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Hamburger moment problem

From Wikipedia, the free encyclopedia
Probability problem

Inmathematics, theHamburgermoment problem, named afterHans Ludwig Hamburger, is formulated as follows: given asequence(m0,m1,m2, ...), does there exist a positiveBorel measureμ (for instance, the measure determined by thecumulative distribution function of arandom variable) on thereal line such that

mn=xndμ(x){\displaystyle m_{n}=\int _{-\infty }^{\infty }x^{n}\,d\mu (x)}?

In other words, an affirmative answer to the problem means that(m0,m1,m2, ...) is the sequence ofmoments of some positive Borel measureμ.

TheStieltjes moment problem,Vorobyev moment problem, and theHausdorff moment problem are similar but replace the real line by[0,+){\displaystyle [0,+\infty )} (Stieltjes and Vorobyev; but Vorobyev formulates the problem in the terms of matrix theory), or abounded interval (Hausdorff).

Characterization

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The Hamburger moment problem is solvable (that is,(mn) is a sequence ofmoments) if and only if the correspondingHankel kernel on the nonnegativeintegers

A=(m0m1m2m1m2m3m2m3m4){\displaystyle A=\left({\begin{matrix}m_{0}&m_{1}&m_{2}&\cdots \\m_{1}&m_{2}&m_{3}&\cdots \\m_{2}&m_{3}&m_{4}&\cdots \\\vdots &\vdots &\vdots &\ddots \end{matrix}}\right)}

ispositive definite, i.e.,

j,k0mj+kcjck¯0{\displaystyle \sum _{j,k\geq 0}m_{j+k}c_{j}{\overline {c_{k}}}\geq 0}

for every arbitrary sequence(cj)j ≥ 0 ofcomplex numbers that are finitary (i.e.,cj = 0 except for finitely many values ofj).

For the "only if" part of the claims simply note that

j,k0mj+kcjck¯=|j0cjxj|2dμ(x){\displaystyle \sum _{j,k\geq 0}m_{j+k}c_{j}{\overline {c_{k}}}=\int _{-\infty }^{\infty }\left|\sum _{j\geq 0}c_{j}x^{j}\right|^{2}\,d\mu (x)},

which is non-negative ifμ{\displaystyle \mu } is non-negative.

We sketch an argument for the converse. LetZ+ be the nonnegative integers andF0(Z+) denote the family of complex valued sequences with finitary support. The positive Hankel kernelA induces a (possibly degenerate)sesquilinear product on the family of complex-valued sequences with finite support. This in turn gives aHilbert space

(H,,){\displaystyle ({\mathcal {H}},\langle \;,\;\rangle )}

whose typical element is anequivalence class denoted by[f].

Leten be the element inF0(Z+) defined byen(m) =δnm. One notices that

[en+1],[em]=Am,n+1=mm+n+1=[en],[em+1]{\displaystyle \langle [e_{n+1}],[e_{m}]\rangle =A_{m,n+1}=m_{m+n+1}=\langle [e_{n}],[e_{m+1}]\rangle }.

Therefore, theshift operatorT onH{\displaystyle {\mathcal {H}}}, withT[en] = [en + 1], issymmetric.

On the other hand, the desired expression

mn=xndμ(x){\displaystyle m_{n}=\int _{-\infty }^{\infty }x^{n}\,d\mu (x)}

suggests thatμ is thespectral measure of aself-adjoint operator. (More precisely stated,μ is the spectral measure for an operatorT¯{\displaystyle {\overline {T}}} defined below and the vector [1], (Reed & Simon 1975, p. 145)). If we can find a "function model" such that the symmetric operatorT ismultiplication byx, then the spectral resolution of aself-adjoint extension ofT proves the claim.

A function model is given by the natural isomorphism fromF0(Z+) to the family ofpolynomials, in one single real variable and complex coefficients: forn ≥ 0, identifyen withxn. In the model, the operatorT is multiplication byx and a densely defined symmetric operator. It can be shown thatT always has self-adjoint extensions. LetT¯{\displaystyle {\overline {T}}} be one of them andμ be its spectral measure. So

T¯n[1],[1]=xndμ(x){\displaystyle \langle {\overline {T}}^{n}[1],[1]\rangle =\int x^{n}d\mu (x)}.

On the other hand,

T¯n[1],[1]=Tn[e0],[e0]=mn{\displaystyle \langle {\overline {T}}^{n}[1],[1]\rangle =\langle T^{n}[e_{0}],[e_{0}]\rangle =m_{n}}.

For an alternative proof of the existence that only usesStieltjes integrals, see also,[1] in particular theorem 3.2.

Uniqueness of solutions

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The solutions form a convex set, so the problem has either infinitely many solutions or a unique solution.

Consider the(n + 1) × (n + 1)Hankel matrix

Δn=[m0m1m2mnm1m2m3mn+1m2m3m4mn+2mnmn+1mn+2m2n]{\displaystyle \Delta _{n}=\left[{\begin{matrix}m_{0}&m_{1}&m_{2}&\cdots &m_{n}\\m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\m_{2}&m_{3}&m_{4}&\cdots &m_{n+2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\m_{n}&m_{n+1}&m_{n+2}&\cdots &m_{2n}\end{matrix}}\right]}.

Positivity ofA means that, for eachn,det(Δn) ≥ 0. Ifdet(Δn) = 0, for somen, then

(H,,){\displaystyle ({\mathcal {H}},\langle \;,\;\rangle )}

is finite-dimensional andT is self-adjoint. So in this case the solution to the Hamburger moment problem is unique andμ, being the spectral measure ofT, has finite support.

More generally, the solution is unique if there are constantsC andD such that, for alln,|mn| ≤CDnn! (Reed & Simon 1975, p. 205). This follows from the more generalCarleman's condition.

There are examples where the solution is not unique; see e.g.[2]

Polynomials

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The Hamburger moment problem is intimately related toorthogonal polynomials on the real line. That is, assume{mn}nN0{\displaystyle \{m_{n}\}_{n\in \mathbb {N} _{0}}} is the moment sequence of some positive measureμ{\displaystyle \mu } onR{\displaystyle \mathbb {R} }. Then for any polynomialp(x)=j=0najxjR,{\displaystyle p(x)=\sum _{j=0}^{n}a_{j}x^{j}\in \mathbb {R} ,} it holds thatp(x)2dμ=(j,k=0najakxj+k)dμ=j,k=0najakmj+k0,nN0,{\displaystyle \int p(x)^{2}d\mu =\int \left(\sum _{j,k=0}^{n}a_{j}a_{k}x^{j+k}\right)d\mu =\sum _{j,k=0}^{n}a_{j}a_{k}m_{j+k}\geq 0,\quad \forall n\in \mathbb {N} _{0},} such that the Hankel matrix is positive semidefinite. This is a necessary condition for a sequence to be a moment sequence and a sufficient condition for the existence of a positive measure.[3]

TheGram–Schmidt procedure gives a basis of orthogonal polynomials in which the operator:T¯{\displaystyle {\overline {T}}} has a tridiagonalJacobi matrix representation. This in turn leads to atridiagonal model of positive Hankel kernels.

An explicit calculation of theCayley transform ofT shows the connection with what is called theNevanlinna class of analytic functions on the left half plane. Passing to the non-commutative setting, this motivatesKrein's formula which parametrizes the extensions of partial isometries.

The cumulative distribution function and the probability density function can often be found by applying the inverseLaplace transform to the moment generating function

m(t)=n=0mntnn!{\displaystyle m(t)=\sum _{n=0}m_{n}{\frac {t^{n}}{n!}}},

provided that this function converges.

Notes

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  1. ^Chihara 1978, p. 56.
  2. ^Chihara 1978, p. 73.
  3. ^Schmüdgen 2017, p. 2.

References

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  • Chihara, T.S. (1978),An Introduction to Orthogonal Polynomials, Gordon and Breach, Science Publishers,ISBN 0-677-04150-0
  • Reed, Michael; Simon, Barry (1975),Fourier Analysis, Self-Adjointness, Methods of modern mathematical physics, vol. 2, Academic Press, pp. 145, 205,ISBN 0-12-585002-6
  • Schmüdgen, Konrad (2017),The Moment Problem, Graduate Texts in Mathematics, vol. 277, Cham: Springer International Publishing,doi:10.1007/978-3-319-64546-9,ISBN 978-3-319-64545-2,ISSN 0072-5285
  • Shohat, J. A.; Tamarkin, J. D. (1943),The Problem of Moments, New York: American mathematical society,ISBN 0-8218-1501-6{{citation}}:ISBN / Date incompatibility (help).
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