Does non-zero correlation imply dependence? We know of the fact that zero correlation does not imply independence. I am interested in whether a non-zero correlation implies dependence - i.e. if $\text{Corr}(X,Y)\ne0$ for some random variables $X$ and $Y$, can we say in general that $f_{X,Y}(x,y) \ne f_X(x) f_Y(y)$?
 A: Here a purely logical proof. If $A\rightarrow B$ then necessarily $\neg B \rightarrow \neg A$, as the two are equivalent. Thus if $\neg B$ then $\neg A$. Now replace $A$ with independence and $B$ with correlation.
Think about a statement "if volcano erupts there are going to be damages". Now think about a case where there are no damages. Clearly a volcano didn't erupt or we would have a condtradicition.
Similarly, think about a case "If independent $X,Y$, then non-correlated $X,Y$". Now, consider the case where $X,Y$ are correlated. Clearly they can't be independent, for if they were, they would also be correlated. Thus conclude dependence.
A: Yes, because
$$\text{Corr}(X,Y)\ne0 \Rightarrow \text{Cov}(X,Y)\ne0$$
$$\Rightarrow E(XY) - E(X)E(Y) \ne 0 $$
$$\Rightarrow \int \int xyf_{X,Y}(x,y)dxdy -\int xf_X(x) dx\int yf_Y(y)dy \ne 0$$
$$\Rightarrow \int \int xyf_{X,Y}(x,y)dxdy -\int \int xyf_X(x) f_Y(y)dxdy \ne 0$$
$$\Rightarrow  \int \int xy \big[f_{X,Y}(x,y) -f_X(x) f_Y(y)\big]dxdy \ne 0$$
which would be impossible if $f_{X,Y}(x,y) -f_X(x) f_Y(y) =0,\;\; \forall \{x,y\}$. So
$$\text{Corr}(X,Y)\ne0 \Rightarrow \exists \{x,y\}:f_{X,Y}(x,y) \ne f_X(x) f_Y(y)$$
Question: what happens with random variables that have no densities?
A: Let $X$ and $Y$ denote random variables such that $E[X^2]$ and $E[Y^2]$
are finite.  Then, $E[XY]$, $E[X]$ and $E[Y]$ all are finite.
Restricting our attention to such random variables, let
 $A$ denote the statement that $X$ and $Y$ are independent random variables
and $B$ the statement that $X$ and $Y$ are uncorrelated random variables,
that is, $E[XY] = E[X]E[Y]$.  Then we know that $A$ implies $B$, that is,
independent random variables are uncorrelated random variables. Indeed,
one definition of independent random variables is that 
$E[g(X)h(Y)]$ equals $E[g(X)]E[h(Y)]$ for all measurable functions $g(\cdot)$
and $h(\cdot)$).  This is usually
expressed as
$$A \implies B.$$
But $A \implies B$ is logically equivalent to $\neg B \implies \neg A$, that is,

correlated random variables are dependent random variables.

If $E[XY]$, $E[X]$ or $E[Y]$ are not finite or do not exist, then it
is not possible to say whether $X$ and $Y$ are uncorrelated or not
in the classical meaning of uncorrelated random variables being those
for which $E[XY] = E[X]E[Y]$. For example,
$X$ and $Y$ could be independent Cauchy random variables (for 
which the mean does not exist). 
Are they uncorrelated random variables in the classical sense?
