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How does the gaussian copula give probability of a sample x? #530

@leanderlindgren

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@leanderlindgren

Problem description

With the gaussian copula, you are modeling a distribution with the cumulative distribution function $\text{F}(\mathbf{x}) = \Phi_R(\Phi^{-1}(\text{G}_1(x_1)), \dots, \Phi^{-1}(\text{G}_d(x_d)))$, where $\text{G}_i$ is the cumulative distribution function of the $i$-th margin. If I want to call the method probability_density(x), then as far as I can tell, it gives me $p(z) = \phi_R(\Phi^{-1}(\text{G}_1(x_1)), \dots, \Phi^{-1}(\text{G}_d(x_d)))$. But doesn't that function give me the probability of $z$, given by $z = (\Phi^{-1}(\text{G}_1(x_1)), \dots, \Phi^{-1}(\text{G}_d(x_d)))$, instead of $p(x)$? Because to get $f(x)$, you take the derivative of $\text{F}(x)$, so you need to apply the chain rule on the right hand side, to get $p(x) = f(x) = \phi_R(\Phi^{-1}(\text{G}_1(x_1)), \dots, \Phi^{-1}(\text{G}_d(x_d))) \prod _{i=1}^{d} \frac{g_i(x)}{\phi(\Phi^{-1}(\text{G}_i(x_i)))}$. Are you accounting for this anywhere? Because in the documentation, it says the method gives probability density values for points in x.

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