Testing High-dimensional Covariance Matrices under the

arXiv:1707.04010v1 [] 13 Jul 2017

Elliptical Distribution and Beyond
Xinxin Yang∗,

Xinghua Zheng†,

Jiaqi Chen‡,

Hua Li§

Abstract
We study testing high-dimensional covariance matrices when data exhibit heteroskedasticity.
The observations are modeled as Yi = ωi Zi , where Zi ’s are i.i.d. p-dimensional random vectors
with mean 0 and covariance Σ, and ωi ’s are random scalars reflecting heteroskedasticity. The
model is an extension of the elliptical distribution, and accommodates several stylized facts of real
data including heteroskedasticity, heavy-tailedness, asymmetry, etc. We aim to test H0 : Σ ∝ Σ0 ,
in the high-dimensional setting where both the dimension p and the sample size n grow to infinity
proportionally. We remove the heteroskedasticity by self-normalizing the observations, and estabP
P
lish a CLT for the linear spectral statistic (LSS) of e
Sn := np ni=1 Yi YTi /|Yi |2 = np ni=1 Zi ZTi /|Zi |2 .
The CLT is different from the existing ones for the LSS of the usual sample covariance matrix
P
Sn := 1n ni=1 Zi ZTi (Bai and Silverstein (2004), Najim and Yao (2016)). Our tests based on the
new CLT neither assume a specific parametric distribution nor involve the fourth moment of Zi .
Numerical studies show that our tests work well even when Zi ’s are heavy-tailed.

Keywords: Elliptical Distribution, Covariance Matrix, High Dimension, Central Limit Theorem,
Self-normalization

1 Introduction
1.1 Tests for high-dimensional covariance matrices
Testing covariance matrices is of fundamental importance in multivariate analysis. There has been a
long history of study on testing: (1) the covariance matrix Σ is equal to a given matrix, or (2) the
∗

Department of Mathematics, The Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong.
xyangai@ust.hk
†
Department of ISOM, The Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong.
xhzheng@ust.hk
‡
Department of Mathematics, Harbin Institute of Technology, Harbin, 150001, China. chenjq1016@hit.edu.cn
§
Department of Applied Statistics, Chang Chun University, Changchun, 130001, China. cculihua@163.com

1

covariance matrix Σ is proportional to a given matrix. To be specific, for a given covariance matrix Σ0 ,
one aims to test either
H0 : Σ = Σ 0

vs.

Ha : Σ = Σ0 , or

(1.1)

H0 : Σ ∝ Σ 0

vs.

Ha : Σ 6∝ Σ0 .

(1.2)

In the classical setting where the dimension p is fixed and the sample size n goes to infinity, the sample
covariance matrix is a consistent estimator, and further inference can be made based on the associated
CLT. Examples include the likelihood ratio tests (see, e.g., Muirhead (1982), Sections 8.3 and 8.4, and
Anderson (2003), Sections 10.7 and 10.8), and the locally most powerful invariant tests (John (1971),
Nagao (1973)).
In the high-dimensional setting, because the sample covariance matrix is inconsistent, the conventional tests may not apply. New methods for testing high-dimensional covariance matrices have been
developed. The existing tests were first proposed under the multivariate normal distribution, then have
been modified to fit more generally distributed data.
• Multivariate normally distributed data. When p/n → y ∈ (0, ∞), Ledoit and Wolf (2002)
show that John’s test for (1.2) is still consistent and propose a modified Nagao’s test for (1.1).
Srivastava (2005) introduces a new test for (1.2) under a more general condition that n = O(pδ )
for some δ ∈ (0, 1]. Birke and Dette (2005) show that the asymptotic null distributions of John’s
and the modified Nagao’s test statistics in Ledoit and Wolf (2002) are still valid when p/n → ∞.
Relaxing the normality assumption but still assuming the kurtosis of data equals 3, Bai et al.
(2009) develop a corrected likelihood ratio test for (1.1) when p/n → y ∈ (0, 1).
• More generally distributed data. Chen et al. (2010) generalize the results in Ledoit and Wolf
(2002) without assuming normality nor an explicit relationship between p and n. By relaxing
the kurtosis assumption, Wang et al. (2013) extend the corrected likelihood ratio test in Bai et al.
(2009) and the modified Nagao’s test in Ledoit and Wolf (2002) for testing (1.1). Along this line,
Wang and Yao (2013) propose two tests by correcting the likelihood ratio test and John’s test for
testing (1.2).

1.2 The elliptical distribution and its applications
The elliptically distributed data can be expressed as
Y = ωZ,
where ω is a positive random scalar, Z is a p-dimensional random vector from N(0, Σ), and further ω
and Z are independent of each other. It is a natural generalization of the multivariate normal distribution, and contains many widely used distributions including the multivariate t-distribution, the symmet2

ric multivariate Laplace distribution and the symmetric multivariate stable distribution. See Fang et al.
(1990) for further details.
One of our motivations of this study arises from the wide applicability of the elliptical distribution
to financial data analysis. The heavy-tailedness of stock returns has been extensively studied, dating
back at least to Fama (1965) and Mandelbrot (1967). Accommodating both heavy-tailedness and flexible shapes makes the elliptical distribution a more admissible candidate for stock-return models than
the Gaussian distribution; see, e.g., Owen and Rabinovitch (1983) and Bingham and Kiesel (2002).
McNeil et al. (2005) state that “elliptical distributions ... provided far superior models to the multivariate normal for daily and weekly US stock-return data" and that “multivariate return data for groups
of returns of similar type often look roughly elliptical." Theoretically, Owen and Rabinovitch (1983)
show that the elliptical distribution extends, among others, the mutual fund separation theorems and
the theory of the capital asset pricing model(CAPM). Chamberlain (1983) finds that the mean-variance
framework implies that asset returns are elliptically distributed. The Markowitz theory for large portfolio optimization has also been studied under the elliptical distribution in El Karoui (2009), El Karoui
(2010) and El Karoui (2013). Recently, Bhattacharyya and Bickel (2014) develop an adaptive procedure for estimating the elliptic density under both low- and high-dimensional settings.

1.3 Performance of the existing tests under the elliptical distribution
Are the existing tests for covariance matrices applicable to the elliptical distribution under the highdimensional setting? Both numerical and theoretical analyses give a negative answer.
We start with a simple numerical study to investigate the empirical sizes of the aforementioned
tests. Consider observations Yi = ωi Zi , i = 1, · · · , n, where
(i) ωi ’s are absolute values of i.i.d. standard normal random variables,
(ii) Zi ’s are i.i.d. p-dimensional standard multivariate normal random vectors, and
(iii) ωi ’s and Zi ’s are independent of each other.
Under such a setting, Yi ’s are i.i.d. random vectors with mean 0 and covariance matrix I. We will test
both (1.1) and (1.2).
For testing (1.1), we examine the tests in Ledoit and Wolf (2002) (LW test), Bai et al. (2009) (BJYZ
test), Chen et al. (2010) (CZZ test) and Wang et al. (2013) (WYMC-LR and WYMC-LW tests). Table 1
reports the empirical sizes for testing H0 : Σ = I at 5% significance level.

3

p/n = 0.5

p/n = 2

p

LW

BJYZ

CZZ

WYMC-LR

WYMC-LW

LW

CZZ

WYMC-LW

100

100%

100%

54.0%

100%

100%

100%

50.2%

100%

200

100%

100%

51.6%

100%

100%

100%

53.0%

100%

500

100%

100%

52.3%

100%

100%

100%

53.3%

100%

Table 1. Empirical sizes of the LW, BJYZ, CZZ, WYMC-LR and WYMC-LW tests for testing H0 : Σ = I at 5% significance level. Data are
generated as Yi = ωi Zi where ωi ’s are absolute values of i.i.d. N(0, 1), Zi ’s are i.i.d. N(0, I), and further ωi ’s and Zi ’s are independent of
each other. The results are based on 10, 000 replications for each pair of p and n.

We observe from Table 1 that the empirical sizes of these tests are far higher than the nominal level
of 5%, suggesting that they are inconsistent for testing (1.1) under the elliptical distribution.
For testing (1.2), we apply the tests proposed by Ledoit and Wolf (2002) (LW test), Srivastava
(2005) (S test), Chen et al. (2010) (CZZ test) and Wang and Yao (2013) (WY-LR and WY-JHN tests).
Table 2 reports the empirical sizes of these tests for testing H0 : Σ ∝ I at 5% significance level.
p/n = 0.5

p/n = 2

p

LW

S

CZZ

WY-LR

WY-JHN

LW

S

CZZ

WY-JHN

100

100%

100%

51.8%

100%

100%

100%

100%

50.2%

100%

200

100%

100%

53.0%

100%

100%

100%

100%

52.3%

100%

500

100%

100%

52.3%

100%

100%

100%

100%

53.5%

100%

Table 2. Empirical sizes of the LW, S, CZZ, WY-LR and WY-JHN tests for testing H0 : Σ ∝ I at 5% significance level. Data are generated as
Yi = ωi Zi where ωi ’s are absolute values of i.i.d. N(0, 1), Zi ’s are i.i.d. N(0, I), and further ωi ’s and Zi ’s are independent of each other. The
results are based on 10, 000 replications for each pair of p and n.

From Table 2, we see again that the empirical sizes of the examined tests are severely distorted.
To conclude, Tables 1 and 2 show that even under such a rather simple setup, the existing tests fail
badly. New tests are therefore needed.
Theoretically, the distorted sizes in Tables 1 and 2 are not unexpected. In fact, denote Sn =
Pn
1 Pn
1 Pn
T
ω
T
2
T
i=1 Zi Zi and Sn = n i=1 Yi Yi = n i=1 ωi Zi Zi . The celebrated Marčenko-Pastur theorem states
that the empirical spectral distribution (ESD) of Sn converges to the Marčenko-Pastur law. However,
Theorem 1 of Zheng and Li (2011) implies that the ESD of Sω
n will not converge to the MarčenkoPastur law except in the trivial situation where ωi ’s are constant. Since all the aforementioned tests
involve certain aspects of the limiting ESD (LSD) of Sω
n , the asymptotic null distributions of the involved test statistics are different from the ones in the usual setting, and consequently the tests are no
1
n

4

longer consistent.

1.4 Our model and aim of this study
In various real situations, the assumption that the observations are i.i.d. is too strong to hold. An important source of violation is (conditional) heteroskedasticity, which is encountered in a wide range
of applications. For instance, in finance, it is well documented that stock returns are (conditionally) heteroskedastic, which motivated the development of ARCH and GARCH models (Engle (1982),
Bollerslev (1986)). In engineering, Yucek and Arslan (2009) explain that the heteroskedasticity of
noise is one of the factors that degrade the performance of target detection systems.
In this paper, we study testing high-dimensional covariance matrices when the data may exhibit
heteroskedasticity. Specifically, we consider the following model. Denote by Yi , i = 1, · · · , n, the
observations, which can be decomposed as
Yi = ωi Zi , where
(i)
(ii)
(iii)
(iv)

(1.3)

ωi ’s are positive random scalars reflecting heteroskedasticity,
Zi ’s can be written as Zi = Σ1/2 e
Zi , where e
Zi consists of i.i.d. standardized random variables,
ωi ’s can depend on each other as well as on {Zi : i = 1, 2, · · · , n} in an arbitrary way, and
ωi ’s do not need to be stationary.

Model (1.3) incorporates the elliptical distribution as a special case. This general model further
possesses several important advantages:
• It can be considered as a multivariate extension of the ARCH/GARCH model, and accommodates the conditional heteroskedasticity in real data. In the ARCH/GARCH model, the volatility
process is serially dependent and depends on past information. Such dependence is excluded
from the elliptical distribution; however, it is perfectly compatible with Model (1.3).
• The dependence of ωi and Zi can feature the leverage effect in financial econometrics, which accounts for the negative correlation between asset return and change in volatility. Various research
has been conducted to study the leverage effect; see, e.g., Schwert (1989), Campbell and Hentschel
(1992) and Aït-Sahalia et al. (2013).
• Furthermore, it can capture the (conditional) asymmetry of data by allowing Zi ’s to be asymmetric. The asymmetry is another stylized fact of financial data. For instance, the empirical study
in Singleton and Wingender (1986) shows high skewness in individual stock returns. Skewness
is also reported in exchange rate returns in Peiro (1999). Christoffersen (2012) documents that
asymmetry exists in standardized returns; see Chapter 6 therein.

5

Since ωi ’s are not required to be stationary, the unconditional covariance matrix may not exist, in
which case there is no basis for testing (1.1). Testing (1.2), however, still makes perfect sense, because
the scalars ωi ’s only scale up or down the covariance matrix by a constant. We henceforth focus on
testing (1.2). As usual, by working with Σ0−1/2 Yi , testing (1.2) can be reduced to testing
H0 : Σ ∝ I vs.

Ha : Σ 6∝ I.

(1.4)

In the following, we focus on testing (1.4), in the high-dimensional setting where both p and n grow to
infinity with the ratio p/n → y ∈ (0, ∞).

1.5 Summary of main results
To deal with heteroskedasticity, we propose to self-normalize the observations. To be specific, we focus
on the self-normalized observations Yi / |Yi |, where | · | stands for the Euclidean norm. Observe that
Yi
Zi
=
,
|Yi | |Zi |

i = 1, 2, · · · , n.

Hence ωi ’s no longer play a role, and this is exactly the reason why we make no assumption on ωi ’s.
There is, however, no such thing as a free lunch. Self-normalization introduces a new challenge in
that the entries of Zi /|Zi | are dependent in an unusual fashion. To see this, consider the simplest case
where Zi ’s are i.i.d. standard multivariate normal random vectors. In this case, the entries of Zi ’s are
i.i.d. random variables from N(0, 1). However, the self-normalized random vector Zi /|Zi | is uniformly
distributed over the p-dimensional unit sphere (known as the Haar distribution on the sphere), and its
p entries are dependent on each other in an unusual way.
To conduct tests, we need some kind of CLTs. Our strategy is to establish a CLT for the linear
spectral statistic (LSS) of the sample covariance matrix based on the self-normalized observations,
specifically,
n
n
p X Yi YTi
p X Zi ZTi
e
Sn =
=
.
(1.5)
n i=1 |Yi |2
n i=1 |Zi |2
When |Yi | or |Zi | = 0, we adopt the convention that 0/0 = 0. Note that e
Sn is not the sample correlation matrix, which normalizes each variable by its standard deviation. Here we are normalizing each
observation by its Euclidean norm.

As we shall see below, our CLT is different from the ones for the usual sample covariance matrix
in Bai and Silverstein (2004) and Najim and Yao (2016). One important advantage of our result is
4 ) = 3 as in Bai and Silverstein (2004), nor the estimation
that applying our CLT requires neither E(Z11
4 ), which is inevitable in Najim and Yao (2016). Based on the new CLT, we propose two tests
of E(Z11
by modifying the likelihood ratio test and John’s test. More tests can be established based on our
4 ) does not
general CLT. Numerical study shows that our proposed tests work well even when E(Z11
exist. Because heavy-tailedness and heteroskedasticity are commonly encountered in practice, such
6

relaxations are appealing in many real applications.
Independently and concurrently, Li and Yao (2017) study high-dimensional covariance matrix test
under a mixture model. There are a couple of major differences between our paper and theirs. First
and foremost, in Li and Yao (2017), the mixture coefficients (ωi ’s in (1.3)) are assumed to be i.i.d. and
drawn from a distribution with a bounded support. Second, Li and Yao (2017) require independence
between the mixture coefficients and the innovation process (Zi ). In our paper, we do not put any
assumptions on the mixture coefficients. As we discussed in Section 1.4, such relaxations allow us
to accommodate several important stylized features of real data, consequently, make our tests more
suitable in many real applications. It can be shown that the test in Li and Yao (2017) can be inconsistent
under our general setting. Furthermore, as we can see from the simulation studies, the test in Li and Yao
(2017) is less powerful than the existing tests in the i.i.d. Gaussian setting and, in general, less powerful
than our tests.
Organization of the paper. The rest of the paper is organized as follows. In Section 2, we state
the CLT for the LSS of e
Sn , based on which we derive the asymptotic null distributions of the modified
likelihood ratio test statistic and John’s test statistic. Section 3 examines the finite-sample performance
of our proposed tests. Section 4 concludes. The proof of the main theorem is given in Appendix A,
with some technical derivations collected in Appendices B – D.
Notation. For any symmetric matrix A ∈ R p×p , F A denotes its ESD, that is,
p

F A (x) =

1X
1 A , for all x ∈ R,
p i=1 {λi ≤x}

where λA
i , i = 1, 2, · · · , p, are the eigenvalues of A and
function f , the associated LSS of A is given by
Z

+∞

1{·} denotes the indicator function. For any
p

1X
f (x)dF (x) =
f (λA
i ).
p i=1
A

−∞

Finally, the Stieltjes transform of a distribution G is defined as
mG (z) =

Z

∞

−∞

1
dG(λ),
λ−z

where supp(G) denotes the support of G.

7

for all z < supp(G),

2 Main results
2.1 CLT for the LSS of e
Sn

As discussed above, we focus on the sample covariance matrix based on the self-normalized Zi ’s,
namely, e
Sn defined in (1.5). Denote by Z = (Z1 , · · · , Zn ).
We now state the assumptions:



2  < ∞;
Assumption A. Z = Zi j p×n consists of i.i.d. random variables with E Z11 = 0 and 0 < E Z11
4  < ∞; and
Assumption B. E Z11
Assumption C. yn := p/n → y ∈ (0, ∞) as n → ∞.
The following proposition gives the LSD of e
Sn .

Proposition 2.1. Under Assumptions A and C, almost surely, the ESD of e
Sn converges weakly to the
standard Marčenko-Pastur law Fy , which admits the density

p
1


 2πxy (x − a− (y))(a+ (y) − x),
py (x) = 

 0,

x ∈ [a− (y), a+ (y)],
otherwise,

and has a point mass 1 − 1/y at the origin if y > 1, where a± (y) = (1 ±

√

y)2 .

Proposition 2.1 is essentially a special case of Theorem 2 in Zheng and Li (2011) but with weaker
moment assumptions, and can be shown by modifying the proof of that theorem. We give the proof in
Appendix B to make this paper self-contained.
Sn shares the same LSD as the usual sample covariance matrix Sn =
According to Proposition 2.1, e
P
1 n
T
2 
n i=1 Zi Zi if one assumes that E Z11 = 1. To conduct tests, we need the associated CLT. The CLTs
for the LSS of Sn have been established in Bai and Silverstein (2004) and Najim and Yao (2016), under
the Gaussian and non-Gaussian kurtosis conditions, respectively. Given that e
Sn and Sn have the same
LSD, one naturally asks whether their LSSs also have the same CLT. The following theorem gives a
negative answer. Hence, an important message is:
Self-normalization does not change the LSD, but it does affect the CLT.
To be more specific, for any function f , define the following centered and scaled LSS:
GeSn ( f ) := p

Z

+∞

−∞

 e

f (x)d F Sn (x) − Fyn (x) .

Theorem 2.2. Suppose that Assumptions A – C hold. Let H denote the set of functions that are
analytic on a domain containing [a− (y)1{0<y<1} , a+ (y)], and f1 , f2 , · · · , fk ∈ H. Then, the random

8



vector GeSn ( f1 ), GeSn ( f2 ), · · · , GeSn ( fk ) converges weakly to a Gaussian vector G( f1 ), G( f2 ), · · · , G( fk )
with mean
1
EG( fℓ ) =
πi

I

ym3 (z)

!

ym2 (z)

!−1

dz
 1−

1 + m(z) 3
1 + m(z) 2
!
!
I
ym2 (z) −2
ym3 (z)
1
dz,
fℓ (z)
−
 1−

2πi C
1 + m(z) 3
1 + m(z) 2

and covariance

fℓ (z)

C

Cov((G( fi ), G( f j )) =

ℓ = 1, 2, · · · , k,

I I

fi (z1 ) f j (z2 )m′ (z1 )m′ (z2 )
2
 dz1 dz2
C2 C1 1 + m(z1 )
1 + m(z2 ) 2
I I
fi (z1 ) f j (z2 )m′ (z1 )m′ (z2 )
1
− 2
dz1 dz2 ,

2π C2 C1
m(z2 ) − m(z1 ) 2

y
2π2

(2.1)

(2.2)
i, j = 1, 2, · · · , k.

Here, C1 and C2 are two non-overlapping contours contained in the domain and enclosing the interval
[a− (y)1{0<y<1} , a+ (y)], and m(z) is the Stieltjes transform of F y := (1 − y)1[0,∞) + yFy .
The proof of Theorem 2.2 is given in Appendix A.
Remark. The second terms on the right-hand sides of (2.1) and (2.2) appeared in equations (1.6) and
(1.7) in Theorem 1.1 of Bai and Silverstein (2004) (in the special case when T = I). The first terms
are new and are due to the self-normalization in e
Sn . It is worth emphasizing that our CLT neither

4
4  as in Najim and Yao (2016).
requires E Z11 = 3 as in Bai and Silverstein (2004), nor involves E Z11

2.2 Tests for the covariance matrix in the presence of heteroskedasticity
Based on Theorem 2.2, we propose two tests for testing (1.4) by modifying the likelihood ratio test and
John’s test. More tests can be established by choosing f in Theorem 2.2 to be different functions.
2.2.1

The likelihood ratio test based on self-normalized observations (LR-SN)

Recall that Sn =

1
n

Pn

T
i=1 Zi Zi .

The likelihood ratio test statistic is

Ln = log |Sn | − p log tr Sn + p log p;

see, e.g., Section 8.3.1 in Muirhead (1982). For the heteroskedastic case, we modify the likelihood

ratio test statistic by replacing Sn with e
Sn . Note that tr e
Sn = p on the event {|Zi | > 0 for i = 1, · · · , n},
which, by Lemma 2 in Bai and Yin (1993), occurs almost surely for all large n. Therefore, we are led

9

to the following modified likelihood ratio test statistic:
e
Sn =
Ln = log e

p
X
i=1

e 
log λSi n .

It is the LSS of e
Sn when f (x) = log(x). In this case, when yn ∈ (0, 1), we have
GeSn (log) =p

Z

+∞

−∞



e
log(x)d F Sn (x) − Fyn (x)

y − 1

e 
n
log λSi n − p
log(1 − yn ) − 1
yn
i=1

y − 1
n
log(1 − yn ) − 1 .
=e
Ln − p
yn

=

p
X

Applying Theorem 2.2, we obtain the following proposition.

Proposition 2.3. Under the assumptions of Theorem 2.2, if yn → y ∈ (0, 1), then



e
Ln − p yny−1
log(1
−
y
)
−
1
− yn − log(1 − yn )/2
n
n
D
−→ N(0, 1),
p
−2yn − 2 log(1 − yn )

(2.3)

The proof of Proposition 2.3 is given in Appendix B.

The convergence in (2.3) gives the asymptotic null distribution of the modified likelihood ratio test
statistic. Since it is derived from the sample covariance matrix based on self-normalized observations,
the test based on (2.3) will be referred to as the likelihood ratio test based on the self-normalized
observations (LR-SN).
2.2.2

John’s test based on self-normalized observations (JHN-SN)

John’s test statistic is given by
2
n 
Sn
T n = tr
 − I − p;
p
1/p tr Sn


Sn and noting again that tr e
Sn = p almost surely for all large n
see John (1971). Replacing Sn with e
lead to the following modified John’s test statistic:
p
2
n 
1 X eSn 2
Ten = tr e
− n − p.
λ
Sn − I − p =
p
yn i=1 i

10

It is related to the LSS of e
Sn when f (x) = x2 . In this case, for any yn ∈ (0, ∞), we have
2

GeSn (x ) = p

Z

+∞

p
 X
e 
x d F (x) − Fyn (x) =
λSi n 2 − p(1 + yn ) = yn Ten .
2

−∞



e
Sn

i=1

Based on Theorem 2.2, we can prove the following proposition.
Proposition 2.4. Under the assumptions of Theorem 2.2, we have
Ten + 1 D
−→ N(0, 1).
2

(2.4)

The proof of Proposition 2.4 is given in Appendix B.
Below we will refer to the test based on (2.4) as John’s test based on the self-normalized observations (JHN-SN).

3 Simulation studies
We now demonstrate the finite-sample performance of our proposed tests. For different values of p and
p/n, we will check both the sizes and powers of the LR-SN and JHN-SN tests.

3.1 I.i.d. Gaussian case
To have a full picture of the performance of our tests, we start with the simplest situation where observations are i.i.d. multivariate normal random vectors. We will compare our proposed tests, LR-SN and
JHN-SN, with the tests mentioned in Section 1.1, namely, LW, S, CZZ and WY-LR, and also the newly
proposed test in Li and Yao (2017) (LY test). In the multivariate normal case, the WY-JHN test reduces
to the LW test.
We start with the size by sampling observations from N(0, I). Table 3 reports the empirical sizes of
these tests for testing H0 : Σ ∝ I at 5% significance level.
p/n = 0.5
p

LW

S CZZ WY-LR

p/n = 2

LY LR-SN JHN-SN

LW

S CZZ

LY JHN-SN

100 4.9% 4.8% 4.9%

4.5% 4.8%

4.6%

5.2%

5.5% 5.5% 5.7% 5.1%

4.9%

200 5.2% 5.0% 5.1%

5.1% 4.8%

5.1%

4.9%

4.6% 4.5% 5.1% 4.8%

4.5%

500 4.9% 5.1% 5.1%

4.8% 5.3%

4.9%

5.2%

5.1% 5.3% 4.9% 5.0%

5.2%

Table 3. Empirical sizes of the LW, S, CZZ, WY-LR, LY, and our proposed LR-SN and JHN-SN tests for testing H0 : Σ ∝ I at 5% significance
level. Observations are i.i.d. N(0, I). The results are based on 10, 000 replications for each pair of p and n.

11

From Table 3, we find that the empirical sizes of all tests are around the nominal level of 5%.
Next, to compare the power, we generate i.i.d. observations from N(0, Σ) under the alternative

with Σ = 0.1|i− j| p×p , and test H0 : Σ ∝ I at 5% significance level. Table 4 reports the empirical
powers.
p/n = 0.5
p

LW

S CZZ WY-LR

p/n = 2
LY LR-SN JHN-SN

LW

S CZZ

LY JHN-SN

100 50.7% 51.3% 50.1% 36.7% 28.1% 35.0%

48.9%

8.4% 8.7% 9.1% 6.3%

8.2%

200 97.3% 97.3% 97.2% 88.0% 79.4% 88.7%

97.0%

18.3% 17.9% 18.1% 11.9%

17.2%

500 100% 100% 100%

100%

70.7% 70.6% 69.8% 43.3%

70.5%

100% 100% 100%

Table 4. Empirical powers of the
WY-LR, LY, and our proposed LR-SN and JHN-SN tests for testing H0 : Σ ∝ I at 5% significance
 LW, S, CZZ,
 
level. Observations are i.i.d. N 0, 0.1|i− j| p×p . The results are based on 10, 000 replications for each pair of p and n.

From Table 4, we find that our proposed LR-SN and JHN-SN tests and the tests mentioned in Section 1.1 have quite good powers especially as the dimension p gets higher, and the powers are roughly
comparable. As in the classical setting, John’s test (JHN-SN) is more powerful than the likelihood ratio
test (LR-SN). The LY test proposed in Li and Yao (2017) is less powerful.
To sum up, while developed under a much more general setup, our tests perform just as well as the
existing ones. Real differences emerge below when we consider more complicated situations, where
existing tests fail while our tests continue to perform well.

3.2 The elliptical case
Now we investigate the performance of our proposed tests under the elliptical distribution. As in
Section 1.3, we take the observations to be Yi = ωi Zi with
(i) ωi ’s being absolute values of i.i.d. standard normal random variables,
(ii) Zi ’s i.i.d. p-dimensional random vectors from N(0, Σ), and
(iii) ωi ’s and Zi ’s independent of each other.
3.2.1

Checking the size

Table 5 completes Table 2 by including the empirical sizes of our proposed LR-SN and JHN-SN tests,
and also the LY test in Li and Yao (2017) for testing H0 : Σ ∝ I at 5% significance level.

12

p/n = 0.5
p

LW

p/n = 2

S CZZWY-LRWY-JHN LYLR-SNJHN-SN

LW

S CZZWY-JHN LYJHN-SN

100 100%100%51.8% 100%

100%4.4% 4.6%

5.2%

100%100%50.2%

100%4.1%

4.9%

200 100%100%53.0% 100%

100%4.5% 5.1%

4.9%

100%100%52.3%

100%4.5%

4.5%

500 100%100%52.3% 100%

100%5.2% 4.9%

5.2%

100%100%53.5%

100%4.7%

5.2%

Table 5. Empirical sizes of the LW, S, CZZ, WY-LR, WY-JHN, LY, and our proposed LR-SN, JHN-SN tests for testing H0 : Σ ∝ I at 5%
significance level. Data are generated as Yi = ωi Zi where ωi ’s are absolute values of i.i.d. N(0, 1), Zi ’s are i.i.d. N(0, I), and further ωi ’s
and Zi ’s are independent of each other. The results are based on 10, 000 replications for each pair of p and n.

Table 5 reveals the sharp difference between the existing tests and our proposed ones: the empirical
sizes of the existing tests are severely distorted, in contrast, the empirical sizes of our LR-SN and JHNSN tests are around the nominal level of 5% as desired. In fact, the sizes of our tests in this table are
exactly the same as in Table 3. The reason is that, with self-normalization in our testing procedure, the
mixing coefficients ωi ’s are removed, and we end up with the same e
Sn as in the previous setting. The
same remark applies to the powers of our tests in Table 6 below. The LY test also yields the right level
of size.
3.2.2

Checking the power

Table 5 shows that the LW, S, CZZ and WY-LR tests are inconsistent under the elliptical distribution,
therefore we exclude them when checking the power. Similarly to Table 4, we will generate observa
tions under the elliptical distribution with Σ = 0.1|i− j| p×p . Table 6 reports the empirical powers of our
proposed tests and the LY test for testing H0 : Σ ∝ I at 5% significance level.
p/n = 0.5

p/n = 2

p

LY

LR-SN

JHN-SN

LY

JHN-SN

100

7.6%

35.0%

48.9%

3.5%

8.2%

200

14.5%

88.7%

97.0%

5.7%

17.2%

500

64.9%

100%

100%

9.0%

70.5%

Table 6. Empirical powers of the LY test and our proposed LR-SN and JHN-SN tests for testing H0 : Σ ∝ I at 5% significance level. Data

are generated as Yi = ωi Zi where ωi ’s are absolute values of i.i.d. N(0, 1), Zi are i.i.d. random vectors from N(0, Σ) with Σ = 0.1|i− j| p×p ,
and further ωi ’s and Zi ’s are independent of each other. The results are based on 10, 000 replications for each pair of p and n.

From Table 6, we find again that the LY test is less powerful.

13

3.3 Beyond elliptical, a GARCH-type case
Recall that in our general model (1.3), the observations Yi admit the decomposition ωi Zi . To examine
the performance of our tests in such a general setup, we simulate data using the following two-step
procedure:
ei , which consists of i.i.d.
1. For each Zi , we first generate another p-dimensional random vector Z
ei j ’s; and with Σ to be specified below, Zi is taken to be Σ1/2 e
standardized random variables Z
Zi . In
ei j ’s are sampled from standardized t-distribution with 4 degrees of freedom,
the simulation below, Z
which is heavy-tailed.

2. For each ωi , inspired by the ARCH/GARCH model, we take ω2i = 0.01+0.85ω2i−1 +0.1|Yi−1 |2 / tr Σ .

3.3.1

Checking the size

We test H0 : Σ ∝ I. Table 7 reports the empirical sizes of our proposed tests and the LY test at 5%
significance level.
p/n = 0.5

p/n = 2

p

LY

LR-SN

JHN-SN

LY

JHN-SN

100

8.2%

5.5%

5.3%

6.8%

5.0%

200

8.5%

5.7%

5.4%

6.8%

5.5%

500

7.6%

5.3%

5.2%

6.6%

5.4%

Table 7. Empirical sizes of the LY test and our proposed LR-SN and JHN-SN tests for testing H0 : Σ ∝ I at 5% significance level. Data are
generated as Yi = ωi Zi with ω2i = 0.01 + 0.85ω2i−1 + 0.1|Yi−1 |2 /p, and Zi consists of i.i.d. standardized t(4) random variables. The results
are based on 10, 000 replications for each pair of p and n.

From Table 7, we find that, for all different values of p and p/n, the empirical sizes of our proposed
tests are around the nominal level of 5%. Again, this is in sharp contrast with the results in Table 2,
where the existing tests yield sizes far higher than 5%.
4 , the
One more important observation is that although Theorem 2.2 requires the finiteness of E Z11
4  does not exist.
simulation above shows that our proposed tests work well even when E Z11

Another observation is that with 10,000 replications, the margin of error for a proportion at 5%
significance level is 1%, hence the sizes of the LY test in Table 7 are statistically significantly higher
than the nominal level of 5%.
3.3.2

Checking the power


To evaluate the power, we again take Σ = 0.1|i− j| p×p and generate data according to the design at the
beginning of this subsection (Section 3.3). Table 8 reports the empirical powers of our proposed tests
14

and the LY test for testing H0 : Σ ∝ I at 5% significance level.
p/n = 0.5

p/n = 2

p

LY

LR-SN

JHN-SN

LY

JHN-SN

100

20.7%

34.4%

47.9%

7.8%

8.7%

200

54.4%

87.8%

96.6%

10.5%

17.6%

500

100%

100%

100%

26.4%

69.9%

Table 8. Empirical powers of the LY test and our proposed LR-SN and JHN-SN tests for testing H0 : Σ ∝ I at 5% significance level. Data
 e
e
are generated as Yi = ωi Zi with ω2i = 0.01 + 0.85ω2i−1 + 0.1|Yi−1 |2 /p, and Zi = 0.1|i− j| 1/2
p×p Zi where Zi consists of i.i.d. standardized t(4)
random variables. The results are based on 10, 000 replications for each pair of p and n.

Table 8 shows again that our tests enjoy a blessing of dimensionality: for a fixed ratio p/n, the
higher the dimension p, the higher the power. Moreover, comparing Table 8 with Table 4, we find that
for each pair of p and n, the powers of our tests are similar under the two designs. Such similarities
show that our tests can not only accommodate conditional heteroskedasticity but also are robust to
heavy-tailedness in Zi ’s. Finally, the LY test is again less powerful.

3.4 Summary of simulation studies
Combining the observations in the three cases, we conclude that
(i) The existing tests, LW, S, CZZ, WY-LR and WY-JHN, work well in the i.i.d. Gaussian setting,
however, they fail badly under the elliptical distribution and our general setup;
(ii) The newly proposed LY test in Li and Yao (2017) is applicable to the elliptical distribution, however, it is less powerful than the existing tests in the i.i.d. Gaussian setting and, in general, less
powerful than ours;
(iii) Our LR-SN and JHN-SN tests perform well under all three settings, yielding the right sizes and
enjoying high powers.

4 Conclusion
We study testing high-dimensional covariance matrices under an extension of the elliptical distribution.
The extended model can feature heteroskedasticity, leverage effect, asymmetry, etc. Under the general
setting, we establish a CLT for the LSS of the sample covariance matrix based on self-normalized
observations. The CLT is different from the existing ones for the usual sample covariance matrix, and
4  = 3 as in Bai and Silverstein (2004), nor involve E Z 4  as in Najim and Yao
it does not require E Z11
11
(2016). Based on the new CLT, we propose two tests by modifying the likelihood ratio test and John’s
15

test, for which explicit CLTs are derived. More tests can be established based on our general CLT.
Numerical studies show that our proposed tests work well regardless of whether the observations are
i.i.d. Gaussian or from the elliptical distribution or feature conditional heteroskedasticity or even when
Zi ’s are heavy-tailed.

Acknowledgements
We thank Zhigang Bao for a suggestion that helps relax the assumptions of Lemma A.3.

Appendix A

Proof of Theorem 2.2

In this section, we prove Theorem 2.2. We first introduce some additional notations. For any symmetric
matrix A, kAk stands for the spectral norm. Denote
Xi =

√

pZi
|Zi |

and

Xi
ri = √ , i = 1, 2, · · · , n.
n

Thus e
Sn = 1/n · XXT with X = (X1 , · · · , Xn ). Recall that mG (z) denotes the Stieltjes transform of
a distribution G with z < supp(G) and Fy the standard Marčenko-Pastur law with index y. Let e
Sn =
T
1/n · X X and define
mn (z) = mFeSn (z),
1 − yn
+ yn m0n (z),
m0n (z) = −
z
1−y
+ ym(z).
m(z) = −
z

mn (z) = mFeSn (z),
m0n (z) = mFyn (z),
m(z) = mFy (z),

Note that e
Sn and e
Sn have the same non-zero eigenvalues, and we have mn (z) = −(1 − yn )/z + yn mn (z).
For any z ∈ C with ℑ(z) , 0, define
D(z) = e
Sn − zI, Di (z) = D(z) − ri rTi , Di j (z) = D(z) − ri rTi − r j rTj for i , j,
−1
−1
ςi (z) = rTi D−1
i (z)ri − n tr Di (z),

−1
−2
τi (z) = rTi D−2
i (z)ri − n tr Di (z),

−1
−1
γi (z) = rTi D−1
i (z)ri − n E tr D1 (z), and

βi (z) =
β̆i j (z) =

1
1+

rTi D−1
i (z)ri

, βi j (z) =

1
1+

rTj D−1
i j (z)r j

, β̆i (z) =

1
1+

n−1 tr D−1
i (z)

(A.1)
(A.2)

,

1
1
1
,
b̄
(z)
=
.
,
b
(z)
=
n
n
−1
−1
1 + n−1 tr Di j (z)
1 + n−1 E tr D1 (z)
1 + n−1 E tr D−1
12 (z)

(A.3)

Throughout the rest of the proofs, K denotes a generic constant whose value does not depend on p

16

and n except possibly on yn and may change from line to line.

A.1 Preliminaries
A.1.1 Truncation, Centralization and Renormalization
4  < ∞, by Lemma 2.2 in Yin et al. (1988), there exists a positive sequence δ  such that
Since E Z11
p


lim pδ8p = ∞, and P Z , b
Z, i.o. = 0,

δ p = o(1),

p→∞

(A.4)

bi j 
b
where b
Z= Z
p×n and Zi j = Zi j 1{|Zi j |≤p1/2 δ p } . Moreover, by enlarging δ p if necessary, we can assume

that δ p also satisfies


4
δ−4
p E |Z11 | 1{|Z11 |>p1/2 δ p } = o(1);


see (1.8) in Bai and Silverstein (2004). Define Z̆ = b
Z−E b
Z and
Zi b
ZTi
pXb
b
,
Sn =
n i=1 |b
Zi |2
n

p X Z̆i Z̆Ti
,
n i=1 |Z̆i |2
n

S̆n =

where b
Zi and Z̆i represent the ith columns of b
Z and Z̆, respectively.

Claim A.1. Under the assumptions of Theorem 2.2, for any f ∈ H, we have
p

Z

+∞

b

f (x)dF Sn (x) = p

−∞

Z

+∞

f (x)dF S̆n (x) + o p (1).

(A.5)

−∞

The proof of Claim A.1 is postponed to Appendix C.
Combining (A.4) and (A.5), we obtain
GeSn ( f ) = p

Z

+∞

−∞

e

f (x)dF Sn (x) = p

Z

+∞

f (x)dF S̆n (x) + o p (1).

−∞

Therefore, to study the asymptotics of GeSn ( f ), it suffices to consider the random matrix S̆n , which is based on
the truncated and centralized entries Z̆i j ’s. Furthermore, because of the self-normalization in S̆n , w.l.o.g. we can

assume that Var Z̆11 = 1.

In the following, for notational ease, we still denote Z̆ by Z. In such a way, for each p, the entries Zi j = Zi(p)
j
are i.i.d. random variables satisfying

(p)
(p) 
(p) 2
Assumption SA. |Z11 | ≤ p1/2 δ p with E Z11 = 0 and E Z11
= 1 for some positive sequence δ p which
satisfies δ p = o(1), lim p→∞ pδ8p = ∞; and
(p) 
Assumption SB. sup p E Z11 4 < ∞.

A.1.2 Elementary estimates
In this section, we give some lemmas which will be used in the sequel.

17

Lemma A.2. [Rank-one perturbation formula] For any z ∈ C with ℑ(z) , 0, any symmetric matrix A ∈ R p×p
and v ∈ R p , we have
vT A + vvT − zI
A + vvT − zI

−1

− A − zI

−1

−1

vT (A − zI)−1
,
1 + vT (A − zI)−1 v
(A − zI)−1 vvT (A − zI)−1
=−
.
1 + vT (A − zI)−1 v

=

(A.6)
(A.7)

Equation (A.6) is a special case of (2.2) in Silverstein and Bai (1995), and (A.7) is a direct result of (A.6).
(p)

(p)

(p)

Lemma A.3. For each p, assume that V1 = V1 , V2 = V2 , · · · , V p = V p are i.i.d. random variables satis
fying V1(p) ≤ p1/2 δ p for some δ p = o(1) and lim p→∞ pδ8p = ∞. Suppose further that EV1(p) = 0, E V1(p) 2 = 1

√
(p)
and sup p E V1 4 < ∞. Let W = pV/|V| with V = (V1 , V2 , · · · , V p ). Then for any (deterministic) symmetric
matrices A, B ∈ C p×p ,



E WT AW − tr A WT BW − tr B

p

X





= E V14 − 3
aii bii − p−1 E V14 − 1 tr A tr B + 2 tr(AB) + o p kAk2 + kBk2 .

(A.8)

i=1

Furthermore, if we denote by ei the p-dimensional vector with a 1 in the ith coordinate and 0’s elsewhere, then



E WT Aei eTi AW · WT Bei eTi BW = o p kAk4 + kBk4 .

(A.9)

Lemma A.4. Under the assumptions of Lemma A.3, for all large p, any (deterministic) symmetric matrix A ∈
C p×p and k ≥ 2, we have
k

E WT AW − tr A ≤K pk−1 δ2k−4
kAkk ,
p

E WT AW − tr A ≤Kδ2p kAk.

(A.10)
(A.11)

Next, we collect some elementary results for the quantities and matrices defined in (A.1) – (A.3). Note that
if λ ≥ 0 and z ∈ C+ := {z ∈ C, ℑ(z) > 0}, we have |λ − z| ≥ ℑ(z) . It follows that

−1
−1
max kD−1 (z)k, kD−1
i (z)k, kDi j (z)k ≤ ℑ(z) ,

for all z ∈ C+ .

(A.12)


Observe also that when λ ≥ 0 and z ∈ C+ , we have ℑ z/(λ − z) ≥ 0. Therefore,
|βi (z)| =

|z|
|z|
,
≤
ℑ(z)
(z)r
|
|z + zrTi D−1
i
i

for all z ∈ C+ .

By the same argument, we can bound the other terms in (A.3):

max |βi (z)|, |βi j (z)|, |β̆i (z)|, |β̆i j (z)|, |bn(z)|, |b̄n(z)| ≤ |z| · ℑ(z)−1 ,

18

for all z ∈ C+ .

(A.13)

Lemma A.5. Under Assumptions SA, SB and C, for any z ∈ C+ ,
E|ς1 (z)|k ≤ K p−1 δ2k−4
ℑ(z)−k ,
p

E|τ1 (z)|k ≤ K p−1 δ2k−4
ℑ(z)−2k for k ≥ 2,
p

(A.14)

E|β̆1 (z) − bn (z)|2 ≤ K p−1 |z|6 ℑ(z)−10 , and

(A.15)

lim bn (z) = lim b̄n (z) = −zm(z).

(A.16)

n→∞

n→∞

√
S(1) = e
Sn −r1 rT1 . The following lemma gives the bounds of the eigenvalues
Recall that a± (y) = (1 ± y)2 . Let e
of e
Sn and e
S(1) .

Lemma A.6. Under the assumptions of Lemma A.5, for any ηl < a− (y)1{0<y<1} and ηr > a+ (y), there holds
 eS∗


e
S∗n
n
P λmin
≤ ηl or λmax
≥ ηr = o n−k for any k > 0,

(A.17)

where e
S∗n can be either e
Sn or e
S(1) .

Lemma A.7. For any α ∈ (0, 1), v0 > n−(1+α) , xl < a− (y)1{0<y<1} and xr > a+ (y), let
Cn = {x ± iv0 : x ∈ [xl , xr ]} ∪ {xl ± iv : v ∈ [n−(1+α) , v0 ]} ∪ {xr ± iv : v ∈ [n−(1+α) , v0 ]}.

(A.18)


Then under the assumptions of Lemma A.5, for any ηl ∈ xl , a− (y)1{0<y<1} and ηr ∈ (a+ (y), xr ), there exists K > 0
such that
n
o


1+α
1 eS(1)
sup kD−1
.
e
1 (z)k, |γ1 (z)|, |β1 (z)| ≤ K 1 + n
(A.19)
Sn
{λmin ≤ηl or λmax ≥ηr }

{z∈Cn }

Furthermore, we have

sup
{n; z∈Cn }

n

o
k
EkD−1
1 (z)k , |bn (z)| = O(1) for k ≥ 1, and

n · sup E|γ1 (z)|k = O δ2k−4
p
{z∈Cn }



for k ≥ 2.

(A.20)
(A.21)

The proofs of Lemmas A.3–A.7 are given in Appendix C.

A.2 Proof of Theorem 2.2
Proof. For any f ∈ H, there exist v0 > 0, xl < a− (y)1{0<y<1} and xr > a+ (y) such that f is analytic on the domain
enclosed by
C := {x ± iv0 : x ∈ [xl , xr ]} ∪ {xl ± iv : v ∈ [0, v0 ]} ∪ {xr ± iv : v ∈ [0, v0]}.

Choose ηl ∈ xl , a− (y)1(0<y<1) and ηr ∈ (a+ (y), xr ). By (A.17) and the Borel-Cantelli lemma, almost surely for
all large n,
e

e

n
n
ηl < λSmin
≤ λSmax
< ηr .

(A.22)

Note that when (A.22) holds, we have
GeSn ( f ) = p

Z

+∞

−∞

I
 e

1
f (z)Mn (z)dz, where
f (x)d F Sn (x) − Fyn (x) = −
2πi C

19

(A.23)


Mn (z) = p mn (z) − m0n (z) .

Recall the definition of Cn in (A.18). For all n large such that n−(1+α) < v0 , define



Mn (z),





bn (z) = 
Mn xl + i · sgn(v)n−(1+α) ,
M







 Mn xr + i · sgn(v)n−(1+α) ,

if z ∈ Cn ,
if z = xl + iv with v ∈ [−n−(1+α) , n−(1+α) ],
if z = xr + iv with v ∈ [−n

−(1+α)

,n

−(1+α)

(A.24)

],

where we make the convention that sgn(0) = 1. By (A.22), almost surely for all large n,
I

C



bn (z)dz ≤ K sup | f (z)| · p |ηl − xl |−1 + |ηr − xr |−1 · n−(1+α) = O n−α .
f (z) Mn (z) − M
{z∈C}

Combining the estimate with (A.23), we get

GSen ( f ) = −

1
2πi

I

bn (z)dz + o p (1).
f (z) M

C

(A.25)

bn (z), we have the following proposition whose proof is given in Section A.3.
For the asymptotics of M

bn (·) converges weakly to a Gaussian process M(·)
Proposition A.8. Under the assumptions of Lemma A.5, M
with mean function
EM(z) = −2

ym3 (z)
1 + m(z)

and covariance function
Cov (M(z1 ), M(z2 )) = −

3

!

1−

ym2 (z)
1 + m(z)

2

!−1

+

ym3 (z)
1 + m(z)

3

!

1−

ym2 (z)
1 + m(z)

2

!−2

2m′ (z1 )m′ (z2 )
2ym′ (z1 )m′ (z2 )
2
+
,


 −
1 + m(z1 ) 2 1 + m(z2 ) 2
m(z2 ) − m(z1 ) 2 (z1 − z2 )2

,

z ∈ C,

(A.26)

z1 , z2 ∈ C.

(A.27)

By Proposition A.8, we have, for any f1 , · · · , fk ∈ H,


1
−
2πi

I

C

bn (z)dz, · · · , − 1
f1 (z) M
2πi

I

C

I
I



1
1
D
b
fk (z) Mn (z)dz −→ −
f1 (z)M(z)dz, · · · , −
fk (z)M(z)dz .
2πi C
2πi C

The right-hand side above is multivariate normal due to the facts that the integrals can be approximated by Riemann sums, which are multivariate normal, and the weak limit of multivariate normal vectors is also multivariate
normal. Furthermore, the mean and variance of the right-hand side follow from (A.26) and (A.27), and are
given by (2.1) and (2.2), respectively. Combining the convergence above with (A.25), we complete the proof of
Theorem 2.2.


A.3 Proof of Proposition A.8
bn (z) as
Proof. To prove Proposition A.8, write M

bn (z) = p mn (z) − Emn (z) + p Emn (z) − m0n (z) =: M
bn,1 (z) + M
bn,2 (z),
M

20

z ∈ C.

bn,1 (z) is random while M
bn,2 (z) is deterministic. The proof of Proposition A.8 is divided into
Observe that M
bn,1 (z). The tightness
three subsections. In Section A.3.1, we prove the finite-dimensional convergence of M
b
bn,2 (·) converges to a
of { Mn,1 (z) : z ∈ C, n ∈ N} is proven in Section A.3.2. In Section A.3.3, we show that M
continuous function uniformly on C.

bn,1 (z)
A.3.1 The finite-dimensional convergence of M

bn,1 (z̄) = M
bn,1 (z), therefore it suffices to show that for any q ∈ N, α j ∈ R and z j ∈ C+ := {z ∈ C, ℑ(z) >
Note that M
P
bn,1 (z j ) converges weakly to a Gaussian random variable, and for any pair
0} for j = 1, · · · , q, the sum qj=1 α j M
bn,1 (z1 ) and M
bn,1 (z2 ) converges to (A.27). We take three steps to prove these
z1 , z2 ∈ C+ , the covariance of M
conclusions.

P
bn,1 (z j ) can be approximated by the sum of a martingale difference
Step I. We first show that the sum qj=1 α j M
√
sequence. Recall that ri = Xi / n. Let Ei (·) denote the conditional expectation with respect to the σ-field
generated by r1 , . . . , ri , and E0 (·) = E(·) stand for the unconditional expectation. Note from the definitions
of βi (z) and β̆i (z) in (A.3) and ςi (z) in (A.2) that
βi (z) = β̆i (z) − βi (z)β̆i (z)ςi (z) = β̆i (z) − β̆2i (z)ςi (z) + βi (z)β̆2i (z)ςi2 (z).
Thus, using the definitions of mn (z), D(z) and Di (z), we have
bn,1 (z) =
M

n  
X



−1
−1
(z)
D
(z)
−
D
(z)
tr Ei D−1 (z) − D−1
−
tr
E
i−1
i
i
i=1

=−

=−
+

n
X
i=1

n
X

i=1
n
X
i=1

=:



Ei − Ei−1 βi (z)rTi D−2
i (z)ri



Ei − Ei−1 β̆i (z)τi (z) − β̆2i (z)ςi (z) · n−1 tr D−2
i (z)
Ei − Ei−1



n
X

Vni,1 (z) +

n
X

Vni,2 (z) ≤ 4

i=1

n
X

β̆2i (z)ςi (z)τi (z)

Vni,2 (z) +

i=1

n
X



−

n
X
i=1

(A.28)



Ei − Ei−1 βi (z)β̆2i (z)ςi2 (z)rTi D−2
i (z)ri

Vni,3 (z),

i=1



where the second equality follows from (A.7) and the third equality uses the fact that Ei −Ei−1 β̆i (z)·1/n tr D−2
i (z) =
0. Applying the Burkholder-Davis-Gundy(BDG) inequality, (A.13), the Cauchy-Schwarz inequality and (A.14)
yields
E

i=1

2

n
X
i=1

2

E β̆2i (z)ςi (z)τi (z) ≤ K

n p
X
i=1


E|ςi (z)|4 E|τi (z)|4 = O δ4p .

(A.29)

Similarly, we can show that
E

n
X
i=1

2

Vni,3 (z) ≤ 4

n
X
i=1

2

E βi (z)β̆2i (z)ςi2 (z)rTi D−2
i (z)ri ≤ K

21

n
X
i=1


E|ςi (z)|4 = O δ4p .

(A.30)

Combining (A.28) with (A.29) and (A.30), we get
q
X
j=1

bn,1 (z j ) =
αjM

q
n X
X

α j Vni,1 (z j ) + o p (1) =:

i=1 j=1

n
X

Yni + o p (1).

i=1

Step II. Here we show that the martingale difference sequence {Yni }ni=1 satisfies condition (ii) of Lemma D.1
in Appendix D. In fact, by (A.13), (A.12) and (A.14), we have,



E|Vni,1 (z)|4 ≤ K E|τi (z)|4 + E|ςi (z)|4 = o n−1 .

Hence for any ε > 0,

q
n
n
 1 X
K XX
E|Yni |4 ≤ 2
|α j |4 E|Vni,1 (z j )|4 = o(1),
E |Yni |2 1{|Yni |≥ε} ≤ 2
ε i=1
ε i=1 j=1

n
X
i=1

namely, condition (ii) in Lemma D.1 holds.
Step III. In this step, we verify condition (i) in Lemma D.1. We have
n
X
i=1

W.l.o.g., we focus on

Pn

i=1

q
n
X

 X
Ei−1 Vni,1 (zℓ )Vni,1 (zm ) .
Ei−1 Yni2 =
αℓ αm

n
X

Ei−1 Vni,1 (z1 )Vni,1 (z2 )

i=1

=

β̆2i (z)ςi (z)

(A.31)

i=1


Ei−1 Vni,1 (z1 )Vni,1 (z2 ) . Note that
β̆i (z)τi (z) −

Thus,

ℓ,m=1

·n

−1

tr D−2
i (z)


d β̆i (z)ςi (z)
.
=
dz



n



∂2 X
Ei−1 Ei β̆i (u1 )ςi (u1 ) Ei β̆i (u2 )ςi (u2 )
∂u2 ∂u1 i=1
(u1 ,u2 )=(z1 ,z2 )
n
X


−
Ei−1 β̆i (z1 )τi (z1 ) − β̆2i (z1 )ςi (z1 ) · n−1 tr D−2
i (z1 ) ·



i=1


=:



Ei−1 β̆i (z2 )τi (z2 ) − β̆2i (z2 )ςi (z2 ) · n−1 tr D−2
i (z2 )


∂2
fn,1 (u1 , u2 )
+ fn,2 (z1 , z2 ).
∂u2 ∂u1
(u1 ,u2 )=(z1 ,z2 )

Denote by E(i) (·) the conditional expectation with respect to the σ-field generated by r1 , · · · , ri−1 , ri+1 , · · · , rn .
By (A.12), (A.13) and (A.11), we have


sup E Ei−1 β̆i (z j )τi (z j ) − β̆2i (z j )ςi (z j ) · n−1 tr D−2
i (z j )
1≤i≤n




 
ς
(z
)
(z
)
·
E
≤ sup E Ei−1 β̆i (z j ) · E(i) τi (z j ) + E Ei−1 β̆2i (z j ) · n−1 tr D−2
j
(i) i j
i
1≤i≤n





≤K sup E E(i) τi (z j ) + E E(i) ςi (z j ) = O n−1 δ2p , j = 1, 2.
1≤i≤n

22



(z
)
, j = 1, 2, are
On the other hand, by (A.12) and (A.13), the terms Ei−1 β̆i (z j )τi (z j ) − β̆2i (z j )ςi (z j ) · n−1 tr D−2
j
i

bounded. Combining the two estimates, we see that fn,2 (z1 , z2 ) = O p δ2p and
n
X
i=1

 
Ei−1 Vni,1 (z1 )Vni,1 (z2 ) =



∂2
fn,1 (u1 , u2 )
+ O p δ2p .
∂u2 ∂u1
(u1 ,u2 )=(z1 ,z2 )

(A.32)

 2

. Let D = {u ∈ C : ṽ/2 < ℑ(u) < k1 }, where k1 > v0 is
It remains to analyze ∂u∂2 ∂u1 fn,1 (u1 , u2 )
(u1 ,u2 )=(z1 ,z2 )

an arbitrary constant and ṽ = min ℑ(z1 ), ℑ(z2 ) . Applying the Cauchy-Schwarz inequality, (A.10), (A.13) and
(A.12), we have for any u1 , u2 ∈ D,
n q
X
2

−1
−1
Ei−1 rTi Ei−1 β̆i (u1 )D−1
| fn,1 (u1 , u2 )| ≤
i (u1 ) ri − n tr Ei−1 β̆i (u1 )Di (u1 ) ·
i=1

q
2

−1
−1
Ei−1 rTi Ei−1 β̆(u2 )D−1
i (u2 ) ri − n tr Ei−1 β̆(u2 )Di (u2 )
q
n q

 K|u1 ||u2 |
KX
2
2
≤
Ei−1 |β̆i (u1 )|2 kD−1
Ei−1 |β̆i (u2 )|2 kD−1
.
i (u1 )k ·
i (u2 )k ≤
n i=1
ṽ4

Similarly, we can show that ∂u∂2 fn,1 (u1 , u2 ) is bounded in n by a constant depending only on u1 , u2 and ṽ. Therefore, by Lemma 2.3 of Bai and Silverstein (2004), if fn,1 (u1 , u2 ) converges to a function f (u1 , u2 ) in probability
for any u1 , u2 ∈ D, then
∂2
P
∂2
fn,1 (u1 , u2 ) −→
f (u1 , u2 ).
∂u1 ∂u2
∂u1 ∂u2
Thus, it suffices to prove the convergence of fn,1 (u1 , u2 ).
To this end, we start with the following decomposition:
fn,1 (u1 , u2 ) =bn (u1 )bn (u2 )

n
X
i=1

+

n
X
i=1




Ei−1 Ei ςi (u1 ) · Ei ςi (u2 )



 

Ei−1 Ei β̆i (u1 )ςi (u1 ) · Ei β̆i (u2 )ςi (u2 ) − Ei bn (u1 )ςi (u1 ) · Ei bn (u2 )ςi (u2 )


=:In,1 (u1 , u2 ) + In,2 (u1 , u2 ).

Using (A.15) and (A.14), one can show E|In,2(u1 , u2 )| = o(1). It follows that
fn,1 (u1 , u2 ) = In,1 (u1 , u2 ) + o p (1).

23

(A.33)

For In,1 (u1 , u2 ), by (A.8) and (A.12), we have
XX



1
4 
−1
−
3
b
(u
)b
(u
)
E
Z
Ei D−1
n
1
n
2
11
i (u1 ) j j Ei Di (u2 ) j j
2
n
i=1 j=1
p

n

In,1 (u1 , u2 ) =

X 1



1
−1
4 
tr Ei D−1
−1
bn (u1 )bn (u2 ) E Z11
i (u1 ) · tr Ei Di (u2 )
2
p
n
i=1
n

−

(A.34)

n
X


2
−1
+ 2 bn (u1 )bn (u2 )
tr Ei D−1
i (u1 ) · Ei Di (u2 ) + o(1)
n
i=1

=:In,11 (u1 , u2 ) + In,12 (u1 , u2 ) + 2In,13 (u1 , u2 ) + o(1).
Next we turn to the studies of In,11 (u1 , u2 ), In,12 (u1 , u2 ) and In,13 (u1 , u2 ).
Term In,11 (u1 , u2 ). By the BDG inequality and (A.7),
sup
{1≤i≤n, 1≤ j≤p}

E Ei D−1
i (u)



jj

 2
2
−1
− Emn (u) ≤ E eT1 D−1
1 (u) − ED (u) e1

n

X
 2
 2
−1
−1
T
−1
(E
−
E
)
D
(u)
−
D
(u)
e1
e
(u)
−
D
(u)
e
≤K E eT1 D−1
+
E
i
i−1
1
i
1
1

≤Kn · E

β1 (z)rT1 D−1
1 (u)e1

·

i=1
2
T −1
e1 D1 (u)r1

= o(1), for u ∈ D,

where the last bound follows from (A.13), (A.9) and (A.12). Further using (A.12) yields
sup
{1≤i≤n, 1≤ j≤p}

E Ei D−1
i (u1 )



jj

Ei D−1
i (u2 )



jj

− Emn (u1 )Emn (u2 ) = o(1),

(A.35)

Note that by Proposition 2.1 and the dominated convergence theorem, we have limn→∞ Emn (u) = m(u) for all u ∈ D.
Therefore, by (A.16) and (A.35),

P 
4 
− 3 y · u1 u2 m(u1 )m(u2 ) · m(u1 )m(u2 ).
In,11 (u1 , u2 ) −→ E Z11

Moreover, by equation (1.2) in Bai and Silverstein (2004), we have um(u) = −(1 + m(u))−1. Hence

ym(u1 )
m(u2 )
P 
4 
− 3 θ(u1 , u2 ), where θ(u1 , u2 ) =
In,11 (u1 , u2 ) −→ E Z11
.
1 + m(u1 ) 1 + m(u2 )
Term In,12 (u1 , u2 ).
Claim A.9. Under the assumptions of Lemma A.5, for any u ∈ D,

−1
−1

n−1
n−1
1/2 
u
−
b̄
(u)
b̄
(u)
E tr D−1
(u)
+
p
u
−
=
O
n
,
and
sup
= O(1).
n
n
1
n
n
n
The proof of Claim A.9 is given in Appendix C.
By Claim A.9 and (A.12), we obtain
sup E
{1≤i≤n}

−1 
−1




n−1
n−1
1
−1
= O n−1/2 .
tr Ei D−1
(u
)
tr
E
D
(u
)
−
y
b̄
(u
)
u
−
b̄
(u
)
u
−
1
i
2
n
2
n
1
n
2
1
i
i
pn
n
n

24

(A.36)

Using further (A.16), we get



P
4 
In,12 (u1 , u2 ) −→ − E Z11
− 1 θ(u1 , u2 ).

Term In,13 (u1 , u2 ).

e i (u) =
Claim A.10. For each i and u ∈ D, let D
have

1
n

P

j,i

(A.37)

Z j ZTj − uI. Under the assumptions of Lemma A.5, we

−1
−1
e −1
e −1
kD
i (u)k ≤ ℑ(u) , and EkD1 (u) − D1 (u)k = o(1).

The proof of Claim A.10 is postponed to Appendix C.
By Claim A.10, we have

−1
e−1
e−1
sup EkEi D−1
i (u1 ) − Ei Di (u1 )k ≤ EkD1 (u1 ) − D1 (u1 )k = o(1),

{1≤i≤n}

which, together with (A.12), implies that
sup E
{1≤i≤n}

 1 

1 
−1
e −1
e −1
tr Ei D−1
tr Ei D
i (u1 ) · Ei Di (u2 ) −
i (u1 ) · Ei Di (u2 ) = o(1).
p
p

(A.38)

Noting further (A.16) and inequality (2.17) in Bai and Silverstein (2004), we see that In,13 (u1 , u2 ) has the same
limit as the term (2.8) in Bai and Silverstein (2004) when T = I, and we get
P

In,13 (u1 , u2 ) −→

Z

0

1

θ(u1 , u2 )
dt;
1 − tθ(u1 , u2 )

(A.39)

see P. 578 in Bai and Silverstein (2004).
Summing up: Plugging (A.34), (A.36), (A.37) and (A.39) into (A.33) yields
P

fn,1 (u1 , u2 ) −→ −2θ(u1 , u2 ) + 2

Z

0

1

θ(u1 , u2 )
dt.
1 − tθ(u1 , u2 )

(A.40)

Combining (A.31) with (A.32) and (A.40), we see that {Yni }ni=1 satisfies condition (i) in Lemma D.1, so we
bn,1 (z) converges, in finite dimension, to a Gaussian process with mean 0 and covariance function
conclude that M
Z 1

θ(u1 , u2 )
∂2 
− 2θ(u1 , u2 ) + 2
dt
∂u1 ∂u2
(u1 ,u2 )=(z1 ,z2 )
0 1 − tθ(u1 , u2 )
2m′ (z1 )m′ (z2 )
2ym′ (z1 )m′ (z2 )
2
,
=−

 +
 −
1 + m(z1 ) 2 1 + m(z2 ) 2
m(z2 ) − m(z1 ) 2 (z1 − z2 )2

z1 , z2 ∈ C,

namely, (A.27) holds.

b
A.3.2 Tightness of M
n,1 (z) : z ∈ C, n ∈ N

We will show the tightness using Lemma D.2 in Appendix D.
We first verify condition (i) in Lemma D.2. Recall the definition of Vni,1 (z) in (A.28). For any fixed z0 ∈ C

25

with ℑ(z0 ) = v0 , by (A.13) and (A.14), we have
E

n
X

n
X

2

Vni,1 (z0 ) =

i=1

i=1

2

E Vni,1 (z0 ) ≤ K

n 
X
2
2
E τi (z0 ) + E ςi (z0 ) = O(1).
i=1

Using further (A.28) – (A.30), we get
n
n
n
 X
X
X
2
2
2
bn,1 (z0 ) 2 ≤ K E
EM
Vni,1 (z0 ) + E
Vni,2 (z0 ) + E
Vn j,3 (z0 ) = O(1).
i=1

i=1

i=1

Hence condition (i) in Lemma D.2 is satisfied.

bn (z) in (A.24). We will verify the
Next, we move to condition (ii) in Lemma D.2. Note the definition of M
Kolmogorov-Chentsov condition:
sup
{n; z1 ,z2 ∈Cn }

bn,1 (z1 ) − M
bn,1 (z2 )
EM
|z1 − z2 |2

2

< ∞.

(A.41)

In fact, for any z1 , z2 ∈ Cn , we have


Ei − Ei−1 tr D−1 (z1 ) − D−1 (z2 )
z1 − z2
n
X

= (Ei − Ei−1 ) tr D−1 (z1 )D−1 (z2 )

bn,1 (z1 ) − M
bn,1 (z2 )
M
=
z1 − z2

Pn

i=1

i=1

=

n
X
i=1

=

n
X
i=1

−
−




−1
(Ei − Ei−1 ) tr D−1 (z1 )D−1 (z2 ) − tr D−1
i (z1 )Di (z2 )

2 
−1
(Ei − Ei−1 ) βi (z1 )βi (z2 ) rTi D−1
i (z1 )Di (z2 )ri

n
X
i=1

n
X
i=1

(A.42)



−1
(Ei − Ei−1 ) βi (z2 )rTi D−2
i (z2 )Di (z1 )ri



−1
(Ei − Ei−1 ) βi (z1 )rTi D−2
i (z1 )Di (z2 )ri

=:Θn,1 (z1 , z2 ) + Θn,2 (z1 , z2 ) + Θn,3 (z1 , z2 ),

where the fourth equality follows from (3.7) in Bai and Silverstein (2004). We thus only need to bound
sup

E|Θn,i (z1 , z2 )|2 fori = 1, 2, 3.

{n; z1 ,z2 ∈Cn }

We start with E|Θn,1 (z1 , z2 )|2 . Note that
βi (z) = bn (z) − bn (z)βi (z)γi (z).

26

(A.43)

Thus,
Θn,1 (z1 , z2 ) =

n
X
i=1


2 

−1
(Ei − Ei−1 ) βi (z1 ) · − bn (z2 )βi (z2 )γi (z2 ) + bn (z2 ) · rTi D−1
i (z1 )Di (z2 )ri

= − bn (z2 )

n
X
i=1


2 
−1
(Ei − Ei−1 ) βi (z1 )βi (z2 )γi (z2 ) rTi D−1
i (z1 )Di (z2 )ri

− bn (z1 )bn (z2 )
+ bn (z1 )bn (z2 )

n
X
i=1

n
X
i=1


2 
−1
(Ei − Ei−1 ) βi (z1 )γi (z1 ) rTi D−1
i (z1 )Di (z2 )ri


2 
2  −1
−1
−1
−1
(z
)
(z
)D
−
n
tr
D
(z
)r
(z
)D
(Ei − Ei−1 ) rTi D−1
2
1
2
i
1
i
i
i
i

=: − bn (z2 )Bn,1(z1 , z2 ) − bn (z1 )bn (z2 )Bn,2(z1 , z2 ) + bn (z1 )bn (z2 )Bn,3(z1 , z2 ).
By the BDG inequality, we have
2

sup
{n; z1 ,z2 ∈Cn }

E Bn,1 (z1 , z2 ) ≤
≤
≤
≤

sup
{n; z1 ,z2 ∈Cn }

sup
{n; z1 ,z2 ∈Cn }

sup
{n; z1 ,z2 ∈Cn }

sup
{n; z1 ,z2 ∈Cn }

−1
Kn E β1 (z1 )β1 (z2 )γ1 (z2 ) rT1 D−1
1 (z1 )D1 (z2 )r1

2 2



4
4
−1
Kn E |β1 (z1 )|2 |β1 (z2 )|2 |γ1 (z2 )|2 kD−1
1 (z1 )k kD1 (z2 )k



Kn E |γ1 (z2 )|2 1 + n1+α 1

e
S(1)
{λmin
<ηl



or

en
λSmax
>ηr }

Kn E |γ1 (z2 )|2 + |γ1 (z2 )|2 · n12(1+α) 1

12 

e
S(1)
{λmin
<ηl

or

en
λSmax
>ηr }



(A.44)

= O(1),

where the third inequality follows from (A.19), and the last step uses (A.21), (A.19) and (A.17). Similarly, one
2
can show sup{n; z1 ,z2 ∈Cn } E Bn,2 (z1 , z2 ) = O(1).
As to E|Bn,3(z1 , z2 )|2 , we have from the BDG inequality that
sup

E|Bn,3(z1 , z2 )|2

{n; z1 ,z2 ∈Cn }

≤K
≤K

sup
{n; z1 ,z2 ∈Cn }

sup
{n; z1 ,z2 ∈Cn }

−1
n · E rT1 D−1
1 (z1 )D1 (z2 )r1

2


2 2
−1
− n−1 tr D−1
1 (z1 )D1 (z2 )


4
−1
−1
−1
−1
n · E rT1 D−1
1 (z1 )D1 (z2 )r1 − n tr D1 (z1 )D1 (z2 )

−1
−1
−1
−1
+ rT1 D−1
1 (z1 )D1 (z2 )r1 − n tr D1 (z1 )D1 (z2 )

4
−1
≤K sup E D−1
1 (z1 )D1 (z2 )

2

·

4
 2
−1
−1
tr
D
(z
)D
(z
)
1
2
1
1
n2

{n; z1 ,z2 ∈Cn }

4 
 2
1+α
−1
−1
−1
−1
1
(z
)
1
+
n
(z
)D
(z
)r
−
n
tr
D
(z
)D
+ n · rT1 D−1
e
S
2
1
2
1
1
en
1
1
(1)
1
1
{λmin
<ηl or λSmax
>ηr }
q
q
−1
−1
8
8
4
4
EkD−1
EkD−1
≤K sup
1 (z1 )k · EkD1 (z2 )k +
1 (z1 )k · EkD1 (z2 )k
{n; z1 ,z2 ∈Cn }


2
4(1+α)
−1
+ n · E kD−1
1
1 (z1 )D1 (z2 )k · n

e
S(1)
{λmin
<ηl

e

n
or λSmax
>ηr }



= O(1),

where the third inequality uses (A.10) and (A.19), the fourth inequality follows from the Cauchy-Schwarz inequality and (A.10) again, and the last step uses (A.20), (A.19) and (A.17).

27

Combining the three bounds above and the boundness of sup{n; z∈Cn } |bn (z)| in (A.20) yields
sup

2

E Θn,1 (z1 , z2 ) = O(1).

(A.45)

{n; z1 ,z2 ∈Cn }

As to Θn,2 (z1 , z2 ), by (A.43), we can rewrite Θn,2 (z1 , z2 ) as
Θn,2 (z1 , z2 ) = − bn (z2 )
+ bn (z2 )

n
X
i=1

n
X
i=1



−1
−1
−2
−1
(Ei − Ei−1 ) rTi D−2
i (z2 )Di (z1 )ri − n tr Di (z2 )Di (z1 )



−1
(Ei − Ei−1 ) βi (z2 )γi (z2 )rTi D−2
i (z2 )Di (z1 )ri .

Similarly to (A.44), by the BDG inequality, (A.10), the Cauchy-Schwarz inequality and Lemma A.7 yields
E|Θn,2 (z1 , z2 )|2

sup
{n; z1 ,z2 ∈Cn }

≤K

sup
{n; z1 ,z2 ∈Cn }

+K

sup


 2
−1
−1
−2
−1
n|bn (z2 )|2 E rT1 D−2
1 (z2 )D1 (z1 )r1 − n tr D1 (z2 )D1 (z1 )

{n; z1 ,z2 ∈Cn }

≤K

sup
{n; z1 ,z2 ∈Cn }

+K

−1
n|bn (z2 )|2 E β1 (z2 )γ1 (z2 )rT1 D−2
1 (z2 )D1 (z1 )r1

2

(A.46)

q
−1
4
8
|bn (z2 )|2 EkD−1
1 (z2 )k · EkD1 (z1 )k

sup
{n; z1 ,z2 ∈Cn }


n|bn (z2 )|2 E |γ1 (z2 )|2 + |γ1 (z2 )|2 · n8(1+α) 1

e
S(1)
{λmin
<ηl

or

en
λSmax
>ηr }



= O(1).

Similarly, we can show that
sup

2

E Θn,3 (z1 , z2 ) = O(1).

(A.47)

{n; z1 ,z2 ∈Cn }

b
Plugging (A.45) – (A.47) into (A.42) yields (A.41). The tightness of M
n,1 (z) : z ∈ C, n ∈ N follows from
Lemma D.2.

bn,2 (·)
A.3.3 Convergence of M

bn,2 (·) converges to a continuous function uniformly on C. Below for each n,
In this subsection, we show that M
we work with Cn .
Define


Rn (z) = −Emn (z) − z −


yn
1
.
+
Emn (z) Emn (z) + 1

(A.48)

We have

1 = Emn (z) − z +

yn
Rn (z) 
=: Emn (z) ̟mn (z).
+
Emn (z) + 1 Emn (z)

(A.49)

On the other hand, by equation (6.1.5) in Bai and Silverstein (2010) and using the fact that H = δ1 in our case,

28

we have
1=

m0n (z)

!
yn
−z+ 0
=: m0n (z) ̟m0n (z).
mn (z) + 1

(A.50)

Combining (A.49) and (A.50) leads to
Emn (z) − m0n (z) =Emn (z) · m0n (z) · ̟m0n (z) − ̟mn (z)

= Emn (z) − m0n (z) ·

bn,2 (z) = n Em (z) − m0 (z). Thus
Note that M
n
n
bn,2 (z) · 1 −
M



yn Emn (z) · m0n (z)

 − m0n (z)Rn (z).
m0n (z) + 1 Emn (z) + 1

!
yn Emn (z) · m0n (z)
= −m0n (z) · nRn (z).


m0n (z) + 1 Emn (z) + 1

(A.51)

For m0n (z), we have from equation (4.2) in Bai and Silverstein (2004) that
lim sup m0n (z) − m(z) = 0.

(A.52)

n→∞ {z∈C}

Proposition A.11. Under the assumptions of Lemma A.5, we have
lim sup

n→∞ {z∈C }
n

and


!−1
!−1
yn Emn (z) m0n (z)
ym2 (z)
− 1−
= 0,
1− 0



mn (z) + 1 Emn (z) + 1
1 + m(z) 2

lim sup nRn (z) − 2

n→∞ {z∈C }
n

ym2 (z)
1 + m(z)

!

3 +

ym2 (z)
1 + m(z)

!

3 1 −

ym2 (z)
1 + m(z)

Moreover, the limiting functions in (A.53) and (A.54) are continuous on C.

2

!−1

= 0.

(A.53)

(A.54)

The proof of Proposition A.11 is given in Appendix C.
Combination of (A.51), (A.52) and Proposition A.11 yields
bn,2 (z) − h(z) −→ 0,
sup M

{z∈Cn }

where h(z) is the function on the right-hand side of (A.26). Note that h(z) is continuous on C. Using further the
bn,2 (z) converges uniformly to h(z) on C.
bn (z) in (A.24), we see that M
definition of M

A.3.4 Conclusion of the proof of Proposition A.8

bn (·) converges weakly to a Gaussian process with
Combining the results in Section A.3.1 – A.3.3, we see that M

mean function (A.26) and covariance function (A.27).

29

Appendix B Proofs of propositions in Section 2
2 −1/2
does not change the definition of e
Sn ,
Proof of Proposition 2.1. Note that scaling the entries of Z by EZ11
2
thus w.l.o.g. we can assume that EZ11 = 1.
√
Sn = 1/n · XXT and Sn = 1/n · ZZT . Let I = {i : |Zi | , 0}.
Let X = (X1 , · · · , Xn ) with Xi = pZi /|Zi |. Then e
Recall that we set Zi /|Zi | = 0 when |Zi | = 0. It follows that

tr e
Sn #I
=
≤ 1.
p
p
On the other hand, by Theorem 3.1 in Yin et al. (1988), we have
lim sup
n→∞

√
tr Sn
≤ lim kSn k = (1 + y)2 , a.s..
n→∞
p

Hence, by Lemma 2.7 in Bai (1999), almost surely for all large n,
 

 
2  e
tr Sn + Sn · tr X − Z X − Z T
2
p n
√
2 K X 
2|Zi | |Zi |2 
1 X 2 p
−1 =
|Zi |
1− √ +
≤K
pn i∈I
|Zi |
n i∈I
p
p
n
K X
2|Zi | |Zi |2 
≤
,
1− √ +
n i=1
p
p

e

L4 (F Sn , F Sn ) ≤

(B.1)

where for any two distributions F and G, L(F, G) denotes the levy distance between them.
Consider the limit of (B.1). By the strong law of large numbers,

1 XX 2
1 X |Zi |2
Zi j = 1, a.s..
= lim
n→∞ np
n→∞ n
p
i=1
i=1 j=1
n

p

n

lim

(B.2)

Furthermore, Lemma 2 in Bai and Yin (1993) implies that for any fixed C > 0,
1 X |Zi |
1X
lim inf
√ ≥ lim inf
n→∞ n
n→∞ n
p
i=1
i=1
n

n

sP

p
j=1

Zi2j 1{|Zi j |≤C}
p

=

q 

E Zi2j 1{|Zi j |≤C} , a.s..

By the arbitrariness of C, we obtain
1 X |Zi |
√ ≥ 1, a.s..
n i=1 p
n

lim inf
n→∞

(B.3)

On the other hand, by the Cauchy-Schwarz inequality and (B.2), almost surely
1 X |Zi |
lim sup
√ ≤ lim sup
p
n→∞
n→∞ n i=1
n

30

v
t

1 X |Zi |2
= 1.
n i=1 p
n

(B.4)

Combining (B.3) with (B.4) yields
1 X |Zi |
√ = 1, a.s..
n→∞ n
p
i=1
n

lim

(B.5)

e

e

By (B.1), (B.2) and (B.5), we obtain that limn→∞ L4 (F Sn , F Sn ) = 0, a.s.. Therefore, almost surely, F Sn converges
weakly to the same limit as F Sn does, which is the standard Marčenko-Pastur law with index y.

Proof of Proposition 2.3. We need to calculate the mean and variance in Theorem 2.2 when f (x) = log(x)
and y ∈ (0, 1).
We start with the mean

EG(log) =

1
πi

I

log(z)

C

1
−
2πi

I

ym3 (z)
1 + m(z)

log(z)

C

3

!

ym3 (z)
1 + m(z)

1−

ym2 (z)
1 + m(z)

!

3 1 −

2

!−1

ym2 (z)
1 + m(z)

dz

2

!−2

(B.6)
dz.

Consider the first term in (B.6). We have from equation (1.2) in Bai and Silverstein (2004) that
z=−

y
1
+
.
m(z) 1 + m(z)

Taking derivative with respect to z and solving for m′ (z) yields

m2 (z) 1 + m(z) 2
.
m (z) =

1 + m(z) 2 − ym2 (z)
′

Therefore,


!
2ym(z)m′ (z) 1 + m(z) − 2ym′ (z)m2 (z)
ym2 (z)
d
=
−
1−

dz
(1 + m(z))2
1 + m(z) 3
!
!−1
ym2 (z)
ym3 (z)
=−2
.
 1−

1 + m(z) 3
1 + m(z) 2

It follows that for any f ∈ H,
1
πi

I

C

f (z)

ym3 (z)
1 + m(z)

!

3 1 −

ym2 (z)
1 + m(z)

2

!−1

31

!
I
ym2 (z)
1
dz = −
f (z) d 1 −

2πi C
1 + m(z) 2
!
I
ym2 (z)
1
′
f (z) 1 −
=
 dz.
2πi C
1 + m(z) 2

(B.7)

Furthermore, by Lemma 3.11 in Bai and Silverstein (2010), for r ∈ [−1, 1],

p

√
yr + i y 1 − r2
,
√
x→0
1 + y + 2 yr
p

√
−1 − yr − i y 1 − r2
√
H2 (r) := lim− m(1 + y + 2 yr + ix) =
.
√
x→0
1 + y + 2 yr
−1 −
√
H1 (r) := lim+ m(1 + y + 2 yr + ix) =

Hence, by (B.7),
!
!−1
I
ym3 (z)
ym2 (z)
1
f (z)
1
−
dz


πi C
1 + m(z) 3
1 + m(z) 2
!
! !
√ Z −1
Z 1
yH12 (r)
yH22 (r)
y
√
√
′
f ′ (1 + y + 2 yr) 1 −
f
(1
+
y
+
2
dr
+
=
yr)
1
−

 dr
πi
1
−1
1 + H1 (r) 2
1 + H2 (r) 2
√ Z
√
4 y 1 ′
√
f (1 + y + 2 yr)r 1 − r2 dr.
=−
π
−1

(B.8)

When f (z) = log(z), routine calculation gives that the integral above equals y. As to the second term of (B.6), it
has been computed in Bai and Silverstein (2004) and equals log(1 − y)/2; see equation (1.21) therein. To sum up,
EG(log) = y +

log(1 − y)
.
2

(B.9)

Next we compute the variance
Var(G(log)) =

I I

log(z1 ) log(z2 )m′ (z1 )m′ (z2 )

 dz1 dz2
C2 C1
1 + m(z1 ) 2 1 + m(z2 ) 2
I I
log(z1 ) log(z2 )m′ (z1 )m′ (z2 )
1
− 2
dz1 dz2 .

2π C2 C1
m(z2 ) − m(z1 ) 2

y
2π2

(B.10)

o
n
We start with the first term. W.l.o.g., we assume that C2 encloses C1 . Let Cim := m : m = m(z), z ∈ Ci for
i ∈ {1, 2}. According to P. 598 in Bai and Silverstein (2004), C2m encloses C1m and C1m encloses −1 and 1/(y − 1).
Hence, by the residue theorem,
y
2π2

I I
C2

C1

log(z1 ) log(z2 )
y
dm(z1 )dm(z2 ) = 2
(1 + m(z1 ))2 (1 + m(z2 ))2
2π

I

C1m

log(z(m))
dm
(1 + m)2

!2

= −2y.

On the other hand, the second term in (B.10) equals −2 log(1−y) according to equation (1.22) in Bai and Silverstein
(2004). Thus,
Var(G(log)) = −2y − 2 log(1 − y).

(B.11)

Combining Theorem 2.2, (B.9) and (B.11), we obtain


D
log(1 − y)
GeSn (log) −→ N y +
, −2y − 2 log(1 − y) .
2

The conclusion in Proposition 2.3 follows.

32



Proof of Proposition 2.4. It suffices to calculate the mean and variance in Theorem 2.2 when f (x) = x2 . We
focus on the case when y < 1. The case when y > 1 can be dealt with similarly. The case y = 1 follows from the
continuity of (2.1) and (2.2) on y.
By (B.8) and equation (1.23) in Bai and Silverstein (2004),
√ Z
√
8 y 1
√
(1 + y + 2 yr) · r · 1 − r2 dr
π
−1
!2
2
1
√   1X 2 i
√ 
+
1− y 4+ 1+ y 4 −
y = −y.
4
2 i=0 i

EG(x2 ) = −

Similarly to (B.10), by the residue theorem and equation (1.24) in Bai and Silverstein (2004), we have
Var(G( f )) =

y
2π2
−

=

I I
C2

z21 z22 m′ (z1 )m′ (z2 )
2 dz1 dz2
C2 C1 m(z2 ) − m(z1 )
!2
z2 (m)
dm
+ 4y(2 + 5y + 2y2 ) = 4y2 .
2
1
(1
+
m)
Cm

1
2π2
I

y
2π2

C1

I I

z21 z22

 dm(z1 ) dm(z2 )
1 + m(z1 ) 2 1 + m(z2 ) 2

The conclusion follows.

Appendix C



Proofs of Claims, Lemmas and Propositions in Appendix A

b11  =
Proof of Claim A.1. Because of the self-normalization in b
Sn and S̆n , w.l.o.g., we can assume that Var Z

bn
n
Var Z̆11 ≡ 1. We first give the bound of λSmin
, λS̆min
and kb
Sn k, kS̆n k. By Lemma 2′ in Bai and Silverstein (2004)
(see P. 601 therein), for any ε > 0 and k > 0, we have,




P max p−1 |Z̆i |2 − p ≥ ε = o n−k .
1≤i≤n

Note that



E |Z11 |4 1{|Z11 |>p1/2 δ p }




b11 = E Z11 1{|Z |>p1/2 δ } ≤
= o p−3/2 δ p .
EZ
11
p
p3/2 δ3p

(C.1)

(C.2)

Thus, by (C.1), for all large n and any ε > 0,





√
√



Zi |2 − p ≥ ε ≤nP |Z̆1 | + |E b
Z1 | ≥ p + pε or |Z̆1 | − |E b
P max p−1 |b
Z1 | ≤ p − pε
1≤i≤n




≤nP p−1 |Z̆1 |2 − p ≥ ε/2 = o n−k , for any k > 0.

33

(C.3)

Applying (C.1), (C.3) and the Borel-Cantelli lemma yields that for any ε ∈ (0, 1), almost surely for all large n,
X
X
1
1
Z̆i Z̆Ti ≤ S̆n ≤
Z̆i Z̆Ti , and
n(1 + ε) i=1
n(1 − ε) i=1
n

n

X
X
1
1
b
b
Sn ≤
Zi b
ZTi ≤ b
Zi b
ZTi .
n(1 + ε) i=1
n(1 − ε) i=1
n

n

√
Recall that a± (y) = (1 ± y)2 . It follows from Theorem 1 and Remark 3 in Bai and Yin (1993), Theorem 3.1 and
Remark 2 in Yin et al. (1988), and the arbitrariness of ε that
n
≤ lim sup kS̆n k ≤ a+ (y), a.s..
a− (y)1{0<y<1} ≤ lim inf λS̆min

n→∞

(C.4)

n→∞

Using further the estimate above (2.1) in Bai and Yin (1993) and Lemma 2.3 in Yin et al. (1988), we obtain
n
≤ lim sup kb
Sn k ≤ a+ (y), a.s..
a− (y)1{0<y<1} ≤ lim inf λSmin

b

n→∞

Next, for any ε > 0, let

n→∞

(C.5)

o
o[n
n


Zi |2 − p ≥ ε
max p−1 |Z̆i |2 − p ≥ ε .
A = max p−1 |b
1≤i≤n

1≤i≤n


By (C.1) and (C.3), we have P(A) = o n−k for any k > 0. Note that for any f ∈ H, f is analytic on a domain
enclosing [a− (y)1{0<y<1} , a+ (y)]. By (C.4) and (C.5), we have, almost surely for all large n,
p

Z

+∞

−∞

b

f (x)dF Sn (x) − p

Z

+∞

−∞

f (x)dF S̆n (x) ≤ C f ·

p
X
i=1

b

λSi n − λS̆i n ,

(C.6)

where
Cf = n

′
o | f (x)| = O p (1).

sup

bn
n
min{λSmin
,λS̆min
}≤x≤max{kb
Sn k,kS̆n k}

By the Cauchy-Schwarz inequality,
p
X
i=1

b
λSi n

− λS̆i n ≤
≤

b=
where X

v
t

p q
X

r

i=1

b
λSi n

−

q

2

λS̆i n ·

v
t

2

p 
X
i=1

b

λSi n + λS̆i n



q
q
2p
b − X̆ X
b − X̆T  · kb
Sn k + kS̆n k,
· tr X
n



√ b b
√
p Z1 /|Z1 |, · · · , b
Zn /|b
Zn | and X̆ = p Z̆1 /|Z̆1 |, · · · , Z̆n /|Z̆n | , and the last step uses the inequality in

34

P. 621 of Bai (1999). Thus, by (C.4), (C.5) and (C.2), we have, almost surely for large n,
p
X
i=1

b
λSi n

−

λS̆i n

· 1 Ac

q
T 
b − X̆ b
X − X̆ · 1Ac
≤K tr X
≤K p ·
≤K p ·
≤K p ·

v
t

v
u
t

q



n
n
 1
X
X
b
Zi − Z̆i b
Zi − Z̆i T
1 2
Z̆i Z̆Ti
−
+
· 1 Ac
|b
Zi |2
|b
Zi | |Z̆i |
i=1
i=1

n
b11
X
p EZ
i=1

bi |2
|Z

2

+

n
 p EZ
b11
1X
· 1 Ac
Z̆i Z̆Ti · n · max
1≤i≤n |b
n i=1
Zi |2 |Z̆i |2

b11 + EZ
b11 · 1Ac = O(δ p ).
p EZ
2

(C.7)

2

2


The conclusion follows by noting further that P A → 0, as n → ∞.



Proof of Lemma A.3. We have




E WT AW − tr A WT BW − tr B


=E WT AW · WT BW − tr A · WT BW − tr B · WT AW + tr A tr B .

(C.8)

Write W = (W1 , · · · , W p )T . Recall that we set V/|V| = 0 when |V| = 0. Thus
p
X
i=1

Wi2 = p1{|V|,0} .

(C.9)

By Lemma 2′ in Bai and Silverstein (2004), we have
E

Hence,


W12

= 1/p · E

p
X
i=1



Wi2 = P(|V| , 0) = 1 + o n−k , for any k > 0.

X



bi j
E tr A · WT BW =E W12 · tr A tr B + E W1 W2 · tr A


= tr A tr B + E W1 W2 · tr A

Similarly,

X
i, j

i, j


bi j + o p kAk2 + kBk2 .

X



ai j + o p kAk2 + kBk2 .
E tr B · WT AW = tr A tr B + E W1 W2 · tr B
i, j

35

(C.10)

(C.11)

(C.12)


As to E WT AW · WT BW , elementary computation yields

p
 

 X


E WT AW · WT BW = E W14 − 3E W12 W22
aii bii + E W12 W22 tr A tr B + 2E W12 W22 tr(AB)
i=1

X
X

 X
aii b jl + 4
ai j bli +
a jl bii
+ E W12 W2 W3 ·
i, j,l

i, j,l

(C.13)

i, j,l

X

 X
  X
ai j blm .
aii bi j + 2
ai j bii + E W1 W2 W3 W4 ·
+ E W13 W2 · 2
i, j

i, j

i, j,l,m

Combining (C.8) with (C.11) – (C.13) gives



E WT AW − tr A WT BW − tr B




W14

=E

− 3E

p
X


W12 W22

i=1




aii bii + E W12 W22 − 1 tr A tr B + 2E W12 W22 tr(AB)

X
X
X 

  X
 
aii bi j + 2
ai j bii
bi j + tr B
ai j + E W13 W2 · 2
− E W1 W2 · tr A
i, j

i, j

i, j

(C.14)

i, j

X
X

 X
 X
ai j blm
aii b jl + 4
ai j bli +
a jl bii + E W1 W2 W3 W4 ·
+ E W12 W2 W3 ·
i, j,l

2

i, j,l

i, j,l,m

i, j,l

2 

+ o p kAk + kBk

.

We turn to analyze the remainder terms.

P
P

Term E W1 W2 · tr A i, j bi j + tr B i, j ai j . Note that
X
i, j

Similarly, |

P

i, j

ai j = 1T A1 − tr A ≤ 2pkAk.

(C.15)

bi j | ≤ 2pkBk. Thus


tr A

X

bi j + tr B

i, j

i, j

Next, for any ε ∈ (0, 1), let
A1 =
We have
E W1 W2
For the first term,



X





ai j ≤ K p2 kAk2 + kBk2 .



1 X 2
Vi − 1 ≥ ε .
p − 2 i,1,2

V1 V2 1 A 1
≤ p· E 2
V1 + V22 + · · · + V p2

V1 V2 1 A 1
p· E 2
V1 + V22 + · · · + V p2

!

(C.16)

!

+p· E

(C.17)

V1 V2 1Ac1
V12 + V22 + · · · + V p2

!

.

!
|V1 V2 |1A1

≤ pP(A1 ) = o p−k , for any k > 0,
≤ p·E 2
2
V1 + V2

(C.18)

(C.19)

where the last estimate is due to Lemma 2′ in Bai and Silverstein (2004). As to the second term, by Taylor

36


expansion, almost surely there exists ζ 2 ∈ 0, V12 + V22 such that


V12 + V22
V12 + V22 2
1
1
=
−
+
2
3 .
V12 + V22 + · · · + V p2 V32 + V42 + · · · + V p2
V32 + V42 + · · · + V p2
ζ 2 + V32 + · · · + V p2

It follows that
p· E
=p · E

V1 V2 1Ac1
V12 + V22 + · · · + V p2
V1 V2 1Ac1

!
!

V1 V2 V12 + V22



1Ac1 !

V1 V2 V12 + V22

2

1Ac1 !

−E
 +E 2

V32 + V42 + · · · + V p2
V32 + V42 + · · · + V p2 2
ζ + V32 + · · · + V p2 3

!

1Ac1
V1 V2 V12 + V22 2 1Ac1


≤
p
·
E
· 2E pδ2p V14 + V24 = O p−1 δ2p ,
≤p · E


3
3
2
2
2
2
2
V3 + · · · + V p
V3 + V4 + · · · + V p

(C.20)

where the second inequality uses the fact that |Vi | ≤ p1/2 δ p . Plugging (C.19) and (C.20) into (C.18) yields

Therefore, by (C.16) and (C.21), we obtain



E W1 W2 = O p−1 δ2p .

X
X 

 
bi j + tr B
ai j = o p kAk2 + kBk2 .
E W1 W2 · tr A
i, j

(C.21)

(C.22)

i, j



P
P
Term E W13 W2 · 2 i, j aii bi j + 2 i, j ai j bii . Note that
X
i, j

aii bi j =

p
X
i=1


aii · eTi B(1 − ei ) ≤ p3/2 kAk · kBk ≤ p3/2 kAk2 + kBk2 .

Thus,
X

 X

2
aii bi j + 2
ai j bii ≤ K p3/2 kAk2 + kBk2 .
i, j

i, j


As to E W13 W2 , for any ε ∈ (0, 1), let

A2 =




1 X 2
Vi − 1 ≥ ε .
p − 1 i,2

37

(C.23)

Similarly to (C.21), we have
E


W13 W2

V13 V2 1A2

2

≤p · E

2

+p · E

!

V13 V2 1Ac2


V12 + V22 · · · + V p2 2
!
!
2V13 V23 1{Ac2 }
V13 V2 1{Ac2 }
−
E
≤K p2 · P(A2 ) + p2 · E


V12 + V32 + · · · + V p2 2
V12 + ζ 2 + V32 + · · · + V p2 3
!

E|V1 |3 E|V2 |3
= O(p−1 ).
≤o p−1 + K p2 ·
p3
V12 + V22 · · · + V p2

2

!

(C.24)

Combining (C.23) and (C.24) yields
X

  X

√
aii bi j + 2
ai j bii = O p kAk2 + kBk2 .
E W13 W2 · 2
i, j

(C.25)

i, j

P
P

 P
Term E W12 W2 W3 ·
i, j,l aii b jl + 4 i, j,l ai j bli + i, j,l a jl bii . We have
X

i, j,l

i, j,l

i=1

aii (1 − ei )T B(1 − ei ) +

p
X

aii

i=1

X

bjj

j,i


≤p2 kAk · kBk + p2 kAk · kBk ≤ p2 kAk2 + kBk2 .

Furthermore,
X

p
X

aii b jl ≤

ai j bli ≤

p
X
i=1

eTi A(1 − ei ) · (1 − ei )T Bei +

p 
X

eTi ABei − aii bii
i=1


≤p kAk · kBk + 2pkAk · kBk ≤ K p kAk2 + kBk2 .
2

Therefore,
X

2

aii b jl + 4

i, j,l

X

ai j bli +

i, j,l


As to E W12 W2 W3 , for any ε ∈ (0, 1), let
A3 =

X

i, j,l





a jl bii ≤ K p2 kAk2 + kBk2 .


1 X 2
Vi − 1 ≥ ε .
p − 2 i,2,3

38

(C.26)

Similarly to (C.21), we have
E


W12 W2 W3

V12 V2 V3 1A3

2

≤p · E

!

V12 V2 V3 1Ac3

2

2 + p · E

!


V12 + V22 + · · · + V p2 2

!
!
2V12 V2 V3 V22 + V32 1Ac3
V12 V2 V3 1Ac3

2
2
−
E
≤K p P A3 + p · E


V12 + V42 + · · · + V p2 2
V12 + V42 + · · · + V p2 3

!
3V12 V2 V3 V22 + V32 2 1Ac3
+E

V12 + ζ 2 + V42 + · · · + V p2 4
!
EV12



· E pδ2p V24 + V34 = O p−1 δ2p .
≤o p−2 + K p2 ·
4
p
V12 + V22 + · · · + V p2

Combining the estimate with (C.26) yields
X
X


 X
aii b jl + 4
ai j bli +
a jl bii = o p kAk2 + kBk2 .
E W12 W2 W3 ·
i, j,l

i, j,l

 P

Term E W1 W2 W3 W4 ·
i, j,l,m ai j blm . Note that
X

i, j,l,m

ai j blm =1T A1 · 1T B1 − tr A tr B − 2 tr AB + 2
−

X

aii b jl + 4

i, j,l

X

ai j bli +

i, j,l

(C.27)

i, j,l

X

p
X
i=1



X

 X
aii bi j + 2
ai j bii
aii bii − 2
i, j

i, j

a jl bii .

i, j,l

By (C.23) and (C.26),
X

i, j,l,m



ai j blm ≤ 2p2 kAk · kBk + 4pkAk · kBk + K p2 kAk2 + kBk2 ≤ K p2 kAk2 + kBk2 .

(C.28)

Moreover, recalling the definition of A1 in (C.17), again similarly to (C.21), we have


2

E W1 W2 W3 W4 ≤ p · E

V1 V2 V3 V4 1 A 1

!

2

V1 V2 V3 V4 1Ac1

!

 +p · E

V12 + V22 + · · · + V p2 2
V12 + V22 + · · · + V p2 2

!
!
2V1 V2 V3 V4 V12 + V22 1Ac1
V1 V2 V3 V4 1Ac1

2
2
−
E
≤K p P A1 + p · E


V32 + V42 + · · · + V p2 2
V32 + V42 + · · · + V p2 3

!
3V1 V2 V3 V4 V12 + V22 2 1Ac1
+E

ζ 2 + V32 + · · · + V p2 4
!


E|V3 V4 |
2
−2 
≤o p + K p ·
· E pδ2p V14 + V24 = O p−1 δ2p .
4
p
Combining the estimate with (C.28) yields


  X
E W1 W2 W3 W4 ·
ai j blm = o p kAk2 + kBk2 .
i, j,l,m

39

(C.29)

Plugging the estimates in (C.22), (C.25), (C.27) and (C.29) into (C.14), we obtain

=




E WT AW − tr A WT BW − tr B



EW14

− 3E


W12 W22

p
X
i=1




aii bii + E W12 W22 − 1 tr A tr B

(C.30)




+ 2E W12 W22 tr AB + o p kAk2 + kBk2 .



Next we relate E W14 to E V14 . For any ε ∈ (0, 1), let
A4 =




1 X 2
Vi − 1 ≥ ε .
p − 1 i,1

Similarly to (C.21), we have


 


 
E W14 − E V14 ≤ E W14 − V14 1A4 + E W14 − V14 1Ac4
!
!
2p2 V16 1Ac4



p2
4
c · EV + E
1
≤p2 P A4 + E V14 · P A4 + E
−
1
A
2
3
1
4
V22 + V32 + · · · + V p2
V22 + V32 + · · · + V p2

≤o(1) + Kε + K

p3 δ2p EV14
p3

.

By the arbitrariness of ε, we obtain


E W14 − E V14 = o(1).


As to E W12 W22 , we have from (C.9) that

(C.31)

 X 2



p2 P |V| , 0 = E
Wi2 = pE W14 + p(p − 1)E W12 W22 .
p

i=1

Thus, by (C.10) and (C.31), we get
E


W12 W22



p2 P(|V| , 0) − pE W14
E V14 − 1
=
=1−
+ o(p−1 ).
p(p − 1)
p

(C.32)

Plugging (C.31) and (C.32) into (C.30) yields


E WT AW − tr A WT BW − tr B

p

X





= E V14 − 3
aii bii − p−1 E V14 − 1 tr A tr B + 2 tr(AB) + o p kAk2 + kBk2 .
i=1

This completes the proof of (A.8).



We now prove (A.9). Let Ã = Aei eTi A =: ãi j and B̃ = Bei eTi B =: b̃i j . Applying (C.13) and the estimates

40

in (C.25), (C.27) and (C.29), we have



E WT ÃW WT B̃W

p




 X

ã j j b̃ j j + E W12 W22 tr Ã tr B̃ + 2E W12 W22 tr ÃB̃ + o p kÃk2 + kB̃k2 .
= E W14 − 3E W12 W22

(C.33)

j=1

Note that

p
X
j=1

kÃk ≤kAk2 ,

kB̃k ≤ kBk2 ,

ã j j b̃ j j ≤kÃk · tr B̃ = kÃk · eTi B2 ei ≤ kAk2 kBk2 ,

tr Ã tr B̃ =eTi A2 ei · eTi B2 ei ≤ kAk2 kBk2 ,
 
tr ÃB̃ =eTi ABei · eTi BAei ≤ kAk2 kBk2 .

Plugging these bounds into (C.33), and further using (C.31) and (C.32) yield the desired estimate in (A.9).

Next we prove Lemma A.4. The following lemma will be used.
Lemma C.1. [Theorem 3 in Rosenthal (1970)] Let U1 , · · · , U p be i.i.d. random variables with EU1 = 0. For
any k ≥ 2, there exists K > 0 such that
E

p
X
i=1

k

Ui ≤ K

p
 X

EUi2

i=1

k/2

+

p
X
i=1


E|Ui |k .

Proof of Lemma A.4. Write


E|WT AW − tr A|k = E |WT AW − tr A|k · 1{|V|=0} + E |WT AW − tr A|k · 1{|V|,0} .

We have




E |WT AW − tr A|k · 1{|V|=0} = E | tr A|k · 1{|V|=0} ≤ pk kAkk P |V| = 0 = kAkk · o(1),

(C.34)

where the last estimate follows from Lemma 2′ in Bai and Silverstein (2004). Furthermore,


 p
k

VT AV − tr A · 1{|V|,0}
E |WT AW − tr A|k · 1{|V|,0} =E
2
|V|
 p


k
k
T
T
≤KE
1
−
1
V
AV
·
+
KE
V
AV
−
tr
A
.
{|V|,0}
|V|2

(C.35)

By Lemma C.1, for any k ≥ 2,
 p

k
−
1
VT AV · 1{|V|,0}
2
|V|
p
p
 X
 k/2 X

k
kAkk ,
≤kAkk E p − |V|2 ≤ KkAkk ·
+
E 1 − Vi2 2
E 1 − Vi2 k ≤ K pk−1 δ2k−4
p
E

i=1

i=1

41

(C.36)


where the last step follows from the fact that |Vi | ≤ p1/2 δ p and E V14 < ∞. On the other hand, if we denote by
A∗ the adjoint matrix of A, then by Lemma 2.7 in Bai and Silverstein (1998), for any k ≥ 2,





k
kAkk .
E VT AV − tr A ≤ K E|V1 |4 tr AA∗ k/2 + E|V1 |2k tr (AA∗ )k/2 ≤ K pk−1 δ2k−4
p

(C.37)

Combining (C.34) – (C.37) gives (A.10).

As to (A.11), by (C.10), (C.15) and (C.21), we get
p
p
X
 X
 X


ai j E W1 W2 −
aii E W12 +
aii = O δ2p kAk .
E WT AW − tr A =
i, j

i=1

i=1


Proof of Lemma A.5. Recall the definitions of ςi (z) and τi (z) in (A.2). Denote by E(i) (·) the conditional expectation with respect to the σ-field generated by r1 , · · · , ri−1 , ri+1 , · · · rn . By (A.10), for any k ≥ 2,

k
E|ς1 (z)|k =E E(1) |ς1 (z)|k ≤ K p−1 δ2k−4
EkD−1
p
1 (z)k , and

2k
EkD−1
E|τ1 (z)|k =E E(1) |τ1 (z)|k ≤ K p−1 δ2k−4
p
1 (z)k .

(C.38)

The bounds in (A.14) then follow from (A.12).
As to (A.15), by (A.13),
2

−1
−1
E β̆1 (z) − bn (z) =E β̆1 (z)bn (z) n−1 tr D−1
1 (z) − n E tr D1 (z)

2

2

−1
≤Kn−2 |z|4 ℑ(z)−4 · E tr D−1
1 (z) − E tr D1 (z) .

Let Ei (·) stand for the conditional expectation with respect to the σ-field generated by r1 , . . . , ri . By the BurkholderDavis-Gundy(BDG) inequality, (A.7) and the Cauchy-Schwarz inequality, for any k ≥ 2,
k

−1
E tr D−1
1 (z) − E tr D1 (z)
n
X
k

−1
=E
Ei − Ei−1 tr D−1
1 (z) − tr D1i (z)
i=2

≤KE

n
X
i=2

tr D−1
1 (z)

−

tr D−1
1i (z)
k

2 k/2

k/2
≤Knk/2 E β12 (z)rT2 D−2
12 (z)r2 ≤ Kn

q

(C.39)

4k
E|β12 (z)|2k · EkD−1
12 (z)k .

Using further the bounds in (A.12) and (A.13), we get (A.15).
Now we show (A.16). We first prove that Eβ1 (z), bn (z) and b̄n (z) converge to the same limit. In fact, by
(A.13) and the Cauchy-Schwarz inequality,
E|β1 (z) − bn (z)| ≤E β1 (z) − β̆1 (z) + E|β̆1 (z) − bn (z)|
1/2

≤|z|2 ℑ(z)−2 E|ς1 (z)| + E|β̆1 (z) − bn (z)|2
1/2
1/2 

+ E|β̆1 (z) − bn (z)|2 .
≤|z|2 ℑ(z)−2 E|ς1 (z)|2

42

Plugging in the bounds in (A.14) and (A.15) yields

E|β1 (z) − bn (z)| ≤ K p−1/2 |z|2 ℑ(z)−3 + |z|3 ℑ(z)−5 .

(C.40)

Furthermore, by (A.13) and Lemma 2.6 in Silverstein and Bai (1995),

−1 2
−3
−1
|bn (z) − b̄n (z)| ≤|z|2 ℑ(z)−2 · n−1 E tr D−1
1 (z) − tr D12 (z) ≤ n |z| ℑ(z) .

(C.41)

Next, using an argument similar to that for equation (2.2) of Silverstein (1995), we have
Eβ1 (z) = −zEmn (z).

(C.42)

By Proposition 2.1 and the dominated convergence theorem, the term Emn (z) converges to m(z). The convergences in (A.16) follow from (C.40) and (C.41).

Proof of Lemma A.6. For any ηl < a− (y)1{0<y<1} and ηr > a+ (y), there exists ε ∈ (0, 1) such that (1 + ε)ηl <
a− (y)1{0<y<1} and (1 − ε)ηr > a+ (y). Note that for e
Sn ,


n



 e
o
en
n
n
n
< (1 + ε)ηl or λSmax
> (1 − ε)ηr + P max p−1 |Zi |2 − p ≥ ε ,
P λSmin
< ηl or λSmax
> ηr ≤ P λSmin
1≤i≤n

and for e
S(1) ,

 n eS(1)


 eS
n e
e
Sn−1
n−1
n−1
n−1 S(1)
P λmin
< ηl or λmax
> ηr .
< ηl or λmax
> ηr = P λmin

The conclusion follows from the estimates (1.9a), (1.9b) and Lemma 2′ in Bai and Silverstein (2004).



Proof of Lemma A.7. For D−1
1 (z), we have


−1
+ |ηr − xr |−1 1
sup kD−1
1 (z)k ≤ |ηl − xl |

{z∈Cn }


≤K 1 + n1+α 1

e
S(1)
{λmin
<ηl

or

e
S

e
S

(1)
(1)
λmin
≥ηl and λmax
≤ηr

en
λSmax
>ηr }


.

+ n1+α 1

e
S

e
S

(1)
(1)
{λmin
<ηl or λmax
>ηr }

(C.43)

Thus, by (A.17), for any k ≥ 1,
 eS(1)

en
k
k(1+α)
sup EkD−1
P λmin
< ηl or λSmax
> ηr = O(1).
1 (z)k ≤ K + Kn

(C.44)

{n; z∈Cn }

Furthermore,

1+α
−1
sup |γ1 (z)| ≤ K sup kD−1
1
1 (z)k + K sup EkD1 (z)k ≤ K 1 + n

{z∈Cn }

{z∈Cn }

{n; z∈Cn }

e
S(1)
{λmin
<ηl

2
T −1
For βi (z), note that by (A.7), rTi D−1 (z)ri − rTi D−1
i (z)ri = −βi (z) ri Di (z)ri . Hence

or

en
λSmax
>ηr }



2
T −1
T −1
T −1
βi (z) = βi (z) − βi (z) rTi D−1
i (z)ri + ri D (z)ri + ri Di (z)ri = 1 − ri D (z)ri ,

43



.

where the second equation holds due to the definition of βi (z). Similarly to (C.43), we can show that

sup kD−1 (z)k ≤ K 1 + n1+α 1

e
S(1)
{λmin
<ηl

{z∈Cn }

or

en
λSmax
>ηr }


.

It follows that

sup |βi (z)| ≤ 1 + sup |rTi D−1 (z)ri | ≤ 1 + yn sup kD−1 (z)k ≤ K 1 + n1+α 1

{z∈Cn }

e
S(1)
{λmin
<ηl

{z∈Cn }

{z∈Cn }

or

en
λSmax
>ηr }


.

(C.45)

We thus complete the proof of (A.19).
We now prove (A.21). Similarly to (C.44) and using the bound in (C.45), we can show that for any k ≥ 1,
sup
{n; z∈Cn }

n

o
k
E|β1 (z)|k , E|β12 (z)|k , EkD−1 (z)kk , EkD−1
12 (z)k = O(1).

(C.46)

For E|γ1 (z)|k , by (C.38), (C.39), (C.44) and (C.46), for any k ≥ 2,


k
k
k
sup E γ1 (z) ≤K sup E γ1 (z) − ς1 (z) + E ς1 (z)

{z∈Cn }

{z∈Cn }



k
k
−1 2k−4
−1
−1
(z)k
(z)
+
K
p
δ
EkD
(z)
−
E
tr
D
≤K sup n−k E tr D−1
p
1
1
1
{z∈Cn }

K
≤ k/2 sup
n {z∈Cn }

The conclusion (A.21) follows.

q
−1 2k−4
4k
E|β12 (z)|2k EkD−1
12 k + K p δ p


=O n−1 δ2k−4
.
p

Finally, we prove (A.20). It remains to show the boundness of sup{n; z∈Cn } |bn (z)|. Note that bn (z) = β1 (z) +
bn (z)β1 (z)γ1 (z). Thus, by the Cauchy-Schwarz inequality,
q
|bn (z)| ≤ E β1 (z) + |bn (z)| E|β1 (z)|2 E|γ1 (z)|2 .
It follows from (C.46) and (A.21) that for all large n,
sup |bn (z)| ≤ sup

{z∈Cn }

{n; z∈Cn }

1−

p

E β1 (z)
E|β1 (z)|2 E|γ1 (z)|2

= O(1).


Proof of Claim A.9. Recall the definition of b̄n (u) in (A.3). By equation (2.9) in Bai and Silverstein (2004), for
any u ∈ C+ ,
−1


n−1
b̄n (u)
− I p + b̄n (u)Ai (u) + Bi (u) + Ci (u) ,
D−1
i (u) = u −
n

44

where
Ai (u) =

X
j,i

Bi (u) =

X
j,i

Ci (u) =
Thus,


r j rTj − n−1 I D−1
i j (u),


βi j (u) − b̄n (u) r j rTj D−1
i j (u), and


b̄n (u) X  −1
Di j (u) − D−1
i (u) .
n j,i

−1

n−1
b̄
(u)
(u)
+
p
u
−
E tr D−1
n
1
n
!−1 


n−1
≤ u−
b̄n (u)
E| tr b̄n (u)A1 (u) | + E| tr B1 (u)| + E| tr C1 (u)| .
n

(C.47)

Now we study the terms in (C.47). For u ∈ D,
u−

n−1
b̄n (u)
n

!−1


1 + n−1 E tr D−1
1 + n−1 E tr D−1
1 + 2p/(nṽ)
12 (u)
12 (u)
= O(1),
≤
=

 ≤
−1
−1
−1
−1
ṽ/2
u 1 + n E tr D12 (u) − (n − 1)/n
ℑ u 1 + n E tr D12 (u)

where the last inequality uses (A.12) and the fact that ℑ (u/(λ − u)) ≥ 0 for any λ ≥ 0 and u ∈ C+ .

Term E| tr b̄n (u)A1 (u) |. Note from (A.10) and (A.12) that


2
2
2
−1 
−1
−1
−1
−1
T −1
.
≤ Kn−1 EkD−1
E rT2 D−1
12 (u)k = O n
12 (u)r2 − n tr D12 (u) = E E(2) r2 D12 (u)r2 − n tr D12 (u)

(C.48)

(C.49)

Thus by (A.13) and the Cauchy-Schwarz inequality,

E tr b̄n (u)A1 (u) ≤ Kn ·

q
2
1/2 
−1
−1
.
E rT2 D−1
12 (u)r2 − n tr D12 (u) = O n

(C.50)

Term E| tr B1 (u)|. By the Cauchy-Schwarz inequality, (A.13) and (A.12), we have


E| tr B1 (u)| ≤ n · E β12 (u) − b̄n (u) rT2 D−1
12 (u)r2
q
q
2
2
≤n · E β12 (u) − b̄n (u)
E rT2 D−1
12 (u)r2
q
2

−1
−1
≤Kn · E β212 (u)b̄2n (u) · rT2 D−1
12 (u)r2 − n E tr D12 (u)
q 

2
−1
−1
−2 tr D−1 (u) − E tr D−1 (u) 2 .
≤Kn · E rT2 D−1
12 (u)r2 − n tr D12 (u) + n
12
12

(C.51)

Similarly to (C.39), using (A.12) and (A.13), we can show

2

−1
E tr D−1
12 (u) − E tr D12 (u) = O(n).

Combining the estimate with (C.49) and (C.51) yields

E tr B1 (u) = O n1/2 .

45

(C.52)

Term E| tr C1 (u)|. By (A.13) and Lemma 2.6 in Silverstein and Bai (1995),

−1
E| tr C1 (u)| ≤ |b̄n (u)| · E tr D−1
12 (u) − D1 (u) = O(1).

(C.53)

Combining (C.47), (C.48), (C.50), (C.52) and (C.53) yields the desired conclusion.

e −1 (u)k ≤ ℑ(u)−1 . As to the differProof of Claim A.10. By the same argument as for (A.12), we have kD
1
e−1
ence D−1
1 (u) − D1 (u), we have
X
1X
T
e −1 (u) ≤ kD−1 (u)k · 1
e −1 (u)k.
D−1
(u)
−
D
Z
Z
−
Xi XTi · kD
i
i
1
1
1
1
n i,1
n i,1

Theorem 1 in Bai and Yin (1993) implies that kn−1

P

i,1

Zi ZTi k = O p (1). Furthermore, by Lemma 2 therein,


1X
1X
1 X
p2
p2
1X
Xi XTi −
Zi ZTi ≤
Zi ZTi = o p (1).
1−
Zi ZTi ≤ max 1 −
·
2
2
{i,1}
n i,1
n i,1
n i,1
n i,1
|Zi |
|Zi |
Using further (A.12), we see that
e −1
D−1
1 (u) − D1 (u) = o p (1).

The conclusion follows from the dominated convergence theorem.



Proof of Proposition A.11. Proof of (A.53). We take two steps to prove (A.53): In Step I we shows that
lim sup Emn (z) − m(z) = 0,

n→∞ {z∈C }
n

(C.54)

and in Step II we prove (A.53).
D
e
Step 1. Recall that e
Sn = 1/n · XT X. By Proposition 2.1, F Sn → F y almost surely. Recall that ηl ∈

xl , a− (y)1{0<y<1} and ηr ∈ (a+ (y), xr ). We have
sup Emn (z) − m(z)
Z
Z 1


1
e
S
c (λ) dF n (λ) .
≤ sup E
1[ηl ,ηr ] (λ) d FeSn (λ) − F y (λ) + sup E
1
λ−z
λ − z [ηl ,ηr ]
{z∈Cn }
{z∈Cn }
{z∈Cn }

The first term converges to 0 by the bounded convergence theorem. As to the second term, we have from (A.17)
that
Z

 e
1
en
n
1[ηl ,ηr ]c (λ)dFeSn (λ) ≤ lim n1+α P λSmin
< ηl or λSmax
> ηr = 0.
lim sup E
n→∞
n→∞ {z∈C }
λ−z
n
Hence (C.54) holds.
Step II. Bai and Silverstein (2004) show that with C0 = {xr ± iv, v ∈ [0, v0 ]} ∪ {xl ± iv, v ∈ [0, v0]}, there exists
a positive number δ such that inf {z∈C0 } |m(z) + 1| > δ; see P. 585 therein. Therefore, by (C.54), for all n large

46

enough,
inf

{z∈C0 ∩Cn }

Emn (z) + 1 ≥

δ
.
2


−1
≤ max{2, 4/v0}.
Moreover, by Lemma 2.3 of Silverstein (1995), for any z with ℑ(z) = ±v0 , we have Emn (z) + 1
Combining the two estimates we get
1
= O(1).
(C.55)
sup
(z)
+1
Em
{n; z∈Cn }
n
Using further (A.52) and (C.54), we have
sup
{n; z∈Cn }

1
= O(1).
m0n (z) + 1

(C.56)

Combining (A.52), (C.54), (C.55) with (C.56) yields
lim sup

n→∞ {z∈C }
n

yn Emn (z) · m0n (z)
ym(z)2
−
= 0.


(1 + m(z))2
m0n (z) + 1 Emn (z) + 1

Finally, by (4.6) in Bai and Silverstein (2004), there exists ξ1 < 1 such that
sup
{z∈C}

ym2 (z)
1 + m(z)

2 ≤ ξ1 < 1.

(C.57)

The conclusion in (A.53) follows. Furthermore, (C.57) and the definition of m(z) implies the continuity of

1 − ym2 (z)/ 1 + m(z) 2 −1 on C.
Proof of (A.54). Recall the definition of Rn (z) in (A.48). Note that Emn (z) = − (1 − yn ) /z + yn Emn (z). We
can rewrite Rn (z) as
Rn (z) =

yn
+ zyn Emn (z).
Emn (z) + 1

Thus, by (C.42) and (A.6), we have
nRn (z) Emn (z) + 1




= tr I + (z − Eβ1 (z)) E tr D−1 (z)




=E tr (D(z) + zI D−1 (z) − Eβ1 (z) · E tr D−1 (z)

n
 X


=E tr
ri rTi · D−1 (z) − Eβ1 (z) · E tr D−1 (z)
i=1




−1
−1
=E nβ1 (z) rT1 D−1
1 (z)r1 − n E tr D (z)



−1
=E nβ1 (z)γ1 (z) + Eβ1 (z) · E tr D−1
1 (z) − tr D (z)

=:Jn (z) + J˜n (z).
We will analyze Jn (z) and J˜n (z), separately.

47

(C.58)

For J˜n (z), by (A.7) and (A.43),


J˜n (z) =Eβ1 (z) · E β1 (z)rT1 D−2
1 (z)r1




T −2
=bn (z)Eβ1 (z) · E rT1 D−2
1 (z)r1 − bn (z)Eβ1 (z) · E β1 (z)γ1 (z)r1 D1 (z)r1 .

For the second term, we have from (A.20), (C.46), Hölder’s inequality and (A.21) that


sup bn (z)Eβ1 (z) · E β1 (z)γ1 (z)rT1 D−2
1 (z)r1

{z∈Cn }

≤K sup

{z∈Cn }

q


1/4 


8 1/4
= O n−1/2 .
E |γ1 (z)|2 · E |β1 (z)|4
· EkD−1
1 (z)k

For the first term, by (A.20), (C.46) and (A.11),


−1
−2
sup bn (z)Eβ1 (z) · E rT1 D−2
1 (z)r1 − n tr D1 (z)

{z∈Cn }


2
−1 
.
≤K sup |E E(1) (τ1 (z) | ≤ Kn−1 δ2p sup EkD−1
1 (z)k = o n
{z∈Cn }

{z∈Cn }

Therefore

J˜n (z) = n−1 bn (z)Eβ1 (z) · E tr D−2
1 (z) + εn (z),

(C.59)

where εn (z) denotes a residual term which satisfies limn→∞ sup{z∈Cn } |εn (z)| = 0. In the following analysis, we will
continue using this notation, whose value may change from line to line.
As to Jn (z), since β1 (z) = bn (z) − b2n (z)γ1 (z) + b2n (z)β1 (z)γ12 (z) by (A.43), we have

Jn (z) = − nb2n (z)Eγ12 (z) + nbn (z)Eγ1 (z) + nb2n (z)E β1 (z)γ13 (z)
2

−1
= − nb2n (z)Eς12 (z) − n−1 b2n (z)E tr D−1
1 (z) − E tr D1 (z)



−1
− 2b2n (z)E ς1 (z) tr D−1
1 (z) − E tr D1 (z)

+ nbn (z)Eγ1 (z) + nb2n (z)E β1 (z)γ13 (z)

(C.60)

=: − nb2n (z)Eς12 (z) + Jn,1 (z) + Jn,2 (z) + Jn,3 (z) + Jn,4 (z).

We turn to study the last four remainder terms.
We first show that
2

−1
sup E tr D−1
1 (z) − E tr D1 (z) = O(1).

(C.61)

{n; z∈Cn }

Firstly, similarly to Lemmas A.6 and A.7, we can prove
−1
−1
T
T
e
e
Claim C.2. Let γ12 = rT1 D−1
12 (z)r1 − n E tr D12 (z) and S(12) = Sn − r1 r1 − r2 r2 . Under the assumptions of
Lemma A.5, for any ηl < a− (y)1{0<y<1} and ηr > a+ (y), we have


 eS(12)

en
P λmin
≤ ηl or λSmax
≥ ηr = o n−k for any k > 0.

48

(C.62)

In addition, there exists K > 0 such that
n
o

1+α
1
sup kD−1
12 (z)k, |β12 (z)|, |γ12 (z)| ≤ K 1 + n

e
S(12)
{λmin
≤ηl

{z∈Cn }

en
λSmax
≥ηr }

or


,

(C.63)

and
sup
{n; z∈Cn }

n

o
k
EkD−1
12 (z)k , |b̄n (z)| = O(1) for k ≥ 1,

n · sup E|γ12 (z)|k ≤ O δ2k−4
p
{z∈Cn }



(C.64)

for k ≥ 2.

(C.65)

We now prove (C.61). By the BDG inequality and (A.7), we have
−1
sup E tr D−1
1 (z) − E tr D1 (z)

{n; z∈Cn }

= sup E
{n; z∈Cn }

= sup

n
X
i=2

n
X

{n; z∈Cn } i=2

≤K sup

{n; z∈Cn }



2


2
−1
(Ei − Ei−1 ) tr D−1
1 (z) − D1i (z)

2




−1
−1
−2
β1i (z) − b̄n (z) · tr D−2
E (Ei − Ei−1 ) β1i (z) rTi D−2
1i (z)
1i (z)ri − n tr D1i (z) + n

(C.66)


2

4
−1
−2
2
2
−1
,
n · E β12 (z) rT2 D−2
(z)
(z)k
(z)r
−
n
tr
D
+
n
·
E
|γ
(z)|
·
|β
(z)
b̄
(z)|
kD
2
12
12
n
12
12
12



where the second equality uses the fact that Ei −Ei−1 b̄n (z) tr D−2
(z)
= 0. For the first term, we have from (C.63)
1i
that

2
−1
−2
sup n · E β12 (z) rT2 D−2
12 (z)r2 − n tr D12 (z)

{n; z∈Cn }


2 
1+α
−1
−2
≤K sup n · E rT2 D−2
1
12 (z)r2 − n tr D12 (z) · 1 + n

e
S(12)
{λmin
<ηl

{n; z∈Cn }

or

en
λSmax
>ηr }


2
−1
4
2(1+α)
−1
−2
1
≤K sup n · E rT2 D−2
12 (z)r2 − n tr D12 (z) + kD12 (z)k · n

2 

e
S(12)
{λmin
<ηl

{n; z∈Cn }

or

(C.67)
en
λSmax
>ηr }



= O(1),

where the last step follows from (A.10), (C.64), (C.63) and (C.62). For the second term, we have

4
sup n · E |γ12 (z)|2 · |β12 (z)b̄n (z)|2 kD−1
12 (z)k

{n; z∈Cn }



≤K sup n · E |γ12 (z)|2 1 + n1+α 1

e
S(12)
{λmin
<ηl

{n; z∈Cn }

e

n
or λSmax
>ηr }


≤K sup n · E |γ12 (z)|2 + |γ12 (z)|2 · n8(1+α) 1

e
S

6 

(C.68)
e

(12)
n
{λmin
<ηl or λSmax
>ηr }

{n; z∈Cn }



= O(1),

where the first inequality uses (C.64) and (C.63), and the last bound follows from (C.65), (C.63) and (C.62).
Plugging (C.67) and (C.68) into (C.66) yields (C.61).
For Jn,1 (z), combining (A.20) and (C.61) yields
2
−1 
−1
sup Jn,1 (z) ≤ sup n−1 |bn (z)|2 E tr D−1
.
1 (z) − E tr D1 (z) = O n

{z∈Cn }

{z∈Cn }

49

(C.69)

As to Jn,2 (z), we obtain from the Cauchy-Schwarz inequality, (A.20), (C.38) and (C.61) that
q


2
−1
sup |Jn,2 (z)| ≤ 2 sup |bn (z)|2 E|ς1 (z)|2 · E tr D−1
= O n−1/2 .
1 (z) − E tr D1 (z)

{z∈Cn }

(C.70)

{z∈Cn }

As to Jn,3 (z), by (A.20) and (A.11),
2
sup |Jn,3 (z)| ≤ sup n|bn (z)| · |Eγ1 (z)| ≤ Kδ2p sup EkD−1
1 (z)k = O δ p .

{z∈Cn }

{z∈Cn }

(C.71)

{z∈Cn }

Finally, for Jn,4 (z), we have

 



−1
2
2
Jn,4 (z) =nb2n (z)E β1 (z)γ12 (z)rT1 D−1
1 (z)r1 − bn (z)E β1 (z)γ1 (z) · E tr D1 (z)




−1
2
2
=nb2n (z)E β1 (z)γ12 (z)rT1 D−1
1 (z)r1 − bn (z)E β1 (z)γ1 (z) tr D1 (z)


+ b2n (z) Cov β1 (z)γ12 (z), tr D−1
1 (z)




=nb2n (z)E β1 (z)γ12 (z)ς1 (z) + b2n (z) Cov β1 (z)γ12 (z), tr D−1
1 (z)

(C.72)

=:Jn,41 (z) + Jn,42 (z).

By the Cauchy-Schwarz inequality,
q 
p

sup |Jn,41 (z)| ≤ sup n · |bn (z)|2 E|ς1 (z)|2 E |β1 (z)|2 |γ1 (z)|4

{z∈Cn }

{z∈Cn }

≤K sup

{z∈Cn }

≤K sup

{z∈Cn }

√

√

n · |bn (z)|2
n·

q

2
EkD−1
1 (z)k

r

E 1 + n1+α 1

e
S(1)
{λmin
<ηl

r 
E |γ1 (z)|4 + |γ1 (z)|4 · n2(1+α) 1

e
S(1)
{λmin
<ηl

or

or

en
λSmax
>ηr }

en
λSmax
>ηr }



2

|γ1 (z)|4



(C.73)

= o(1),

where the second inequality follows from (C.38) and (A.19), the third inequality uses (A.20), and the last
step uses (A.21), (A.19) and (A.17). Similarly, we have from the Cauchy-Schwarz inequality, (C.46), (A.21)
and (C.61) that
sup |Jn,42 (z)| ≤ sup K E |β1 (z)|4

{z∈Cn }

{z∈Cn }

1/4

E |γ1 (z)|8

1/4 

−1
E tr D−1
1 (z) − E tr D1 (z)

2 1/2


= o n−1/4 .

(C.74)

Plugging (C.73) and (C.74) into (C.72) , we obtain
sup |Jn,4 | = o(1).

{z∈Cn }

Combining (C.60), (C.69) – (C.71) and (C.75) yields
Jn (z) = −nb2n (z)Eς12 (z) + εn (z).

50

(C.75)

Moreover, by (A.8) and (A.20),
− nb2n (z)Eς12 (z)
p 


X


2
 2
4 
−2
−1
4 
−1
= − n−1 b2n (z)E E Z11
−3
(z)
+ εn (z).
(z)
+
2
tr
D
(z)
−
1
tr
D
D−1
E
Z
−
p
1
1
1
11
ii
i=1

Thus, by (A.20) and (C.61), we have
p
4 
 1X


−1 2
 2 E Z11
2
4 
−3 ·
sup Jn (z) + b2n (z) E Z11
(z)
E D−1
−
bn (z) · n−1 E tr D−1
1
1 (z)
ii
n i=1
yn
{z∈Cn }

+ 2b2n (z) · n−1 E tr D−2
1 (z) = o(1).

(C.76)

Plugging (C.59) and (C.76) into (C.58), and applying (C.55), we obtain
nRn (z)

P p  −1  2 

4
EZ11
− 3 b2n (z) · n−1 i=1
E D1 (z) ii


2
4
EZ11
− 1 b2n (z) · n−1 E tr D−1
1 (z)
=−
+

Emn (z) + 1
yn Emn (z) + 1


bn (z)Eβ1 (z) · n−1 E tr D−2
2b2n (z) · n−1 E tr D−2
1 (z)
1 (z)
+
+ εn (z).
−
Emn (z) + 1
Emn (z) + 1

(C.77)

The remaining task is to study (C.77).
We start with Eβ1 (z) and bn (z). Combining (C.42) and (C.54) yields
Eβ1 (z) = −zEmn (z) = −zm(z) + εn (z).

(C.78)

By the Cauchy-Schwarz inequality, (A.20), (C.46) and (A.21), we obtain
sup |bn (z) − Eβ1 (z)| = sup |bn (z)| E |β1 (z)γ1 (z)| ≤ sup K

{z∈Cn }

{z∈Cn }

{z∈Cn }

q
q
E |β1 (z)|2 E |γ1 (z)|2 = o(1).

Therefore, by (C.78),
bn (z) = −zEmn (z) + εn (z) = −zm(z) + εn (z).

(C.79)



p×p
−1
Next we relate D−1
, let diag(A) = diag a11 , a22 , · · · , a pp .
1 (z) to D . For any matrix A = ai j ∈ C

51

By (A.7), the Cauchy-Schwarz inequality, (C.46) and (A.20), we have
p
p
X

 2 X  −1  2
E D−1
E D (z) ii
1 (z) ii −

sup
{z∈Cn }

i=1

i=1



= sup E tr diag
{z∈Cn }

D−1
1 (z)



−1
+ D−1 (z) · D−1
1 (z) − D (z)

 −1
−1
−1
≤ sup E β1 (z)rT1 D−1
1 (z) · diag D1 (z) + D (z) · D1 (z)r1
{z∈Cn }

≤K sup

{z∈Cn }

≤K sup

{z∈Cn }

q

 −1
−1
2
−1
E|β1 (z)|2 · EkD−1
1 (z) · diag D1 (z) + D (z) · D1 (z)k

r

E D−1
1 (z)

6

+

q

8
−1
4
EkD−1
1 (z)k · EkD (z)k = O(1).

Therefore,
1 X  −1  2 1 X  −1  2
E D1 (z) ii =
E D (z) ii + εn (z).
n i=1
n i=1
p

p

Similarly, we can show that
 1



1
−1
E tr D−1
1 (z) = E tr D (z) + εn (z),
n
n

 1


1
−2
E tr D−2
1 (z) = E tr D (z) + εn (z).
n
n

(C.80)
(C.81)

Thus, we only need to focus on D−1 (z).
Pp 
 2
Now we analyze n−1 i=1 E D−1 (z) ii . By the Cauchy-schwarz inequality, we get
p

1 X  −1  2
 2
E D (z) ii − yn E D−1 (z) 11
{z∈Cn } n i=1


  

 
≤yn · sup E D−1 (z) 11 − E D−1 (z) 11 · D−1 (z) 11 + E D−1 (z) 11

sup

{z∈Cn }

≤yn · sup

{z∈Cn }

≤K · sup

{z∈Cn }

r

E

v
t

E

√
≤K n · sup

{z∈Cn }

√
≤K n · sup

{z∈Cn }

eT1



D−1 (z)

n
X
i=1

q

−

ED−1 (z)

 2
e1 ·

r


 2
E eT1 D−1 (z) + ED−1 (z) e1

 2 p
EkD−1 (z)k2
eT1 (Ei − Ei−1 ) D−1 (z) − D−1
i (z) e1 ·

T −1
E β1 (z)rT1 D−1
1 (z)e1 e1 D1 (z)r1

r


E 1 + n1+α 1

e
S(1)
{λmin
<ηl

or

en
λSmax
>ηr }

2

2

T −1
· rT1 D−1
1 (u)e1 e1 D1 (u)r1

r 

√
2
T −1
−1
4
2(1+α) 1
≤K n · sup
E rT1 D−1
1 (z)e1 e1 D1 (z)r1 + kD1 (z)k · n

e
S(1)
{λmin
<ηl

{z∈Cn }

2

or

en
λSmax
>ηr }



= o(1),

where the fourth inequality uses (A.7) and (C.46), the fifth inequality follows from (A.19), and the last step

52

uses (A.9), (A.20), (A.19) and (A.17). It follows from (C.54) that
1 X  −1  2
E D (z) ii − yn m2 (z) = o(1).
sup
{z∈Cn } n i=1
p

(C.82)

As to n−1 E tr D−1 (z), we have from (C.54) that
sup n−1 E tr D−1 (z) − yn m(z) = o(1).

(C.83)

{z∈Cn }

Finally, we turn to n−1 E tr D−2 (z). Similarly to equation (4.13) in Bai and Silverstein (2004), rewrite D−1 (z)
as
D−1 (z) = −

I − bn (z)A(z) − B(z) − C(z)
,
z − bn (z)

where
A(z) =

n
X
i=1


ri rTi − n−1 I D−1
i (z),

B(z) =

n
X
i=1

X

1
−1
C(z) = bn (z)
D−1
i (z) − D (z) .
n
i=1
n


βi (z) − bn (z) ri rTi D−1
i (z), and

Thus
tr D−2 (z) = −




tr D−1 (z) − bn (z) tr A(z)D−1 (z) − tr B(z)D−1 (z) − tr C(z)D−1 (z)
.
z − bn (z)

(C.84)

Similarly to (C.52) and (C.53), we can show that


sup E tr B(z)D−1 (z) = O n1/2 and

{z∈Cn }



sup E tr C(z)D−1 (z) = O 1 .

{z∈Cn }

Furthermore, by (C.55) and (C.79),
sup
{n; z∈Cn }

1
= O(1).
z − bn (z)

(C.85)

Plugging the above bounds into (C.84) yields

n−1 E tr D−1 − n−1 bn (z)E tr A(z)D−1 (z)
1
−2
= o(1).
sup E tr D (z) +
z − bn (z)
{z∈Cn } n
We next rewrite


tr A(z)D−1 (z) = An,1 (z) + An,2 (z) + An,3 (z),

53

(C.86)

(C.87)

where
An,1 (z) = tr

n
X
i=1

An,3 (z) =

1
tr
n



−1
−1
,
ri rTi D−1
(z)
D
(z)
−
D
(z)
i
i

n
X
i=1

An,2 (z) = tr

i=1





n 
X


−1 −2
,
ri rTi D−2
(z)
−
n
D
(z)
i
i

−1
−1
D−1
i (z) Di (z) − D (z) .

For An,2 (z), by (A.11) and (A.20), we have
2
−1 2 
sup |n−1 EAn,2 (z)| = sup Eτ1 (z) ≤ K sup n−1 δ2p EkD−1
1 (z)k = O n δ p .

{z∈Cn }

{z∈Cn }

(C.88)

{z∈Cn }

Moreover, applying (A.7), the Cauchy-Schwarz inequality, (C.46) and (A.20) yields



−1
−1
sup n−1 EAn,3 (z) ≤ sup n−1 · E tr D−1
1 (z) D (z) − D1 (z)

{z∈Cn }

{z∈Cn }

= sup n−1 · E β1 (z)rT1 D−3
1 (z)r1
{z∈Cn }

≤K sup n−1 ·
{z∈Cn }

q

(C.89)

q
−1 
6
.
E|β1 (z)|2 EkD−1
1 (z)k = O n

Now we consider An,1 (z). We have from (A.7) and (A.43) that
T −1
n−1 EAn,1 (z) = − E β1 (z)rT1 D−2
1 (z)r1 · r1 D1 (z)r1






T −1
T −2
T −1
= − bn (z)E rT1 D−2
1 (z)r1 · r1 D1 (z)r1 + bn (z)E β1 (z)γ1 (z) · r1 D1 (z)r1 · r1 D1 (z)r1



−1
−2
T −1
−1
= − n−2 bn (z)E tr D−2
1 (z) · tr D1 (z) − bn (z)E τ1 (z) · r1 D1 (z)r1 − n bn (z)E ς1 (z) · tr D1 (z)


T −1
+ bn (z)E β1 (z)γ1 (z) · rT1 D−2
1 (z)r1 · r1 D1 (z)r1

−1
=: − n−2 bn (z)E tr D−2
1 (z) · tr D1 (z) + An,11 (z) + An,12 (z) + An,13 (z).

Applying (A.20), the Cauchy-Schwarz inequality and (C.38) yields
sup |An,11 (z)| ≤ K sup

{z∈Cn }

{z∈Cn }

q
−1/2
2
E|τ1 (z)|2 · EkD−1
).
1 (z)k = O(n

Similarly, we have sup{z∈Cn } |An,12 (z)| = O(n−1/2 ). As to An,13 (z), by (A.20), the Cauchy-Schwarz inequality, (C.46) and (A.21), we get
sup |An,13 (z)| ≤ K sup E|β1 (z)|2

{z∈Cn }

{z∈Cn }

1/2

E|γ1 (z)|4

54

1/4

12
EkD−1
1 (z)k

1/4

= o(n−1/4 ).

Finally, we have from (A.20), the Cauchy-Schwarz inequality and (C.61) that




−1
−1
−2
−1
−1
sup − n−2 bn (z)E tr D−2
1 (z) · tr D1 (z) + bn (z)E n tr D1 (z) · E n tr D1 (z)

{z∈Cn }



−1
≤K sup n−2 · Cov tr D−2
1 (z), tr D1 (z)
{z∈Cn }

≤K sup n−2 ·
{z∈Cn }

≤K sup n−1 ·
{z∈Cn }

q
2
2
−1
−1
−2
E tr D−2
1 (z) − E tr D1 (z) · E tr D1 (z) − E tr D1 (z)

q
−1 
4
.
EkD−1
1 (z)k = O n

Combining the estimates above with (C.80) and (C.81), we obtain


sup n−1 EAn,1 (z) + bn (z)E n−1 tr D−2 (z) · E n−1 tr D−1 (z) = o(1).

(C.90)

{z∈Cn }

Plugging (C.88), (C.89) and (C.90) into (C.87) yields



n−1 E tr A(z)D−1 (z) = −bn (z)E n−1 tr D−2 (z) · E n−1 tr D−1 (z) + εn (z).

(C.91)

Recall that zm(z) = −(1 + m(z))−1 . Plugging (C.91) into (C.86), and further using (C.79) and (C.83) yield



E n−1 tr D−1 (z)
b2n (z)E n−1 tr D−2 (z) · E n−1 tr D−1 (z)
n E tr D (z) = −
−
+ εn (z)
z − bn (z)
z − bn (z)


y

2 2
−1
−2
=
2 1 + z m (z)E n tr D (z) + εn (z),
2
z 1 + m(z)
−1

−2



where the second inequality also uses (A.20) and (C.85). Solving for n−1 E tr D−2 (z) and using (C.57) yield


n−1 E tr D−2 (z) =

y
z2 1 + m(z)

2

!

1−

ym2 (z)

1 + m(z)

2

!−1

+ εn (z).

(C.92)

Combining (C.77) – (C.79) with (C.54), (C.82), (C.83) and (C.92), we obtain
lim sup nRn (z) − 2

n→∞ {z∈C }
n

ym2 (z)
1 + m(z)

3

!

ym2 (z)

+

1 + m(z)

3

!

1−

ym2 (z)
1 + m(z)

2

!−1

= 0.

Moreover, by the definition of m(z) and (C.57), one can show that the limiting function of nRn (z) is continuous

on C. We thus complete the proof of Proposition A.11.

Appendix D

Two Lemmas

n
is a martingale differLemma D.1. [Theorem 35.12 in Billingsley (1995)] Suppose that for each n, {Yni }ℓi=1
ence sequence having second moments with respect to an increasing σ-field sequence {Fn,i }. If the following
assumptions hold:
Pn
 P
(i) ℓi=1
E Yni2 |Fn,i−1 −→ σ2 ; and

55

(ii) for each ε > 0,
Then we have

Pℓn

i=1


E |Yni |2 I{|Yni |≥ε} −→ 0, as n −→ ∞.
ℓn
X
i=1


D
Yni −→ N 0, σ2 .

Lemma D.2. [Theorem 12.3 of Billingsley (1968)] The set, { Xn (t)| t ∈ [0, 1], n ∈ N}, is tight if it satisfies the
following two conditions:
(i) the set { Xn (0)| n ∈ N} is tight;
(ii) there exist constants γ ≥ 0, α > 1 and a non-decreasing continuous function h(·) on [0, 1] such that
P (|Xn (t2 ) − Xn (t1 )| ≥ λ) ≤

1
|h(t2 ) − h(t1 )|α ,
λγ

for all t1 , t2 ∈ [0, 1], n and all λ > 0.

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