Moments and weak convergence of probability measures

Solution 1:

Partial answer, further details need to be added.

The result may be not true if we don't assume that $\mu$ is determined by its moments, and it's not automatic since we have counter-examples (a famous one is given by Feller, using the density of a log-normal law.

But if $\mu$ is determined by its moments, we can get the result. Since $$\mu_n\{x:|x|\geq A\}\leq \frac 1{A^2}\int x^2d\mu_n(x)$$ is bounded, the sequence $\{\mu_n\}$ is tight. Therefore, we can extract a converging (in law) subsequence, to $\nu$. What we have to show is that for this subsequence $\{\mu_{n_k}\}$, we have for all $p\geq 0$ and $m>0$ that $$\lim_{k\to\infty}\int_{|x|\leq m} x^pd\mu_{n_k}(x)=\int_{|x|\leq m} x^pd\nu(x).$$ Then we will get that $\mu$ and $\nu$ have the same moments.

Solution 2:

Take two distinct probability measures $\mu$ and $\nu$ with the same moments, as in this MathOverflow answer. Now consider the tight sequence $\mu,\nu,\mu,\nu,\mu\dots.$

Solution 3:

Edit: this answer has been expanded upon further consideration.

The answer is yes if $\mu$ is the only measure with the moments $\int x^k d\mu$. Otherwise, there are examples with the lognormal distribution that show the result can be false (see Durrett's probability book). So assume that $\mu$ is the only measure with the aforementioned moments. In the comments, we concluded that for a compact set the answer is true. When we have arbitrary support, the key notion that is required is tightness. It is a fact that the cdf's $F_n$ converge to a cdf $F$ iff they are tight. Thus, it is sufficient to check that the $\mu_n$ are tight which will imply what you desire by the Levy-Cramer Continuity Theroem. To check for tightness, I refer you to the following result (which can be found in more detail in Durrett's probability book):

If there is a $g\geq 0$ such that $g(x)\rightarrow\infty$ for $|x|\rightarrow\infty$ and

$L:=\sup_n \int g(x)dF_n <\infty$

then $F_n$ is tight. The proof is trivial by considering for large enough $A$:

$1-F_n(A)-F_n(-A) \leq \frac{L}{\inf_{|x|\geq M}g(x)}$