I recently learned that two independent random variables $X$ and $Y$ must have a covariance of $0$. That means that the correlation between them is also $0$.

However, apparently, the converse is not true. 2 random variables $X$ and $Y$ can have a correlation of $0$, yet still be dependent. I don't understand why this is. Doesn't a correlation of $0$ imply that the random variables do not affect each other?


Correlation is a measure of linear dependence, so it kind of gives you an indication of how the two variables are related linearly. It doesn't capture however more complicated behaviour.

Therefore if you have $X$ and $X^2$ with $X \sim N(0,1)$, then

$$\operatorname{Cov}(X, X^2) = E(X^3) - E(X)E(X^2) = 0$$

but the two random variables are clearly dependent.

  • 4
    $\begingroup$ It's one of the reasons why when you're working with statistical data, you should always study data plots. Claims that there's no statistical relationship when there's an obvious "smile" on a scatter plot can be very misleading. $\endgroup$ – johnny Dec 2 '12 at 20:32
  • $\begingroup$ I am not sure whether you can see this or not, however, how can we guarantee $E(X^3) = 0$ i.e., 3rd moment of $X$ $\endgroup$ – kurtkim Sep 11 '15 at 1:21
  • $\begingroup$ @kurtkim The distribution of $X^3$ is symmetric around 0 $\endgroup$ – Stockfish Jul 22 '20 at 14:36

No, it doesn't imply that. Correlation is just a one-dimensional measure, whereas dependence can take many forms. For instance, the indicator variable of the event that a normally distributed random variable is within one standard deviation of the mean is uncorrelated with the random variable itself, but is clearly not independent of it.


The definition of covariance between two random variables $X,Y$ is $Cov(X,Y)=E(XY)-E(X)E(Y)$. Zero covariance means that $E(X,Y)=E(X)E(Y)$. But the independency condition is much stricter: $E(g(X)h(Y))=E(g(X)) E(h(Y))$ for any functions $g,h$. To give some intuition on the difference, I will assume that $X,Y$ are smooth (infinately differentiable) random variables and Taylor expand both functions. I will be expanding around zero in order for the presentation to be clearer. Doing so, gives $$g(X)=\sum_n\dfrac{1}{n!}g^{(n)}(0)X^n$$ where $g^{(n)}$ is the $n-$th derivative of $g$, and similarly for $h(Y)$. Then using the linearly of $E(\ .)$, namely $E(aX+bY)=aE(x)+bE(Y)$ for $a,b$ constants, we get $$E(g(X)h(Y))=\sum_{n,m}\dfrac{1}{n!m!}g^{(n)}(0)h^{(m)}(0)E(X^nY^m)$$ Now, if $Cov(X,Y)=0$, then the $n,m=1$ term simplifies through $E(X,Y)=E(X)E(Y)$ but we can say nothing about terms containing $E(X^nY^m)$. This shows you that covariance measures independence on a "linear level", whereas dependence of two random variables is much more general that their correlation.

Remark for applications: For $x,y$ ($x,y$ being the realizations of $X,Y$) sufficiently small, then we might be able to approximately treat them as independent (although they are not), which is what we do in any system in which we care about understanding its behaviour when $x,y$ are small (we do this a lot in physics). More concretely, if we can neglect all powers $n,m>1$, then we can approximate uncorrelation as independence.

So, as you can see, in order to have independence between $X,Y$, $E(X^nY^m)=E(X^n)E(Y^m),\ \forall n,m$. If this is the case, we can write the second Taylor expansion as $$E(g(X)h(Y))=\sum_{n}\dfrac{1}{n!}g^{(n)}(0)E(X^n)\sum_m\dfrac{1}{m!}h^{(m)}(0)E(Y^m)$$ and using linearity and the Taylor expansions of $g(X)$ and $h(Y)$, we arrive at the final result: $$E(g(X)h(Y))=E(g(X))E(h(Y))$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.