On Wed, Nov 14, 2012 at 1:22 PM, 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.
That's true. I'll just drop points that are too far away. > >> 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. Heh, so, know any nice introductory articles (preferably with lots of pictures) about the multivariate distribution? Thanks! :)
