In my textbook for my measure theory class (Marek Capinski and Ekkehard Kopp: Measure, Integral and Probability), in the chapter on the independence of two random variables, the author(s) make the following claim after defining the covariance of two random variables and correlation:
As we can see $\operatorname{Cov}(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$, so clearly independent random variables $X$ and $Y$ are uncorrelated; it is sufficient to take $f(x)=x-\mathbb{E}(X)$ and $g(x)=x-\mathbb{E}(Y)$ in theorem $5.18$.
Where theorem $5.18$ states:
The random variables $X$ and $Y$ are independent if and only if $\mathbb{E}(f(X)g(Y))=\mathbb{E}(f(X))\mathbb{E}(g(Y))$ holds for all Borel measurable bounded functions $f,g$.
My question:
Since the author did not provide a proof, i'm not understanding how using $f(x)=x-\mathbb{E}(X)$ and $g(x)=x-\mathbb{E}(Y)$ in theorem $5.18$ shows that independent random variables $X$ and $Y$ are uncorrelated. If any one could explain or provide insight, it would be a great help.
$\endgroup$1 Answer
$\begingroup$Suppose two random variables are independent. Then (choose identity functions) $E(XY)=E(X)E(Y)$.
It remains to note that
$$Cov(X,Y)=E(XY)-E(X)E(Y).$$
EDIT: Using the given functions, we have
$$Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(X-E(X))E(Y-E(Y))=0$$
$\endgroup$ 3
