Independence does not imply Zero Correlation If I take $X$ to be a degenerate random variable, i.e. $X=1$ WP1 and $Y=X$ defined over the singleton sample space $\Omega=\{1\}$. Then
$$\mathbb{P}(X=1|Y=1)=1=\mathbb{P}(X=1)$$
i.e. I'd assume they're independent. But, we have
$$\rho_{XY}=\frac{\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)}{\sigma_X\sigma_Y}=\frac{0}{0}$$
which is undefined, not zero. Where am I being imprecise here/what's the misunderstanding? Thanks!

Independence implies the correlation can't be any number other than zero, but as this example and @Dave's example of independent Cauchy variables shows, the correlation might not be a number at all. So the usual statement is slightly inexact.
The point-mass example is worse, in a way.  In the Cauchy example it's still true that the correlation is zero for, eg, all bounded functions $f(X)$, $g(Y)$  whereas in the point-mass example there's no way to get a well-defined correlation.
