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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?

develarist
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  • My initial guess would have been $\Sigma = \begin{bmatrix} 1 & \rho \ \rho & 1 \end{bmatrix}$ but the numerator in the exponential doesn't seem to match up... – angryavian Nov 23 '20 at 01:37
  • How about $\Sigma=\left[\begin{array}{cc} \sigma^{2} & \rho \sigma^{2} \ \rho \sigma^{2} & \sigma^{2} \end{array}\right]$? – develarist Nov 23 '20 at 01:38
  • Also, $$\frac{\det(\Sigma_{0})}{\det(\Sigma_{1})} = \frac{\sigma_{11}\sigma_{22}-\sigma_{12}^{2} }{ \sigma_{11}\sigma_{22}} = 1-\frac{\sigma_{12}^{2}}{\sigma_{11}\sigma_{22}} = 1-\rho^{2}$$ where $\Sigma_{0}$ is the covariance matrix of the multivariate Gaussian in question and $\Sigma_{1}$ is the covariance matrix of the marginal product measure – develarist Nov 23 '20 at 01:47

1 Answers1

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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,

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

RRL
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  • why does your $\Sigma$ look like a correlation matrix? I thought it was supposed to be a covariance matrix – develarist Nov 24 '20 at 05:00
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    As per the starting point of your linked answer $\Phi$ is a standard univariate normal distribution function (variance is $1$) and $\Sigma$ is the correlation matrix which is also a covariance matrix when the marginals have unit variance. Furthermore the desired form of $c(u_1,u_2;\rho)$ does not depend on $\sigma_i \neq 1$. – RRL Nov 24 '20 at 06:16