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


3 Answers 3


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?

  • 2
    $\begingroup$ Alecos, I have a dumb question. What does the fancy arrow mean in, e.g., line 1? I imagine something like "imply," but I'm uncertain. $\endgroup$
    – Sycorax
    Aug 27, 2014 at 13:56
  • 2
    $\begingroup$ @user777 You mean $\Rightarrow$? Indeed, it means "implies". $\endgroup$ Aug 27, 2014 at 13:58
  • $\begingroup$ The reason to only use the implication arrow in informal argument: is the implication arrow left or right associative? $\endgroup$
    – kasterma
    Aug 27, 2014 at 14:33
  • $\begingroup$ \implies produces $\implies$ which looks better than \rightarow which produces $\Rightarrow$. $\endgroup$ Sep 3, 2017 at 18:00

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?

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    $\begingroup$ The nice thing about this answer is that it applies whether or not the random variables in question admit a density function, as opposed to other answers on this thread. This is true due to the fact that expectations can be defined with Stieltjes integrals using the CDF, with no mention of the density. $\endgroup$
    – ahfoss
    Aug 27, 2014 at 15:52

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.

  • $\begingroup$ If you will read my answer carefully, you will see that I too used the argument that you have made in your answer, namely that $A \implies B$ is the same as $\neq B \implies \neg A$. $\endgroup$ Jul 24, 2017 at 16:40
  • $\begingroup$ @DilipSarwate Edited to reflect that. $\endgroup$
    – Tony
    Jul 24, 2017 at 17:00

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