Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Adapted concentration of bounded complex resolvent quadratic forms

Statement

On a common probability space, for each positive integer n, let Xn∈Cn×Nn have entries from one common iid law with mean zero, unit second absolute moment, and absolute value at most K, where K is independent of n. Let Tn∈Cn×n be deterministic Hermitian positive semidefinite with ∥Tn∥op≤M, and let xn∈Cn be deterministic with ∥xn∥2=1. Assume Nn/n→κ∈(0,∞). For fixed z=u+iη with η>0, set An:=1NnTn1/2XnXn∗Tn1/2,Rn(z):=(An−zI)−1. Then xn∗Rn(z)xn−E[xn∗Rn(z)xn]→0 almost surely. Moreover, for every t>0 there is a constant C=C(K,M,z,κ) such that, for all sufficiently large n, P ⁣{∣xn∗Rn(z)xn−E[xn∗Rn(z)xn]∣>t}≤Ct4Nn2.

Statement references

Bai, Miao, and Pan, §2, equations (2.1)–(2.3), printed pp. 12–13 (full paper PDF).

Silverstein, Theorem 1.1 proof, PDF pp. 3–8 (full paper PDF).

Facts & Assumptions

Given: The common probability space, bounded iid entries, deterministic PSD Tn, unit vectors xn, and fixed z∈C+ in the Statement.

[F1]

The entries are mutually independent with the stated common law (Identical distribution and IID families).

[F2]

Independent columns have their product joint law (Independent random elements have product joint law).

[F3]

Fubini permits the finite product-integral conditional averages used for the column filtration (Fubini's theorem for L^1 functions on a sigma-finite product).

[F4]

The Hermitian rank-one resolvent update and nonzero denominator are given by Hermitian rank-one resolvent update and trace-norm bound.

[F5]

A summable sequence of probabilities has probability zero of occurring infinitely often (First Borel-Cantelli lemma for events).

[F6]

A finite-dimensional Hermitian PSD matrix has nonnegative real eigenvalues and a unitary eigenbasis (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

[F7]

For a nonnegative random variable Y and a>0, P(Y>a)≤E[Y]/a (Markov's inequality for random variables).

[F8]

The operator norm is the maximum of ∥Av∥2 over unit vectors in the domain (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

Proof

Given: The hypotheses and notation in the Statement.

1.1givenF1algebra

We first prove a linear-form bound. If ζi are independent centered complex variables with ∣ζi∣≤K and ai∈C, expansion of E∣∑iaiζi∣8 has zero contribution whenever an index occurs exactly once. Each remaining equality pattern partitions the eight positions into blocks of sizes at least two. For a fixed pattern, the absolute sum is at most K8∏B∑i∣ai∣∣B∣≤K8(∑i∣ai∣2)4, since ∑i∣ai∣r≤(∑i∣ai∣2)r/2 for r≥2. There are finitely many such patterns, so E∣∑iaiζi∣8≤CK(∑i∣ai∣2)4.

2.1givenF2F4F6F8step 1.1algebra

Write sn,j=Tn1/2Xn,⋅j/Nn, An,j=An−sn,jsn,j∗, and Rn,j=(An,j−zI)−1. By [F4], Un,j:=xn∗(Rn−Rn,j)xn=−(xn∗Rn,jsn,j)(sn,j∗Rn,jxn)1+sn,j∗Rn,jsn,j. Diagonalizing An,j as in [F6] gives Im⁡(z(1+sn,j∗Rn,jsn,j))≥η, hence the denominator has modulus at least η/∣z∣. Conditional on all columns except j, each factor in the numerator is a linear form in the bounded independent entries of column j, with coefficient-vector norm at most M/(ηNn). Step 1.1 and Cauchy–Schwarz therefore give E∣Un,j∣4≤CNn−4.

3.1givenF2F3step 2.1algebra

Let Fn,j=σ(Xn,⋅1,…,Xn,⋅j), and let En,j be the finite-product average over the remaining columns. By [F2]–[F3], xn∗Rnxn−E[xn∗Rnxn]=∑j=1Nn(En,j−En,j−1)xn∗Rnxn. The leave-one-column resolvent Rn,j depends on all columns except j, so its product-integral averages under En,j and En,j−1 agree. Therefore the jth summand is Dn,j:=(En,j−En,j−1)Un,j. Integrating over column j shows En,j−1Dn,j=0, so these are martingale differences. Conditional Jensen gives E∣Dn,j∣4≤16E∣Un,j∣4≤CNn−4.

4.1step 3.1algebra

Put Sk=∑j≤kDn,j, ak=E∣Sk∣4, vk:=En,k−1∣Dn,k∣2, and wk:=En,k−1∣Dn,k∣4. Expanding ∣Sk−1+Dn,k∣4, the term linear in Dn,k vanishes after conditional integration; ∣Re⁡(S‾k−1Dn,k)∣≤∣Sk−1∣∣Dn,k∣ and 4∣Sk−1∣∣Dn,k∣3≤2∣Sk−1∣2∣Dn,k∣2+2∣Dn,k∣4 give ak≤ak−1+8E(∣Sk−1∣2vk)+3Ewk. By Cauchy–Schwarz and conditional Jensen, E(∣Sk−1∣2vk)≤ak−1Evk2≤ak−1Ewk≤Cak−1/Nn2, while Ewk≤C/Nn4. Induction gives ak≤C0k2/Nn4 for a fixed sufficiently large C0 and k≤Nn, hence E∣SNn∣4≤C0/Nn2.

5.1givenF5F7step 4.1algebra∎

Apply [F7] to Y=∣SNn∣4 and a=t4 to obtain the stated C/(t4Nn2) tail bound. Since Nn/n→κ>0, the probabilities are summable in n. All deviations lie on the common probability space in the Statement, so [F5] proves almost-sure convergence for this fixed z.

Source notes

Bai, Miao, and Pan, §2, equations (2.1)–(2.3), printed pp. 12–13 (full paper PDF).

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