On Wed, Nov 14, 2012 at 12:33 PM, Sean Owen <srowen@gmail.com> 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 <dangeorge.filimon@gmail.com>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!
>>
