Timeline for Intuition of law of iterated logarithm?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2023 at 19:52 | comment | added | RegularGraph | This is likely implicit in some of the comments, but the reason why it is enough to do Borel-Cantelli on the geometric sequence as George Lowther suggests is that the events $S_n/\sqrt n > \sqrt{A \log \log n}$ are correlated for close by values of $n$. If none of these event holds along the geometric sequence, then it is likely that none hold along all the integers. | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Jan 29, 2016 at 20:59 | comment | added | A.S. | @Anthony I just saw an alternative, seemingly more intuitive approach: $X_t=e^{-t}B(e^{2t})$ is standard O-U process and because it's correlations drop off quickly (exponentialy unlike Brownian motion), we can apply Borel Cantelli directly to the sum of probability of its deviations and ignore the dependencies. The core of the transform is the same - but operates on more explicit probabilistic objects. | |
| Jan 28, 2016 at 17:07 | comment | added | Anthony Quas | @A.S.: right. I guess George Lowther's comment says it right. One should imagine one is doing Borel-Cantelli along geometrically spaced times. Along these times, the summability works out. | |
| Jan 28, 2016 at 14:07 | comment | added | A.S. | @Anthony But $\mathbb P(N>\sqrt{C\log\log n})$ is not summable over naturals for any C. But $\mathbb P(N>\sqrt{(2+\epsilon)\log n})$ is. So it's subtler. | |
| Jan 12, 2016 at 10:58 | answer | added | Lan GAO | timeline score: 0 | |
| S Apr 20, 2014 at 8:35 | history | suggested | Davide Giraudo |
added two tags.
|
|
| Apr 20, 2014 at 8:28 | review | Suggested edits | |||
| S Apr 20, 2014 at 8:35 | |||||
| Apr 20, 2014 at 8:11 | answer | added | Bjørn Kjos-Hanssen | timeline score: 17 | |
| Jun 1, 2013 at 2:31 | comment | added | user16557 | George: That is a very curious property that I find surprising. I would have thought that convergence would depend on the asymptotics of $K=O(f(n))$ and not on the constant coefficient of $K$. | |
| Jun 1, 2013 at 0:52 | comment | added | George Lowther | Actually, for Borel-Cantelli to work in this situation you need to sample time points along a geometric progression, not linearly spaced. i.e., for any $q > 1$, $$ \sum_{n\in\lbrace q^r\colon r\in\mathbb{N}\rbrace}\mathbb{P}\left(\frac {S_n}{\sqrt{n}} > \sqrt{A\log\log n}\right) $$ is finite for $A > 2$ and infinite for $A < 2$. This follows from $\mathbb{P}(S_n/\sqrt{n} > K)\sim (2\pi)^{-1/2}K^{-1}\exp(-K^2/2)$. | |
| May 31, 2013 at 13:30 | comment | added | George Lowther | @unknown: that's clearly false, but I think it's true if the summand is divided by n. | |
| May 31, 2013 at 7:38 | comment | added | user16557 | @Anthony: Are you saying that with $S_n$ having mean $0$ and variance $n$ that it is in fact true that $$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n}} > \sqrt{(2+\epsilon)\log{\log{n}}}\right) < \infty$$ and $$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n}} > \sqrt{(2-\epsilon)\log{\log{n}}}\right) = \infty$$ | |
| May 31, 2013 at 6:40 | comment | added | Mark Lewko | To supplement Anthony's comment slightly: recall that the normal distribution in the central limit theorem (with variance $1$) is $\frac{1}{ \sqrt{2\pi}} e^{\frac{x^2}{2 }} $. Roughly speaking, it's the 2 in the denominator of the exponent that ultimately gives rise to the $\sqrt{2}$ in the law of the iterated logarithm. | |
| May 31, 2013 at 6:09 | comment | added | Anthony Quas | The proof goes via lots of Borel-Cantelli. Heuristically if you believe the central limit theorem, $S_n$ should be normal with mean 0 and variance $n$, so that $S_n/\sqrt n$ is approximately $N(0,1)$. The appearance of the $\sqrt{2\log\log n}$ is roughly because $\mathbb P(N>\sqrt{2\log\log n})$ is on the cusp of summability. So that $\mathbb P(N>\sqrt{2.000001\log\log n})$ is summable, so happens finitely many times (this is the easier part), whereas $\mathbb P(N>\sqrt{1.999999\log\log n})$ is not summable and so [quite a lot of annoying technical details skipped] happens infinitely often. | |
| May 31, 2013 at 4:29 | history | asked | user16557 | CC BY-SA 3.0 |