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Suppose the distribution of an $n$-dimensional random vector $X$ is characterized by a Gaussian copula $C_R$ with correlation matrix $R$ and a set of marginal $\{(F_{X_i}, f_{X_i})\}_{i=1}^n$ (pairs of a distribution function and a density function). I would like to derive the probability density of $X$. Wikipedia says that the density of the copula is as follows:

$$ f_{C_R}(u) = \frac{1}{\sqrt{\det{R}}}\exp\left(-\frac{1}{2} \begin{pmatrix}\Phi^{-1}(u_1)\\ \vdots \\ \Phi^{-1}(u_n)\end{pmatrix}^T \cdot \left(R^{-1}-I\right) \cdot \begin{pmatrix}\Phi^{-1}(u_1)\\ \vdots \\ \Phi^{-1}(u_n)\end{pmatrix} \right) $$

Given this information, I decided to apply a change-of-variables transformation and, thereby, express the density of $X$ in terms of the density of $U \sim C_R$. I arrived at the following formula:

$$ f_X(x) = \left| \prod_{i=1}^n f_{X_i}(x_i) \right| \, \, f_{C_R}\left( \begin{pmatrix}F_{X_1}(x_1)\\ \vdots \\ F_{X_n}(x_n)\end{pmatrix} \right) $$

The coefficient before $f_{C_R}$ is the absolute value of the determinant of the Jacobian of the inverse of the transformation from $U$ to $X$, which is just an application of $\{F^{-1}_{X_i}\}$ to the corresponding elements of $U$.

Since I do not have a proper background, I am not entirely sure should that all the steps that I have taken are legitimate. I, therefore, would be grateful if somebody could confirm that the above computations are correct.

I also wonder why $f_{C_R}$ differs so much from the density of a centered multivariate Gaussian vector:

$$ f_Y(y) = \frac{1}{\sqrt{(2\pi)^n \det{\Sigma}}} \exp\left(-\frac{1}{2}y^\mathrm{T}\Sigma^{-1}y \right). $$

How did $-I$ in $f_{C_R}$ come about? And why does not $f_{C_R}$ have $\sqrt{(2\pi)^n}$? It would be nice if somebody could explain this.

Thank you!

Regards, Ivan

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The Gaussian copula (with correlation matrix $R$): $$ C_R(u)=\Phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R). $$

Then, letting $\zeta:=(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n))^{\top}$, $$ c_R(u)=\frac{\phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R)}{\phi(\Phi^{-1}(u_1))\cdots \phi(\Phi^{-1}(u_1))} \\=(2\pi)^{-n/2}\lvert R \rvert^{-1/2}\frac{-\exp(\zeta^{\top}R^{-1}\zeta/2)}{\prod_{i=1}^n (2\pi)^{-1/2}\exp(-\zeta_i^2/2)}=\lvert R \rvert^{-1/2}\frac{\exp(-\zeta^{\top}R^{-1}\zeta/2)}{\exp(-\zeta^{\top}I_n\zeta/2)}. $$


Here, $\Phi_n(\cdot;R),\phi_n(\cdot;R)$ are zero-mean multivariate normal CDF and PDF with covariance matrix $R:n\times n$.

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  • $\begingroup$ That section on Wikipedia you referred to also contains a general formula for the density of a multivariate distribution given its copula and marginals, which I totally missed: $$f(x_1,\dots x_d)= c(F_1(x_1),\dots F_d(x_d))\cdot f_1(x_1)\cdot\dots\cdot f_d(x_d).$$It seems to be exactly what I derived except for the redundant $|\cdot|$. $\endgroup$ – Ivan May 3 '16 at 4:38
  • $\begingroup$ Thank you for the explanation! $\endgroup$ – Ivan May 3 '16 at 4:46

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