This is a constructivist question.

Please provide a bi-variate distribution or density/mass function of two absolutely continuous/discrete (but not mixed-type) random variables, which (may) have covariance zero and still be dependent.
But mind the constraints: The density/mass function should not have branches, i.e. each has to be a single mathematical expression for all the support of the distribution, where it is not zero. The same should hold for the marginal distributions. (In a parametric context, I guess the ideal will be to have a parameter governing covariance, and some other parameter(s) governing any other form of dependence, but "hard-wired" zero-covariance is ok for starters). And it would be preferable to avoid exotic supports.

Notes: It may be the case that this is trivially answered using the concept of Copulas. But we do want the marginals here. So any Copula-based answer I hope will be kind enough to explicitly provide also the marginals (or an example of marginals, if the answer provides a general mechanism of distributions-generation). Also, situations like e.g. a Bernoulli whose probability parameter is modeled as random and drawn from some other distribution, are not of interest here.

I am asking this and placing all these restrictions because to date, I find all examples of "zero covariance does not imply independence" absolutely true but artificial. I am thinking about situations where your "everyday" statistics practitioner (not necessarily a high-profile academic/professional) will be called upon to model -and most likely he will tend to try first some "smooth" representation. Does he have the theoretical modeling tools that will permit him to conclude "zero-covariance together with dependence"?

Naturally, I did try but have not yet found any such example -but this only attests to my knowledge gaps, not to whether such bi-variate distributions are hard to find. Meaning, any literature references are also welcome.

  • 2
    $\begingroup$ $X\sim\mathcal{N}(0,1)$ and $Y=X*(-1)^\xi$, where $\xi\in Bernoulli(1/2)$. $X$ and $Y$ are dependent but uncorrelated. $\endgroup$ – Aksakal Jan 6 '15 at 21:07
  • $\begingroup$ @Aksakal That belongs to the "well-known and artificial examples". What is their joint distribution? $\endgroup$ – Alecos Papadopoulos Jan 6 '15 at 21:15
  • $\begingroup$ These constraints seem highly artificial, even to practitioners. They would rule out things like Gamma distributions, for instance (which are not necessarily smooth at $0$ and require a "branch" to describe the negative part). Since copulas automatically apply to any marginal, your objections to their use are hard to fathom. You might have gotten lost in the abstractions. If you have a particular statistical problem where you would like a model that incorporates bivariate distributions having no necessary relationship between covariance and independence, then why not provide that context? $\endgroup$ – whuber Jan 6 '15 at 21:24
  • 2
    $\begingroup$ What is the point you are making here? Is it that in practice if there's no correlation, then there's no dependence? Otherwise, what's the problem with "artificial" distributions? $\endgroup$ – Aksakal Jan 6 '15 at 21:25
  • 3
    $\begingroup$ It should be more specific, because it's unclear why don't you like the well-known example to which I referred. You called "artificial", yet you don't seem to have any particular phenomenon in mind. It's shown many times that dependence does not always manifest in correlation. It's an important observation for anyone who models anything $\endgroup$ – Aksakal Jan 6 '15 at 21:36

Among spherically symmetric distributions, i.e. distributions of the form $f(\boldsymbol x) = \varphi(\|\boldsymbol x\|)$ where $f$ is the density with respect to Lebesgue measure, it can be shown that the coordinates always have zero correlation, essentially due to the fact that the distribution is invariant under orthogonal transformations. This family includes, for example, the multivariate normal distribution, but also things like the multivariate t distribution.

It can be shown that, in fact, the multivariate normal is the only spherically symmetric distribution with independent components. Now, spherical symmetry is in fact a very reasonable property that a statistician might expect a distribution to have in some settings. Hence, inferring independence of components from no correlation makes very strong assumptions about the tails of the distribution in this case!

See for example this paper. See this report for more on the properties of spherically symmetric distributions.

  • $\begingroup$ Thanks for contributing. The main point of my question is that I would like to be able to infer exactly the opposite ("zero-covariance but still dependent"), in a tangible way. I am looking into the references now. $\endgroup$ – Alecos Papadopoulos Jan 6 '15 at 21:07
  • $\begingroup$ @MartijnWeterings True. But I was very young four years ago... $\endgroup$ – Alecos Papadopoulos Jan 21 '19 at 14:21

The last row of the first image of https://en.wikipedia.org/wiki/Correlation_and_dependence provide exemple where the Pearson correlation is 0 but there is strong non-linear dependance between X and Y.

The relation $\rho=0$ is quite strong in general but easily obtainable considering discrete and / or symetrical distributions. That's why you usually end up with discrete / symmetrical distribution as exemples. The article provide an exemple of an assymetrical distribution (but discrete). A sum of three gaussian centered on the three points should yield the same result.

In the real life the symmetry could arise from the set-up of the problem as evoked above. More rarely it could arise from an assymetrical setup, that just happen to give a 0 correlation.


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.