{t=1}^T F_{kt}^2\right)^{1/2}\left(\frac 1 T\sum_{s=1}^T F_{ks}^2\right)^{1/2}, \end{equation} and, by Assumption (ref)(b), using (ref) and again Cauchy-Schwarz inequality, \begin{align} \max_{k=1,\ldots, r}\frac{1}{T}\sum_{t=1}^T\sum_{s=1}^T\mathbb{E}[F_{kt}F_{ks}]& =\max_{k=1,\ldots, r}\mathbb{E}\left[\frac{1}{T}\sum_{t=1}^T\sum_{s=1}^TF_{kt}F_{ks}\right]\le\max_{k=1,\ldots, r}\mathbb{E}\left[\left\vert \frac{1}{T}\sum_{t=1}^T\sum_{s=1}^TF_{kt}F_{ks}\right\vert\right]\nonumber\\ &\le \max_{k=1,\ldots, r} \mathbb{E}\left[\left(\frac 1 T\sum_{t=1}^T F_{kt}^2\right)^{1/2}\left(\frac 1 T\sum_{s=1}^T F_{ks}^2\right)^{1/2}\right]\nonumber\\ &\le \max_{k=1,\ldots, r} \left(\mathbb{E}\left[\left(\frac 1 T\sum_{t=1}^T F_{kt}^2\right)\right] \right)^{1/2} \left(\mathbb{E}\left[\left(\frac 1 T\sum_{s=1}^T F_{ks}^2\right)\right] \right)^{1/2}\nonumber\\ &\le \max_{k=1,\ldots, r} \left(\frac 1 T\sum_{t=1}^T \mathbb{E}[F_{kt}^2] \right)^{1/2} \left(\frac 1 T\sum_{s=1}^T \mathbb{E}[F_{ks}^2] \right)^{1/2}\nonumber\\ &=\max_{k=1,\ldots, r} \frac{1}{T}\sum_{t=1}^T \mathbb{E}[F_{kt}^2]\le \max_{t=1,\ldots,T}\max_{k=1,\ldots, r}\mathbb{E}[F_{kt}^2]\nonumber\\ &= \max_{k=1,\ldots, r} \bm\eta_k^\prime \bm\Gamma_F\bm\eta_k \le \Vert\bm\Gamma^F\Vert\le M_F, \end{align} since $M_F$ is independent of $t$ and where $\bm\eta_k$ is an $r$-dimensional vector with one in the $k$th entry and zero elsewhere. And, because of Lemma (ref)(ii) \begin{equation} \max_{t,s=1,\ldots,T}\frac 1{n^2T}\sum_{j=1}^n\sum_{\ell=1}^n\left\vert \mathbb{E}[\xi_{jt}\xi_{\ell s}]\right\vert\le \frac{M_\xi}{nT}, \end{equation} since $M_\xi$ is independent of $n$ and $t$. By substituting (ref) and (ref) into (ref), \begin{equation} \mathbb{E}\left[\left\Vert\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}{\bm\lambda}_j^\prime \right\Vert^2\right] \le \frac{r^2M_FM_\Lambda^2M_\xi}{nT}. \end{equation} By substituting (ref) into (ref), we prove part (a). For part (b), for any $i=1,\ldots,n$, because of Lemma (ref)(i), \begin{equation} \left\Vert\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime\sum_{j=1}^n\bm\lambda_j\bm\lambda_j^\prime\right\Vert= \left\Vert\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime(\bm\Lambda^\prime\bm\Lambda)\right\Vert\le \left\Vert\frac 1{T} \sum_{t=1}^T \xi_{it}\mathbf F_t\right\Vert\,\left\Vert \frac{\bm\Lambda}{\sqrt n} \right\Vert^2\le \left\Vert\frac 1{T} \sum_{t=1}^T \xi_{it}\mathbf F_t\right\Vert M_\Lambda^2. \end{equation} Then, by Assumptions (ref) and (ref)(b) and using (ref) \begin{align} \mathbb{E}\left[\left\Vert\frac 1{T} \sum_{t=1}^T \xi_{it}\mathbf F_t\right\Vert^2\right]&=\frac 1{T^2} \sum_{j=1}^r \mathbb{E}\left[\left(\sum_{t=1}^T \xi_{it}F_{jt}\right)^2\right]\nonumber\\ &\le \frac{r}{T^2}\max_{j=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[\xi_{it}F_{jt}\xi_{is}F_{js}]= \frac{r}{T^2}\max_{j=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[F_{jt}F_{js}]\,\mathbb{E}[\xi_{it}\xi_{is}]\nonumber\\ &\le \left\{\frac{r}{T}\max_{j=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[F_{jt}F_{js}]\right\} \left\{\frac 1T \max_{t,s=1,\ldots, n} \left\vert\mathbb{E}[\xi_{it}\xi_{is}]\right\vert\right\}\nonumber\\ &\le rM_F\frac{ M_{ii}}T \max_{t,s=1,\ldots, n}\rho^{|t-s|} \le \frac{rM_F M_{\xi}}T, \end{align} since $M_\xi$ is independent of $i$. Or, equivalently, by Lemma (ref)(iii) and Cauchy-Schwarz inequality \begin{align} \mathbb{E}\left[\left\Vert\frac 1{T} \sum_{t=1}^T \xi_{it}\mathbf F_t\right\Vert^2\right] &\le \frac{r}{T^2}\max_{j=1,\ldots, r}\sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[F_{jt}F_{js}]\,\mathbb{E}[\xi_{it}\xi_{is}]\nonumber\\ &\le \left\{\frac{r}{T}\max_{j=1,\ldots, r}\max_{t,s=1,\ldots, n} \vert\mathbb{E}[F_{jt}F_{js}]\vert\right\} \left\{\frac 1T \sum_{t=1}^T\sum_{s=1}^T\left\vert\mathbb{E}[\xi_{it}\xi_{is}]\right\vert\right\}\nonumber\\ &\le \frac r T \max_{j=1,\ldots, r}\max_{t,s=1,\ldots, n} \mathbb{E}[F_{jt}^2] \frac{M_\xi(1+\rho)}{1-\rho} \le \frac{r M_F M_{3\xi}}{T}, \end{align} since $M_F$ is independent of $t$ and $M_{3\xi}$ is independent of $i$. Notice that $M_\xi\le M_{3\xi}$. Notice that (ref) is a special case of Lemma (ref)(i). By substituting (ref), or (ref), into (ref), we prove part (b). For part (c), for any $i=1,\ldots,n$, because of Assumption (ref)(a), \begin{align} \left\Vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} \bm\lambda_j^\prime\right\Vert &=\left\{\sum_{k=1}^r \left(\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} \lambda_{jk}\right)^2\right\}^{1/2}\le \sqrt rM_\Lambda\left\vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} \right\vert\nonumber\\ &\le\sqrt r M_\Lambda\left\{ \left\vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\} \right\vert+\left\vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\mathbb{E}[\xi_{it}\xi_{jt}] \right\vert\right\}. \end{align} Then, by Assumption (ref)(b), \begin{align} \left\vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\mathbb{E}[\xi_{it}\xi_{jt}] \right\vert\le \frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\left\vert\mathbb{E}[\xi_{it}\xi_{jt}]\right\vert\le \max_{t=1,\ldots,T}\frac 1n\sum_{j=1}^n\left\vert\mathbb{E}[\xi_{it}\xi_{jt}]\right\vert\le\frac 1n \sum_{j=1}^n M_{ij} \le \frac{M_\xi}n, \end{align} since $M_\xi$ is independent of $i$ and $t$. Moreover, by Assumption (ref)(c), \begin{equation} \mathbb{E}\left[\left\vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\} \right\vert^2\right]\le \frac{K_\xi}{nT}. \end{equation} By substituting (ref) and (ref) into (ref), we prove part (c). For part (d), for any $i=1,\ldots,n$, because of Assumption (ref)(a) \begin{align} \left\Vert\frac 1{nT}{\bm\lambda}_i^\prime\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}(\widehat{\bm\lambda}_j^\prime-{\bm\lambda}_j^\prime\bm{\mathcal H})\right\Vert &\le M_\Lambda\left\Vert\frac 1{nT}\sum_{t=1}^T\sum_{j=1}^n\mathbf F_t\xi_{jt}(\widehat{\bm\lambda}_j^\prime-{\bm\lambda}_j^\prime\bm{\mathcal H}) \right\Vert\nonumber\\ &= M_\Lambda\left\Vert\frac {\bm F^\prime\bm \Xi (\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H} )}{nT} \right\Vert\le M_\Lambda \left\Vert\frac {\bm F^\prime\bm \Xi}{\sqrt n T} \right\Vert\, \left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert. \end{align} Then, by using Lemma (ref) and Proposition (ref) in (ref), we prove part (d). For part (e), for any $i=1,\ldots,n$, \begin{align} \left\Vert\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime\sum_{j=1}^n\bm\lambda_j(\widehat{\bm\lambda}_j^\prime-{\bm\lambda}_j^\prime\bm{\mathcal H})\right\Vert&= \left\Vert\frac 1{nT} \sum_{t=1}^T \xi_{it}\mathbf F_t^\prime\bm\Lambda^\prime(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H})\right\Vert\nonumber\\ &\le \left\Vert\frac 1{T} \sum_{t=1}^T \xi_{it}\mathbf F_t\right\Vert\,\left\Vert \frac{\bm\Lambda}{\sqrt n} \right\Vert \, \left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert . \end{align} By substituting part (ii), Lemma (ref)(i), and part (a) of Proposition (ref) into (ref), we prove part (e). Finally, for part (f), for any $i=1,\ldots,n$, let $\bm\zeta_i=(\xi_{i1}\cdots \xi_{iT})^\prime$, then \begin{align} \left\Vert\frac 1{nT} \sum_{t=1}^T\sum_{j=1}^n\xi_{it}\xi_{jt} (\widehat{\bm\lambda}_j^\prime-{\bm\lambda}_j^\prime\bm{\mathcal H})\right\Vert &= \left\Vert\frac{\bm\zeta_i^\prime\bm\Xi\left(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}\right)}{nT} \right\Vert \le \left\Vert\frac{\bm\zeta_i^\prime\bm\Xi}{\sqrt n T}\right\Vert\, \left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert. \end{align} Then, by the $C_r$-inequality with $r=2$, \begin{align} \left\Vert\frac{\bm\zeta_i^\prime\bm\Xi}{\sqrt n T}\right\Vert^2&= \left\Vert\frac 1 {\sqrt n T} \sum_{t=1}^T \xi_{it} \bm\xi_t^\prime\right\Vert^2 =\frac 1{n} \sum_{j=1}^n \left(\frac 1T\sum_{t=1}^T \xi_{it}\xi_{jt}\right)^2\nonumber\\ & =\frac 1{n} \sum_{j=1}^n \left(\frac 1T\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\}+\frac 1T\sum_{t=1}^T \mathbb{E}[\xi_{it}\xi_{jt}]\right)^2\nonumber\\ &\le\frac 2{n} \sum_{j=1}^n\left\{ \left(\frac 1T\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\}\right)^2+\left(\frac 1T\sum_{t=1}^T \mathbb{E}[\xi_{it}\xi_{jt}]\right)^2\right\} \end{align} By taking the expectation of (ref), and because of Assumption (ref)(c) and Lemma (ref)(v), \begin{align} \mathbb{E}\left[\left\Vert\frac{\bm\zeta_i^\prime\bm\Xi}{\sqrt n T}\right\Vert^2\right]&=\frac 2{n} \sum_{j=1}^n \mathbb{E}\left[ \left(\frac 1T\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\}\right)^2 \right]+\frac 2{n} \sum_{j=1}^n \left(\frac 1T\sum_{t=1}^T \mathbb{E}[\xi_{it}\xi_{jt}]\right)^2\nonumber\\ &\le 2\max_{j=1,\ldots, n}\mathbb{E}\left[ \left(\frac 1T\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\}\right)^2 \right]+\frac{2}{n}\sum_{i=1}^n\left(\frac 1T\sum_{t=1}^T \mathbb{E}[\xi_{it}\xi_{jt}]\right)^2\nonumber\\ &\le \frac{2K_\xi}{T} + \frac{2}{nT^2}\sum_{i=1}^n\sum_{t=1}^T \mathbb{E}[\xi_{it}\xi_{jt}]\sum_{s=1}^T\mathbb{E}[\xi_{is}\xi_{js}]\nonumber\\ &\le \frac{2K_\xi}{T} + \max_{t=1,\ldots ,T}\frac{2}{n}\sum_{i=1}^n \left\vert\mathbb{E}[\xi_{it}\xi_{jt}]\right\vert \max_{s=1,\ldots,T}\max_{i,j=1,\ldots, n} \mathbb{E}[\xi_{is}\xi_{js}]\nonumber\\ &\le \frac{2K_\xi}{T} + \frac{2}n \sum_{i=1}^n M_{ij} \max_{i,j=1,\ldots, n} \bm\varepsilon_i^\prime\bm\Gamma^\xi\bm\varepsilon_j\le \frac{2K_\xi}{T}+ \frac{2M_\xi}{n}\Vert \bm\Gamma^\xi\Vert\le \frac{2K_\xi}{T}+ \frac{2M_\xi M_{2\xi}}{n}, \end{align} since $K_\xi$ is independent of $j$ and $M_\xi$ is independent of $i$, $j$, $t$, and $s$ and where $\bm\varepsilon_i$ is an $n$-dimensional vector with one in the $i$th entry and zero elsewhere. By substituting (ref) and part (a) of Proposition (ref) into (ref), we prove part (f). This completes the proof. \end{proof} \begin{prop} Under Assumptions (ref) through (ref), \begin{compactenum}[(a)] • $ \left\Vert \bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right)-\bm Q_0\right\Vert = o(1)$, as $n\to\infty$; • $\left\Vert \bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right)-\left(\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}\right)\right\Vert = o_{\mathrm P}(1)$, as $n,T\to\infty$. \end{compactenum} where $\bm Q_0=\bm V_0\bm{\mathcal J}_0\bm\Upsilon_0^\prime (\bm\Gamma^F)^{-1/2}$, and where $\bm{\mathcal J}_0$ is an $r\times r$ diagonal matrix with entries $\pm 1$, $\bm\Upsilon_0$ is the $r\times r$ matrix having as columns the normalized eigenvectors of $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$, and $\bm V_0$ is the $r\times r$ matrix of corresponding eigenvalues sorted in descending order. \end{prop} \begin{proof} Start with part (a). From (ref) and (ref) \begin{equation} \bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right) = \mathbf J(\mathbf M^\chi)^{1/2} \mathbf V^{\chi\prime}\bm\Lambda(\bm\Lambda^\prime\bm\Lambda)^{-1} \left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right) =\mathbf J\left(\frac{\mathbf M^\chi}n\right)^{1/2} \frac{\mathbf V^{\chi\prime}\bm\Lambda}{\sqrt n}=\mathbf J\left(\frac{\mathbf M^\chi}n\right) \mathbf K^{-1}(\bm\Gamma^F)^{-1/2}. \end{equation} Then, from (ref) \begin{equation} \mathbf V^{\chi} = \bm\Lambda (\bm\Gamma^F)^{1/2}\mathbf K ({\mathbf M^\chi})^{-1/2} \end{equation} thus, from (ref) and (ref) \begin{align} \mathbf J\left(\frac{\mathbf M^\chi}n\right) \mathbf K^{-1} &=\mathbf J \left(\frac{\mathbf M^\chi}n\right) (\mathbf M^\chi)^{-1/2}{\mathbf V^{\chi\prime}\bm\Lambda}(\bm\Gamma^F)^{1/2}\nonumber\\ &= \mathbf J\left(\frac{\mathbf M^\chi}n\right) ({\mathbf M^\chi})^{-1}\mathbf K^\prime(\bm\Gamma^F)^{1/2}\bm\Lambda^\prime \bm\Lambda(\bm\Gamma^F)^{1/2}\nonumber\\ &= \mathbf J\mathbf K^\prime(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime \bm\Lambda}n(\bm\Gamma^F)^{1/2}. \end{align} And, from (ref) \begin{align} \mathbf J\mathbf K\mathbf K^\prime\mathbf J &= \mathbf J (\bm\Gamma^F)^{-1/2} (\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\mathbf V^\chi(\mathbf M^\chi)^{1/2} (\mathbf M^\chi)^{1/2}\mathbf V^{\chi\prime}\bm\Lambda(\bm\Lambda^\prime\bm\Lambda)^{-1}(\bm\Gamma^F)^{-1/2} \mathbf J\nonumber\\ &=\mathbf J (\bm\Gamma^F)^{-1/2} (\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\bm\Gamma^\chi\bm\Lambda(\bm\Lambda^\prime\bm\Lambda)^{-1}(\bm\Gamma^F)^{-1/2} \mathbf J\nonumber\\ &=\mathbf J (\bm\Gamma^F)^{-1/2} (\bm\Lambda^\prime\bm\Lambda)^{-1}\bm\Lambda^\prime\bm\Lambda\bm\Gamma^F\bm\Lambda^\prime\bm\Lambda(\bm\Lambda^\prime\bm\Lambda)^{-1}(\bm\Gamma^F)^{-1/2} \mathbf J\nonumber\\ &= \mathbf J (\bm\Gamma^F)^{-1/2} \bm\Gamma^F(\bm\Gamma^F)^{-1/2} \mathbf J=\mathbf I_r. \end{align} Therefore, the columns of $\mathbf J\mathbf K$ are the normalized eigenvectors of $(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime \bm\Lambda}n(\bm\Gamma^F)^{1/2}$ with eigenvalues $\frac{\mathbf M^\chi}n$ (notice that $\mathbf J\left(\frac{\mathbf M^\chi}n\right)=\left(\frac{\mathbf M^\chi}n\right)\mathbf J$). Moreover, by Assumption (ref)(a) \begin{equation} \lim_{n\to\infty} \left\Vert(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime \bm\Lambda}n(\bm\Gamma^F)^{1/2}-(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}\right\Vert=0. \end{equation} Letting, $\bm V_0$ be the matrix of eigenvalues of $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$ sorted in descending order, from (ref) we also have (this is proved also in Lemma (ref)(i)) \begin{equation} \lim_{n\to\infty} \left\Vert\frac{\mathbf M^\chi}n - \bm V_0\right\Vert=0. \end{equation} Let $\bm\Upsilon_0$ be the normalized eigenvectors of $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$ Hence, by continuity of eigenvectors, and since the eigenvalues $\frac{\mathbf M^\chi}n$ are distinct because of Assumption (ref), from yu15, by Lemma (ref)(iii) and using (ref) it follows that \begin{equation} \lim_{n\to\infty} \left\Vert\mathbf J\mathbf K-\bm\Upsilon_0\bm{\mathcal J}_0\right\Vert \le \lim_{n\to\infty}\frac{2^{3/2}\sqrt r \left\Vert(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime \bm\Lambda}n(\bm\Gamma^F)^{1/2}-(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}\right\Vert}{\mu_r(\bm V_0)}=0, \end{equation} where $\bm{\mathcal J}_0$ is an $r\times r$ diagonal matrix with entries $\pm 1$, which is in general different from $\mathbf J$. Finally, since $\bm\Upsilon_0$ is an orthogonal matrix, we have $\bm\Upsilon_0^{-1}=\bm\Upsilon_0^\prime$, so from (ref) \begin{equation} \lim_{n\to\infty} \left\Vert\mathbf K^{-1}\mathbf J-\bm{\mathcal J}_0\bm\Upsilon_0^\prime\right\Vert =0. \end{equation} By using (ref) and (ref) into (ref) and since $\mathbf K^{-1}\mathbf J=\mathbf J\mathbf K^{-1}$, we have \begin{equation}\nonumber \lim_{n\to\infty}\left\Vert\bm{\mathcal H}^\prime\left(\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right) -\bm V_0\bm{\mathcal J}_0\bm\Upsilon_0^\prime(\bm\Gamma^F)^{-1/2}\right\Vert=0. \end{equation} By defining $\bm Q_0=\bm V_0\bm{\mathcal J}_0\bm\Upsilon_0^\prime(\bm\Gamma^F)^{-1/2}$, we prove part (a). Part (b) follows from Proposition (ref) and Lemma (ref)(i), and since \begin{equation} \left\Vert \frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}-\bm {\mathcal H}^\prime\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right\Vert \le \left\Vert \frac{\widehat{\bm\Lambda}^\prime-\bm {\mathcal H}^\prime\bm\Lambda^\prime}{\sqrt n}\right\Vert\,\left\Vert \frac{\bm\Lambda}{\sqrt n}\right\Vert = o_{\mathrm P}(1). \end{equation} This completes the proof. \end{proof} \begin{prop} Under Assumptions (ref) through (ref), as $n,T\to\infty$, $\min(n,\sqrt{nT},T)\left\Vert \frac{(\bm\Lambda\widehat{\mathbf H}-\widehat{\bm\Lambda})^\prime\widehat{\bm\Lambda}}{n}\right\Vert=O_{\mathrm {P}}(1)$, where $\widehat{\mathbf H}$ is defined in (ref). \end{prop} \begin{proof}We have, \begin{align} \left\Vert \frac{(\bm\Lambda\widehat{\mathbf H}-\widehat{\bm\Lambda})^\prime\widehat{\bm\Lambda}}{n}\right\Vert&\le \left\{\left\Vert \frac{(\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H})^\prime\bm\Lambda\widehat{\mathbf H}}{n}\right\Vert+\left\Vert \frac{(\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H})^\prime (\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H})}{n}\right\Vert\right\}\nonumber\\ &=\left\{\bm I +\bm{II}\right\},\;\text{say.} \end{align} First, consider $\bm I$ in (ref). From (ref) \begin{align} \bm I=&\,\frac {(\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H})^\prime \bm\Lambda \widehat{\mathbf H}}n\nonumber\\ =&\, \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\bm{\mathcal H}^\prime\frac{\bm\Lambda^\prime\bm\Lambda}{n} \frac{\bm F^\prime\bm\Xi\bm\Lambda \widehat{\mathbf H}}{ nT} +\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\bm{\mathcal H}^\prime\frac{\bm\Lambda^\prime\bm\Xi^\prime\bm F}{nT}\frac{\bm\Lambda^\prime\bm\Lambda \widehat{\mathbf H}}{n} + \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\bm{\mathcal H}^\prime \frac{\bm\Lambda^\prime\bm\Xi^\prime\bm\Xi\bm\Lambda \widehat{\mathbf H}}{n^2T}\nonumber\\ &+\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H})^\prime}{\sqrt n}\frac{\bm \Xi^\prime\bm F\bm\Lambda^\prime}{nT}\frac{\bm\Lambda \widehat{\mathbf H}}{\sqrt n} +\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H})^\prime}{\sqrt n}\frac{\bm\Lambda\bm F^\prime\bm\Xi}{nT}\frac{\bm\Lambda \widehat{\mathbf H}}{\sqrt n}\nonumber\\ &+\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{(\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H})^\prime}{\sqrt n}\frac{\bm\Xi^\prime\bm \Xi}{nT}\frac{\bm\Lambda \widehat{\mathbf H}}{\sqrt n} = \bm I_a+\bm I_b+\bm I_c+\bm I_d+\bm I_e+\bm I_f, \;\text{say.} \end{align} Then, because of (ref) and (ref) in the proof of Proposition (ref)(a), \begin{align} \mathbb{E}\left[\left\Vert \frac{\bm F^\prime\bm\Xi\bm\Lambda}{ nT}\right\Vert^2\right] =\frac 1{n^2T^2}\sum_{k=1}^r\sum_{h=1}^r \mathbb{E}\left[\left(\sum_{t=1}^T F_{kt}\sum_{j=1}^n \xi_{jt}\lambda_{jh}\right)^2\right] \le \frac{r^2M_FM_\Lambda^2M_\xi}{nT}. \end{align} Therefore, by Assumption (ref)(a), Lemma (ref)(iv), (ref)(i), (ref)(i), and using (ref) , we get \begin{align} \Vert \bm I_a\Vert \le \left\Vert \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert \bm{\mathcal H}\right\Vert\,\left\Vert\frac{\bm\Lambda^\prime\bm\Lambda}{n}\right\Vert\, \left\Vert\frac{\bm F^\prime\bm\Xi\bm\Lambda }{ nT}\right\Vert\,\left\Vert \widehat{\mathbf H}\right\Vert=O_{\mathrm P}\left(\frac 1{\sqrt {nT}}\right),\\ \Vert \bm I_b\Vert\le \left\Vert \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert \bm{\mathcal H}\right\Vert\,\left\Vert \frac{\bm\Lambda^\prime\bm\Xi^\prime\bm F}{nT}\right\Vert \,\left\Vert \frac{\bm\Lambda^\prime\bm\Lambda}n \right\Vert\,\left\Vert \widehat{\mathbf H}\right\Vert =O_{\mathrm P}\left(\frac 1{\sqrt {nT}}\right). \end{align} Moreover, because of Lemma (ref)(i), (ref)(iii), (ref)(iv), (ref)(i), and (ref)(i), \begin{align} \Vert \bm I_c\Vert \le \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert\bm{\mathcal H}\right\Vert\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert\, \left\Vert\frac{\bm\Lambda^\prime\bm\Xi^\prime\bm\Xi }{n^{3/2}T}\right\Vert\, \left\Vert\widehat{\mathbf H}\right\Vert=O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}\right)\right). \end{align} Similarly, because of Proposition (ref)(a), Lemma (ref)(i), (ref)(i), (ref)(iv), (ref)(iv), and (ref)(i), \begin{align} \Vert \bm I_d\Vert &\le \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert \frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert \,\left\Vert \frac{\bm \Xi^\prime\bm F\bm}{\sqrt nT}\right\Vert\, \left\Vert \frac{\bm\Lambda^\prime \bm\Lambda }{n} \right\Vert\, \left\Vert \widehat{\mathbf H}\right\Vert= O_{\mathrm P}\left(\max\left(\frac{1}{n\sqrt {T}},\frac 1{T}\right)\right), \\ \Vert \bm I_e\Vert &\le \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert \frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert \,\left\Vert \frac{\bm \Xi^\prime\bm F\bm}{\sqrt nT}\right\Vert\, \left\Vert \frac{\bm\Lambda }{\sqrt n} \right\Vert^2\, \left\Vert \widehat{\mathbf H}\right\Vert= O_{\mathrm P}\left(\max\left(\frac{1}{n\sqrt {T}},\frac 1{T}\right)\right), \end{align} and, because of Proposition (ref)(a), Lemma (ref)(i), (ref)(iii), (ref)(iv), and (ref)(i), \begin{align} \Vert \bm I_f\Vert \le \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\,\left\Vert \frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert \,\left\Vert \frac{\bm \Xi^\prime \bm\Xi}{nT}\right\Vert\, \left\Vert \frac{\bm\Lambda }{\sqrt n} \right\Vert\, \left\Vert \widehat{\mathbf H}\right\Vert = O_{\mathrm P}\left(\max\left(\frac 1{n^2},\frac{1}{n\sqrt {T}},\frac 1{T}\right)\right). \end{align} By using (ref), (ref), (ref), (ref), (ref), and (ref) into (ref) \begin{equation} \Vert\bm{I}\Vert = O_{\mathrm P}\left(\max\left(\frac 1 {n},\frac 1{\sqrt{nT}},\frac 1T\right)\right). \end{equation} Second, consider $\bm {II}$ in (ref). From Proposition (ref)(b), \begin{equation} \Vert\bm{II}\Vert \le \frac 1 {n}\left\Vert \widehat{\bm\Lambda}-{\bm\Lambda}\widehat{\mathbf H}\right\Vert^2 = O_{\mathrm P}\left(\max\left(\frac{1}{n^2},\frac 1{T}\right)\right). \end{equation} And, by using (ref) and (ref) in (ref), because of Lemma (ref)(ii), and Lemma (ref)(i), we complete the proof. \end{proof} \begin{prop} Under Assumptions (ref) through (ref), and Assumption (ref), if $\sqrt T/n\to0$ and $\sqrt n/T\to 0$, as $n,T\to\infty$, $\min(\sqrt{n},\sqrt T)\Vert \widehat{\mathbf H}-\bm J\Vert = o_{\mathrm P}(1)$; where $\bm J$ is a diagonal $r\times r$ matrix with entries $\pm 1$. \end{prop} \begin{proof} From Proposition (ref) and by using (ref) \begin{equation} \frac{\widehat{\bm\Lambda}^\prime\bm\Lambda\widehat{\mathbf H}}{n}=\frac{\widehat{\bm\Lambda}^\prime(\bm\Lambda\widehat{\mathbf H}-\widehat{\bm\Lambda}+\widehat{\bm\Lambda})}{n}=\frac{\widehat{\mathbf M}^x}{n}+O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} Or, equivalently, \begin{equation} \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}=\widehat{\mathbf H}^{-1}+O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} Moreover, by Assumption (ref) and using (ref) in (ref) we have \begin{equation} \widehat{\mathbf H}^\prime = \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}=\widehat{\mathbf H}^{-1}+O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} Therefore, because of (ref), as $n,T\to\infty$, $\widehat{\mathbf H}$ is an $r\times r$ orthogonal matrix thus it has eigenvalues $\pm 1$. Moreover, because of Proposition (ref) \begin{equation} \frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}= \frac{(\widehat{\bm\Lambda}-\bm\Lambda\widehat{\mathbf H}+\bm\Lambda\widehat{\mathbf H})^\prime\bm\Lambda}{n} =\frac{\widehat{\mathbf H}^\prime\bm\Lambda^\prime\bm\Lambda}{n}+ O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} Thus, from (ref) and (ref) and by Lemma (ref)(iv), \begin{equation} \widehat{\mathbf H}^\prime = \left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{\widehat{\bm\Lambda}^\prime\bm\Lambda}{n}=\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\frac{\widehat{\mathbf H}^\prime\bm\Lambda^\prime\bm\Lambda}{n}+ O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} And from (ref) it follows that \begin{equation} \left(\frac{\widehat{\mathbf M}^x}{n}\right)\widehat{\mathbf H}^\prime = \widehat{\mathbf H}^\prime\frac{\bm\Lambda^\prime\bm\Lambda}{n}+ O_{\mathrm P}\left(\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)\right). \end{equation} So, because of (ref), as $n,T\to\infty$, the columns of $\widehat{\mathbf H}$ are the eigenvectors of $\frac{\bm\Lambda^\prime\bm\Lambda}{n}$ with eigenvalues $\frac{\widehat{\mathbf M}^x}{n}$. The eigenvectors are normalized since $\widehat{\mathbf H}$ is orthogonal, as $n,T\to\infty$. Moreover, under Assumption (ref), $\frac{\bm\Lambda^\prime\bm\Lambda}{n}$ is diagonal, so, as $n,T\to\infty$, $\widehat{\mathbf H}$ must be diagonal with eigenvalues $\pm 1$. By noticing that, if $\sqrt T/n\to0$ and $\sqrt n/T\to 0$, then $\max\left(\frac 1n,\frac 1{\sqrt {nT}}, \frac 1T\right)=o_{\mathrm P}\left(\max\left(\frac 1{\sqrt {n}}, \frac 1{\sqrt T}\right)\right)$, we complete the proof. \end{proof} \setcounter{equation}{0} \section{Auxiliary lemmata} \begin{lem} Under Assumptions (ref) and (ref): \begin{compactenum} • for all $n\in\mathbb N$ and $T\in\mathbb N$, $\frac 1{nT}\sum_{i,j=1}^n\sum_{t,s=1}^T \vert\mathbb{E}_{}[\xi_{it}\xi_{js}]\vert \le M_{1\xi}$, for some finite positive real $M_{1\xi}$ independent of $n$ and $T$; • for all $n\in\mathbb N$ and $t\in\mathbb Z$, $\frac 1{n}\sum_{i,j=1}^n \vert\mathbb{E}_{}[\xi_{it}\xi_{jt}]\vert \le M_{2\xi}$, for some finite positive real $M_{2\xi}$ independent of $n$ and $t$; • for all $i\in\mathbb N$ and $T\in\mathbb N$, $\!\frac 1{T}\sum_{t,s=1}^T \vert\mathbb{E}_{}[\xi_{it}\xi_{is}]\vert \le M_{3\xi}$, $\!$for some finite positive real $M_{3\xi}$ independent of $i$ and $T$; • for all $j=1,\ldots,r$, $\underline C_j\!\le \lim\inf_{n\to\infty} \frac{ \mu_{j}^\chi}n\le\lim\sup_{n\to\infty} \frac{\mu_{j}^\chi}n\le\! \overline C_j$, $\!\!$ for some finite positive reals $\underline C_j$ and $\overline C_j$; • for all $n\in\mathbb N$, $\mu_{1}^\xi \le M_{2\xi}$, where $M_\xi$ is defined in part (ii); • for all $j=1,\ldots,r$, $\underline C_j\le \lim\inf_{n\to\infty} \frac{ \mu_{j}^x}n\le\lim\sup_{n\to\infty} \frac{\mu_{j}^x}n\le \overline C_j$, and for all $n\in\mathbb N$, $\mu_{r+1}^x \le M_\xi$, where $M_\xi$ is defined in Assumption (ref)(b). \end{compactenum} \end{lem} \begin{proof} Using Assumptions (ref)(a) and (ref)(b), we have: \begin{align} \frac 1{nT}\sum_{i,j=1}^n\sum_{t,s=1}^T \vert\mathbb{E}_[\xi_{it}\xi_{js}]\vert &=\frac 1{n}\sum_{i,j=1}^n\sum_{k=-(T-1)}^{T-1} \left(1-\frac{\vert k\vert}{T}\right) \vert\mathbb{E}_[\xi_{it}\xi_{j,t-k}]\vert\nonumber\\ &\le \frac 1n\sum_{i=1}^n \sum_{k=-\infty}^{\infty} \rho^{|k|}\sigma_i^2 + \max_{j=1,\ldots, n}\sum_{j=1, j\ne i}^n\sum_{k=-\infty}^{\infty} \rho^{\vert k\vert} M_{ij}\nonumber\\ &\le \frac{C_\xi^\prime(1+\rho)}{1-\rho}+\frac{M_\xi(1+\rho)}{1-\rho}. \nonumber \end{align} Similarly, \begin{align} \frac 1{n}\sum_{i,j=1}^n \vert\mathbb{E}_[\xi_{it}\xi_{jt}]\vert &\le \frac 1n\sum_{i=1}^n \sigma_i^2 + \max_{j=1,\ldots,n}\sum_{j=1, j\ne i}^n M_{ij}\le C_\xi^\prime+ M_\xi,\nonumber \end{align} and \begin{align} \frac 1{T}\sum_{t,s=1}^T \vert\mathbb{E}_[\xi_{it}\xi_{is}]\vert &= \sum_{k=-(T-1)}^{T-1} \left(1-\frac{\vert k\vert}{T}\right) \vert\mathbb{E}_[\xi_{it}\xi_{i,t-k}]\vert\le \sum_{k=-\infty}^{\infty} \rho^{\vert k\vert} \sigma_i^2 \le \frac{C_\xi^\prime (1+\rho)}{1-\rho}.\nonumber \end{align} Defining, $M_{1\xi}=\frac{(C_\xi^\prime+M_\xi)(1+\rho)}{1-\rho}$ and $M_{2\xi}=C_\xi^\prime+M_\xi$, and $M_{3\xi}=\frac{C_\xi^\prime(1+\rho)}{1-\rho}$, we prove parts (i), (ii), and (iii). For part (iv), by MK04, for all $j=1,\ldots, r$, we have \begin{equation} \frac{\mu_r(\bm\Lambda^\prime\bm\Lambda)}n\mu_j(\bm\Gamma^F) \le \frac{\mu_{j}^\chi}n \le \frac{\mu_j(\bm\Lambda^\prime\bm\Lambda)}n\ \mu_1(\bm\Gamma^F). \end{equation} The proof then follows from Assumption (ref)(a) which, by continuity of eigenvalues, implies that, for any $j=1,\ldots, r$, as $n\to\infty$ \[ \lim_{n\to\infty}\frac{\mu_j(\bm\Lambda^\prime\bm\Lambda)}n = \mu_j(\bm\Sigma_\Lambda). \] with $$ 0<m_\Lambda^2\le \mu_r(\bm\Sigma_\Lambda)\le\mu_1(\bm\Sigma_\Lambda)\le M_\Lambda^2<\infty ,$$ and by Assumption (ref)(b) and Assumption (ref)(c) which imply $$ 0<m_F\le \mu_r(\bm\Gamma^F)\le\mu_1(\bm\Gamma^F)\le M_F<\infty. $$ For part (v), by Assumption (ref)(b): \begin{align} \Vert\bm\Gamma^\xi\Vert\le \max_{i=1,\ldots,n}\sum_{j=1}^n \vert \mathbb{E}[\xi_{it}\xi_{jt}]\vert \le \max_{i=1,\ldots, n}\sigma_i^2+ \max_{i=1,\ldots,n}\sum_{j=1,j\ne i}^n M_{ij}\le C_\xi^\prime + M_{\xi}.\nonumber \end{align} Part (vi) follows from parts (iv) and (v) and Weyl's inequality. This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), for all $t=1,\ldots, T$ and all $n,T\in\mathbb N$ \begin{compactenum}[(i)] • $\Vert\frac{\bm\Lambda}{\sqrt n}\Vert=O(1)$; • $\Vert \mathbf F_t\Vert = O_{\mathrm {ms}}(1)$ and $\Vert\frac{\bm F}{\sqrt T}\Vert=O_{\mathrm {ms}}(1)$. \end{compactenum} \end{lem} \begin{proof} By Assumption (ref)(a), which holds for all $n\in\mathbb N$, \begin{equation} \sup_{n\in\mathbb N}\left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^2\le \sup_{n\in\mathbb N}\left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^2_F =\sup_{n\in\mathbb N}\frac 1n\sum_{j=1}^r\sum_{i=1}^n \lambda_{ij}^2 \le\sup_{n\in\mathbb N} \max_{i=1,\ldots, n} \Vert\bm\lambda_i\Vert^2\le M_\Lambda^2,\nonumber \end{equation} since $M_\Lambda$ is independent of $i$. This proves part (i). By Assumption (ref)(b), which holds for all $T\in\mathbb N$ because of stationarity (see also part (i) of Lemma (ref)), \begin{align} \sup_{T\in\mathbb N}\max_{t=1,\ldots ,T}\mathbb{E}[\Vert\mathbf F_t\Vert^2]&= \sup_{T\in\mathbb N} \max_{t=1,\ldots ,T}\sum_{j=1}^r \mathbb{E}[F_{jt}^2]\le r\sup_{T\in\mathbb N}\max_{t=1,\ldots ,T}\max_{j=1,\ldots,r} \mathbb{E}[F_{jt}^2] \nonumber\\ &\le r\sup_{T\in\mathbb N}\max_{t=1,\ldots ,T}\max_{j=1,\ldots,r} \bm\eta_j^\prime\bm\Gamma^F\bm\eta_j\le r \Vert\bm\Gamma^F\Vert\le r M_F, \end{align} since $M_F$ is independent of $t$ and where $\bm\eta_j$ is an $r$-dimensional vector with one in the $j$th entry and zero elsewhere. Therefore, from (ref): \begin{align} \sup_{T\in\mathbb N}\mathbb{E}\left[\left\Vert\frac{\bm F}{\sqrt T}\right\Vert^2\right]\le\sup_{T\in\mathbb N} \mathbb{E}\left[\left\Vert\frac{\bm F}{\sqrt T}\right\Vert^2_F\right] =\sup_{T\in\mathbb N}\frac 1T\sum_{j=1}^r\sum_{t=1}^T \mathbb{E}[[\bm F]_{tj}^2] \le\sup_{T\in\mathbb N} \max_{t=1,\ldots, T} \mathbb{E}[\Vert\mathbf F_t\Vert^2] \le r M_F.\nonumber \end{align} This proves part (ii) and it completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), for all $n,T\in\mathbb N$ $\sqrt T\left\Vert\frac{\bm F^\prime\bm \Xi}{\sqrt n T} \right\Vert = O_{\mathrm{ms}}(1)$. \end{lem} \begin{proof} By Assumption (ref), Lemma (ref)(iii) and Cauchy-Schwarz inequality \begin{align} \mathbb{E}\left[ \left\Vert\frac {\bm F^\prime\bm \Xi}{\sqrt n T} \right\Vert^2\right] &=\mathbb{E}\left[ \left\Vert\frac 1{\sqrt nT}\sum_{t=1}^T \mathbf F_t\bm\xi_t^\prime \right\Vert^2\right]\le \mathbb{E}\left[ \left\Vert\frac 1{\sqrt nT}\sum_{t=1}^T \mathbf F_t\bm\xi_t^\prime \right\Vert_F^2\right]\nonumber\\ &=\frac 1{nT^2}\sum_{j=1}^r\sum_{i=1}^n \mathbb{E}\left[\left(\sum_{t=1}^T F_{jt}\xi_{it}\right)^2\right]\nonumber\\ &\le\frac r{T^2}\max_{j=1\ldots, r} \max_{i=1\ldots, n} \sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[\xi_{it}F_{jt}\xi_{is}F_{js}]\nonumber\\ &=\frac r{T^2}\max_{j=1\ldots, r} \max_{i=1\ldots, n} \sum_{t=1}^T\sum_{s=1}^T \mathbb{E}[F_{jt}F_{js}]\, \mathbb{E}[\xi_{it}\xi_{is}]\nonumber\\ &\le \left\{\frac{r}{T}\max_{j=1,\ldots, r}\max_{t,s=1,\ldots, n} \vert\mathbb{E}[F_{jt}F_{js}]\vert\right\} \left\{\frac 1T \sum_{t=1}^T\sum_{s=1}^T\left\vert\mathbb{E}[\xi_{it}\xi_{is}]\right\vert\right\}\nonumber\\ &\le \frac r T \max_{j=1,\ldots, r}\max_{t,s=1,\ldots, n} \mathbb{E}[F_{jt}^2] \frac{M_\xi(1+\rho)}{1-\rho}\le\frac{rM_F M_{3\xi}}T,\nonumber \end{align} since $M_F$ is independent of $t$ and $s$ and $M_{3\xi}$ is independent of $i$. This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), for all $n,T\in\mathbb N$ \begin{compactenum}[(i)] • $\min(n,\sqrt T)\left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT} \right\Vert = O_{\mathrm{ms}}(1)$; • $\min(n,\sqrt {nT})\left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T} \right\Vert = O_{\mathrm{ms}}(1)$; • $\left\Vert\frac{\bm \Lambda^\prime\bm \Lambda}{n} \right\Vert = O(1)$. \end{compactenum} \end{lem} \begin{proof} For part (i), first notice that, by Lemmas (ref)(v) and (ref)(ii) \begin{equation} \left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT} \right\Vert \le \left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT}-\frac{\bm\Gamma^\xi}{n} \right\Vert +\left\Vert\frac{\bm\Gamma^\xi}{n}\right\Vert = \left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT}-\frac{\bm\Gamma^\xi}{n} \right\Vert +\frac{\mu_1^\xi}{n}=O_{\mathrm P}\left(\frac 1{\sqrt T}\right)+O\left(\frac 1{n}\right). \end{equation} Similarly, for part (ii) by Lemmas (ref)(i) and (ref)(v) \begin{equation} \left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T} \right\Vert \le \left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T}-\frac{\bm\Lambda^\prime\bm\Gamma^\xi}{n^{3/2}} \right\Vert +\left\Vert \frac{\bm\Lambda}{\sqrt n}\right\Vert\,\left\Vert\frac{\bm\Gamma^\xi}{n}\right\Vert= \left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T}-\frac{\bm\Lambda^\prime\bm\Gamma^\xi}{n^{3/2}} \right\Vert +O\left(\frac 1{n}\right). \end{equation} Then, because of Assumption (ref)(c), \begin{align} \mathbb{E}\left[\left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T}-\frac{\bm\Lambda^\prime\bm\Gamma^\xi}{n^{3/2}} \right\Vert ^2\right]&\le \mathbb{E}\left[\left\Vert\frac{\bm\Lambda^\prime\bm \Xi^\prime\bm \Xi}{n^{3/2}T}-\frac{\bm\Lambda^\prime\bm\Gamma^\xi}{n^{3/2}} \right\Vert ^2_F\right]\nonumber\\ &= \sum_{k=1}^r\sum_{j=1}^n\mathbb{E}\left[\left\vert\frac 1{n^{3/2}T} \sum_{i=1}^n \sum_{t=1}^T \left\{\lambda_{ik}\xi_{it}\xi_{jt}-\lambda_{ik}\mathbb{E}[\xi_{it}\xi_{jt}]\right\} \right\vert^2 \right]\nonumber\\ &\le \frac{rM_\Lambda n}{n^2T} \max_{j=1,\ldots, n} \mathbb{E}\left[\left\vert\frac 1{\sqrt{nT}} \sum_{i=1}^n \sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\mathbb{E}[\xi_{it}\xi_{jt}]\right\} \right\vert^2 \right]\le \frac{rM_\Lambda K_\xi}{nT}, \end{align} since $K_\xi$ is independent of $j$. By using (ref) into (ref), we prove part (ii). Part (iii) follows from Lemma (ref)(i) since \[ \left\Vert\frac{\bm \Lambda^\prime\bm \Lambda}{n} \right\Vert\le \left\Vert\frac{\bm \Lambda}{\sqrt n} \right\Vert^2\le M_\Lambda^2. \] This completes the proof. \end{proof} \begin{lem} $\,$ \begin{compactenum}[(i)] • Under Assumption (ref), for all $T\in\mathbb N$, $\sqrt T\left\Vert\frac{\bm F^\prime\bm F}{T} -\bm\Gamma^F\right\Vert = O_{\mathrm{ms}}(1)$; • Under Assumption (ref), for all $n,T\in\mathbb N$, $\sqrt T \left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT}-\frac{\bm\Gamma^\xi}{n} \right\Vert= O_{\mathrm{ms}}(1)$. \end{compactenum} \end{lem} \begin{proof} Part (i) is direct consequence of Assumption (ref)(c-ii), since \begin{align} \mathbb{E}\left[\left\Vert\frac{\bm F^\prime\bm F}{T}-\bm\Gamma^F\right\Vert^2\right]&= \mathbb{E}\left[\left\Vert \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right\Vert^2\right]\le \mathbb{E}\left[\left\Vert \frac 1T\sum_{t=1}^T \left\{\mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right\}\right\Vert^2_F\right]\nonumber\\ &=\frac 1{T^2} \sum_{j=1}^r\sum_{k=1}^r\mathbb{E}\left[\left(\sum_{t=1}^T \left\{F_{jt}F_{kt}-\mathbb{E}[F_{jt}F_{kt}] \right\}\right)^2\right] \le \frac {r^2C_F}{T},\nonumber \end{align} since $C_F$ is independent of $j$ and $k$. For part (ii), by Assumption (ref)(c), letting $\gamma_{ij}^\xi$ be the $(i,j)$the entry of $\bm\Gamma^\xi$, we have \begin{align} \mathbb{E}\left[\left\Vert\frac{\bm \Xi^\prime\bm \Xi}{nT}-\frac{\bm\Gamma^\xi}{n} \right\Vert ^2\right]&=\mathbb{E}\left[\left\Vert \frac 1{nT} \sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime -\frac{\bm\Gamma^\xi}n \right\Vert^2\right]\le \mathbb{E}\left[\left\Vert \frac 1{nT} \sum_{t=1}^T\left\{\bm\xi_t\bm\xi_t^\prime -\bm\Gamma^\xi\right\} \right\Vert^2_F\right]\nonumber\\ &=\frac 1{n^2T^2}\sum_{i,j=1}^n\mathbb{E}\left[\left(\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\gamma_{ij}^\xi\right\}\right)^2\right]\nonumber\\ &\le \max_{i,j=1,\ldots, n}\frac 1{T^2} \mathbb{E}\left[\left(\sum_{t=1}^T \left\{\xi_{it}\xi_{jt}-\gamma_{ij}^\xi\right\}\right)^2\right] \le \frac {K_\xi}{T},\nonumber \end{align} since $K_\xi$ is independent of $i$ and $j$. This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), for all $n,T\in\mathbb N$ \begin{compactenum} • $\frac {\sqrt T} n\Vert \widehat{\bm\Gamma}^x-\bm\Gamma^x\Vert = O_{\mathrm {ms}}(1)$; • $\frac {\min(n,\sqrt T)} n\Vert \widehat{\bm\Gamma}^x-\bm\Gamma^\chi\Vert = O_{\mathrm {ms}}(1)$; • $\frac {\min(n,\sqrt T)} n \Vert \widehat{\mathbf M}^x-\mathbf M^\chi\Vert=O_{\mathrm {ms}}(1)$; • as $n\to\infty$, ${\min(n,\sqrt T)} \Vert\widehat{\mathbf V}^x-\mathbf V^\chi\mathbf J\Vert = O_{\mathrm {ms}}(1)$, where $\mathbf J$ is $r\times r$ diagonal with entries $\pm 1$. \end{compactenum} \end{lem} \begin{proof} For part (i), from Lemma (ref)(i), (ref)(ii) and (ref)(i): \begin{align} \mathbb{E}\left[\left\Vert\frac 1n\left( \widehat{\bm\Gamma}^x-\bm\Gamma^x\right)\right\Vert^2\right] & = \mathbb{E}\left[\left\Vert \frac 1n\left\{ \bm\Lambda\left( \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right)\bm\Lambda^\prime + \frac 1T\sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime-\bm\Gamma^\xi\right\}\right\Vert^2 \right]\nonumber\\ & \le \mathbb{E}\left[\left\Vert \frac 1n\left\{ \bm\Lambda\left( \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right)\bm\Lambda^\prime \right\}\right\Vert^2\right] +\mathbb{E}\left[\left\Vert \frac 1n\left( \frac 1T\sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime-\bm\Gamma^\xi\right)\right\Vert^2 \right]\nonumber\\ &\le \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^4 \, \mathbb{E}\left[\left\Vert \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right\Vert^2\right]+\mathbb{E}\left[\left\Vert \frac 1n\left( \frac 1T\sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime-\bm\Gamma^\xi\right)\right\Vert^2 \right]\nonumber\\ &\le \frac{M_\Lambda^4 r^2 C_F+K_\xi}{T}.\nonumber \end{align} This proves part (i). Part (ii), follows from \begin{align} \mathbb{E}\left[\left\Vert\frac 1n\left( \widehat{\bm\Gamma}^x-\bm\Gamma^\chi\right)\right\Vert^2\right] &\le \mathbb{E}\left[\left\Vert\frac 1n\left( \widehat{\bm\Gamma}^x-\bm\Gamma^x\right)\right\Vert^2\right] +\left\Vert\frac {\bm\Gamma^\xi}n\right\Vert^2 \le \frac{M_\Lambda^4K_F+K_\xi}{T}+ \frac{\mu_1^\xi}{n^2}\le \frac{M_\Lambda^4K_F+K_\xi}{T}+\frac{M_{2\xi}}{n^2}\nonumber\\ &\le M_1\max \left(\frac 1{n^2},\frac 1T\right),\;\text{say,}\nonumber \end{align} because of part (i) and Lemma (ref)(v) and where $M_1$ is a finite positive real independent of $n$ and $T$. For part (iii), for any $j=1,\ldots, r$, because of Weyl's inequality and part (ii), it holds that \begin{equation} \vert\widehat\mu_j^x-\mu_j^\chi \vert\le \mu_1(\widehat{\bm\Gamma}^x-\bm\Gamma^\chi) =\left\Vert\widehat{\bm\Gamma}^x-\bm\Gamma^\chi\right\Vert. \end{equation} Hence, from (ref) and part (ii), \begin{align} \mathbb{E}\left[ \left\Vert\frac 1n\left( \widehat{\mathbf M}^x-\mathbf M^\chi\right)\right\Vert^2\right] &\le \mathbb{E}\left[ \left\Vert\frac 1n\left( \widehat{\mathbf M}^x-\mathbf M^\chi\right)\right\Vert^2_F\right] = \frac 1{n^2} \sum_{j=1}^r \mathbb{E}\left[\left(\widehat\mu_j^x-\mu_j^\chi \right)^2\right]\nonumber\\ &\le \frac{r}{n^2} \mathbb{E}\left[\left(\widehat\mu_1^x-\mu_1^\chi \right)^2\right]\le r\mathbb{E}\left[\left\Vert\frac 1n\left(\widehat{\bm\Gamma}^x-\bm\Gamma^\chi\right)\right\Vert^2\right]\le \frac{rM_\Lambda^4K_F+K_\xi}{T}+\frac{rM_\xi}{n^2}\nonumber\\ &\le rM_1\max\left(\frac 1{n^2},\frac 1T\right). \end{align} This proves part (iii). For part (iv), because of yu15, which is a special case of Davis Kahn Theorem, there exists an $r\times r$ diagonal matrix $\mathbf J$ with entries $\pm 1$ such that \begin{equation} \Vert\widehat{\mathbf V}^x-\mathbf V^\chi\mathbf J\Vert\le \frac{2^{3/2}\sqrt r\left\Vert \widehat{\bm\Gamma}^x-\bm\Gamma^\chi\right\Vert }{\min(\vert \mu_{0}^\chi-\mu_{1}^\chi \vert,\vert \mu_{r}^\chi-\mu_{r+1}^\chi \vert)}, \end{equation} where $\mu_0^\chi=\infty$. This holds provided the eigenvalues $\mu_j^\chi$ are distinct as required by Assumption (ref). Therefore, from (ref), part (ii) and Lemma (ref)(iv), and since $\mu_{r+1}^\chi =0$, as $n\to\infty$ \begin{align} \min(n^2,T) \mathbb{E}\left[\Vert\widehat{\mathbf V}^x-\mathbf V^\chi\mathbf J\Vert^2\right]&\le \frac {\min(n^2,T) 2^3 \frac r{n^2} \mathbb{E}\left[\left\Vert\widehat{\bm\Gamma}^x-\bm\Gamma^\chi \right\Vert^2\right]} {\frac 1{n^2}\left\{\min(\vert \mu_{0}^\chi-\mu_{1}^\chi \vert,\vert \mu_{r}^\chi-\mu_{r+1}^\chi \vert)\right\}^2} \le \frac{8rM_1}{\underline C_r^2}= M_2,\;\text{say,}\nonumber \end{align} where $M_2$ is a finite positive real independent of $n$ and $T$. This proves part (iv). This completes the proof. \end{proof} \begin{lem} Let $\bm\varepsilon_i$ be the $n$-dimensional vector with one in entry $i$ and zero elsewhere. Under Assumptions (ref) through (ref), for all $i=1,\ldots, n$ and $T\in\mathbb N$ and as $n\to\infty$ \begin{compactenum}[(i)] • $\frac {\min(\sqrt n,\sqrt T)}{\sqrt n} \Vert\bm\varepsilon^\prime_i (\widehat{\bm\Gamma}^x-\bm\Gamma^\chi)\Vert = O_{\mathrm {ms}}(1)$; • $\sqrt n\Vert\mathbf v_i^\chi\Vert = O(1)$; • ${\min(\sqrt n,\sqrt T)} \sqrt n \Vert\widehat{\mathbf v}^{x\prime}_i-\mathbf v^{\chi\prime}\mathbf J\Vert = O_{\mathrm {ms}}(1)$. \end{compactenum} \end{lem} \begin{proof} First notice that \begin{align} \max_{i=1,\ldots, n}\mathbb{E}\left[\left\Vert \frac 1{\sqrt n}\bm\varepsilon^\prime_i\left( \sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime -\bm\Gamma^\xi\right) \right\Vert^2\right] &\le\max_{i=1,\ldots, n}\frac 1{n}\sum_{j=1}^n\mathbb{E}\left[\left(\frac 1T\sum_{t=1}^T \xi_{it}\xi_{jt}-[\bm\Gamma^\xi]_{ij}\right)^2\right]\nonumber\\ &\le \max_{i,j=1,\ldots, n} \mathbb{E}\left[\left(\frac 1T\sum_{t=1}^T \xi_{it}\xi_{jt}-[\bm\Gamma^\xi]_{ij}\right)^2\right] \le \frac {K_\xi}{T}, \end{align} since $K_\xi$ is independent of $i$ and $j$. Therefore, from Lemma (ref)(i) and Lemma (ref)(i), and using (ref), \begin{align} \max_{i=1,\ldots, n} \mathbb{E}\left[\left\Vert\frac 1{\sqrt n}\bm\varepsilon^\prime_i\left( \widehat{\bm\Gamma}^x-\bm\Gamma^x\right)\right\Vert^2\right] =&\, \max_{i=1,\ldots, n} \mathbb{E}\left[\left\Vert \frac 1{\sqrt n}\left\{ \bm\lambda_i^\prime \left( \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right)\bm\Lambda^\prime +\bm\varepsilon^\prime_i\left( \frac 1T\sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime-\bm\Gamma^\xi\right)\right\}\right\Vert^2 \right]\nonumber\\ \le&\,\max_{i=1,\ldots, n}\Vert \bm\lambda_i\Vert^2\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^2 \, \mathbb{E}\left[\left\Vert \frac 1T\sum_{t=1}^T \mathbf F_t\mathbf F_t^\prime-\bm\Gamma^F\right\Vert^2\right]\nonumber\\ &+\max_{i=1,\ldots, n}\mathbb{E}\left[\left\Vert \frac 1{\sqrt n}\bm\varepsilon^\prime_i\left( \frac 1T\sum_{t=1}^T\bm\xi_t\bm\xi_t^\prime-\bm\Gamma^\xi\right)\right\Vert^2 \right]\nonumber\\ \le&\, \frac{M_\Lambda^4 r^2 C_F+K_\xi}{T}, \end{align} since $M_\Lambda$ and $C_F$ are independent of $i$. Then, following the same arguments as Lemma (ref)(ii), because of (ref) and Lemma (ref)(v): \begin{align} \max_{i=1,\ldots, n}\mathbb{E}\left[\left\Vert\frac 1{\sqrt n}\bm\varepsilon^\prime_i\left( \widehat{\bm\Gamma}^x-\bm\Gamma^\chi\right)\right\Vert^2\right]& \le\max_{i=1,\ldots, n} \mathbb{E}\left[\left\Vert\frac 1{\sqrt n}\bm\varepsilon^\prime_i\left( \widehat{\bm\Gamma}^x-\bm\Gamma^x\right)\right\Vert^2\right] +\max_{i=1,\ldots, n}\left\Vert\bm\varepsilon^\prime_i\frac {\bm\Gamma^\xi}{\sqrt n}\right\Vert^2\nonumber\\ &\le \frac{M_\Lambda^4r^2C_F+K_\xi}{T}+\max_{i=1,\ldots, n} \Vert \bm\varepsilon_i\Vert^2\, \left\Vert\frac{\bm\Gamma^\xi}{\sqrt n}\right\Vert^2\nonumber\\ &=\frac{M_\Lambda^4r^2C_F+K_\xi}{T}+ \frac{\mu_1^\xi}{n} \le \frac{M_\Lambda^4r^2C_F+K_\xi}{T}+ \frac{M_{2\xi}}{n}\le M_1\max\left(\frac 1T,\frac 1n\right),\; \text{say,}\nonumber \end{align} since $\Vert \bm\varepsilon_i\Vert=1$ and where $M_1$ is a finite positive real independent of $n$ and $T$ defined in Lemma (ref)(ii). This proves part (i). For part (ii), notice that for all $i=1,\ldots, n$ we must have: \begin{equation} \mathbb{V}\mathrm{ar}(\chi_{it}) =\bm\lambda_i^\prime\bm\Gamma^F\bm\lambda_i\le \Vert\bm\lambda_i\Vert^2\,\left\Vert\bm\Gamma^F\right\Vert \le M_\Lambda^2 M_F, \end{equation} which is finite for all $i$ and $t$. So, since by Lemma (ref)(iv) \[ \liminf_{n\to\infty}\max_{i=1,\ldots, n}\mathbb{V}\mathrm{ar}(\chi_{it})=\liminf_{n\to\infty}\max_{i=1,\ldots, n}\sum_{j=1}^r \mu_j^\chi [\mathbf V^\chi]_{ij}^2\ge \liminf_{n\to\infty}\mu_r^\chi \max_{i=1,\ldots, n}\sum_{j=1}^r [\mathbf V^\chi]_{ij}^2 \ge \underline C_r n \max_{i=1,\ldots, n}\Vert\mathbf v_i^\chi\Vert^2, \] then, because of (ref), we must have, as $n\to\infty$, \[ \underline C_r n \max_{i=1,\ldots, n}\Vert\mathbf v_i^\chi\Vert^2\le M_\Lambda^2 M_F \] which implies that, as $n\to\infty$, \[ n \max_{i=1,\ldots, n}\Vert\mathbf v_i^\chi\Vert^2 \le M_V, \] for some finite positive real $M_V$ independent of $n$. This proves part (ii). Finally, using the same arguments in Lemma (ref)(iv), from (ref), part (i) and Lemma (ref)(iv), and since $\mu_0^\chi=\infty$ and $\mu_{r+1}^\chi =0$, as $n\to\infty$ \begin{align} \max_{i=1,\ldots, n} \min(n,T)\mathbb{E}\left[\Vert\sqrt n(\widehat{\mathbf v}_i^{x\prime}-\mathbf v_i^{\chi\prime}\mathbf J)\Vert^2\right] &=\max_{i=1,\ldots, n}\min(n,T)\mathbb{E}\left[\Vert\sqrt n\bm\varepsilon_i^\prime(\widehat{\mathbf V}^x-\mathbf V^\chi\mathbf J)\Vert^2\right]\nonumber\\ &\le\max_{i=1,\ldots, n}\ \frac {\min(n,T) 2^3 \frac r{n^2} n \mathbb{E}\left[\left\Vert\bm\varepsilon_i^\prime(\widehat{\bm\Gamma}^x-\bm\Gamma^\chi) \right\Vert^2\right]} {\frac 1{n^2}\left\{\min(\vert \mu_{0}^\chi-\mu_{1}^\chi \vert,\vert \mu_{r}^\chi-\mu_{r+1}^\chi \vert)\right\}^2}\nonumber\\ &=\max_{i=1,\ldots, n}\ \frac {\min(n,T) 2^3 \frac r{n} \mathbb{E}\left[\left\Vert\bm\varepsilon_i^\prime(\widehat{\bm\Gamma}^x-\bm\Gamma^\chi) \right\Vert^2\right]} {\frac 1{n^2}\left\{\min(\vert \mu_{0}^\chi-\mu_{1}^\chi \vert,\vert \mu_{r}^\chi-\mu_{r+1}^\chi \vert)\right\}^2}\nonumber\\ &\le \frac{8r M_1}{\underline C_r^2}= M_2,\nonumber \end{align} where $M_2$ is a finite positive real independent of $n$ and $T$ defined in Lemma (ref)(iv). This proves part (iii) and completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), for all $n,T\in\mathbb N$, \begin{compactenum}[(i)] • $\Vert\frac{{\mathbf M}^\chi}{n}\Vert =O(1)$; • $\Vert(\frac{{\mathbf M}^\chi}{n})^{-1}\Vert=O(1)$; • $\Vert\frac{\widehat{\mathbf M}^x}{n}\Vert=O_{\mathrm{P}}(1)$; • $\Vert(\frac{\widehat{\mathbf M}^x}{n})^{-1}\Vert=O_{\mathrm{P}}(1)$. \end{compactenum} \end{lem} \begin{proof} Parts (i) and (ii) follow directly from Lemma (ref)(iv), indeed, \[ \left\Vert\frac{{\mathbf M}^\chi}{n}\right\Vert =\frac{\mu_1^\chi}{n}\le \overline C_1, \] and \[ \left\Vert\left(\frac{{\mathbf M}^\chi}{n}\right)^{-1}\right\Vert = \frac n{\mu_r^\chi}\le \frac 1{\underline C_r}. \] Both statements hold for all $n\in\mathbb N$ since the eigenvalues are an increasing sequence in $n$. For part (iii), because of part (i) and Lemma (ref)(iii), \begin{align} \left\Vert\frac{\widehat{\mathbf M}^x}{n}\right\Vert\le \left\Vert\frac{{\mathbf M}^\chi}{n}\right\Vert+ \left\Vert\frac{\widehat{\mathbf M}^x}{n}-\frac{{\mathbf M}^\chi}{n}\right\Vert\le \overline C_1+O_{\mathrm P}\left(\max\left(\frac1{\sqrt n},\frac 1{\sqrt T}\right)\right).\nonumber \end{align} For part (iv) just notice that, because of Lemma (ref)(iii) and part (ii), then $\frac{\widehat{\mathbf M}^x}{n}$ is positive definite with probability tending to one as $n,T\to\infty$. This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), \begin{compactenum}[(i)] • $\Vert\frac{\mathbf M^\chi}{n}-\bm V_0 \Vert=o(1)$, as $n\to\infty$ and $\Vert \bm V_0\Vert = O(1)$; • $\Vert\frac{\widehat{\mathbf M}^x}{n}-\bm V_0 \Vert=o_{\mathrm {ms}}(1)$, as $n,T\to\infty$; • $\Vert(\frac{\mathbf M^\chi}{n})^{-1}-\bm V_0^{-1} \Vert=o(1)$, as $n\to\infty$ and $\Vert \bm V_0^{-1}\Vert = O(1)$ • $\Vert(\frac{\widehat{\mathbf M}^x}{n})^{-1}-\bm V_0^{-1} \Vert=o_{\mathrm {ms}}(1)$, as $n,T\to\infty$, \end{compactenum} where $\bm V_0$ is $r\times r$ diagonal with entries the eigenvalues of $\bm\Sigma_\Lambda\bm\Gamma^F$ sorted in descending order. \end{lem} \begin{proof} For part (i), first notice that the $r$ non-zero eigenvalues of $\frac{\bm\Gamma^\chi}{n}$ are also the $r$ eigenvalues of $(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime\bm\Lambda}{n}(\bm\Gamma^F)^{1/2}$ which in turn are also the entries of $\bm V_0$. The proof follows from continuity of eigenvalues and since, because of Assumption (ref)(a), as $n\to\infty$, \[ \left\Vert(\bm\Gamma^F)^{1/2}\frac{\bm\Lambda^\prime\bm\Lambda}{n}(\bm\Gamma^F)^{1/2} -(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2} \right\Vert=o(1).\nonumber \] Part (ii) is a consequence of part (i) and Lemma (ref)(iii). For part (iii) notice that \[ \left\Vert\left(\frac{\mathbf M^\chi}{n}\right)^{-1}-\bm V_0^{-1} \right\Vert \le\left \Vert\left(\frac{\mathbf M^\chi}{n}\right)^{-1}\right\Vert\, \left\Vert\frac{\mathbf M^\chi}{n}-\bm V_0 \right\Vert \, \left\Vert \bm V_0^{-1} \right\Vert, \] then the proof follows from part (i), Lemma (ref)(ii), and since $\bm V_0$ is positive definite since $\bm\Sigma_\Lambda$ and $\bm\Gamma^F$ are positive definite by Assumptions (ref)(a) and (ref)(b), respectively. This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), as $n\to\infty$, \begin{compactenum}[(i)] • $\Vert{\mathbf K}\Vert=O(1)$; • $\Vert{\mathbf K}^{-1}\Vert=O(1)$. \end{compactenum} \end{lem} \begin{proof} For part (a), from (ref) in the proof of Proposition (ref)(a) it follows that \begin{equation} \left\Vert\mathbf K-\mathbf J\bm\Upsilon_0\right\Vert =o(1), \end{equation} where $\bm\Upsilon_0$ is the $r\times r$ matrix of normalized eigenvectors of $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$ and $\mathbf J$ is a diagonal matrix with entries $\pm 1$. Part (i) follows from the fact that $\Vert \mathbf J\bm\Upsilon_0\Vert = O(1)$, since $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$ is finite. Likewise part (ii) follows from the fact that $\mathbf J$ is obviously positive definite and $\bm\Upsilon_0$ is also positive definite because the eigenvalues of $(\bm\Gamma^F)^{1/2}\bm\Sigma_\Lambda(\bm\Gamma^F)^{1/2}$ are distinct by Assumption (ref). This completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), as $n\to\infty$, $\Vert{\bm{\mathcal H}}\Vert=O(1)$. \end{lem} \begin{proof} From (ref) and (ref) in the proof of Proposition (ref) $\bm{\mathcal H}=(\bm\Gamma^F )^{1/2}\mathbf K\mathbf J$. Then, the proof follows immediately from Assumption (ref)(b), Lemma (ref)(i), and since $\mathbf J$ is obviously finite and positive definite. \end{proof} \begin{lem} Under Assumptions (ref) through (ref), as $n,T\to\infty$, $\Vert\widehat{\mathbf H}\Vert=O_{\mathrm {P}}(1)$. \end{lem} \begin{proof} From (ref), by Proposition (ref)(a), Lemma (ref)(v), (ref)(iv) (ref)(i), we have \begin{align} \left\Vert\widehat{\mathbf H}\right\Vert &\le \left\Vert\frac{\bm F^\prime\bm F}{T}\right\Vert\, \left\Vert\frac{\bm\Lambda^\prime\widehat{\bm\Lambda}}{n}\right\Vert\, \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\le \left\Vert\frac{\bm F}{\sqrt T}\right\Vert^2\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert\, \left\Vert\frac{\widehat{\bm\Lambda}}{\sqrt n}\right\Vert\,\left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\nonumber\\ &\le\left\Vert\frac{\bm F}{\sqrt T}\right\Vert^2\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert^2\, \left\Vert\bm{\mathcal H}\right\Vert\, \left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert+ \left\Vert\frac{\bm F}{\sqrt T}\right\Vert^2\, \left\Vert\frac{\bm\Lambda}{\sqrt n}\right\Vert\, \left\Vert\frac{\widehat{\bm\Lambda}-\bm\Lambda\bm{\mathcal H}}{\sqrt n}\right\Vert\,\left\Vert\left(\frac{\widehat{\mathbf M}^x}{n}\right)^{-1}\right\Vert\nonumber\\ &= O_{\mathrm P}(1)+ O_{\mathrm P}\left(\max\left(\frac 1{\sqrt n},\frac 1{\sqrt T}\right)\right),\nonumber \end{align} which completes the proof. \end{proof} \begin{lem} Under Assumptions (ref) through (ref) and letting $\bm\Sigma^\xi$ be the $n\times n$ diagonal matrix with diagonal entries the diagonal entries of $\bm\Gamma^\xi$, as $n\to\infty$, $$ n^{-1}\left\Vert\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\left\{{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1}{\bm\Lambda}\right\}-\mathbf I_r\right\Vert= O\left(1\right). $$ \end{lem} \begin{proof} First notice that for any two invertible matrices $\bm K$ and $\bm H$ we have \begin{align} (\bm H+ \bm K)^{-1}&= (\bm H + \bm K)^{-1}- \bm K^{-1} + \bm K^{-1}= (\bm H+ \bm K)^{-1}(\bm K - (\bm H + \bm K))\bm K^{-1}+ \bm K^{-1}\nonumber\\ &= (\bm H + \bm K)^{-1}(-\bm H)\bm K^{-1} + \bm K^{-1}= \bm K^{-1}- (\bm H + \bm K)^{-1}\bm H\bm K^{-1}. \end{align} Then, setting $\bm K=\bm \Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda $ and $\bm H=\mathbf I_r$ from (ref) we have \begin{align} \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}=\{\bm \Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda \}^{-1}-\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\{\bm \Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda \}^{-1} . \end{align} which implies \begin{equation} \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\{\bm \Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda \} = \mathbf I_r-\{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}. \end{equation} Then, by Weyl's inequality: \begin{equation} \nu_r(\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}) \ge 1+\nu_r({\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda})\ge \nu_r({\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}). \end{equation} From, (ref), we have \begin{align} \Vert \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\Vert&=\frac 1{\nu_r(\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda})}\le \frac 1{\nu_r({\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda})}. \end{align} Now, the $r$ eigenvalues of ${\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}$ are also the $r$ non-zero eigenvalues of $\bm \Lambda\bm \Lambda^\prime \bm ({\bm\Sigma}^\xi)^{-1}$, and the $r$ non-zero eigenvalues of $\bm \Lambda\bm \Lambda^\prime$ are the $r$ eigenvalues of $\bm \Lambda^\prime\bm \Lambda$. Therefore, because of MK04 and Proposition (ref)(a), and Lemma (ref)(iv): \begin{align} \nu_r({\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda})&\ge \nu_r (\bm \Lambda\bm \Lambda^\prime)\nu_r(({\bm\Sigma}^\xi)^{-1})=\frac {\nu_r (\bm \Lambda^\prime\bm \Lambda)}{\nu_1({\bm\Sigma}^\xi)} = \frac {\mu_r^\chi}{\max_{i=1,\ldots, n}\sigma_i^2}\ge \frac{n\underline C_r}{C_\xi}. \end{align} By substituting (ref) into (ref) we have \begin{equation} \Vert \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\Vert\le \frac{C_\xi}{n\underline C_r}. \end{equation} Therefore, from (ref) and (ref), \[ \left\Vert \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\{\bm \Lambda^\prime(\bm\Sigma^\xi)^{-1}\bm\Lambda \} - \mathbf I_r\right\Vert =\left\Vert \{\mathbf I_r+{\bm\Lambda}^\prime({\bm\Sigma}^\xi)^{-1} {\bm\Lambda}\}^{-1}\right\Vert\le \frac{C_\xi}{n\underline C_r}, \] which completes the proof. \end{proof} $\,$