Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
4  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.