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Concentration of the normalized complex Hermitian resolvent trace

Statement

On a common probability space, for each positive integer n, let Yn,1,…,Yn,Nn be independent random vectors in Cn, let Bn=∑j=1NnYn,jYn,j∗, and assume Nn/n→κ∈(0,∞). Fix z=u+iη with u∈R and η>0, and define Mn(z):=1ntr⁡(Bn−zI)−1. Then for every t>0, P{∣Mn(z)−EMn(z)∣>t}≤4exp⁡ ⁣(−t2n2η24Nn), and Mn(z)−EMn(z)→0 almost surely.

Statement references

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

Bai and Silverstein, Chapter 3, §3.3, printed pp. 53–55 (full book PDF).

Facts & Assumptions

Given: The common probability space, independent complex columns Yn,j, the normalized resolvent trace, and z=u+iη with η>0.

[F1]

Adding one positive rank-one term changes the unnormalized Hermitian resolvent trace by at most 1/η (Hermitian rank-one resolvent update and trace-norm bound).

[F2]

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

[F3]

Fubini's theorem permits integration in either order on the finite product spaces used to define the conditional averages (Fubini's theorem for L^1 functions on a sigma-finite product).

[F4]

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

[F5]

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

Proof

Given: The hypotheses and notation of the Statement.

1.1givenF1algebra

Replacing one column Yn,j by any other vector removes one positive rank-one term and adds another. Applying [F1] twice shows that tr⁡(Bn−zI)−1 changes by at most 2/η, so the real and imaginary parts of Mn(z) each have coordinate oscillation at most cn:=2/(nη).

2.1givenF2F3step 1.1construct

Let f be either real part of Mn(z) and use [F2]–[F3] to form its product-integral Doob averages Fj over the first j columns, integrating the remaining columns against their marginal laws. Boundedness ∣tr⁡(Bn−zI)−1∣≤n/η makes these integrals finite. Then Dj:=Fj−Fj−1 has conditional mean zero and lies, conditionally on the first j−1 columns, in an interval of length at most cn by step 1.1.

3.1F5step 2.1algebra

If a real random variable D has mean zero and lies in an interval of width c, then log⁡EeλD has second derivative equal to a variance under an exponentially tilted law, at most c2/4; the variance bound follows from E[(D−a)(b−D)]≥0 for interval endpoints a≤D≤b. Since the log moment and its first derivative vanish at 0, integrating twice gives EeλD≤eλ2c2/8. Apply this conditional estimate to each Dj, iterate, and use [F5] with λ=4t/(Nncn2) to obtain each one-sided tail P{f−Ef≥t}≤e−2t2/(Nncn2). The same bound applies to the lower tail and to the imaginary part. If a complex number has modulus greater than t, one of its real or imaginary parts has magnitude greater than t/2, so the union bound and cn=2/(nη) give the displayed probability inequality.

4.1givenF4step 3.1algebra∎

Since Nn/n→κ∈(0,∞), for every fixed t>0 the step 3.1 bound is at most 4e−atn for all sufficiently large n and some at>0. The probabilities are summable. The deviations lie on the common probability space in the Statement, so [F4] applies; use t=1/m, m≥1, and intersect the resulting countable probability-one events to conclude Mn(z)−EMn(z)→0 almost surely.

Source notes

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

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