Thank you! I think I did just the way you described. Pierre-Edouard Portier
On Wed, Jan 11, 2017 at 12:25:56PM -0500, Brian Schott wrote: > Your example has 0 valued off-diagonal elements in sq, so there is no > covariance and 2 independent gaussians can be used. > > But it your sg had non-zero off-diagonals then according to a very old > reference, if you can find lower triangular matrix c such that sg -: c+/ . > *|:c then you can generate X=(c+/ . *Z)+mu where Z are standard normal and > your desired result is X . > > I am assuming you have access to a standard normal random generator. One, > which I have not used is the verb normalrandom in stats/base/random.ijs . > > > -- > (B=) <-----my sig > Brian Schott > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
