How Many Moments Uniquely Define a Distribution with Finite Support? Simple question, but one to which I could not find the exact answer elsewhere.  How many moments of a discrete probability distribution with finite support are required to uniquely identify the exact probability mass function?  Suppose we know that the distribution has support on at most $N$ points within a bounded interval (for my purposes the interval is $[0, 1]$), but we do not know the points.
Is it the case that the distribution is uniquely identified by some number of moments? My hypothesis is that it may be the first $2N-1$ moments. Since we have to identify $N$ mass points and their $N$ individual probabilities, one might think we need $2N$ equations and each moment gives us one equation, plus the restriction that the probabilities sum to $1$.  But these equations are not linear in the mass points, so it's not immediately obvious to me that we're identified.
I am aware of the Hausdorff Moment Problem, so I know that an infinite sequence of moments uniquely identifies any bounded distribution, but I am particularly interested in further restricting the domain to distributions with finite support. Any references would also be appreciated!
Thanks!
 A: Let $F$ be the distribution supported on the numbers $x_1 \lt x_2 \lt \ldots \lt x_n$ that assigns probabilities $p_i \gt 0$ to each $x_i.$  By definition, its (raw) moment of degree $k$ is
$$\mu_k = \sum_{i=1}^n p_i x_i^k.$$
I will begin with a series of observations about this situation, each of interest in its own right.  A basic tool is the sequence of vectors $\mathbf{x}_k = (x_1^k, x_2^k, \ldots, x_n^k)$ for $k=0, 1, \ldots,n-1.$  Writing $\mathbf{p} = (p_1,p_2,\ldots, p_n),$ each moment can be expressed as a vector product 
$$\mu_k = \sum_{i=1}^n p_i x_i^k = \mathbf{p}\, \mathbf{x}_k^\prime.$$


*

*The collection $\{\mathbf{x}_0,\mathbf{x}_1, \ldots, \mathbf{x}_{n-1}\}$ is linearly independent.  To show this, assume the contrary: that is, let coefficients $c_k$ not all zero be such that $$\sum_{k=0}^{n-1} c_k \mathbf{x}_k = \mathbf{0}.\tag{1}$$  Written out component-by-component,  $(1)$ asserts that for each $i=1,2,\ldots, n,$ $$\sum_{k=0}^{n-1} c_k x_i^k = 0.$$  That exhibits each $x_i$ as a root of the polynomial $c(T)=c_{n-1}T^{n-1}+c_{n-2}T^{n-2}+\cdots + c_0.$ Such a polynomial has at most $\operatorname{deg}(c)\le n-1$ distinct roots, contradicting the distinctness of the $n$ $x_i.$

*All moments are determined by the first $n$ moments $\mu_0,\mu_1,\ldots,\mu_{n-1}.$  The preceding result shows the vectors $\mathcal{X} = \{\mathbf{x}_k,k=0,1,\ldots, n-1\},$ are a basis for $\mathbb{R}^n.$  Therefore for any $m,$ $\mathbf{x}_m$ is a linear combination of the $\mathbf{x}^k,$ $k=0,1,\ldots,n-1;$ that is, there exist coefficients $\,_ma_k$ (determined solely by the $x_i$) for which $$\mathbf{x}_m = \,_ma_0\mathbf{x}_0 + \,_ma_1\mathbf{x}_1 + \cdots + \,_ma_{n-1}\mathbf{x}_{n-1}.$$  Consequently $$\mu_m = \mathbf{p}\,\mathbf{x}_m^\prime = \mathbf{p}\,\sum_{i=0}^{n-1}\,_ma_k \mathbf{x}_k^\prime  = \sum_{i=0}^{n-1}\,_ma_k \mathbf{p}\,\mathbf{x}_k^\prime= \sum_{i=0}^{n-1}\,_ma_k \mu_k.$$

*The numbers $x_i$ and the first $n$ moments determine $\mathbf{p}.$  Indeed, the first $n$ moments are the coefficients of $\mathbf{p}$ in the basis dual to $\mathcal X.$

*The first $n$ moments of $F$ determine, and are determined by, the distribution shifted by a constant $\lambda.$  This is the distribution supported on $x_1-\lambda, x_2-\lambda, \ldots, x_n-\lambda$ with probabilities $p_i.$  The demonstration is straightforward: use the Binomial theorem to expand $(x_i-\lambda)^k$ in terms of $x_i^0, x_i^1, \ldots, x_i^k.$
Part of the question is whether there exist $n^\prime,$ a positive probability vector $\mathbf{q},$ and support points $y_1\lt y_2\lt \ldots \lt y_{n^\prime},$ determining a distribution $G$ having the same moments as $F.$  Suppose there is.  Shift both distributions by $\lambda=-\min(x_1,y_1),$ simplifying the situation to distributions with nonnegative support.  By taking $m$ arbitrarily large, the largest support points eventually dominate the moments: $$q_{n^\prime} y_{n^\prime}^m \approx \mu_m \approx p_n x_n^m$$ This is possible only when $q_{n^\prime}=p_n$ and $y_{n^\prime} = x_n.$  Continuing inductively, we conclude $n=n^\prime,$ $\mathbf{q}=\mathbf{p},$ and $\mathbf{x}_1=\mathbf{y}_1:$ that is, $G=F.$
Finally, how many moments need to be known to determine $\mathbf{p}$ and $\mathbf{x}$?  Consider the map $f:\mathbb{R}^n\times \mathbb{R}^n\approx \mathbb{R}^{2n}\to\mathbb{R}^{2n}$ defined by $$f(\mathbf{p}^\prime, \mathbf{x}^\prime) = (\mathbf{p}\mathbf{x}_0^\prime, \mathbf{p}\mathbf{x}_1^\prime, \ldots, \mathbf{p}\mathbf{x}_{2n-1}^\prime)^\prime.$$ Its derivative is the $2n\times 2n$ matrix
$$Df(\mathbf{p}^\prime, \mathbf{x}^\prime) = \pmatrix{1 & \cdots & 1 & 0 & \cdots & 0 \\
x_1 & \cdots & x_n & p_1 & \cdots & p_n \\
x_1^2 & \cdots & x_n^2 & 2p_1x_1 & \cdots & 2p_n x_n \\
\vdots & \cdots & \vdots & \vdots & \cdots & \vdots \\
x_1^{2n-1} & \cdots & x_n^{2n-1} & (2n-1)p_1x_1^{2n-2} & \cdots & (2n-1)p_nx_n^{2n-2}}$$
with a Vandermonde-like structure, enabling us to obtain a simple formula for its determinant, 
$$\operatorname{Det}\left(Df(\mathbf{p}^\prime, \mathbf{x}^\prime)\right) = -(p_1p_2\cdots p_n)^{2n} \left(\prod_{1\le i\lt j \le n}(x_i-x_j)\right)^4.$$
Because none of the $p_i$ is zero and all the $x_i$ are distinct, this is nonzero.  The Inverse Function Theorem implies $f$ is locally invertible: that is, provided $\mathbf{\mu}=(\mu_0,\mu_1,\ldots,\mu_{2n-1})$ is in the range of $f$, there exists an inverse $f^{-1}\subset\mathbb{R}^n\times \mathbb{R}^n$ in a neighborhood of $\mathbf{\mu}.$  That is, 

The first $2n$ moments $\mu_0,\mu_1,\ldots,\mu_{2n-1}$ determine a discrete set of solutions $(\mathbf{p},\mathbf{x})$ corresponding to those moments.

As we have already shown, all such solutions correspond to the same distribution: they differ only by permuting the indexes $1,2,\ldots, n$ of the variables.
