Does anyone know if are there some assumptions for Copula method? I heard from someone that the data should be i.i.d (independent and identically distributed). Let's say, if I want to capture the dependence structure between two variables. I have to use marginal distribution to transform data to rank space, then I can fit a theoretical Copula function. Is there an assumption for the marginals that the data should be i.i.d? If so, why? And does one apply Copula method if the time series are not i.i.d?
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2 Answers
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Yes, you need to have independent observations to apply conventional copulas. If you have serial correlation in data, look for "autocopulas", e.g. here "Autocopulas: investigating the interdependence structure of stationary time series" by Rakonczai et al.
The reason is that you're looking for a wrong joint. Consider this, you're trying to recover the joint distribution $F(X,Y)$ with copulas, where $X,Y$ are random variables. However, in case of first order serial correlation in observations (autocorrelation) the joint distribution is $F(X_t,X_{-1},Y_t,Y_{t-1})$ not the one you're trying to recover.
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1$\begingroup$ if the marginals are non-i.i.d., then transforming them to rank space would shuffle the original data out of order right, removing their dependence and make them i.i.d.? how are autocopulas different w.r.t. this ranking step? $\endgroup$ Commented Sep 1, 2020 at 0:22
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Copulas are based on Sklar's theorem and this theorem does not require independency. Besides, there exists an independence copula such that: $$F(x_1, \ldots , x_d ) = \Pr(X_1 ≤ x_1, \ldots , X_d ≤ x_d )$$ $$= \Pr(X_1 ≤ x_1)\cdots\Pr(X_d ≤ x_d ) = F_1(x_1)\cdots F_d(x_d)$$ and $$ F(x_1, \ldots , x_d ) $$ $$= C(F_1(x_1), \ldots , F_d(x_d))$$
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6$\begingroup$ I believe the nature of this question is different from the spirit of your answer, Dennis: in this context, the "Copula method" must refer to some procedure to estimate multivariate distributions from data. $\endgroup$– whuber ♦Commented May 28, 2018 at 12:56