Analysis of the Limiting Spectral
Distribution of Large Dimensional
Random Matrices
by
Jack W.Silverstein*
and
Sang-Il Choi
Department of Mathematics,Box8205
North Carolina State University
Raleigh,North Carolina27695-8205
Summary
Results on the analytic behavior of the limiting spectral distribution of matrices of
sample covariance type,studied in Marˇc enko and Pastur[2]and Yin[8],are derived.
Through an equation defining its Stieltjes transform,it is shown that the limiting distri-
bution has a continuous derivative away from zero,the derivative being analytic wherever
it is positive,and resembles
1.Introduction.For N =1,2,...let M N =
1N →c >0as N →∞,and T n is n ×
n Hermitian,independent of X N .For any square matrix A with real eigenvalues,let F A denote the spectral distribution of A ,that is,the empirical distribution function of the eigenvalues of A .Assume F T n D −→H ,a.s.,where H is a non-random probability distribution function (p.d.f.).Then it is known that,with probability one,F M N D −→F ,a non-random p.d.f.depending on H and c ,if either:1)T n is diagonal ([2],[4]),or 2)T n ≥0(T n is non-negative definite),and H has moments of all order satisfying Carleman’s sufficiency condition (ensuring only one p.d.f.having these moments)([8]).
This result has direct bearing on multivariate statistical applications when the vector dimension and sample size are both large but have the same order of magnitude (see [6]
for an application to array signal processing).Indeed,when T n ≥0and E X 111=0,the
matrix 1
N T n X ∗N X N ,and for any Hermitian T n
the spectra of this latter matrix and M N differ by |n −N |zero eigenvalues.From this it is
a simple matter to verify F M N =(1−n N
F 1N T n X ∗N X N D −→F 0≡(1−1
c F.(1.1)
Important to applications is the behavior of F and its dependence on H and c .The purpose of this paper is to derive certain fundamental properties,the most important being the analyticity of F .
Under condition 2)it is shown in [8]that F 0has moments of all order satisfying Carleman’s sufficiency condition,and are explicitly expressed.From the moments,F 0has been derived in two cases:when T n =I n (the n ×n identity matrix),that is,when H =1[1,∞)([1]),and when T n =(1
N
T n X ∗N X N i.p.−→T n for n fixed and N →∞).Further analysis relying on the moment expressions appear extremely difficult.How-ever,the approach taken in [2],[4]under condition 1)(where H is arbitrary)leads to a characterization of F most suitable to analysis.It uses the Stieltjes transforms of mea-sures,that is,for z ∈D ≡{z ∈:Im z >0}and p.d.f.G on ,the Stieltjes transform m G of G is the analytic function mapping D into itself defined by m G (z )= dG (λ)
G{[a,b]}=lim
η→0+
1
1+λm −1
.
(1.3)
It follows that:
On D,m F(z)has a unique inverse,given by
z F(m)=−1
1+λm
m∈m F(D).
(1.4)
Although it appears unlikely a general form for F exists,quite a bit can be inferred from this representation.We show how the above qualitative properties can be derived.
From(1.4)wefind for all m∈m F(D)z F(m)=−1
m −c
m
).Thus,m H is deter-
mined by F or F0(via z F)and c on the open set{m:−1
z +cm F
(z).Us-
ing the fact that|m G(z)|≤1
z
as c→0.From
(1.4)it follows that zm F
0(z)=−1+
λm F(z)
1+λm F(z)
≤|m F(z)|
1−λ/z
dH(z)=zm H(z).Thus we
get ii).
We can also use(1.4)to show F{0},the mass F places at0,is max(0,1−c(1−H{0})). Consider a sequence{T n}from1)satisfying F T n D−→H,and F T n{0}→H{0}.Then it is a simple matter to verify F M N{0}≥max(0,N−(n−nF T n{0})
1+λm F(iy)
.If F{0}>0,then, as y↓0,|m F(iy)|→∞,and,since yRe m F(iy)→0and yIm m F(iy)→F{0},we have
Re m F(iy)
|1+λm F(iy)|2≤1+
Re m F(iy)
π
Im m0(x).The density f is analytic 2
(possesses a power series expansion)for every x =0for which f (x )>0.Moreover,for these x ,πf (x )is the imaginary part of the unique m ∈D satisfying
x =−11+λm
.(1.6)Obviously the theorem reveals much of the analytic behavior of F in general,and its dependence on H .For example,when H is discrete with finite support,m 0(x )is the root of a polynomial with coefficients depending linearly on x ,making f algebraic in nature.The theorem also shows how to determine F .For some H (1.6)can be solved explicitly,for example,in the above 2cases,or when H has support on at most 3distinct points in (the degree of the resulting polynomial being at most 4).If no way of solving (1.6)is apparent,then a simple numerical scheme can be applied.
It is remarked here that,even if F {0}=0,it is still possible for f not to exist at 0.For example,for the case H =1[1,∞)and c =1,f (x )=1(0,4)14−x
m ∈S c H }.Section 5shows,for most cases of x 0∈∂S F ,that f (x ),for
x ∈S F ,resembles πIm m G (x 0).
Proof.Fix >0.Let δ>0be s.t.12whenever |x −x 0|<δ,y ∈(0,δ).Let x 1<x 2be continuity points of G s.t.x 1<x 2and |x i −x 0|<δ,i =1,2.From (1.2),we can choose y ∈(0,δ)s.t. G (x 2)−G (x 1)−12(x 2−x 1).For any x ∈[x 1,x 2],we have |x −x 0|<δ.Thus G (x 2)−G (x 1)πIm m G (x 0) ≤1π x 2x 1Im m G (x +iy )dx +1π
(Im m G (x +iy )−Im m G (x 0)) dx < .It follows that G is continuous at x 0,and for any sequence {x n }of continuity points of G converging to x 0
lim n →∞G (x n )−G (x 0)π
Im m G (x 0).(2.1)Let {x n }be any real sequence satisfying x n ↓x 0.For each n choose continuity