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In order to draw a sample from an N-dimensional Gaussian copula, we draw N independent standard Gaussian random variables, form a vector, and multiply it by an appropriate matrix (Cholesky and such). Each independent standard Gaussian can be computed by drawing a sample from the uniform distribution on [0, 1] followed by an application of the inverse of the standard Gaussian CDF.

Question: Given an arbitrary N-dimensional copula and N independent samples uniformly distributed on [0, 1], is it possible to demonstrate that there is always a transformation which allows one to obtain a sample from the copula using only the uniform samples?

Update: I’m aware of the Rosenblatt transformation. The inverse of this transformation might be an answer to my question. I don’t have deep knowledge in this area, and I’m looking for a person who could consolidate all ideas and describe the whole process step by step.

Thank you.

Regards, Ivan

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  • $\begingroup$ What kind of "transformation" are you looking for? I am puzzled, because according to the definitions of copula I know (such as in Nelsen's book), it is defined as a distribution $F$ on $[0,1]^N$, so one can "sample" from it with a uniform vector $(x_1,\ldots,x_N)$ merely by computing $F(x_1,\ldots,x_N)$. $\endgroup$ – whuber Dec 10 '14 at 17:03
  • $\begingroup$ @whuber, I’m looking for a transformation that would turn N independent uniformly distributed random variables into N dependent uniformly distributed random variables according to the copula. $F(x_1, \dots, x_N)$ is just a single value, isn’t it? Or may be I’m missing your point. Can you please elaborate a bit? Thank you. $\endgroup$ – Ivan Dec 10 '14 at 20:29
  • $\begingroup$ Are you perhaps asking the multidimensional generalization of the question at stats.stackexchange.com/questions/124865/…? $\endgroup$ – whuber Dec 10 '14 at 20:57
  • $\begingroup$ @whuber, thank you for your time. It’s certainly related, but not quite. I’m trying to find a way to correlate independent variables according to a given copula. Please have a look at the update in the original post. $\endgroup$ – Ivan Dec 11 '14 at 9:30
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    $\begingroup$ Yes, I think so. $\endgroup$ – whuber Dec 11 '14 at 15:15

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