On Wed, Nov 14, 2012 at 12:33 PM, Sean Owen <[email protected]> wrote: > In the univariate case, there is no max/min possible value. We just have > the variance to say how unlikely a value is that is far from the > distribution mean, though any value is possible. Same in multivariate, so I > don't think you could say the distribution fits strictly inside a sphere.
Right, I probably want a modified version in my case where I normalize the distances somehow. > The distribution will only be symmetrical and not 'elongated' if the > variances are the same, which is the case I think you're talking about. > > Ted I am also confused by the naming in this class. What I'd imagine is the > vector of means is called "offset". The variances come in to the picture > via a matrix called "mean". (That's not the covariance matrix right? might > expect that from an API perspective but I don't think that's how it is > used.) And the parameter for the case where all variances are the same is > "radius". So "radius" in this case is the variance of the ith component of the vector, since the covariance matrix is diagonal? > > On Wed, Nov 14, 2012 at 8:32 AM, Dan Filimon > <[email protected]>wrote: > >> Hi, >> >> I'm familiar with the basic univariate normal distribution but am >> having trouble understanding how the Mahout multivariate normal >> distribution works. >> >> Specifically, what does the radius of the distribution stand for? >> What I'm imagining (at lest for 3 dimensions) is that all points would >> fit into a sphere centered in the mean with the given radius and that >> they would be normally distributed inside. >> >> This however doesn't seem to be the case (unless my tests are broken). >> >> What am I missing? >> Thanks! >>
