If we have two random variables $X,Y$ with support on $A=\{a_1,\dots,a_n\}$ such that $\mathbb{E}[X^k]=\mathbb{E}[Y^k]$ for all $1\le k\le n$, does this imply that $\mathbb{E}[X^k]=\mathbb{E}[Y^k]$ for all $k\in\mathbb{N}$?
It seems like this should be true to me since you only have $n-1$ 'degrees of freedom' in creating such random variables but I'm not sure how I would go about proving this. I've thought about trying to create a degree $n-1$ polynomial with coefficients related to $p_X(a_i)-p_Y(a_i)$ and showing that the $a_i$ are zeros to show $X=Y$ in distribution but I haven't been able to make this work out. Any help would be appreciated, thanks!
Let $p_j := P(X=a_j)$ and $q_j := P(Y=a_j)$. Suppose we have $\sum_{j=1}^n p_j a_j^k = \sum_{j=1}^n q_j a_j^k$ for all $1 \le k \le \color{blue}{n-1}$. These relationships (along with the "zero moment" condition $\sum_{j=1}^n p_j = \sum_{j=1}^n q_j$) can be written as $$A p = A q$$ where $p=(p_1, \ldots, p_n)$ and $q=(q_1, \ldots, q_n)$ are column vectors containing the probabilities associated with $X$ and $Y$, and where $A$ is [the transpose of] an $n \times n$ Vandermonde matrix induced by the $a_1, \ldots, a_n$. If the $a_i$ are distinct, you can show the matrix $A$ is invertible (e.g. by looking at the formula for the determinant of a Vandermonde matrix), and thus $p=q$.