#### Central Limit Theorem for Linear Eigenvalue Statistics for a Tensor Product Version of Sample Covariance Matrices

Mathematics Subject Classification
Central Limit Theorem for Linear Eigenvalue Statistics for a Tensor Product Version of Sample Covariance Matrices
A. Lytova 0 1
Mn 0 1
m 0 1
k 0 1
0 Institute of Mathematics and Informatics, Opole University , 45 052 Opole , Poland
1 Department of Mathematical and Statistical Sciences, University of Alberta , Edmonton, AB T6G 2G1 , Canada
For k, m, n ∈ N, we consider nk × nk random matrices of the form m
Random matrices; Sample covariance matrices; Central Limit Theorem; Linear eigenvalue statistics
1 Introduction: Problem and Main Result
Y = y(1) ⊗ · · · ⊗ y(k) ∈ (Rn)⊗k ,
where y(1),…, y(k) are i.i.d. copies of a normalized isotropic random vector y =
(y1, . . . , yn) ∈ Rn,
E{y j } = 0, E{yi y j } = δi j n−1, i, j ∈ [n],
where we use the notation j for k-multiindex:
j = { j1, . . . , jk }, j1, . . . , jk ∈ [n].
m m
For every m ∈ N, let {Yα}α=1 be i.i.d. copies of Y , and let {τα}α=1 be a collection of
real numbers. Consider an nk × nk real symmetric random matrix corresponding to a
normalized isotropic random vector y,
We suppose that
Mn = Mn,m,k = Mn,m,k (y) =
m → ∞
and m/nk → c ∈ (0, ∞) as n → ∞.
Note that Mn,m,k can be also written in the form
Mn,m,k = Bn,m,k Tm BnT,m,k ,
Bn,m,k = (Y1 Y2 . . . Ym ), Tm = {ταδαβ }αm,β=1.
Such matrices with Tm ≥ 0 (not necessarily diagonal) are known as sample
covariance matrices. The asymptotic behavior of their spectral statistics is well studied when
all entries of Yα are independent. Much less is known in the case when columns Yα
have dependence in their structure.
The model constructed in (1.3) appeared in the quantum information theory and was
introduced to random matrix theory by Hastings (see [3,14,15]). In [3], it was studied
as a quantum analog of the classical probability problem on the allocation of p balls
among q boxes (a quantum model of data hiding and correlation locking scheme). In
particular, by combinatorial analysis of moments of n−k Tr Mnp, p ∈ N, it was proved
that for the special cases of random vectors y uniformly distributed on the unit sphere
in Cn or having Gaussian components, the expectations of the Normalized Counting
Measures of eigenvalues of the corresponding matrices converge to the Marchenko–
Pastur law [17]. The main goal of the present paper is to extend this result of [3] to
a wider class of matrices Mn,m,k (y) and also to prove the Central Limit Theorem for
linear eigenvalue statistics in the case k = 2.
Let {λl(n)}ln=k 1 be the eigenvalues of Mn counting their multiplicity, and introduce
their Normalized Counting Measure (NCM) Nn, setting for every ⊂ R
Nn( ) = Card{l ∈ [nk ] : λl(n) ∈
m ,
Likewise, define the NCM σm of {τα}α=1
σm ( ) = Card{α ∈ [m] : τα ∈
In the case k = 1, there are a number of papers devoted to the convergence of the
NCMs of the eigenvalues of Mn,m,1 and related matrices (see [1,6,12,17,20,27] and
references therein). In particular, in [20] the convergence of NCMs of eigenvalues of
Mn,m,1 was proved in the case when corresponding vectors {Yα}α are “good vectors”
in the sense of the following definition.
Definition 1.1 We say that a normalized isotropic vector y ∈ Rn is good, if for every
n × n complex matrix Hn which does not depend on y, we have
Var{(Hny, y)} ≤ ||Hn||2δn, δn = o(1), n → ∞,
where ||Hn|| is the Euclidean operator norm of Hn.
Following the scheme of the proof proposed in [20], we show that despite the fact
that the number of independent parameters, kmn = O(nk+1) for k ≥ 2, is much less
than the number of matrix entries, n2k , the limiting distribution of eigenvalues still
obeys the Marchenko–Pastur law. We have:
Theorem 1.2 Fix k ≥ 1. Let n and m be positive integers satisfying ( 1.4), let {τα}α be
real numbers satisfying (1.7), and let y be a good vector in the sense of Definition 1.1.
Then there exists a nonrandom measure N of total mass 1 such that the NCMs Nn
of the eigenvalues of Mn (1.3) converge weakly in probability to N as n → ∞. The
Stieltjes transform f of N ,
f (z) =
z = 0,
is the unique solution of the functional equation
z f (z) = c − 1 − c
Nn[ϕ] =
j=1
is the linear eigenvalue statistic of Mn corresponding to a bounded continuous test
function ϕ : R → C, then we have in probability
This can be viewed as an analog of the Law of Large Numbers in probability theory
for (1.11). Since the limit is nonrandom, the next natural step is to investigate the
fluctuations of Nn[ϕ]. This corresponds to the question of validity of the Central
Limit Theorem (CLT). The main goal of this paper is to prove the CLT for the linear
eigenvalue statistics of the tensor version of the sample covariance matrix Mn,m,2
defined in (1.3).
There are a considerable number of papers on the CLT for linear eigenvalue statistics
of sample covariance matrices Mn,m,1 (1.5), where all entries of the matrix Bn,m,1 are
independent (see [4,7–9,11,16,18,19,21,25] and references therein). Less is known
in the case where the components of vector y are dependent. In [13], the CLT was
proved for linear statistics of eigenvalues of Mn,m,1, corresponding to some special
class of isotropic vectors defined below.
Definition 1.3 The distribution of a random vector y ∈ Rn is called unconditional if
n n
its components {y j } j=1 have the same joint distribution as {±y j } j=1 for any choice
of signs.
Definition 1.4 We say that normalized isotropic vectors y ∈ Rn, n ∈ N, are very good
if they have unconditional distributions, their mixed moments up to the fourth order
do not depend on i, j, n, there exist n-independent a, b ∈ R such that as n → ∞,
a2,2 := E{yi2 y 2j} = n−2 + an−3 + O(n−4), i = j,
κ4 := E{y 4j} − 3a2,2 = bn−2 + O(n−3),
and for every n × n complex matrix Hn which does not depend on y,
E{|(Hny, y)◦|4} ≤ C ||Hn||4n−2.
Here and in what follows we use the notation ξ ◦ = ξ − E{ξ }.
An important step in proving the CLT for linear eigenvalue statistics is the
asymptotic analysis of their variances Var{Nn[ϕ]} := E{|Nn◦[ϕ]|2}, in particular, the proof
of the bound
Var{Nn[ϕ]} ≤ Cn||ϕ||2H,
where || . . . ||H is a functional norm and Cn depends only on n. This bound determines
the normalization factor in front of Nn◦[ϕ] and the class H of the test functions for
which the CLT, if any, is valid. It appears that for many random matrices normalized
so that there exists a limit of their NCMs, in particular for sample covariance matrices
Mn,m,1, the variance of the linear eigenvalue statistic corresponding to a smooth
enough test function does not grow with n, and the CLT is valid for Nn◦[ϕ] itself
without any n-dependent normalization factor in front. Consider the test functions
ϕ : R → R from the Sobolev space Hs , possessing the norm
||ϕ||s2 =
(1 + |t |)2s |ϕ(t )|2dt, ϕ(t ) =
The following statement was proved in [13] (see Theorem 1.8 and Remark 1.11):
and let y be a very good vector in the sense of Definition 1.4. Consider matrix
Mn,m,1(y) (1.3) and the linear statistic of its eigenvalues Nn[ϕ] (1.11)
corresponding to a test function ϕ ∈ Hs , s > 2. Then {Nn◦[ϕ]}n converges in distribution to a
Gaussian random variable with zero mean and the variance V [ϕ] = limη↓0 Vη[ϕ],
where
L(z1, z2) − L(z1, z2) (ϕ(λ1) − ϕ(λ2))2dλ1dλ2
Lemma 1.6 Let {τα}α be a collection of real numbers satisfying (1.7) and (1.18), and
let y be a normalized isotropic vector having an unconditional distribution, such that
a2,2 = n−2 + O(n−3), κ4 = O(n−2).
Consider the corresponding matrix Mn (1.3) and a linear statistic of its eigenvalues
Nn[ϕ]. Then for every ϕ ∈ Hs , s > 5/2, and for all sufficiently large m and n, we
have
Var{Nn[ϕ]} ≤ C nk−1||ϕ||s2,
where C does not depend on n and ϕ.
It follows from Lemma 1.6 that in order to prove the CLT (if any) for linear
eigenvalue statistics of Mn, one needs to normalize them by n−(k−1)/2. To formulate our
main result we need more definitions.
Definition 1.7 We say that the distribution of a random vector y ∈ Rn is
permutationally invariant (or exchangeable) if it is invariant with respect to the permutations
of entries of y.
Definition 1.8 We say that normalized isotropic vectors y ∈ Rn, n ∈ N, are the
CLTvectors if they have unconditional permutationally invariant distributions and satisfy
the following conditions:
(i) their fourth moments satisfy (1.13)–(1.14),
(ii) their sixth moments satisfy conditions
a2,2,2 := E{yi2 y 2j yk2} = n−3 + O(n−4),
a2,4 := E{yi2 y 4j} = O(n−3), a6 := E{yi6} = O(n−3),
(iii) for every n × n matrix Hn which does not depend on y,
E{|(Hny, y)◦|6} ≤ C ||Hn||6n−3.
It can be shown that a vector of the form y = x/n1/2, where x has i.i.d. components
with even distribution and bounded twelfth moment is a CLT-vector as well as a vector
uniformly distributed on the unit ball in Rn or a properly normalized vector uniformly
distributed on the unit ball Bnp = x ∈ Rn : nj=1 |x j | p ≤ 1 in lnp (see [13],
Section 2 for k = 1).
The main result of the present paper is:
Theorem 1.9 Let m and n be positive integers satisfying (1.4) with k = 2, and let
m
{Cτoαn}sαi=d1erbemaatsreicteosf Mreanl,mnu,2m(yb)er(s1.u3n)icfoorrmrelsypboonudnindgedtoinCαLTa-nvdecmtorasndy s∈atRisnfy.iInfgN(1n.[7ϕ)].
are the linear statistics of their eigenvalues (1.11) corresponding to a test function
ϕ ∈ Hs , s > 5/2, then {n−1/2Nn◦[ϕ]}n converges in distribution to a Gaussian random
variable with zero mean and the variance V [ϕ] = limη↓0 Vη[ϕ], where
and f is given by (1.10).
V [ϕ] =
a−
where a± = (1 ± √c)2 and am = 1 + c.
(ii) We can replace the condition of the uniform boundedness of τα with the
condition of uniform boundedness of eighth moments of the Normalized Counting
Measures σn, or take {τα}α being real random variables independent of y with
common probability law σ having finite eighth moment. In general, it is clear from
(1.23) that it should be enough to have second moments of σn being uniformly
bounded in n.
(iii) If in (1.23) a + b + 2 = 0, then to prove the CLT one needs to renormalize linear
eigenvalue statistics. In particular, it can be shown that if y in the definition of
Mn,m,k (y) is uniformly distributed on the unit sphere in Rn, then a + b + 2 = 0
and under additional assumption m/n = c + O(n−1) the variance of the linear
eigenvalue statistic corresponding to a smooth enough test function is of the order
O(nk−2) (cf 1.20).
The paper is organized as follows. Section 3 contains some known facts and
auxiliary results. In Sect. 4, we prove Theorem 1.2 on the convergence of the NCMs
of eigenvalues of Mn,m,k . Sections 5 and 7 present some asymptotic properties of
bilinear forms (H Y, Y ), where Y is given by (1.1) and H does not depend on Y . In
Sect. 6, we prove Lemma 1.6. In Sect. 8, the limit expression for the covariance of the
resolvent traces is found. Section 9 contains the proof of the main result, Theorem 1.9.
2 Notations
Here and in what follows
gn(z) = n−k γn(z), fn(z) = E{gn(z)}.
j=1
so that for the nonbold Latin and Greek indices the summations are from 1 to n and
from 1 to m, respectively. For α ∈ [m], let
Thus the upper index α indicates that the corresponding function does not depend on
Yα. We use the notations Eα{. . .} and (. . .)◦α for the averaging and the centering with
respect to Yα, so that (ξ )◦α = ξ − Eα{ξ }.
In what follows we also need functions (see (4.5) below)
Writing O(n− p) or o(n− p) we suppose that n → ∞ and that the coefficients in
the corresponding relations are uniformly bounded in {τα}α, n ∈ N, and z ∈ K . We
use the notation K for any compact set in C \ R.
Given matrix H , ||H || and ||H ||H S are the Euclidean operator norm and the
HilbertSchmidt norm, respectively. We use C for any absolute constant which can vary from
place to place.
3 Some Facts and Auxiliary Results
We need the following bound for the martingales moments, obtained in [10]:
Proposition 3.1 Let {Sm }m≥1 be a martingale, i.e., ∀m, E{Sm+1 | S1, . . . , Sm } = Sm
and E{|Sm |} < ∞. Let S0 = 0. Then for every ν ≥ 2, there exists an absolute constant
Cν such that for all m = 1, 2 . . .
Lemma 3.2 Let {ξα}α be independent random variables assuming values in Rnα and
having probability laws Pα, α ∈ [m], and let : Rn1 × . . . × Rnm → C be a Borel
measurable function. Then for every ν ≥ 2, there exists an absolute constant Cν such
that for all m = 1, 2 . . .
− E{ }|ν } ≤ Cν mν/2−1
Proof This simple statement is hidden in the proof of Proposition 1 in [25]. We give its
proof for the sake of completeness. For α ∈ [m], denote E≥α = Eα . . . Em . Applying
Proposition 3.1 with S0 = 0, Sα = E≥α+1{ } − E{ }, Sm = − E{ }, we get
By the Ho¨ lder inequality
which implies (3.2).
− E{ }|ν } ≤ Cν mν/2−1
E{|E≥α+1{ } − E≥α { }|ν }.
Lemma 3.3 Fix ≥ 2 and k ≥ 2. Let y ∈ Rn be a normalized isotropic random
vector (1.2) such that for every n × n complex matrix H which does not depend on y,
we have
E{|( H y, y)◦| } ≤ || H || δn , δn = o(1), n →
∞.
Then there exists an absolute constant C such that for every nk × nk complex matrix
H which does not depend on y, we have
E{|(HY, Y )◦| } ≤ C k /2||H|| δn ,
Proof It follows from (3.2) that
E{|(HY, Y )◦| } ≤ C k /2−1
E{|(HY, Y )◦j | },
j =1
(HY, Y ) =
Hp, qYpYq = ( H ( j )y( j ), y( j )),
where H ( j ) is an n × n matrix with the entries
(H ( j))st =
This and (3.3) yield
= E j {|( H ( j )y( j ), y( j ))◦| } ≤ || H ( j )|| δn .
|| H ( j )|| ≤ ||H||
i = j
E j {|(HY, Y )◦j | } ≤ ||H||
E{||y(i)||2 }δn ≤ C ||H|| δn.
i = j
This and (3.5) lead to (3.4), which completes the proof of the lemma.
The following statement was proved in [20].
Proposition 3.4 Let Nn be the NCM of the eigenvalues of Mn = α ταYαYα T , where
m m
{Yα}α=1 ∈ R p are i.i.d. random vectors and {τα}α=1 are real numbers. Then
Var{Nn( )} ≤4m/ p2, ∀
⊂ R,
Var{gn(z)} ≤4m/( p| z|)2, ∀z ∈ C \ R.
Also, we will need the following simple claim:
Claim 3.5 If h1, h2 are bounded random variables, then
Var{h1h2} ≤ C Var{h1} + Var{h2} .
4 Proof of Theorem 1.2
Theorem 1.2 essentially follows from Theorem 3.3 of [20] and Lemma 3.3; here we
give a proof for the sake of completeness. In view of (3.6) with p = nk , it suffices
to prove that the expectations N n = E{Nn} of the NCMs of the eigenvalues of Mn
converge weakly to N . Due to the one-to-one correspondence between nonnegative
measures and their Stieltjes transforms (see, e.g., [2]), it is enough to show that the
Stieltjes transforms of N n,
λ − z
converge to the solution f of (1.10) uniformly on every compact set K ⊂ C \ R, and
that
fn(z) =
In [20], it is proved that the solution of (1.10) satisfies (4.1), so it is enough to show
that
fn (z) ⇒ f (z), z ∈ K , (4.2)
n→∞
where we use the double arrow notation for the uniform convergence. Assume first
that all τα are bounded:
Since Mn − Mn = ταYαYαT , the rank one perturbation formula
α
= − 1 + τα(GαYα, Yα)
τα((Gα)2Yα, Yα) Bα
γn − γnα = − 1 + τα(GαYα, Yα) = − Aα .
It follows from the spectral theorem for the real symmetric matrices that there exists
a nonnegative measure mα such that
| Aα| ≥ | Aα| = |τα|| z|
where we use ||G|| ≤ | z|−1. Let us show that
It follows from (1.2) that
| Aα−1| ≤ 1 + |τα| · ||Yα||2/| z|,
|Eα{ Aα}|−1, |E{ Aα}|−1 ≤ 4(1 + |τα|/| z|).
Eα{ Aα} = 1 + τα gnα(z), E{ Aα} = 1 + τα fnα(z).
where Nnα is the counting measure of the eigenvalues of Mnα. For every η ∈ R \ {0},
consider
< n−k
n−k
|Eα{ Aα}| ≥ | Eα{ Aα}| = |τα||η|n−k
zgn(z) = −1 + n−k Tr Mn G =
− 1 + mn−k
− n−k
This and the identity
z fn(z) = − 1 + n−k
− n−k
rn(z) =n−k
It follows from the Schwarz inequality that
|E{ A◦α Aα−1}| ≤ E{| A◦α|2}1/2E{| Aα−2|}1/2.
Note that since E{||Yα|| = 1}, we have by (1.8) E{||Yα||4} ≤ C . This and (4.8) imply
that E{| Aα−2|} is uniformly bounded in |τα| ≤ L and z ∈ K . We also have
A◦α = ( Aα)◦α + τα(gnα)◦ = τα (GαYα, Yα)◦α + (gnα)◦ ,
By (1.4) and (3.7) with p = nk , Var{gnα} ≤ C n−k | z|−2. It follows from (1.8) and
Lemma 3.3 with H = Gα and = 2 that
Eα{|(GαYα, Yα)◦α|2} ≤ C2k||Gα||2δn ≤ C2k| z|−2δn.
|rn| ≤ C (kδn + n−k )1/2.
uniformly in |τα| ≤ L and z ∈ K . Hence
z fn(z) = (−1 + mn−k ) − n−k
It follows from (4.5) and (4.7) that
| fn(z) − fnα(z)| ≤ n−k | z|−1.
This and (4.9) imply that |1+τα fn(z)|−1 is uniformly bounded in |τα| ≤ L and z ∈ K .
Hence, in (4.15) we can replace fnα with fn (the corresponding error term is of the
order O(n−k )) and pass to the limit as n → ∞. Taking into account (1.7) we get that
the limit of every convergent subsequence of { fn(z)}n satisfies (1.10). This finishes
the proof of the theorem under assumption (4.3).
Consider now the general case and take any sequence {σn} = {σm(n)} satisfying
(1.7). For any L > 0, introduce the truncated random variables
τα, |τα| < L ,
0, otherwise.
L
rank(Mn − Mn ) ≤ Card{α ∈ [m] : |τα| ≥ L}.
i → ∞. If NnLi is the NCM of the eigenvalues of MnLi and N nLi is its expectation,
then the mini-max principle implies that for any interval ⊂ R:
|N n( ) − N nLi ( )| ≤
uniformly on K . It follows from the first part of the proof that
where cLi = cσ [−Li , Li ] → c as Li → ∞. Since N (R) = 1, there exists C > 0,
such that
Hence we have for all sufficiently big Li :
Thus |τ /(1 + τ f Li (z))| ≤ | f Li (z)|−1 ≤ 2/C < ∞, z ∈ K . This allows us to pass
to the limit Li → ∞ in (4.17) and to obtain (1.10) for f , which completes the proof
of the theorem.
Remark 4.1 It follows from the proof that in the model we can take k depending on n
such that
k → ∞
5 Variance of Bilinear Forms
Lemma 5.1 Let Y be defined in (1.1–1.2), where y has an unconditional distribution
and satisfies (1.19). Then for every symmetric nk ×nk matrix H which does not depend
on y and whose operator norm is uniformly bounded in n, there is an absolute constant
C such that
nVar{(H Y, Y )} ≤ C n−k ||H ||2H S ≤ C ||H ||2.
If additionally y satisfies (1.13–1.14), then we have
nVar{(H Y, Y )} = ka|n−k Tr H |2
+ n−2k+1
+ O(n−1),
i=1 j,p
where j( pi ) = { j1, . . . , ji−1, pi , ji+1, . . . , jk }.
Proof Since y has an unconditional distribution, we have
E{|(H Y, Y )|2} =
i=1
For W ⊂ [k], W c = [k] \ W , denote
For every fixed W, j, s, we have
(W, j, s, p, q) =
i∈W c
∈W
(W, j, s, p, q) = O(n−k−|W |).
Indeed, the number of pairs for which (W, j, s, p, q) = 0 does not exceed 2|W |nk−|W |
(the number of choices of indices pi = qi for i ∈/ W equals to nk−|W |; all other indices
p , q ( ∈ W ) must satisfy { p , q } = { j , s } and, therefore, can be chosen in at
most two ways each). Since a2,2, wi = O(n−2), (5.4) follows.
For every fixed W ,
|Hj, s||Hp, q| (W, j, s, p, q) ≤
(W, j, s, p, q)/2
= O(n−k−|W |)||H ||2H S.
Var{(H Y, Y )} =
r=0 |W |=r j,s,p,q
This and (1.19) imply that
i=1
T0 :=
(a2,2δ ji si δ pi qi ) = a2,2| Tr H |2.
k
n T0 − n−2k | Tr H |2 ≤ C n−k ||H ||2H S,
and by (1.13),
and by (1.13)
T1 : =
i=1 j,p
i=1 j,p
The term corresponding to
Hj, s H p, q wi (j, s, p, q)
=i
a2k,2 Hj, j( pi ) H p, p( ji ) + a2k,−21κ4 Hj, j H p, pδ pi ji ,
Proof The proof follows the scheme proposed in [25] (see also Lemma 3.2 of [13]).
For q = 1, 2, by (3.2) we have
Also it follows from (5.5) that the terms corresponding to W : |W | ≥ 2 are less than
C n−k−2||H ||2H S. Summarizing (5.6–5.8), we get (5.1) and (5.2) and complete the
proof of the lemma.
6 Proof of Lemma 1.6
Lemma 6.1 Let {τα}α be a collection of real numbers satisfying (1.7), (1.18), and let y
be a normalized isotropic vector having an unconditional distribution and satisfying
(1.19). Consider the corresponding matrix Mn (1.3) and the trace of its resolvent
γn(z) = Tr(Mn − z I )−1. We have
Var{γn(z)} ≤ C nk−1| z|−6.
E{|γn◦(z)|4} ≤ C n2k−2| z|−12.
Applying (4.5), (4.7), and (4.9) we get
E{|(γn)◦α|2q } = E{|γn − γnα − Eα{γn − γnα}|2q }
≤ C E ABαα − EEαα{{ABαα}} 2q = C E E(αB{αA)α◦α} − ABαα · E(αA{αA)α◦α} 2q
≤ C (1 + |τα|/| z|)2q E Eα{|(Bα)◦α|2q } + Eα{|( Aα)◦α|2q }/| z|2q .
(6.4)
Here by (5.1)
nτα−2Eα{|(Bα)◦α|2} ≤ C n−k ||(Gα)2||2H S ≤ | z|−4.
This and (6.3–6.4) lead to (6.1). Also it follows from (1.15) and Lemma 3.3 that
Eα{|(Bα)◦α|4}, Eα{|( Aα)◦α|4}/| z|4 ≤ C τα4| z|−8n−2,
which leads to (6.2).
Proof of Lemma 1.6 The proof of (1.20) is based on the following inequality obtained
in [25]: for ϕ ∈ Hs (see 1.17),
Var{Nn[ϕ]} ≤ Cs ||ϕ||s2
≤ C n−k−1
τα2(1 + η−2τα2)E{||(Gα)2||2H S + η−2||Gα||2H S}.
By the spectral theorem for the real symmetric matrices,
where Nnα is the expectation of the counting measure of the eigenvalues of Mnα. We
have
n−k
Summarizing, we get
Var{Nn[ϕ]} ≤ C nk−1||ϕ||s2
dηe−ηη2s−6 ≤ C nk−1||ϕ||s2
provided that s > 5/2. This finishes the proof of Lemma 1.6.
7 Case k = 2: Some Preliminary Results
From now on we fix k = 2 and consider matrices Mn = Mn,m,2. For every j =
{ j1, j2} = j1 j2,
j=1
Lemma 7.1 Under conditions of Theorem 1.9,
Eα{|( Aα)◦α| p} ≤C (τα/| z|) pn− p/2,
Eα{|(Bα)◦α| p} ≤C (τα/| z|2) pn− p/2,
E{| A◦α| p}, E{|Bα◦| p} = O(n− p/2), 2 ≤ p ≤ 6.
Eα{|( Aα)◦α|6} ≤ C (τα/| z|)6n−3,
and by the Ho¨lder inequality we get the first estimate in (7.1). Analogously one can
get the second estimate in (7.1). Also we have by (6.1)
E{|(gnα)◦| p} ≤ | z|2− pE{|(gnα)◦|2} = O(n−3), p ≥ 2,
which together with (4.13) and (7.1) leads to (7.2).
It follows from (5.2) with k = 2 that
nVar{(H Y, Y )} =2a|n−2 Tr H |2
+ 2n−3
+ bn−3
Hj, j1 p2 H p, p1 j2 + Hj, p1 j2 H p, j1 p2
Hj, j H p, p(δ p1 j1 + δ p2 j2 ) + O(n−1).
Consider an n × n matrix of the form
Gs, p =
Since G =
||G|| ≤
We define functions
Similarly, we introduce the matrix
gn(1)(z1, z2) :=n−3
gn(2)(z1, z2) :=n−3
His, is (z1)H js, js (z2) = n−3
Gss (z1)Gss (z2).
Gi, j =
and define functions
It follows from (7.3) that
=2a(gnα(z))2 + 2(gn(1)(z, z) + gn(1)(z, z))
+ b(gn(2)(z, z) + gn(2)(z, z)) + O(n−1).
Var{gn(i)}, Var{gn(i)} =O(n−2),
lim E{gn(i)(z1, z2)} = nl→im∞ E{gn(i)(z1, z2)} = f (z1) f (z2),
n→∞
where f is the solution of (1.10).
Proof We prove the lemma for gn(1); the cases of gn(2), gn(2), and gn(2) can be treated
similarly. Without loss of generality we can assume that in the definitions of G and
gn(1), H = G. It follows from (3.2) that
Var{gn(1)} ≤
gn(1) − gn(1)α =n−3 Tr(G(z1) − Gα(z1))G(z2)
+ n−3 Tr Gα(z1)(G(z2) − Gα(z2)) =: Sn(1) + Sn(2).
and to get (7.7), it is enough to show that
Consider Sn(1). It follows from (4.4) that
E{|Sn( j)|2} = O(n−4), j = 1, 2.
Since for x , ξ ∈ Rn and an n × n matrix D
Di j xi ξ j ≤ ||D|| · ||x || · ||ξ ||,
This and following from (1.2) and (1.22) bound
z1 fn(1)(z1, z2) = − fn(z2) + n−3
− n−3
= − fn(z2) + Tn(1) + Tn(2).
j,p α
τα2E Yαj(H Aα(αz(1z)1Y)α) j1 p2 · (H α(z2)Yα)Apα((Hz2α)(z2)Yα)p1 j2
imply (7.9) for j = 1. The case j = 2 can be treated similarly. So we get (7.7) for
gn(1).
(YαYαT H )j, q = Aα−1Yαj(H αYα)q.
This and the resolvent identity yield
Hj, q(z1) = −z1−1δj, q + z1−1
It follows from (1.2) that
This and (4.12) yield
Tn(1) = n−5
rn = n−3
Treating rn we note that
By the Ho¨lder inequality, (4.8), and (7.13)
≤ C n−3
τα2E{||Yα||4| Aα(z1)|−1| Aα(z2)|−1} = O(n−1).
Eα{Yαj(H αYα) j1 p2 } = n−2 Hjα, j1 p2 .
E A◦α(z1) Yαj(H α(z1)Yα) j1 p2 Hpα, p1 j2 (z2) .
Aα(z1)
n−1
Hence, by the Schwarz inequality, (4.8), (4.9), (7.2), and (7.13)
|rn| ≤ C n−2
≤ C n−2
E{| A◦α|2}1/2E{| Aα|−2||Yα||4}1/2 = O(n−1/2).
Also one can replace fnα and H α with fn and G (the error term is of the order O(n−1)).
Hence,
z1 fn(1)(z1, z2) = − fn(z2) + fn(1)(z1, z2)n−2
This, (1.4), (1.7), and (1.10) lead to
f (1)(z1, z2) = f (z2) c
−1
= f (z1) f (z2)
and finishes the proof of the lemma.
It follows from Lemmas 5.1 and 7.2 that under conditions of Theorem 1.9
lim nτα−2E{ A◦α(z1) Aα(z2)} = 2(a + b + 2) f (z1) f (z2),
n→∞
where f is the solution of (1.10).
Lemma 7.3 Under conditions of Theorem 1.9
Var{Eα{( A◦α) p}} = O(n−4), p = 2, 3.
Proof Since τα, α ∈ [m], are uniformly bounded in α and n, then to get the desired
bounds it is enough to consider the case τα = 1, α ∈ [m]. By (4.13), we have
Eα{( A◦α)2} = Eα{(H Yα, Yα)◦α2} + (gnα)◦2,
It also follows from (7.6) and Lemmas 6.1 and 7.2 that
Var{Eα{(H Yα, Yα)◦α2}} = O(n−4),
Var{Eα{(H Yα, Yα)◦α3}} = O(n−4).
Var Eα{(H Yα, Yα)3} − gnα3
= O(n−4).
Hi, j Hp, q Hs, t (i, j, p, q, s, t),
k=1
(i, j, p, q, s, t) =
− 3Eα{(H Yα, Yα)} · Eα{(H Yα, Yα)◦α2}
= Eα{(H Yα, Yα)3} − gnα3 − 3gnα · Eα{(H Yα, Yα)◦α2}.
+ O(n−4),
and by (1.21)
(i, j, p, q, s, t) = O(n−6).
Also, due to the unconditionality of the distribution, contains only even moments.
Thus in the index pairs i, j, p, q, s, t ∈ [n]2, every index (both on the first positions and
on the second positions) is repeated an even number of times. Hence, there are at most 6
independent indices: ≤ 3 on the first positions (call them i, j, k) and ≤ 3 on the second
positions (call them u, v, w). For every fixed set of independent indices, consider maps
from this set to the sets of index pairs {i, j, p, q, s, t}. We call such maps the index
schemes. Let | | be the cardinality of the corresponding set of independent indices.
For example,
is an index scheme with 5 independent indices (i, j on the first positions and u, v, w on
the second positions). The inclusion–exclusion principle allows to split the expression
(7.19) into the sums over fixed sets of independent indices of cardinalities from 2 to
6 with the fixed coefficients depending on a2,2,2, a2,4, and a6 in front of every such
sum. We have
=2
S , S =
: | |=
where the last sum is taken over the set of independent indices of cardinality , is an
index scheme constructing pairs {i, j, p, q, s, t} from this set, and ( ) is a certain
expression, depending on , a2,2,2, a2,4, and a6. For example,
S2 = F (a2,2,2, a2,4, a6)
where F (a2,2,2, a2,4, a6) can be found by using the inclusion–exclusion formulas. As
to ( ) in (7.21), the only thing we need to know is that
( ) = O(n−6),
and that in the particular case of
we have by (1.21)
Tr : {i, j, k ; u, v, w} → {(i, u), (i, u); ( j, v), ( j, v); (k, w), (k, w)},
=2
= O(n−4).
Hence to get (7.18) it suffices to consider terms with 5 and 6 independent indices and
show that
= O(n−4).
Consider S5. In this case we have exactly 5 independent indices. By the symmetry
we can suppose that there are two first independent indices, i, j , and three second
independent indices, u, v, w, and that we have i on four places and j on two places.
Thus, S5 is equal to the sum of terms of the form
S5 =O(n−6)
S5 =O(n−6)
Here we suppose that there are some fixed indices on the dot places, which are different
from explicitly mentioned ones. Note that S5 has a single “external” pairing with
respect to j . While estimating the terms, our argument is essentially based on the
simple relations
and on the observation that the more the mixing of matrix entries we have the lower
order of sums we get. Let V ⊂ Rn be the set of vectors of the form
ξ = {ξ j }nj=1 = {H··, j·}nj=1 or ξ = {H··, ·u }un=1,
and let W be the set of n × n matrices of the form
It follows from (7.24) that
∀ξ ∈ V ||ξ || = O(1) and ∀D ∈ W
||D|| = O(1).
|H··, j· H··, j·| =O(1),
|H··, ·u H··, ·u | = O(1),
Hi·, j· Hi·, ·· H··, j· =O(1), and
Hi·, ·u Hi·, ·· H··, ·u = O(1).
In particular, by (7.24) and (7.25), we have for S5
|S5| ≤ O(n−6)
|Hi·, j· Hi·, j·| = O(n−2),
so that Var{S5} = O(n−4). Consider S5 . Note that if in S5 we have a single “external”
pairing with respect to at least one index on the second positions, then similar to S5,
the variance of this term is of the order O(n−4). So we are left with the terms of the
form
S5 = O(n−6)
Hiu, iu Hiv, iv H jw, jw.
It follows from (7.5) that
S5 = O(n−1) · gnα(z) · gn(2)(z, z).
Now (3.8), (6.1), and (7.7) imply that
Var{S5 } ≤ C n−2(Var{gn } + Var{gn(2)}) = O(n−4).
α
Summarizing we get Var{S5} = O(n−4).
Consider S6 and show that Var{S6 − gα3
n } = O(n−4). In this case we have 6
independent indices, i, j, k for the first positions and u, v, w for the second positions.
Suppose that we have two single external pairing with respect to two different first
indices and consider terms of the form
S6 =O(n−6)
S6 =O(n−6)
S6 = O(n−6)
If the second indices in Hk·, k· are not equal, then we get the expression of the form
S6 = O(n−6)
It follows from (7.26) that S6 = O(n−2); hence Var{S6 } = O(n−4). If the second
indices in Hk·, k· in (7.27) are equal, then we get the expressions of three types:
O(n−6)
O(n−6)
O(n−6)
Hiu, jv Hiu, jv Hkw, kw =gnαn−4
(Hiu, jv)2 = O(n−2),
Hiu, jv Hiv, ju Hkw, kw =gnαn−4
Hiu, jv Hiv, ju = O(n−2),
Hiu, ju Hiv, jv Hkw, kw =O(n−1)gn(1)(z, z)gnα(z),
where we used (7.24) to estimate the first two expressions, so that their variances are
of the order O(n−4). It also follows from (3.8), (6.1), and (7.7) that the variance of
the third expression is of the order O(n−4). Hence, Var{S6 } = O(n−4). It remains
to consider the term without external pairing, which corresponds to
2 3
Hiu, iu H jv, jv Hkw, kw = (a2,2,2) γn
(see (7.22)). Summarizing we get
= O(n−2)Var{gnα3} + O(n−4) = O(n−4),
where we used (1.21) and (6.1). This leads to (7.23) and completes the proof of the
lemma.
8 Covariance of the Resolvent Traces
Lemma 8.1 Suppose that the conditions of Theorem 1.9 are fulfilled. Let
C (z1, z2) = 2(a + b + 2)c
f (z1) f (z2)
(1 + τ f (z1))2 (1 + τ f (z2))2
Proof For a convergent subsequence {Cni }, denote
C (z1, z2) := nil→im∞ Cni (z1, z2).
We will show that for every converging subsequence, its limit satisfies (8.1). Applying
the resolvent identity, we get (see (4.11))
E{ Aα−1(z1)(γn − γnα)◦(z2)} =: Tn(1) + Tn(2).
Consider Tn(1). Iterating (4.12) four times, we get
=: Sn(1) + Sn(2) + Sn(3) + Sn(4).
It follows from (4.9), (6.1), and (7.15) that Sn(i) = O(n−1/2), i = 2, 3. Also, by (4.8)
we have
where by the Schwarz inequality, (6.2), (7.1), and (7.13)
≤ E{|γnα◦(z2)|}/|z1| + E{|γn◦(z1)|}/|z2| = O(n1/2).
1 τα
= Cn(z1, z2) n2z1 α (1 + τα fnα(z1))2 + O(n−1/2).
and we have
Summarizing, we get
Consider now Tn(2) of (8.2). By (4.5),
1 ∂ E{ A◦α(z1) A◦α(z2)}
E{ Aα−1(z1)(Bα/ Aα)◦(z2)} = − E{ Aα(z1)}2 ∂ z2 E{ Aα(z2)}
+ O(n−3/2).
For shortness let for the moment Ai = Aα(zi ), i = 1, 2, B2 = Bα(z2). Iterating (4.12)
with respect to A1 and A2 two times we get
E{(1/ A1)◦(B2/ A2)◦}
E{ A1}2E{ A2}2
−E{ A1◦ B2}E{ A2} + E{B2}E{ A1◦ A2}
E{ A1}2E{ A2}2
E{ A1}2E{ A2}2
Applying (1.22), (7.13), and using bounds (4.7), (4.8), (4.9) for |B2/ A2|, | Ai |−1,
|E{ Ai }|−1, i = 1, 2, one can show that the terms containing at least three centered
factors A◦1, A◦2, B2◦ are of the order O(n−3/2). This implies that
E{(1/ A1)◦(B2/ A2)◦} =
E{ A1}2E{ A2}2
Returning to the original notations and taking into account that
D = 2(a + b + 2).
It follows from (7.14) and (8.4–8.5) that
Denote for the moment
This and (8.2–8.3) yield
Note that by (1.10),
(1 + τ f (z1))2 ∂ z2 1 + τ f (z2)
Dc
C (z1, z2) = c τ (1+τ f (z1))−2dσ (τ )−z1
(1+τ f (z1))2 ∂ z2 1 + τ f (z2)
which completes the proof of the lemma.
9 Proof of Theorem 1.9
τ dσ (τ ) f (z)
(1 + τ f (z))2 − z = f (z)
C (z1, z2) = Dc
f (z1) f (z2)
(1 + τ f (z1))2 (1 + τ f (z2))2
and “∗” denotes the convolution. We have
Denote for the moment the characteristic function (9.1) by Zn[ϕ], to make explicit
its dependence on the test function. Take any converging subsequence {Zn j [ϕ]} ∞j=1
Without loss of generality assume that the whole sequence {Zn j [ϕη]} converges as
n j → ∞. By (1.20), we have
|Zn j [ϕ] − Zn j [ϕη]| ≤ |x |n−1/2 Var{Nn j [ϕ] − Nn j [ϕη]}
lim Zn j [ϕ] = lηi↓m0 n j →∞
lim Zn j [ϕη].
n j →∞
Thus it suffices to find the limit of
as n → ∞. It follows from (9.2) – (9.3) that
This allows to write
ϕ(μ) γn(z)dμ, z = μ + i η.
Yn(z, x ) = n−1/2E{γn(z)eη◦n (x )}.
Since |Yn(z, x )| ≤ 2n−1/2Var{γn(z)}1/2, it follows from the proof of Lemma 1.6 that
for every η > 0 the integrals of |Yn(z, x )| over μ are uniformly bounded in n. This
and the fact that ϕ ∈ L2 together with Lemma 9.1 below show that to find the limit of
integrals in (9.7) it is enough to find the pointwise limit of Yn(μ + i η, x ). We have
ϕ(λ1) (γn − γnα)◦(z1)dλ1
ϕ(λ1) (γn − γnα)◦(z1)dλ1
i x eηαn
E{ Aα−n1(z)(eηn − eηαn )◦(x )} = √
ϕ(λ1) (γn − γnα)◦(z1)dλ1
where z j = λ j + i η, j = 1, 2, and
Using the argument of the proof of the Lemma 8.1, it can be shown that Rn =
O(n−5/2). Hence,
1
Yn(z, x ) = − zn1/2
Treating the r.h.s. similarly to Tn(1) and Tn(2) of (8.2), we get
Yn(z, x ) =
where C (z, z1) is defined in (8.1). It follows from (9.7) and (9.8) that
E{eηαn (x )( Aα−1(z))◦ (γn − γnα)◦(z1)}dλ1 + O(n−1).
(see (1.23)) and finally
Taking into account (9.5), we pass to the limit η ↓ 0 and complete the proof of the
theorem.
It remains to prove the following lemma.
Lemma 9.1 Let g ∈ L 2(R) and let {hn } ⊂ L 2(R) be a sequence of complex-valued
functions such that
and hn → h a.e. as n →
∞,
where |h(x )| ≤ ∞
g(x )hn (x )dx →
g(x )h(x )dx as n →
∞.
Proof According to the convergence theorem of Vitali (see, e.g., [24]), if ( X, F , μ)
is a positive measure space and
then F ∈ L 1(μ) and limn→∞
that g(x ) = 0, x ∈ R, and take
{Fn }n is uniformly integrable,
∞,
|F (x )| ≤ ∞
Fn = ghn /|g|2,
F = gh/|g|2.
|g(x )|2dx < ∞,
∞,
|F (x )| ≤ ∞
Hence, the conditions of Vitali’s theorem are fulfilled and we get
|hn − h||g|dx = 0,
Acknowledgements The author would like to thank Leonid Pastur for an introduction to the problem and
for fruitful discussions.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
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