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May 26, 2021 at 17:31 comment added Iosif Pinelis @RajeshD : Substitute $Kn^{-1/m}\ln^{1/m}n$ for $\varepsilon$, where $K=c^{1/m}C$. Then the upper bound in (2) will go to $0$, which implies $\zeta_n=O_P(n^{-1/m}\ln^{1/m}n)$.
May 26, 2021 at 6:49 comment added Rajesh D @losif : It would be helpful if you could expand the last statement of your answer, how (2) implies the final asymptote in probability.
May 23, 2021 at 2:09 comment added Iosif Pinelis @RajeshD : Right now, I do not have good ideas about the lower bound, but I will will have this in mind.
May 22, 2021 at 6:54 vote accept Rajesh D
May 22, 2021 at 6:39 comment added Rajesh D for a lower bound, we need a different inequality on the probabilities though.
May 22, 2021 at 4:30 comment added Rajesh D @losif : Yes. one last thing, can we derive a lower bound as well. As I see there exist another constant $c$ such that $N_{\epsilon}>= (c/\epsilon)^m$(as covering number is bounded from below as well). Similar is the case for $|B_0(\epsilon)|$ as $|B_0(\epsilon)|>= (c\epsilon)^m$. Appreciate your advice on the lower bound, and if its true, please add to the answer if there exists or not a lower bound.
May 21, 2021 at 17:27 comment added Iosif Pinelis @RajeshD : All right. So, to have a closure, does this answer look satisfactory to you?
May 20, 2021 at 15:48 comment added Rajesh D Thank you very much Losif. It is helpful.
May 20, 2021 at 15:32 comment added Iosif Pinelis @RajeshD : You can get such a bound but only with an extra log factor, as is now detailed.
May 20, 2021 at 15:31 history edited Iosif Pinelis CC BY-SA 4.0
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May 20, 2021 at 14:38 comment added Rajesh D Thank you for the valuable answer. My expectation was the asymptotic $\zeta_n \sim n^{-1/m}$. (don't know how to formulate this in terms of probability. Can this asymptote still be derived from your answer, is what I am trying to figure out. It is already known that $\gamma_n \sim n^{-1/m}$ if I am not wrong, but I don't remember the source.
May 20, 2021 at 14:20 comment added Iosif Pinelis @RajeshD : I have added details.
May 20, 2021 at 13:24 history edited Iosif Pinelis CC BY-SA 4.0
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May 20, 2021 at 9:16 comment added Rajesh D @losifPinelis: Thank you for the answer. I have understood the inequlities on probabilities that you had mentioned. But I am not able to quantify as its not apparent to me how I can use Dudley's entropy. Sorry I am new to probability theory. Appreciate some help on how I can use Dudley's. It seems to be defined for Gaussian processes. I have no clue how I can apply.
Apr 27, 2021 at 20:23 history edited Iosif Pinelis CC BY-SA 4.0
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Apr 27, 2021 at 19:54 history bounty ended Rajesh D
Apr 27, 2021 at 16:49 history edited Iosif Pinelis CC BY-SA 4.0
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Apr 27, 2021 at 16:39 history answered Iosif Pinelis CC BY-SA 4.0