Timeline for Does a Central Limit Theorem imply a series is $O(\sqrt{N})$?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2017 at 3:36 | answer | added | JGWang | timeline score: 2 | |
| Feb 14, 2017 at 0:01 | vote | accept | André LeClair | ||
| Feb 14, 2017 at 0:01 | vote | accept | André LeClair | ||
| Feb 14, 2017 at 0:01 | |||||
| Feb 13, 2017 at 23:38 | comment | added | Christian Remling | @AndréLeClair: The party seems over here by now, but one more comment on your question above: even for a deterministic sequence $a_n=O(n^{1/2+\epsilon})$ for all $\epsilon>0$ will not imply that $a_n=O(n^{1/2})$ because the implied constant may depend on $\epsilon$ (trivial example: $a_n=n^{1/2}\log n$). | |
| Feb 13, 2017 at 23:24 | answer | added | John Dawkins | timeline score: 6 | |
| Feb 13, 2017 at 22:50 | vote | accept | André LeClair | ||
| Feb 14, 2017 at 0:01 | |||||
| Feb 13, 2017 at 22:46 | answer | added | Nate Eldredge | timeline score: 8 | |
| Feb 13, 2017 at 22:39 | comment | added | André LeClair | Thank you Nate. I've received some very good answers on this site on questions I am just not an expert on. This is very helpful since for my purposes I only need $O(\sqrt{N})$ up to logs anyway, since the latter don't spoil the convergence of a series I am trying to establish. | |
| Feb 13, 2017 at 22:34 | comment | added | Nate Eldredge | In your notation, LIL says that $X_N$ is $O((N \log \log N)^{1/2})$ with probability $1$, and moreover this is sharp. | |
| Feb 13, 2017 at 22:20 | comment | added | Nate Eldredge | But the latter statement is implied by the LIL. LIL is substantially harder to prove than CLT, but at least in your iid finite-variance setting, it's a classical result. | |
| Feb 13, 2017 at 22:18 | comment | added | Nate Eldredge | Watch the quantifiers. It's clear that CLT implies $O(N^{1/2+\epsilon})$ with high probability. That is, for every $\epsilon$ and $C$ we have $P(|X_N| \le C N^{1/2+\epsilon}) \to 1$ as $N \to \infty$. When you say "$O(N^{1/2+\epsilon"})$ with probability 1" I read that as "for $P$-almost every $\omega$, there exists $C = C(\omega)$ such that $|X_N(\omega)| \le C N^{1/2+\epsilon}$ for all $N$". The latter statement is a lot stronger. You could perhaps get it from a version of CLT with good bounds on the convergence rate, via Borel-Cantelli, but I don't think it's trivial. | |
| Feb 13, 2017 at 21:30 | comment | added | André LeClair | The CLT would seem to imply that it is of $O(N^{1/2+\epsilon})$ with probability =1 for any $\epsilon >0$. So what is subtle? | |
| Feb 13, 2017 at 21:09 | comment | added | André LeClair | Christian, can you clarify? I'm interpreting your answer that it IS $O(\sqrt{N})$ with probability 1. Is that right? | |
| Feb 13, 2017 at 21:06 | comment | added | André LeClair | I am getting two conflicting answers. I better look up the iterated logarithm. | |
| Feb 13, 2017 at 21:04 | comment | added | Christian Remling | Yes, it does; law of the iterated logarithm again. | |
| Feb 13, 2017 at 21:01 | comment | added | André LeClair | Let me refine the statement: Does the CLT imply it is $O(\sqrt{N^{1/2+\epsilon})$ for any $\epsilon > 0$? | |
| Feb 13, 2017 at 20:58 | comment | added | Nate Eldredge | It's not even $O(\sqrt{N})$ with probability 1. Do you know about the law of the iterated logarithm? | |
| Feb 13, 2017 at 20:57 | comment | added | Robert Israel | As a mathematical physicist, I must protest: we would indeed be careful with such distinctions. Maybe not a theoretical physicist. | |
| Feb 13, 2017 at 20:53 | history | asked | André LeClair | CC BY-SA 3.0 |