Timeline for Minimum distance distribution between N random points in a cube and the origin
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2015 at 14:44 | history | edited | Michael Lugo | CC BY-SA 3.0 |
improved formatting of code
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| Aug 1, 2011 at 9:34 | vote | accept | Thomas | ||
| Jul 31, 2011 at 15:46 | comment | added | Thomas | In fact, your result is confirmed by a paper dealing with any dimension "The average distance of the n-th neighbour in a uniform distribution of random points" by P. Bhattacharyya, B. K. Chakrabarti and A. Chakraborti I computed it for $D=3$ and it gives almost exaclty the same result as you. | |
| Jul 30, 2011 at 17:30 | comment | added | Thomas | Impressive! I tried your proposition $E(D) = 0.554N^{-1/3}$ and I get almost exactly the same function as I got with my own simulations. The major error appears for small numbers. As you mentioned, for large $N$, the probability that the closest point lies at a distance $< 0.5$ is high.This is not the case for small $N$. That is why we got a little imprecision. Nevertheless, it is a very nice way of thinking the problem. Thank you very much! P.S. I finally computed the distributiion of $D_i$ but it involves nasty $\arctan$ so I think I will stick with your proposition. Thomas | |
| Jul 29, 2011 at 0:37 | comment | added | Michael Lugo | They're not more complicated, but you have to keep track of factors of $N^{1/3}$ which is a bit of a nuisance. | |
| Jul 28, 2011 at 16:44 | comment | added | Alekk | the computations are not more complicated if one computes exactly what $\mathbb{P}(D>rN^{\frac{1}{3}})$ is, right? | |
| Jul 28, 2011 at 15:04 | history | edited | Michael Lugo | CC BY-SA 3.0 |
changed integrand
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| Jul 28, 2011 at 14:54 | history | answered | Michael Lugo | CC BY-SA 3.0 |