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I know the Nadaraya-Watson kernel regression. What is new to me is the Kernel ridge regression from scitkit-learn's KernelRidge with kernel='rbf'. It mentions that the kernel trick is used, which is not done in the Nadaraya-Watson kernel regression, so it seems to me they are different concepts. Am I right, or are they the same afterall?

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migrated from stackoverflow.com Nov 9 '17 at 21:17

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Yeah, you are right. You practically replace the square matrix $X^TX$ with a Kernel $K$ when you estimate your coefficients.

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    $\begingroup$ Can you elaborate? Kernel-Trick based methods are considered to be non-parametric, are they not? $\endgroup$ – Make42 Nov 15 '17 at 9:41
  • $\begingroup$ I dropped it since it seems to the distinction seems to be confusing. Parametric regression models, I would argue, establish a clear dependency of Y on X given by the coefficients (dimension of coefficients is finite). This is also what kernelized ridge regression does, since your $beta$ captures this. Non-parametric models do not establish this dependency in the same way. Reading your comment I am not sure if you still can say that KRR is parametric, because the dependency Y on X is not so clear anymore due to the kernel. $\endgroup$ – Simon Dirmeier Nov 15 '17 at 10:00
  • $\begingroup$ With "beta", do you mean the factor before the regularization sum (attributing for the "ridge" in the name)? I would say this is a model-hyperparameter, but not a "standard" model-parameter (which a coefficient would be). Or do you mean a kernel-parameter, like a kernel scale parameter? Btw, my professor considers methods non-parametric if there is an arbitrary number of parameters. $\endgroup$ – Make42 Nov 15 '17 at 12:20
  • $\begingroup$ No, with beta I mean the coefficients you are estimating. The estimator for beta for classical ridge is $beta = (X^T X + alpha I)^{-1} X^T y$ which you can also solve like this: $beta = X^T (XX^T + alpha I)^{-1} y $. Beta is p-dimensional, as the number of covariates. In kernel ridge regression you replace $X X^T$ with a kernel. So, I would argue that this is still a parametric approach, if we take the definition of a parametric model being one that has coefficients with of finite dimensionality (p). However, since you operate in some high-dimensional feature space now, I might be wrong. $\endgroup$ – Simon Dirmeier Nov 15 '17 at 14:55

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