Project Report on Random Toeplitz andHankel Matrices

Alok Bakshi

May 11, 2011

Abstract

Using method of moments, Bryc, Dembo, and Jiang (2006) showedthat for both Toeplitz and Hankel matrices (denoted by T and H re-

spectively), the corresponding empirical spectral distribution µ(n−12Tn)

and µ(n−12Hn) weakly converges to non-random symmetric probabil-

ity measures γT and γH with unbounded support, and moreover γTand γH don’t depend upon the distribution of input sequences. Theyconjectured that both measures are absolutely continuous and in par-ticular γT has a smooth density. Thereafter Sen, and Virag (2011)gave a partial solution to the conjecture about Toeplitz matrix, prov-ing that γT is absolutely continuous with bounded density. In theirproof, rather than using method of moments, they realized the spec-trum of Gaussian Toeplitz matrix as diagonal matrix (consisting ofindependent Gaussian random variables) conjugated by the determin-istic projection matrix. The next key step then was to use the spectralaveraging technique, which was developed by Combes, Hislop, andMourre (1996) in different context.

In this project, we undertook the task of adapting the techniqueto Hankel matrix. Building upon the work of Bose and Hazra, whereusing reverse circulant matrices they realized the spectrum of GaussianHankel matrix as random Diagonal matrix (consisting of independentGaussian and Rayleigh distributed random variables) conjugated by(nondeterministic) projection matrix we seek out to find a suitablevariant of spectral averaging technique to apply in this particular case.

1

1 Empirical and limiting spectral distribu-

tions

Given an n× n matrix An, its empirical spectral measure µn is defined as

µn =1

n

n∑i=1

δλi

where (λi)ni=1 are the eigenvalues of An and δ is the dirac delta function.

For symmetric real matrices, with which we will be concerned here, all theeigenvalues are real and so µn is a measure on real line R. The correspondingempirical distribution function (ESD) is given by

FAn(x) =1

n

n∑i=1

Iλi≤x

For a sequence of matrices (An)∞n=1, the limiting spectral distribution(LSD) is defined as the weak limit of FAn with the implicit assumption thatsuch a limit exists. In real symmetric case, all ESD and therefore LSD toois a distribution on R.

2 Spectrum for random symmetric Toeplitz

Matrix

A random symmetric Toeplitz matrix is defined as

Tn = ((x|i−j|))0≤i,j≤n−1

where the input sequence (xi)∞i=0 consists of independent identically dis-

tributed random variables with Var(x0) = 1.Since γT doesn not depend upon the distribution of individual entries,

so without loss of generality we can assume xi to be independent standardGaussian distributed random variables. For the sake of convenience in com-putation, we will work with Ton , which has

√2x0 on its diagonal entries and

2

rest of the entries are same as that of T. Using Mallow’s metric we see

W 22 (F n−1/2Tn , F n−1/2Ton) ≤ 1

n

n∑i=1

(λi(n− 1

2Tn)− λi(n−12Ton)))2 ≤ 1

nTr(n−

12Tn − n−

12Ton)2

≤ 1

n2

n−1∑i=0

(√

2− 1)2x20 → 0 as n→∞

so that Ton and Tn will have same limiting spectral distribution.A n×n random symmetric Toeplitz matrix Ton is n×n principal submatrix

of 2n× 2n symmetric circulant matrix, which is defined as

C2n = ((yn−|n−|i−j||))0≤i,j≤2n−1, where yj =

xj 0 < j < n

x2n−j n < j < 2n− j√2xj j ∈ {0, n}

Thus, we do have

Q2nC2nQ2n =

[Ton 0n0n 0n

], where Q2n =

[In 0n0n 0n

](1)

The circulant matrix C2n can be diagonalized as (2n)−12C2n = U2nD2nU

∗2n,

the unitary matrix U2n being given by

U2n(j, k) =1√2n

exp

(2πijk

2n

), where 0 ≤ j, k ≤ 2n− 1

and the diagonal matrix D2n = diag(d0, d1, . . . , d2n−1), where

dj =1√2n

2n−1∑k=0

yk exp

(2πijk

2n

)=

1√2n

(√

2x0 + (−1)j√

2xn + 2n−1∑k=1

xk cos

(2πjk

2n

))Since yj = y2n−j, and all the dj are real so for n < j < 2n we observe

that dj = d2n−j. Furthermore it can be seen that (dj)0≤j≤n are independent

mean zero Gaussian random variables, where Var(dj) =

{1 0 < j < n2 j ∈ {0, n}

Defining P2n = U∗2nQ2nU2n, we see that P 22n = P2n = P ∗2n or P2n is a

Hermitian projection matrix. Moreover P2n is a deterministic project matrix(since its entries don’t depend upon (xi)) with P2n(j, j) = 1

2∀j. Thus

(2n)−12U∗2nQ2nC2nQ2nU2n = P2nD2nP2n (2)

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so that spectrum of Q2nC2nQ2n is realized as the diagonal matrix D2n,conjugated by the projection matrix P2n. But from equation (1), we seethat (counting multiplicities) eigenvalues of Q2nC2nQ2n precisely consist ofall eigenvalues of Ton and zero (n times). Thus the above procedure gives usa handle on the spectrum of the Toeplitz matrix.

3 Spectrum for random Hankel Matrix

A random Hankel matrix is defined as

Hn = ((xi+j))0≤i,j≤n−1

where the input sequence (xi)∞i=0 consists of independent identically dis-

tributed random variables with Var(x0) = 1.Since γH doesn not depend upon the distribution of individual entries, so

again without loss of generality we can assume xi to be independent standardGaussian distributed random variables. Furthermore Hankel matrix can berealized as the principal minor of reverse circulant matrix. Reverse circulantmatrix is related to circulant matrix via the following relation

R2n = T2nC2n, where T2n =

1 0 0 . . . 0 00 0 0 . . . 0 10 0 0 . . . 1 0

...0 0 1 . . . 0 00 1 0 . . . 0 0

In words, in reverse circulant matrix, shifting each row to the left instead

of right, we get the next row. Thus we have

Q2nR2nQ2n =

[Hn 0n0n 0n

], where Q2n =

[In 0n0n 0n

](3)

So as in the case of Toeplitz matrix, we need to analyze the eigenvalues ofR2n. But unlike its relative C2n, the description is bit complicated. Similarto the case of symmetric Circulant matrix, for circulant matrix we have thespectral decomposition namely, (2n)−

12C2n = U2nD2nU

∗2n, where (denoting

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primitive 2n root of unity by ω) U2n is the following matrix

U2n =

1 1 1 . . . 1 11 ω ω2 . . . ω2n−2 ω2n−1

1 ω2 ω4 . . . ω2(2n−2) ω2(2n−1)

...1 ω2n−2 ω(2n−2)2 . . . ω(2n−2)(2n−2) ω(2n−2)(2n−1)

1 ω2n−1 ω(2n−1)2 . . . ω(2n−1)(2n−2) ω(2n−1)(2n−1)

Interestingly, we have the following relation

U2nU2n = U∗2nU∗2n = T2n where R2n = T2nC2n

Therefore, we have

(2n)−12R2n = (2n)−

12T2nC2n = T2nU2nD2nU

∗2n = U2nT2nD2nU

∗2n

= U2nM2nU∗2n where M2n = T2nD2n

Now to get hold of eigenvalues of reverse circulant matrix (2n)−12R2n, it

suffices to get the spectral decomposition of M2n. Concretely, we have

M2n =

d0 0 0 . . . 0 00 0 0 . . . 0 d2n−10 0 0 . . . d2n−2 0

...0 0 d2 . . . 0 00 d1 0 . . . 0 0

Here dj are the eigenvalues of the circulant matrix C2n. Denoting the firstrow of R2n as (yk)0≤k≤2n−1 we have

dj =1√2n

2n−1∑k=0

yk exp

(2πijk

2n

)

=1√2n

2n−1∑k=0

yk cos

(2πjk

2n

)+ i

1√2n

2n−1∑k=0

yk sin

(2πjk

2n

)= Aj,n + iBj,n = rj exp(−iθj)

5

Where rj =√A2j,n +B2

j,n and θj = tan−1(Bj,nAj,n

) When (yk)0≤2n−1 are inde-

pendent standard Gaussian random variables then (rj)1≤j≤n−1, (θj)1≤j≤n−1,d0 and dn are independent random variables, where rj follow Rayleigh dis-tribution and θj follow uniform distribution over [−π, π].

Let Ij =

{ej j ∈ {0, n}

1√2[sgn(n− j) exp(iθj)ej + e2n−j] otherwise

We define L2n = [I0, I1, I2, . . . , I2n−1]. Concretely we have

L2n =

1 0 0 . . . 0 00 1√

2exp(iθ1) 0 . . . 0 − 1√

2exp(iθ1)

0 0 1√2

exp(iθ2) . . . − 1√2

exp(iθ2) 0...

0 0 1√2

. . . 1√2

0

0 1√2

0 . . . 0 1√2

Moreover letD†2n = [d0, r1, r2, . . . , rn−1, dn,−rn−1, . . . ,−r1]. By result of Boseand Mitra(2002), D†2n consist of all the eigenvalues of reverse circulant matrixR2n and in fact M2n = L2nD2nL

∗2n

So we have

(2n)−12R2n = U2nT2nD2nU

∗2n

= U2nM2nU∗2n

= U2nL2nD†2nL

∗2nU

∗2n

= V2nD†2nV

∗2n where V2n = U2nL2n

Being a product of two unitary matrices, V2n is also a unitary matrix.Defining hermitian projection matrix PH

2n = V ∗2nQ2nV2n, we get the relation

(2n)−12Q2nR2nQ2n = Q2nV2nD

†2nV

∗2nQ2n

= V2nPH2nD

†2nP

H2nV

∗2n

From equation (3), we see that (counting multiplicities) eigenvalues of

(2n)−12R2n precisely consists of eigenvalues of (2n)−

12H2n and zero (n times).

Thus spectrum of (2n)−12H2n is realized as the diagonal matrix D†2n conju-

gated by the non-deterministic projection matrix PH2n.

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4 Spectral Properties

Given a finite measure ν on R, its Cauchy-Stieltjes transformation is

S(z, ν) =

∫R

1

x− zν(dx), where z ∈ C, =(z) > 0

In the special case, when ν is the empirical measure of a n×n symmetricmatrix A, we have

S(z, µ(A)) =1

n

n∑i=1

1

λi − z

=1

nTr(A− zI)−1 =

1

n

n−1∑j=0

〈ej, (A− zI)−1ej〉

We can recover measure back from its Stieltjes transformation by thefollowing inversion formula

ν{(x, y)} = limδ↓0

1

π

∫ y

x

=(S(E + iδ; ν))dE

where x < y are the continuity points of ν.Moreover putting a bound on =(S(z; ν)) in the upper half complex plane

will entail that ν is absolutely continuous with respect to the Lebesgue mea-sure. This follows from the inversion formula, as

supz:=(z)>0

=(S(z; ν) ≤ K =⇒ ν{(x, y)} ≤ K

πλ{(x, y)} =⇒ dν

dλ≤ K

π

Sen and Virag(2011) used the same approach for proving the absolutecontinuity of γT . To put a bound on Stieltjes transform, first we need torewrite it in suitable format.

Let (ej)0≤j≤2n−1 be the coordinate vectors of R2n, then for z ∈ C and=(z) > 0, we next prove the following two results

S(z;Eµ(n−12T0

n)) =

√2

n

2n−1∑j=0

E〈Pej, (PDP − zI)−1Pej〉

S(z;Eµ(n−12Hn)) =

√2

n

2n−1∑j=0

E〈PHej, (PHD†PH − zI)−1PHej〉

7

and these thus give us relation between the Stieltjes transformation of Toeplitz/Hankelmatrix and the corresponding representation of it as diagonal matrix conju-gated by the projection matrix. We prove the first of these results, the secondone is proved by following the same procedure. So, we have

S(z, µ(n−12Ton)) =

1

n

n−1∑j=0

〈ej, (n−12QCQ− zI)−1ej〉

=

√2

n

n−1∑j=0

〈ej, (QUDU∗Q− zI)−1ej〉

=

√2

n

n−1∑j=0

〈U∗ej, (PDP − zI)−1U∗ej〉

Moreover we have

2n−1∑j=0

P eje∗j P = P

(2n−1∑j=0

eje∗j

)P = P 2 = P = U∗QU

= U∗

(n−1∑j=0

eje∗j

)U =

n−1∑j=0

U∗ eje∗j U

8

Thus we do the following manipulation

S(z, µ(n−12Ton)) =

√2

n

n−1∑j=0

〈U∗ej, (PDP − zI)−1U∗ej〉

=n−1∑j=0

〈ej, U(PDP − zI)−1U∗ej〉 =n−1∑j=0

U(PDP − zI)−1U∗(j, j)

=n−1∑j=0

2n−1∑j1,j2=0

U(j, j1)(PDP − zI)−1(j1, j2)U∗(j2, j)

=2n−1∑j1,j2=0

(PDP − zI)−1(j1, j2)

(n−1∑j=0

U∗(j2, j)U(j, j1)

)

=2n−1∑j1,j2=0

(PDP − zI)−1(j1, j2)

(n−1∑j=0

U∗eje∗jU

)(j2, j1)

=2n−1∑j1,j2=0

(PDP − zI)−1(j1, j2)

(2n−1∑j=0

Peje∗jP

)(j2, j1)

=2n−1∑j1,j2=0

(PDP − zI)−1(j1, j2)

(2n−1∑j=0

P (j2, j)P (j, j1)

)

=2n−1∑j=0

2n−1∑j1,j2=0

P (j, j1)(PDP − zI)−1(j1, j2)P (j2, j)

=2n−1∑j=0

P (PDP − zI)−1P (j, j) =2n−1∑j=0

〈Pej, (PDP − zI)−1Pej〉

Thus taking expectation of both sides, we get

S(z;Eµ(n−12T0

n)) =

√2

n

2n−1∑j=0

E〈Pej, (PDP − zI)−1Pej〉

Next we state and prove the spectral averaging technique by Combes et.al. (1996) and afterwards a simple application of the above theorem will giveus a bound on the Stieltjes transformation of γT , thus proving the absolutecontinuity of γT . Later we will see the difficulties in using the same approachfor Hankel matrix.

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5 Spectral averaging technique by Combes

et. al. (1996)

Let Hλ, λ ∈ R be a C2 family of self adjoint operators such that D(Hλ) =D0 ⊂ H ∀λ ∈ R, and such that (Hλ − zI)−1 is twice strongly differentiablein λ for all z, =(z) 6= 0. Assume that there exists a finite positive constantc0, and a positive bounded self adjoint operator B such that,on D0

Hλ =dHλ

dλ≥ c0B

2

Also assume that Hλ is linear in λ, i.e. Hλ = d2Hλdλ2

= 0. Then for all E ∈ Rand twice continuously differentiable function g such that g, g′, g′′ ∈ L1(R)and for all φ ∈ H, we have

supδ>0

∣∣∣∣∫Rg(λ)〈φ,B(Hλ − (E + iδ)I)−1Bφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1) ‖φ‖2

(4)

The vanishing second derivative hypothesis can be relaxed by just requir-ing that |Hλ| ≤ c1Hλ, as it was done in Combes (1996) paper. But restrictivehypothesis is satisfied in our application, and so we don’t need to prove theresult in most general form, whose proof is considerably harder than thisparticular case.

This result is trivially true for φ = 0, and so dividing both sides by ‖φ‖2,we can assume without loss of generality that ‖φ‖ = 1.

Since 〈φ,B(Hλ−(E+iδ)I)−1Bφ〉 = 〈φ,B(Hλ − (E − iδ)I)−1Bφ〉, so prov-ing the above inequality (where the supremum is taken in the upper halfcomplex plane) will imply the same inequality where the supremum is takenin the lower half complex plane and vice versa. In particular we do have

supδ<0

∣∣∣∣∫Rg(λ)〈φ,B(Hλ − (E + iδ)I)−1Bφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1) ‖φ‖2

⇓

supδ>0

∣∣∣∣∫Rg(λ)〈φ,B(Hλ − (E + iδ)I)−1Bφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1) ‖φ‖2

In the proof, we will take supremum over the negative half complex plane.

10

Finally we observe that we don’t need to have supremum over the wholehalf plane but taking supremum over the strip δ : 0 < δ < 1, will suffice. Thisfollows since in the case of when δ ≥ 1, we can let Gλ = Hλ

δand D = B√

δ, so

that Gλ ≥ c0D2 and thefore

sup0<δ<1

∣∣∣∣∫Rg(λ)〈φ,D(Gλ − (E + iδ)I)−1Gφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1) ‖φ‖2

⇓

supδ>0

∣∣∣∣∫Rg(λ)〈φ,B(Hλ − (E + iδ)I)−1Bφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1) ‖φ‖2

Now, for ε > 0 and 0 < δ < 1, we make the following definitions

R(λ, ε, δ) = (Hλ − (E − iδ)I + iεHλ)−1 and K(λ, ε, δ) = BR(λ, ε, δ)B

Now, keeping in mind the inequality Hλ ≥ c0B2, we have

−=〈φ,K(λ, ε, δ)φ〉 = 〈φ,BR(λ, ε, δ)∗(δI + εHλ)R(λ, ε, δ)Bφ〉 ≥ c0ε‖K(λ, ε, δ)φ‖2

But by Cauchy Schwartz inequality and the fact that ‖φ‖ = 1, we also have−=〈φ,K(λ, ε, δ)φ〉 ≤ ‖K(λ, ε, δ)φ‖. So we get the following inequality

‖K(λ, ε, δ)φ‖ ≤ 1

c0ε(5)

Now we make another definition

F (ε, δ) =

∫Rg(λ)〈φ,K(λ, ε, δ)φ〉dλ (6)

Inequality (5) gives us the following bound

F (ε, δ) ≤ 1

c0ε‖g‖1 (7)

Differentiating F (ε, δ) with respect to ε, and using integration by parts andthe fact that Hλ = 0, we get

idF (ε, δ)

dε=

∫Rg(λ)BR(λ, ε, δ)HλR(λ, ε, δ)Bφ〉dλ

= −∫Rg(λ)

d

dλ〈φ,K(λ, ε, δ)φ〉dλ

11

Therefore from (7) and using integration by parts, we have∣∣∣∣dF (ε, δ)

dε

∣∣∣∣ =

∣∣∣∣∫Rg′(λ)〈φ,K(λ, ε, δ)φ〉dλ

∣∣∣∣ ≤ 1

c0ε‖g′‖1 (8)

Integration of the above inequality, yields the following improved bound onF (ε, δ)

|F (ε, δ)| ≤ c−10 ‖g′‖1| log(ε)|+ |F (1, δ)| ≤ c−10 ‖g′‖1| log(ε)|+ c−10 ‖g‖1 (9)

If we define the function F (ε, δ) =∫R g′(λ)〈φ,K(λ, ε, δ)φ〉dλ then following

the same procedure for F (ε, δ) instead of F (ε, δ), we get

|F (ε, δ)| ≤ c−10 ‖g′‖1| log(ε)|+ |F (1, δ)| ≤ c−10 ‖g′′‖1| log(ε)|+ c−10 ‖g′‖1which implies that∣∣∣∣dF (ε, δ)

dε

∣∣∣∣ ≤ c−10 ‖g′′‖1| log(ε)|+ c−10 ‖g′‖1

Finally integrating the above inequality again, we get

|F (ε, δ)| ≤ c−10 (‖g′′‖1 + ‖g′‖1) + |F (1, δ)| ≤ c−10 (‖g′′‖1 + ‖g′‖1 + ‖g‖1)(10)

We note that R(λ, ε, δ) → (Hλ − (E − iδ)I)−1 as ε → 0+, therefore bydominated convergence theorem we get the following result

sup0<δ<1

∣∣∣∣∫Rg(λ)〈φ,B(Hλ − (E − iδ)I)−1Bφ〉dλ

∣∣∣∣ ≤ c−10 (‖g‖1 + ‖g′‖1 + ‖g′′‖1)

(11)

and as discussed earlier, the general result follows from the aforementionedinequality.

6 Absolute Continuity for γT

Let’s take

Ej =

{eje∗j + e2n−je

∗2n−j 1 ≤ j < n

eje∗j j ∈ {0, n}

Bj =

{Peje

∗jP or Pe2n−je

∗2n−jP 1 ≤ j < n

Peje∗jP j ∈ {0, n}

12

Now we takeHλ = P (D+(λ−dj)Ej)P , so thatHλ is a random self adjoint op-erator, whose randomness comes from {d0, d1, . . . , dj−1, dj+1, . . . , dn}. More-over by definition, we have Hλ = 0. We can verify that Hλ = PEjP ≥ Bj =P (j, j)−1B2

j = 2B2j , so that in theorem we can use c0 = 2 and P (j, j) = 1

2.

Now for the choice of g, let Z denote the standard Gaussian random variable.Then we take

g =

{density of Z 0 < j < n

density of√

2Z j ∈ {0, n}

From the definition, we have ‖g‖1 = 1, ‖g′‖1 ≤√

2π, ‖g′′‖1 ≤ 2.

Thus plugging in φ = ej or e2n−j and taking expectation with respect to{d0, d1, . . . , dj−1, dj+1, . . . , dn} (noticing that integral is equivalent to takingexpectation with respect to dj, and by virtue of independence of (dk) we canbreak up expectation in this form) we finally get

supz:=(z)>0

P (j, j)2∣∣E〈Pej, (PDP − zI)−1Pej〉∣∣ ≤ c−10 (‖g′′‖1 + ‖g′‖1 + ‖g‖1)

Therefore, we get the following bound on S(z;Eµ(n−12T0

n))

∣∣∣S(z;Eµ(n−12T0

n))∣∣∣ ≤ √2

n

2n−1∑j=0

∣∣E〈Pej, (PDP − zI)−1Pej〉∣∣≤√

2

n× 2n× 8

≤ 16√

2

So as a consequence of bound, we know that γT is absolutely continuouswith respect to Lebesgue measure.

7 Absolute Continuity for γH

The same approach can be used for Hankel matrix. Though projection matrixin this case PH is non-deterministic, still it does not poses major problemsas we still have P (j, j) = 1

2. However because of the structure of diagonal

matrix D†, we are forced to choose Ej as eje∗j−e2n−je∗2n−j for 1 ≤ j < n, and

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that means Hλ is no longer bounded below since Hλ = PEjP is not even apostive semidefinite matrix.

Our task in the project was to study the proof of spectral averagingtechnique and to check whether it could be suitably modified so that we nolonger need the condition on lower bound of Hλ. As of now, we have notmade any progress. But it looks feasible as we never really need the full forceof Combes’s result. Firstly we only need the result for finite dimensionaloperator, and moreover g in our case is not only smooth but is a rapidlydecreasing function (or belongs to the Schwartz class).

References

[1] Arup Bose Lecture notes on Random Matrices.

[2] Arup Bose and Rajat Hazra Some remarks about Hankel Matrices.

[3] Bryc, W. Dembo, and A. Jiang, Spectral measure of large random Hankel,Markov and Toeplitz matrices 2006: Ann. Probab. 34, 1–38.

[4] Arnab Sen, and Balint Virag, Absolute continuity of the limiting eigen-value distribution of the random Toeplitz matrix 2011: Electronic Com-munications in Probability 61, 706–711.

[5] A. Bose, and J. Mitra, Limiting spectral distribution of a special circulant2002: Statistics and Probability Letters, 60 (1). pp. 111–120

[6] J. M. Combes, P. D. Hislop, and E. Mourre, Spectral averaging, perturba-tion of singular spectra, and localization 1996: Trans. Amer. Math. Soc.348, 4883–4894.

[7] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to ran-dom matrices, Cambridge University Press, 2010.

[8] Patrick Billingsley, Probability and Measure, Wiley Series in Probabilityand Mathematical Statistics, 1995.

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