I'm talking about the case here where covariances are 0. The marginals in each dimension are independent and are normally distributed. Right?
What is that matrix connecting the univariate normals to multivariate BTW? 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. Yes that is right about sampling, you can't sample the radius uniformly! Very important detail to overlook. On Wed, Nov 14, 2012 at 5:38 PM, Ted Dunning <[email protected]> wrote: > On Wed, Nov 14, 2012 at 3:53 AM, Sean Owen <[email protected]> wrote: > > > The wikipedia article is a fine intro. If your covariances are 0, there's > > not much to know at all. The multivariate normal is just several > univariate > > normals, independent in each dimension. > > > > Not quite. The multivariate normal is several univariate normals > multiplied by a matrix. > > > > If you want a uniform distribution over a unit sphere, that's different, > > but you're actually also on the right track. You don't need to sample and > > discard, just pick your point as above and normalize to a length randomly > > chosen in (0,radius]. 90% sure that's correct off the top of my head. > > > > In high dimensions, this doesn't work because you get vastly too much mass > near the origin. You need to sample the radius from a distribution biased > toward larger values. The idea is that each shell of radius r and > thickness dr needs to have probability according to the volume of the shell > which is proportional to r^(d-1). If every shell has equal mass, then the > inner shells are much too dense. The desired cumulative distribution is > proportional to r^d so you can sample r by using the inverse method > > u ~ Uniform(0,1) > r = Math.pow(u, 1/d) > > See https://dl.dropbox.com/u/36863361/spherical-sampling.png for a picture > of 2-d sampling which is the first interesting case. For higher > dimensional cases, to get a comparable picture, you need to take a slice of > data near a 2-d plane since simply projecting to the x-y plane will give > very biased results. >
