Timeline for Largest deviations for uniform order statistics

Current License: CC BY-SA 4.0

11 events

when toggle format	what		by	license	comment
Jun 19, 2018 at 12:28	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 1 character in body
Jun 18, 2018 at 20:55	comment	added	Iosif Pinelis		@fedja : I suspected that something like this should be true. It seems that the argument in the above "binary" counterexample can be rather easily modified to show that, with any gap in the distribution, the probability in question will not go to $0$. In the other direction, the matter seems significantly more difficult.
Jun 18, 2018 at 16:37	comment	added	fedja		@Gericault The necessary and sufficient condition seems to be that there should be no gaps (intervals of $0$ probability) in the distribution.
Jun 17, 2018 at 16:32	comment	added	Iosif Pinelis		@Gericault : Indeed, 8 needed to be changed to 16 (done now). Also, added a paragraph at the end.
Jun 17, 2018 at 16:31	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 275 characters in body
Jun 17, 2018 at 16:10	comment	added	Gericault		Great, thanks ! Indeed, I was thinking about a good assumption on the density of $X_1$, I think the bounded from below can be relaxed, still ensuring the inverse of the cumulative is Lipschitz (allowing the density to go to 0 at the edges of [a,b] for instance). PS : it does not matter but I think after using $(a-b)^4 \leq 8(a^4 + b^4)$, you get $\mu_4(Y^{(i)}) \leq 16L^4\mu_4(X^{(i)})$, right ?
Jun 17, 2018 at 12:54	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 13 characters in body
Jun 17, 2018 at 4:57	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 515 characters in body
Jun 17, 2018 at 4:14	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 115 characters in body
Jun 17, 2018 at 3:38	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 1236 characters in body
Jun 17, 2018 at 2:40	history	answered	Iosif Pinelis	CC BY-SA 4.0