Moment problem

Existence

A sequence of numbers ${\displaystyle m_n}$ is the sequence of moments of a measure ${\displaystyle \mu }$ if and only if a certain positivity condition is fulfilled; namely, the Hankel matrices ${\displaystyle H_n}$,

${\displaystyle (H_n{)}_{ij}=m_{i+j},}$

should be positive semi-definite. This is because a positive-semidefinite Hankel matrix corresponds to a linear functional ${\displaystyle \mathrm{\Lambda }}$ such that ${\displaystyle \mathrm{\Lambda }(x^n)=m_n}$ and ${\displaystyle \mathrm{\Lambda }(f^2)\ge 0}$ (non-negative for sum of squares of polynomials). Assume ${\displaystyle \mathrm{\Lambda }}$ can be extended to ${\displaystyle \mathbb{R}[x{]}^{\ast }}$. In the univariate case, a non-negative polynomial can always be written as a sum of squares. So the linear functional ${\displaystyle \mathrm{\Lambda }}$ is positive for all the non-negative polynomials in the univariate case. By Haviland's theorem, the linear functional has a measure form, that is ${\displaystyle \mathrm{\Lambda }(x^n)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{n}d\mu }$. A condition of similar form is necessary and sufficient for the existence of a measure ${\displaystyle \mu }$ supported on a given interval ${\displaystyle [a,b]}$.

One way to prove these results is to consider the linear functional ${\displaystyle \phi }$ that sends a polynomial

${\displaystyle P(x)=\sum _k{a}_k{x}^k}$

to

${\displaystyle \sum _k{a}_k{m}_k.}$

If ${\displaystyle m_k}$ are the moments of some measure ${\displaystyle \mu }$ supported on ${\displaystyle [a,b]}$, then evidently

Vice versa, if (1) holds, one can apply the M. Riesz extension theorem and extend ${\displaystyle \phi }$ to a functional on the space of continuous functions with compact support ${\displaystyle C_c([a,b])}$), so that

By the Riesz representation theorem, (2) holds iff there exists a measure ${\displaystyle \mu }$ supported on ${\displaystyle [a,b]}$, such that

${\displaystyle \phi (f)=\int fd\mu }$

for every ${\displaystyle f\in C_c([a,b])}$.

Thus the existence of the measure ${\displaystyle \mu }$ is equivalent to (1). Using a representation theorem for positive polynomials on ${\displaystyle [a,b]}$, one can reformulate (1) as a condition on Hankel matrices.

Uniqueness (or determinacy)

The uniqueness of ${\displaystyle \mu }$ in the Hausdorff moment problem follows from the Weierstrass approximation theorem, which states that polynomials are dense under the uniform norm in the space of continuous functions on ${\displaystyle [0,1]}$. For the problem on an infinite interval, uniqueness is a more delicate question. There are distributions, such as log-normal distributions, which have finite moments for all the positive integers but where other distributions have the same moments.

Formal solution

When the solution exists, it can be formally written using derivatives of the Dirac delta function as

${\displaystyle d\mu (x)=\rho (x)dx,\rho (x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}{\delta }^{(n)}(x)m_n}$.

The expression can be derived by taking the inverse Fourier transform of its characteristic function.

Variations

An important variation is the truncated moment problem, which studies the properties of measures with fixed first k moments (for a finite k). Results on the truncated moment problem have numerous applications to extremal problems, optimisation and limit theorems in probability theory.

Probability

The moment problem has applications to probability theory. The following is commonly used:

Theorem (Fréchet-Shohat)—If $\mu$ is a determinate measure (i.e. its moments determine it uniquely), and the measures ${\mu }_n$ are such that $${\displaystyle \mathrm{\forall }k\ge 0\underset{n\to \mathrm{\infty }}{lim}{m}_k\left[{\mu }_n\right]=m_k[\mu ],}$$ then ${\mu }_n\to \mu$ in distribution.

By checking Carleman's condition, we know that the standard normal distribution is a determinate measure, thus we have the following form of the central limit theorem:

Corollary—If a sequence of probability distributions ${\nu }_n$ satisfy $${\displaystyle m_{2k}[{\nu }_n]\to \frac{(2k)!}{2^{k}k!};m_{2k+1}[{\nu }_n]\to 0}$$ then ${\nu }_n$ converges to $N(0,1)$ in distribution.