Lvy-Khintchine Random Matrices Paul Jung University of Alabama - - PowerPoint PPT Presentation

Lévy-Khintchine Random Matrices

Paul Jung University of Alabama Birmingham Western States Mathematical Physics Meeting February 15, 2016

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α-Stable limit laws (heavy tails)

Finite variance is important in the CLT: If P(|Xi| > Cn

1 α ) ∼ 1

n for 0 < α < 2 then for all x

P

• X1 + X2 + · · · + Xn

√n

• < x
• → 0

On the otherhand, P

X1 + X2 + · · · + Xn

n1/α < x

• → F(x)

where F(x) is an α-Stable cdf. When α = 1, F(x) is the Cauchy distribution function.

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α-Stable limit laws (heavy tails)

Finite variance is important in the CLT: If P

• |Xi|/n

1 α > Cn 1 α /n 1 α

• ∼ 1

n for 0 < α < 2 then for all x

P

• X1 + X2 + · · · + Xn

√n

• < x
• → 0

On the otherhand, P

X1 + X2 + · · · + Xn

n1/α < x

• → F(x)

where F(x) is an α-Stable cdf. When α = 1, F(x) is the Cauchy distribution function.

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3 Different normalizations

CLT: EX 2

i = σ2

• n
• i=1

1 √nXi ∼ N(0, σ2). Poisson:

• P(Xi = 1) = 1/n

P(Xi = 0) = 1 − 1/n

• n
• i=1

Xi ∼ Poisson (λ) . α-Stable: P(|Xi| > Cn

1 α ) ∼ 1

n

• n
• i=1

1 n1/α Xi ∼ Stableα(σ).

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Wigner’s theorem (1955, 58)

Matrix An has i.i.d. entries up to symmetry An(i, j) = An(j, i) and such that EAn(i, j)2 < ∞ Normalized empirical measure of eigenvalues ej ∈ R converges 1 n

n

• j=1

δej = ESDn = ⇒ 1 2π

• 4 − x2 dx

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Normalization for Wigner matrices

1 n

n

• j=1

δej = ESDn = ⇒ 1 2π

• 4 − x2 dx

What is the correct scaling? Tightness of probability measures: Second Moment(ESDn) = 1 n Tr(A2

n) = 1

n

• i,j

An(i, j)2 ≃ nAn(i, j)2. So we need An(i, j)2 = O

1

n

• .

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Normalization for Wigner matrices

1 n

n

• j=1

δej = ESDn = ⇒ 1 2π

• 4 − x2 dx

What is the correct scaling? Tightness (in expectation) of random probability measures: E(Second Moment(ESDn)) = E1 n Tr(A2

n) = nEAn(i, j)2.

So we need EAn(i, j)2 = O

1

n

• .

Instead of normalizing, change the distribution as n varies: An(i, j) ∼ Bernoulli (λ/n) so that EAn(i, j)2 = λ/n.

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3 Different normalizations for random matrices

Wigner: EAn(i, j)2 ∼ 1 n

• An(i, j) ∼

1 √nN(0, σ2). Sparse: EAn(i, j)2 ∼ 1 n

• An(i, j) ∼ Poisson

λ

n

• .

Heavy-tailed: P(|An(i, j)| > Cn

1 α ) ∼ 1

n

• An(i, j) ∼

1 n1/α Stableα(σ).

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Infinite divisibility

X (symmetric) is infinitely divisible if for every n X d = An(1, 1) + An(1, 2) + · · · + An(1, n) (i.i.d.) and it is determined by (σ2, ν) satisfying, when X d = −X, EeiθX = exp

• −1

2σ2θ2 +

• R

(eiθx − 1)ν(dx)

• 7/18

Existence of the LSD

Suppose each An has i.i.d. entries up to self-adjointness such that for each i: limn→∞

n

j=1 An(i, j) d

= ID(σ2, ν).

• J. (2015)

With probability 1, ESDn weakly converges to a symm.

• prob. meas. µ∞.

µ∞ is the expected spectral measure for vector δroot of a self-adjoint operator on L2(G). Wigner matrices: G = N Sparse matrices: G is a Poisson Galton-Watson tree

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ESD of heavy-tailed and gamma random matrices

Histogram of Cauchy

Cauchy Frequency

• 40
• 20

20 40 50 100 150 200 250

Histogram of Gamma

Gamma Frequency

• 4
• 2

2 4 50 100 150 200 250 300 350

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Proof sketch: Existence of the LSD

(1) As rooted graphs, Erdős-Rényi(λ/n) locally converge to a branching process with Poisson(λ) offspring distribution. (2) Bordenave-Lelarge (2010) If Gn[1] ⇒ G∞[1], then for all z ∈ C+, (An − zI)−1

11 → (A∞ − zI)−1 11 := δ1, (A∞ − zI)−1δ1,

where A∞ is an adjacency operator on L2(G∞). (3) E(An − zI)−1

11 = E1

n Tr(An − zI)−1 =

• 1

x − z dE(ESDn).

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Local weak limits of Erdős-Rényi graphs

Fixing an offspring k with depth 2 or more: P(1 is connected to k) = λ/n → 0. Local weak convergence to a Poiss(λ) branching process

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Topology for rooted weighted graphs (locally-finite)

Such graphs form a Polish space so weak convergence makes sense.

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Aldous’ Poisson weighted infinite tree

For heavy-tailed matrices, the weights 1/|An(i, j)| are arrivals of a Poisson process of intensity |x|α−1dx.

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The Poisson weighted infinite skeleton tree

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The cords to infinity: σ2 > 0

Interpret the weights as the lengths of edges. Thus, v and ∞v are infinitely far apart, but have infinitely many parallel edges between. Conductance of each parallel edge is zero, but their collective effective conductance is σ.

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Corollary

We handle infinite variance in the Gaussian domain of attraction.

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Schur complement formula for the limiting operators

Corollary (J. 2014) For z ∈ C+, Rjj(z) d = (A∞ − zI)−1

11

satisfies R00(z) d = −

 z + σ2R11(z) +

• j≥2

a2

j Rjj(z)

 

−1

where {aj} are arrivals of an independent Poisson(ν) process. Proof follows from the resolvent identity.

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Thanks for your attention!

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