Moments of products of independent random variables: $E[X^mY^n]$ Part II This is a follow up question from here.
Let $X$ and $Y$ denote two real-valued bounded random variables. Then all joint moments exist and uniquely define their joint probability $P(X,Y)$. 
Given for all $m,n\in\mathbb{N}\cup\{0\}$ we have $E[X^m Y^n]=a_m b_n$ with $a_m, b_n \in \mathbb{R}$. Does it follow that $X$ and $Y$ are independent random variables and, thus, $a_m=E[X^m]$, $b_n=E[Y^n]$ and $E[X^m Y^n]=E[X^m]E[Y^n]$?
If yes, is it a straight forward result? 
Thanks for your comments / suggestions.

Assuming that the equation $EX^{m}Y^{n}=a_mb_n$ holds for all $n, m \geq 0$ and $a_0b_0=1$ the answer is YES. In this case $a_mb_0=EX^{m}$ and $b_na_0=EY^{n}$ so we get  $EX^{m}Y^{n}=EX^{m}EY^{n}$.
From the fact that the moments uniquely determine the joint distribution it follows that $X$ and $Y$ are independent. The reason is if $U$ and $V$ are independent with the same distributions as $X$ and $Y$ respectively then  $(U,V)$ and $(X,Y)$ have the same moments. The uniqueness of the  joint distribution now implies that $(U,V)$ has same distribution as $(X,Y)$ which implies that $X$ and $Y$ are independent. 
