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I want to generate a random vector $z$ of dimension $k+m$ with some given correlation matrix $\Sigma$, such that the first $k$ elements of the vector are distributed normally and the last $m$ elements follow the Gamma distribution with some given parameters $a,b$.

Hier is suggested (applied to this case) to generate a normal r.v. Z as $N(0,\Sigma)$ and then solve $G_{[a,b]}(Y_i)=\Phi_{[0,\Sigma]}(Z_i), i\geq m$ and replace the last m elements of Z with Ys, however it's not guaranteed that the vector $(Z_1, ... , Z_k, Y_{k+1}, ..., Y_{k+m})$ will still have the correlation matrix $\Sigma$.

Is there some nice copula out there that does the job or some other approach?

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2 Answers 2

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Look here for multivariate distribution on the positive quadrant ... quite difficult. http://xianblog.wordpress.com/tag/multivariate-analysis/ I have been thinking about this for weeks and months in the context of credit risk (modelling default intensities jointly) and I think this does not work.

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  • $\begingroup$ Assume that this mixture of distribution is given and I can't escape into log-normal and to some extent tolerant towards negative values. Simulating into positive quadrant with given covariance is also interesting, thanks for this! $\endgroup$
    – Max Li
    Jan 13, 2013 at 20:22
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In Oracle Crystal Ball (or in a few other Excel based MC simulation add-ins), we can do this without much hassle: define $k$ normal distribution and $m$ gamma distributions. Define (or load) the correlation matrix $\Sigma$ and then generate the random variates. Each trial run would give you one random vector $z$ of dimension $k+m$.

We use normal copula to generate correlated random numbers from risk distributions in Oracle Crystal Ball.

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  • $\begingroup$ can you elaborate on "and then generate the random variates"? Do you generate k normals and m gammas and the put it into a copula? $\endgroup$
    – Max Li
    Jan 16, 2013 at 15:51
  • $\begingroup$ We generate streams of uniforms, put them through a normal copula to rotate according to the correlation value, and then use inverse PDF to go back to the distribution. $\endgroup$
    – Samik R
    Jan 17, 2013 at 21:51
  • $\begingroup$ from these 3 steps, they are correlated with the desired correlation matrix after the 2nd step, but not after the final 3rd step. correct me if I'm wrong $\endgroup$
    – Max Li
    Jan 20, 2013 at 18:57
  • $\begingroup$ We get streams of uniform numbers with the desired correlation after 2nd step, and then using the inverse PDF, we retain the correlation structure after 3rd step. $\endgroup$
    – Samik R
    Jan 21, 2013 at 19:37

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