2]$ by the sum of $n$ iid uniform random variables from $[-1,1]$

8 events

when toggle format	what		by	license	comment
Jun 16, 2020 at 22:39	comment	added	dohmatob		I'm interested with how far one can go with the given data. Obviously, the tighter the bound, the better. One bound is not as good as another...
Jun 16, 2020 at 21:43	comment	added	Pat Devlin		Even still, you're cutting it kind of close! :-) What kind of lower bound would you be happy with? Anything that converges to 1?
Jun 16, 2020 at 21:35	comment	added	dohmatob		Take $C= 1/100$ if you like.
Jun 16, 2020 at 20:11	comment	added	Pat Devlin		Naturally, we'd need $\varepsilon > 2^{-n}$ or there won't be enough different sums to land in the $1/\varepsilon$ intervals needed. Can we assume $\varepsilon > e^{-n/100}$ if we feel like? Otherwise, if we have no control over $C$, then $\varepsilon$ might be much too small for this to be possible.
Jun 16, 2020 at 19:25	comment	added	dohmatob		You may assume $\varepsilon \ge e^{-Cn}$, for some fixed constant $C>0$, independent of $n$.
Jun 16, 2020 at 19:22	comment	added	Pat Devlin		What sort of dependence does $\varepsilon$ have on $n$? [If $n \varepsilon$ is bounded, this is going to be much different than the case where say $\varepsilon n > 10 \log(n)$.]
Jun 16, 2020 at 16:59	history	edited	dohmatob	CC BY-SA 4.0	added 516 characters in body
Jun 16, 2020 at 16:52	history	asked	dohmatob	CC BY-SA 4.0