Timeline for Supermartingales and convergence
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2011 at 3:06 | comment | added | Ori Gurel-Gurevich | To add another example to that of Anthony: If $X_n=-\infty$ with probability $p_n$ and 0 otherwise and $\sum p_n < \infty$ then $X_n = 0$ for all $n$ with positive probability, but the expectation you wrote is $-\infty$. | |
| Feb 18, 2011 at 0:04 | answer | added | James Zhao | timeline score: 3 | |
| Feb 17, 2011 at 15:52 | vote | accept | Colin Reid | ||
| Feb 17, 2011 at 15:47 | comment | added | Anthony Quas | Sounds like the ans are yes to the first and no to the second. Yes: by the decomposition it's a martingale minus something. The mg can't go to $+\infty$ by @Camomille's ans so nor can the supermg. No: If you define $X_n$=(position of a r.w. at time $n$) - $\log n$ then you get $c_n\approx 1/n$ but $X_n\not\to-\infty$. | |
| Feb 17, 2011 at 15:36 | answer | added | camomille | timeline score: 4 | |
| Feb 17, 2011 at 15:00 | comment | added | Steve Huntsman | en.wikipedia.org/wiki/Azuma%27s_inequality | |
| Feb 17, 2011 at 14:45 | history | asked | Colin Reid | CC BY-SA 2.5 |