[SciPy-User] scipy.interpolate.rbf: how is "smooth" defined?
Mon Aug 30 13:15:28 CDT 2010
On Mon, Aug 30, 2010 at 12:45:27PM -0400, firstname.lastname@example.org 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.
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