Does non-zero correlation imply dependence?

Question

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 $f_{X,Y}(x,y) \ne f_X(x) f_Y(y)$?

Answer 1

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?

Answer 2

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?

Answer 3

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.