Timeline for Is there a probabilistic (not analytical) argument for why the sum of independent Poissons is Poisson?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2017 at 21:22 | vote | accept | conv3d | ||
| Feb 14, 2017 at 19:21 | history | edited | whuber♦ | CC BY-SA 3.0 |
added 9 characters in body; edited tags; edited title; edited tags
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| Feb 14, 2017 at 19:18 | answer | added | whuber♦ | timeline score: 7 | |
| Nov 15, 2015 at 1:39 | vote | accept | conv3d | ||
| Mar 2, 2017 at 21:22 | |||||
| Nov 4, 2015 at 2:14 | comment | added | Glen_b | Because the superposition of the two processes (which count of events is the sum of the two Poisson counts) satisfies all the original conditions for the Poisson process. | |
| Nov 4, 2015 at 2:10 | comment | added | conv3d | Really? What about the poisson process allows for the sums to be poisson processes? | |
| Nov 4, 2015 at 1:25 | comment | added | Glen_b | If $\lambda_1\neq \lambda_2$, they're not iid, and if $\lambda_1=\lambda_2$, why distinguish them? But you can argue directly from the Poisson process itself (without needing the two Poissons to be identically distributed) | |
| Nov 4, 2015 at 0:29 | answer | added | Alex R. | timeline score: 2 | |
| Nov 3, 2015 at 23:27 | review | First posts | |||
| Nov 3, 2015 at 23:56 | |||||
| Nov 3, 2015 at 23:25 | history | asked | conv3d | CC BY-SA 3.0 |