Statement about independence of random variables

If I have 2 random variables $$\xi, \eta$$ and $$\forall n,m \ \mathbb E\xi^n\eta^m=\mathbb E\xi^n \mathbb E\eta^m$$, does this imply that $$\xi,\eta$$ are independent? How to show it?

The answer is no. Indeed, let $$X:=\xi$$ and $$Y:=\eta$$. Let $$U$$ and $$V$$ be any independent random variables (r.v.'s) with different distributions but with the same finite moments of all orders: $$EU^m=EV^m=:\mu_m$$ for all natural $$m$$. A standard example of the distributions of such r.v.'s $$U$$ and $$V$$ is given in the answer by saz.

Let the cumulative distribution function (cdf) $$F_{X,Y}$$ of the random pair $$(X,Y)$$ be the half-and-half mixture of the cdf's $$F_{U,V}$$ and $$F_{V,U}$$, so that $$\begin{equation*} F_{X,Y}(x,y)=\frac{F(x)G(y)+G(x)F(y)}2 \tag{1} \end{equation*}$$ for all real $$x,y$$, where $$F$$ and $$G$$ are the cdf's of $$U$$ and $$V$$, respectively. Then for the cdf's $$F_X$$ and $$F_Y$$ one has $$F_X=F_Y=\frac{F+G}2$$ and hence $$\begin{equation*} 4[F_X(x)F_Y(y)-F_{X,Y}(x,y)]=[F(x)-G(x)][F(y)-G(y)]\ne0 \end{equation*}$$ for some real $$x,y$$, so that $$X$$ and $$Y$$ are not independent.

However, $$\begin{equation*} EX^mY^n=\tfrac12\,EU^m\, EV^n+\tfrac12\,EV^m\,EU^n=\mu_m\mu_n=EX^m\,EY^n \tag{2} \end{equation*}$$ for all natural $$m,n$$.

Details on (2), in response to comments by the OP: To simplify the matter, here we may use the fact that in the mentioned standard example of the distributions of r.v.'s $$U$$ and $$V$$, these r.v.'s actually have probability density functions (pdf's), say $$f$$ and $$g$$. Then one may rewrite (1) as $$\begin{equation*} f_{X,Y}(x,y)=\frac{f(x)g(y)+g(x)f(y)}2 \tag{1a} \end{equation*}$$ for all real $$x,y$$, where $$f_{X,Y}$$ is the joint pdf of $$(X,Y)$$. So, $$\begin{multline*} EX^mY^m=\int_{-\infty}^\infty\int_{-\infty}^\infty x^my^n f_{X,Y}(x,y)\,dx\,dy \\ =\frac12\,\int_{-\infty}^\infty x^m f(x)\,dx\ \int_{-\infty}^\infty y^n g(y)\,dy+\frac12\,\int_{-\infty}^\infty x^m g(x)\,dx\ \int_{-\infty}^\infty y^n f(y)\,dy \\ =\tfrac12\,EU^m\, EV^n+\tfrac12\,EV^m\,EU^n =\mu_m\mu_n=EX^m\,EY^n. \tag{2a} \end{multline*}$$

• Can you explain me please why $E X^mY^n=1/2EU^mEV^n+1/2EV^mEU^n$? @Iosif Pinelis – user157895564 Mar 13 at 18:10
• @user157895564 : One way to do this is to use (1) to write the following: $EX^mY^m=\int_{-\infty}^\infty\int_{-\infty}^\infty x^my^n dF_{X,Y}(x,y) =\tfrac12\,\int_{-\infty}^\infty x^m dF(x)\int_{-\infty}^\infty y^n dG(y)+\tfrac12\,\int_{-\infty}^\infty x^m dG(x)\int_{-\infty}^\infty y^n dF(y)=\tfrac12\,EU^m\, EV^n+\tfrac12\,EV^m\,EU^n$. – Iosif Pinelis Mar 13 at 19:03
• Why second equality is correct? @Iosif Pinelis – user157895564 Mar 13 at 19:46
• @user157895564 : This equality immediately follows by (1) and the definitions of the Lebesgue--Stieltjes integrals with respect to $dF_{X,Y}$, $dF$, and $dG$. To simplify this, we may use the fact that in the cited example of the distributions of r.v.'s $U$ and $V$, these r.v.'s actually have pdf's. I have added this detail to the answer. – Iosif Pinelis Mar 13 at 21:08
• But it doesn't work, isn't it? $(F_{X,Y})'_x=f_{X,Y}$ only if $G=g,F=f$, but $f$ is always summable, but $F$ is not. Or I'm missed something? – user157895564 Mar 14 at 4:45