Question about independence

Question

First of all is true that given $X,Y$ two random variables indenpendent; $(X,Y)\in D\subset \mathbb{R}^2$ then $\text{Cov}(X,Y)=0$?

I tried to prove it and this is my solution:

If $D=[a,b]\times [c,d]$ is trivial.

But for a general $D$?

Answer 1

You don't need to know what $D$ is. By definition of independence, $f(x, y) = g(x)h(y)$. Then

$\begin{align} E[XY] & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy\, f(x, y)\, dx\,dy\\ & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy\, g(x)h(y)\, dx\,dy\\ & = \int_{-\infty}^{\infty} x\, g(x)\, dx \times \int_{-\infty}^{\infty} y\, h(y)\, dy\\ & = E[X]E[Y] \end{align}$

$\therefore \boxed{\text{Cov}[X, Y] = E[X, Y] - E[X]E[Y] = 0}$

Answer 2

In general: if $X$ and $Y$ are rv's with $\text{Var}X<\infty$ and $\text{Var}Y<\infty$ then: $$\mathbb{E}XY=\int xydF_{X,Y}\left(x,y\right)$$ If $X$ and $Y$ are independent then: $$F_{X,Y}\left(x,y\right)=F_{X}\left(x\right)F_{Y}\left(y\right)$$ This leads to: $$\int xydF_{X,Y}\left(x,y\right)=\int\int xydF_{X}\left(x\right)dF_{Y}\left(y\right)=\int xdF_{X}\left(x\right)\times\int ydF_{Y}\left(y\right)=\mathbb{E}X\times\mathbb{E}Y$$