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Can someone explain to me why the CopulaDistribution function with a "Multinormal" kernal asks for the covariance matrix instead of the correlation matrix. Just by looking at the formula for copula, it uses correlation matrix. So I was wondering why it is so for mathematica.

Edit: So what difference would it make if I use correlation matrix instead of covariance matrix?

Edit: After some research, I realised that for this particular case, the theoretical input is in fact the correlation matrix. However, using this correlation matrix as the input, I simulate 10000 vector probabilities. When I calculate the correlation of these results, I do not end up with the correlation matrix that I started with (in fact it is well off). Does anyone have an idea as to why this may be the case. Surely the simulated correlation matrix should line up with the input correlation?.

Jim
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  • Allowing for the Covariance is more general than using a Correlation matrix. You can assume unit variance and they are equivalent. – Andy Ross Dec 08 '14 at 14:10
  • So, if I understand what you're saying, is that using a Correlation matrix instead of a covariance matrix should not have too much impact on the final result given that I do not have a covariance matrix to use – Jim Dec 08 '14 at 22:08
  • Correlation is expressed as a unitless number, whereas covariance is expressed in units that are the xunitsyunits. The two statistics differ by the product of the standard deviations in the x and y dimensions: Correlation = Covariance/(sigmaX sigmaY). Relying on the covariance is mathematically more parsimonious than relying on the correlation in that the covariance generalizes to the variance in one dimension. – David G. Stork Dec 08 '14 at 22:49
  • Anyone else offer any input? I am really stumped by this. (The second edit) – Jim Dec 11 '14 at 00:11

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