Timeline for Probability of a point on a unit sphere lying within a cube
Current License: CC BY-SA 4.0
16 events
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| Jun 28, 2022 at 10:37 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 28, 2022 at 10:00 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 17, 2022 at 10:59 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
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| Jun 21, 2010 at 2:56 | vote | accept | L.Z. Wong | ||
| Jun 21, 2010 at 2:56 | vote | accept | L.Z. Wong | ||
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| Jun 20, 2010 at 8:42 | answer | added | Gideon Schechtman | timeline score: 16 | |
| Jun 19, 2010 at 12:52 | comment | added | Steve Huntsman | I didn't see that there were other special cases. But I'd try i-e as soon as Maple or thumbing through Gradshteyn and Ryzhik came up empty. | |
| Jun 19, 2010 at 9:33 | history | edited | L.Z. Wong | CC BY-SA 2.5 |
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| Jun 19, 2010 at 8:54 | history | edited | L.Z. Wong | CC BY-SA 2.5 |
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| Jun 19, 2010 at 8:42 | comment | added | L.Z. Wong | As regards the caps mentioned by Steve, I have considered that option. In three dimensions, we need only consider the cases you've mentioned (and the trivial cases where the probabilities are 1 or 0). However, it seems that for larger n, in each interval $\frac{1}{\sqrt{k+1}} < d < \frac{1}{\sqrt{k}}, k < n$, we get a different special case. Would inclusion-exclusion still help for these cases? | |
| Jun 19, 2010 at 8:33 | comment | added | L.Z. Wong | Thanks for the comments. I am not familiar with the methods mentioned by Thorny and Scott, and will read about them first and report back on the results. I should perhaps add that this question was first posed to me by a software engineer who's been using experimental estimates on large n. His method involves randomly selecting large numbers of points on an n-sphere and then checking the proportion of points that fulfill the conditions, and this sounds a lot like (the little I know of) the Monte-Carlo method. If this is indeed the best method, I guess there isn't much more to be said. | |
| Jun 19, 2010 at 3:59 | comment | added | Steve Huntsman | For $d > 1/\sqrt{2}$ you "just" need to figure out how much a bunch of spherical caps contribute to the total surface area. For $d < 1/\sqrt{2}$ there are a bunch of identical disconnected regions of the cube outside the sphere, and at first glance it seems like computing their areas is the harder part of the problem. But here it seems like inclusion-exclusion should render this doable. | |
| Jun 19, 2010 at 1:46 | comment | added | Ryan O'Donnell | I agree with the above questions: do you think of d as fixed and n quite large (in which case the probability will go to 1 quite rapidly even for small d)? Or is n considered quite small (in which case one could find a numerical answer, but probably no simple closed form)? | |
| Jun 18, 2010 at 14:02 | comment | added | S. Carnahan♦ | Have you considered approximating with a Monte-Carlo method? | |
| Jun 18, 2010 at 11:10 | comment | added | Thorny | What kind of accuracy are you aiming for? Approximating the surface measure of the sphere by a standard Gaussian would give results not worse than deviating ~1/n^{1/2} in d, and would probably be much better than that in most cases. | |
| Jun 18, 2010 at 10:50 | history | asked | L.Z. Wong | CC BY-SA 2.5 |