On Wed, Nov 14, 2012 at 3:22 AM, Sean Owen <[email protected]> wrote: > > Right, I probably want a modified version in my case where I normalize > > the distances somehow. > > > > You can divide the result by any scalar you want and it will still have > non-zero probability of being farther than any given distance d from the > mean. Of course, it gets very very unlikely as the distance increases. >
Cosine normalization gives symmetric results on a sphere. You can also use a rejection method to drop all points outside the sphere. > > So "radius" in this case is the variance of the ith component of the > > vector, since the covariance matrix is diagonal? > > > > > I don't think that's a covariance matrix in general given how it is > applied. The argument in question is used as a standard deviation rather > than (co)variance from what I can see. > Standard deviation is just the 1-d version of covariance. If you want a spherically symmetric multi-variate normal, then a diagonal covariance does the trick.
