Derivation of bivariate Gaussian copula density

bivariate-distributionscopulacorrelationexponential functiongaussian

The multivariate Gaussian copula density, derived here, is

$$c(u_1,\ldots,u_n;\Sigma)=|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}(\Sigma^{-1}-I)x\right)$$
where $\Sigma$ is the covariance matrix, and $x=[\Phi^{-1}(u_1),\ldots,\Phi^{-1}(u_n)]^{\top}$.

The bivariate Gaussian copula density, based on the pair-wise correlation coefficient $\rho$, is
$$
c\left(u_{1}, u_{2} ; \rho\right)=\frac{1}{\sqrt{1-\rho^{2}}} \exp \left\{-\frac{\rho^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho x_{1} x_{2}}{2\left(1-\rho^{2}\right)}\right\}
$$

What is the derivation of the second formula from the first?

Best Answer

Note that with standard normal marginals

$$\Sigma=\left[\begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array}\right],\,\, |\Sigma| = 1 - \rho^2$$

and

$$\Sigma^{-1}= \frac{1}{1- \rho^2}\left[\begin{array}{cc} 1 & -\rho \\ -\rho & 1 \end{array}\right], \,\, \Sigma^{-1}-I= \frac{1}{1- \rho^2}\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho & \rho^2 \end{array}\right]$$

Hence,

$$- \frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x} = \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho & \rho^2 \end{array}\right]\left[\begin{array}{cc} x_1 \\ x_2 \end{array}\right] \\= \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2x_1 -\rho x_2 \\ -\rho x_1 + \rho^2 x_2 \end{array}\right] \\= -\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)},$$

and, thus,

$$|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x}\right) = \frac{1}{\sqrt{1- \rho^2}} \exp \left\{-\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)} \right\}$$