Covariance and Independence

By definition, independence implies zero Covariance. The converse is not true.

Easy example: Let $X$ be a random variable that is $-1$ or $+1$ with probability 0.5. Then let $Y$ be a random variable such that $Y=0$ if $X=-1$, and $Y$ is randomly $-1$ or $+1$ with probability 0.5 if $X=1$.

Clearly $X$ and $Y$ are highly dependent (since knowing $Y$ allows me to perfectly know $X$), but their covariance is zero: They both have zero mean, and

$$\begin{array}{rl}\mathbb{E}[XY]=& (-1)\cdot 0\cdot P(X=-1)\\ & +1\cdot 1\cdot P(X=1,Y=1)\\ & +1\cdot (-1)\cdot P(X=1,Y=-1)\\ =& 0.\end{array}$$

Or more generally, take any distribution $P(X)$ and any $P(Y|X)$ such that $P(Y=a|X)=P(Y=-a|X)$ for all $X$ (i.e., a joint distribution that is symmetric around the $x$ axis), and you will always have zero covariance. But you will have non-independence whenever $P(Y|X)\ne P(Y)$; i.e., the conditionals are not all equal to the marginal. Or ditto for symmetry around the $y$ axis.)