Timeline for Renewal process: domination by product measure
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 8 hours ago | 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 17 at 16:04 | 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. | |
| Apr 19 at 15:05 | 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. | |
| Dec 21, 2023 at 15:04 | 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 23, 2023 at 14:04 | 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. | |
| Apr 25, 2023 at 13:09 | 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. | |
| Mar 26, 2023 at 12:21 | answer | added | jdods | timeline score: 0 | |
| May 25, 2011 at 19:09 | comment | added | James Martin | I could have said "dominated" again: (the distribution of) a random variable U is stochastically bounded above by (the distribution of) another random variable V if you can find a coupling of the two such that U≤V with probability 1. Or in this case, more simply, $P(U>x)<P(V>x)$ for all $x$. | |
| May 25, 2011 at 14:03 | comment | added | fedja | Looks like an interesting thing to think of but can somebody, please, remind me what "stochastically bounded" means? | |
| May 25, 2011 at 8:03 | history | asked | James Martin | CC BY-SA 3.0 |