Timeline for Rate of convergence in the Law of Large Numbers
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2015 at 15:56 | vote | accept | Anthony Quas | ||
| Apr 16, 2015 at 8:43 | comment | added | Christian Chapman | @AnthonyQuas Perhaps you will be interested in Sanov's theorem: en.wikipedia.org/wiki/Sanov%27s_theorem | |
| Apr 16, 2015 at 8:17 | comment | added | user5810 | You'll also need $\: |\mathbb{E}X_1\hspace{-0.02 in}| < \infty \:$ in order to reach the conclusions you claim from the laws of large numbers. $\;\;\;$ | |
| Apr 16, 2015 at 7:27 | answer | added | ofer zeitouni | timeline score: 17 | |
| Apr 16, 2015 at 6:33 | answer | added | coudy | timeline score: 5 | |
| Apr 16, 2015 at 6:31 | comment | added | Anthony Quas | So it doesn't seem likely that $\mathbb E|X|^\alpha<\infty$ will be sufficient for the averages to converge to the $\alpha$-stable law: after all if this condition holds, then so does $\mathbb E|X|^\beta<\infty$ for $\beta<\alpha$. As far as I could tell (without claiming to really understand it), it looked as though the things proved to be in the domain of attraction of the $\alpha$-stable law had infinite $\alpha$th moments (the $\alpha$-stable law certainly has this property). | |
| Apr 16, 2015 at 6:14 | history | edited | Anthony Quas | CC BY-SA 3.0 |
sharpen question
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| Apr 16, 2015 at 2:27 | comment | added | Nate Eldredge | Related to your other questions: mathoverflow.net/questions/73647/… | |
| Apr 16, 2015 at 2:25 | answer | added | Igor Rivin | timeline score: 5 | |
| Apr 16, 2015 at 2:21 | comment | added | Nate Eldredge | I don't know an exact reference either, but I suspect that with a little work you can show that a distribution satisfying $E|X|^\alpha < \infty$ will satisfy the K-G hypotheses. | |
| Apr 16, 2015 at 2:19 | comment | added | Anthony Quas | Sorry @NateEldredge for the misunderstanding. So yes. I hope that a K-G stable law limit theorem would do the trick. It's just that I did not find a reference for such a theorem that didn't have a bunch of extra conditions (about slowly varying densities etc). If you know where I can find a statement of the type you're mentioning, that will probably resolve (at least the first part of) my issue. | |
| Apr 16, 2015 at 2:06 | comment | added | Nate Eldredge | Right. In the infinite second moment case, by "CLT-type result" I'm talking about a Kolmogorov-Gnedenko style stable law limit theorem giving the weak convergence of $(S_n - \mu n)/n^{1/\alpha}$ (typically the weak limit is a stable law, not the normal distribution). | |
| Apr 16, 2015 at 2:05 | comment | added | Anthony Quas | @Nate: Don't you need a second moment condition to apply CLT? | |
| Apr 16, 2015 at 1:55 | comment | added | Nate Eldredge | Maybe I'm confused, but doesn't a CLT-type result give you exactly what you want? In the finite-variance case, the classical CLT says that $(S_n - \mu n)/n^{1/2}$ converges weakly, so as a result $E_n/n^{1/2 + \epsilon} \to 0$ weakly and hence in probability (standard result). So a CLT-type result telling you that $(S_n - \mu n)/ n^{1/\alpha}$ converges in distribution would imply that $E_n / n^{1/\alpha + \epsilon} \to 0$ in probability. | |
| Apr 15, 2015 at 23:59 | answer | added | Taha | timeline score: 1 | |
| Apr 15, 2015 at 23:51 | history | asked | Anthony Quas | CC BY-SA 3.0 |