Let $P$ be the joint distribution of two random variables $X$ and $Y$, that both have support on $(0,1)$ (I am also interested in the case where $X$ takes values on $k$-dimensional simplex, but I would be happy to start with the simple case).


Now, suppose that for all $n,m \in \mathbb{N}$, we have:


$E[X^nY^m] = E[X^n]E[Y^m]$


Is this a sufficient condition for independence of $X$ and $Y$? Are there other conditions I would need?


I suspect that I may need some knowledge of higher order moment problems: does anyone have any suggestions for useful references?



Sufficient conditions for independence based on moments
By your assumptions, for all polynomial $Q$ and $R$, you have :
$$
E(Q(X)R(Y))=E(Q(X))E(R(Y)).
$$
Then, as $X$ and $Y$ are bounded, by Stone-Weierstrass theorem, you get the same equality for all continuous maps $Q$ and $R$. Using for example results about the Fourier transform you get independence of $X$ and $Y$.


You can avoid quite easily the use of Fourier transform if you prefer. The aim is to prove that $(X,Y)$ has the same law as $(X',Y')$ where $X'$ and $Y'$ are independent and have the same law as $X$ and $Y$, respectively. Your assumption is $E(X^nY^m)=E(X'^nY'^m)$ for all $n,m$ and Stone-Weistrass gives you $E(f(X,Y))=E(f(X',Y'))$ for all continuous $f$. The result follows.


When the random variable are bounded this is quite easy. 


