Can two random variables $X,Y$ be dependent and such that $E(XY)=E(X)E(Y)$? Can someone define independence of two random variables with this "product rule", or are there any counterexamples?
 A: Independence implies uncorrelation
We say that two random variables are uncorrelated if $E[XY]  = E[X]E[Y]$.
If $X$ and $Y$ are independent and discrete (this can also be extended to continuous random variables), then
$$p_{X,Y}(x,y) = p_X(x)p_Y(y)\qquad \forall (x,y)$$
Using this in the definition of $E[XY]$ and doing some rearrangements of terms, we have
\begin{align}
E[XY] &= \sum_x\sum_y xyp_{X,Y}(x,y)\\
&= \sum_x\sum_y xyp_X(x)p_Y(y)\\
&= \sum_x\sum_y [xp_X(x)][yp_Y(y)]\\
&= \sum_xxp_X(x)\sum_yyp_Y(y)\\
&= E[X]E[Y]
\end{align}
Therefore, if $X$ and $Y$ are independent, then they are uncorrelated.
$\\$
Uncorrelation does not imply independence (in general)
In general$^1$, it is false to say that if $X$ and $Y$ are uncorrelated, then they are independent. Let's see a counterexample. Let $X$ and $Y$ be random variables taking values in $\{(0,1), (1,0), (-1,0), (0,-1)\}$ with equal probability $0.25$. Then, $XY = 0$ with probability $1$, and therefore $E[XY] = 0$.
By other hand,
$\displaystyle p_X(x) = \left\{
\begin{array}{ll}
\frac{1}{4} & x=-1\\
\frac{1}{2} & x=0\\
\frac{1}{4} & x=1\\
\end{array}
\right.
$ $\hspace{1cm}$ and $\hspace{1cm}$ $p_Y(y) = \left\{
\begin{array}{ll}
\frac{1}{4} & y=-1\\
\frac{1}{2} & y=0\\
\frac{1}{4} & y=1\\
\end{array}
\right.
$
which implies that $E[X]=E[Y]=0$. Then, $E[XY] = E[X]E[Y]$ and the random variables are uncorrelated.
However,
$$p_X(1)p_Y(1) =  \frac{1}{4}\frac{1}{4} \neq p_{X,Y}(1,1) = 0$$
which shows that $X$ and $Y$ are not independent.
We can conclude then that independence is a sufficient but not a necessary condition for uncorrelation.

$^1$ The only two exceptions I am aware of are the jointly Gaussian variables and two Bernoulli random variables...any other?
