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comparing to svm...

The claim: "RVM avoids the set of free parameters of the SVM (that usually require cross-validation-based post-optimizations)" does not make sense to me.

First, RVM also uses a regularization-like a priori distribution for weights. Usually the distribution is simply assumed as a Gaussian distribution, which have already made a questionable assumption. The width of the Gaussian distribution is also unknown, and should be estimated by cross-validation. (Sure you can estimate the parameter from the training data, which is also applicable for the parameter of SVM)

Second, the most important and the most influential parameter of SVM is not the weight of regularization, but the parameter of the kernel. RVM does not touch this issue. — Preceding unsigned comment added by Eyounx (talkcontribs) 03:41, 3 April 2011 (UTC)[reply]

Stub???

This article is a stub that tell very little about RVMs. No practitioner could just read this and understand what an RVM is, even with some background in SVMs, unless they have a lot of additional background on Gaussian processes since the only vaguely elucidating comment is how they are equivalent to a particular sort of Gaussian process.

This should certainly have the "THIS ARTICLE IS A STUB" disclaimer and be soliciting for additional information and sections to be added. — Preceding unsigned comment added by 99.109.19.38 (talk) 06:20, 7 May 2012 (UTC)[reply]

Yah, I think that is implicitly understood by all. Sadly most articles in math on WP are stub class, so no surprise here... linas (talk) 22:52, 14 July 2012 (UTC)[reply]

Enough information in the references

There's enough information in the references already, anyone wants to help gather it all? thisbugisonfire (talk) 14:09, 24 April 2016 (UTC)[reply]

"... an identical functional form to SVM, but provides probabilistic classification." Lifted verbatim from Lavecchia, A 2015 Drug Discovery Today