Suppose there are two random variables $x,y$, with nonzero means. Their covariance is $$Cov(x,y)=E(xy)-E(x)E(y).$$ If they are independent, $E(xy)=E(x)E(y)$ and hence $Cov(x,y)=0$.
However, is it possible that $Cov(x,y)\neq0$ due to $E(xy)=0$ and $E(x)E(y)\neq0$? I can't think of/visualize an example.
(I'm asking because I'm trying to gain some intuition behind the algebra of the least squares formula, where $\beta=E[(X'X)]^{-1}E[X'y]$ in the population. I'm well aware that the estimator $\hat{\beta}$ ultimately involves $(X'MX)^{-1}$ and $X'MY$ when an intercept is included, where $M$ is the demeaning matrix.)
Example: Suppose $Y$ and $Z$ satisfy (i) $\Bbb E[Y]=1$, $\Bbb E[Z]=0$, (ii) $Var[Y]=Var[Z]=1$, and (iii) $Cov[Y,Z]=\rho>0$. Define $X=Z-\rho$. Then $\Bbb E[XY]=\Bbb E[ZY]-\Bbb E[\rho Y]=Cov[Z,Y]-\rho=0$, but $Cov[X,Y]=Cov[Z,Y]=\rho>0$.