[SciPy-User] scipy.interpolate.rbf: how is "smooth" defined?
josef.pktd@gmai...
josef.pktd@gmai...
Mon Aug 30 13:38:42 CDT 2010
On Mon, Aug 30, 2010 at 2:15 PM, Gael Varoquaux
<gael.varoquaux@normalesup.org> wrote:
> On Mon, Aug 30, 2010 at 12:45:27PM -0400, josef.pktd@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.
Maybe you find out before me, with all the other things I don't find
time for, Ridge, informative Bayesian priors and Tychonov will have to
wait.
Josef
>
>> 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
> _______________________________________________
> SciPy-User mailing list
> SciPy-User@scipy.org
> http://mail.scipy.org/mailman/listinfo/scipy-user
>
More information about the SciPy-User
mailing list