Timeline for Uniform convergence of averages for stationary ergodic process
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2019 at 13:43 | vote | accept | zhoraster | ||
| Jun 30, 2019 at 13:42 | answer | added | zhoraster | timeline score: 0 | |
| Jun 28, 2019 at 9:50 | vote | accept | zhoraster | ||
| Jun 30, 2019 at 13:43 | |||||
| Jun 28, 2019 at 8:42 | answer | added | Yuval Peres | timeline score: 2 | |
| Jun 24, 2019 at 16:01 | comment | added | zhoraster | @D.Thomine, the problem is that the only viable bound I can obtain so far is for the variance, which is of order $1/n$. This gives, through Chebyshev's inequality, the estimate of the same order for the probability, so $R_n\gg n$ is, unfortunately, impossible. | |
| Jun 23, 2019 at 21:19 | comment | added | D. Thomine | I'd look into concentration of measure. If the $X_t$ are close to being i.i.d. and the tails of $X_t$ are nice enough, you can get bounds such as $\mathbb{P} (|(2n)^ {-1}\int X_t - \mathbb{E} (X_0)| > \varepsilon) \leq C(\varepsilon, n)$, whence $\mathbb{P} ( \sup |(2n)^ {-1}\int X_t - \mathbb{E} (X_t)| > \varepsilon) \leq R_n C(\varepsilon, n)$. Then you only need to find $R_n$ such that $\lim R_n C(\varepsilon, n) = 0$ for all $\varepsilon$ to get convergence in distribution. | |
| Jun 23, 2019 at 16:00 | history | edited | zhoraster | CC BY-SA 4.0 |
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| Jun 23, 2019 at 15:49 | history | asked | zhoraster | CC BY-SA 4.0 |