# Do mixed moments determine joint distribution?

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 $$\left\lbrace\mathbb{E}\left[\prod_{i=1}^n X_i^{n_i}\right]\,:\,n_i\in[N_i],i=1,2,\ldots,n\right\rbrace.$$

• Do you know how to prove this in the $n=1$ dimensional case? Commented Dec 17, 2019 at 16:48
• 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/… Commented Dec 17, 2019 at 17:01

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.