Timeline for VC dimension, fat-shattering dimension, and other complexity measures, of a class BV functions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 17 at 1:03 | comment | added | ABIM | @BenoîtKloeckner Are their bounds in the literature for the covering number is o-BV functions for positive p? | |
| Mar 2, 2022 at 18:30 | comment | added | Benoît Kloeckner | @TomTheQuant Sorry, this goes a bit too far back and I do not have a reference on hands. | |
| Feb 28, 2022 at 20:08 | comment | added | ABIM | @BenoîtKloeckner Do you have a link for the $C^k$ case you mention when $\mathcal{X}$ is a complete Riemannian manifold; esp. Euclidean space? | |
| Feb 20, 2021 at 9:02 | history | edited | Benoît Kloeckner | CC BY-SA 4.0 |
Corrected a typo in BV ($1/p$ plays the role of the Hölder exponent)
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| Feb 18, 2021 at 22:49 | vote | accept | dohmatob | ||
| Aug 1, 2018 at 12:19 | comment | added | Benoît Kloeckner | @dohmatob: yes if $\varepsilon$ is fixed and $\alpha\to 0$, but then it amounts to looking only at constant functions, and more importantly it says nothing if $\varepsilon$ goes to $0$ as well (and the limits cannot be exchanged here). | |
| Aug 1, 2018 at 11:56 | comment | added | dohmatob | Thanks, I see the big picture. One last take: In the limit $\alpha \rightarrow 0^+$, $\mathcal H_\alpha$ "approaches" the set of constant functions from $\mathcal X$ to $[0,1]$ (as all the functions in it are then essentially constants). Thus in this limit, one would then expect the covering number of $\mathcal H_\alpha$ to be essentially that of the interval $[0,1]$ (independent of the geometry of $\mathcal X$), namely $1/\epsilon$. No ? | |
| Aug 1, 2018 at 11:18 | comment | added | Benoît Kloeckner | @dohmatob: no, your patch would not fix anything. $1/2$ has no particular role, it could be replaced by any value. You do need to use the geometry of $\mathcal{X}$ in some way: any condition that is invariant by bijective map will give you the same complexity for $[0,1]$ or $[0,1]^{124643234}$ - hence either trivial or infinite. | |
| Aug 1, 2018 at 11:03 | comment | added | dohmatob | Thanks for the generous response (upvoted). (A) What if we impose $\alpha \ll 1$ and $|\bar{h} - 1/2| \gg \alpha$ for all $h$ (see the of definition of $\mathcal H_\alpha$) ? This will sure diminish the richness of $\mathcal H_\alpha$, right ? Can anything be salvaged from the wreckage ? (B) What minimal conditions could be imposed on the $\bar{h}$'s so as to have something +ve to say about the claim ? | |
| Aug 1, 2018 at 10:29 | history | answered | Benoît Kloeckner | CC BY-SA 4.0 |