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From Dan Filimon <dangeorge.fili...@gmail.com>
Date Wed, 14 Nov 2012 11:18:53 GMT
```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

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!
>>

```
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