Convergence rates of spectral distributions of large sample covariance matrices Z. D. BAI
Baiqi M IAO y
Jianfeng YAO SAMOS-MATISSE
Dept. of Stat. & Appl. Prob.
Dept. of Stat. & Fin.
Natl. U. of Singapore
U. of Sci. & Tech. of China
Universit´e Paris I
[email protected]
[email protected]
[email protected]
March 17, 2000
Abstract In this paper, we improve known results on the convergence rates of spectral distributions of large dimensional sample covariance matrices of size p n. Depending on the limiting value y of the ratio p=n and by using the tool of Stieltjes transforms, we first prove that the expected 1 spectral distribution converges to the limiting Marˇcenko-Pastur distribution at a rate of O(n, 2 ) 1 , for y 2 = f0; 1g, and of O (n 4 ) for y = 1, under the assumption that the entries have a finite 8-th order moment. Furthermore, the rates for both the convergence in probability and the almost sure convergence are investigated.
AMS 1991 subject classification: 60F99. Keywords and phrases: Convergence rate, random matrix, spectral distribution, Manˇchenko-Pastur distribution
Supported in part by the NUS...TO BE CLARIFIED y Supported in part by the National Foundation of Natural Science of China.
1
1 Introduction The spectral analysis of large dimensional random matrices has been actively developed in the last decades since the initial contributions of Wigner (1955, 1958), see the review by Bai (1999) and the references therein. Various limiting distributions were discovered including the Wigner semicircular law (Wigner, 1955), the Marˇcenko-Pastur law (Marˇcenko and Pastur, 1967) and the circular law (Bai,1997). Let A be an n n symmetric matrix, and 1 n be the eigenvalues of A. The spectral distribution F A of A is defined as
F A (x) = n1 number of elements in fk : k xg : Let Xp = (xij )pn be a p n observation matrix whose entries are mutually independent and have a common mean zero and variance 1. The entries of Xp may depend on n but we suppress the index n for simplicity. In this paper, we consider the sample covariance matrix Sp = n,1Xp X0p. Assume that the ratio p=y of sizes tends to a positive limit y as n ! 1. Under suitable moment conditions on the entries xij ’s, it is known that the empirical spectral distribution (ESD) Fp := F S converges to the following Marˇcenko-Pastur distribution Fy with index y with density 8 1 p(x , a)(b , x) ; if a < x < b < 2xy 0 Fy (x) = : 0; otherwise, p p where a = (1 , y )2 , b = (1 + y )2 . p
An important question arose here is the problem of the convergence rates. However, no significant progress had been made before the introduction of a novel and powerful tool, namely the Stieltjes transforms, by Bai (1993a,1993b). Using this methodology, Bai (1993b) proved that the expected ESD, E Fp converges to Fy at a rate of O(n,1=4) and O(n,5=48) according to y 6= 1 or y = 1, respectively. In a further work by Bai et al. (1997), these rates are also established for the convergence in probability of the ESD Fp itself. In this work, we prove the following theorems which give a significant improvement of these rates. The following conditions will be used.
E xij
(C.2)
sup E jxij j8 < 1 ,
(C.3) (C.2’)
= 0,
E x2ij
= 1, 1 i p; 1 j n,
(C.1)
i;j;n
X ij
E x8ij I(jxij j"pn)
= o(n2) , for any " > 0.
sup E jxij jk < 1 , for all integer k 1. i;j;n
1 Throughout the text, we use the notation Zn = Op(an ) if the sequence (a, n Zn ) is tight, and Zn op (an ) when a,n 1 Zn tends to 0 in probability. Let be kf k = supx jf (x)j.
8 , > < O(n ); kEFp , Fy k = > : O(n, );
Theorem 1.1 Assume that the conditions C.1-2-3 are satisfied. Then
2
1 2
if
0 : Op(n, ); 2 5
if
0 : o(n,
);
if
y 6= 1;
);
if
y = 1:
2 5+ 2 9+
> 0 and almost surely,
It is worth noticing that the convergence rates given above for the case 0 < y < 1 also apply to the case y > 1, since the last case can be reduced to the first case by interchanging the roles of row and column sizes p and n. The proofs of these main results will be given in Section 4. To simplify their presentation, we first establish several intermediate results in Section 3 after the introduction of some necessary notations and preliminary consequences in Section 2.
2
Definitions and easy consequences
A
A
Throughout the paper, the transpose of a possibly complex matrix is denoted by T , and its conjugate by . For each fixed p; n and k = 1; : : :; p, let us denote by k = (xk1; : : :; xkn )T the k-th row of p arranged as a column vector, p (k) be the (p , 1) n sub-matrix obtained from p by deleting its k-th row. Let us define
X
A
X
k := n1 Xp(k)xk ;
Sk
:= n1 Xp(k)XTp (k)
Bk
x
X
:= n1 XTp (k)Dk Xp (k);
B := n1 XTp DXp Dk := (Sk , zIp, ), ; D := (S , zIp), ; ,k := Dk Dk ; k := Dk Sk Dk : Here Im is the m-dimensional identity matrix and z a complex number with a positive imaginary part. Following Bai (1993b), the Stieltjes transform of the spectral distribution Fp of the sample covariance matrix Sp is defined for z = u + iv with v > 0, by Z1 1 1
mp (z ) =
1
,1 x , z
1
dFp (x);
and it is well-known that
mp(z) = 1p tr(S , z Ip ),1 : Similarly, the Stieltjes transform of the spectral distribution Fp
k of the sub-matrix S satisfies k
( )
mpk
( )
(z ) =
Z1
1 dF (k) (x) = 1 tr(S , z I ),1 : p,1 p p,1 k ,1 x , z 3
(2.1)
Lastly, the Stieltjes transform of the (limiting) Marˇcenko-Pastur distribution Fy is
m(z) := =
Z1
1 dF (x) y x , 8,1 z >> , y + z , 1 , p(1 , y , z)2 , 4y ; < 2yz
0 < y < 1;
p >> : , z , z , 4z ; 2z
(2.2)
y = 1:
2
p
Here the square root z is the one with a positive imaginary part. Bai (1993b) also provided the following bounds for m(z ) which will play a key role in next derivations :
8 1 + 3py >> p < y(1 , y) ; m(z ) > >: p2 ; v
x
0 < y < 1; (2.3)
y = 1:
y
Lemma 2.1 Let = (x1 ; : : : ; xn )T and = (y1; :::; yn)T be independent real random vectors with independent elements. Suppose that for all 1 j n, E xj = Eyj = 0, E jxj j2 = E jyj j2 = 1, E jxj j4 L < 1, and that is an n n complex symmetric matrix. Let k = maxj n ( E jxj jk ; E jyj jk ). Then
A
x Ayj = tr(AA ) ; ) + jtrAj ; E jxT Axj (L , 1)tr(AA ); E jxT Ax , trAj (L , 1)(trAA )k + (Ltr(AA ))k E jxT Ax , trAj k dk k tr(AA
(i). E j T (ii). (iii). (iv).
2
2
2
2
2
4
stant dk depending on k only.
for k
2 and some positive con-
Lemma 2.1 can be proved in an elementary way and is stated in Bai et al. (1997). Lemma 2.2 Let G1 and G2 be probability distribution functions and z each positive integer m,
= u + iv , v > 0.
2 Z 1 1 d(G (x) , G (x)) v m kG , G k: m j x , z j ,1 1
2
4
1
2
Then for
Proof.
Let be G
:= G1 , G2 . We have, by integration by parts,
Z 1 1 dG ,1 j x ,Z z jm 1 1 G (x)d jx , zjm = , ,1 Z Re z 1 Z1 1 = , G (x)d jx , z jm + G (x)d , jx , zjm ,1 Re z (Z Re z Z 1 1 ) 1 ( )
( )
kGk
=
( )
(
,1
d jx , zjm +
Re z kG k jx , zjm ,1 1
( )
Re(z)
d , jx , z jm
1 !) , jx , zjm = kGk v2m : Re z 1
+
( )
We will need the following auxiliary variables.
n X 1 "k = , n (x2kj , 1) + n1 (xk0 Bk xk , EtrB); j =1 n X
"k = , n1 (x2kj , 1) + n1 (xk0 Bk xk , trBk ); j =1 1 "ek = n (trBk , E trBk ) = nz (trDk , E trDk ); k = n1 E (trBk , trB) = nz E (trDk , trD) , n1 ; n X 1 k = , (x2kj , 1) + z , 1 + 1 xk 0Bk xk ;
n j=1
n
k = z , 1 + n1 trBk ; = z , 1 + n1 trB
We summarize below some inequalities which will be used in the derivations. Let = k E Fp F k and M := supi;j;n E jxij j4. For fixed (n; p) and 1 k p, we define the -algebra F (k) ( i : i = 1; : : : ; p ; i 6= k) and Fk = ( i : i = 1; : : : ; p ; i > k). Notice that Fk F (k).
x
x
,
=
(i). (Lemma 3.3 of Bai (1993a)) :
j(p , 1)Fpk (x) , pFp(x)j 1: ( )
(2.4)
Z 1 k (x)] d [ pF ( x ) , ( p , 1) F p v, : p jtrD , trDk j = x , z ,1
(ii). ((3.11) of Bai (1993a)) :
( )
(iii). ((4.7) of Bai (1993a)) :
mp(z) = , p1 5
p X
1:
k=1 k
1
(2.5)
(2.6)
(iv). (Lemma 2.2 of Bai et al. (1997)) :
E jmp(z ) , E (mp (z ))j2 p,1 v ,2 :
(2.7)
(v). (from j kj Im( k) = v (1 + n,1 tr k )) :
j kj, (1 + n, trk ) v, :
(2.8)
j kj Im( k) = v(1 + n1 Tk Dk Dk k ):
(2.9)
j1 + n1 Tk Dk k j 1 + n1 Tk Dk Dk k :
(2.10)
1
1
1
(vi).
(vii). 2
S
Let kj , j = 1; 2; : : : ; p , 1, be the eigenvalues of k which can be decomposed in a diagonal form on a basis of orthonormal and real eigenvectors. Let be a complex matrix having the ` `0 for some integers `; `0 and factors ; equal to one of the matrices product form = f k ; k ; k ; k g. An important feature that we will frequently use in the sequel is that such a matrix can be decomposed into a diagonal form on the same basis of the eigenvectors of k . Moreover, the eigenvalues of can be straightforwardly expressed in term of the kj ’s. In particular, we have the following
L MN D D B B L L
L M N
S
Lemma 2.3 Assume that jz j T where T
tr(,k )`
1. Then for all integers ` 1
1 `,
1
tr,k ; v2 T `,1 ` tr(k ) v 2 trk :
(2.11) (2.12)
Proof. (i) The inequality (2.11) follows from
tr(,k )` =
p, X 1
j =1
1
, jkj , zj ` v 2
p,1 1 `,1) X 2 j =1 jkj , z j
2(
= v ,2(`,1) tr,k :
(ii) For the inequality (2.12), we have
p,1 ` X ` tr(k ) = j ,kj zj2` : j =1 kj The conclusion follows from that The function '() := ,1 j , z j2 defined on (0; 1) is convex and has an unique minimum of value ' satisfying 2 2 p ' = 2( u2 + v 2) , u = 2 jzjv+ u vT :
6
Lemma 2.4 For the Marˇcenko-Pastur distribution Fy , we have
Zb
8 p v, ; 1 dF (x) < ,y y : jzj, v, = ; jx , zj y 1 )
(1
2
a
0 , () is decreasing and then () () ; and for < , v , () is increasing and then () ( , v ). Hence () reaches its maximum only for some 2 (max( , v; 0); ). Now suppose that 2 ( , v; ), it follows from (2.14) that
Z +v + v , u p 1=4 () 22yy udu n u+a hp p = 2(y 3=4),1 ( + v + a) ( + v , )
r !#
r
,pa arctan +a v , arctan a
Notice that ,
pa arctan px is convex , we get a
)
h i , 13 ( + v)3=2 , 3=2 :
!
r + v a p + v , p ; arctan , arctan a a a +v +a p p and by setting = + v , , we have p () 23=4 f(a + + v )( , a + a + v ) , (( + + 13 2 )g y p 2 2 3 2 = y3=4 + 3 : p1
Let c2
r
= 21+pyy . Since + v c,2 a, we have
p
p pa +c pv ; p ( + v + ) p 1 p (pa +2cpv)v : ( + v + ) 2
3
Hence
p
7 2(1 + y ) p 1 2 () 23=4 3(pa7+c pv ) v 2 = 3y v + (1 , py ) v : y Case y
=1:
Here a = 0 and
() =
Z v + v , ur4 , u 2" ur du # r 4, 1 Z v 4,u +
d() = d 2
+
u ,
du :
But (4 , u)=u is decreasing for u > 0, thus () is decreasing for 0. Hence
() (0) =
Z v v , ur4 , u 0
2
Combining these two cases proves the lemma. 8
2 3=2 u du v :
(2.15)
3
Intermediate lemmas
In this section, we establish some more technical lemmas. Let `
= supi;j;n fE jxij j` g.
Lemma 3.1 For each ` > 1=2 with 4` < 1, there exist positive constants c` independent of n and v , such that for all n; v satisfying nv T , we have
E j"k j2` F (k)
1 ` , ` c`n 1 + trk n
and
(3.1)
!
("k )2` (k) ,` ,` j j` F c` n v :
E
(3.2)
k
0 ` 1 n X E j"k j ` jF k = EB A @ , n1 (xkj , 1) + n1 (x0k Bk xk , trBk ) F k C j 9 8 ` > >< X n = k ` , , ` 0 ` 2 n > E (xkj , 1) + E jxkBk xk , trBk j F > ; : j
Proof.
We have
2
2
( )
2
( )
=1
2
2
1
2
2
2
=1
:= A + B :
For the first term A, by the Burkholder inequality, we get
` n X E (xkj , 1) j 2n 3` 2n 3 X X c` E 4 (xkj , 1) 5 c`n`, E 4 (xkj , 1) `5 c` ` n` : 2
2
=1
2
2
1
j =1
2
2
j =1
For the second term B , we first notice that
,
4
tr Bk Bk = trBk + ztrk ;
and
1 jtrB j = y + z trD 1 + T 2 : n k n k nv
Hence
1 T k Bk 2 + n trk T 1 + n trk :
1 tr ,B
n Therefore by Lemma 2.1,
E jx0kBk xk , trBk j2` F (k)
c`( ` 4
` ,` ,` k Bk ) c` T n
+ M ` )(trB
9
` 1 1 + n trk :
( )
Combining the bounds for A and B proves the first conclusion. The second conclusion immediately follows by taking into account the inequality (2.8). Lemma 3.2 If n,1=2 each k n, (i).
v < 1, then there are positive constants C ; C 1
2
such that for large n and
j E tr(Dk Dk )j C p v+ v : 1
(ii). E j"k j2 Proof.
(i).
2
C n1 1 + jzj v+ v : Recall that = k E Fp , Fy k. By Lemma 2.2, 2
2
2
2 Z 1 1 d( E Fp (x) , Fy (x)) v : ,1 jx , z j 2
2
Application of Lemmas 2.1 and (2.4) yields that
Z1 1 k d[ E Fp (x)] j E tr(Dk Dk )j = (p , 1) j x , z j ,1 Z 1 1 k jx , zj d[(p , 1) E Fp (x) , p EFp (x)] ,1 Z Z 1 1 1 1 d[ E Fp(x) , Fy (x)] + p dFy (x) +p j x , z j j x , z j ,1 ,1 Z 1 1 2 2 v + p v + p jx , zj dFy (x) : ,1 ( )
2
( )
2
2
2
2
2
2
p
By Lemma 2.4, the last term is bounded by C3pv ,1 or C3p(jz j v),1 according to 0 < y < 1 or y = 1. Taking into account thep condition v pnp 1, we have for large n, pv 2C3 for the first case and for the second one, since v v jz j, p v 2C3. The conclusion (i) follows in both cases. (ii). The conclusion follows from (i), (3.1) and the fact
trBk Bk = tr(Ip,1 + zDk )(I + z Dk ) 2(p + jz j2trDk Dk ) : p Let us define vy = v for 0 < y < 1 and vy = v for y = 1. p p Lemma 3.3 Assume jz j T with T 2, and nv 6 2T (M + 2). Then for large n and a positive constants C1 , p X k=1
E (j kj,1 ) C1n( + vy )v ,1 :
Proof. First notice that from the definition of "k , we have ( k ),1
= k,1 (1 + k,1 "k ). By (2.5),
j k , j = n1 j , 1 + z(trDk , trD)j n1 (1 + jvzj ) 2nvT :
10
(3.3)
Taking account of (2.6) and (3.2), we obtain
p X
E (j k j,1)
k X X X p p p p X 1 1 1 1 1 1 , E , + E ( , ) + E ( , ) + E k j j j j k k k k k k k k p p p X j k , j X j"k j X j"k j =1
1
2
=1
=1
=1
=1
2
j kj + k E j k jj kj + p E jmp(z)j p p p X ( E (j"k j jF k )) = X E (j"k j jF k ) 4T X E (j kj, ) + E + E + p E jmp (z )j nv j j j v j k k k k k p p X X 4 T 2 MT = , = , , ( nv + (2MT ) n v ) E (j k j ) + nv E (j kj, ) + p E jmp(z )j k k k=1
E
j jj kj + k
E
2
2
=1
=1 2
1
2
( )
1 2
2
2
=1
2
=1
1 2
1 2
( )
=1
1
1
1
2
2
=1
p X
(2T (2 + M )v, n, + (2MT ) = n, = v, ) 2
1
1 2
1 2
1
+pj E (mp (z )) , m(z )j + pjm(z )j
2[2T (2 + M )]
k=1
=1
E (j k j,1) + p E jmp (z ) , E (mp(z ))j
p = v ,1 n,1=2 X E (j j,1 ) + ppv ,1 + 2pv ,1 + pjm(z )j: k k=1
1 2
Since 2[2T (2 + M )]1=2v ,1 n,1=2
p X k=1
< 1=3, we find
E (j kj,1 )
Notice that for large n, 12 yn for m(z ) given in Eq. (2.3).
3 ,ppv ,1 + 2pv ,1 + pjm(z )j : 2
p yn. The conclusion follows by taking into account the bounds 3 2
D
D
D
D P
Lemma 3.4 Let zk = E (tr jFk,1) , E (tr jFk ). Then tr , E tr = pk=1 zk and (zk ) is a martingale difference with respect to (Fk ), k = p; p , 1; : : : ; 0. Moreover, we have the following formula for zk
zk = f E (ak jFk,1) , E ( ak jFk )g , E ( bk jFk,1) ; with
(1 + T Dk 2 k ) Tk Dk 2 k , n1 tr[(I + zDk )Dk ] " k k ak = ; bk = : k k k Proof.
D
D E [(trD , trDk )jFk, ] , E [(trD , trDk )jFk ]:
Since E (tr k jFk,1) = E (tr k jFk ), we have
zk =
1
11
(3.4)
On the other hand, 1 T 2 trD , trDk = , 1 + n k Dk k k 1 + n1 tr[(I + z Dk )Dk ] "k (1 + Tk Dk 2 k ) Tk Dk 2 k , n1 tr[(I + z Dk )Dk ] = , + , k k k k 1 + 1 tr[(I + z D )D ] = , n k k + ak , bk : k
The conclusion follows from
k
! Fk, =
1 + n1 tr[(I + z Dk )Dk ]
E and
1
! Fk ;
1 + n1 tr[(I + z Dk )Dk ]
E
k
Tk Dk 2 k F (k) = n1 tr[(I + zDk )Dk ] : Lemma 3.5 For each ` > 1=2 with 4` < 1, there exist positive constants c` and L0 independent of n and v, such that for all n; v satisfying L0 n,1=2 v < 1, E jmp (z ) , Emp(z )j2` c` n,2` v ,4` ( + vy )` : E
Proof. In the proof of this lemma, c` and c`;0 will be used to denote universal positive constants which may depend on the moments up to order ` of underlying variables and may represent different values at different appearance, even in one expression. Recall that we have
p X 1 mp (z ) , Emp (z ) = p [trD , E trD] = zk ; k=1
where the (zk ) are defined in Lemma 3.4. We have
E
j zk j
` F
2
k
` = E [ E ( ak jFk, ) , E ( ak jFk )] , E ( bk jFk, ) Fk o n 2 `, E [ E (ak jFk, ) , E (ak jFk )] ` + [ E ( bk jFk, )] ` Fk o n 2 `, E [ E (ak jFk, )] ` + [ E (bk jFk, )] ` Fk o n 2 `, E (ak ) ` Fk + E (bk ) ` Fk : 2
1
2
1
2
1
2
1
1
2
1 1
2
1
1
2
2
2
2
Note that by (2.9) and (2.10), jak j v ,1 j"k = k j. Hence by Lemma 3.1
E
jak j ` F k v1` E 2
( )
2
" ` ! k F k c`; n,` v, `j kj,` : k 2
On the other hand, by Lemma 2.1 and assuming ` 1,
E
( )
0
3
jbk j ` F k c`; (n k), `( ` + M `) tr(I + zDk )(I + zDk )Dk Dk ` : 2
( )
0
2
4
12
Since from (2.8) and (2.12), it holds that
j kj, tr(I + zDk )(I + zDk )Dk Dk j kj, trk nTv, ; 1
1
we obtain
E Therefore for all ` 1,
E
n
o
3
h
i
h
i
jbk j ` Fk c`; n,` v, ` E j kj,` Fk : 2
0
jzk j ` Fk 2
3
c`; n,` v, ` E j kj,` Fk c`; n,` v, ` E j kj, Fk 3
0
4 +1
0
Applying Lemma 3.3 gives for ` 1
n X
Case ` = 1 :
2
k=1
1
(3.5)
E jzk j2` c`;0 n,`+1 ( + vy )v ,4` :
(3.6)
Since that fzk g is a martingale difference sequence, the above inequality yields
E jmp (z ) ,
p X , 2 E mp (z )j = n E jzk j2 c1;0n,2 ( + vy )v ,4 : k=1 2
(3.7)
The lemma is proved in this case. Case 12 < ` < 1 : By applying the Burkholder inequality for martingales (see Burkholder (1973)) and using the the concavity of the function x` , we find
E jmp (z ) , E mp (z )j2` ` p , 2` 2 c`p E jzk j c`n,2` k=1
!
X
" Xp E(
k=1
#`
jzk j ) c`n, ` ( + vy )v, 2
2
4
` ;
where the last step follows from the previous case ` = 1. The lemma is then proved in this case. Case ` > 1 : We proceed by induction in this general case. First, by another Burkholder inequality for martingales, we have
8p 2 then nv 2 > 2 and that
, , , j j , j k j j , ( k), j = jtrDpj ,jj trDj k j pv1 1
1
1
1
2
(3.10)
min(j j,1; j kj,1 )
k (this comes from (2.5) and j k j,1 v ,1 min(j j,1 ; j k j,1 )). This yields j kj,1 j j,1 + p,1v,2j kj,1 2j j,1 and
X X X p p p jp , j ( k), + j( k), , , j ( k), + v, j j, k k k X p p p X X , , , , 2 ( k ) 2 (( k ) , k ) + 2 k k k k p X j"k j 1
1
1
=1
1
=1
1
2
2 k=1 j k jj k j
=1
+ 2pjmp(z )j:
Therefore, by applying Lemma 3.1 and if we choose L0 have
p X k=1
E j k j,`
c`;
0
1
1
1
=1
=1
2
=1
1
2
1
> (2c`;0)1=` so that c`;0n,` v ,2` < 1=2, we
! p j ` X j " v,` E k + p E jmp (z)j` 2
j k j ` 2
! p X , ` , 2` , ` ` c`;0 n v E j k j + p E jmp (z )j k=1 2c`;0 p E jmp (z )j` : k=1
From the above inequality and (3.10), we get
I2 c`n,2` v,3` hEjmp (z )j` i c`n,2` v,3` Ejmp (z) , Emp (z)j` + j Emp (z) , m(z)j` + jm(z)j` h i c`n,2` v,3` Ejmp (z) , Emp (z)j` + ( + vy )`v,` :
(3.11)
It can be readily checked that the ratio of the upper bound for I1 , Eq. (3.9), over the second term from the last inequality is bounded by a constant (for both cases 0 < y < 1 and y = 1), namely
1 `,1 ( + vy )n,3`+1 v ,4` = 1; n,2`v ,3` ( + vy )` v,` n( + vy ) 14
because nv 2
1 and v 1. Therefore by (3.8) and (3.11), it follows that E jmp (z ) , E mp (z )j ` c`n, ` v, ` Ejmp (z) , E mp (z)j` + c`n, ` ( + vy )`v, `: 2
2
3
2
4
(3.12)
Finally by the induction hypothesis, we find
E jmp (z ) , E mp (z )j2` c`n,2` v,3` n,` v,2`( + vy )`=2 + c`n,2` ( + vy )`v,4` `=2 1 ,2` ` ,4` = n2v 2( + vy ) + 1 c`n ( + vy ) v 2c`n,2` ( + vy )`v,4` :
h
"
i
#
The proof of Lemma 3.5 is complete. Remark 3.1. such that
Application of Lemma 3.5 to the case `
= 1 gives that there is some constant c1 > 0
D , E trDj c ( + vy )v, : (3.13) It is also worth noticing that if we substitute D for any Dk with k n, Lemma 3.5 as well as the E jtr
2
4
1
above consequence (3.13) are still valid, with slightly different constants c` ’s.
4
Proofs
Suppose that G is a function of bounded variation. The Stieltjes transform g of G is defined as
g (z ) = where z
Z1
,1
1 dG(x); x,z
= u + iv and v > 0. Our main tool is the following proposition (Bai (1993a)).
4.1 Let G be a distribution function and H be a function of bounded variation satisfying RProposition jG(x) , H (x)j dx < 1. Denote their Stieltjes transforms by g(z) and h(z), respectively. Then "Z A 1 2 Z
kG , H k (1 , )(2 , 1)
Z 1 + sup v
x
jyj2va
,A
jg(z) , h(z)j du + v
#
jG(x) , H (x)j dx
jxj>B
jH (x + y) , H (x)j dy ;
where the constants A > B , and a are restricted by
Z 1
=
1 du > 1 ; 2 juja u + 1 2
and
= (A , B4B)(2 , 1) 2 (0; 1):
15
Denote the Stieltjes transform of Fp and Fy by mp (z ) and m(z ), respectively. By Proposition 4.1 with A = 25, B = 5 and Lemma 2.5, we have, for some constant c > 0,
"Z A
k E Fp , F k c
#
Z 1 j E mp(z) , m(z)j du + v
j E Fp (x) , e(x)j dx+ vy ;
jxj>5
,A
(4.1)
where e(x) = 1 for x > 0 and e(x) = 0 otherwise. In the sequel, for brevity, c will be an universal constant which is not related to the estimation of the order. Since it is already proved in Bai (1993a) that = k E Fp , Fy k = O(n,1=4), will be treated as of order n,1=4 .
4.1
Proof of Theorem 1.1
We will estimate the first two terms on the right hand side of (4.1) with various choices of v , subject to v ' n,1=2 . We begin with the the second term. Let p be the largest eigenvalue of p and recall p that b = (1 + y )2 . Yin, Bai and Krishnaiah [1988] proved that under the conditions C.1-2-3, one 1 can find two sequences (p) and (mp) satisfying p ! 0 and m, p log n ! 0 such that
S
E (p )m
(b + p)m : p
(4.2)
Notice that
1 , Fp (x) If xg ; p
Take B
for x 0 :
(4.3)
= 5, we get for all t > 0
Z1 B
E jFp (x) , Fy (x)j dx
Z1
B
P (p x) dx
Z 1 b + p m
p
B
B
dx = o(n,t ) :
Thus the second term of the equation (3.1) can be neglected. Therefore what remains is to estimate the order of the first term of (4.1). By Eq. (3.14) of Bai [1993b],
mp(z ) =
Z1 0
p 1 dF (x) = 1 trD = , 1 X 1 x,z p p p : k=1 k
Let us define p such that
mp(z) = , z + y , 1 +1 yz E m (z ) + p = , E1 + p : p
Since
1 = 1 1 , "k ; k E k it is easy to see that
!
p p p 2 X X X " 1 1 1 " 1 1 k p = p = ( E )2 p "k , p k : E k k=1 k=1 k=1 k 16
Now
j E pj
1
pj E j
p X 2
1
(j E "k j + j E "k j) 2
k
k=1 p X
4 j E ("k + "~k ) + k j + E1 E "2k , ( E1 )2 E "3k + ( E1 )2 E ( "k ) k
= pj E j2 k"=1 X p p p X X 1 1 1 2 3 j E ("k + "~k ) + k j + E E "k + ( E )2 E "k pj E j2
# " E ( k ) k
k=1
p X + ( E1 )
2
=
k=1
k=1
4
k=1 j E j,2 [I0 + I1 + I2 + I3] :
We will estimate each of Ii ’s to obtain a bound on j E p j (cf. (4.4)). Since that E ("k (2.5), we have
I0 = 1p
p X
p X
k=1
k=1
1 jk j = pn
+ "~k ) = 0, by
j E trDk , E trDj 1=(nv) Cv v:
Here and hereafter, the symbol Cv denotes a positive constant which may be made arbitrarily small p by choosing nv large. From Lemma 3.2, Remark 3.1 and noticing that v vy , we have
1
p X
1
p X
I1 pj E j E j"k j = pj E j ( E j"k j2 + E j"~k j2 + jk j2) k=1 k=1 c( + vy ) Cv ( + vy ) c 1 + v + v 1 y y j E j n + nv2 + n2v4 + n2v2 j E jnv2 j E j ; 1
2
p X
1
p X
I2 = pj E j2 j E "k j pj E j2 ( E j"k j2 + E j"k j4 ): k=1 k=1 Now
3
p p X 1X 27 4 E j"k j ( E j"k j4 + E j"~k j4 + jk j4) =^ c(I21 + I22 + I23):
p k=1
p
k=1
Since
trBk Bk = tr(Ip,1 + z Dk )(I + zDk ) 2(p + jz j2trDk Dk ) ; We have from the proof of Lemma 3.1,
E j"k j4
cn, 1 + n, E(trBk Bk ) cn, 1 + n, E(trDk Dk ) : 2
2
17
2
2
2
2
Now
DD
E (tr( k k ))2
= v ,2 E (Im(tr(Dk )))2 2v,2[v,2 + E (Im(tr(D)))2] = 2v ,4 + 2p2v ,2 E (Im(mp(z )))2 2v,4 + 4p2v,2j E mp(z)j2 + 4p2v,2 E jmp(z) , E mp(z)j2 cp2v,4( + vy )2 + cv,6( + vy );
where the second inequality follows from (2.5) and the last step follows from Lemma 3.5 and j E mp (z )j j E mp(z) , m(z)j + jm(z)j v,1(2 + y vy ) with y := (1 + 3py)=[py(1 , y)] for 0 < y < 1 and y := 2 for y = 1. Thus
I21 c n,2 + n,4 v,4 + n,2v ,4 ( + vy )2 + n,4 v,6 ( + vy ) Cv [vy2 + 2] :
D
D as in Lemma 3.5 and applying (2.4), one can show that for some
Also, considering k instead of L0 such that for all L0 n,1=2 v
< 1, I22 c( + vy )2n,4 v,8 Cv [vy2 + 2]: Since jk j jz j(nv ),1, we have I23 jz j4 (nv ),4, and hence, p X , 1 p E j"k j4 k=1
c(I + I + I ) 21
22
23
c[Cv( + vy ) + Cv ( + vy ) + (nv), ] Cv( + vy ): Consequently, for some constant Cv > 0, + vy ) + Cv (v + ) Cv ( + vy ) ; I cj ( E j nv j E j y j E j 2
2
2
2
2
and
I3 pvj 1E j2
2
4
2
2
2
2
2
2
p X k=1
2
E j"k j4
Cv 2 2 v j E j2 ( + vy ):
Summing up the above results, we obtain
j E pj j E1 j [I + I + I + I ] " # + v C + v + v j E vj v + j E jy + j E j y + vj E j y : 2
0
1
2
3
2
2
2
2
2
(4.4)
On the other hand, by Lemma 2.2 and (4.1), we have
1 = j, E + E [m (z ) , m(z )] + m(z )j j E j + 2 + y vy : p p p j E j v 18
(4.5)
Note that the estimates (4.4) and (4.5) are valid for all L0 n,1=2 v < 1. As proved in Bai (1993b (see Eq. (3.39)-(3.40) there), there is a constant c such that for every v > 0
ZA
,A
j E mp(z) , m(z)j du < cv
provided that supu j E n j v (here and hereafter, supu refers to supjujA ). Thus, if supu j E p j v , in view of (4.1), we can find a positive constant c1 such that
< c1 vy :
(4.6)
Part (i) of Theorem 1.1 :
< y < 1 and vy = v . Write M0 = (1 + 2c1 + y ) and select L > L0 such that when Ln,1=2 v < 1, we have
In this part, 0
Cv,1 > M02[1 + (1 + c1)M0 + (2 + c1 + c21)M02] : The proof will be complete once we have shown that for all large n and Ln,1=2
v < 1,
sup j E p j v:
(4.7)
u
It is proved in Bai (1993b) that (4.7) holds for all large n and c2n,1=4 a constant, and hence < c1 v . Applying these to (4.5), we have
v < 1, where c > 0 is 2
j E j, v + 2=v + y < M : 1
(4.8)
0
This means that for all large n and c2 n,1=4 v < 1, both (4.7) and (4.8) hold. Now letting v decrease to Ln,1=2 , since supu j E p j is continuous in v , one of the following cases must hold: Case 1. supu j E p j < v is true for all Ln,1=2 v < 1; Case 2. There is a v 2 [Ln,1=2; c2n,1=4 ) such that supu j E p j = v and j E j,1 M0 ; Case 3. There is a v 2 [Ln,1=2; c2n,1=4 ) such that supu j E p j < v and j E j,1 = M0 . The theorem then follows if Case 1 is true. Thus to complete the proof of the theorem, it suffices to show that Cases 2 and 3 are impossible. Note that in either Cases 2 or 3, we have < c1v by (4.6). If Case 3 happens, then there exist v0 2 [Ln,1=2; c2n,1=4 ) and u0 , such that j E (z0 )j,1 = M0 , where z0 = u0 + iv0. Then, by (4.5), we have
j E (z )j, 2c + y + v < 2c + y + 1 = M ; 0
1
1
0
1
0
which leads to a contradiction to the equality assumption. If Case 2 happens, then there exist v0 [Ln,1=2 ; c2n,1=4 ) and u0, such that j E p (z0 )j = v0 , where z0 = u0 + iv0 . From (4.4) we have
2
j E p(z )j v Cv M [1 + c M + (1 + c )M ] < v : 0
0
2 0
1
0
2 1
2 0
0
This is also a contradiction to the equality assumption. The proof of Theorem 1.1 is complete for the case 0 < y < 1. 19
Part (ii) of Theorem 1.1 : When y
= 1, Fp (x) and Fy (x) satisfy the following conditions:
Rx
Fp(0) = Fy (0) ;
Z1 0
Rx
xdFp (x) =
Z1 0
xdFy (x) = 1 :
Thus F~p (x) = 0 tdFp (t) and F~y (x) = 0 tdFy (t) are two distributions and F~y (x) satisfies the Lipschitz condition, i.e. there exists a constant L > 0 for any x and y such that
jF~y (x0) , F~y (x)j Ljx0 , xj
(4.9)
Therefore there is a constant c1 such that
1 sup Z
v
x
juj2 v
jF~y (x + u) , F~y (x)j du c v 1
According to the definition of F~p (x) and F~y (x) it follows for any > 0 and every t > 0 that
Z1
Z 1 4+
4+
jEF~p (x) , F~y (x)j dx = o(n,t) E jF~p (x) , F~y (x)j dx
= o(n,t )
Let m ~ p (z ) and m ~ (z ) denote the Stieltjes transform of F~p (x) and F~y (x) respectively , then
m~ p (z ) = 1 + zmp (z );
m~ (z ) = 1 + zm(z )
~ = kF~p (x) , F~y (x)k and The proof of the Theorem 1.1, part (i) can be applied to the estimations of k E jm ~ p (z ) , m~ (z )j . Therefore there is a constant c~ > 0 , when 1=2 v c~n,1=2 it is followed that sup E jzp j < v; u E jzmp (z ) , zm(z )j = E jm ~ p (z ) , m ~ (z )j < v: By (4.1) and Lemma 2.5, there is a constant c2 , such that
Z
jEmp (z) , m(z)j du + c pv Zjuj jEzmp (z) , zm(z)j p v = du + c jzj jZuj p du + c pv v log cv + c pv: v u +v juj p Since v log cv3 < v when v is small enough, it is followed that p < (c + 1) v:
2
25
2
25
25
2
2
2
2
The proof of Theorem 1.1, part (ii) is complete.
.
20
3
2
(4.10) (4.11)
4.2
Proof of Theorem 1.2
By Chebyshev inequality, it suffices to show that
(
2 O (n, 5 ); E jjFp , Fy jj = O(n, 92 );
Case 0 < y
