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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.

Its_me
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  • If $X$ and $Y$ are independent then $EX^{m}Y^{n}=EX^{m}EY^{n}$ for all $n,m \geq 0$. So a reasonable question in the converse direction is to assume that this equation holds for $n, m \geq 0$ an ask for independence in which case the converse is true.But if you are keen on not assuming the equation for $n=0$ and $m=0$ it may be very hard to construct a counter-example. – Kavi Rama Murthy Jan 22 '20 at 12:32

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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.

Kavi Rama Murthy
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  • That "the moments uniquely determine the joint distribution" is not a condition, afa I understand it follows from the bounded support of X and Y and also holds for dependent X and Y. – Its_me Jan 22 '20 at 11:53
  • @Its_me Answer edited. I had not noticed that your variables were bounded. Actual my earlier answer shows that this works even for certain unbounded variables. – Kavi Rama Murthy Jan 22 '20 at 11:56