Timeline for splitting exponential random variable into independent components
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2015 at 5:16 | vote | accept | Gourab Mukherjee | ||
| Oct 1, 2015 at 5:16 | vote | accept | Gourab Mukherjee | ||
| Oct 1, 2015 at 5:16 | |||||
| Sep 30, 2015 at 20:25 | vote | accept | Gourab Mukherjee | ||
| Oct 1, 2015 at 5:16 | |||||
| Sep 30, 2015 at 20:25 | vote | accept | Gourab Mukherjee | ||
| Sep 30, 2015 at 20:25 | |||||
| Sep 30, 2015 at 10:23 | answer | added | Anton Muratov | timeline score: 3 | |
| Sep 30, 2015 at 8:16 | vote | accept | Gourab Mukherjee | ||
| Sep 30, 2015 at 20:25 | |||||
| Sep 29, 2015 at 23:55 | answer | added | Ori Gurel-Gurevich | timeline score: 4 | |
| Sep 29, 2015 at 21:14 | comment | added | usul | It's a bit weird because your example uses two independent variables to produce another two... but if we can we use two independent exponentials $X,Y$, then I think the minimum $M = \min\{X,Y\}$ and the absolute difference $Z = |X-Y|$ should be independent by the memoryless property (and both are exponential). | |
| Sep 29, 2015 at 20:48 | comment | added | Anthony Quas | With a uniform random variable, this is trivial by picking odd and even binary digits ($g$ and $h$ are not continuous though). It's not hard to transform exponential to uniform. | |
| Sep 29, 2015 at 19:49 | history | asked | Gourab Mukherjee | CC BY-SA 3.0 |