I am trying to create a joint distribution that has a specific copula (e.g. Clayton) and whose marginals come from a mixture distribution (e.g. the mixture of two Gaussian distributions). My idea was to first sample from the copula and then transform the marginals using the inverse of the marginal CDFs.

Since I am working with mixed marginals, I don't have an analytic expression of the inverse CDF. My first workaround was to try this method:

Say I have a mixture of two marginals, marginal 1 has inverse CDF $P_1$ and marginal 2 has inverse CDF $P_2$. I also have copula samples $C$. For each of the two dimensions of the copula samples I follow this approach:

  1. Draw samples from a uniform distribution
  2. Use the uniform samples to decide which inverse CDF ($P_1$ or $P_2$) to use for transforming the copula sample dimension.

However, the resulting joint distribution does not seem to have the same copula structure.

Is there a way to have an analytical expression of the inverse of the mixed CDF? I am now calculating it by numerically inverting the CDF, but this takes quite long and I would prefer an analytical expression. If there is no analytical expression, is there another workaround that preserves the copula structure of the joint distribution?

  • 1
    $\begingroup$ I am missing something: a copula is intended to preserve marginals, however they are defined. Is this a practical issue where computing the cdf is impossible? In the Gaussian mixture case this is not an issue. $\endgroup$
    – Xi'an
    Jan 23 at 18:23
  • $\begingroup$ My issue is that I don't have the inverse of the Gaussian mixture CDF, which I would like to use to transform the marginals from being uniform to being from the Gaussian Mixture Distribution. The workaround I am using at the moment, with first transforming from one distribution and then the other seems to change the copula, probably because I am disregarding the overlap between the two Gaussians that form the mixture. $\endgroup$
    – bossemel
    Jan 23 at 19:08
  • $\begingroup$ One can write down the density of the random variable though, meaning simulation algorithms like Gibbs sampling are available. $\endgroup$
    – Xi'an
    Jan 23 at 21:02
  • $\begingroup$ In what sense do you not "have the inverse"? Practically any specification of a Gaussian mixture provides the data needed to invert its distribution function. $\endgroup$
    – whuber
    Jan 23 at 22:06
  • $\begingroup$ What do you mean by data exactly? I am now numerically inverting the CDF of the mixture. Is there a more direct way? $\endgroup$
    – bossemel
    Jan 24 at 23:57

Your Answer

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

Browse other questions tagged or ask your own question.