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I have the intuition that a sample of size $n$ should be able to be described uniquely by just N moments. But I don't know if that's true.

The idea is that, first, it is obvious that a sample of $n = 1$ can be uniquely describe by just the sample mean. For $n = 2$, knowing the first two (uncentered) moments is also enough because (if I did my maths correctly)

$$ X_1 = 2\overline{X} - X_2 $$ and

$$ M_2 = 1/2X_2^2 - 2M_1X_2+2M_1^2 $$ From which I can solve for $X_2$ and then $X_1$ by knowing $M_1$ and $M_2$.

It seems intuitive that this must hold on for $n > 2 $. The generalization for a sample of size $n$ and $m$ moments would be

\begin{matrix} x_1 + x_2 + ... + x_n = nM_1\\ x_1^2 + x_2^2 + ... + x_n^2 = nM_2 \\ \vdots \\ x_1^m + x_2^m + ... + x_n^m = nM_m \end{matrix}

It these were linear equations, they would only have an unique solution for $n = m$ if the system described a full rank matrix. But they aren't, so... no dice.

From an information theory standpoint it seems to me (again, intuitively) that you shouldn't need more than $N$ numbers to represent any other $N$ numbers.

Note that I'm not trying to estimate the population distribution by the sample moments. I'm talking about describing a specific vector of $n$ values.

Also note that I assume that the order of samples does not matter so that $(x_1, x_2)$ and $(x_2, x_2)$ would be treated as the same solution.

Any hints on this problem?

Elio Campitelli
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    It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $\left(x_1\to \frac{1}{2} \left(1-\sqrt{3}\right),x_2\to \frac{1}{2} \left(\sqrt{3}+1\right)\right)$ is one solution and $\left(x_1\to \frac{1}{2} \left(\sqrt{3}+1\right),x_2\to \frac{1}{2} \left(1-\sqrt{3}\right)\right)$ is another. – JimB Dec 10 '18 at 06:26
  • Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like). – Elio Campitelli Dec 10 '18 at 14:25

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We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n. From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).

(Summing up a discussion with Elio via twiiter)

Pablo De Napoli
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