Timeline for Doob Martingale: Where is the catch?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2016 at 11:19 | comment | added | user63957 | Your sequence of partial sums is not a martingale. | |
| Sep 4, 2016 at 5:48 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 5, 2016 at 5:47 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 6, 2016 at 5:23 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 6, 2016 at 4:59 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 7, 2016 at 4:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 9, 2015 at 13:35 | comment | added | David White | Thanks for editing! This is a useful post and I hope it sticks around | |
| Oct 10, 2015 at 0:45 | comment | added | Sriram S | If you would like, please add a solution. Otherwise, I have added a resolution to my post. thanks once again! The nice thing is that this gives the same answer as another paper I am reading by Jansen that uses the chromatic number of the dependence graph. So I am much happier! | |
| Oct 10, 2015 at 0:42 | history | edited | Sriram S | CC BY-SA 3.0 |
added 815 characters in body
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| Oct 9, 2015 at 20:57 | comment | added | alpoge | (I'm certainly no expert so the bound above is probably not optimal --- e.g. the constant in the exponential is probably suboptimal.) | |
| Oct 9, 2015 at 20:55 | comment | added | alpoge | Sorry, I keep indexing wrong! Yeah it's as I said in the comment I deleted --- the bound on the increments doesn't hold for Y_1 - Y_0. The general bound is Prob.(Y_k - Y_0\geq t)\leq \exp(-t^2/(2\sum_{i=1}^k ||Y_i - Y_{i-1}||_\infty^2)). Applying this you get a denominator that is order n^2, rather than order n. | |
| Oct 9, 2015 at 20:00 | comment | added | Sriram S | Not quite, $Y_0 = 0$ and $Y_n = (n+1) X_0$ right? But between your previous comment that you just modified, and Edgar's comment below, I think I can see a mistake. $Y_1$ is already $(n+1) X_0$. So the difference $| Y_{i} - Y_{i-1} | = 0$ for $i \geq 2$ but $|Y_{1} - Y_{0}| = 2 (n+1)$. The result will be different. Is that the mistake? | |
| Oct 9, 2015 at 19:59 | comment | added | alpoge | Azuma gives you a bound on the probability that Y_n - Y_0 is large, no? (Here it's 0.) | |
| Oct 9, 2015 at 17:55 | answer | added | Gerald Edgar | timeline score: 3 | |
| Oct 9, 2015 at 17:02 | review | First posts | |||
| Oct 9, 2015 at 17:08 | |||||
| Oct 9, 2015 at 17:02 | history | asked | Sriram S | CC BY-SA 3.0 |