[SciPy-User] scipy.interpolate.rbf: how is "smooth" defined?

[Gael Varoquaux]

On Mon, Aug 30, 2010 at 12:45:27PM -0400, josef.pktd at gmail.com wrote:
> If you use this with moment/normal equations for penalized
> least-squares, do you have to multiply also `dot(x,y)/n_samples` by
> `alpha/delta` ?

I do not know, and I do not want to reply with something stupid. If you
find out the answer, I would be interest.

> In the normalization of RBF, smooth would be `beta/delta * mu /
> (alpha/delta)`  or maybe additionally also divided by n_samples or
> something like this.

I believe it would be 'beta/delta * mu / (alpha/delta)', but I am not
sure whether or not the multiplication with n_samples should be applied,
as I haven't really looked at the exact formulation of the problem in the
context of RBFs (sorry, no time). All I know is that the trace of the
covariance matrix estimate should be kept constant as much as possible.

Gaël