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Sklar's theorem states that: for any given n-variate distribution function $F$ with uniform margins $F_1,...,F_n$ then there exist copula $C$, such that: $F(x_1,..,x_n) =C(F_1(x_1),...,F_n(x_n))$, then if all margins is continuous, then copula is unique and defined as:

$$C(u_1,...,u_n)=F(F^{-1}(u_1),...,F^{-1}(u_n))\qquad\qquad (1)$$

Hence, since in copula we need to transform all the margins to uniform, then we need to go back to the original data using the inverse as in $(1)$.

However, I think we do not need to have the inverse of copula because based on Sklar's theorem we can write the density of $F$ (in the bivariate case) as follows:

$$f(x_1,x_2)=c(F(_1(x_1),F_2(x_2)) \, f_1 \,f_2 $$

where $f_1$ and $f_2$ are the densities of the margins and can be any types of margins. Hence because of that, we do not need to have the inverse of the copula. Is that correct? Any help please?

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You don't invert the copula. The transformations (back and forth) are of the marginals.

The conversion to marginal uniformity is achieved by the probability integral transform $U_i=F_{i}(X_i)$. Note that each variable is transformed by its own cdf.

Similarly you transform the uniform margins of the copula back to the original scale via $X_i=F_{i}^{-1}(U_i)$.

Note that the expression you have at the end is for the bivariate density of the original variables, not the variables themselves -- the question was about transforming the variables ("go back to the original data"). That expression will be for the density of the back-transformed variables.

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  • $\begingroup$ Oh, it is simple and great answer. I think the transformation can be done using any type of transformation such as empirical cdf of each margins. $\endgroup$ – user130885 Sep 28 '16 at 2:32
  • $\begingroup$ It is the case that a variety of approaches are used in practice including some based on empirical cdfs. $\endgroup$ – Glen_b -Reinstate Monica Sep 28 '16 at 2:39
  • $\begingroup$ Yes, I meant that we should back to original data but not to invert copula itself. I think probability integral transformation is used when we know the CDF, as I understand from its definition in Wikipedia. Sorry, I really new to this stuff. $\endgroup$ – user130885 Sep 28 '16 at 2:43

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