Harmonic Means of Wishart Matrices Hi! Im Asad Lodhia, Im a postdoc - - PowerPoint PPT Presentation

Harmonic Means of Wishart Matrices Hi! I’m Asad Lodhia, I’m a postdoc at the University of Michigan. Feel free to email me at alodhia@umich.edu Based on joint work with Keith Levin and Liza Levina. Thanks to the organizers for the invitation! I hope you enjoy it.

What’s a Harmonic Mean? ake two positive definite matrices P1 and P2

• P−1

1

+ P−1

2

2 −1

he AMHM inequality:

• P−1

1

+ P−1

2

2 −1

• P1 + P2

2

• the operator norm is smaller...

What’s a Harmonic Mean? Take two positive definite matrices P1 and P2

• P−1

1

+ P−1

2

2 −1

he AMHM inequality:

• P−1

1

+ P−1

2

2 −1

• P1 + P2

2

• the operator norm is smaller...

What’s a Harmonic Mean? Take two positive definite matrices P1 and P2

• P−1

1

+ P−1

2

2 −1

The AMHM inequality:

• P−1

1

+ P−1

2

2 −1

• P1 + P2

2

• the operator norm is smaller...

What’s a Harmonic Mean? Take two positive definite matrices P1 and P2

• P−1

1

+ P−1

2

2 −1

The AMHM inequality:

• P−1

1

+ P−1

2

2 −1

• P1 + P2

2

So the operator norm is smaller...

The Wishart Ensemble et X be P × N, i.i.d complex standard normals P < N:

• P

N − γ

• ≤ K

P 2,

for some K > 0 and γ ∈ (0, 1). he matrix W = XX∗

N

has limiting spectral measure

ργ(x) :=

• ((1 + √γ)2 − x)(x − (1 − √γ)2)

2γπx

his is invertible with probability 1.

The Wishart Ensemble Let X be P × N, i.i.d complex standard normals P < N:

• P

N − γ

• ≤ K

P 2,

for some K > 0 and γ ∈ (0, 1). he matrix W = XX∗

N

has limiting spectral measure

ργ(x) :=

• ((1 + √γ)2 − x)(x − (1 − √γ)2)

2γπx

his is invertible with probability 1.

The Wishart Ensemble Let X be P × N, i.i.d complex standard normals P < N:

• P

N − γ

• ≤ K

P 2,

for some K > 0 and γ ∈ (0, 1). The matrix W = XX∗

N

has limiting spectral measure

ργ(x) :=

• ((1 + √γ)2 − x)(x − (1 − √γ)2)

2γπx

his is invertible with probability 1.

The Wishart Ensemble Let X be P × N, i.i.d complex standard normals P < N:

• P

N − γ

• ≤ K

P 2,

for some K > 0 and γ ∈ (0, 1). The matrix W = XX∗

N

has limiting spectral measure

ργ(x) :=

• ((1 + √γ)2 − x)(x − (1 − √γ)2)

2γπx

This is invertible with probability 1.

Covariance Estimation he matrix W is an estimate of the Covariance Matrix (in this case I). he MP-Law show’s W isn’t good in this high-d setting. uantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. s there something closer to I in operator norm? ptimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Covariance Estimation The matrix W is an estimate of the Covariance Matrix (in this case I). he MP-Law show’s W isn’t good in this high-d setting. uantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. s there something closer to I in operator norm? ptimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Covariance Estimation The matrix W is an estimate of the Covariance Matrix (in this case I). The MP-Law show’s W isn’t good in this high-d setting. uantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. s there something closer to I in operator norm? ptimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Covariance Estimation The matrix W is an estimate of the Covariance Matrix (in this case I). The MP-Law show’s W isn’t good in this high-d setting. Quantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. s there something closer to I in operator norm? ptimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Covariance Estimation The matrix W is an estimate of the Covariance Matrix (in this case I). The MP-Law show’s W isn’t good in this high-d setting. Quantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. Is there something closer to I in operator norm? ptimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Covariance Estimation The matrix W is an estimate of the Covariance Matrix (in this case I). The MP-Law show’s W isn’t good in this high-d setting. Quantitatively

lim

P,N→∞ W − I → γ + 2√γ

a.s. Is there something closer to I in operator norm? Optimizing the Frobenius Norm has been done. (Ledoit, Pech´ e, Wolf)

Bear with me here... et’s imagine I have multiple matrices {Xi}n

i=1 all P × N.

uppose we form the average A = 1

n

n

• i=1

Wi, here Wi =

XiX∗ i

N

ust line up the Xi, side by side

lim

P,N→∞ A − I → γ

n + 2 γ n

a.s. loser now.

Bear with me here... Let’s imagine I have multiple matrices {Xi}n

i=1 all P × N.

uppose we form the average A = 1

n

n

• i=1

Wi, here Wi =

XiX∗ i

N

ust line up the Xi, side by side

lim

P,N→∞ A − I → γ

n + 2 γ n

a.s. loser now.

Bear with me here... Let’s imagine I have multiple matrices {Xi}n

i=1 all P × N.

Suppose we form the average A = 1

n

n

• i=1

Wi, here Wi =

XiX∗ i

N

ust line up the Xi, side by side

lim

P,N→∞ A − I → γ

n + 2 γ n

a.s. loser now.

Bear with me here... Let’s imagine I have multiple matrices {Xi}n

i=1 all P × N.

Suppose we form the average A = 1

n

n

• i=1

Wi, here Wi =

XiX∗ i

N

Just line up the Xi, side by side

lim

P,N→∞ A − I → γ

n + 2 γ n

a.s. loser now.

Bear with me here... Let’s imagine I have multiple matrices {Xi}n

i=1 all P × N.

Suppose we form the average A = 1

n

n

• i=1

Wi, here Wi =

XiX∗ i

N

Just line up the Xi, side by side

lim

P,N→∞ A − I → γ

n + 2 γ n

a.s. Closer now.

Alternative Means ake their Harmonic Mean: H = n(W−1

1

+ · · · + W−1

n )−1,

esult: the limiting ESD is

n 2πγx

• (e+ − x)(x − e−)

where

e± = 1 − γ + 2γ n ± 2 γ n

• 1 − γ + γ

n

lso:

lim

P,N→∞ H − I = 1 − e− = γ − 2γ

n + 2 γ n

• 1 − γ + γ

n

Alternative Means Take their Harmonic Mean: H = n(W−1

1

+ · · · + W−1

n )−1,

esult: the limiting ESD is

n 2πγx

• (e+ − x)(x − e−)

where

e± = 1 − γ + 2γ n ± 2 γ n

• 1 − γ + γ

n

lso:

lim

P,N→∞ H − I = 1 − e− = γ − 2γ

n + 2 γ n

• 1 − γ + γ

n

Alternative Means Take their Harmonic Mean: H = n(W−1

1

+ · · · + W−1

n )−1,

Result: the limiting ESD is

n 2πγx

• (e+ − x)(x − e−)

where

e± = 1 − γ + 2γ n ± 2 γ n

• 1 − γ + γ

n

lso:

lim

P,N→∞ H − I = 1 − e− = γ − 2γ

n + 2 γ n

• 1 − γ + γ

n

Alternative Means Take their Harmonic Mean: H = n(W−1

1

+ · · · + W−1

n )−1,

Result: the limiting ESD is

n 2πγx

• (e+ − x)(x − e−)

where

e± = 1 − γ + 2γ n ± 2 γ n

• 1 − γ + γ

n

Also:

lim

P,N→∞ H − I = 1 − e− = γ − 2γ

n + 2 γ n

• 1 − γ + γ

n

Figure of the ESD vs LSD P = 500, N = 1000 and n = 2

Improved Operator Norm Estimate e have the a.s. result

lim

P,N→∞ H − I <

lim

P,N→∞ A − I

for n ≤ n∗(γ). ndeed this is equivalent to

γ − 2γ n + 2 γ n

• 1 − γ + γ

n < γ n + 2 γ n

• r n = 2 it’s always true for γ ∈ (0, 1)
• 2γ
• 1 − γ

2 < γ 2 +

• 2γ

Improved Operator Norm Estimate We have the a.s. result

lim

P,N→∞ H − I <

lim

P,N→∞ A − I

for n ≤ n∗(γ). ndeed this is equivalent to

γ − 2γ n + 2 γ n

• 1 − γ + γ

n < γ n + 2 γ n

• r n = 2 it’s always true for γ ∈ (0, 1)
• 2γ
• 1 − γ

2 < γ 2 +

• 2γ

Improved Operator Norm Estimate We have the a.s. result

lim

P,N→∞ H − I <

lim

P,N→∞ A − I

for n ≤ n∗(γ). Indeed this is equivalent to

γ − 2γ n + 2 γ n

• 1 − γ + γ

n < γ n + 2 γ n

• r n = 2 it’s always true for γ ∈ (0, 1)
• 2γ
• 1 − γ

2 < γ 2 +

• 2γ

Improved Operator Norm Estimate We have the a.s. result

lim

P,N→∞ H − I <

lim

P,N→∞ A − I

for n ≤ n∗(γ). Indeed this is equivalent to

γ − 2γ n + 2 γ n

• 1 − γ + γ

n < γ n + 2 γ n

For n = 2 it’s always true for γ ∈ (0, 1)

• 2γ
• 1 − γ

2 < γ 2 +

• 2γ

Error Comparison for n = 2 as a function of γ

Is this Just an Identity Matrix Fact? nswer: No, but general Covar is tricky. ubmultiplicative bound

• √

ΣH √ Σ − Σ

• ≤
• √

ΣA √ Σ − Σ

• ΣΣ−1H − I

A − I

• if we have

lim sup

P,N→∞

ΣΣ−1H − I A − I < 1

then

lim sup

P,N→∞

• √

ΣH √ Σ − Σ

• √

ΣA √ Σ − Σ

• < 1

Is this Just an Identity Matrix Fact? Answer: No, but general Covar is tricky. ubmultiplicative bound

• √

ΣH √ Σ − Σ

• ≤
• √

ΣA √ Σ − Σ

• ΣΣ−1H − I

A − I

• if we have

lim sup

P,N→∞

ΣΣ−1H − I A − I < 1

then

lim sup

P,N→∞

• √

ΣH √ Σ − Σ

• √

ΣA √ Σ − Σ

• < 1

Is this Just an Identity Matrix Fact? Answer: No, but general Covar is tricky. Submultiplicative bound

• √

ΣH √ Σ − Σ

• ≤
• √

ΣA √ Σ − Σ

• ΣΣ−1H − I

A − I

• if we have

lim sup

P,N→∞

ΣΣ−1H − I A − I < 1

then

lim sup

P,N→∞

• √

ΣH √ Σ − Σ

• √

ΣA √ Σ − Σ

• < 1

Is this Just an Identity Matrix Fact? Answer: No, but general Covar is tricky. Submultiplicative bound

• √

ΣH √ Σ − Σ

• ≤
• √

ΣA √ Σ − Σ

• ΣΣ−1H − I

A − I

So if we have

lim sup

P,N→∞

ΣΣ−1H − I A − I < 1

then

lim sup

P,N→∞

• √

ΣH √ Σ − Σ

• √

ΣA √ Σ − Σ

• < 1

Condition on the Condition Number! he ratio

c := ΣΣ−1 = λmax(Σ) λmin(Σ)

is the condition number of Σ

• r γ = 1

2 the condition number can be below

c < 5 4

• 4

3 ≈ 1.44337567 . . .

and the result still holds.

• re on general Σ later.

Condition on the Condition Number! The ratio

c := ΣΣ−1 = λmax(Σ) λmin(Σ)

is the condition number of Σ

• r γ = 1

2 the condition number can be below

c < 5 4

• 4

3 ≈ 1.44337567 . . .

and the result still holds.

• re on general Σ later.

Condition on the Condition Number! The ratio

c := ΣΣ−1 = λmax(Σ) λmin(Σ)

is the condition number of Σ For γ = 1

2 the condition number can be below

c < 5 4

• 4

3 ≈ 1.44337567 . . .

and the result still holds.

• re on general Σ later.

Condition on the Condition Number! The ratio

c := ΣΣ−1 = λmax(Σ) λmin(Σ)

is the condition number of Σ For γ = 1

2 the condition number can be below

c < 5 4

• 4

3 ≈ 1.44337567 . . .

and the result still holds. More on general Σ later.

Applications: Data Splitting uppose T = nN is my total observations, define

Γ := lim

P,T→∞

P T = γ n ∈

• 0, 1

2

• hen

lim

P,T→∞ A − I = Γ + 2

√ Γ

and

lim

P,T→∞ H − I = (n − 2)Γ +

√ Γ

• 1 − (n − 1)Γ

he argmin is 2! If T is at least twice P split your data in two and take the harmonic mean.

Applications: Data Splitting Suppose T = nN is my total observations, define

Γ := lim

P,T→∞

P T = γ n ∈

• 0, 1

2

• hen

lim

P,T→∞ A − I = Γ + 2

√ Γ

and

lim

P,T→∞ H − I = (n − 2)Γ +

√ Γ

• 1 − (n − 1)Γ

he argmin is 2! If T is at least twice P split your data in two and take the harmonic mean.

Applications: Data Splitting Suppose T = nN is my total observations, define

Γ := lim

P,T→∞

P T = γ n ∈

• 0, 1

2

• Then

lim

P,T→∞ A − I = Γ + 2

√ Γ

and

lim

P,T→∞ H − I = (n − 2)Γ +

√ Γ

• 1 − (n − 1)Γ

he argmin is 2! If T is at least twice P split your data in two and take the harmonic mean.

Applications: Data Splitting Suppose T = nN is my total observations, define

Γ := lim

P,T→∞

P T = γ n ∈

• 0, 1

2

• Then

lim

P,T→∞ A − I = Γ + 2

√ Γ

and

lim

P,T→∞ H − I = (n − 2)Γ +

√ Γ

• 1 − (n − 1)Γ

The argmin is 2! If T is at least twice P split your data in two and take the harmonic mean.

Proof of Results (Techniques) eed Xi to be P × N complex gaussian entries and

• P

N − γ

• ≤ K

P 2

for some K. hen Wi are asymptotically free and Q is a non-commutative polynomial (result due to Donati-Martin and Capitaine)

lim

P,N→∞ Q(W1, . . . , Wn) = Q(p1, . . . , pn)F

the variables pj are freely independent non-commutative Free Poisson Random Variables.

Proof of Results (Techniques) Need Xi to be P × N complex gaussian entries and

• P

N − γ

• ≤ K

P 2

for some K. hen Wi are asymptotically free and Q is a non-commutative polynomial (result due to Donati-Martin and Capitaine)

lim

P,N→∞ Q(W1, . . . , Wn) = Q(p1, . . . , pn)F

the variables pj are freely independent non-commutative Free Poisson Random Variables.

Proof of Results (Techniques) Need Xi to be P × N complex gaussian entries and

• P

N − γ

• ≤ K

P 2

for some K. Then Wi are asymptotically free and Q is a non-commutative polyno- mial (result due to Donati-Martin and Capitaine)

lim

P,N→∞ Q(W1, . . . , Wn) = Q(p1, . . . , pn)F

the variables pj are freely independent non-commutative Free Poisson Random Variables.

Free Probability Calculation he pi can be thought of as bounded linear operators on some Hilbert space whose spectrum is the MP-law with parameter γ here is a unit vector e0 in the hilbert space such that

ν(pk

j) := e0, pk je0 =

• xkρMP,γ(dx),

ree independence means

ν n

• l=1
• Ql(pl) − ν[Ql(pl)]
• = 0

Free Probability Calculation The pi can be thought of as bounded linear operators on some Hilbert space whose spectrum is the MP-law with parameter γ here is a unit vector e0 in the hilbert space such that

ν(pk

j) := e0, pk je0 =

• xkρMP,γ(dx),

ree independence means

ν n

• l=1
• Ql(pl) − ν[Ql(pl)]
• = 0

Free Probability Calculation The pi can be thought of as bounded linear operators on some Hilbert space whose spectrum is the MP-law with parameter γ There is a unit vector e0 in the hilbert space such that

ν(pk

j) := e0, pk je0 =

• xkρMP,γ(dx),

ree independence means

ν n

• l=1
• Ql(pl) − ν[Ql(pl)]
• = 0

Free Probability Calculation The pi can be thought of as bounded linear operators on some Hilbert space whose spectrum is the MP-law with parameter γ There is a unit vector e0 in the hilbert space such that

ν(pk

j) := e0, pk je0 =

• xkρMP,γ(dx),

Free independence means

ν n

• l=1
• Ql(pl) − ν[Ql(pl)]
• = 0

More Proof Ingredients ree random variables have an explicit way of computing their laws (Voiculescu). et µ be a measure compactly supported on the positive reals,

mµ(z) = µ(dx) z − x Kµ(mµ(z)) = mµ(Kµ(z)) = z.

efine µ1 ⊞ µ2 as the measure such that

Rµ(z) := Kµ(z) − 1 z Rµ1⊞µ2(z) = Rµ1(z) + Rµ2(z)

More Proof Ingredients Free random variables have an explicit way of computing their laws (Voiculescu). et µ be a measure compactly supported on the positive reals,

mµ(z) = µ(dx) z − x Kµ(mµ(z)) = mµ(Kµ(z)) = z.

efine µ1 ⊞ µ2 as the measure such that

Rµ(z) := Kµ(z) − 1 z Rµ1⊞µ2(z) = Rµ1(z) + Rµ2(z)

More Proof Ingredients Free random variables have an explicit way of computing their laws (Voiculescu). Let µ be a measure compactly supported on the positive reals,

mµ(z) = µ(dx) z − x Kµ(mµ(z)) = mµ(Kµ(z)) = z.

efine µ1 ⊞ µ2 as the measure such that

Rµ(z) := Kµ(z) − 1 z Rµ1⊞µ2(z) = Rµ1(z) + Rµ2(z)

More Proof Ingredients Free random variables have an explicit way of computing their laws (Voiculescu). Let µ be a measure compactly supported on the positive reals,

mµ(z) = µ(dx) z − x Kµ(mµ(z)) = mµ(Kµ(z)) = z.

Define µ1 ⊞ µ2 as the measure such that

Rµ(z) := Kµ(z) − 1 z Rµ1⊞µ2(z) = Rµ1(z) + Rµ2(z)

More Proof Ingredients e add the R-transforms and work backwards to get the spectrum. e are studying is H which we want to say (more on this) converges to h := n(p−1

1

+ · · · + p−1

n )−1

f we compute

nRp−1(z) =

n

• i=1

Rp−1

i (z)

we can obtain the Stieltjest transform of

nh−1

More Proof Ingredients We add the R-transforms and work backwards to get the spectrum. e are studying is H which we want to say (more on this) converges to h := n(p−1

1

+ · · · + p−1

n )−1

f we compute

nRp−1(z) =

n

• i=1

Rp−1

i (z)

we can obtain the Stieltjest transform of

nh−1

More Proof Ingredients We add the R-transforms and work backwards to get the spectrum. We are studying is H which we want to say (more on this) converges to h := n(p−1

1

+ · · · + p−1

n )−1

f we compute

nRp−1(z) =

n

• i=1

Rp−1

i (z)

we can obtain the Stieltjest transform of

nh−1

More Proof Ingredients We add the R-transforms and work backwards to get the spectrum. We are studying is H which we want to say (more on this) converges to h := n(p−1

1

+ · · · + p−1

n )−1

If we compute

nRp−1(z) =

n

• i=1

Rp−1

i (z)

we can obtain the Stieltjest transform of

nh−1

More Proof Ingredients rick due to Edelman and Rao ach mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

his means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

f you directly invert to get Kp−1

j (z) you are doing more work than you

need to. lug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients Trick due to Edelman and Rao ach mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

his means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

f you directly invert to get Kp−1

j (z) you are doing more work than you

need to. lug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients Trick due to Edelman and Rao Each mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

his means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

f you directly invert to get Kp−1

j (z) you are doing more work than you

need to. lug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients Trick due to Edelman and Rao Each mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

This means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

f you directly invert to get Kp−1

j (z) you are doing more work than you

need to. lug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients Trick due to Edelman and Rao Each mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

This means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

If you directly invert to get Kp−1

j (z) you are doing more work than you

need to. lug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients Trick due to Edelman and Rao Each mpj(z) satisfies a quadratic

γzmpj(z)2 + mpj(z)(1 − z − γ) + 1 = 0.

This means each p−1

j

satisfies a quadratic

γz2mp−1

j (z)2 − mp−1 j (z)[z(1 + γ) − 1] + 1 = 0.

If you directly invert to get Kp−1

j (z) you are doing more work than you

need to. Plug in z = Kp−1

j (w) and then plug in the R-transform.

More Proof Ingredients

• u get the quadratic

γzRp−1(z)2 + (γ − 1)Rp−1(z) + 1 = 0.

anipulate some more and you get

γz n mh(z)2 + (1 − γ − z)mh(z) + 1 = 0.

alculation is quick but how to justify and why does the operator norm converge?

More Proof Ingredients You get the quadratic

γzRp−1(z)2 + (γ − 1)Rp−1(z) + 1 = 0.

anipulate some more and you get

γz n mh(z)2 + (1 − γ − z)mh(z) + 1 = 0.

alculation is quick but how to justify and why does the operator norm converge?

More Proof Ingredients You get the quadratic

γzRp−1(z)2 + (γ − 1)Rp−1(z) + 1 = 0.

Manipulate some more and you get

γz n mh(z)2 + (1 − γ − z)mh(z) + 1 = 0.

alculation is quick but how to justify and why does the operator norm converge?

More Proof Ingredients You get the quadratic

γzRp−1(z)2 + (γ − 1)Rp−1(z) + 1 = 0.

Manipulate some more and you get

γz n mh(z)2 + (1 − γ − z)mh(z) + 1 = 0.

Calculation is quick but how to justify and why does the operator norm converge?

More Proof Ingredients nly have non-commutative polynomials in Wi. se Neumann series to approximate W−1

i

and then H. se operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! nd AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) e prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

More Proof Ingredients Only have non-commutative polynomials in Wi. se Neumann series to approximate W−1

i

and then H. se operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! nd AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) e prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

More Proof Ingredients Only have non-commutative polynomials in Wi. Use Neumann series to approximate W−1

i

and then H. se operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! nd AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) e prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

More Proof Ingredients Only have non-commutative polynomials in Wi. Use Neumann series to approximate W−1

i

and then H. Use operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! nd AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) e prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

More Proof Ingredients Only have non-commutative polynomials in Wi. Use Neumann series to approximate W−1

i

and then H. Use operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! And AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) e prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

More Proof Ingredients Only have non-commutative polynomials in Wi. Use Neumann series to approximate W−1

i

and then H. Use operator norm bounds on smallest singular value of Xi due to Rudel- son and Vershynin — applies for subgaussian! And AMHM inequality: P(H > t) ≤ P(A > t) ≤ nP(W1 > t) We prove that there exists a κ > 0: P({max (Wi, W−1

i , H, H−1) > κ})

is summable in P .

General Covar Fixed Point Equation air of fixed point equation for e = lim

√ ΣH √ Σ − Σ me(z) =

• F(dx)

z − γx{ 1

n(zme(z) − 1)Sh−1(zme(z) − 1) + 1 nzme(z) − 1},

and

γ nzSh−1(z)2 + γ(1 + z) n Sh−1(z) − γSh−1(z) − 1 = 0

dentify the edge for any F ?

General Covar Fixed Point Equation Pair of fixed point equation for e = lim

√ ΣH √ Σ − Σ me(z) =

• F(dx)

z − γx{ 1

n(zme(z) − 1)Sh−1(zme(z) − 1) + 1 nzme(z) − 1},

and

γ nzSh−1(z)2 + γ(1 + z) n Sh−1(z) − γSh−1(z) − 1 = 0

dentify the edge for any F ?

General Covar Fixed Point Equation Pair of fixed point equation for e = lim

√ ΣH √ Σ − Σ me(z) =

• F(dx)

z − γx{ 1

n(zme(z) − 1)Sh−1(zme(z) − 1) + 1 nzme(z) − 1},

and

γ nzSh−1(z)2 + γ(1 + z) n Sh−1(z) − γSh−1(z) − 1 = 0

Identify the edge for any F ?

General Covar Fixed Point Equation Pair of fixed point equation for e = lim

√ ΣH √ Σ − Σ me(z) =

• F(dx)

z − γx{ 1

n(zme(z) − 1)Sh−1(zme(z) − 1) + 1 nzme(z) − 1},

and

γ nzSh−1(z)2 + γ(1 + z) n Sh−1(z) − γSh−1(z) − 1 = 0

Identify the edge for any F ?

Acknowledgments and Advertisement Anna Maltsev is looking for a PhD Student at Queen Mary University

• f London. Ask me more questions

Thanks to Alice Guionnet, Alan Edelman and Jinho Baik for helpful comments and suggestions.