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I'm wondering if I can get the best of both worlds of the Gaussian and T copulas (and if not, why not?). A property of the Gaussian copula that I like is that if the off-diagonal entry into the correlation matrix is 0, X and Y are independent. This is not true with the T copula. Because correlation increases in the tails, having 0 correlation in your correlation matrix does not guarantee independence under the T-copula. Is there a way I can get tail dependence (increased correlation for extreme values) for positively correlated variables, while getting no tail dependence (and independence) for uncorrelated variables?

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  • $\begingroup$ how about plug Student t marginals into Gaussian copula? $\endgroup$
    – Aksakal
    Jul 25, 2018 at 17:17
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    $\begingroup$ Regardless of the marginals you use, the Gaussian copula will have no tail dependence. $\endgroup$
    – jntrcs
    Jul 25, 2018 at 18:02
  • $\begingroup$ The questions is a bit ambiguous (i.e. correlation vs independence). For what I understood, the following might provide an answer/some hints: For large degrees of freedom, the t copula convergences to the Gaussian copula, hence one could tune that parameter - see www.copulatheque.org for a quick visualization. Another option would be to pick a copula family that actually includes the product copula (e.g. Gumbel), or to build a "new" family by mixing the product copula with the t copula in a convex manner for the cost of an additional parameter. $\endgroup$
    – Ben
    Jul 31, 2018 at 6:05
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    $\begingroup$ You'll have to describe what's vague about my use of correlation and independence--I thought I was using standard statistical definitions. I would really like to look at that link you shared, but the app is crashing every time I click anything. I've done a bit more research on the T-copula. The problem is even if cor(X, Y)=0, if X is conditioned on Y, and y is an extreme value, x will be in the extreme upper or extreme lower of it's distribution, it will not be in the middle. Likewise if Y is in the middle, X will not be in its extremes. $\endgroup$
    – jntrcs
    Jul 31, 2018 at 19:17
  • $\begingroup$ You can try using Gaussian copula with Student t marginals. $\endgroup$
    – Aksakal
    Jun 22, 2020 at 19:16

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The best way to think of the Gaussian and t copulas is as a single family with two parameters: the correlation value and the degrees of freedom. There is one combination of those parameters that is equivalent to the independence copula, namely, $\rho=0$ and $\nu = \infty$ (where $\rho$ is the off-diagonal term in the correlation matrix and $\nu$ is the degrees of freedom). Any other combination will give you either correlation, tail dependence, or both.

So, to do what you want to do, which is to create a one-parameter family that has the independence copula as a limiting case, one easy way to do it is to take a curve in the $\rho -\nu$ plane that passes through $(0, \infty)$ and come up with a convenient parameterization for the position along that curve. There are infinitely many such curves, and which one you want depends on what kind of relationship you want to have between correlation and tail dependence. One simple prescription is to have a parameter $\theta \in [0,1]$ and let $\rho = \theta$ and $\nu = 1/\theta$. Then you get dependence coefficients that look something like this:

plot of dependence coefficients

If that's not exactly what you are looking for, you could play around with the parameterization in various ways to get the kind of relationship between $\rho$ and $\lambda$ that you want, but the thing to remember is that there is no single, natural relationship between the two. In the end, whatever relationship you come up with is a choice, which should be motivated by the needs of whatever problem you are planning to use the copulas to solve.

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