# Testing Alpha in High Dimensional Linear Factor Pricing Models with
Dependent Observations

Huifang Ma
School of Statistics and Data Science, KLMDASR, LEBPS, and LPMC,
Nankai University

Long Feng
School of Statistics and Data Science, KLMDASR, LEBPS, and LPMC,
Nankai University

Zhaojun Wang
School of Statistics and Data Science, KLMDASR, LEBPS, and LPMC,
Nankai University

Jigang Bao
Tsinghua Shenzhen International Graduate School, Tsinghua University
Shenzhen Wukong Investment Management Co. Ltd

###### Abstract

In this study, we introduce three distinct testing methods for testing alpha
in high dimensional linear factor pricing model that deals with dependent
data. The first method is a sum-type test procedure, which exhibits high
performance when dealing with dense alternatives. The second method is a max-
type test procedure, which is particularly effective for sparse alternatives.
For a broader range of alternatives, we suggest a Cauchy combination test
procedure. This is predicated on the asymptotic independence of the sum-type
and max-type test statistics. Both simulation studies and practical data
application demonstrate the effectiveness of our proposed methods when
handling dependent observations.

Keywords: asymptotic independence, Cauchy combination test, high-dimensional
linear factor model, time series.

##  1 Introduction

This study focuses on the detection of alpha in Linear Factor Pricing Models
(LFPM), particularly when the quantity of securities significantly outweighs
the temporal dimension of the return series. LFPM is a cornerstone in the
field of finance. Inspired by the Arbitrage Pricing Theory (Ross, 1976), LFPM
elucidates the relationship between security returns and market factors. LFPM
manifests in various forms, such as the renowned single-factor model, namely
the Sharpe-Lintner Capital Asset Pricing Model (CAPM) (Sharpe, 1964; Lintner,
1965), the Fama-French three-factor model (Fama and French, 1993), and the
Fama-French five-factor model (Fama and French, 2015). Each factor in these
models typically carries substantial economic significance and pricing
capability.

When the number of securities $N$ is fixed, Gibbons et al. (1989) introduced
an exact multivariate F-test under the assumption of joint normality. This
classic GRS test procedure was later expanded by MacKinlay and Richardson
(1991), Zhou (1993), and Beaulieu et al. (2007). In today’s financial markets,
the number of securities often exceeds a thousand, which may surpass the
length of the considered time period. Therefore, it’s not appropriate to
assume a fixed number of securities. To address this, recent advancements have
focused on devising tests that accommodate situations where the number of
securities $N$ exceeds the time periods $T$. For instance, Pesaran and
Yamagata (2012, 2017, 2023) introduced a sum-type test statistic that replaces
the sample covariance matrix with an identity matrix in the traditional
F-test, performing well under dense alternatives. Lan et al. (2018) proposed a
random projection procedure which admits the covariance matrix of the
idiosyncratic term to be nonsparse. For sparse alternatives, Gungor and Luger
(2013) and Feng et al. (2022b) proposed max-type test statistic. Fan et al.
(2015) proposed a power enhancement procedure to strengthen the test power
under the sparse alternatives. Yu et al. (2023) utilized the thresholding
covariance estimator of Fan et al. (2013) to propose a novel Wald-type test
and proposed a new Cauchy combination test which combinates the Wald-type test
and max-type test. Xia and Zhang (2023) consider $L$-statistics for handling
the sparse alternatives. For heavy-tailed distributions, Liu et al. (2023)
proposed a robust spatial sign-based nonparametric test procedure. Zhao et al.
(2022) extended Liu et al. (2023)’s results by considering a class of weighted
spatial-sign tests and choosing an optimal test INST.

In addition, Some literatures considered some more general LFPMs. For example,
Gagliardini et al. (2016) and Ma et al. (2020) established two sum-type
testing procedures which allow the existence of time-variation coefficients.
Ma et al. (2023) introduce a maximum-type test that performs well in scenarios
where the alternative hypothesis is sparse and an adaptive test by combining
the maximum-type test and Ma et al. (2020)’s sum-type test. Zhao (2023) also
proposed a robust spatial-sign-based nonparametric test procedure for the
time-varying factor pricing model. In addressing observed non-traded factors,
a substantial body of literature exists that estimates the risk premiums
associated with these factors using a two-pass regression methodology. This
approach, pioneered by Fama and MacBeth (1973), is also featured in recent
works such as those by Pesaran and Smith (2021b) and Pesaran and Smith
(2021a). When it comes to unobserved factors, numerous methods have been
developed to estimate these factors and their corresponding loadings,
primarily through principal component analysis. For example, Giglio et al.
(2021) devised a multiple testing procedure that operates on the residuals
remaining after the influence of the estimated latent factors and loadings has
been removed.

Most existing studies do not take into account the dependence structure among
observations, instead requiring the assumption of identical and independent
distribution for random errors. Only the works of Ma et al. (2020) and Ma et
al. (2023) have extended their testing procedures under the assumption of
martingale difference. In the financial market, observations often exhibit
some form of dependence structure, particularly in daily and weekly data. As
demonstrated in real data applications, the residuals of the traditional
linear factor pricing model are time-dependent, making the assumption of
identical and independent distribution unsuitable for random error. In high-
dimensional settings, Ayyala et al. (2017) proposed a sum-type test procedure
for the $M-$dependent stationary Gaussian process, with corrections to the
proofs of Ayyala et al. (2017)’s main theorems provided by Cho et al. (2019).
For the max-type test procedure, significant efforts have been made towards
Gaussian approximation for high-dimensional time series, as seen in the works
of Zhang and Wu (2017), Zhang and Cheng (2018), and Chang et al. (2024). An
overview of these efforts can be found in Chernozhukov et al. (2023). To the
best of our knowledge, no literature exists that considers the high-
dimensional alpha test of linear factor pricing models with dependent
observations.

In this study, we initially introduce a sum-type test statistic, which is
derived by simply summing the squares of the estimated alphas for each
security. We compute the expectation and asymptotic variance of this sum-type
test statistic and affirm its asymptotic normality under the assumption of an
$MA(\infty)$ process. Notably, we relax the assumptions of normal distribution
and $M$-dependence as presented in Ayyala et al. (2017). Subsequently, we
provide a consistent estimator of the expectation and asymptotic variance and
propose a sum-type test procedure that performs exceptionally well under dense
alternatives, i.e., when many alphas of the securities are non-zero. In cases
where only a few alphas of the securities are non-zero, we propose a max-type
test statistic, which is the maximum of the square of estimated alphas divided
by its corresponding long-run variances. For dependent observations, we
demonstrate that the limit null distribution of the proposed max-type test
statistic also follows a Gumbel distribution and establish its consistency
under sparse alternatives.

It’s important to note that the underlying truth in real applications is
typically unknown, as the density or sparsity is contingent on the properties
of the securities involved. Recent literature has demonstrated that the
maximum-type test statistic is asymptotically independent of the sum-type test
statistic, leading to the proposal of a corresponding combination test in
numerous high-dimensional problems. These include high-dimensional mean
testing problems by Xu et al. (2016), He et al. (2021) and Feng et al. (2023),
cross-sectional independence test in high-dimensional panel data models by
Feng et al. (2022a), testing of high-dimensional covariance matrix by Yu et
al. (2022), high-dimensional change point inference by Wang and Feng (2023)
and testing alpha of high dimensional time-varying factor model by Ma et al.
(2023). All the aforementioned papers affirm the asymptotic independence
between the sum-type test statistic and the max-type test statistic, based on
the assumption of identical and independent distribution. However, this
assumption may not be directly applicable in our context. In this study, we
are the first to establish the asymptotic independence between the sum-type
test statistic and the max-type test statistic under the assumption of time-
dependence. We then introduce a Cauchy combination test procedure, which is
capable of handling both dense and sparse alternatives simultaneously.
Furthermore, we provide a power analysis of the proposed Cauchy combination
test procedure. The advantages of our proposed tests over existing methods are
demonstrated through extensive Monte Carlo experiments and an empirical
application.

In summary, there are three main contributions of this paper:

  * 1.

We have relaxed the assumption of Gaussian distribution and $M$-dependence as
presented in Ayyala et al. (2017), and have established the asymptotic
normality of the sum-type test statistic in the presence of dependent
observations.

  * 2.

For the max-type test, we have also established the limit null distribution
and its consistency under sparse alternatives, assuming time-dependent
observations.

  * 3.

We are the first to establish the asymptotic independence between the sum-type
test statistic and the max-type test statistic under the assumption of time-
dependence. We have proposed a Cauchy combination test procedure, which
performs very well under neither dense nor sparse alternatives.

The remainder of this paper is structured as follows. In Sections 2, we
present the sum-type test, the max-type test procedures and the Cauchy
combination test procedure. Section 3 showcases some simulation studies. An
empirical application is provided in Section 4. Section 5 contains some
discussions. All technical details are included in the Appendix.

##  2 Test Procedures

###  2.1 Sum-type test

Assume that there are a total of $N$ securities and each security has $T$
observations of its excess return that are regularly collected from different
time periods. Let $y_{it}$ be the excess return of security
$i\in\\{1,\cdots,N\\}$ at time $t\in\\{1,\cdots,T\\}$. Let
${\bm{f}}_{t}=(f_{t1},\cdots,f_{tp})^{\top}\in\mathcal{R}^{p}$ denote the
$p$-vector of observed factors at time $t$, which represents the excess return
over the excess return on the $p$ observed factors. In this paper, we assume
that the number of factors $p$ is fixed, which depends on the specific multi-
factor model studied.

Assume that security returns are governed by the following multi-factor model:

$\displaystyle
y_{it}=\alpha_{i}+\bm{\beta}_{i}^{\top}{\bm{f}}_{t}+\varepsilon_{it},~{}i\in\\{1,\cdots,N\\},~{}t\in\\{1,\cdots,T\\},$
(1)

where $\alpha_{i}$ is a scalar representing the security specific intercept,
$\bm{\beta}_{i}=(\beta_{i1},\cdots,\beta_{ip})^{\top}\in\mathcal{R}^{p}$ is a
vector of multiple regression betas of security $i$ with respect to the $p$
factors, and the $\varepsilon_{it}$’s are the corresponding idiosyncratic
error terms. Note that for the classic CAPM (Sharpe, 1964; Lintner, 1965), the
single common factor is the market risk, while for the Fama and French three-
factor model (Fama and French, 1993), the common factors are the market risk,
SMB and HML, where SMB and HML measure the historic excess returns of small-
cap stocks over big-cap stocks and of value stocks over growth stocks,
respectively.

Stacking by cross-sectional observations, we can write the above multi-factor
model as

$\displaystyle{\bf y}_{t}=\bm{\alpha}+{\bf
B}{\bm{f}}_{t}+\bm{\varepsilon}_{t},~{}t\in\\{1,\cdots,T\\},$ (2)

where ${\bf y}_{t}=(y_{1t},\cdots,y_{Nt})^{\top}\in\mathcal{R}^{N}$,
$\bm{\alpha}=(\alpha_{1},\cdots,\alpha_{N})^{\top}\in\mathcal{R}^{N}$, ${\bf
B}=(\bm{\beta}_{1},\cdots,\bm{\beta}_{N})^{\top}\in\mathcal{R}^{N\times p}$
and
$\bm{\varepsilon}_{t}=(\varepsilon_{1t},\cdots,\varepsilon_{Nt})^{\top}\in\mathcal{R}^{N}$.
In addition, there are some notation to be used in later sections: ${\bf
y}_{i\cdot}=(y_{i1},\cdots,y_{iT})^{\top}\in\mathcal{R}^{T}$,
$\bm{\varepsilon}_{i\cdot}=(\varepsilon_{i1},\cdots,\varepsilon_{iT})^{\top}\in\mathcal{R}^{T}$,
for each $i\in\\{1,\cdots,N\\}$; ${\bf Y}=({\bf y}_{1},\cdots,{\bf
y}_{N})\in\mathcal{R}^{T\times N}$,
$\bm{\varepsilon}=(\bm{\varepsilon}_{1},\cdots,\bm{\varepsilon}_{N})\in\mathcal{R}^{T\times
N}$, ${\bf F}=({\bm{f}}_{1},\cdots,{\bm{f}}_{T})^{\top}\in\mathcal{R}^{T\times
p}$.

To evaluate the marginal return associated with an additional strategy that is
not explained by existing factors, many researchers employ the specification
test for a factor model by testing

$\displaystyle
H_{0}:\bm{\alpha}=\mathbf{0}~{}~{}\text{versus}~{}~{}H_{1}:\bm{\alpha}\not=\mathbf{0}.$
(3)

This test has a special explanation in the seminal single-factor model CAPM.
Specifically, if a single factor or a particular portfolio is mean-variance
efficient, i.e. it minimizes variance for a given level of expected return,
then $E(y_{it})=\bm{\beta}_{i}^{\top}E({\bm{f}}_{t})$ must be satisfied with
$\bm{\beta}_{i}=\mathrm{cov}(y_{it},{\bm{f}}_{t})\mathrm{var}^{-1}({\bm{f}}_{t})\in\mathcal{R}^{p}$
standing for the systematic risk on security $i$, which is equivalent to
$\bm{\alpha}=\mathbf{0}$ under the distributional assumption given by (1).
Thus, in CAPM the above test in (3) is actually a test of the mean-variance
efficiency.

Let $\bm{1}_{T}=(1,\cdots,1)^{\top}\in\mathcal{R}^{T}$, ${\bf I}_{T}$ be the
$T\times T$ identity matrix and ${\bf M}_{{\bf F}}={\bf I}_{T}-{\bf F}({\bf
F}^{\top}{\bf F})^{-1}{\bf F}^{\top}$;
$\hat{\bm{\alpha}}=(\hat{\alpha}_{1},\cdots,\hat{\alpha}_{N})^{\top}$ be the
Ordinary Least Squares (OLS) estimator of, where

$\hat{\alpha}_{i}=\mathbf{y}_{i\cdot}^{\top}\left(\frac{{\bf M}_{{\bf
F}}\bm{1}_{T}}{\bm{1}_{T}^{\top}{\bf M}_{{\bf F}}\bm{1}_{T}}\right).$

And
$\hat{{\bm{\Sigma}}}=\frac{1}{T}\sum_{t=1}^{T}\hat{\bm{\varepsilon}}_{t}\hat{\bm{\varepsilon}}_{t}^{\top}$
with
$\hat{\bm{\varepsilon}}_{t}=(\hat{\bm{\varepsilon}}_{1t},\cdots,\hat{\bm{\varepsilon}}_{Nt})^{\top}$,
where $\hat{\bm{\varepsilon}}_{it}$ is the OLS residual from the regression of
$y_{it}$ on an intercept and ${\bm{f}}_{t}$, i.e.

$\hat{\bm{\varepsilon}}_{i\cdot}=(\hat{\varepsilon}_{i1},\cdots,\hat{\varepsilon}_{iT})^{\top}={\bf
M}_{{\bf F}}(\mathbf{y}_{i}-\hat{\alpha}_{i}).$

To handle dense alternatives, we propose a sum-type test, named the SUM test,
with the test statistic constructed as

$\displaystyle T_{{\text{Sum}}}=\hat{\bm{\alpha}}^{\top}\hat{\bm{\alpha}}.$
(4)

Note that, if the error term is identical and independent distributed, Pesaran
and Yamagata (2023) also construct a similar sum-type test statistic

$\displaystyle T_{{\rm PY}}$
$\displaystyle=\frac{N^{-1/2}\sum_{i=1}^{N}\\{t_{i}^{2}-v/(v-2)\\}}{v/(v-2)\sqrt{2(v-1)/(v-4)\\{1+(N-1)\tilde{\rho}_{MT}^{2}\\}}},$
$\displaystyle t_{i}^{2}$
$\displaystyle=\frac{\hat{\alpha}_{i}^{2}(\bm{1}_{T}^{\top}{\bf M}_{{\bf
F}}\bm{1}_{T})}{v^{-1}\hat{\bm{\varepsilon}}_{i\cdot}^{\top}\hat{\bm{\varepsilon}}_{i\cdot}},$

where $v=T-p-1$ and
$\tilde{\rho}_{MT}=2/\\{N(N-1)\\}\sum_{i=2}^{N}\sum_{j=1}^{i-1}\tilde{\rho}^{2}_{ij}$
is the corresponding correlation estimator of
$\rho^{2}=2/\\{N(N-1)\\}\sum_{i=2}^{N}\sum_{j=1}^{i-1}\rho^{2}_{ij}$ with
$\tilde{\rho}_{ij}$ denoting the multiple testing estimator of the correlation
$\rho_{ij}=\sigma_{ij}/(\sigma_{ii}^{1/2}\sigma_{jj}^{1/2})$ (Bailey et al.,
2019). However, when the errors $\bm{\varepsilon}_{t}$ are time-dependent, a
bias term that cannot be ignored appears in $T_{{\rm PY}}$. That is, under the
null hypothesis, the expectation of $T_{{\rm PY}}$ is not zero, i.e.,
$E(T_{{\rm PY}})\not=0$. Furthermore, the asymptotic variance of $T_{{\rm
PY}}$ will no longer be one. Therefore, $T_{{\rm PY}}$ may not perform well
with dependent observations.

In our proposed test, we allow both $N$ and $T$ to be large, and allow
$\bm{\varepsilon}_{t}$’s be dependent linear processes. We need the conditions
for the error terms and factors given below.

  * (C1)

(i) $\\{{\bm{f}}_{1},\cdots,{\bm{f}}_{T}\\}$ is an strictly stationary. (ii)
${\bm{\Sigma}}_{{\bm{f}}}\equiv\mathrm{cov}({\bm{f}}_{t})$ is strictly
positive definite. (iii) ${\bm{f}}_{t}^{\top}{\bm{f}}_{t}$ is bounded for all
$t$.

Under Condition (C1), we have $T^{-1}\bm{1}_{T}^{\top}{\bf M}_{{\bf
F}}\bm{1}_{T}\mathop{\rightarrow}\limits^{p}\omega$, $T^{-1}{\bf
F}^{\top}\bm{1}_{T}\mathop{\rightarrow}\limits^{p}E({\bm{f}}_{t})=:{\bm{\mu}}_{{\bm{f}}}$,
$T^{-1}{\bf F}^{\top}{\bf
F}\mathop{\rightarrow}\limits^{p}\mu_{{\bm{f}}}\mu_{{\bm{f}}}^{\top}+{\bm{\Sigma}}_{{\bm{f}}}=:{\bf\Lambda}_{{\bm{f}}}$.
Define
$e_{t}:=\frac{1}{\omega}(1-{\bm{f}}_{t}^{\top}{\bf\Lambda}_{{\bm{f}}}^{-1}{\bm{\mu}}_{{\bm{f}}})$.
Because ${\bm{f}}_{t}$ is strictly stationary, so $e_{t}$ is also strictly
stationary. Define ${\bm{X}}_{t}:=\bm{\varepsilon}_{t}e_{t}$. We consider the
following dependent model for ${\bm{X}}_{t}$:

  * (C2)

${\bm{X}}_{1},\cdots,{\bm{X}}_{T}\in\mathbb{R}^{N}$ follow an $MA(\infty)$
process of the form (Zhang et al., 2018)

$\displaystyle{\bm{X}}_{t}=\bm{\Sigma}^{1/2}\sum_{k=0}^{\infty}b_{k}{\bm{z}}_{t-k}.$
(5)

(i) $\\{{\bm{z}}_{t}\\}$ is a sequence of $N$-dimensional random vectors such
that, if the coordinates of ${\bm{z}}_{t}$ are $\left\\{z_{it}\right\\},$ then
the two dimensional array $\left\\{z_{it}:1\leq i\leq N,t\geq 1\right\\}$ of
variables are i.i.d. satisfying the moment conditions
$\mathbb{E}z_{it}=0,\mathbb{E}\left|z_{it}\right|^{2}=1$ and
$\mathbb{E}\left|z_{it}\right|^{4}=\nu_{4}<\infty$.

(ii) $\sum_{k=0}^{\infty}|b_{k}|<\infty$, $\sum_{k=0}^{\infty}b_{k}=s\neq 0$
and $b_{k}=o(T^{-5})$, i.e. $\lim_{T\to\infty}T^{5}b_{T}=0$.

Unlike the $M-$dependent Gaussian stationary assumption in Ayyala et al.
(2017), we allow the error term can follows a $MA(\infty)$ process and
$z_{it}$ could be not normal distributed. Under Condition (C2), ${\bm{X}}_{t}$
has mean zero and auto-covariance structure given by
$\Gamma_{h},h\in\mathbb{Z}$, i.e.
$\mathbb{E}({\bm{X}}_{t}{\bm{X}}_{t+h}^{\top})=:\Gamma_{h}=a_{h}\bm{\Sigma}$
for $h=0,1,2,\cdots$ and $\Gamma_{h}=\Gamma_{-h}^{\top}$ with
$a_{h}:=\sum_{k=0}^{\infty}b_{k}b_{k+h}$. Next, we give some conditions of
structure of the auto-covariance matrix.

  * (C3)

There exist two positive constants $M_{0}$ and $M_{1}$ such that
$\|\bm{\Sigma}\|_{2}\leq M_{0}$ and $\mathrm{tr}(\bm{\Sigma})/N\geq M_{1}$,
and $\mathrm{tr}^{2}(\bm{\Sigma})/(T\mathrm{tr}(\bm{\Sigma}^{2}))=o(1)$. Here
$\|\cdot\|_{2}$ stands for either the spectral norm of a matrix or the
Euclidean norm of a vector.

  * (C4)

${\bf\Omega}_{T}:=T\mathrm{cov}(\bar{{\bm{X}}}_{T})=\Gamma_{0}+2\sum_{h=1}^{T}(1-\frac{h}{T})\Gamma_{h}\to\Gamma_{0}$,
where $\Gamma_{0}$ is the spectral matrix evaluated at the zero frequency.

  * (C5)

$N=O(T)$, and $N\to\infty$ as $T\to\infty$.

Condition (C3) reflects Assumption (A3) as stated in Zhang et al. (2018),
suggesting that the correlation among these variables should not be overly
high. Condition (C4) is a standard assumption made for multivariate time
series. In this study, we permit the number of securities $N$ to be comparable
to the length of the time period $T$, which is generally sufficient for most
empirical studies.

We propose to use a standardized version of $T_{{\text{Sum}}}$ as the test
statistic. From simple mathematical derivation, we have

$\displaystyle\hat{\bm{\alpha}}=\bm{\alpha}+\frac{1}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}\eta_{t}$

with $\bm{\eta}=(\eta_{t},\cdots,\eta_{T})^{\top}=\frac{{\bf M}_{{\bf
F}}\bm{1}_{T}}{T^{-1}\bm{1}_{T}^{\top}{\bf M}_{{\bf F}}\bm{1}_{T}}$. Define

$\displaystyle\tilde{T}_{{\text{Sum}}}=\tilde{\bm{\alpha}}^{\top}\tilde{\bm{\alpha}}~{}~{}\text{with}~{}~{}\tilde{\bm{\alpha}}=\bm{\alpha}+\frac{1}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}e_{t}.$
(6)

In the appendix, we will show the difference between $T_{{\text{Sum}}}$ and
$\tilde{T}_{{\text{Sum}}}$ are negligible. So, we only need to consider the
mean and variance of $\tilde{T}_{{\text{Sum}}}$ under $H_{0}$, given below:

$\displaystyle\mathbb{E}(\tilde{T}_{{\text{Sum}}})=\mathbb{E}(\bar{{\bm{X}}}_{T}^{\top}\bar{{\bm{X}}}_{T})=\frac{1}{T}\left\\{\mathrm{tr}(\Gamma_{0})+2\sum_{h=1}^{T}\left(1-\frac{h}{T}\right)\mathrm{tr}(\Gamma_{h})\right\\}=:\mu_{T},$
$\displaystyle\mathrm{var}(\tilde{T}_{{\text{Sum}}})\approx\frac{2}{T^{2}}\mathrm{tr}({\bf\Omega}_{T}^{2})=:\sigma_{T}^{2},$

We use
$\mu_{T}^{\prime}=\frac{1}{T}\\{\mathrm{tr}(\Gamma_{0})+2\sum_{h=1}^{M}(1-\frac{h}{T})\mathrm{tr}(\Gamma_{h})\\}$
replace $\mu_{T}$, where $M=\lceil\min(N,T)^{1/8}\rceil$. The reasons for such
substitution will be explained later.

First, we give the asymptotic normality of the test statistic
$T_{{\text{Sum}}}$.

###### Theorem 1

Under Conditions (C1)-(C5) hold and assuming $M=\lceil\min(N,T)^{1/8}\rceil$.
Under $H_{0}$, we have
$(T_{{\text{Sum}}}-\mu_{T}^{\prime})/\sigma_{T}\mathop{\rightarrow}\limits^{d}N(0,1)$
as $\min(N,T)\to\infty$.

Next, we consider the following alternative hypothesis:

  * (C6)

$\bm{\alpha}^{\top}(\Gamma_{h}\Gamma_{-h})^{1/2}\bm{\alpha}=o\\{(M+1)^{-1}T^{-1}\mathrm{tr}({\bf\Omega}_{T}^{2})\\}$,
$h=0,\cdots,M$.

###### Theorem 2

Under Conditions (C1)-(C6) hold and assuming $M=\lceil\min(N,T)^{1/8}\rceil$,
we have
$(T_{{\text{Sum}}}-\mu_{T}^{\prime})/\sigma_{T}\mathop{\rightarrow}\limits^{d}N\left(\frac{\bm{\alpha}^{\top}\bm{\alpha}}{\sqrt{2T^{-2}\mathrm{tr}({\bf\Omega}_{T}^{2})}},1\right)$
as $\min(N,T)\to\infty$.

It is worth noting that $(T_{{\text{Sum}}}-\mu_{T}^{\prime})/\sigma_{T}$ is
usually unknown since it involves population parameters. Thus it cannot be
used as a test statistic in practice. We therefore need to find consistent
estimators of $\mu_{T}^{\prime}$ and $\sigma_{T}^{2}$. Since
$\hat{\varepsilon}_{it}\mathop{\rightarrow}\limits^{p}\varepsilon_{it}$ and
$\eta_{t}-e_{t}=o_{p}(1)$, the following formula gives an estimator of the
auto-covariance matrix $\Gamma_{h}$:

$\displaystyle\hat{\Gamma}_{h}=\frac{1}{T-h}\sum_{t=1}^{T-h}\hat{\bm{\varepsilon}}_{t+h}\hat{\bm{\varepsilon}}_{t}^{\top}\eta_{t+h}\eta_{t}.$

Then, we consider the plug-in estimator of $\mu_{T}^{\prime}$:

$\displaystyle\hat{\mu}_{T}=\frac{1}{T-p-1}\left\\{\mathrm{tr}(\hat{\Gamma}_{0})+2\sum_{h=1}^{M}\left(1-\frac{h}{T-p-1}\right)\mathrm{tr}(\hat{\Gamma}_{h})\right\\}.$

We next find a ratio consistent estimator of $\sigma_{T}^{2}$. For testing
unit root, Zhang et al. (2018) proposed an estimator for
$a_{0}^{2}\mathrm{tr}(\bm{\Sigma}^{2})$ and prove that the estimator is
suitable for some special correlation structure. Their main idea was to
construct a special set of subscripts such that only pairwise correlations
exist between the four samples in each component of the estimator. Similarly,
we consider the following estimator for
$a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2})=\mathrm{tr}(\Gamma_{h_{1}}\Gamma_{h_{2}})$:

$\displaystyle\widehat{\mathrm{tr}(\Gamma_{h_{1}}\Gamma_{h_{2}})}\equiv
S_{h_{1},h_{2}}=\frac{\sum_{t=1}^{[T/2]-h_{2}}\sum_{s=t+[T/2]}^{T-h_{2}}\hat{\bm{\varepsilon}}_{t}^{\top}\hat{\bm{\varepsilon}}_{s}\hat{\bm{\varepsilon}}_{t+h_{1}}^{\top}\hat{\bm{\varepsilon}}_{s+h_{2}}\eta_{t}\eta_{s}\eta_{t+h_{1}}\eta_{s+h_{2}}}{(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})}.$

Accordingly,

$\displaystyle\hat{\sigma}_{T}^{2}=\frac{2}{T^{2}}\left(S_{0,0}+2\sum_{r=1}^{M}S_{0,r}+2\sum_{r=1}^{M}S_{r,0}+4\sum_{r=1}^{M}\sum_{s=1}^{M}S_{r,s}\right).$

Next, we show that the above two estimators are consistent.

###### Theorem 3

Assume the conditions in Theorems 1 and 2 hold. Then as $\min(N,T)\to\infty$,
$\hat{\mu}_{T}-\mu_{T}^{\prime}=o_{p}(\sigma_{T})$ and
$\hat{\sigma}_{T}^{2}/\sigma_{T}^{2}\mathop{\rightarrow}\limits^{p}1$.

According to Theorems 1-3, the $p$-value associated with $T_{{\text{Sum}}}$ is

$p_{{\text{Sum}}}=1-\Phi\left(\frac{T_{{\text{Sum}}}-\hat{\mu}_{T}}{\hat{\sigma}_{T}}\right),$

where $\Phi(\cdot)$ is the cumulative distribution function (CDF) of
${N}(0,1)$. If the $p$-value is below some pre-specified significant level,
say $\gamma\in(0,1)$, then we rejected the null hypothesis.

According to Theorem 1 and 2, the power function of $T_{{\text{Sum}}}$ is

$\displaystyle\beta_{{\text{Sum}}}(\bm{\alpha})=\Phi\left(-z_{\gamma}+\frac{T\bm{\alpha}^{\top}\bm{\alpha}}{\sqrt{2\mathrm{tr}({\bf\Omega}_{T}^{2})}}\right)$
(7)

where $z_{\gamma}$ is the upper $\gamma$ quantile of $N(0,1)$. Note that the
power function $\beta_{{\text{Sum}}}(\bm{\alpha})$ has the similar form of the
power function of Chen and Qin (2010)’s test for the high dimensional two
sample mean problem. If the error term $\bm{\varepsilon}_{t}$ are identical
and independent distributed and the variances of each
$\varepsilon_{it},i=1,\cdots,N$ are equal, $\beta_{{\text{Sum}}}$ are the same
as the power function of $T_{{\rm PY}}$.

###  2.2 Max-type test

The power function (7) indicates that the sum-type test $T_{{\text{Sum}}}$ can
achieve strong performance when a significant number of alphas are nonzero.
Conversely, if only a limited number of alphas are nonzero, $T_{{\text{Sum}}}$
may not exhibit sufficient power. In scenarios involving sparse alternatives,
numerous studies have shown that the max-type test procedure is highly
effective. For instance, Cai et al. (2014) demonstrated its utility in high-
dimensional two-sample problems, Feng et al. (2022b) applied it to high-
dimensional alpha testing in linear factor pricing models, and Ma et al.
(2023) used it for high-dimensional alpha testing in time-varying factor
pricing models.

Consequently, we proposed the following max-type test

$\displaystyle T_{{\text{Max}}}=\max_{1\leq i\leq
N}\frac{T\hat{\alpha}_{i}^{2}}{\hat{\sigma}_{i}},$

where $\hat{\sigma}_{i}=\sum_{h\in\mathcal{M}}\omega(h/M)\hat{\phi}_{i,h}$,
$\hat{\phi}_{i,h}=\frac{1}{T-h}\sum_{t=h+1}^{T}\hat{\varepsilon}_{it}\hat{\varepsilon}_{i,t-h}\eta_{t}\eta_{t-h}$
and $\mathcal{M}=\\{0,\pm 1,\dots,\pm M\\}$. Here $\omega(\cdot)$ is a
weighted function. For simplicity, we use the plain estimate in the article,
i.e. $\omega(\cdot)=1$. Here $\hat{\sigma}_{i}$ is the long-run variance
estimator.

Next, we will demonstrate that when $(N,T)\to\infty$,
$T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}$ has a type I extreme value
distribution. To proceed, we first introduce some conditions.

Let ${\bf D}_{0}\equiv\mathrm{diag}(\sigma_{1}^{0},\dots,\sigma_{N}^{0})$ be
the diagonal matrix of the long-run covariance of ${\bm{X}}_{t}$,
${\bf\Omega}:=\Gamma_{0}+2\sum_{h=0}^{\infty}\Gamma_{h}$, and
$\bm{R}_{0}:={\bf D}_{0}^{-1/2}{\bf\Omega}{\bf D}_{0}^{-1/2}$. Let $[{\bf
A}]_{ij}$ denote the $(i,j)$-th element of matrix ${\bf A}$.

  * (C7)

$\max_{1\leq i\leq N}\sum_{j=1}^{N}[\bm{\Sigma}^{1/2}]_{ij}^{4}<C$ for some
positive constant $C$.

  * (C8)

$\max_{1\leq i<j\leq N}[\bm{R}_{0}]_{ij}\leq r<1$ and $\max_{1\leq j\leq
N}\sum_{i=1}^{N}[\bm{R}_{0}]_{ij}^{2}\leq c$ for some constants $r,c>0$.

  * (C9)

$M^{1+a}T^{1/q-1/2}\log^{2}(N)\log^{3/2}(NT)=o(1)$ and
$M^{qa^{2}+(q+7)a+2}T^{-qa/2}\log^{2+5qa/2}(N)\log^{7a}(NT)=o(1)$ for some
$q\geq 4$ and $a>1/2-1/q$.

Condition (C7) and (C8) are similar to the conditions in Lemma 6 in Cai et al.
(2014), which assume the correlations between those variables are not very
large. In the special case with $M\asymp T^{1/8}$, Condition (C9) becomes
$\log(N)=o(T^{\tau})$ where
$\tau=\frac{2}{5}(\frac{1}{2}-\frac{1}{q}-\frac{1+a}{8})\wedge\frac{-qa^{2}+(3q-7)a-2}{16+(56+20q)a}$
with $q\geq 4$ and $a>1/2-1/q$.

###### Theorem 4

Under Conditions (C1)-(C2) and (C7)-(C9), and assuming
$M=\lceil\min(N,T)^{1/8}\rceil$, we have as $\min(N,T)\to\infty$,

$\displaystyle P_{H_{0}}\left(T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}\leq
x\right)\to\exp\left\\{-\frac{1}{\sqrt{\pi}}\exp\left(-\frac{x}{2}\right)\right\\}\equiv
F(x).$

Here, $\mathbb{P}_{H_{0}}$ denotes the probability measure under the null
hypothesis $H_{0}$. According to the limiting null distribution derived in
Theorem 4, we can easily obtain the $p$-value associated with
$T_{{\text{Max}}}$, namely,

$p_{{\text{Max}}}=1-F(T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}).$

Again, small values of $p_{{\text{Max}}}$ indicate rejecting the null
hypothesis. Next, we turn to the power analysis of the proposed max-type test.

###### Theorem 5

Under Conditions (C1)-(C2) and (C7)-(C9), and assuming
$M=\lceil\min(N,T)^{1/8}\rceil$, we have as $\min(T,N)\to\infty$,

$\displaystyle\inf_{\bm{\alpha}\in\mathcal{A}(c)}P(\Phi_{\gamma}=1)\to 1,$

where $\Phi_{\gamma}\equiv I[T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}\geq
q_{\gamma}]$,
$q_{\gamma}\equiv-\log(\pi)-2\log\left\\{\log(1-\gamma)^{-1}\right\\}$, $c=4$
and

$\displaystyle\mathcal{A}(c)\equiv\left\\{\bm{\alpha}=(\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{N})\in\mathcal{R}^{N}:\max_{1\leq
i\leq N}\alpha_{i}/\sqrt{\sigma_{i}^{0}}\geq c\sqrt{\log(N)/{T}}\right\\}.$
(8)

Theorem 5 demonstrates that if at least one alpha is significantly large, the
suggested max-type test maintains consistency. As showed in Theorem 3 in Feng
et al. (2022b), the rate $\sqrt{\log(N)/T}$ is minimax rate optimal, which
means any $\gamma$-level test is unable to detect the alternative hypothesis
in probability uniformly over

$\left\\{\bm{\alpha}\in\mathcal{R}^{N}:\sum_{i=1}^{N}I\left(\alpha_{i}\neq
0\right)=k_{N}\right\\}\bigcap\\{\bm{\alpha}\in\left.\mathcal{R}^{N}:\max_{1\leq
i\leq N}\left|\alpha_{i}\right|>c_{4}\sqrt{\log(N)/T}\right\\}$

where $k_{N}=O(N^{r}),r<1/4$. Hence the order of the lower bound $\max_{1\leq
i\leq N}\alpha_{i}/\sqrt{\sigma_{i}^{0}}\geq c\sqrt{\log(N)/{T}}$ cannot be
improved.

###  2.3 Cauchy combination test

In real-world scenarios, it’s often unclear whether the intercept vector is
sparse or dense. To accommodate different alternative behaviors, we integrate
the Sum-test and Max-test. The crucial point is that these two test statistics
are asymptotically independent under certain mild conditions and the null
hypothesis.

###### Theorem 6

Under Conditions (C1)-(C9) and assuming $M=\lceil\min(N,T)^{1/8}\rceil$, under
$H_{0}$, we have as $\min(N,T)\to\infty$,

$\mathbb{P}_{H_{0}}\left(\frac{T_{{\text{Sum}}}-\hat{\mu}_{T}}{\hat{\sigma}_{T}}\leq
x,T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}\leq y\right)\to\Phi(x)F(y).$

According to Theorem 6, we suggest combining the corresponding $p$-values by
using Cauchy Combination Method (Liu and Xie, 2020), to wit,

$p_{CC}=1-G[0.5\tan\\{(0.5-p_{{\text{Max}}})\pi\\}+0.5\tan\\{(0.5-p_{{\text{Sum}}})\pi\\}],$

where $G(\cdot)$ is the CDF of the the standard Cauchy distribution. If the
final $p$-value is less than some pre-specified significant level
$\gamma\in(0,1)$, then we reject $H_{0}$.

Next, we analyze the power of the adaptive testing procedure. We consider the
following sequence of alternative hypotheses, to wit,

$H_{1,NT}:||\bm{\alpha}||_{0}=o[N/\log^{2}\\{\log(N)\\}]~{}~{}\text{and}~{}~{}||\bm{\alpha}||_{2}=O\\{T^{-1/2}\mathrm{tr}^{1/4}({\bf\Omega}_{T}^{2})\\}.$
(9)

In fact, the asymptotic independence between the maximum-type and sum-type
statistics still hold under the hypotheses given in (9).

###### Theorem 7

Under the same condition as Theorem 6, we have as $(N,T)\to\infty$, under
$H_{1,NT}$,

$\displaystyle\mathbb{P}\left(\frac{T_{{\text{Sum}}}-\hat{\mu}_{T}}{\hat{\sigma}_{T}}\leq
x,T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}\leq y\right)$
$\displaystyle\qquad\to\mathbb{P}\left(\frac{T_{{\text{Sum}}}-\hat{\mu}_{T}}{\hat{\sigma}_{T}}\leq
x\right)\mathbb{P}\left(T_{{\text{Max}}}-2\log(N)+\log\\{\log(N)\\}\leq
y\right).$

Long et al. (2023) show that the power of Cauchy combination-based test would
be powerful than that of the test based on
$\min\\{p_{{\text{Max}}},p_{{\text{Sum}}}\\}$ (referred to as the minimal
$p$-value combination), say $\beta_{M\wedge
S,\gamma}=P(\min\\{{p}_{{\text{Max}}},{p}_{{\text{Sum}}}\\}\leq
1-\sqrt{1-\gamma})$. Obviously,

$\displaystyle\beta_{CC,\gamma}\geq\beta_{M\wedge S,\gamma}$
$\displaystyle\geq
P(\min\\{{p}_{{\text{Max}}},{p}_{{\text{Sum}}}\\}\leq\gamma/2)$
$\displaystyle=\beta_{{{\text{Max}}},\gamma/2}+\beta_{{{\text{Sum}}},\gamma/2}-P({p}_{{\text{Max}}}\leq\gamma/2,{p}_{{\text{Sum}}}\leq\gamma/2)$
$\displaystyle\geq\max\\{\beta_{{{\text{Max}}},\gamma/2},\beta_{{{\text{Sum}}},\gamma/2}\\}.$
(10)

On the other hand, under $H_{1}$ in (9), we have

$\displaystyle\beta_{CC,\gamma}\geq\beta_{{{\text{Max}}}\wedge
S,\gamma}\geq\beta_{{{\text{Max}}},\gamma/2}+\beta_{{{\text{Sum}}},\gamma/2}-\beta_{{{\text{Max}}},\gamma/2}\beta_{{{\text{Sum}}},\gamma/2}+o(1),$
(11)

due to the asymptotic independence entailed by Theorem 7. For a small
$\gamma$, the difference between $\beta_{{{\text{Max}}},\gamma}$ and
$\beta_{{{\text{Max}}},\gamma/2}$ should be small, and the same fact applies
to $\beta_{S,\gamma}$. Consequently, by (2.3)–(11), the power of the Cauchy
combination test would be no smaller than or even significantly larger than
that of either max-type or sum-type test.

Next, we present some specific cases to demonstrate the power performance of
each test. For simplicity, we assume that $\sigma_{i}^{0}=1,i=1,\cdots,N$ and
$\mathrm{tr}({\bf\Omega}_{T}^{2})=c_{O}N$. Additionally,
$\alpha_{i}=a,i=1,\cdots,s$ and $\alpha_{i}=0$ for $i=s+1,\cdots,N$. Let
$\xi_{\gamma}:=\lim\beta_{{\text{Max}},\gamma}$,
$\eta_{\gamma}:=\lim\beta_{{\text{Sum}},\gamma}$. Table 1 summarizes the
asymptotic powers of these tests under different configurations of the change
magnitude $a$ and the sparsity level $s$. Here we provide a detailed analysis
as follows:

  * (I)

$Ta^{2}/\log N\to\infty$. Then, $\xi_{\gamma}=1$ by Theorem 5 and
$\beta_{CC,\gamma}=1$ by (2.3).

  * (II)

$a=C\sqrt{(\log N)/T}$. If $C\geq 4$, $\xi_{\gamma}=1$ by Theorem 5. Next, we
consider three case for $s$.

    * (i)

$s=o(N^{1/2}/\log N)$. Now
$T\bm{\alpha}^{T}\bm{\alpha}/\sqrt{2\mathrm{tr}({\bf\Omega}_{T}^{2})}\to 0$,
so $\eta_{\gamma}=\gamma$ by (7). And
$\beta_{CC,\gamma}\geq\xi_{\gamma/2}+\eta_{\gamma/2}-\xi_{\gamma/2}\eta_{\gamma/2}$
by (11);

    * (ii)

$s\sim N^{1/2}/\log N$; Both $T_{{\text{Sum}}}$ and $T_{{\text{Max}}}$ possess
a certain level of power. And
$\beta_{CC,\gamma}\geq\xi_{\gamma/2}+\eta_{\gamma/2}-\xi_{\gamma/2}\eta_{\gamma/2}$
could yield higher power than either sum-type or max-type procedure.

    * (iii)

$N^{-1/2}s\log N\to\infty$. Now
$T\bm{\alpha}^{T}\bm{\alpha}/\sqrt{2\mathrm{tr}({\bf\Omega}_{T}^{2})}\to\infty$,
so $\eta_{\gamma}=1$ by (7) and $\beta_{CC,\gamma}=1$ by (2.3).

  * (III)

$a=o\\{\sqrt{(\log N)/T}\\}$. The max-type test would not have any power, i.e.
$\xi_{\gamma}=\gamma$. Specially, if $N^{-1/2}Tsa^{2}\to\infty$, we have
$T\bm{\alpha}^{T}\bm{\alpha}/\sqrt{2\mathrm{tr}({\bf\Omega}_{T}^{2})}\to\infty$,
so, similar to II-(iii), $\eta_{\gamma}=1$ and $\beta_{CC,\gamma}=1$.

[ht]  Case $a$ $s$ $\lim\beta_{{\text{Max}},\gamma}$
$\lim\beta_{{\text{Sum}},\gamma}$ $\lim\beta_{CC,\gamma}$ I $Ta^{2}/\log
N\to\infty$ any $1$ $\eta_{\gamma}$ $1$ II–(i) $a\sim\sqrt{(\log N)/T}$
$s=o(N^{1/2}/\log N)$ $\xi_{\gamma}$ $\gamma$
$\geq\xi_{\gamma/2}+\gamma/2(1-\xi_{\gamma/2})$ II–(ii) $s\sim N^{1/2}/\log N$
$\xi_{\gamma}$ $\eta_{\gamma}$
$\geq\xi_{\gamma/2}+\eta_{\gamma/2}-\xi_{\gamma/2}\eta_{\gamma/2}$ II–(iii)
$N^{-1/2}s\log N\to\infty$ $\xi_{\gamma}$ $1$ $1$ II’ $a=C\sqrt{(\log N)/T}$*
any $1$ $\eta_{\gamma}$ $1$ III $a=o\\{\sqrt{(\log N)/T}\\}$ any $\gamma$
$\eta_{\gamma}$ $\geq\eta_{\gamma/2}+\gamma/2(1-\eta_{\gamma/2})$ III’
$a=o\\{\sqrt{(\log N)/T}\\}$ $N^{-1/2}Tsa^{2}\to\infty$ $\gamma$ $1$ $1$

Table 1: Asymptotic powers of the max-type and sum-type tests, and the Cauchy
combination test, respectively.

  * •

∗ $C>0$ is a large enough constant

##  3 Simulation

In this section, we compare our proposed methods with three test procedures
for the independent case: the sum-type test procedure proposed by Pesaran and
Yamagata (2017) (abbreviated as PY), the max-type test procedure proposed by
Feng et al. (2022b) (abbreviated as FLM) and the combination test procedure
proposed by Feng et al. (2022b) (abbreviated as COM).

This simulation is designed to mimic the commonly used Fama-French three-
factor model, where the factors ${\bm{f}}_{t}$ have strong serial correlation
and heterogeneous variance. Specifically, we consider the example studied in
Feng et al. (2022b). The response $Y_{it}$ are generated according to the LFPM
in (1) with $p=3$:

$\displaystyle
Y_{it}=\alpha_{i}+\sum_{j=1}^{p}\beta_{ij}f_{tj}+\varepsilon_{it},$

where the three factors, $f_{t1}$, $f_{t2}$ and $f_{t3}$, are the Fama-French
three factors (Market factor, SMB, HML). We generate each factor from an
autoregressive conditional heteroskedasticity process and the GARCH(1,1) model
respectively. Specifically

$\displaystyle f_{t1}=$ $\displaystyle
0.53+0.06f_{t-1,1}+h_{t1}^{1/2}\zeta_{t1},~{}\mbox{Market factor},$
$\displaystyle f_{t3}=$ $\displaystyle
0.19+0.19f_{t-1,2}+h_{t2}^{1/2}\zeta_{t2},~{}\mbox{SMB factor},$
$\displaystyle f_{t3}=$ $\displaystyle
0.19+0.05f_{t-1,3}+h_{t3}^{1/2}\zeta_{t3},~{}\mbox{HML factor},$

where for $j=1,2,3$, $\zeta_{tj}$ are generated independently from a standard
normal distribution, and the variance term $h_{tj}$ is generated as follows:

$\displaystyle h_{t1}$
$\displaystyle=0.89+0.85h_{t-1,1}+0.11\zeta_{t-1,1}^{2},~{}\mbox{Market
factor},$ $\displaystyle h_{t2}$
$\displaystyle=0.62+0.74h_{t-1,2}+0.19\zeta_{t-1,2}^{2},~{}\mbox{SMB},$
$\displaystyle h_{t3}$
$\displaystyle=0.80+0.76h_{t-1,3}+0.15\zeta_{t-1,3}^{2},~{}\mbox{HML}.$

The above process is simulated over the periods
$t\in\\{-49,\cdots,0,1,\cdots,T\\}$ with the initial values $f_{-50,j}=0$ and
$h_{-50,j}=1$ for any $j\in\\{1,2,3\\}$, and the generated data that belong to
the periods $\\{1,\cdots,T\\}$ is extracted as the simulation data.

The three groups of coefficients corresponding to the three factors,
$\beta_{i1}$, $\beta_{i2}$ and $\beta_{i3}$, are generated independently from
$U(0.2,2)$, $U(-1,1.5)$ and $U(-1.5,1.5)$, respectively. Then, we set
$\bm{\alpha}=\bm{0}$ under the null hypothesis.

The error $\bm{\varepsilon}_{t}$ are generated from
$\bm{\varepsilon}_{t}=\sum_{h=0}^{M}A_{h}{\bm{z}}_{t-h}$, where
$A_{0},\cdots,A_{M}$ are $N\times N$ matrices which determine the
autocovariance structure. And ${\bm{z}}_{t}={\bm{\Sigma}}^{1/2}\bm{\zeta}_{t}$
where $\bm{\zeta}_{t}=(\zeta_{t1},\cdots,\zeta_{tN})$ are independently and
identically distributed from $N(0,1)$ and $t(3)$. Here we consider $A_{0}={\bf
I}_{N}$,

$\displaystyle A_{h}(i,j)$
$\displaystyle=\left\\{\begin{array}[]{cl}h^{-1}\phi_{1}&\text{if}~{}i=i,h\geq
1\\\ \frac{\phi_{1}}{h(i-j)^{2}}&\text{if}~{}1\leq|i-j|\leq\omega N,\\\
0&\text{if}~{}|i-j|>\omega N.\end{array}\right.\text{if}~{}~{}h=1,2.$
$\displaystyle{\bm{\Sigma}}(i,j)$
$\displaystyle=\left\\{\begin{array}[]{cl}1&\text{if}~{}i=j,\\\
\phi_{2}&\text{if}~{}i=i,h\geq 1\\\
\frac{\phi_{2}}{(i-j)^{2}}&\text{if}~{}1\leq|i-j|\leq\omega N,\\\
0&\text{if}~{}|i-j|>\omega N.\end{array}\right.$

For $h>2$, we simply set $A_{h}=e^{-2h}$. We set $T=400,800$, $N=250,500$ and
$(\omega,\phi_{1},\phi_{2})=(0.9,0.6,0.4)$. And we consider two case for $M$:
(i) independent case, $M=0$; (ii) 2-dependent case, $M=2$; (iii)
$\infty$-dependent case, $M=T-1$. Table 2 presents the empirical dimensions of
each test. In the case of independence, all tests effectively manage the
empirical dimensions in most scenarios. However, in the dependent case, the
empirical dimensions of the PY, FLM, and COM tests exceed the nominal level,
which is expected as these tests do not account for the error terms’ dependent
structure. On the other hand, our proposed test procedures - SUM, MAX, CC -
demonstrate strong performance in terms of empirical dimensions, particularly
when dealing with large sample sizes.

For power comparison, we generate $\alpha_{i},i\in S$ independently from
$U(0,\sqrt{s^{-1}c_{M}\log(N)/T})$ where each element in $S$ is uniformly and
randomly drawn from $\\{1,\cdots,N\\}$ and $|S|=s$. And
$c_{0}=12,c_{2}=80,c_{\infty}=90$ for a fair power comparison. Figure 1
illustrates the power curves of each test under varying levels of sparsity in
the independent case where $M=0$. It was observed that the SUM and MAX tests
perform comparably to the PY and FLM tests, respectively. The CC test
surpasses the COM test in all scenarios due to the superior efficiency of the
Cauchy combination test procedure over the minimal value procedure (Long et
al., 2023). As the PY, FLM, and COM tests are unable to control the empirical
sizes in the dependent case, we only consider our proposed three test
procedures for $M=2,\infty$. Figure 2-3 depicts the power curves of the SUM,
MAX, and CC tests under varying sparsity. The SUM test outperforms the MAX
test when signals are very dense, while the MAX test excels when signals are
very sparse. These observations align with intuitive expectations and are
corroborated by several other studies, such as Feng et al. (2023).
Furthermore, our proposed CC test performs similarly to the SUM test under
very dense alternatives and the MAX test under very sparse alternatives. When
the signals are neither very dense nor sparse, the CC test exhibits the best
performance, underscoring the superiority of the Cauchy combination test
procedure.

Table 2: Empirical sizes of different tests. | $M=0$ | $M=2$ | $M=\infty$
---|---|---|---
| $T=400$ | $T=800$ | $T=400$ | $T=800$ | $T=400$ | $T=800$
$N$ | 250 | 500 | 250 | 500 | 250 | 500 | 250 | 500 | 250 | 500 | 250 | 500
| $\zeta_{ti}\sim N(0,1)$
PY | 6.1 | 4.6 | 5.8 | 4.6 | 100 | 100 | 100 | 100 | 100 | 100 | 100 | 100
FLM | 4.8 | 5.8 | 3.6 | 5.4 | 97.3 | 99.6 | 95.8 | 99.5 | 100 | 100 | 100 | 100
COM | 6.1 | 5.2 | 4.9 | 5.2 | 100 | 100 | 100 | 100 | 100 | 100 | 100 | 100
SUM | 6.2 | 5.1 | 5.1 | 5.7 | 5.7 | 5.5 | 5.7 | 5.8 | 5.6 | 6.1 | 4.9 | 5.8
MAX | 7.3 | 6.7 | 5.2 | 6.7 | 6.7 | 6.6 | 4.1 | 6.2 | 6.3 | 6.5 | 5.4 | 5.9
CC | 6.7 | 6.0 | 5.6 | 6.2 | 6.6 | 5.8 | 6.0 | 5.7 | 5.4 | 6.2 | 5.7 | 5.0
| $\zeta_{ti}\sim t(3)$
PY | 6.7 | 5.3 | 5.8 | 5.2 | 100 | 100 | 100 | 100 | 100 | 100 | 100 | 100
FLM | 4.2 | 4.5 | 3.4 | 4.8 | 96.9 | 99.8 | 95.5 | 99.7 | 100 | 100 | 100 | 100
COM | 5 | 4.4 | 4.8 | 5.3 | 100 | 100 | 100 | 100 | 100 | 100 | 100 | 100
SUM | 6.4 | 5.7 | 5.8 | 6.3 | 5.6 | 6.3 | 5.7 | 5.4 | 6.0 | 5.8 | 5.4 | 5.3
MAX | 5.2 | 5.8 | 2.9 | 6.1 | 5.9 | 6.4 | 3.1 | 5.4 | 6.5 | 5.9 | 4.7 | 5.1
CC | 6.7 | 6.5 | 4.8 | 6.1 | 6.1 | 5.5 | 5.2 | 4.6 | 5.7 | 5.5 | 5.1 | 4.8
Figure 1: Power comparison of different methods with different sparsity under
independent case $M=0$ with $(T,N)=(400,500)$. Figure 2: Power comparison of
different methods with different sparsity under 2-dependent case $M=2$ with
$(T,N)=(400,500)$. Figure 3: Power comparison of different methods with
different sparsity under $\infty$-dependent case $M=\infty$ with
$(T,N)=(400,500)$.

Continuing with the same settings, we examine the power performance of the
tests involved across varying signal strengths. For the alternative
hypothesis, we contemplate three scenarios for the sparsity level $s=2,15,50$.
The nonzero $\alpha_{i}$s are drawn from $U(0,\sqrt{\delta\log N/T})$. Here we
only consider $\zeta_{ti}\sim N(0,1)$. The outcomes would be analogous if we
consider $\zeta_{ti}\sim t(3)$. Figures 4-6 depict the power curves of our
proposed three methods for $M=0,2,\infty$, respectively. We observe that the
power of each test increases as the signal strength escalates, demonstrating
the consistency of each test. Moreover, we find that the Cauchy combination
tests outperform the other methods, particularly when the sparsity level is
moderate.

Figure 4: Power comparison of different methods with different signal strength
under independent case $M=0$ with $(T,N)=(400,500)$. Figure 5: Power
comparison of different methods with different signal strength under
2-dependent case $M=2$ with $(T,N)=(400,500)$. Figure 6: Power comparison of
different methods with different signal strength under $\infty$-dependent case
$M=\infty$ with $(T,N)=(400,500)$.

##  4 Real Data Application

In this study, we focus on the stocks listed in the S&P 500 index, a globally
recognized benchmark for U.S. equities. We have compiled weekly returns for
all the securities in the S&P 500 index from January 14, 2005, to November 24,
2023, excluding any missing values. This gives us a total of $T=888$
observations over this extensive period. As the composition of the index
changes over time, we have limited our analysis to $N=393$ securities that
were consistently part of the S&P 500 index throughout the entire period. The
time series data for the risk-free rate of return and other factors were
sourced from Ken French’s data library web page. We have chosen the one-month
U.S. treasury bill rate as the risk-free rate, denoted as $r_{ft}$. As a proxy
for the market return $r_{mt}$, we used the value-weighted return on all
stocks listed on the NYSE, AMEX, and NASDAQ, as provided by CRSP. We
calculated the Small Minus Big (SMB) factor, $\text{SMB}_{t}$, as the average
return on three small portfolios minus the average return on three big
portfolios. Similarly, the High Minus Low (HML) factor, $\text{HML}_{t}$, is
calculated as the average return on two value portfolios minus the average
return on two growth portfolios. Both these factors are based on the stocks
listed on the NYSE, AMEX, and NASDAQ.

We use the Fama-French three-factor model to describe the above panel data.
Recall that the Fama-French three-factor model is

$\displaystyle
Y_{it}=r_{it}-r_{ft}=\alpha_{i}+\beta_{i1}(r_{mt}-r_{ft})+\beta_{i2}SMB_{t}+\beta_{i3}HML_{t}+\epsilon_{it},$
(12)

for each $i\in\\{1,\cdots,N\\}$ and $t\in\\{\tau,\cdots,\tau+T^{\prime}-1\\}$,
where $r_{mt}-r_{ft}$ is also referred to as the market factor. We are
interested in testing

$\displaystyle
H_{0}:\alpha_{1}=\dots\alpha_{N}=0,\text{versus}~{}H_{1}:\exists~{}i\in\\{1,\cdots,N\\}\text{
s.t. }\alpha_{i}\neq 0.$ (13)

Initially, we examine the presence of time series dependence in the residual
sequences of the dataset. We apply the Box-Pierce test, a conventional
autocorrelation test, to the residuals of each security under the Fama-French
three-factor model with the total sample sizes. The histogram of p-values from
the Box-Pierce test for the U.S. datasets is shown in Figure 7. We notice a
significant number of $p$-values below 0.1, leading us to believe that some
residual sequences may exhibit autocorrelation. Furthermore, we employ the
max-type high-dimensional white noise test procedure, as described in Feng et
al. (2022c), to verify if the residuals are white noise. The $p$-value of this
test is 2e-11, indicating some time-dependence among $\epsilon_{it}$.
Consequently, test procedures based on the i.i.d assumption may not perform
well and could yield inaccurate results. To analyze this dataset, we apply
three test procedures proposed in this paper - SUM, MAX, and CC, which are
designed to handle such time-dependence.

We next examine the hypothesis (13) using a rolling window procedure with a
window length of $T=260$ (equivalent to 5 years). The p-values for each alpha
test procedure over the time period are depicted in Figure 8. Our observations
reveal that for the initial seven years and the final six years, nearly all
tests do not reject the null hypothesis. This suggests that the three observed
factors can effectively account for the stock returns. However, during the
five-year period in the middle, the CC test frequently rejects the null
hypothesis, leading us to infer that the markets were inefficient during these
periods. Interestingly, there are instances over the long time period where
the MAX or SUM test does not reject the null hypothesis. This demonstrates
that our CC test procedure is more robust and efficient under various sparsity
alternatives.

Figure 7: Histogram of $p$-values of the Box-Pierce of residuals of Fama-
French three factor model for U.S.’s datasets. Figure 8: $p$-values of alpha
tests for U.S.’s datasets with Fama-French three factor model.

##  5 Conclusion

Initially, we introduced the sum-type and max-type tests for a high-
dimensional linear pricing model with dependent observations. These tests
perform well against both dense and sparse alternatives. To broaden the
application, we introduced a Cauchy combination test that combines the sum-
type and max-type tests. Both simulation studies and real data applications
have shown the necessity of considering the time-dependence of the error term.
Ignoring this could lead to significant size distortion in methods based on
the i.i.d assumption.

Furthermore, given the importance of allowing for time-variation in the risk-
exposure coefficients under LFPMs, our future work will focus on developing
sum-type, max-type, and combination test procedures that allow for such time-
variation. We plan to use a more complex and powerful technical framework to
establish the asymptotic theory, as opposed to the current one.

##  6 Appendix

###  6.1 Proof of Theorems 1-3

Let $M=\lceil\min(N,T)^{1/8}\rceil$. We can define a M-dependent approximation
sequence for $\\{\bm{{\bm{X}}}_{t}\\}$:

$\displaystyle\bm{\gamma}_{t}:=\mathbb{E}(\bm{{\bm{X}}}_{t}|{\bm{z}}_{t-M},\cdots,{\bm{z}}_{t})=\bm{\Sigma}^{1/2}\sum_{k=0}^{M}b_{k}{\bm{z}}_{t-k}.$

By replacing ${\bm{X}}_{t}$ in $\tilde{T}_{{\text{Sum}}}$ with
$\bm{\gamma}_{t}$, we define
$T_{{\text{Sum}}}^{NG}:=\bar{\bm{\gamma}}_{T}^{\top}\bar{\bm{\gamma}}_{T}$.
Let $a_{h,M}:=\sum_{k=0}^{M-h}b_{k}b_{k+h}$, then
$\Gamma_{h,M}:=\mathbb{E}(\bm{\gamma}_{t}\bm{\gamma}_{t+h}^{\top})=a_{h,M}\bm{\Sigma}$,
and
${\bf\Omega}_{T,M}:=T\mathbb{E}(\bar{\bm{\gamma}}_{T}\bar{\bm{\gamma}}_{T}^{\top})=\sum_{h\in\mathcal{M}}(1-\frac{|h|}{T})\Gamma_{h,M}$,
where $\mathcal{M}=\\{0,\pm 1,\dots,\pm M\\}$.

Let $\\{\bm{\delta}_{t}\\}$ be a Gaussian sequence which is independent of
$\\{\bm{\gamma}_{t}\\}$ and preserves the same auto-covariance structure as
$\\{\bm{\gamma}_{t}\\}$, i.e.
$\mathbb{E}(\bm{\delta}_{t}\bm{\delta}_{t+h}^{\top})=\Gamma_{h,M}$. Similarly,
define
$T_{{\text{Sum}}}^{G}:=\bar{\bm{\delta}}_{T}^{\top}\bar{\bm{\delta}}_{T}$.

Firstly, we establish some useful results in the following lemmas.

###### Lemma 1

Suppose $z_{it}$ are independent and identically distributed random variables
satisfying $\mathbb{E}(z_{it})=0$, $\mathbb{E}(z_{it}^{2})=1$ and
$\mathbb{E}(z_{it}^{4})=\mu_{4}<\infty$. Let
$W=\sum_{t=0}^{\infty}a_{t}{\bm{z}}_{t}$, where
${\bm{z}}_{t}=(z_{1t},\dots,z_{Nt})^{\top}$ and
$\sum_{i=0}^{\infty}|a_{i}|<\infty$. Then, there exists a positive constant
$\tau_{1}\geq 3$ such that for all $N\times N$ positive semi-definite matrix
$B$ and absolute convergent $\\{a_{i}\\}_{i\geq 0}$,

$\displaystyle\mathbb{E}[(W^{\top}BW)^{2}]\leq\tau_{1}[\mathbb{E}(W^{\top}BW)]^{2}.$

Proof. Let $[B]_{ij}$ denote the $(i,j)$-th element of $B$. Since $B$ is
positive semi-definite, we can decompose $B$ as $B=D\Lambda D^{\top}$, where
$D$ is an orthogonal matrix and
$\Lambda=\mathrm{diag}(\lambda_{1},\dots,\lambda_{N})$ with
$\lambda_{1}\geq\dots\geq\lambda_{N}\geq 0$ being the eigenvalues of $B$.
Define $\tilde{{\bm{z}}}_{t}=D^{\top}{\bm{z}}_{t}$, then

$\displaystyle\mathbb{E}(\tilde{z}_{it}^{2})=\mathbb{E}\left\\{\left(\sum_{j=1}^{N}[D]_{ji}z_{jt}\right)^{2}\right\\}=1,~{}~{}\text{and}$
$\displaystyle\mathbb{E}(\tilde{z}_{it}^{4})=\mathbb{E}\left\\{\left(\sum_{j=1}^{N}[D]_{ji}z_{jt}\right)^{4}\right\\}=\sum_{j=1}^{N}[D]_{ij}^{4}(\mu_{4}-3)+3\leq\mu_{4}.$

Thus,

$\displaystyle\mathbb{E}(W^{\top}BW)=\sum_{r=1}^{N}[B]_{rr}\mathbb{E}\left\\{\left(\sum_{t=0}^{\infty}a_{t}z_{rt}\right)^{2}\right\\}=\mathrm{tr}(B)\sum_{t=0}^{\infty}a_{t}^{2},$

and

$\displaystyle\mathbb{E}[(W^{\top}BW)^{2}]=$ $\displaystyle\sum_{1\leq
r_{1},r_{2},r_{3},r_{4}\leq
N}[B]_{r_{1}r_{2}}[B]_{r_{3}r_{4}}\mathbb{E}\left\\{\prod_{l=1}^{4}\left(\sum_{t=0}^{\infty}a_{t}z_{r_{l},t}\right)\right\\}$
$\displaystyle=$
$\displaystyle\sum_{r=1}^{N}[B]_{rr}^{2}\mathbb{E}\left\\{\left(\sum_{t=0}^{\infty}a_{t}z_{it}\right)^{4}\right\\}$
$\displaystyle+\sum_{1\leq r\neq s\leq
N}(2[B]_{rs}^{2}+[B]_{rr}[B]_{ss})\mathbb{E}\left\\{\left(\sum_{t=0}^{\infty}a_{t}z_{rt}\right)^{2}\left(\sum_{t=0}^{\infty}a_{t}z_{st}\right)^{2}\right\\}$
$\displaystyle=$
$\displaystyle\sum_{r=1}^{N}[B]_{rr}^{2}\left(\sum_{t=0}^{\infty}a_{t}^{4}\mu_{4}+3\sum_{t\neq
s\geq 0}a_{t}^{2}a_{s}^{2}\right)+\sum_{1\leq r\neq s\leq
N}(2[B]_{rs}^{2}+[B]_{rr}[B]_{ss})\left(\sum_{t=0}^{\infty}a_{t}^{2}\right)^{2}$
$\displaystyle\leq$
$\displaystyle\mathrm{tr}^{2}(B)\left(\sum_{t=0}^{\infty}a_{t}^{4}\mu_{4}+3\sum_{t\neq
s\geq
0}a_{t}^{2}a_{s}^{2}\right)+(2\mathrm{tr}(B^{2})+\mathrm{tr}^{2}(B))\left(\sum_{t=0}^{\infty}a_{t}^{2}\right)^{2}$
$\displaystyle\leq$
$\displaystyle(\mu_{4}+6)\left(\sum_{t=0}^{\infty}a_{t}^{2}\right)^{2}\mathrm{tr}^{2}(B).$

Then, taking $\tau_{1}=\mu_{4}+6$ completes the proof.

###### Lemma 2

Under Conditions (C1)-(C5), for all $0\leq k_{1},k_{2},k_{3},k_{4}\leq T$,
there exists a positive constant $\tau_{2}$ such that

$\displaystyle|\mathbb{E}\\{(\bm{{\bm{X}}}_{k_{1}}^{\top}\bm{{\bm{X}}}_{k_{2}}-\mathrm{tr}(\Gamma_{|k_{1}-k_{2}|}))(\bm{{\bm{X}}}_{k_{3}}^{\top}\bm{{\bm{X}}}_{k_{4}}-\mathrm{tr}(\Gamma_{|k_{3}-k_{4}|}))\\}|\leq\tau_{2}\mathrm{tr}(\Omega_{T}^{2}),~{}~{}\text{and}$
$\displaystyle|\mathbb{E}\\{(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}-\mathrm{tr}(\Gamma_{|k_{1}-k_{2}|,M}))(\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}}-\mathrm{tr}(\Gamma_{|k_{3}-k_{4}|,M}))\\}|\leq\tau_{2}\mathrm{tr}(\Omega_{T,M}^{2}).$

Proof. To prove the first result, since

$\displaystyle|\mathbb{E}\\{(\bm{{\bm{X}}}_{k_{1}}^{\top}\bm{{\bm{X}}}_{k_{2}}-\mathrm{tr}(\Gamma_{|k_{1}-k_{2}|}))(\bm{{\bm{X}}}_{k_{3}}^{\top}\bm{{\bm{X}}}_{k_{4}}-\mathrm{tr}(\Gamma_{|k_{3}-k_{4}|}))\\}|$
$\displaystyle\leq$
$\displaystyle[\mathbb{E}\\{(\bm{{\bm{X}}}_{k_{1}}^{\top}\bm{{\bm{X}}}_{k_{2}}-\mathrm{tr}(\Gamma_{|k_{1}-k_{2}|}))^{2}\\}\mathbb{E}\\{(\bm{{\bm{X}}}_{k_{3}}^{\top}\bm{{\bm{X}}}_{k_{4}}-\mathrm{tr}(\Gamma_{|k_{3}-k_{4}|}))^{2}\\}]^{1/2},$

it suffices to show that for all $h\geq 0$,

$\displaystyle\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h}-\mathrm{tr}(\Gamma_{h}))^{2}\\}\leq\tau_{2}\mathrm{tr}(\Omega_{T}^{2}).$

We split $\bm{{\bm{X}}}_{h}$ into two independent parts as

$\displaystyle\bm{{\bm{X}}}_{h}=\bm{\Sigma}^{1/2}\sum_{k=0}^{\infty}b_{k+h}{\bm{z}}_{-k}+\bm{\Sigma}^{1/2}\sum_{k=0}^{h-1}b_{k}{\bm{z}}_{h-k}=:\bm{{\bm{X}}}_{h,(1)}+\bm{{\bm{X}}}_{h,(2)},$

then

$\displaystyle\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h}-\mathrm{tr}(\Gamma_{h}))^{2}\\}=\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h,(1)}-\mathrm{tr}(\Gamma_{h}))^{2}\\}+\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h,(2)})^{2}\\}.$

By the independence of ${\bm{z}}_{it}$’s, we have

$\displaystyle\mathbb{E}\\{({\bm{z}}_{r}\bm{\Sigma}{\bm{z}}_{s})^{2}\\}$
$\displaystyle=\mathbb{E}\\{\mathrm{tr}(\bm{\Sigma}{\bm{z}}_{s}{\bm{z}}_{s}^{\top}\bm{\Sigma}{\bm{z}}_{r}{\bm{z}}_{r}^{\top})\\}=\mathrm{tr}(\bm{\Sigma}^{2}),~{}~{}\text{and}$
$\displaystyle\mathbb{E}\\{({\bm{z}}_{k}^{\top}\bm{\Sigma}{\bm{z}}_{k}-\mathrm{tr}(\bm{\Sigma}))^{2}\\}$
$\displaystyle=\mathbb{E}\left[\left\\{\sum_{1\leq i,j\leq
N}(z_{ik}z_{jk}-\mathbb{I}_{(i=j)})(\bm{\Sigma})_{ij}\right\\}^{2}\right]$
$\displaystyle=\sum_{i=1}^{N}[\bm{\Sigma}]_{ii}^{2}(\mu_{4}-1)+2\sum_{i\neq
j}[\bm{\Sigma}]_{ij}^{2}\leq(\mu_{4}+1)\mathrm{tr}(\bm{\Sigma}^{2}).$

Hence, we have

$\displaystyle\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h,(2)})^{2}\\}=\left(\sum_{k=0}^{h-1}b_{k}^{2}\right)\left(\sum_{k=0}^{\infty}b_{k}^{2}\right)\mathrm{tr}(\bm{\Sigma}^{2}),$

and

$\displaystyle\mathbb{E}\\{(\bm{{\bm{X}}}_{0}^{\top}\bm{{\bm{X}}}_{h,(1)}-\mathrm{tr}(\Gamma_{h}))^{2}\\}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}b_{k}^{2}b_{k+h}^{2}\mathbb{E}\\{({\bm{z}}_{r}\bm{\Sigma}{\bm{z}}_{s})^{2}\\}+\sum_{r\neq
s\geq
0}b_{r}b_{s+h}b_{s}b_{r+h}\mathbb{E}({\bm{z}}_{r}^{\top}\bm{\Sigma}{\bm{z}}_{s}{\bm{z}}_{s}^{\top}\bm{\Sigma}{\bm{z}}_{r})+\sum_{r\neq
s\geq
0}b_{r}^{2}b_{s+h}^{2}\mathbb{E}({\bm{z}}_{r}^{\top}\bm{\Sigma}{\bm{z}}_{s}{\bm{z}}_{r}^{\top}\bm{\Sigma}{\bm{z}}_{s})$
$\displaystyle\leq$
$\displaystyle\sum_{k=0}^{\infty}b_{k}^{2}b_{k+h}^{2}(\mu_{4}+1)\mathrm{tr}(\bm{\Sigma}^{2})+\sum_{r\neq
s\geq
0}(b_{r}b_{s+h}b_{s}b_{r+h}+b_{r}^{2}b_{s+h}^{2})\mathrm{tr}(\bm{\Sigma}^{2})$
$\displaystyle\leq$
$\displaystyle\left\\{\sum_{k=0}^{\infty}b_{k}^{2}b_{k+h}^{2}(\mu_{4}+1)+\left(\sum_{k=0}^{\infty}b_{k}^{2}\right)\left(\sum_{k=0}^{\infty}b_{k+h}^{2}\right)+\left(\sum_{k=0}^{\infty}b_{k}b_{k+h}\right)^{2}\right\\}\mathrm{tr}(\bm{\Sigma}^{2})$
$\displaystyle\leq$
$\displaystyle(\mu_{4}+3)\left(\sum_{k=0}^{\infty}b_{k}^{2}\right)^{2}\mathrm{tr}(\bm{\Sigma}^{2}).$

These results above, together with the fact that
$\mathrm{tr}({\bf\Omega}_{T}^{2})=\sum_{h_{1},h_{2}\in\mathcal{T}}(1-\frac{|h_{1}|}{T})(1-\frac{|h_{2}|}{T})a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2})$,
imply the desired result.

Let $b_{k,M}=b_{k}\mathbb{I}_{(k\leq M)}$ and repeat the above proof process,
then we get the second result.

###### Lemma 3

Let $\\{b_{n}\\}_{n\geq 0}$ be a sequence of numbers. If there exists a
constant $k>1$ such that $b_{n}=o(n^{-k})$, then

$\displaystyle\sum_{m=n}^{\infty}b_{m}=o(n^{-k+1}).$

Proof. For all $\epsilon>0$, there exists $N_{0}\in\mathbb{N}$ such that for
all $n>N_{0}$, $n^{k}b_{n}<\epsilon$. Then for all $n>N_{0}$,

$\displaystyle
n^{k-1}\sum_{m=n}^{\infty}b_{m}=\sum_{l=1}^{\infty}\sum_{m=ln}^{(l+1)n-1}\frac{(nl)^{k}}{nl^{k}}b_{m}\leq\sum_{l=1}^{\infty}\sum_{m=ln}^{(l+1)n-1}\frac{1}{nl^{k}}m^{k}b_{m}\leq\epsilon\sum_{l=1}^{\infty}\frac{1}{l^{k}}.$

Next, we prove Theorem 1 in the following three parts: first, we prove the
asymptotic normality of $T_{{\text{Sum}}}^{G}$, then we use some Gaussian
approximation approach to prove the asymptotic normality of
$T_{{\text{Sum}}}^{NG}$. Finally, we use $T_{{\text{Sum}}}^{NG}$ to
approximate $T_{{\text{Sum}}}$ and complete the proof of Theorem 1.

###### Lemma 4

Under Conditions (C1)-(C5), we have

$\displaystyle\frac{T_{{\text{Sum}}}^{G}-T^{-1}\mathrm{tr}({\bf\Omega}_{T,M})}{\sqrt{2T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\mathop{\rightarrow}\limits^{d}N(0,1).$

Proof. Let $\lambda_{1}\geq\dots\geq\lambda_{N}$ be the eigenvalues of
$\bm{\Sigma}$. By Condition (C3), we have $\lambda_{1}^{2}\leq M_{0}$ and
$\mathrm{tr}(\bm{\Sigma})\geq NM_{1}$. Furthermore, using the fact that
$\mathrm{tr}^{2}(\bm{\Sigma})\leq N\mathrm{tr}(\bm{\Sigma}^{2})$, we have

$\displaystyle\frac{M^{4}\mathrm{tr}(\bm{\Sigma}^{4})}{\mathrm{tr}^{2}(\bm{\Sigma}^{2})}\leq\frac{N^{1/2}\sum_{i=1}^{N}\lambda_{i}^{4}}{(N^{-1}\mathrm{tr}^{2}(\bm{\Sigma}))^{2}}\leq\frac{N^{1/2}NM_{0}^{2}}{(NM_{1}^{2})^{2}}=N^{-1/2}M_{0}^{2}M_{1}^{-4}=o(1),$

which implies that

$\displaystyle\mathrm{tr}(\Gamma_{a,M}\Gamma_{b,M}\Gamma_{c,M}\Gamma_{d,M})=o\\{M^{-4}\mathrm{tr}^{2}({\bf\Omega}_{T,M}^{2})\\},\forall
a,b,c,d\in\mathcal{M}.$

Hence, recalling that
$T_{{\text{Sum}}}^{G}=\bar{\bm{\delta}}_{T}^{\top}\bar{\bm{\delta}}_{T}$ and
$\mathbb{E}(T_{{\text{Sum}}}^{G})=T^{-1}\mathrm{tr}({\bf\Omega}_{T,M})$, and
according to equation (10) in the proof of theorem 2.1 in Cho et al. (2019),
we have

$\displaystyle\frac{T_{{\text{Sum}}}^{G}-T^{-1}\mathrm{tr}({\bf\Omega}_{T,M})}{\sqrt{\mathrm{var}(T_{{\text{Sum}}}^{G})}}\mathop{\rightarrow}\limits^{d}N(0,1).$

Next, we analyze the variance of $T_{{\text{Sum}}}^{G}$. Let
$\tilde{\bm{\delta}}:=(\bm{\delta}_{1}^{\top},\dots,\bm{\delta}_{T}^{\top})^{\top}\in\mathcal{R}^{TN}$
and $\tilde{\bm{\Sigma}}:=\sum_{h\in\mathcal{M}}D_{h}\otimes\Gamma_{h,M}$,
where $D_{h}$ is an $T\times T$ matrix with $[D_{h}]_{ts}=1$ if $t-s=h$ and
all remaining elements equaling zero. Let $\mathbbm{1}\in\mathcal{R}^{T\times
T}$ be the all $1$’s matrix and ${\bf I}_{N}$ be the $N\times N$ identity
matrix. Here we use a property about the fourth moment of multivariate
Gaussian distribution: if ${\bm{X}}\in\mathcal{R}^{k}$ is Gaussian with
$\mathbb{E}({\bm{X}})=\bm{0}$ and $\mathrm{var}({\bm{X}})=\bm{\Sigma}$, then

$\displaystyle\mathrm{cov}({\bm{X}}^{\top}{\bf A}{\bm{X}},{\bm{X}}^{\top}{\bf
B}{\bm{X}})=2\mathrm{tr}({\bf A}\bm{\Sigma}{\bf B}\bm{\Sigma}).$

Then, we have

$\displaystyle\mathrm{var}(T_{{\text{Sum}}}^{G})=$
$\displaystyle\mathrm{var}(T^{-2}\tilde{\bm{\delta}}^{\top}(\mathbbm{1}\otimes{\bf
I}_{N})\tilde{\bm{\delta}})$ $\displaystyle=$ $\displaystyle
2T^{-4}\mathrm{tr}((\mathbbm{1}\otimes{\bf
I}_{N})\tilde{\bm{\Sigma}}(\mathbbm{1}\otimes{\bf I}_{N})\tilde{\bm{\Sigma}})$
$\displaystyle=$ $\displaystyle
2T^{-4}\sum_{a\in\mathcal{M}}\sum_{b\in\mathcal{M}}\mathrm{tr}(\mathbbm{1}D_{a}\mathbbm{1}D_{b})\mathrm{tr}(\Gamma_{a,M}\Gamma_{b,M})$
$\displaystyle=$ $\displaystyle
2T^{-4}\sum_{a\in\mathcal{M}}\sum_{b\in\mathcal{M}}(T-|a|)(T-|b|)\mathrm{tr}(\Gamma_{a,M}\Gamma_{b,M})$
$\displaystyle=$ $\displaystyle 2T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2}).$

In summary, the proof is finished.

###### Lemma 5

Under Conditions (C1)-(C5), we have
$T_{{\text{Sum}}}^{G}-T_{{\text{Sum}}}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Proof. Let
$\mathcal{F}_{ts}^{G}:=T^{-2}(\bm{\delta}_{t}^{\top}\bm{\delta}_{s}-\mathrm{tr}(\Gamma_{|t-s|,M}))$,
then $\sum_{1\leq t,s\leq
T}\mathcal{F}_{ts}^{G}=\bar{\bm{\delta}}_{T}^{\top}\bar{\bm{\delta}}_{T}-T^{-1}\mathrm{tr}({\bf\Omega}_{T,M})$.
For any $T$, choose $\zeta\in(0,1)$ and $c>0$ such that $w_{T}=cT^{\zeta}>M$
and $T=w_{T}q_{T}+r_{T}$, where $0\leq r_{T}<w_{T}$. For $1\leq t,s\leq
q_{T}$, define

$\displaystyle
B_{ts}^{G}:=\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{l=(s-1)w_{T}+1}^{sw_{T}-M}\mathcal{F}_{kl}^{G},$
$\displaystyle
D_{ts}^{G}:=\sum_{k=(t-1)w_{T}+1}^{tw_{T}}\sum_{l=(s-1)w_{T}+1}^{sw_{T}}\mathcal{F}_{kl}^{G}-B_{ts}^{G},$
$\displaystyle
F^{G}:=\sum_{(k,l)\in\\{1,\dots,T\\}^{2}-\\{1,\dots,q_{T}w_{T}\\}^{2}}\mathcal{F}_{kl}^{G}.$
(A.1)

Then, $\sum_{1\leq t,s\leq T}\mathcal{F}_{ts}^{G}=\sum_{1\leq t,s\leq
q_{T}}(B_{ts}^{G}+D_{ts}^{G})+F^{G}$. For non-Gaussian sequence
$\\{\bm{\delta}_{t}\\}_{t=1}^{T}$, we can define $B_{ts}^{NG}$, $D_{ts}^{NG}$
and $F^{NG}$ similarly. We use the unmarked symbol $B_{ts}$, $D_{ts}$ and $F$
when we do not emphasize the difference between Gaussian situation and non-
Gaussian situation. Define

$\displaystyle S_{1}:=\sum_{1\leq t<s\leq
q_{T}}B_{ts},~{}~{}S_{2}:=\sum_{t=1}^{q_{T}}B_{tt},~{}~{}S_{3}:=\sum_{1\leq
t<s\leq q_{T}}D_{ts},~{}~{}S_{4}:=F,$

and $\Delta S_{i}=S_{i}^{G}-S_{i}^{NG}$. Then,
$T_{{\text{Sum}}}^{G}-T_{{\text{Sum}}}^{NG}=\Delta S_{1}+\Delta S_{2}/2+\Delta
S_{3}/2+\Delta S_{4}/2$. Therefore, it suffices to show that $\Delta
S_{i}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$ for
$i=1,2,3,4$.

Step 1. Show that

$\displaystyle\Delta
S_{1}=o_{p}\left\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\right\\}.$
(A.2)

Let $\mathcal{C}_{b}^{3}(\mathcal{R})$ be the class of bounded functions with
continuous derivatives up to order $3$. For
$f\in\mathcal{C}_{b}^{3}(\mathcal{R})$, let $f^{(i)}$ denote the $i$-th
derivative of $f$, $i=1,2,3$. Define

$\displaystyle\bm{\zeta}_{t}:=\frac{1}{w_{T}-M}\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\bm{\delta}_{k}~{}~{}\text{and}~{}~{}\bm{\xi}_{t}:=\frac{1}{w_{T}-M}\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\bm{\gamma}_{k}.$

Then, $\\{\bm{\zeta}_{t}\\}_{t=1}^{q_{T}}$ and
$\\{\bm{\xi}_{t}\\}_{t=1}^{q_{T}}$ are both independent series with
$\mathbb{E}(\bm{\zeta}_{t})=\bm{0}=\mathbb{E}(\bm{\xi}_{t})$ and
$\mathrm{var}(\bm{\zeta}_{t})=\mathrm{var}(\bm{\xi}_{t})=\frac{1}{w_{T}-M}{\bf\Omega}_{w_{T}-M}$,
where
${\bf\Omega}_{w_{T}-M}=\sum_{h\in\mathcal{M}}(1-\frac{|h|}{w_{T}-M})\Gamma_{h,M}$.
Then for $t<s$, we can rewrite $B_{ts}^{G}$ and $B_{ts}^{NG}$ as

$\displaystyle
B_{ts}^{G}=\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{l=(s-1)w_{T}+1}^{sw_{T}-M}T^{-2}\bm{\delta}_{k}^{\top}\bm{\delta}_{l}=\frac{(w_{T}-M)^{2}}{T^{2}}\bm{\zeta}_{t}^{\top}\bm{\zeta}_{s},~{}~{}\text{and}$
$\displaystyle
B_{ts}^{NG}=\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{l=(s-1)w_{T}+1}^{sw_{T}-M}T^{-2}\bm{\gamma}_{k}^{\top}\bm{\gamma}_{l}=\frac{(w_{T}-M)^{2}}{T^{2}}\bm{\xi}_{t}^{\top}\bm{\xi}_{s}.$

Define

$\displaystyle W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}):=\sum_{1\leq t<s\leq
q_{T}}B_{ts}^{G}/\sqrt{2T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}=\sum_{1\leq
t<s\leq q_{T}}u(\bm{\zeta}_{t},\bm{\zeta}_{s}),$ (A.3)

where
$u(\bm{\zeta}_{t},\bm{\zeta}_{s})=\frac{(w_{T}-M)^{2}}{T\sqrt{2\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\bm{\zeta}_{t}^{\top}\bm{\zeta}_{s}$.
Since $\zeta_{t}$ and $\bm{\xi}_{t}$ are linear combinations of ${\bm{z}}$ and
$\mathbb{E}(z_{it}^{4})<\infty$, using the results of Lemma 1, it is easy to
check that they have the following properties: for $1\leq t,s\leq q_{T}$,

  * •

$\forall\bm{a}\in\mathcal{R}^{N},~{}~{}\mathbb{E}(u(\bm{\zeta}_{t},\bm{a}))=\mathbb{E}(u(\bm{a},\bm{\zeta}_{t}))=0,~{}~{}\mathbb{E}(u(\bm{\xi}_{t},\bm{a}))=\mathbb{E}(u(\bm{a},\bm{\xi}_{t}))=0$.

  * •

$\forall\bm{a},\bm{b}\in\mathcal{R}^{N}$,

$\displaystyle\mathbb{E}(u(\bm{a},\bm{\xi}_{t})u(\bm{b},\bm{\xi}_{t}))=\mathbb{E}(u(\bm{a},\bm{\zeta}_{t})u(\bm{b},\bm{\zeta}_{t})),$
$\displaystyle\mathbb{E}(u(\bm{a},\bm{\xi}_{t})u(\bm{\xi}_{t},\bm{b}))=\mathbb{E}(u(\bm{a},\bm{\zeta}_{t})u(\bm{\zeta}_{t},\bm{b})),$
$\displaystyle\mathbb{E}(u(\bm{\xi}_{t},\bm{a})u(\bm{b},\bm{\xi}_{t}))=\mathbb{E}(u(\bm{\zeta}_{t},\bm{a})u(\bm{b},\bm{\zeta}_{t})).$

  * •

Let $c_{ts}^{2}=\mathbb{E}(u(\bm{\xi}_{t},\bm{\xi}_{s})^{2})$ and
$\rho_{0}=\max(\tau_{1},3)$, then

$\displaystyle\max(\mathbb{E}(u(\bm{\xi}_{t},\bm{\xi}_{s})^{4}),\mathbb{E}(u(\bm{\xi}_{t},\bm{\zeta}_{s})^{4}),\mathbb{E}(u(\bm{\zeta}_{t},\bm{\zeta}_{s})^{4}))\leq\rho_{0}c_{ts}^{4}<\infty.$

It is known that a sequence of random variables $\\{Z_{n}\\}_{n=1}^{\infty}$
converges weakly to a random variable $Z$ if and only if for every
$f\in\mathcal{C}_{b}^{3}(\mathbb{R})$,
$\mathbb{E}(f(Z_{n}))\to\mathbb{E}(f(Z))$, see, e.g. Pollard (1984). Hence, it
suffices to show that

$\displaystyle|\mathbb{E}f(W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}))-\mathbb{E}f(W(\bm{\xi}_{1},\dots,\bm{\xi}_{q_{T}}))|\to
0,$

for every $f\in\mathcal{C}_{b}^{3}(\mathbb{R})$ as $(N,T)\to\infty$. Define

$\displaystyle
W_{k}:=W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{k-1},\bm{\xi}_{k},\dots,\bm{\xi}_{q_{T}})~{}~{}\text{for}~{}~{}k=1,\dots,q_{T}+1,~{}~{}\text{and}$
$\displaystyle W_{k,0}:=\sum_{1\leq t<s\leq
k-1}u_{T}(\bm{\zeta}_{t},\bm{\zeta}_{s})+\sum_{k+1\leq t<s\leq
q_{T}}u_{T}(\bm{\xi}_{t},\bm{\xi}_{s})+\sum_{1\leq t\leq k-1,k+1\leq s\leq
q_{T}}u_{T}(\bm{\zeta}_{t},\bm{\xi}_{s}).$ (A.4)

Then,

$\displaystyle|\mathbb{E}f(W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}))-\mathbb{E}f(W(\bm{\xi}_{1},\dots,\bm{\xi}_{q_{T}}))|\leq\sum_{k=1}^{q_{T}}|\mathbb{E}f(W_{k})-\mathbb{E}f(W_{k+1})|.$

By Taylor’s expansion, we have

$\displaystyle
f(W_{k})-f(W_{k,0})=\sum_{i=1}^{2}\frac{1}{i!}f^{(i)}(W_{k,0})(W_{k}-W_{k,0})^{i}+O(|W_{k}-W_{k,0}|^{3}),~{}~{}\text{and}$
$\displaystyle
f(W_{k+1})-f(W_{k,0})=\sum_{i=1}^{2}\frac{1}{i!}f^{(i)}(W_{k,0})(W_{k+1}-W_{k,0})^{i}+O(|W_{k+1}-W_{k,0}|^{3}).$

According to the properties of $u(\bm{\xi}_{t},\bm{\zeta}_{s})$, we have

$\displaystyle\mathbb{E}(W_{k}-W_{k,0}|\bm{\zeta}_{1},\dots,\bm{\zeta}_{k-1},\bm{\xi}_{k+1},\dots,\bm{\xi}_{q_{T}})=\sum_{t=1}^{k-1}\mathbb{E}(u(\bm{\zeta}_{t},\bm{\xi}_{k})|\bm{\zeta}_{t})+\sum_{s=k+1}^{q_{T}}\mathbb{E}(u(\bm{\xi}_{k},\bm{\xi}_{s})|\bm{\xi}_{s})=0,$
$\displaystyle\mathbb{E}(W_{k+1}-W_{k,0}|\bm{\zeta}_{1},\dots,\bm{\zeta}_{k-1},\bm{\xi}_{k+1},\dots,\bm{\xi}_{q_{T}})=\sum_{t=1}^{k-1}\mathbb{E}(u(\bm{\zeta}_{t},\bm{\zeta}_{k})|\bm{\zeta}_{t})+\sum_{s=k+1}^{q_{T}}\mathbb{E}(u(\bm{\zeta}_{k},\bm{\xi}_{s})|\bm{\xi}_{s})=0,$
$\displaystyle\mathbb{E}[(W_{k}-W_{k,0})^{2}|\bm{\zeta}_{1},\dots,\bm{\zeta}_{k-1},\bm{\xi}_{k+1},\dots,\bm{\xi}_{q_{T}}]=\mathbb{E}[(W_{k+1}-W_{k,0})^{2}|\bm{\zeta}_{1},\dots,\bm{\zeta}_{k-1},\bm{\xi}_{k+1},\dots,\bm{\xi}_{q_{T}}].$
(A.5)

Furthermore,

$\displaystyle\mathbb{E}\left[\sum_{i=1}^{2}\frac{1}{i!}f^{(i)}(W_{k,0})(W_{k}-W_{k,0})^{i}\right]=\mathbb{E}\left[\sum_{i=1}^{2}\frac{1}{i!}f^{(i)}(W_{k,0})(W_{k+1}-W_{k,0})^{i}\right].$

Thus, there exists a positive constant $C$ such that

$\displaystyle|\mathbb{E}f(W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}))-\mathbb{E}f(W(\bm{\xi}_{1},\dots,\bm{\xi}_{q_{T}}))|\leq$
$\displaystyle
C\sum_{k=1}^{q_{T}}(\mathbb{E}|W_{k}-W_{k,0}|^{3}+\mathbb{E}|W_{k+1}-W_{k,0}|^{3})$
$\displaystyle\leq$ $\displaystyle
C\sum_{k=1}^{q_{T}}[(\mathbb{E}|W_{k}-W_{k,0}|^{4})^{3/4}+(\mathbb{E}|W_{k+1}-W_{k,0}|^{4})^{3/4}].$

Next, we derive the upper bounds for $\mathbb{E}|W_{k}-W_{k,0}|^{4}$ and
$\mathbb{E}|W_{k+1}-W_{k,0}|^{4}$. According to the properties of
$u(\bm{\xi}_{t},\bm{\zeta}_{s})$, we have

$\displaystyle\mathbb{E}|W_{k}-W_{k,0}|^{4}$ $\displaystyle=$
$\displaystyle\sum_{t=1}^{k-1}\mathbb{E}(u(\bm{\zeta}_{t},\bm{\xi}_{k})^{4})+\sum_{s=k+1}^{q_{T}}\mathbb{E}(u(\bm{\xi}_{k},\bm{\xi}_{s})^{4})+6\sum_{t_{1}=1}^{k-1}\sum_{t_{2}=1}^{k-1}\mathbb{E}(u(\bm{\zeta}_{t_{1}},\bm{\xi}_{k})^{2}u(\bm{\zeta}_{t_{2}},\bm{\xi}_{k})^{2})$
$\displaystyle+6\sum_{s_{1}=k+1}^{q_{T}}\sum_{s_{2}=k+1}^{q_{T}}\mathbb{E}(u(\bm{\xi}_{k},\bm{\xi}_{s_{1}})^{2}u(\bm{\xi}_{k},\bm{\xi}_{s_{2}})^{2})+6\sum_{t=1}^{k-1}\sum_{s=k+1}^{q_{T}}\mathbb{E}(u(\bm{\zeta}_{t},\bm{\xi}_{k})^{2}u(\bm{\xi}_{k},\bm{\xi}_{s})^{2})$
$\displaystyle\leq$
$\displaystyle\rho_{0}\left(\sum_{t=1}^{k-1}c_{tk}^{4}+\sum_{s=k+1}^{q_{T}}c_{ks}^{4}+6\sum_{t_{1}=1}^{k-1}\sum_{t_{2}=1}^{k-1}c_{t_{1},k}^{2}c_{t_{2},k}^{2}+6\sum_{s_{1}=k+1}^{q_{T}}\sum_{s_{2}=k+1}^{q_{T}}c_{k,s_{1}}^{2}c_{k,s_{2}}^{2}+6\sum_{t=1}^{k-1}\sum_{s=k+1}^{q_{T}}c_{tk}^{2}c_{ks}^{2}\right)$
$\displaystyle\leq$ $\displaystyle
3\rho_{0}\left(\sum_{t=1}^{k-1}c_{tk}^{2}+\sum_{s=k+1}^{q_{T}}c_{ks}^{2}\right)^{2}.$
(A.6)

Similarly, we get $\mathbb{E}|W_{k+1}-W_{k,0}|^{4}\leq
3\rho_{0}(\sum_{t=1}^{k-1}c_{tk}^{2}+\sum_{s=k+1}^{q_{T}}c_{ks}^{2})^{2}$.
Thus,

$\displaystyle|\mathbb{E}f(W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}))-\mathbb{E}f(W(\bm{\xi}_{1},\dots,\bm{\xi}_{q_{T}}))|=O\left\\{\sum_{k=1}^{q_{T}}\left(\sum_{t=1}^{k-1}c_{tk}^{2}+\sum_{s=k+1}^{q_{T}}c_{ks}^{2}\right)^{3/2}\right\\}.$

Since $\\{\bm{\xi}_{t}\\}_{t=1}^{q_{T}}$ has the same first two moment as
$\\{\bm{\zeta}_{t}\\}_{t=1}^{q_{T}}$, we have

$\displaystyle c_{ts}^{2}=$
$\displaystyle\frac{(w_{T}-M)^{2}}{2T^{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\mathbb{E}[(\bm{\xi}_{t}^{\top}\bm{\xi}_{s})^{2}]$
$\displaystyle=$
$\displaystyle\frac{1}{2T^{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\sum_{k_{1}=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{k_{2}=(s-1)w_{T}+1}^{sw_{T}-M}\sum_{k_{3}=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{k_{4}=(s-1)w_{T}+1}^{sw_{T}-M}\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})$
$\displaystyle=$
$\displaystyle\frac{1}{2T^{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\mathrm{tr}\left\\{\left(\sum_{k_{1}=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{k_{2}=(t-1)w_{T}+1}^{tw_{T}-M}\Gamma_{k_{1}-k_{2},M}\right)^{2}\right\\}$
$\displaystyle=$
$\displaystyle\frac{(w_{T}-M)^{2}\mathrm{tr}({\bf\Omega}_{w_{T}-M}^{2})}{2T^{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\{1+o(1)\\}.$

Furthermore,

$\displaystyle|\mathbb{E}f(W(\bm{\zeta}_{1},\dots,\bm{\zeta}_{q_{T}}))-\mathbb{E}f(W(\bm{\xi}_{1},\dots,\bm{\xi}_{q_{T}}))|=O\left[q_{T}\left\\{q_{T}\frac{(w_{T}-M)^{2}\mathrm{tr}({\bf\Omega}_{w_{T}-M}^{2})}{2T^{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\right\\}^{3/2}\right]=o(1).$

In summary, (6.1) is proved.

Step 2. Show that

$\displaystyle\Delta
S_{2}=\sum_{t=1}^{q_{T}}B_{tt}^{G}-\sum_{t=1}^{q_{T}}B_{tt}^{NG}=o_{p}\left\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\right\\}.$
(A.7)

By the classical CLT, it holds that
$\sum_{t=1}^{q_{T}}B_{tt}/\sqrt{q_{T}\mathrm{var}(B_{11})}\mathop{\rightarrow}\limits^{d}N(0,1)$.
Since
$\mathrm{var}(B_{11})=(w_{T}-M)^{2}T^{-4}\mathrm{tr}({\bf\Omega}_{w_{T}-M}^{2})$,
we have

$\displaystyle\frac{\sum_{t=1}^{q_{T}}B_{tt}}{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}=\frac{1}{\sqrt{q_{T}}}\sqrt{\frac{q_{T}^{2}(w_{T}-M)^{2}T^{-4}\mathrm{tr}({\bf\Omega}_{w_{T}-M}^{2})}{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\frac{\sum_{t=1}^{q_{T}}B_{tt}}{\sqrt{q_{T}\mathrm{var}(B_{11})}}=o_{p}(1).$

Step 3. Show that

$\displaystyle\Delta S_{3}=\sum_{1\leq t,s\leq q_{T}}D_{ts}^{G}-\sum_{1\leq
t,s\leq
q_{T}}D_{ts}^{NG}=o_{p}\left\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\right\\}.$
(A.8)

Define

$\displaystyle
L_{ts}:=\sum_{k=tw_{T}-M+1}^{tw_{T}}\sum_{l=(s-1)w_{T}+1}^{sw_{T}-M}\mathcal{F}_{kl},$
$\displaystyle
R_{ts}:=\sum_{k=(t-1)w_{T}+1}^{tw_{T}-M}\sum_{l=sw_{T}-M+1}^{sw_{T}}\mathcal{F}_{kl},$
$\displaystyle
C_{ts}:=\sum_{k=tw_{T}-M+1}^{tw_{T}}\sum_{l=sw_{T}-M+1}^{sw_{T}}\mathcal{F}_{kl}.$

Then, $D_{ts}=L_{ts}+R_{ts}+C_{ts}=2L_{ts}+C_{ts}$. Define

$\displaystyle\mathcal{L}_{1}=\sum_{t=1}^{q_{T}-1}\sum_{s=t+2}^{q_{T}}L_{ts},~{}~{}\mathcal{L}_{2}=\sum_{t=2}^{q_{T}}\sum_{s=1}^{t-1}L_{ts},$
$\displaystyle~{}~{}\mathcal{L}_{3}=\sum_{t=1}^{q_{T}}L_{tt},~{}~{}\mathcal{L}_{3}=\sum_{t=1}^{q_{T}}L_{t,t+1}$
$\displaystyle\mathcal{C}_{1}=\sum_{t=1}^{q_{T}}\sum_{s=t+1}^{q_{T}}C_{ts},~{}~{}$
$\displaystyle\mathcal{C}_{2}=\sum_{t=1}^{q_{T}}C_{tt}.$

Then,
$S_{3}=2\mathcal{L}_{1}+2\mathcal{L}_{2}+2\mathcal{L}_{3}+2\mathcal{L}_{4}+2\mathcal{C}_{1}+\mathcal{C}_{2}$.

Next, we analyze the second moment of $\mathcal{L}_{i}$’s and
$\mathcal{C}_{i}$’s. For $k_{1},k_{2},k_{3},k_{4}\in\\{1,\dots,T\\}^{4}$, by
the M-dependence of $\bm{\gamma}_{t}$’s,
$\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})\neq
0$ only if $k_{(2)}-k_{(1)}\leq M$ and $k_{(4)}-k_{(3)}\leq M$, where
$k_{(1)},\dots,k_{(4)}$ are the sorted $k_{1},\dots,k_{4}$. Accordingly,

$\displaystyle\mathbb{E}\\{(\mathcal{L}_{1}^{NG})^{2}\\}$ $\displaystyle=$
$\displaystyle\sum_{t_{1}=1}^{q_{T}-1}\sum_{s_{1}=t_{1}+2}^{q_{T}}\sum_{t_{2}=1}^{q_{T}-1}\sum_{s_{2}=t_{2}+2}^{q_{T}}\mathbb{E}(L_{t_{1}s_{1}}^{NG}L_{t_{2}s_{2}}^{NG})$
$\displaystyle=$
$\displaystyle\sum_{t_{1}=1}^{q_{T}-1}\sum_{s_{1}=t_{1}+2}^{q_{T}}\sum_{t_{2}=1}^{q_{T}-1}\sum_{s_{2}=t_{2}+2}^{q_{T}}\sum_{k_{1}=t_{1}w_{T}-M+1}^{t_{1}w_{T}}\sum_{k_{2}=(s_{1}-1)w_{T}+1}^{s_{1}w_{T}-M}\sum_{k_{3}=t_{2}w_{T}-M+1}^{t_{2}w_{T}}\sum_{k_{4}=(s_{2}-1)w_{T}+1}^{s_{2}w_{T}-M}\frac{\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})}{T^{4}}$
$\displaystyle=$
$\displaystyle\frac{1}{T^{4}}\sum_{t=1}^{q_{T}-1}\sum_{s=t+2}^{q_{T}}\mathbb{E}\left\\{\left(\sum_{k=(t-1)w_{T}-M+1}^{tw_{T}}\sum_{l=(s-1)W_{T}+1}^{sw_{T}-M}\bm{\gamma}_{k}\top\bm{\gamma}_{l}\right)^{2}\right\\}$
$\displaystyle=$
$\displaystyle\frac{1}{T^{4}}\frac{(q_{T}-1)(q_{T}-2)}{2}M(w_{T}-M)\mathrm{tr}({\bf\Omega}_{M}{\bf\Omega}_{w_{T}-M})$
$\displaystyle\leq$
$\displaystyle\frac{1}{T^{4}}\frac{(q_{T}-1)(q_{T}-2)}{2}M(w_{T}-M)\\{\mathrm{tr}({\bf\Omega}_{M}^{2})\mathrm{tr}({\bf\Omega}_{w_{T}-M}^{2})\\}^{1/2}$
$\displaystyle=$ $\displaystyle
O\\{q_{T}MT^{-3}\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\}=o\\{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\}.$

Here, we explain the third equation as follows. Since $s_{1}\geq t_{1}+2$,
which implies $k_{2}-k_{1}\geq M$,
$\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})\neq
0$ only if $t_{1}=t_{2}$ or $t_{1}=s_{2}$ or $t_{1}=s_{2}-1$. If
$t_{1}=t_{2}$, then
$\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})\neq
0$ only if $s_{1}=s_{2}$. If $t_{1}=s_{2}$ or $t_{1}=s_{2}-1$, then $t_{2}\leq
s_{2}-2\leq t_{1}-1$, which implies
$k_{(2)}-k_{(1)}=\min(k_{1}-k_{3},k_{4}-k_{3})>M$. Therefore,
$\mathbb{E}(\bm{\gamma}_{k_{1}}^{\top}\bm{\gamma}_{k_{2}}\bm{\gamma}_{k_{3}}^{\top}\bm{\gamma}_{k_{4}})\neq
0$ only if $t_{1}=t_{2}$ and $s_{1}=s_{2}$. Then, we get
$\mathcal{L}_{1}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
Similarly, we get
$\mathcal{L}_{2}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{C}_{1}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{L}_{1}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{L}_{2}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$
and
$\mathcal{C}_{1}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

$\displaystyle\mathbb{E}\\{(\mathcal{L}_{3}^{NG})^{2}\\}=$
$\displaystyle\sum_{t_{1}=1}^{q_{T}}\sum_{t_{2}=1}^{q_{T}}\mathbb{E}(L_{t_{1}t_{1}}^{NG}L_{t_{2}t_{2}}^{NG})$
$\displaystyle=$
$\displaystyle\sum_{t_{1}=1}^{q_{T}}\sum_{t_{2}=1}^{q_{T}}\sum_{k_{1}=t_{1}w_{T}-M+1}^{t_{1}w-T}\sum_{k_{2}=(t_{1}-1)w_{T}+1}^{t_{1}w_{T}-M}\sum_{k_{3}=t_{2}w_{T}-M+1}^{t_{2}w-T}\sum_{k_{4}=(t_{2}-1)w_{T}+1}^{t_{2}w_{T}-M}\mathbb{E}(\mathcal{F}_{k_{1}k_{2}}^{NG}\mathcal{F}_{k_{3}k_{4}}^{NG})$
$\displaystyle=$
$\displaystyle\sum_{t=1}^{q_{T}}\mathbb{E}\left\\{\left(\sum_{k=tw_{T}-M+1}^{tw_{T}}\sum_{l=(t-1)w_{T}+1}^{tw_{T}-M}\mathcal{F}_{kl}^{NG}\right)^{2}\right\\}$
$\displaystyle\leq$ $\displaystyle
T^{-4}q_{T}M^{2}(w_{T}-M)^{2}\tau_{2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})=o(T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})).$

Here, the third equation follows from the independence of
$\mathcal{F}_{k_{1}k_{2}}$ and $\mathcal{F}_{k_{3}k_{4}}$, and the fourth
inequality follows from Lemma 2. Then, we get
$\mathcal{L}_{3}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
Similarly, we get
$\mathcal{L}_{4}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{C}_{2}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{L}_{3}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$,
$\mathcal{L}_{4}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$
and
$\mathcal{C}_{2}^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
In summary, we get (A.8).

Step 4. Show that

$\displaystyle\Delta
S_{4}=F^{G}-F^{NG}=o_{p}\left\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\right\\}.$
(A.9)

Define
$\mathcal{A}=\\{q_{T}w_{T}+1,\dots,T\\}\times\\{1,\dots,T\\}\times\\{q_{T}w_{T}+1,\dots,T\\}\times\\{1,\dots,T\\}$,
$\mathcal{A}_{1}=\\{(k_{1}.k_{2},k_{3},k_{4})\in\\{1,\dots,T\\}^{4}:|k_{1}-k_{3}|\geq
3M,\min(k_{1},k_{3})>q_{T}w_{T}\\}$ and
$\mathcal{A}_{2}=\\{(k_{1}.k_{2},k_{3},k_{4})\in\\{1,\dots,T\\}^{4}:|k_{1}-k_{3}|\geq
3M,\min(k_{1},k_{3})>q_{T}w_{T},\min(|k_{1}-k_{2}|,|k_{1}-k_{4}|,|k_{3}-k_{2}|,|k_{3}-k_{4}|)\leq
M\\}$. Then, for all
$(k_{1}.k_{2},k_{3},k_{4})\in\mathcal{A}_{1}-\mathcal{A}_{2}$, we have
$\mathbb{E}(\mathcal{F}_{k_{1}k_{2}}\mathcal{F}_{k_{3}k_{4}})=0$. Since
$\mathcal{A}_{2}\subset\\{q_{T}w_{T}+1,\dots,T\\}^{4}$,
$|\mathcal{A}_{2}|\leq(w_{T}+M)^{4}$. Similarly, we have
$|\mathcal{A}-\mathcal{A}_{1}|\leq 6MTw_{T}^{2}$. Accordingly, we have

$\displaystyle\mathbb{E}\\{(F^{NG})^{2}\\}=$
$\displaystyle\mathbb{E}\left\\{\left(2\sum_{k=1}^{q_{T}w_{T}}\sum_{l=q_{T}w_{T}+1}^{T}\mathcal{F}^{NG}+\sum_{k=q_{T}w_{T}+1}^{T}\sum_{l=q_{T}w_{T}+1}^{T}\mathcal{F}^{NG}\right)^{2}\right\\}$
$\displaystyle\leq$ $\displaystyle
4\sum_{k_{1}=1}^{T}\sum_{k_{2}=q_{T}w_{T}+1}^{T}\sum_{k_{3}=1}^{T}\sum_{k_{4}=q_{T}w_{T}+1}^{T}|\mathbb{E}(\mathcal{F}_{k_{1}k_{2}}\mathcal{F}_{k_{3}k_{4}})|$
$\displaystyle=$ $\displaystyle
4\left(\sum_{(k_{1},k_{2},k_{3},k_{4})\in\mathcal{A}_{1}}+\sum_{(k_{1},k_{2},k_{3},k_{4})\in\mathcal{A}_{2}}\right)|\mathbb{E}(\mathcal{F}_{k_{1}k_{2}}\mathcal{F}_{k_{3}k_{4}})|$
$\displaystyle\leq$ $\displaystyle
4\\{6MTw_{T}^{2}+(w_{T}+M)^{4}\\}T^{-4}\mathrm{tr}(\Omega_{T,M}^{2})=o(T^{-2}\mathrm{tr}(\Omega_{T,M}^{2})).$

Thus, $F^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
Similarly, we get
$F^{G}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$. Then,
$\Delta S_{4}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

In summary, (A.2) and (A.7)-(A.9) yield Lemma 5.

###### Lemma 6

Under Conditions (C1)-(C5), we have
$\mathbb{E}(T_{{\text{Sum}}}^{NG})-\mu_{T}^{\prime}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Proof. By the definitions and the fact that $\mathrm{tr}^{2}(\bm{\Sigma})\leq
N\mathrm{tr}(\bm{\Sigma}^{2})$, we have

$\displaystyle|\mu_{T}^{\prime}-\mathbb{E}(T_{{\text{Sum}}}^{NG})|=$
$\displaystyle\left|\frac{1}{T}\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\mathrm{tr}(\Gamma_{h}-\Gamma_{h,M})\right|$
$\displaystyle=$
$\displaystyle\left|\frac{1}{T}\mathrm{tr}(\bm{\Sigma})\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\sum_{k=M-h+1}^{\infty}b_{k}b_{k+h}\right|$
$\displaystyle\leq$
$\displaystyle\frac{2}{T}\\{N\mathrm{tr}(\bm{\Sigma}^{2})\\}^{1/2}\left(\sum_{k=0}^{\infty}|b_{k}|\right)\left(\sum_{k=M+1}^{\infty}|b_{k}|\right)$
$\displaystyle=$ $\displaystyle
o\\{N^{1/2}T^{-1}\mathrm{tr}^{1/2}(\bm{\Sigma}^{2})M^{-4}\\}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\},$

where the last equality is because

$\displaystyle\sum_{k=M}^{\infty}|b_{k}|=o(M^{-4})$ (A.10)

due to Condition (C2) (ii) and Lemma 3.

###### Lemma 7

Under Conditions (C1)-(C5), we have
$\tilde{T}_{{\text{Sum}}}-T_{{\text{Sum}}}^{NG}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Proof. By the definitions, we have

$\displaystyle\tilde{T}_{{\text{Sum}}}-T_{{\text{Sum}}}^{NG}=$
$\displaystyle\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}({\bm{X}}_{t}^{\top}{\bm{X}}_{s}-\bm{\gamma}_{t}^{\top}\bm{\gamma}_{s})$
$\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}({\bm{X}}_{t}-\bm{\gamma}_{t})^{\top}({\bm{X}}_{s}-\bm{\gamma}_{s})+\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\gamma}_{t}^{\top}({\bm{X}}_{s}-\bm{\gamma}_{s}).$

Define
$\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}=\\{(k_{1},k_{2},k_{3},k_{4}):t_{1}-M\leq
k_{1}\leq t_{1},k_{2}\leq s_{1}-M-1,t_{2}-M\leq k_{3}\leq t_{2},k_{4}\leq
s_{2}-M-1\\}$, $\mathcal{S}_{1}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{2}\neq
k_{3}=k_{4}\\}$, $\mathcal{S}_{2}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{3}\neq
k_{2}=k_{4}\\}$, $\mathcal{S}_{3}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{4}\neq
k_{2}=k_{3}\\}$ and
$\mathcal{S}_{4}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{2}=k_{3}=k_{4}\\}$.
Then,

$\displaystyle\mathrm{var}\left\\{\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\gamma}_{t}^{\top}({\bm{X}}_{s}-\bm{\gamma}_{s})\right\\}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\end{subarray}}b_{t_{1}-k_{1}}b_{s_{1}-k_{2}}b_{t_{2}-k_{3}}b_{s_{2}-k_{4}}\mathbb{E}\left\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}}-\mathbb{I}_{(k_{1}=k_{2})}\mathrm{tr}(\bm{\Sigma}))({\bm{z}}_{k_{3}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{4}}-\mathbb{I}_{(k_{3}=k_{4})}\mathrm{tr}(\bm{\Sigma}))\right\\}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{1}\end{subarray}}b_{t_{1}-k_{1}}b_{s_{1}-k_{1}}b_{t_{2}-k_{3}}b_{s_{2}-k_{3}}\mathbb{E}\left\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{1}}-\mathrm{tr}(\bm{\Sigma}))({\bm{z}}_{k_{3}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{3}}-\mathrm{tr}(\bm{\Sigma}))\right\\}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{2}\end{subarray}}b_{t_{1}-k_{1}}b_{s_{1}-k_{2}}b_{t_{2}-k_{1}}b_{s_{2}-k_{2}}\mathbb{E}\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}})^{2}\\}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{3}\end{subarray}}b_{t_{1}-k_{1}}b_{s_{1}-k_{2}}b_{t_{2}-k_{2}}b_{s_{2}-k_{1}}\mathbb{E}\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}})({\bm{z}}_{k_{2}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{1}})\\}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{4}\end{subarray}}b_{t_{1}-k_{1}}b_{s_{1}-k_{1}}b_{t_{2}-k_{1}}b_{s_{2}-k_{1}}\mathbb{E}\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{1}}-\mathrm{tr}(\bm{\Sigma}))^{2}\\}$
$\displaystyle\leq$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{2}\end{subarray}}|b_{t_{1}-k_{1}}b_{s_{1}-k_{2}}b_{t_{2}-k_{1}}b_{s_{2}-k_{2}}|\mathrm{tr}(\bm{\Sigma}^{2})$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{3}\end{subarray}}|b_{t_{1}-k_{1}}b_{s_{1}-k_{2}}b_{t_{2}-k_{2}}b_{s_{2}-k_{1}}|\mathrm{tr}(\bm{\Sigma}^{2})$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},s_{1},t_{2},s_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}_{t_{1}s_{1}t_{2}s_{2}}\cap\mathcal{S}_{4}\end{subarray}}|b_{t_{1}-k_{1}}b_{s_{1}-k_{1}}b_{t_{2}-k_{1}}b_{s_{2}-k_{1}}|\tau_{1}\mathrm{tr}^{2}(\bm{\Sigma})$
$\displaystyle\leq$ $\displaystyle T^{2}\left\\{\left(\sum_{t_{1},t_{2}\geq
0}|b_{t_{1}}b_{t_{2}}|\right)\left(\sum_{s_{1},s_{2}\geq
M}|b_{s_{1}}b_{s_{2}}|\right)+\left(\sum_{t_{1}\geq 0,s_{2}\geq
M}|b_{t_{1}}b_{s_{2}}|\right)\left(\sum_{t_{2}\geq 0,s_{1}\geq
M}|b_{t_{2}}b_{s_{1}}|\right)\right\\}\mathrm{tr}(\bm{\Sigma}^{2})$
$\displaystyle+\tau_{1}NT\sum_{t_{1},t_{2}\geq 0,s_{1}1,s_{2}\geq
M}|b_{t_{1}}b_{t_{2}}b_{s_{1}}b_{s_{2}}|\mathrm{tr}(\bm{\Sigma}^{2})=o\\{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\},$

where the first inequality comes from Lemma 1 and the last equation comes from
(A.10). Hence, we get
$T^{-2}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\gamma}_{t}^{\top}({\bm{X}}_{s}-\bm{\gamma}_{s})=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
Similarly, we can prove
$T^{-2}\sum_{t=1}^{T}\sum_{s=1}^{T}({\bm{X}}_{t}-\bm{\gamma}_{t})^{\top}({\bm{X}}_{s}-\bm{\gamma}_{s})=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
This completes the proof.

###### Lemma 8

Under Conditions (C1)-(C5), we have
$T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})=\sigma_{T}^{2}\\{1+o(1)\\}$.

Proof. Recalling the definitions, we have

$\displaystyle{\bf\Omega}_{T}-{\bf\Omega}_{T,M}=\left[\left\\{a_{0}+2\sum_{h=1}^{T}\left(1-\frac{h}{T}\right)a_{h}\right\\}-\left\\{a_{0,M}+2\sum_{h=1}^{M}\left(1-\frac{h}{T}\right)a_{h,M}\right\\}\right]\bm{\Sigma}=:B_{0}\bm{\Sigma},$

and

$\displaystyle|B_{0}|\leq$ $\displaystyle
2\left|\sum_{h=0}^{T}\left(1-\frac{h}{T}\right)a_{h}-\sum_{h=0}^{M}\left(1-\frac{h}{T}\right)a_{h,M}\right|$
$\displaystyle=$ $\displaystyle
2\left|\sum_{h=M+1}^{T}\left(1-\frac{h}{T}\right)\sum_{k=0}^{\infty}b_{k}b_{k+h}+\sum_{h=0}^{M}\left(1-\frac{h}{T}\right)\sum_{k=M-h+1}^{\infty}b_{k}b_{k+h}\right|$
$\displaystyle\leq$ $\displaystyle
4\left(\sum_{k=0}^{\infty}|b_{k}|\right)\left(\sum_{l=M+1}^{\infty}|b_{l}|\right)=o(M^{-4})$

due to (A.10). Then, using the fact that

$\displaystyle\sigma_{T}^{2}-T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})=$
$\displaystyle
T^{-2}\\{\mathrm{tr}({\bf\Omega}_{T}^{2})-\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\}$
$\displaystyle=$ $\displaystyle
T^{-2}\mathrm{tr}\\{({\bf\Omega}_{T}-{\bf\Omega}_{T,M})^{2}\\}+2T^{-2}\mathrm{tr}\\{{\bf\Omega}_{T,M}({\bf\Omega}_{T}-{\bf\Omega}_{T,M})\\},$

we have $T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})=\sigma_{T}^{2}\\{1+o(1)\\}$.

###### Lemma 9

Under Conditions (C1)-(C6), we have
$T_{{\text{Sum}}}-\tilde{T}_{{\text{Sum}}}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Proof. Since $|1-{\bm{f}}_{t}^{\top}({\bf F}^{\top}{\bf F})^{-1}{\bf
F}^{\top}\bm{1}_{T}|\leq 1+\sum_{k=1}^{p}|f_{tk}|\cdot\|({\bf F}^{\top}{\bf
F}/T)^{-1}\|_{\infty}\cdot\|{\bf F}^{\top}\bm{1}_{T}/T\|_{\infty}\leq
Cp(1+\|{\bm{f}}_{t}\|)\leq C^{\prime}$ for some positive constants
$C,C^{\prime}$, we have $\bm{1}_{T}^{\top}{\bf M}_{\bf F}\bm{1}_{T}/T=O(1)$
and $\eta_{t}=O(1)$. Define $\zeta_{t}:=\eta_{t}-e_{t}$. By the definition,

$\displaystyle\zeta_{t}=$ $\displaystyle\frac{T}{\bm{1}_{T}^{\top}{\bf M}_{\bf
F}\bm{1}_{T}}\\{1-{\bm{f}}_{t}^{\top}({\bf F}^{\top}{\bf F})^{-1}{\bf
F}^{\top}\bm{1}_{T}\\}-\frac{1}{\omega}(1-{\bm{f}}_{t}^{\top}{\bf\Lambda}_{{\bm{f}}}^{-1}{\bm{\mu}}_{{\bm{f}}})$
$\displaystyle=$ $\displaystyle\frac{T}{\bm{1}_{T}^{\top}{\bf M}_{\bf
F}\bm{1}_{T}}\\{{\bm{f}}_{t}^{\top}{\bf\Lambda}_{{\bm{f}}}^{-1}{\bm{\mu}}_{{\bm{f}}}-{\bm{f}}_{t}^{\top}({\bf
F}^{\top}{\bf F})^{-1}{\bf
F}^{\top}\bm{1}_{T}\\}+\left(\frac{T}{\bm{1}_{T}^{\top}{\bf M}_{\bf
F}\bm{1}_{T}}-\frac{1}{\omega}\right)(1-{\bm{f}}_{t}^{\top}{\bf\Lambda}_{{\bm{f}}}^{-1}{\bm{\mu}}_{{\bm{f}}})$

By Condition (C1), we have $T^{-1}\bm{1}_{T}^{\top}{\bf M}_{\bf
F}\bm{1}_{T}=\omega+O_{p}(T^{-1/2})$, $T^{-1}{\bf
F}^{\top}\bm{1}——T={\bm{\mu}}_{{\bm{f}}}+O_{p}(T^{-1/2})$, and $T^{-1}{\bf
F}^{\top}{\bf F}={\bf\Lambda}_{{\bm{f}}}+O_{p}(T^{-1/2})$. Additionally,
$\|{\bm{f}}_{t}\|$ is bounded, so

$\displaystyle\zeta_{t}=O_{p}(T^{-1/2})~{}~{}\text{and}~{}~{}e_{t}=O_{p}(1).$
(A.11)

By the definitions, we have

$\displaystyle T_{{\text{Sum}}}-\tilde{T}_{{\text{Sum}}}=$
$\displaystyle(\tilde{\bm{\alpha}}+\frac{1}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}\zeta_{T})^{\top}(\tilde{\bm{\alpha}}+\frac{1}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}\zeta_{T})-\tilde{\bm{\alpha}}^{\top}\tilde{\bm{\alpha}}$
$\displaystyle=$
$\displaystyle\frac{2}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}^{\top}\tilde{\bm{\alpha}}\zeta_{t}+\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\varepsilon}_{t}^{\top}\bm{\varepsilon}_{s}\zeta_{t}\zeta_{s}$
$\displaystyle=$
$\displaystyle\frac{2}{T}\sum_{t=1}^{T}\bm{\varepsilon}_{t}^{\top}{\bm{\alpha}}\zeta_{t}+\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\varepsilon}_{t}^{\top}\bm{\varepsilon}_{s}\zeta_{t}e_{s}+\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\varepsilon}_{t}^{\top}\bm{\varepsilon}_{s}\zeta_{t}\zeta_{s}$
$\displaystyle=:$ $\displaystyle A_{1}+A_{2}+A_{3}.$

We consider the first term $A_{1}$. $\mathbb{E}(A_{1})=0$ and

$\displaystyle E(A_{1}^{2})=$ $\displaystyle
E\left(\frac{4}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\alpha}^{\top}\bm{\varepsilon}_{t}\bm{\varepsilon}_{s}^{\top}\bm{\alpha}\zeta_{t}\zeta_{s}\right)$
$\displaystyle=$ $\displaystyle
E\left(\frac{4}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\alpha}^{\top}{\bm{X}}_{t}{\bm{X}}_{s}^{\top}\bm{\alpha}\zeta_{t}\zeta_{s}e_{t}^{-1}e_{s}^{-1}\right)$
$\displaystyle=$ $\displaystyle
O(T^{-3})\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\alpha}^{\top}\Gamma_{t-s}\bm{\alpha}$
$\displaystyle=$ $\displaystyle
O(T^{-3}\bm{\alpha}^{\top}\Omega_{T}\bm{\alpha})=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\},$

by Condition (C6) (see the detail proof of this term in the proof of Theorem
2). Hence, $A_{1}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.
Similarly, we have

$\displaystyle
E(A_{2})=\frac{2}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}E({\bm{X}}_{t}^{\top}{\bm{X}}_{s}\zeta_{t}e_{t}^{-1})=O(T^{-3/2})\mathrm{tr}({\bf\Omega}_{T})=o\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\},$

by condition (C2) (iii). And
$\mathrm{var}(A_{2})=O\\{T^{-3}\mathrm{tr}({\bf\Omega}_{T}^{2})\\}=o\\{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\}$.
Taking the same procedure, we have
$E(A_{3})=O\\{T^{-2}\mathrm{tr}({\bf\Omega}_{T})\\}=o\\{\sqrt{T^{-2}\mathrm{tr}(\Omega_{T,M}^{2})}\\}$
and
$\mathrm{var}(A_{3})=O\\{T^{-4}\mathrm{tr}({\bf\Omega}_{T}^{2})\\}=o\\{T^{-2}\mathrm{tr}(\Omega_{T,M}^{2})\\}$.
Here we complete the proof.

####  6.1.1 Proof of Theorem 1

Proof. Note that

$\displaystyle\frac{T_{{\text{Sum}}}-\mu_{T}^{\prime}}{\sigma_{T}}=$
$\displaystyle\sqrt{\frac{2T^{-2}\mathrm{tr}(\Omega_{T,M}^{2})}{\sigma_{T}^{2}}}\frac{1}{\sqrt{2T^{-2}\mathrm{tr}(\Omega_{T,M}^{2})}}\\{(T_{{\text{Sum}}}-\tilde{T}_{{\text{Sum}}})+(\tilde{T}_{{\text{Sum}}}-T_{{\text{Sum}}}^{NG})$
$\displaystyle+(T_{{\text{Sum}}}^{NG}-T_{{\text{Sum}}}^{G})+(T_{{\text{Sum}}}^{G}-\mathbb{E}(T_{{\text{Sum}}}^{G}))+(\mathbb{E}(T_{{\text{Sum}}}^{G})-\mu_{T}^{\prime})\\}.$
(A.12)

According to Lemmas 4-9, we can easily obtain the result by Slutsky’s Theorem.

####  6.1.2 Proof of Theorem 2

Proof. According to the proof of Theorem 1, we have

$\displaystyle
T_{{\text{Sum}}}=\bar{\bm{\gamma}}_{T}^{\top}\bar{\bm{\gamma}}_{T}+\bm{\alpha}^{\top}\bm{\alpha}+\bar{\bm{\gamma}}_{T}^{\top}\bm{\alpha}+o_{p}(\sigma_{T}).$

Thus, to proof the result, we only need to show that
$\mathbb{E}\\{(\bar{\bm{\gamma}}_{T}^{\top}\bm{\alpha})^{2}\\}=\frac{1}{T}\bm{\alpha}^{\top}{\bf\Omega}_{T,M}\bm{\alpha}=o(\sigma_{T}^{2})$.
Note that
${\bf\Omega}_{T,M}=\sum_{h\in\mathcal{M}}(1-\frac{|h|}{T})\Gamma_{h,M}$. For
each $h$, we consider the singular value decomposition of $\Gamma_{h,M}$ as
$\Gamma_{h,M}={\bf U}\Lambda{\bm{V}}^{\top}$, where $\Lambda$ is the diagonal
matrix with positive square root of eigenvalues of
$\Gamma_{h,M}\Gamma_{h,M}^{\top}$. It yields that
$\Gamma_{h,M}\Gamma_{-h,M}={\bf U}\Lambda^{2}{\bf U}^{\top}$ which gives ${\bf
U}\Lambda{\bf U}^{\top}=(\Gamma_{h,M}\Gamma_{-h,M})^{1/2}$. Similarly, we have
${\bm{V}}\Lambda{\bm{V}}^{\top}=(\Gamma_{-h,M}\Gamma_{h,M})^{1/2}$. Thus,

$\displaystyle\bm{\alpha}^{\top}\Gamma_{h,M}\bm{\alpha}=$
$\displaystyle\bm{\alpha}^{\top}{\bf
U}\Lambda{\bm{V}}^{\top}\bm{\alpha}=(\Lambda^{1/2}{\bf
U}^{\top}\bm{\alpha})^{\top}(\Lambda^{1/2}{\bm{V}}^{\top}\bm{\alpha})$
$\displaystyle\leq$
$\displaystyle\sqrt{\bm{\alpha}^{\top}(\Gamma_{h,M}\Gamma_{-h,M})^{1/2}\bm{\alpha}}\sqrt{\bm{\alpha}^{\top}(\Gamma_{-h,M}\Gamma_{h,M})^{1/2}\bm{\alpha}},$

where the inequality comes from the Cauchy-Schwarz inequality and

$\displaystyle(\Lambda^{1/2}{\bf
U}^{\top}\bm{\alpha})^{\top}(\Lambda^{1/2}{\bf
U}^{\top}\bm{\alpha})=\bm{\alpha}^{\top}{\bf U}\Lambda{\bf
U}^{\top}\bm{\alpha}=\bm{\alpha}^{\top}(\Gamma_{h,M}\Gamma_{-h,M})^{1/2}\bm{\alpha},$
$\displaystyle(\Lambda^{1/2}{\bm{V}}^{\top}\bm{\alpha})^{\top}(\Lambda^{1/2}{\bm{V}}^{\top}\bm{\alpha})=\bm{\alpha}^{\top}{\bm{V}}\Lambda{\bm{V}}^{\top}\bm{\alpha}=\bm{\alpha}^{\top}(\Gamma_{-h,M}\Gamma_{h,M})^{1/2}\bm{\alpha}.$

Then, the upper bound on $\bm{\alpha}^{\top}{\bf\Omega}_{T,M}\bm{\alpha}$ is
given by

$\displaystyle\bm{\alpha}^{\top}{\bf\Omega}_{T,M}\bm{\alpha}=$
$\displaystyle\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\bm{\alpha}^{\top}\Gamma_{h,M}\bm{\alpha}$
$\displaystyle\leq$
$\displaystyle\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\sqrt{\bm{\alpha}^{\top}(\Gamma_{h,M}\Gamma_{-h,M})^{1/2}\bm{\alpha}}\sqrt{\bm{\alpha}^{\top}(\Gamma_{-h,M}\Gamma_{h,M})^{1/2}\bm{\alpha}}.$

Thus,

$\displaystyle\frac{T^{-1}\bm{\alpha}^{\top}{\bf\Omega}_{T,M}\bm{\alpha}}{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}$
$\displaystyle\leq$
$\displaystyle\frac{1}{{T^{-1}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\sqrt{\bm{\alpha}^{\top}(\Gamma_{h,M}\Gamma_{-h,M})^{1/2}\bm{\alpha}}\sqrt{\bm{\alpha}^{\top}(\Gamma_{-h,M}\Gamma_{h,M})^{1/2}\bm{\alpha}}$
$\displaystyle\lesssim$
$\displaystyle\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\sqrt{\frac{\bm{\alpha}^{\top}(\Gamma_{h}\Gamma_{-h})^{1/2}\bm{\alpha}}{{T^{-1}\mathrm{tr}({\bf\Omega}_{T}^{2})}}}\sqrt{\frac{\bm{\alpha}^{\top}(\Gamma_{-h}\Gamma_{h})^{1/2}\bm{\alpha}}{{T^{-1}\mathrm{tr}({\bf\Omega}_{T}^{2})}}}\to
0,$

by Condition (C6). This, together with the result of Lemma 8, implies the
desired result.

###### Lemma 10

Under Conditions (C1)-(C5), we have
$\hat{\mu}_{T}-\mu_{t}^{\prime}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Proof. Define

$\displaystyle\hat{\mu}_{T}^{\prime}:=\frac{1}{T}\left\\{\mathrm{tr}(\hat{\Gamma}_{0})+2\sum_{h=1}^{M}\left(1-\frac{h}{T}\right)\mathrm{tr}(\hat{\Gamma}_{h})\right\\}.$

Note that

$\displaystyle\hat{\bm{\varepsilon}}_{t}=\bm{y}_{t}-\hat{\bm{\alpha}}-\hat{{\bf
B}}{\bm{f}}_{t}=\bm{\varepsilon}_{t}+(\bm{\alpha}-\hat{\bm{\alpha}})+({\bf
B}-\hat{{\bf B}}){\bm{f}}_{t},$ (A.13)

where $\hat{{\bf B}}=(\hat{\bm{\beta}}_{1},\dots,\hat{\bm{\beta}}_{N})^{\top}$
with $\hat{\bm{\beta}}_{i}=({\bf F}^{\top}{\bf F})^{-1}{\bf
F}^{\top}(\bm{y}_{i.}-\alpha_{i})$. Then,

$\displaystyle\hat{\mu}_{T}^{\prime}=$
$\displaystyle\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\hat{\bm{\varepsilon}}_{t}^{\top}\hat{\bm{\varepsilon}}_{s}\eta_{t}\eta_{s}\mathbb{I}_{(|t-s|\leq
M)}$ $\displaystyle=$
$\displaystyle\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}\\{\bm{\varepsilon}_{t}^{\top}\bm{\varepsilon}_{s}+(\bm{\varepsilon}_{t}+\bm{\varepsilon}_{s})^{\top}(\bm{\alpha}-\hat{\bm{\alpha}})+\bm{\varepsilon}_{t}^{\top}({\bf
B}-\hat{{\bf B}}){\bm{f}}_{s}+\bm{\varepsilon}_{s}^{\top}({\bf B}-\hat{{\bf
B}}){\bm{f}}_{t}$
$\displaystyle+(\bm{\alpha}-\hat{\bm{\alpha}})^{\top}(\bm{\alpha}-\hat{\bm{\alpha}})+(\bm{\alpha}-\hat{\bm{\alpha}})^{\top}({\bf
B}-\hat{{\bf B}})({\bm{f}}_{t}+{\bm{f}}_{s})$
$\displaystyle+{\bm{f}}_{t}^{\top}({\bf B}-\hat{{\bf B}})^{\top}({\bf
B}-\hat{{\bf B}}){\bm{f}}_{s}\\}\eta_{t}\eta_{s}\mathbb{I}_{(|t-s|\leq M)}.$

Define
$\tilde{\mu}_{T}=T^{-2}\sum_{t=1}^{T}\sum_{s=1}^{T}\bm{\varepsilon}_{t}^{\top}\bm{\varepsilon}_{s}\eta_{t}\eta_{s}\mathbb{I}_{(|t-s|\leq
M)}$ and
$\check{\mu}_{T}=T^{-2}\sum_{t=1}^{T}\sum_{s=1}^{T}{\bm{X}}_{t}^{\top}{\bm{X}}_{s}\mathbb{I}_{(|t-s|\leq
M)}$.

Step 1. According to the proof of Theorem 3 in Fan et al. (2011), we have

$\displaystyle\max_{1\leq i\leq
N}|\hat{\alpha}_{i}-\alpha_{i}|=O_{p}\\{\sqrt{\log(N)/T}\\}=\max_{1\leq i\leq
N}\|\hat{\bm{\beta}}_{i}-\bm{\beta}_{i}\|,~{}~{}\text{and}$
$\displaystyle\max_{1\leq i\leq
N}\left|T^{-1}\sum_{t=1}^{T}\varepsilon_{it}\right|=O_{p}\\{\sqrt{\log(N)/T}\\}=\max_{1\leq
i\leq N}\left\|T^{-1}\sum_{t=1}^{T}\varepsilon_{it}{\bm{f}}_{t}\right\|,$
(A.14)

which, together with the fact that
$\mathrm{tr}({\bf\Omega}_{T,M}^{2})\lesssim\mathrm{tr}(\bm{\Sigma}^{2})=O(N)$
by Condition (C3), imply that

$\displaystyle\frac{\hat{\mu}_{T}^{\prime}-\tilde{\mu}_{T}}{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\leq
C\frac{T^{-2}\cdot TM\cdot N\log N/T}{\sqrt{T^{-2}N}}=CMN^{1/2}T^{-1}\log N\to
0$

with probability tending to one, i.e.
$\hat{\mu}_{T}^{\prime}-\tilde{\mu}_{T}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Step 2. Note that $\mathbb{E}(\check{\mu}_{T})=\mu_{T}^{\prime}$ and

$\displaystyle\check{\mu}_{T}=\frac{1}{T}\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{T}\right)\mathrm{tr}(\hat{\Gamma}_{h,M}),$

where
$\hat{\Gamma}_{h,M}=\frac{1}{T-h}\sum_{t=1}^{T-h}{\bm{X}}_{t+h}^{\top}{\bm{X}}_{t}$
and $\hat{\Gamma}_{-h,M}=\hat{\Gamma}_{h,M}^{\top}$ for $h\geq 0$. Then,

$\displaystyle\mathrm{var}(\check{\mu}_{T})=$
$\displaystyle\frac{1}{T^{4}}\sum_{h_{1}\in\mathcal{M}}\sum_{h_{2}\in\mathcal{M}}\mathrm{cov}((T-|h_{1}|)\mathrm{tr}(\hat{\Gamma}_{h_{1},M}),(T-|h_{2}|)\mathrm{tr}(\hat{\Gamma}_{h_{2},M}))$
$\displaystyle\lesssim$
$\displaystyle\frac{4}{T^{4}}\sum_{h_{1}=0}^{M}\sum_{h_{2}=0}^{M}\mathrm{cov}((T-h_{1})\mathrm{tr}(\hat{\Gamma}_{h_{1},M}),(T-h_{2})\mathrm{tr}(\hat{\Gamma}_{h_{2},M})).$
$\displaystyle\mathrm{cov}((T-h_{1})\mathrm{tr}(\hat{\Gamma}_{h_{1},M}),(T-h_{2})\mathrm{tr}(\hat{\Gamma}_{h_{2},M}))$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\sum_{t_{1}=1}^{T-h_{1}}\sum_{t_{2}=1}^{T-h_{2}}\\{{\bm{X}}_{t_{1}}^{\top}{\bm{X}}_{t_{1}+h_{1}}-\mathbb{E}({\bm{X}}_{t_{1}}^{\top}{\bm{X}}_{t_{1}+h_{1}})\\}\\{{\bm{X}}_{t_{2}}^{\top}{\bm{X}}_{t_{2}+h_{2}}-\mathbb{E}({\bm{X}}_{t_{2}}^{\top}{\bm{X}}_{t_{2}+h_{2}})\\}\right]$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq t_{1}\leq
T-h_{1},1\leq t_{2}\leq T-h_{2}\\\ k_{1}\leq t_{1},k_{2}\leq
t_{1}+h_{1},k_{3}\leq t_{2},k_{4}\leq
t_{2}+h_{2}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}+h_{1}-k_{2}}b_{t_{2}-k_{3}}b_{t_{2}+h_{2}-k_{4}}$
$\displaystyle\qquad\times\mathbb{E}\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}}-\mathbb{I}_{(k_{1}=k_{2})}\mathrm{tr}(\bm{\Sigma}))({\bm{z}}_{k_{3}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{4}}-\mathbb{I}_{(k_{3}=k_{4})}\mathrm{tr}(\bm{\Sigma}))\\}.$

Consider the value of
$\rho_{k_{1}k_{2}k_{3}k_{4}}:=\mathbb{E}\\{({\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}}-\mathbb{I}_{(k_{1}=k_{2})}\mathrm{tr}(\bm{\Sigma}))({\bm{z}}_{k_{3}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{4}}-\mathbb{I}_{(k_{3}=k_{4})}\mathrm{tr}(\bm{\Sigma}))\\}$.

$\rho_{k_{1}k_{2}k_{3}k_{4}}\begin{cases}=\mathrm{tr}(\bm{\Sigma}^{2}),&k_{1}=k_{3}=r,k_{2}=k_{4}=s,r\neq
s,\\\ =\mathrm{tr}(\bm{\Sigma}^{2}),&k_{1}=k_{4}=r,k_{2}=k_{3}=s,r\neq s,\\\
\leq\tau_{1}\mathrm{tr}^{2}(\bm{\Sigma}),&k_{1}=k_{2}=k_{3}=k_{4}=r,\\\
=0,&\text{otherwise}.\end{cases}$

So we just have to look at the coefficient of $\rho_{k_{1}k_{2}k_{3}k_{4}}$.
Denote it as $C_{k_{1}k_{2}k_{3}k_{4}}$. Then,

$\displaystyle\sum_{t_{1}=1}^{T-h_{1}}\sum_{t_{2}=1}^{T-h_{2}}|C_{rsrs}|\leq\sum_{\begin{subarray}{c}1\leq
t_{1}\leq T-h_{1},1\leq t_{2}\leq T-h_{2}\\\ r\leq t_{1}\wedge
t_{2},s\leq(t_{1}+h_{1})\wedge(t_{2}+h_{2})\end{subarray}}|b_{t_{1}-r}b_{t_{1}+h_{1}-s}b_{t_{2}-r}b_{t_{2}+h_{2}-s}|.$

Consider the number of $|b_{k_{1}}b_{k_{2}}b_{k_{3}}b_{k_{4}}|$ appears in the
right hand side. This can be equivalently written as

$\displaystyle
t_{1}-r=k_{1},~{}~{}t_{1}+h_{1}-s=k_{1},~{}~{}t_{2}-r=k_{3},~{}~{}t_{2}+h_{2}-s=k_{4}.$

Fix $t_{1},k_{1},k_{2},k_{3},k_{4}$, $(r,s,t_{2})$ can be uniquely represented
by $t_{1},k_{1},k_{2},k_{3},k_{4}$ or not exist. Therefore,
$|b_{k_{1}}b_{k_{2}}b_{k_{3}}b_{k_{4}}|$ appears at most $T-h_{1}$ times.
Hence,

$\displaystyle\sum_{t_{1}=1}^{T-h_{1}}\sum_{t_{2}=1}^{T-h_{2}}|C_{rsrs}|\leq
T\left(\sum_{k=0}^{\infty}|b_{k}|\right)^{4}.$

Similarly, we get

$\displaystyle\sum_{t_{1}=1}^{T-h_{1}}\sum_{t_{2}=1}^{T-h_{2}}|C_{rssr}|\leq
T\left(\sum_{k=0}^{\infty}|b_{k}|\right)^{4},~{}~{}\sum_{t_{1}=1}^{T-h_{1}}\sum_{t_{2}=1}^{T-h_{2}}|C_{rrrr}|\leq\left(\sum_{k=0}^{\infty}|b_{k}|\right)^{4}.$

These above results, together with the fact that
$\mathrm{tr}^{2}(\bm{\Sigma})\leq N\mathrm{tr}(\bm{\Sigma}^{2})$, yiled that

$\displaystyle\mathrm{var}(\check{\mu}_{T})=O(M^{2}(T^{-3}+NT^{-4})\mathrm{tr}(\bm{\Sigma}^{2}))=o\\{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})\\},$

which implies that
$\check{\mu}_{T}-\mu_{T}^{\prime}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

Step 3. By (A.11), we have
$1-\frac{\eta_{t}\eta_{s}}{e_{t}e_{s}}=\frac{e_{t}(e_{s}-\eta_{s})+\eta_{s}(e_{t}-\eta_{t})}{e_{t}e_{s}}=O_{p}(T^{-1/2})$.
Furthermore, we have

$\displaystyle\check{\mu}_{T}-\tilde{\mu}_{T}=\frac{1}{T^{2}}\sum_{t=1}^{T}\sum_{s=1}^{T}{\bm{X}}_{t}^{\top}{\bm{X}}_{s}\left(1-\frac{\eta_{t}\eta_{s}}{e_{t}e_{s}}\right)\mathbb{I}_{(|t-s|\leq
M)}=o_{p}\\{T^{-1/2}\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}.$

Step 4. Recalling the definitions of $\hat{\mu}_{T}$ and
$\hat{\mu}_{T}^{\prime}$, we have

$\displaystyle\hat{\mu}_{T}-\hat{\mu}_{T}^{\prime}=\frac{p+1}{T-p-1}\hat{\mu}_{T}^{\prime}-\sum_{h=1}^{M}\frac{2(p+1)h}{T(T-p-1)^{2}}\mathrm{tr}(\hat{\Gamma}_{h}).$

According to the results of Step 1-3 and Lemma 6, we have

$\displaystyle\frac{p+1}{T-p-1}\hat{\mu}_{T}^{\prime}/\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}=O\left\\{\frac{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M})}{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\right\\}=O(T^{-1}N^{1/2})=o(1),$

and

$\displaystyle\sum_{h=1}^{M}\frac{2(p+1)h}{T(T-p-1)^{2}}\mathrm{tr}(\hat{\Gamma}_{h})/\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}=O\left\\{\frac{MT^{-3}\mathrm{tr}({\bf\Omega}_{T,M})}{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}}\right\\}=O(MT^{-2}N^{1/2})=o(1),$

which yield that
$\hat{\mu}_{T}-\hat{\mu}_{T}^{\prime}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

In summary, we have
$\hat{\mu}_{T}-\mu_{t}^{\prime}=o_{p}\\{\sqrt{T^{-2}\mathrm{tr}({\bf\Omega}_{T,M}^{2})}\\}$.

###### Lemma 11

Under Conditions (C1)-(C5), we have
$\hat{\sigma}_{T}^{2}/\sigma_{T}^{2}\mathop{\rightarrow}\limits^{p}1$.

Proof. Define

$\displaystyle\tilde{\sigma}_{T}^{2}:=\frac{2}{T^{2}}\left(\tilde{S}_{0,0}+2\sum_{r=1}^{M}\tilde{S}_{0,r}+2\sum_{r=1}^{M}\tilde{S}_{r,0}+4\sum_{r=1}^{M}\sum_{s=1}^{M}\tilde{S}_{r,s}\right),$

where

$\displaystyle\tilde{S}_{h_{1},h_{2}}:=\frac{\sum_{t=1}^{[T/2]-h_{2}}\sum_{s=t+[T/2]}^{T-h_{2}}{\bm{X}}_{t}^{\top}{\bm{X}}_{s}{\bm{X}}_{t+h_{1}}^{\top}{\bm{X}}_{s+h_{2}}}{(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})},$

which can be expanded in terms of $z_{it}$’s, because

$\displaystyle{\bm{X}}_{t}^{\top}{\bm{X}}_{s}{\bm{X}}_{t+h_{1}}^{\top}{\bm{X}}_{s+h_{2}}=\sum_{k_{1}\leq
t,k_{2}\leq s,k_{3}\leq t+h_{1},k_{4}\leq
s+h_{2}}b_{t-k_{1}}b_{s-k_{2}}b_{t+h_{1}-k_{3}}b_{s+h_{2}-k_{4}}{\bm{z}}_{k_{1}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{2}}{\bm{z}}_{k_{3}}^{\top}\bm{\Sigma}{\bm{z}}_{k_{4}}.$

Next, write $\tilde{S}_{h_{1},h_{2}}$ a sum of the terms involving the high
order of $z_{it}$ and the terms involving the low order of $z_{it}$.
Specifically, write
$\tilde{S}_{h_{1},h_{2}}=\tilde{S}_{h_{1},h_{2},H}+\tilde{S}_{h_{1},h_{2},L}$.

Step 1. Here

$\displaystyle(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})\tilde{S}_{h_{1},h_{2},H}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t\leq[T/2]-h_{1}\\\ t+[T/2]\leq s\leq T-h_{2}\end{subarray}}\sum_{1\leq
i,j\leq
N}[\bm{\Sigma}]_{ii}[\bm{\Sigma}]_{ij}\\{(\sum_{\begin{subarray}{c}r\leq
t\wedge s\wedge t+h_{1}\\\ k\leq
s+h_{2}\end{subarray}}b_{t-r}b_{s-r}b_{t+h_{1}-r}b_{s+h_{1}-k}$
$\displaystyle+\sum_{\begin{subarray}{c}r\leq t\wedge s\wedge s+h_{2}\\\ k\leq
t+h_{1}\end{subarray}}b_{t-r}b_{s-r}b_{t+h_{1}-k}b_{s+h_{1}-r}+\sum_{\begin{subarray}{c}r\leq
t\wedge t+h_{1}\wedge s+h_{2}\\\ k\leq
s\end{subarray}}b_{t-r}b_{s-k}b_{t+h_{1}-r}b_{s+h_{1}-r}$
$\displaystyle+\sum_{\begin{subarray}{c}r\leq s\wedge t+h_{1}\wedge s+h_{2}\\\
k\leq
t\end{subarray}}b_{t-k}b_{s-r}b_{t+h_{1}-r}b_{s+h_{1}-r})z_{ir}^{3}z_{jk}+\sum_{r\leq
t\wedge t+h_{1}\wedge
s+h_{2}}b_{t-r}b_{s-r}b_{t+h_{1}-r}b_{s+h_{1}-r}z_{ir}^{3}z_{jr}\\}.$

Thus, by Condition (C2) (ii) and the fact that
$(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})\geq T^{2}/9$ when $T$ is large
enough, we have

$\displaystyle\mathbb{E}|\tilde{S}_{h_{1},h_{2},H}|=o(N^{2}T^{-2}).$

Furthermore, since
$\mathrm{tr}(\bm{\Sigma}^{2})\geq\mathrm{tr}^{2}(\bm{\Sigma})/N\geq
M_{1}^{2}N$ due to Condition (C3), we have

$\displaystyle\frac{\mathbb{E}\left|\tilde{S}_{0,0,H}+2\sum_{r=1}^{M}\tilde{S}_{0,r,H}+2\sum_{r=1}^{M}\tilde{S}_{r,0,H}+4\sum_{r=1}^{M}\sum_{s=1}^{M}\tilde{S}_{r,s,H}\right|}{\left(a_{0}^{2}+4\sum_{r=1}^{M}a_{0}a_{r}+4\sum_{r=1}^{M}\sum_{s=1}^{M}a_{r}a_{s}\right)\mathrm{tr}(\bm{\Sigma}^{2})}=o(NT^{-2})=o(1).$

Step 2. Note that

$\displaystyle(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})\mathbb{E}(\tilde{S}_{h_{1},h_{2},L})$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t\leq[T/2]-h_{1}\\\ t+[T/2]\leq s\leq T-h_{2}\end{subarray}}\\{\sum_{1\leq
i,j\leq N}[\bm{\Sigma}]_{ij}^{2}(\sum_{\begin{subarray}{c}r\leq t\wedge
t+h_{1}\\\ k\leq s\wedge
s+h_{2}\end{subarray}}b_{t-r}b_{s-k}b_{t+h_{1}-r}b_{s+h_{1}-k}+\sum_{\begin{subarray}{c}r\leq
t\wedge s+h_{2}\\\ k\leq s\wedge
t+h_{1}\end{subarray}}b_{t-r}b_{s-k}b_{t+h_{1}-k}b_{s+h_{1}-r})$
$\displaystyle+\sum_{1\leq i,j\leq
N}[\bm{\Sigma}]_{ii}[\bm{\Sigma}]_{jj}\sum_{\begin{subarray}{c}r\leq t\wedge
s\\\ k\leq t+h_{1}\wedge
s+h_{2}\end{subarray}}b_{t-r}b_{s-r}b_{t+h_{1}-k}b_{s+h_{1}-k}$
$\displaystyle+(\sum_{1\leq i\neq j\leq
N}[\bm{\Sigma}]_{ii}[\bm{\Sigma}]_{jj}+2\sum_{1\leq i\neq j\leq
N}[\bm{\Sigma}]_{ij}^{2})\sum_{r\leq t\wedge s\wedge t+h_{1}\wedge
s+h_{2}}b_{t-r}b_{s-r}b_{t+h_{1}-r}b_{s+h_{1}-r}\\}.$

Then, by Conditions (C2) (ii) again, we have

$\displaystyle(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})\mathbb{E}(\tilde{S}_{h_{1},h_{2},L}-a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2}))$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
t\leq[T/2]-h_{1}\\\ t+[T/2]\leq s\leq
T-h_{2}\end{subarray}}\\{a_{s+h_{2}-t}a_{t+h_{1}-s}\mathrm{tr}(\bm{\Sigma}^{2})+a_{t-s}a_{s-t+h_{2}-h_{1}}\mathrm{tr}^{2}(\bm{\Sigma})$
$\displaystyle+(\sum_{1\leq i\neq j\leq
N}[\bm{\Sigma}]_{ii}[\bm{\Sigma}]_{jj}+2\sum_{1\leq i\neq j\leq
N}[\bm{\Sigma}]_{ij}^{2})\sum_{r\leq t\wedge s\wedge t+h_{1}\wedge
s+h_{2}}b_{t-r}b_{s-r}b_{t+h_{1}-r}b_{s+h_{1}-r}\\}$ $\displaystyle=$
$\displaystyle o(N^{2}T^{-1}).$

Hence,
$\mathbb{E}\\{(\tilde{S}_{h_{1},h_{2},L}-a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2}))/\mathrm{tr}(\bm{\Sigma}^{2})\\}=o(NT^{-3})=o(1)$.
Similarly, we can also verify that
$\mathrm{var}\\{(\tilde{S}_{h_{1},h_{2},L}-a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2}))/\mathrm{tr}(\bm{\Sigma}^{2})\\}=o(NT^{-2}+N^{-1})=o(1)$.
Thus,
$(\tilde{S}_{h_{1},h_{2},L}-a_{h_{1}}a_{h_{2}}\mathrm{tr}(\bm{\Sigma}^{2}))/\mathrm{tr}(\bm{\Sigma}^{2})=o_{p}(1)$,
which implies that

$\displaystyle\frac{\tilde{S}_{0,0,L}+2\sum_{r=1}^{M}\tilde{S}_{0,r,L}+2\sum_{r=1}^{M}\tilde{S}_{r,0,L}+4\sum_{r=1}^{M}\sum_{s=1}^{M}\tilde{S}_{r,s,L}}{\left(a_{0}^{2}+4\sum_{r=1}^{M}a_{0}a_{r}+4\sum_{r=1}^{M}\sum_{s=1}^{M}a_{r}a_{s}\right)\mathrm{tr}(\bm{\Sigma}^{2})}=1+o_{p}(1).$

In summary, we get
$\tilde{\sigma}_{T}^{2}/\sigma_{T}^{2}\mathop{\rightarrow}\limits^{p}1$.

According to the proof of Lemma 9,

$\displaystyle|\tilde{S}_{h_{1},h_{2}}-S_{h_{1},h_{2}}|=$
$\displaystyle\left|\frac{\sum_{t=1}^{[T/2]-h_{1}}\sum_{s=t+[T/2]}^{T-h_{2}}{\bm{X}}_{t}^{\top}{\bm{X}}_{s}{\bm{X}}_{t+h_{1}}^{\top}{\bm{X}}_{s+h_{2}}}{(T-h_{2}/2-\frac{3}{2}[T/2]+1/2)([T/2]-h_{2})}\left(1-\frac{\eta_{t}\eta_{s}\eta_{t+h_{1}}\eta_{s+h_{2}}}{e_{t}e_{s}e_{t+h_{1}}e_{s+h_{2}}}\right)\right|$
$\displaystyle\leq$ $\displaystyle CT^{-1}|\tilde{S}_{h_{1},h_{2}}|,$

which implies that
$\hat{\sigma}_{T}^{2}/\tilde{\sigma}_{T}^{2}=1+O_{p}(T^{-1})=1+o_{p}(1)$. In
summary, we get
$\hat{\sigma}_{T}^{2}/\sigma_{T}^{2}\mathop{\rightarrow}\limits^{p}1$.

####  6.1.3 Proof of Theorem 3

Proof. According to Lemmas 10-11, we can easily obtain the result.

###  6.2 Proof of Theorems 4-5

Let ${\bf D}\equiv\mathrm{diag}\\{\sigma_{1},\cdots,\sigma_{N}\\}$ be the
diagonal matrix of the long-run covariance of $\bm{\gamma}_{t}$,
${\bf\Omega}_{M}:=\Gamma_{0,M}+2\sum_{h=0}^{M}\Gamma_{h,M}$. Define
$\bm{R}:={\bf D}^{-1/2}{\bf\Omega}_{M}{\bf D}^{-1/2}$. For a $d$-dimension
vector $\bm{Y}=(Y_{1},\cdots,Y_{d})^{\top}$, define
$\|\bm{Y}\|_{\infty}=\max_{1\leq i\leq d}|Y_{i}|$. For $X\in\mathcal{R}$,
define $||X||_{q}=E[|X|^{q}]^{1/q}$.

###### Lemma 12

Let $(Z_{1},\cdots,Z_{N})^{\top}$ be a zero mean multivariate normal random
vector with covariance matrix $\Sigma=(\sigma_{ij})_{N\times N}$ and diagonal
$\sigma_{ii}=1$ for $1\leq i\leq N$. Suppose that $\max_{1\leq i<j\leq
N}|\sigma_{ij}|\leq r<1$ and $\max_{1\leq j\leq
N}\sum_{i=1}^{N}\sigma_{ij}^{2}\leq c$ for some $r$ and $c$. Then for any
$x\in\mathcal{R}$ as $N\to\infty$,

$\displaystyle P\left(\max_{1\leq i\leq
N}Z_{i}^{2}-2\log(N)+\log\\{\log(N)\\}\leq
x\right)\to\exp\left\\{-\frac{1}{\sqrt{\pi}}\exp\left(-\frac{x}{2}\right)\right\\}.$

Proof. See Lemma 6 in Cai et al. (2014).

###### Lemma 13

Suppose $\left\\{x_{i}\right\\}_{i=1}^{n}$ is a p-dimensional M-dependent
sequence. Let $n=(N+M)r,$ where $N\geq M$ and $N,M,r\rightarrow+\infty$ as
$n\rightarrow+\infty$. Define the block sums

$\displaystyle
A_{ij}=\sum_{l=iN+(i-1)M-N+1}^{iN+(i-1)M}x_{lj},~{}~{}\text{and}~{}~{}B_{ij}=\sum_{l=i(N+M)-M+1}^{i(N+M)}x_{lj}.$

It is not hard to see that $\left\\{A_{ij}\right\\}_{i=1}^{r}$ and
$\left\\{B_{ij}\right\\}_{i=1}^{r}$ with $1\leq j\leq p$ are two sequences of
i.i.d random variables. Let
$V_{nj}=\sqrt{\sum_{i=1}^{r}(A_{ij}^{2}+B_{ij}^{2})}$. Assume that there exist
$a_{j},b_{j}>0$ such that

$\displaystyle P\left(\sum_{i=1}^{n}x_{ij}>a_{j}\right)\leq
1/4,~{}~{}P\left(V_{nj}^{2}>b_{j}^{2}\right)\leq 1/4.$

Then we have

$\displaystyle P\left(\left|\sum_{i=1}^{n}x_{ij}\right|\geq
x\left(a_{j}+b_{j}+V_{nj}\right)\right)\leq 8\exp\left(-x^{2}/8\right)$

for any $1\leq j\leq p.$ In particular, we can choose
$b_{j}^{2}=4\mathbb{E}V_{nj}^{2}$ and
$a_{j}^{2}=2b_{j}^{2}=8\mathbb{E}V_{nj}^{2}$.

Proof. See Lemma A.1 in Zhang and Cheng (2018).

###### Lemma 14

Under Conditions (C1)-(C2) and (C7)-(C9), we have

$\displaystyle\mathbb{P}\left(T\left\|{\bf
D}^{-1/2}\bar{\bm{\gamma}}_{T}\right\|_{\infty}^{2}-2\log(N)+\log\\{\log(N)\\}\right)\to\exp\left\\{-\frac{1}{\sqrt{\pi}}\exp\left(-\frac{x}{2}\right)\right\\}.$

Proof. Recalling the definition of
$\bm{\gamma}_{t}:=\mathbb{E}(\bm{{\bm{X}}}_{t}|{\bm{z}}_{t-M},\cdots,{\bm{z}}_{t})$
in the proof of Theorem 1, we can write it as
$\bm{\gamma}_{t}=G(\mathcal{F}^{t})$, where
$\mathcal{F}^{t}=({\bm{z}}_{t-M},\dots,{\bm{z}}_{t-1},{\bm{z}}_{t})$ and
$G=(g_{1}(\cdot),\dots,g_{N}(\cdot))^{\top}$ is an $\mathcal{R}^{N}$-valued
measurable function. Define the functional dependence measure

$\displaystyle\iota_{t,q,i}:=\|\gamma_{it}-g_{i}(\mathcal{F}^{t,\\{0\\}})\|_{q},$

where
$\mathcal{F}^{t,\\{0\\}}=({\bm{z}}_{t-M},\dots,{\bm{z}}_{k-1},{\bm{z}}_{k}^{\prime},{\bm{z}}_{k+1},\dots,{\bm{z}}_{t})$
is a coupled version of $\mathcal{F}^{t}$ with ${\bm{z}}_{k}$ in
$\mathcal{F}^{t}$ is replaced by ${\bm{z}}_{k}^{\prime}$, and
${\bm{z}}_{t},{\bm{z}}_{t}^{\prime}$ are i.i.d. random elements. To account
for the dependence in the process $\\{\bm{\gamma}_{i.}\\}_{i=1}^{N}$, we
define the dependence adjusted norm

$\displaystyle\|\iota_{i.}\|_{q,a}:=\sup_{m\geq
0}(m+1)^{a}\sum_{t=m}^{\infty}\iota_{t,q,i}.$

To account for high dimensionality, we define the uniform and the overall
dependence adjusted norm of $\\{\bm{\gamma}_{t}\\}_{t=1}^{T}$:

$\displaystyle\psi_{q,a}:=\max_{1\leq i\leq
N}\|\iota_{i.}\|_{q,a},~{}~{}\text{and}~{}~{}\Upsilon_{q,a}:=\left(\sum_{i=1}^{N}\|\iota_{i.}\|_{q,a}^{q}\right)^{1/q}.$

Additionally, define the $\mathcal{L}^{\infty}$ functional dependence measure
and its corresponding dependence adjusted norm of
$\\{\bm{\gamma}_{t}\\}_{t=1}^{T}$:

$\displaystyle\omega_{t,q}:=\|\|\bm{\gamma}_{t}-G(\mathcal{F}^{t,\\{0\\}})\|_{\infty}\|_{q}~{}~{}\text{and}~{}~{}\||\gamma_{.}|_{\infty}\|_{q,a}:=\sup_{m\geq
0}(1+m)^{a}\sum_{t=m}^{\infty}\omega_{t,q}.$

Following the same notations as Zhang and Wu (2017), define the following
quantities

$\displaystyle\Theta_{q,a}:=\Upsilon_{q,a}\wedge\\{\||\gamma_{.}|_{\infty}\|_{q,a}\log^{3/2}(N)\\},~{}~{}$
$\displaystyle~{}~{}L_{1}:=\\{\psi_{2,a}\psi_{2,0}\log^{2}(N)\\}^{1/a},$
$\displaystyle W_{1}:=(\psi_{3,0}^{6}+\psi_{4,0}^{4})\log^{7}(NT),~{}~{}$
$\displaystyle~{}~{}W_{2}:=\psi_{2,a}^{2}\log^{4}(NT),$ $\displaystyle
N_{1}:=\\{T/\log(N)\\}^{q/2}/\Theta_{q,a}^{q},~{}~{}$
$\displaystyle~{}~{}N_{2}:=T\log^{-2}(T)\psi_{2,a}^{-2}.$

Since $\\{\bm{\gamma}\\}_{t=1}^{T}$ is M-dependent and $\|\gamma_{it}\|_{q}$
is bounded for any positive integrate $q$ due to Condition (C7), we have

$\displaystyle\Theta_{q,a}=O\\{M^{1+a}\log^{2}(N)\\},~{}~{}$
$\displaystyle~{}~{}L_{1}=O[\\{M^{2+a}\log^{2}(N)\\}^{1/a}],$ $\displaystyle
W_{1}=O\\{M^{6}\log^{7}(NT)\\},~{}~{}$
$\displaystyle~{}~{}W_{2}=O\\{M^{2a+2}\log^{4}(NT)\\},$ $\displaystyle
N_{1}=O[\\{T\log^{-5}(N)M^{-2a-2}\\}^{q/2}],~{}~{}$
$\displaystyle~{}~{}N_{2}=O\\{T\log^{-2}(N)M^{-2a-2}\\}.$

Furthermore, by Condition (C9), we have
$\Theta_{q,a}T^{1/q-1/2}\log^{3/2}(N)=o(1)$ and
$L_{1}\max(W_{1},W_{2})=o(1)\min(N_{1},N_{2})$ for some $q\geq 4$ and
$a>1/2-1/q$. So by Theorem 3.2 in Zhang and Wu (2017), we have

$\displaystyle\sup_{x\geq 0}\left|\mathbb{P}\left(T\left\|{\bf
D}^{-1/2}\bar{\bm{\gamma}}_{T}\right\|_{\infty}^{2}\geq
x\right)-\mathbb{P}\left(\|\bm{Z}\|_{\infty}^{2}\geq x\right)\right|\to 0,$
(B.1)

where $\bm{Z}\sim N(\bm{0},\bm{R})$.

By Condition (C8) and Lemma 12, we have

$\displaystyle\mathbb{P}(\|\bm{Z}\|_{\infty}^{2}-2\log(N)+\log\\{\log(N)\\}\leq
x)\to F(x).$ (B.2)

In summary, the desired result comes from (B.1) and (B.2).

###### Lemma 15

Under Conditions (C1)-(C2) and (C7)-(C9), we have

$\displaystyle T\left\|{\bf
D}^{-1/2}(\bar{{\bm{X}}}_{T}-\bar{\bm{\gamma}}_{T})\right\|_{\infty}^{2}=\max_{1\leq
i\leq
N}\frac{1}{T\sigma_{i}}\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})\right\\}^{2}=o_{p}(1).$

Proof. Let $[\bm{\Sigma}^{1/2}]_{(i)}$ be the $i$-th row of
$\bm{\Sigma}^{1/2}$, then $X_{it}-\gamma_{it}=\sum_{k\leq
t-M+1}b_{t-k}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k}$. Using the independence of
$z_{it}$’s, we have

$\displaystyle\mathrm{var}\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})\right\\}=$
$\displaystyle\sum_{t_{1}=1}^{T}\sum_{t_{2}=1}^{T}\sum_{k_{1}\leq
t_{1}-M+1}\sum_{k_{2}\leq
t_{2}-M+1}b_{t_{1}-k_{1}}b_{t_{2}-k_{2}}\mathbb{E}([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{2}})$
$\displaystyle=$
$\displaystyle\sum_{t_{1}=1}^{T}\sum_{t_{2}=1}^{T}\sum_{k\leq(t_{1}\wedge
t_{2})-M+1}b_{t_{1}-k}b_{t_{2}-k}[\bm{\Sigma}]_{ii}$ $\displaystyle\leq$
$\displaystyle
T^{2}\left(\sum_{k=M+1}^{\infty}|b_{k}|\right)^{2}[\bm{\Sigma}]_{ii}=o(T^{2}M^{-8})$

due to (A.10). Thus, $T\left\|{\bf
D}_{0}^{-1/2}(\bar{{\bm{X}}}_{T}-\bar{\bm{\gamma}}_{T})\right\|_{\infty}^{2}=o_{p}(T^{-1}\cdot
T^{2}M^{-8})=o_{p}(1)$.

###### Lemma 16

Under Conditions (C1)-(C2) and (C7)-(C9), we have

$\displaystyle\max_{1\leq i\leq
N}\frac{T(\hat{\alpha_{i}}-\tilde{\alpha}_{i})^{2}}{\sigma_{i}}=\max_{1\leq
i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}X_{it}(\zeta_{t}/e_{t}-1)\right\\}^{2}=O_{p}\\{\sqrt{T^{-1}\log(N)}\\}.$

Proof. Note that

$\displaystyle\max_{1\leq i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}X_{it}(\zeta_{t}/e_{t}-1)\right\\}^{2}\leq$
$\displaystyle\max_{1\leq i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})(\zeta_{t}/e_{t}-1)\right\\}^{2}$
$\displaystyle\quad+\max_{1\leq i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}\gamma_{it}(\zeta_{t}/e_{t}-1)\right\\}^{2}.$

For the first term on the right side of the above inequality, by Cauchy-
Schwarz inequality, we have

$\displaystyle\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})(\zeta_{t}/e_{t}-1)\right\\}^{2}\leq\sum_{t=1}^{T}(X_{it}-\gamma_{it})^{2}\cdot\sum_{s=1}^{T}(\zeta_{s}/e_{s}-1)^{2}.$

By (A.11), we have $\sum_{s=1}^{T}(\zeta_{s}/e_{s}-1)^{2}=O_{p}(1)$. Next,
following the proof of Lemma 7, we analyze
$\sum_{t=1}^{T}(X_{it}-\gamma_{it})^{2}$.

$\displaystyle\mathrm{var}\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})^{2}\right\\}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq
T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{2}}b_{t_{2}-k_{3}}b_{t_{2}-k_{4}}\mathbb{E}\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{2}}-\mathbb{I}_{(k_{1}=k_{2})}[\bm{\Sigma}]_{ii})$
$\displaystyle\qquad\times([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{3}}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{4}}-\mathbb{I}_{(k_{3}=k_{4})}[\bm{\Sigma}]_{ii})\\}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq
T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{1}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{1}}b_{t_{2}-k_{3}}b_{t_{2}-k_{3}}\mathbb{E}[\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}})^{2}-[\bm{\Sigma}]_{ii}\\}\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{3}})^{2}-[\bm{\Sigma}]_{ii}\\}]$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{2}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{2}}b_{t_{2}-k_{1}}b_{t_{2}-k_{2}}\mathbb{E}\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{2}})^{2}\\}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{3}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{2}}b_{t_{2}-k_{2}}b_{t_{2}-k_{1}}\mathbb{E}\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}}[\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{2}})^{2}\\}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{4}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{1}}b_{t_{2}-k_{1}}b_{t_{2}-k_{1}}\mathbb{E}[\\{([\bm{\Sigma}^{1/2}]_{(i)}{\bm{z}}_{k_{1}})^{2}-[\bm{\Sigma}]_{ii}\\}^{2}]$
$\displaystyle=$ $\displaystyle\left(\sum_{\begin{subarray}{c}1\leq
t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{2}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{2}}b_{t_{2}-k_{1}}b_{t_{2}-k_{2}}+\sum_{\begin{subarray}{c}1\leq
t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{3}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{2}}b_{t_{2}-k_{2}}b_{t_{2}-k_{1}}\right)[\bm{\Sigma}]_{ii}^{2}$
$\displaystyle+\sum_{\begin{subarray}{c}1\leq t_{1},t_{2}\leq T\\\
(k_{1},k_{2},k_{3},k_{4})\in\mathcal{S}\bigcap\mathcal{S}_{4}\end{subarray}}b_{t_{1}-k_{1}}b_{t_{1}-k_{1}}b_{t_{2}-k_{1}}b_{t_{2}-k_{1}}\left(\mu_{4}\sum_{j=1}^{N}[\bm{\Sigma}^{1/2}]_{ij}^{4}+5[\bm{\Sigma}]_{ii}^{2}\right)$
$\displaystyle\lesssim$ $\displaystyle\sum_{t_{1},t_{2}\geq
0}|b_{t_{1}}b_{t_{2}}|\sum_{s_{1},s_{2}\geq
M}|b_{s_{1}}b_{s_{2}}|+\sum_{t_{1}\geq 0,s_{2}\geq
M}|b_{t_{1}}b_{s_{2}}|\sum_{t_{2}\geq 0,s_{1}\geq
M}|b_{t_{2}}b_{s_{1}}|+\frac{N}{T}\sum_{t_{1},t_{2}\geq
0}\sum_{s_{1},s_{2}\geq M}|b_{t_{1}}b_{t_{2}}b_{s_{1}}b_{s_{2}}|$
$\displaystyle=$ $\displaystyle o(M^{-8}),$

where $\mathcal{S}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1},k_{2}\leq
t_{1}-M+1,k_{3},k_{4}\leq t_{2}-M+1\\}$,
$\mathcal{S}_{1}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{2}\neq k_{3}=k_{4}\\}$,
$\mathcal{S}_{2}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{3}\neq k_{2}=k_{4}\\}$,
$\mathcal{S}_{3}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{4}\neq k_{2}=k_{3}\\}$
and $\mathcal{S}_{4}=\\{(k_{1},k_{2},k_{3},k_{4}):k_{1}=k_{2}=k_{3}=k_{4}\\}$.
Therefore, $\sum_{t=1}^{T}(X_{it}-\gamma_{it})^{2}=o_{p}(M^{-4})$.
Furthermore, we get $\max_{1\leq i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}(X_{it}-\gamma_{it})(\zeta_{t}/e_{t}-1)\right\\}^{2}=o_{p}(T^{-1}M^{-4})$.

For the second term, by Cauchy-Schwarz inequality again, we have

$\displaystyle\left\\{\sum_{t=1}^{T}\gamma_{it}(\zeta_{t}/e_{t}-1)\right\\}^{2}\leq\sum_{t=1}^{T}\gamma_{it}^{2}\cdot\sum_{s=1}^{T}(\zeta_{t}/e_{t}-1)^{2}.$

Let $T=(M^{*}+M)r$ where $M^{*}\geq M$ and $M^{*},M,r\to\infty$ as
$T\to\infty$. Define the block sums

$\displaystyle
A_{it}=\sum_{l=(t-1)(M^{*}+M)+1}^{tM^{*}+(t-1)M}\gamma_{il}^{2},~{}~{}\text{and}~{}~{}B_{it}=\sum_{l=tM^{*}+(t-1)M+1}^{t(M^{*}+M)}\gamma_{il}^{2}.$

Since $\\{\bm{\gamma}_{t}\\}_{t=1}^{T}$ is M-dependent,
$\\{A_{il}\\}_{l=1}^{r}$ and $\\{B_{il}\\}_{l=1}^{r}$ with $1\leq i\leq N$ are
two sequence of i.i.d. random variables. Let
$V_{iT}=\\{\sum_{l=1}^{r}(A_{il}^{2}+B_{il}^{2})\\}^{1/2}$. Then, by Lemma 13,
we have

$\displaystyle\mathbb{P}\left(\sum_{t=1}^{T}\gamma_{it}^{2}\geq
x\\{(2+2\sqrt{2})\sqrt{\mathbb{E}(V_{iT}^{2})}+V_{iT}\\}\right)\leq
8\exp(-x^{2}/8).$

Note that $(2+2\sqrt{2})\sqrt{\mathbb{E}(V_{iT}^{2})}+V_{iT}\leq CT^{1/2}$ for
some positive constant $C$, then we have

$\displaystyle\mathbb{P}\left(\max_{1\leq i\leq
N}\sum_{t=1}^{T}\gamma_{it}^{2}\geq
Cx\right)\leq\sum_{i=1}^{N}\mathbb{P}\left(\sum_{t=1}^{T}\gamma_{it}^{2}\geq
Cx\right)\leq 8N\exp(-\frac{x^{2}}{8T})\to 0$

by setting $x=10\sqrt{T\log(N)}$. Hence, $\max_{1\leq i\leq
N}\sum_{t=1}^{T}\gamma_{it}^{2}=O_{p}\\{\sqrt{T\log(N)}\\}$ and

$\displaystyle\max_{1\leq i\leq
N}T^{-1}\sigma_{i}^{-1}\left\\{\sum_{t=1}^{T}\gamma_{it}(\zeta_{t}/e_{t}-1)\right\\}^{2}=O_{p}\\{\sqrt{T^{-1}\log(N)}\\}.$

In summary, we have

$\displaystyle\max_{1\leq i\leq
N}\frac{T(\hat{\alpha_{i}}-\tilde{\alpha}_{i})^{2}}{\sigma_{i}}=o_{p}(T^{-1}M^{-4})+O_{p}\\{\sqrt{T^{-1}\log(N)}\\}=O_{p}\\{\sqrt{T^{-1}\log(N)}\\}.$

###### Lemma 17

Under Conditions (C1)-(C2) and (C7)-(C9), we have

$\displaystyle\max_{1\leq i\leq
N}|\hat{\sigma}_{i}-\sigma_{i}|=O_{p}(T^{-\delta})$

for sufficiently some $\delta>0$.

Proof. According to (A.13), we can rewrite $\hat{\phi}_{i,h}$ as

$\displaystyle\hat{\phi}_{i,h}=$
$\displaystyle\frac{1}{T-h}\sum_{t=h+1}^{T}\\{\varepsilon_{it}\varepsilon_{i,t-h}+(\varepsilon_{it}+\varepsilon_{i,t-h})(\alpha_{i}-\hat{\alpha}_{i})+\varepsilon_{it}(\bm{\beta}_{i}-\hat{\bm{\beta}}_{i})^{\top}{\bm{f}}_{t-h}+\varepsilon_{i,t-h}(\bm{\beta}_{i}-\hat{\bm{\beta}}_{i})^{\top}{\bm{f}}_{t}$
$\displaystyle+(\alpha_{i}-\hat{\alpha}_{i})^{2}+(\alpha_{i}-\hat{\alpha}_{i})(\bm{\beta}_{i}-\hat{\bm{\beta}}_{i})^{\top}({\bm{f}}_{t}+{\bm{f}}_{t-h})+(\bm{\beta}_{i}-\hat{\bm{\beta}}_{i})^{\top}{\bm{f}}_{t}(\bm{\beta}_{i}-\hat{\bm{\beta}}_{i})^{\top}{\bm{f}}_{t-h}\\}\eta_{t}\eta_{t-h}.$

Define
$\tilde{\phi}_{i,h}:=\frac{1}{T-h}\sum_{t=h+1}^{T}\varepsilon_{it}\varepsilon_{i,t-h}\eta_{t}\eta_{t-h}$,
$\phi_{i,h}^{NG}:=\frac{1}{T-h}\sum_{t=h+1}^{T}\gamma_{it}\gamma_{i,t-h}$ and
$\phi_{i,h}:=\mathbb{E}(\gamma_{i0}\gamma_{ih})$, then
$\sigma_{i}=\sum_{h\in\mathcal{M}}\phi_{i,h}$ and
$\hat{\sigma}_{i}-\sigma_{i}=\sum_{h\in\mathcal{M}}(\hat{\phi}_{i,h}-\phi_{i,h})$.
By (6.1.2), we have

$\displaystyle\sqrt{T}\max_{1\leq i\leq
N}\left\|\hat{\phi}_{i,h}-\tilde{\phi}_{i,h}\right\|_{q/2}=O\\{\log(N)T^{-1/2}\\}=o(1).$

By Lemma E.5 in Jirak (2015), we have for any positive integrate $q>4$,

$\displaystyle\sqrt{T}\max_{1\leq i\leq
N}\left\|\phi_{i,h}^{NG}-\phi_{i,h}\right\|_{q/2}=O(1).$

Following the proof of Lemma 7, we can prove
$\frac{1}{T-h}\sum_{t=h+1}^{T}X_{it}X_{i,t-h}-\phi_{i,h}^{NG}=o_{p}(M^{-4})$,
which together with (A.11), yields that

$\displaystyle\sqrt{T}\max_{1\leq i\leq
N}\left\|\phi_{i,h}^{NG}-\tilde{\phi}_{i,h}\right\|_{q/2}=o(T^{1/2}M^{-4})=o(1).$

In summary, we have

$\displaystyle\sqrt{T}\max_{1\leq i\leq
N}\left\|\hat{\phi}_{i,h}-\phi_{i,h}\right\|_{q/2}=O(1).$

It follows that for large enough $N,T,$ we have

$\displaystyle\mathbb{P}\left(\max_{1\leq i\leq
N}|\hat{\sigma}_{i}-\sigma_{i}|\geq T^{-\delta}\right)\leq$
$\displaystyle\sum_{i=1}^{N}\mathbb{P}\left(\left|\sum_{h\in\mathcal{M}}(\hat{\phi}_{i,h}-\phi_{i,h})\right|\geq
T^{-\delta}\right)$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{N}\sum_{h=1}^{M}\mathbb{P}\left(\left|\hat{\phi}_{i,h}-\phi_{i,h}\right|\geq\frac{T^{-\delta}}{2M}\right)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{N}\sum_{h=1}^{M}\left(\frac{T^{-\delta}}{2M}\right)^{-q/2}\left\|\hat{\phi}_{i,h}-\phi_{i,h}\right\|_{q/2}^{q/2},$

which is bounded by $NT^{\delta q/2}M^{q/2+1}T^{-q/4}\lesssim T^{-C}$, $C>0$
for sufficiently small $\delta>0$ and large $q$.

####  6.2.1 Proof of Theorem 4

Proof. Note that

$\displaystyle\left|\max_{1\leq i\leq
N}\frac{T(\hat{\alpha}_{i}-\alpha_{i})^{2}}{\hat{\sigma}_{i}}-\max_{1\leq
i\leq N}\frac{T(\hat{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}\right|$
$\displaystyle\leq$ $\displaystyle\max_{1\leq i\leq
N}\frac{T(\hat{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}\cdot\max_{1\leq i\leq
N}\left|\frac{\sigma_{i}}{\hat{\sigma}_{i}}-1\right|$ $\displaystyle\leq$
$\displaystyle 2\left\\{\max_{1\leq i\leq
N}\frac{T(\hat{\alpha}_{i}-\tilde{\alpha}_{i})^{2}}{\sigma_{i}}+\max_{1\leq
i\leq
N}\frac{T(\tilde{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}\right\\}\cdot\max_{1\leq
i\leq N}\left|\frac{\sigma_{i}}{\hat{\sigma}_{i}}-1\right|,$ (B.3)
$\displaystyle\left|\max_{1\leq i\leq
N}\frac{T(\hat{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}-\max_{1\leq i\leq
N}\frac{T(\tilde{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}\right|$
$\displaystyle\leq$ $\displaystyle\max_{1\leq i\leq
N}\frac{T(\hat{\alpha}_{i}-\tilde{\alpha}_{i})^{2}}{\sigma_{i}}+2\left\\{\max_{1\leq
i\leq
N}\frac{T(\tilde{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}\cdot\max_{1\leq
i\leq
N}\frac{T(\hat{\alpha}_{i}-\tilde{\alpha}_{i})^{2}}{\sigma_{i}}\right\\}^{1/2},$
(B.4)

and

$\displaystyle\max_{1\leq i\leq
N}\frac{T(\tilde{\alpha}_{i}-\alpha_{i})^{2}}{\sigma_{i}}=T\left\|{\bf
D}^{-1/2}\bar{{\bm{X}}}_{T}\right\|_{\infty}^{2}\leq T\left\|{\bf
D}^{-1/2}\bar{\bm{\gamma}}_{T}\right\|_{\infty}^{2}+T\left\|{\bf
D}^{-1/2}(\bar{{\bm{X}}}_{T}-\bar{\bm{\gamma}}_{T})\right\|_{\infty}^{2}.$
(B.5)

By Lemmas 14-17, we have

$\displaystyle P\left(\max_{1\leq i\leq