Fluctuations of linear statistics of half-heavy-tailed random matrices

Introduction

Let $A=[a_{ij}]$ be an $N\times N$ Hermitian random matrix whose entries are i.i.d. and let $\lambda_{1},\ldots,\lambda_{N}$ be its eigenvalues. It is well known that if the entries of $A$ are duly renormalized, then for any continuous bounded test function $\varphi$, the random variable

has a deterministic limit, which is equal to the integral of $f$ with respect to the limit spectral distribution of $A$, namely the semicircle law when the entries have at least a second moment and different distributions depending on $\alpha$ if the entries are heavy-tailed with exponent $\alpha\in(0,2)$ (see ). The rate of convergence of the random variables of (1) to its limit is not usually $\frac{1}{\sqrt{N}}$, as i.i.d. $\lambda_{i}$’s would give. In particular, if the entries of $A$ have a fourth moment, then the fluctuations of $\frac{1}{N}\operatorname{Tr}\varphi(A)$ around its expectation have order $\frac{1}{N}$ (see ). On the other hand, if the entries are heavy-tailed with exponent $\alpha\in(0,2)$ or Bernoulli with parameter of order $N^{-1}$, then the fluctuations of $\frac{1}{N}\operatorname{Tr}\varphi(A)$ around its expectation have order $N^{-1/2}$ . This difference of order in the fluctuations is due to the fact that when the entries of $A$ have enough moments, the eigenvalues of $A$ fluctuate very little, as studied by Erdös, Schlein, Yau, Tao, Vu and their co-authors, who analyzed their rigidity in e.g. . On the other hand, the heavier the tails the more similar to a sparse matrix the (renormalized) matrix $A$ is, and the more independently its eigenvalues behave.

Viewed in the light of concentration inequalities for linear spectral functionals of random matrices, random matrices with half-heavy tailed entries interpolate between two extreme regimes, as shown in Table 1 :

is bounded in probability and explains why the order of the fluctuations of (1) cannot be larger than $N^{-1/2}$,

is bounded in probability. It explains why the order of the fluctuations of functionals as (1) cannot be larger than $N^{-1}$ in the case of matrices with independent Log-Sobolev entries.

Equation (2) shows that the case $\alpha<2$ corresponds to the largest possible fluctuations order in (1). On the other hand, (3) proves that in the Gaussian case (this has been extended by Bai et. al. to the case $\alpha\geq 4$), the actual order is $N^{-1}$ (to be more precise, (3) only gives an upper-bound for this order, but one can easily check, using, for example, $\varphi(\lambda)=\lambda$ or $\varphi(\lambda)=\lambda^{2}$, that $N^{-1}$ is actually the right order). The case $2<\alpha<4$ is an intermediate case, where concentration inequalities neither allow to guess the order of the fluctuations, nor allow to extend fluctuation results from a first class of test functions to a wider classer (as was done for example in ).

Main result

Let us consider a random real symmetric or Hermitian matrix

where one of two conditions holds: either

(real case) $x_{ij}$’s, $1\leq i\leq j$, are i.i.d. real random variables with mean and variance $1$ such that for a certain $\alpha\in(2,4)$ and a certain $c>0$, as $x\to+\infty$,

This theorem proves Gaussian convergence for any random variable of the form

where $\varphi$ is a function of the type

The remainder of the paper consists of the proof of Theorem 2.1. In Section 3.1, we truncate the random variables appropriately, and centralize in the real case (in the complex case, centralization is automatic due to our assumption of symmetry). In Section 3.2, we restate our problem in terms of a martingale approach and cite relevant martingale convergence theorem. In Section 3.3, we show that off-diagonal terms of the resolvent can be neglected in further calculations. Lastly, in Section 3.4 we show that the diagonal terms of the resolvent yield the desired formula for the covariance, using a lemma proved in Section 3.5 that allows us to approximate the diagonal elements of the resolvent by the Stietjes transform of the spectral measure.

Proof of Theorem 2.1

where the $X_{i}$’s are independent Bernoulli r.v. with parameters

and 1 is added for the rank 1 perturbation of shifting each entry by $\mu_{N}/\sqrt{N}$. In order to upper-bound $\operatorname{rank}(B-A)$ with high probability thanks to Bennett’s inequality or Lemma 5.7 in , we compute the sum of these parameters:

Thus, by Bennett’s inequality or Lemma 5.7 in , as soon as $\alpha\beta>1$, we know that $\operatorname{rank}(B-A)$ has order at most $N^{(2-\alpha\beta)_{+}}$ (i.e. for any $\varepsilon>0$, $N^{-(2-\alpha\beta)_{+}-\varepsilon}\operatorname{rank}(B-A)$ tends in probability to zero). So one can replace $A$ by $B$ as long as

Furthermore we want to renormalize our new truncated centered random variables to have variance 1, so we let

and thus $\sigma_{N}=1+O(N^{\beta(2-\alpha)})$. We keep the variance of the entries as $\sigma_{N}^{2}/N$ throughout.

In the complex case, subtracting the mean from each matrix entry is no longer a rank 1 perturbation, so this argument will no longer work. This is the reason why, in the complex case, we only consider random variables which are symmetric so that we can truncate and still retain a mean.

Let $\epsilon>0$, $\beta=\frac{1}{\alpha}+\frac{1}{4}+\epsilon$ and

in the complex case. Then there is a constant $C$ depending only on the distribution of the $x_{ij}$’s such that

Recalling that $\mu_{N}\to 0$ we get from (4) that for a certain constant $C>0$,

Since $\mu_{N}\rightarrow 0$, the random variable $x_{ij}-\mu_{N}$ will also satisfy (4) for large $N$, and therefore (iv) holds for the shifted entry $x_{ij}-\mu_{N}$. ∎

for $\epsilon>0$ that can be chosen as small as needed. By a slight abuse of notation, we still denote this random variable by $a_{ij}$ and we henceforth assume the conclusions of Lemma 3.1 to be true for the $a_{ij}$’s.

2. Martingale approach

converges in distribution to a certain Gaussian distribution. We will use Theorem 4.3 for $M(\varphi,N)$, with $M_{N}(N)=M(\varphi,N)$ and $\mathcal{F}_{k}(N):=\sigma(x_{i,j}\,;\,i\leq k\textrm{ and }j\leq k).$

Note first that by the interlacing property between the spectrums of $A$ and $A^{(k)}$, when $\varphi$ has finite total variation, we have

As a consequence, $|Y_{k}|\leq\frac{\|\varphi\|_{\operatorname{TV}}}{N^{1-\alpha/4}}$ and the $L(\varepsilon,N)$ of Theorem 4.3 is null for $N$ large enough.

have finite deterministic limits that agree with the limit covariance of Theorem 2.1.

converges in probability to a deterministic constant which agrees with the limit covariance of Theorem 2.1.

hence we shall prove that for any $u\in(0,1)$, as $N\to\infty$ and $k\to\infty$ with $k/N\to u$, we have

with $C(z,z^{\prime})$ the function defined in Theorem 2.1.

Note also that for $G^{(k)}(z):=(z-A^{(k)})^{-1}$, by (35),

where $\mathbf{a}_{k}$ is the $k$th column of $A$ without the diagonal term.

3. Removing the off-diagonal terms

where for a matrix $M$, $M_{\operatorname{diag}}$ denotes the diagonal matrix obtained from $M$ by setting all its non-diagonal entries to zero. Then

so that the argument of the $\log$ cannot vanish.

hence by the Cauchy inequalities for holomorphic functions, it suffices to prove that uniformly on $k,z,z^{\prime}$ (as $z,z^{\prime}$ stay at a macroscopic distance from the real line) we have

We also define $\eta_{k}^{\prime}$ and $\varepsilon_{k}^{\prime}$ in the same way with $z$ instead of $z^{\prime}$.

From now on, $C$ will denote a finite constant (that will change from line to line) depending uniformly in $k,z,z^{\prime}$ as $z,z^{\prime}$ stay at any positive distance away from the real line.

Inequality (i) follows from Lemma 4.1, which allows to claim that

To make subsequent calculations less cumbersome to write, we introduce the notation

and correspondingly $J_{k}^{\prime},J_{k,\operatorname{diag}}^{\prime}$ with $z^{\prime}$ instead of $z$. Let furthermore

By Cauchy-Schwarz and Lemma 3.3 we get that

We note that for $2<\alpha<4$, it can be ensured that $\epsilon$ is small enough to obtain the desired inequality:

Let us partition the space of matrices as follows. We define the events

and note that since $|1-\eta_{k}J_{k,\operatorname{diag}}|$ and $|1+\eta_{k}J_{k}|$ are reciprocals, we have that

Taking the expectation of $|\varepsilon_{k}|^{2}$, and using that $|J_{k}|$ and $|J_{k,\operatorname{diag}}|$ are absolutely bounded we get that

where the last inequality follows by the Lemma 3.3. Here the $\epsilon$ of (11) is chosen small enough. By Cauchy-Schwartz, it proves that $|T_{3}|\leq CN^{-1}$.

Let us now treat $T_{1}$ and $T_{2}$. We have

The same bound holds for $T_{1}$. This concludes the proof of Proposition 3.2.∎

4. Computation of the limit

By what precedes, to prove (13), it suffices to prove that the random variables

where $M_{t}=\max_{B(z,\Im z/2)}|\delta(z,t)|$. Let $z_{t}$ be the maximizer of $\delta(w,t)$ on $B(z,\Im z/2)$. Then

Let $0<\gamma<(\alpha-2)/(2\alpha)<1/2$. Then we split the above integral into two parts:

To get a bound on $\delta_{1}$ we use Lemma 4.5 for each $t$ with

where we used the fact that $\gamma<1/2$.

To get a bound on $\delta_{2}$, we can use the Taylor series expansion to check that for $|x|\leq 1/2$

so that, as $\gamma<(\alpha-2)/(2\alpha)$ implies that $\gamma\alpha-\alpha+2<-\alpha/2+1$ and for some constants $C_{1},C_{2}$,

To get a bound on $\delta_{3}$ we do a dyadic decomposition of the integral. We integrate on $[2^{k},2^{k+1}]$ with $k$ such that $2^{k}<N^{\gamma}$. This yelds

Noting that $\sum_{k}2^{k\alpha/2}e^{-|C\Im z|2^{k}}$ is convergent we get that

where $o(1)$ is for the convergence in probability.

This equation, together with (30) and Lemma 3.4, imply (27). This concludes the proof.

5. Concentration of the diagonal terms of the resolvent

As $|G(z)_{jj}-G_{\operatorname{sc}}(z)|\leq 2(\Im z)^{-1}$, it suffices to prove the result for any value of $p$. By the Schur complement formula (see [3, Th. 11.4]), we know that

where $G^{(j)}(z)=(z-A^{(j)})^{-1}$ and $A^{(j)}$ is the matrix obtained after removing the $j$th row and the $j$th column of $A$.

We recall that the denominator is equal to

where $\eta_{j}$ is as in (18) and $E=\frac{1}{N}\sum G_{jj}^{(j)}\left(|x_{jj}|^{2}-1\right)$.

We will show that for a certain choice of $0<p<1$ and a certain choice of $\epsilon_{0}>0$,

Recall that here $x_{jj}$ has been truncated at $N^{\frac{1}{\alpha}+\frac{1}{4}+\epsilon}$. In the proof of this Lemma we will truncate the variables further at $N^{\beta_{1}}.$ Since each variable is truncated, independence of variables is retained. We let $\Omega_{k}$ be event that for $k$ of $j$’s, we have that $|x_{jj}|\geq N^{\frac{1}{4}-\delta}$. Then, by Jensen’s inequality and Lemma 4.1,

Choosing $p$ and $\beta_{1}$ in such a way that $\beta_{1}(4-\alpha)-1<0$ and $-\alpha\beta_{1}+\left(\frac{1}{4}+\frac{1}{\alpha}+\epsilon\right)p<0$ finishes the proof of (32).

Again, using Jensen’s Inequality we obtain that

Lastly, using that $|\operatorname{Tr}G^{(j)}(z)-\operatorname{Tr}G(z)|\leq C$ (see Lemma 4.6), the denominator of the RHS of (31) can be written

We conclude by using the fact that $\displaystyle\frac{1}{z-G_{\operatorname{sc}}(z)}=G_{\operatorname{sc}}(z)$.∎

Appendix

Let $\mathbf{a}=(a_{1},\ldots,a_{N})^{T}$ be a column vector whose entries are i.i.d., centered and satisfy (ii) and (iii) of Lemma 3.1. Then for any deterministic matrix $G$, the random variables

Direct computations, the second one using that $a_{j}$ is centred for the first one and that $|a_{j}|^{2}-\frac{1}{N}$ is centred for the second one. ∎

We shall sometimes use this lemma after removal of the $k$th row and column of $B$ and of the $k$th entry of $\mathbf{a}$, but it suffices to apply the lemma with the matrix deduced from $B$ by setting its $k$th row and column to zero.

2. CLT for martingales

Let $(\mathcal{F}_{k})_{k\geq 0}$ be a filtration such that $\mathcal{F}_{0}=\{\emptyset,\Omega\}$ and let $(M_{k})_{k\geq 0}$ be a square-integrable complex-valued martingale starting at zero with respect to this filtration. For $k\geq 1$, we define the random variables

Let now everything depend on a parameter $N$, so that $\mathcal{F}_{k}=\mathcal{F}_{k}(N),M_{k}=M_{k}(N),Y_{k}=Y_{k}(N),v=v(N),\tau=\tau(N),L(\varepsilon)=L(\varepsilon,N),\ldots$

Then we have the convergence in distribution

To apply this theorem, we shall use the following lemma.

3. A lemma about large products and the exponential function

Let $u_{i}$, $i=1,\ldots,N$, be some complex numbers and set

There is a universal constant $R>0$ (independent of $N$ and of $M$) such that

Let $L(z)$ be defined on $B(0,1)$ by $\log(1+z)=z+z^{2}L(z)$ and $R>0$ be such that on $B(0,R)$, $|L(z)|\leq 1.$ If $\frac{M}{N}\leq R$, we have

Since for any $z$, $|e^{z}-1|\leq|z|e^{|z|},$ the conclusion follows. ∎

4. Linear algebra

Let $H_{k}$ be the submatrix of $H$ obtained by removing its $k$-th row and $k$-th column and set $G_{k}:=(z-H_{k})^{-1}$. Let also $\mathbf{a}_{k}$ be the $k$-th column of $H$ where the $k$-th entry has been removed. Then

References