Timeline for Lipschitzness of posterior mean of Gaussian process?
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| Aug 26 at 10:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 16 at 16:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 18, 2023 at 13:09 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 3, 2023 at 16:01 | comment | added | Banach | Can it be independent of $X$? Imagine $x_1 = x_2 = x_3 = ... = x_n$. My line of attack would be to assume something on filling and separation distance of points in $X$ and use, say, Teckentrup (2020, arxiv.org/abs/1909.00232 ) that bounds uniformly $|\mu - f|$. | |
| Jan 3, 2023 at 15:43 | answer | added | ccriscitiello | timeline score: 0 | |
| Jan 2, 2023 at 21:00 | history | tweeted | twitter.com/StackStats/status/1610018107149492231 | ||
| Jan 2, 2023 at 19:54 | history | edited | ccriscitiello | CC BY-SA 4.0 |
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| Jan 2, 2023 at 19:53 | comment | added | ccriscitiello | Agreed, that is the important part. I will put it in italics. | |
| Jan 2, 2023 at 19:35 | comment | added | Yves | Yes a path $f$ lies in the RKHS with probability zero but the Kriging mean $\mu$ is always in the RKHS. Again, the important part of the question is for me independent of $X$ and maybe you should emphasize this in the question e.g. using italics? | |
| Jan 2, 2023 at 19:25 | comment | added | ccriscitiello | @Yves Thanks for the response! As far as I see it, Comment (3) doesn't quite give the answer because $\|\mu\|_K = \alpha^T K(X,X) \alpha = y^T K(X, X)^{-1} y$, but it is not clear how to bound this quantity (independently of $X$). For the exponential kernel $k(x,y) = \exp(-|x-y|)$, we don't expect the sample paths to be Lipschitz, so of course we don't expect the kriging mean to Lipschitz either. However, for kernels which are sufficiently smooth, the sample paths will be Lipschitz whp and so it is reasonable to expect the kriging mean to be as well (with a similar Lipschitz constant). | |
| Jan 2, 2023 at 18:56 | comment | added | Yves | As for the Lipschitz condition, you gave the answer (yes) in comment (3)! However for the important part independent of $X$ I think that the answer is no. Think of the exponential kernel: the smoothness of the paths is that of Brownian paths. When the design $X$ becomes dense the kriging mean is very rough. | |
| Jan 2, 2023 at 17:54 | history | edited | ccriscitiello | CC BY-SA 4.0 |
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| S Jan 2, 2023 at 17:45 | review | First questions | |||
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| S Jan 2, 2023 at 17:45 | history | asked | ccriscitiello | CC BY-SA 4.0 |