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Let $\vec{X}:=(X_1,X_2,\cdots,X_n)$ be an $n$-dimensional random vector where $X_i$ takes values in $[N_i]:=\{0,1,\ldots,N_i\}$ for each $i=1,2,\ldots,n$. Show that the distribution of $\vec{X}$ is uniquely determined by $$\left\lbrace\mathbb{E}\left[\prod_{i=1}^n X_i^{n_i}\right]\,:\,n_i\in[N_i],i=1,2,\ldots,n\right\rbrace.$$

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  • $\begingroup$ Do you know how to prove this in the $n=1$ dimensional case? $\endgroup$ Commented Dec 17, 2019 at 16:48
  • $\begingroup$ Yes! I can write the distribution as a solution to a system of linear equations which has a unique solution. For higher dimension, this method in principle should work. But I don't know if the system will have unique solution. For 1 dimension, this question gives the answer: stats.stackexchange.com/questions/416186/… $\endgroup$ Commented Dec 17, 2019 at 17:01

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Here is a hint, the details of which you should be able to fill in.

In the $n=1$ dimensional case the matrix of the linear transformation sending a probability distribution to its vector of moments is a Vandermonde matrix, and hence invertible. In the $n=2$ dimensional case the corresponding matrix is the Kronecker product $A\otimes B$ of the two $n=1$ matrices corresponding to the marginal distributions. Since thee Kronecker product of invertible matrices is invertible, and the moments determine the distribution here, too. And for $n>2$ the same method works, too.

You have to know that the $n=1$ matrices are Vandermonde matrices and have to know why they are invertible in your case. You also need to know the $\otimes$ contruction. (Maybe its easiest to work out the matrix if $n=2$ and $N_1=N_2=2$ or $3$, and see what's going on with $\otimes$.) And then you have to know that the Kronecker product of invertibles is invertible.

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