Assume ${\bf x} \in \mathbb{R}^n$ denotes a real-valued and bounded random variable. Then, the moments of $\bf x$ uniquely define its distribution.
Assume we have not only one distribution, but a sequence indexed by $n\in\mathbb{N}$. Let $M_{k,n}$ denote the $k$-th mixed moment of the $n$-th distribution.
If the limit
$~\lim_{n\to\infty} M_{k,n}=M_{k}$
exists for all $k$, can we conclude that the $M_{k}$ are the moments of a distribution?
Best Answer
Here is the 1-d version of proposition that the question is looking for, which I believe to be correct (but idk for sure):
Proposition: Suppose that $X_n \in \mathbb{R}$ is a sequence of random variables such that:
Then there is a random variable $X$ with $E[X^k] = m_k$.
Proof:
First, we observe that the sequence of random variables $X_n$ is tight, using: Tightness of a sequence of random variables with bounded mean and variance .Since you are assuming the random variables are universally bounded, they are tight.
By Prokohorov's theorem (in particular the corollary for measures in $\mathbb{R}^m$), tightness implies that there is a subsequence $X_{n_k}$ which converges in distribution to some random variable $X$. We can pass to this subsequence, since this does not effect the two hypothesis.
Now, for any $k \geq 1$, let $f_k(x)$ be any function that agrees with $x^k$ on $[-N, N]$ and which is bounded and continuous. For instance, $f(x)$ might linearly interpolate so that it is zero on $(\infty, -N - 1) \cup (N + 1, \infty)$.
Since we have $X_n \to X$ weakly, we have from the Portmanteau theorem that $E [ f_k(X_n) ] \to E[f(X)]$. But since $|X_n| \leq N$ and $|X| \leq N$ a.s. (this is a consequence of weak convergence, e.g. using the CDF), we have that $f_k(x) = x^k$ on $[-N, N]$, which means that $E[X_n^k] \to E[X^k]$. By the second assumption, this implies that $E[X^k] = m_k$, which is the desired conclusion. QED
Hope this helps!
I think it should be relatively straightforward to adapt to $\mathbb{R}^n$.
We don't have to worry about tightness because of your assumption about the universal boundedness:
You just want to make sure you that your assumption about moments again implies tightness -- I'm not completely sure what mixed moments you are controlling, but if you get control over the first two moments of $||X_n||_2$ (or $L_p$ norms for some $p$), you can prove that the $||X_n||_2$ are tight, which implies that $X_n$ are.(I'm not sure that controlling moments like $E[X_1^k X_2^k]$ is enough; e.g. its possible that $X_1$ blows up a lot for small $X_2$, and maybe mass can escape to infinity that way? But if you know that all moments of the form $E[X_i^k]$ have limits I think you are in good shape.)