Timeline for Does this quantity related to independence have a name?
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2012 at 21:25 | vote | accept | Michael McGowan | ||
| Dec 7, 2012 at 19:49 | history | post merged (destination) | |||
| Nov 3, 2012 at 22:49 | answer | added | RichardN | timeline score: 6 | |
| Nov 3, 2012 at 14:09 | vote | accept | Piotr Migdal | ||
| Nov 3, 2012 at 23:20 | |||||
| Oct 23, 2012 at 9:53 | answer | added | conjugateprior | timeline score: 8 | |
| Oct 22, 2012 at 19:26 | answer | added | Piotr Migdal | timeline score: 14 | |
| Oct 22, 2012 at 19:18 | comment | added | Piotr Migdal | @Procrastinator Sure, I just didn't want to steal your rep. But if you are fine with it, I'll do. :) | |
| Oct 22, 2012 at 19:16 | comment | added | user10525 | @PiotrMigdal Thanks for the kind offer. I would prefer to see your own answer. Maybe including how you came up with this question and how that quantity can be useful. | |
| Oct 22, 2012 at 19:07 | comment | added | Piotr Migdal | @Procrastinator Thx! Could you make it an answer, as the link point to: "Then, at least in the environmental, medical and life sciences literature, P(A∩B)/(P(A)P(B)) is called the observed to expected ratio (abbreviation o/e)." | |
| Oct 22, 2012 at 18:56 | comment | added | Bitwise | Go for "Migdal Probability" ;) | |
| Oct 22, 2012 at 18:55 | comment | added | user10525 | This question was asked in Mathematics: About joint probability divided by the product of the probabilities. | |
| Feb 8, 2012 at 13:49 | comment | added | naught101 | This SE could do with some more "quite silly" questions. It's very intimidating, even for someone who enjoyed basic undergrad level statistics. +1 for silliness | |
| Jan 7, 2011 at 14:59 | vote | accept | Michael McGowan | ||
| Dec 7, 2012 at 21:25 | |||||
| Jan 7, 2011 at 14:23 | comment | added | vqv | I think it depends on the context. Note that $$Q = \frac{\Pr(A|B)}{\Pr(A)} = \frac{\Pr(B|A)}{\Pr(B)}$$ so that $\Pr(A|B) = Q \Pr(A)$ and $\Pr(B|A) = Q \Pr(B)$. This form has more of a Bayesian inference flavor. | |
| Jan 7, 2011 at 8:54 | answer | added | Yorgos | timeline score: 11 | |
| Jan 7, 2011 at 3:33 | answer | added | Kenneth Cabrera | timeline score: 0 | |
| Jan 6, 2011 at 19:50 | history | asked | Michael McGowan | CC BY-SA 2.5 |