Gaussian Copula and the addition of an Identity matrix - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T07:35:50Zhttp://mathoverflow.net/feeds/question/96007http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/96007/gaussian-copula-and-the-addition-of-an-identity-matrixGaussian Copula and the addition of an Identity matrix Moe2012-05-04T18:40:41Z2012-08-26T16:06:43Z
<p>When I was looking at the Gaussian Copula Example @ <a href="http://en.wikipedia.org/wiki/Copula_(probability_theory" rel="nofollow">http://en.wikipedia.org/wiki/Copula_(probability_theory</a>)</p>
<p>I realized the Gaussian Copula is stated as follow
\begin{equation}
C^{Gauss}_\Sigma (u) = \frac{1}{\sqrt\det{\Sigma}} \exp{\Bigg ( -\frac{1}{2} \begin{pmatrix} \Phi^{-1}(u_1) \ \dots \ \Phi^{-1}(u_d)\end{pmatrix}^T. (\Sigma^{-1} - I).\begin{pmatrix} \Phi^{-1}(u_1) \ \dots \ \Phi^{-1}(u_d)\end{pmatrix} \Bigg) } \end{equation}
where $\Sigma$ is the correlation matrix, $\Phi^{-1}$ is the inverse cumulative distribution function of a standard normal and $I$ is the identity matrix.</p>
<p>The question is, why is there an identity matrix in the exponential form?</p>
<p>Thank you</p>
http://mathoverflow.net/questions/96007/gaussian-copula-and-the-addition-of-an-identity-matrix/96009#96009Answer by mike for Gaussian Copula and the addition of an Identity matrix mike2012-05-04T18:55:43Z2012-05-04T18:55:43Z<p>It's the jacobian.</p>
http://mathoverflow.net/questions/96007/gaussian-copula-and-the-addition-of-an-identity-matrix/105548#105548Answer by mathtick for Gaussian Copula and the addition of an Identity matrix mathtick2012-08-26T16:06:43Z2012-08-26T16:06:43Z<p>What's a good reference for this derivation (online). Wikipedia only has an advertisement for someone's book as a reference. </p>
<p>I find this presentation very confusing, for example I had missing that this was the density (small c) as opposed to the CDF and was obvious not making any sense of it. I can imagine other readers doing the same thing.</p>