LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Uniqueness of the deformed Marchenko–Pastur fixed point

Statement

Let $c > 0$ , let $H$ be a Borel probability measure on $[0, \infty)$ , and fix $z \in C$ with $Im z > 0$ . There is at most one $q \in C$ satisfying $Im q > 0$ , $u : = - (1 - c) / z + c q$ with $Im u > 0$ , and $q = \int_{[0, \infty)} \frac{1}{t (1 - c - c z q) - z} d H (t) .$

Statement references

Silverstein, Theorem 1.1 and equation (1.4), PDF pp. 2–3 (full paper PDF).

Bai, Miao, and Pan, equation (1.1), PDF pp. 3–4 (full paper PDF).

Facts & Assumptions

Given: $c > 0$ , a Borel probability measure $H$ on $[0, \infty)$ , $z \in C^{+}$ , and two candidate solutions $q_{1}, q_{2}$ satisfying the Statement.

[A1]

The aspect ratio $c$ is positive.

[A2]

$H$ is a countably additive Borel measure on $[0, \infty)$ with total mass one (Probability measures and probability spaces).

[A3]

The spectral parameter is $z = u + i η$ with $u \in R$ and $η > 0$ .

[A4]

Each candidate $q$ has positive imaginary part, $- (1 - c) / z + c q$ has positive imaginary part, and $q$ satisfies the fixed-point equation in the Statement.

Proof

1.1givenA1A2A3A4algebra

For either candidate $q$ , set $u = - (1 - c) / z + c q \in C^{+}$ . Since $t \geq 0$ , $1 + t u \neq 0$ for every $t$ , and the fixed-point equation becomes $q = - z^{- 1} \int (1 + t u)^{- 1} d H (t)$ . Substituting this into $u = - (1 - c) / z + c q$ and using $(1 + t u)^{- 1} = 1 - t u / (1 + t u)$ gives $z = - \frac{1}{u} + c \int_{[0, \infty)} \frac{t}{1 + t u} d H (t) .$ All integrals are finite because $∣ t / (1 + t u) ∣ \leq 1 / Im u$ .

2.1A1A2A3A4step 1.1algebra

If $u_{1} \neq u_{2}$ are the transforms of two solutions, subtracting their equations in step 1.1 and dividing by $u_{1} - u_{2}$ yields $\frac{1}{u_{1} u_{2}} = c \int_{[0, \infty)} \frac{t^{2}}{(1 + t u_{1}) (1 + t u_{2})} d H (t) .$ For either solution $u$ , taking imaginary parts in step 1.1 gives $Im z = Im u (\frac{1}{∣ u ∣^{2}} - c \int_{[0, \infty)} \frac{t^{2}}{∣ 1 + t u ∣^{2}} d H (t)) > 0,$ so $c ∣ u ∣^{2} \int t^{2} ∣ 1 + t u ∣^{- 2} d H (t) < 1$ .

3.1A1A2A3A4step 2.1algebra∎

Taking absolute values in the identity of step 2.1 and applying Cauchy–Schwarz gives $1 \leq {(c ∣ u_{1} ∣^{2} \int \frac{t^{2}}{∣ 1 + t u_{1} ∣^{2}} d H (t))}^{1 / 2} {(c ∣ u_{2} ∣^{2} \int \frac{t^{2}}{∣ 1 + t u_{2} ∣^{2}} d H (t))}^{1 / 2} < 1,$ contradicting step 2.1. Hence $u_{1} = u_{2}$ , and the relation $u = - (1 - c) / z + c q$ with $c > 0$ gives $q_{1} = q_{2}$ .

Source notes

Silverstein, Theorem 1.1 and equation (1.4), PDF pp. 2–3 (full paper PDF).

Depends on

Probability measures and probability spaces

Used by

Dependency tree · two levels

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Sources

J. W. Silverstein, Strong Convergence of the Empirical Distribution of Eigenvalues of Large Dimensional Random Matrices (standard reference, not scraped)
Z. D. Bai, B. Q. Miao, and G. M. Pan, On Asymptotics of Eigenvectors of Large Sample Covariance Matrix (standard reference, not scraped)