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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?

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Yes because all continuous functions $f$, $g$ can be approximated by polynomials on compact sets, so you get $E(f(X)g(Y))=Ef(X)Eg(Y)$, which is equivalent to the independence. It all gets trickier when the supports are unbounded but it is another story. – fedja May 29 '11 at 14:50
up vote 6 down vote accepted

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.

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