Erroneous Argument for uncorrelated implies Independence I've been working on the problem where for a bivariate normal random variable (X,Y), uncorrelated implies Independence. However, I realized that I didn't use the bivariate normal assumption, so there is something wrong with my argument. Can someone point out the problem with the following:
Starting from the first direction. From $Corr(X,Y) = 0$, we can deduce that $E[XY] = E[X]E[Y]$.
This means that $$\int_{\mathbb{R}^2} xy dF_{X,Y}(x,y) = \int x dF_X(x) \cdot \int y dF_y(y)$$
$$ = \int \int xy dF_X(x) dF_Y(y)$$
$$\implies \int \int xy f_{X,Y}(x,y) dx dy = \int \int xy f_X(x)f_Y(y)dx dy$$
thanks in advance

Note: you can safely work with the pdf here since we know it exists
For $(\Rightarrow)$, when replacing $\text{cor}(X,Y)=0$ in the expression of $f_{X,Y}(x,y)$, do you see that the joint density factors into two products, one involving only $x$ and the other involving only $y$?
In general, this implication is false, i.e. there are random vectors which are uncorrelated but their density doesn't factor. The multivariate $t$ is one of these.
