> 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. > 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.
