On Wed, Nov 14, 2012 at 9:48 AM, Sean Owen <[email protected]> wrote: > I'm talking about the case here where covariances are 0. The marginals in > each dimension are independent and are normally distributed. Right? >
Yes. With no covariance, all of the axes are independent. > What is that matrix connecting the univariate normals to multivariate BTW? > http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Drawing_values_from_the_distribution > I don't know what it is theoretically. It's not the covariance matrix, > which might be a reader's first guess on looking at the constructor. > Should I fix the constructor to accept covariance? The SVD or Cholesky decomposition required is pretty quick.
