Truncated Sieltjes r-atomic moment problem

Question

Given a certain $n \in \mathbb{N}$, can i construct a discrete positive random variable $X$ that fullfill the following conditions :

$$\forall k \in \{1,...,n\}, \mathbb{E}(X^k) = \mu_k$$

for some $\mu_k$ given (parameters). We suppose that there exists such a random variable (in facts, the $\mu$'s come from a random variable that i want to estimate).

For exemple, for $n=1$, it is enough to set a dirac in $\mu_1$. for $n=2$, take a radom variable with 2 atoms $\mu_1 +x$ and $\mu_1 -x$ with equal probability and choose x such that :

$$\mu_1^2 + x^2 = \mu_2$$

giving $x = \sqrt{\mu_2 - \mu_1^2}$.

What about $n \ge 3$ ?

Edits :

1° Yes, it is possible (prooved to be possible by many diffrent papers, notably Tchakaloff's theorems). 2° Yes, there exists some algorithm to do it, but i did not found a propper one yet.

Answer 1

The Full answer about the existence question is given in the paper of Curto-Fialkow, which we can resume (For the case of Hamburger moment problem) in two different cases, but first we consider the notation : We put $A_k$ is the matrix $(\mu_{i+j})_{i,j\leq k}$ and $v_{k}^l$ is the vector $(\mu_{k+j})_{j\leq l}$.

So the first case is when $n=2k+1$ is odd.

Then we get a positive measure $\iff$ we get a discrete measure $\iff$ $A_k$ is positive semi-definite and $v_{k+1}^k$ is in the range of $A_k$.

The second case is when $n=2k$ is even.

Then we get a positive measure $\iff$ we get a discrete measure $\iff$ $A_k$ is positive semi-definite and $rank(A_k)=\max\{p\leq k\;; \; \det(A_p)>0\}$.

For the fact that every positive measure can be represented as an atomic measures on the space of polynomials of degree $p$, is wildly known in dimension 1 as Gaussian quadrature (Or As Richter-Tchakaloff's theorems if the dimension is bigger then one, for more detail, please check the resent book of K. Schmudgen The moment problem) which state that :

Let $\mu$ be a positive measure supported on $I\subset R$, we assume that $\int_I x^nd\mu(x)<\infty$ for all $n$, then for every fixed $p\in\mathbb{N}$, we can find a set of points of $I$, $(x_i)_{1\leq i\leq p}$ called nodes, and a sequence of positive numbers $(w_i)_{1\leq i\leq p}$ called weights, such that : $$\int_I p(x)d\mu(x)=\sum_{i=1}^p w_i p(x_i)\qquad \qquad \forall p \in \mathbb{R}_{2p-1}[X]$$ Furthemore $(x_i)_i$ are exactely the zeros of the $p^{th}$ orthogonal polynomial $P_p(x)= \det \begin{bmatrix} \mu_0 & \mu_1 & \mu_2 &\cdots & \mu_p \\ \mu_1 & \mu_2 & \mu_3 &\cdots & \mu_{p+1} \\ &&\vdots&& \vdots \\ \mu_{p-1} &\mu_p& \mu_{p+1} &\cdots &\mu_{2p-1}\\ 1 & x & x^2 & \cdots & x^p \end{bmatrix}$

So to give an answer to your second question : If $n=2k+1$, is odd, then the atomic measure will be concentrated on the zero set of $P_{k+1}(x)=\det \begin{bmatrix} \mu_0 & \mu_1 & \mu_2 &\cdots & \mu_{k+1} \\ \mu_1 & \mu_2 & \mu_3 &\cdots & \mu_{k+2} \\ &&\vdots&& \vdots \\ \mu_{k} &\mu_{k+1}& \mu_{k+2} &\cdots &\mu_{2k+1}\\ 1 & x & x^2 & \cdots & x^{k+1} \end{bmatrix}$

and after you can get your weights (for example by resolving a Vandermonde system ...) For the even case $n=2k$ you should extend your sequence to an odd one, and then use the same result.

Answer 2

Certainly. Let $P$ be the set of values $X$ can take. The idea is to treat the the system of $n$ equations $\mathbb{E}(X^k)=\mu_k$ for $k=1, 2, \cdots, n$ as "simultaneous equations" and solve for $\mathbb{P}(X=i)$ for each $i \in P$.

Recall that $$\mathbb{E}(X^k)=\sum_{i \in P}\mathbb{P}(X=i)i^k$$

Seeing that we are constructing $X$, we only need to define $P$ and the probability distribution on $P$, so for any $n$, let $j$ be one more than the maximum of $\{\mu_1, \mu_2, \dots, \mu_n\}$, and set $P=\{0, 1, \dots, j\}$. Then just solve the simultaneous equations (there may be infinitely many solutions).

Answer 3

For general sequences of $\mu_k$, the answer is of course no; necessarily the moment matrix $M$ where $M_{i,j}=\mathbb{E}[X^{i+j}]$ must be positive semi-definite (see also the Hamburger moment problem, for instance). But in the case you describe, where the $\mu_k$, $k=1,\ldots,n$ are induced moments from some nonnegative random variable, the answer is yes, you can indeed find a discrete, nonnegative random variable with the same moments. In fact, one can make the support have size $n$. However, this result is not obvious at all, and is not really explicitly constructive.

For a reference, see for instance Theorem 5.9 of these notes by Monique Laurent. This essentially says that if the moments are generated by some appropriate Borel measure, then there is a finite, atomic measure that induces the same moments. The idea is to look at the closed convex cone generated by the moment sequences for all elements of the support of the initial random variable, then argue that the desired moment sequence lies in the relative interior. Then Caratheodory's Theorem will imply the moment sequence is a conic combination of finitely many elements in the support, giving the desired discrete random variable. Technically, the result there holds for arbitrary measures (not necessarily probability measures), but by inspecting the zeroth moment condition, the result specializes to probability measures (as in the resulting atomic measure will also be a probability measure).