Timeline for Nearest neighbor for planar Poisson is normally distributed
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 21, 2015 at 14:34 | answer | added | Carlo Beenakker | timeline score: 1 | |
| Sep 20, 2015 at 18:02 | history | edited | zhoraster | CC BY-SA 3.0 |
added 633 characters in body
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| Sep 20, 2015 at 18:01 | comment | added | zhoraster | @Michael, maybe I was not very careful writing this. I don't claim that the distance is normally distributed, but rather that the neighbor is. The neighbor is a two dimensional vector, whose length has Rayleigh distribution with cdf $1-e^{-ax^2}$. But the vector does have normal distribution. I copied the paragraph from my MathSE post in order to clarify. | |
| Sep 20, 2015 at 17:49 | comment | added | Michael | The $e^{-x^2}$ function is associated with the Gaussian density. In your link, the $e^{-x^2}$ is associated with a CDF function. I do not think it is correct to call your distance "normally distributed," even if you overlook the non-negative issue. | |
| Sep 20, 2015 at 17:41 | comment | added | Michael | Normal random variables can be negative, while distances cannot... | |
| Sep 20, 2015 at 6:13 | comment | added | Anthony Quas | This is not really an answer, but a possibly related reference. Terry Soo and co-authors have some papers where they study "Poisson thinning". That is, they construct for fixed $\mu<\lambda$ a deterministic measurable map which sends an instance of a Poisson process with rate $\lambda$ to an instance of a Poisson process with rate $\mu$. This is all about "extracting randomness" from PPs; maybe the fact that you mention is used somewhere. (NB: of course non-deterministic Poisson thinning is simple) | |
| Sep 20, 2015 at 5:37 | history | asked | zhoraster | CC BY-SA 3.0 |