I have a growing interest and respect for the subject of copulas initially thanks to comments made on stats.SE by kjetil-b-halvorsen. The most interesting to me right now is the following:
"[T]he Pearson correlation is not a functional of the copula[.]"
My understanding is that if two random variables are stochastically independent, then their Pearson correlation will be zero. The converse does not hold, which I have believed to be the case in the sense that correlation only quantifies dependence in a narrow sense. The (possibly unique depending on continuity) copula between a joint cumulative distribution and its marginals seems to specify the type of dependence. It surprises me that there wouldn't be a mapping from the copula to the Pearson correlation function (of a given collection of random variables) that "loses information" about the exact dependence, leaving only a measure of dependence in the more narrow linear sense.
Is there a (dis)proof of this assertion?
Best Answer
I believe the remark simply means that you can have different joint distributions with the same copula but different Pearson correlation. So the copula alone does not determine the correlation; the correlation also depends on the marginals.
Take, for example, the bivariate copula that has density $c(u,v) = 2$ when $u,v$ are both $\le 1/2$ or both $>1/2$, and zero otherwise. If you take $U$ and $V$ both uniform over $(0,1)$ with this joint distribution, you have $\rho_{U,V} = 0.75$. Then let $X=U^3$ and $Y=V^3$; now $(X,Y)$ has still the same copula but $\rho_{X,Y} \approx 0.595$.
Here's a more drastic example where $\rho_{U,V}$ and $\rho_{X,Y}$ have different signs. Here the copula density is $3$ in the three small squares, and the nonlinear transformation is again $(X,Y)=(U^3,V^3)$.
On the other hand, if you know the copula and the marginals, then you know the joint distribution so surely the correlation is then determined.