the moment problems

The Moment Problems

Abhishek Bhardwaj

Laboratoire d’analyse et d’architectures des systemes

January 27, 2021

Abhishek Bhardwaj (LAAS) LAAS Talk January 27, 2021 :

The Classical Moment Problem

Given a sequence of real numbers (an)n∈N0 , the Classical MomentProblems entails finding a measure µ such that

an =

∫Kxn dµ(x),

where K is R (Hamburger), [0,+∞) (Stieltjes) or [0, 1] (Hausdorff).


Classical Solution

Each of these can be solved by considering the Hankel matrix associatedto the moments,

H =

a0 a1 a2 . . .

a1 a2. . . . . .

a2. . .

. . . . . .... . . . . . .

. . .

(or a variation of it).


Extending to higher dimensions

We extend the problem R2. Given a multiindexed sequence β ≡ (βij)i ,j∈Nin R, does there exist a positive Borel measure µ on R2 such that

βij =

∫R2

x iy j dµ.

We call such a measure µ, a representing measure for β.

Given β, the Riesz functional of β, Lβ : R[x , y ]→ R is defined as

Lβ

∑i ,j∈N

fi ,jxiy j

=∑i ,j∈N

fi ,jβi ,j .


Solution!

A complete solution to the moment problem, is given by the followingtheorem,

Theorem (Riesz-Haviland)

Given β ≡ (βij)i ,j∈N, there exists a representing measure µ for β if andonly if

Lβ(f ) ≥ 0

for all polynomials f which are non-negative on R2.

...or is it?


The Truncated Moment Problem

Stochel’s Theorem

β ≡ (βij) has representing measure µ if and only if for each n ∈ N, thetruncation of β toβ(2n) : β00, β01, β10, . . . , β0,2n, β1,2n−1, . . . , β2n−1,1, β2n,0, has arepresenting measure.

Given a multiindexed sequence β ≡ β(2n) : β00, β01, β10, . . . , β0,2n, β2n,0,when does β have a representing measure?

Theorem (Bayer-Teichmann)

If a truncated sequence β(2n) has a representing measure µ, then it hasanother representing measure ν which is finitely atomic withsupp(ν) ⊆ supp(µ), i.e., there is a set {(x1, y1), . . . , (xk , yk)} such that

βi ,j =k∑`=1

λ`xi`y

j` .


The Truncated Moment Problem

The solution to this problem comes by analyzing the column relations inthe associated multivariate Hankel matrix,

M(n) =

1 X Y . . . Xn . . . Yn

1 β00 β10 β01 . . . βn,0 . . . β0,n

X β10 β20 β11 . . . βn+1,0 . . . β1,n

Y β01 β11 β02 . . . βn,1 . . . β0,n+1

......

......

. . ....

......

Xn βn,0 βn+1,0 βn,1 . . . β2n,0 . . . βn,n...

......

......

......

...Yn β0,n β1,n β0,n+1 . . . βn,n . . . β0,2n

.


Notation and Definitions

Given a multivariate Hankel matrix, M(n) and a polynomialp =

∑pijx

iy j ∈ R[x , y ]n, we let p(X,Y) :=∑

pijXiYj .

A column relation in M(n) is always described as p(X,Y) = 0 for somep ∈ R[x , y ]n.

For example, in the following M(1), we have that p(X,Y) = X + Y = 0.

M(1) =

1 1 −11 2 −2−1 −2 2

.


Notation and Definitions

We define the algebraic variety of β by

V(β) :=⋂

p∈R[x ,y ]n, p(X,Y)=0

Z(p)

where Z(p) is the zero set of p.

We say that the matrix M(n) is recursively generated if wheneverp(X,Y) = 0, and deg(pq) ≤ n, then (pq)(X,Y) = 0.


Classic Results

Theorem (Curto & Fialkow)

If µ is a representing measure for β, then for p ∈ R[x , y ]n,

p(X,Y) = 0⇔ supp µ ⊆ Z(p).


If β admits a representing measure then M(n) is positive semi-definite,recursively generated and satisfies rank M(n) ≤ card V(β).


Classic Results


β admits a representing measure if and only if M(n) ≥ 0 admits anextension to M(n + k) which then admits a rank preserving extensionM(n + k + 1).


β admits a rank M(n)-atomic measure if and only if M(n) ≥ 0 admits anextension to M(n + 1) which is rank preserving.

Using this theorem we are trying to build a flat extension, i.e., anextension of the form

M(n + 1) =

(M(n) BB∗ C

)such that B =M(n)W and C = B∗W for some matrix W .


The Problem

Of course, the extension B and C have to satisfy some constraints.Namely,

B =

Xn+1 . . . Yn+1 1 βn+1,0 . . . β0,n+1

......

. . ....

Xn β2n+1,0 . . . βn,n+1

and

C =

Xn+1 . . . Yn+1 Xn+1 β2n+2,0 . . . βn+1,n+1

......

. . ....

Yn+1 βn+1,n+1 . . . β0,2n+2


Applications!

The Truncated Moment Problem:

Probability theory

Markov Chains

Mathematical Finance

Bounds on Linear PDE’s

Minimum covering ellipsoid problem

The Truncated Tracial Moment Problem:

Control theory and operations research

Quantum information science

Statistical quantum mechanics


The Truncated Tracial Moment Problem

Let 〈X ,Y 〉 denote the free monoid generated by X and Y .

Two words v ,w ∈ 〈X ,Y 〉 are cyclically equivalent (w ∼ v) if there arewords u1, u2 ∈ 〈X ,Y 〉 such that w = u1u2 and v = u2u1.

For a word w ∈ 〈X ,Y 〉 let w∗ be the reverse of w .

A sequence of numbers (λw ) indexed by words w ∈ 〈X ,Y 〉 satisfying

λw = λv whenever w ∼ v , λw = λw∗ for all w ,

and λφ = 1 is called a (normalized) tracial sequence.



Given n ∈ N and A,B ∈ SRn×n, the sequence given by

λw := Tr(w(A,B)) =1

ttr(w(A,B))

is a tracial sequence.

We are interested in the converse.



For which sequence λ ≡ (λw )≤k , where the length of w is at most k , dothere exist

N ∈ Nµi ≥ 0, with

∑Ni µi = 1

n ∈ NA(i),B(i) ∈ SRn×n

such that,

λw =N∑i

µiTr(w(A,B))?

When this is true, we call λ a truncated tracial moment sequence.



Theorem (Burgdorf & Klep)

Let λ ≡ (λw )≤4 be a truncated tracial sequence. If the associatedmultivariate Hankel matrix if positive definite, then λ is a truncated tracialmoment sequence.

This theorem solves the problem when Rank(MNC (2)) = 7.


Singular MNC - Rank 4

Suppose the Hankel matrix MNC (2) of rank 4. Then the set{1,X,Y,XY} is a basis for the column space of MNC (2). Write

X2 = a11 + b1X + c1Y + d1XY,YX = a21 + b2X + c2Y + d2XY,Y2 = a31 + b3X + c3Y + d3XY,

where aj , bj , cj , dj ∈ R for j = 1, 2, 3.



Theorem (Rank 4 Classification)

The following statements are true:

1 d1 = d3 = 0, d2 = −1.

2 λ admits a nc measure ⇔ MNC (2) has a flat extension to a momentmatrix MNC (3) ⇔ MNC (2) is positive semidefinite andc1 = b3 = 0, b2 = c3, c2 = b1.

3 The minimal measure is a unique atom (X ,Y ) ∈ (SR2×2)2 given by

X =

√a1 +b2

14

0

0 −√

a1 +b2

14

+b1

2· I2,

Y =

√a3 +

c23

4·(

a2

12

√4− a2

12

√4− a2 − a

2

)+

c3

2· I2,

where a = 4a2+2b1c3√(4a1+b2

1)(4a3+c23 )

.


Singular MNC

Proposition (Canonical Reductions)

Suppose the matrix MNC (2) is of rank 5 or 6. If λ admits a nc measure,then:

1 If MNC (2) is of rank 5, then it satisfies XY + YX = 0 and one of thefollowing relations:

Basic relation 1 X2 + Y2 = 1,Basic relation 2 Y2 = 1,Basic relation 3 Y2 − X2 = 1,Basic relation 4 Y2 = X2.

2 IfMNC (2) is of rank 6, then it satisfies one of the following relations:

Basic relation 1 Y2 = 1− X2,Basic relation 2 Y2 = 1 + X2,Basic relation 3 XY + YX = 0,Basic relation 4 Y2 = 1.



Theorem (Rank 5)

Suppose MNC (2) is of rank 5 satisfying the relations XY + YX = 0 andX2 + Y2 = 1. Then λ admits a nc measure if and only if

|λY | < 1− |λX |, |λX | < λX 2 < 1− |λY |, c ≤ λX 4 < λX 2 , (1)

where

c =−λ2

X 2 − |λX |+ 2λX 2 |λX |+ |λYλX |−1 + |λY |+ |λX |

.



Suppose MNC (2) is of rank 6 satisfying the relation Y2 = 1− X2. LetL(a, b, c , d , e) be the following linear matrix polynomial

a λX λY b c c a− bλX b c λX 3 λX 2Y λX 2Y λX − λX 3

λY c a− b λX 2Y λX − λX 3 λX − λX 3 λY − λX 2Y

b λX 3 λX 2Y d e e b − dc λX 2Y λX − λX 3 e b − d b − d c − ec λX 2Y λX − λX 3 e b − d b − d c − e

a− b λX − λX 3 λY − λX 2Y b − d c − e c − e a− 2b + d

,

where a, b, c , d , e ∈ R.



Theorem (Rank 6 to LMI)

λ admits a nc measure if and only if there exist a, b, c , d , e ∈ R such that

1 L(a, b, c , d , e) � 0,

2 MNC (2)− L(a, b, c , d , e) � 0,

3 (MNC (2)− L(a, b, c , d , e)){1,X,Y,XY} � 0,

4 rank(L(a, b, c , d , e)) ≤ cardVL, where VL is the variety associated tothe moment matrix L(a, b, c, d , e).


Beyond Quadratic Varieties - Elliptic Moment Problem

Suppose that MNC (n) is positive semi-definite, recursively generated andin the column space of MNC (n) we have only the relations

Y2 = X3 + aX + b1, (2)

and RG relations via (2), i.e.,V(MNC (n)) = {(X ,Y ) ∈ (SRt×t)2 : Y 2 = X 3 + aX + bIs}. We call sucha moment matrix Elliptic-pure


Elliptic Moment Problem - Results

Theorem (Tracial to Commutative)

Let MNC (n) be a positive semi-definite, recursively generated momentmatrix which satisfies Y2 = X3 + aX + b1. If a ≥ 0 and MNC (n) has arepresenting measure µ, then MNC (n) is a commutative moment matrix,and has a commutative representing measure µcm.

Theorem (Commutative Flat Extension)

Let n ≥ 3. Suppose that MNC (n) is positive and Elliptic-pure. A flatextension MNC (n + 1) of MNC (n) exists if the quartic polynomial Q(θ)has a real root.


Thank you.


the moment problems

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