I have a D-dimensional space of volume V, and I uniformly sample it $P$ times by randomly positioning points / throwing darts / etc. I also randomly position some number, $N$, of non-overlapping $D$-spheres, for the same value $D$ as the dimension of the space, with volume $v_{sphere}$. So we have circles for $D$ = 2, regular spheres for $D$ = 3, higher dimensional spheres for greater values of $D$. What is the probability for finding a certain number, $r$, of my $P$ points/darts/etc. in a given $D$-sphere?
Probability distribution for the number of points in a D-sphere when uniformly sampling a D-dimensional space
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0How are you sampling the $D$-spheres? Do you know something about their distribution? – 2011-07-27
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0@Srivatsan, the idea is that the D-spheres, and the points that sample them, occur with uniform probability across the D-dimensional volume. – 2011-07-27
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0Added the criterion that the spheres are non-overlapping, and assigned them a volume $v_{sphere}$. – 2011-07-27
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0Well, there is no uniform distribution over the $D$-dimensional space. For instance, how would you pick a uniformly random circle in the plane? You can perhaps consider a bounding box, and sample uniformly from the box. Even then, in your question, it doesn't make that much sense. – 2011-07-27
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0@Srivatsan, as long as the D-spheres are non-overlapping and the points are chosen with uniform probability across the space's volume, why would the positioning of the spheres matter? Wouldn't any distribution serve equally well? – 2011-07-27
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0@B.M.: It doesn't even matter that they don't overlap, since the question just asks about one of them individually. – 2011-07-27
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0@B.M.: However, note that Srivatsan referred to "the $D$-dimensional space". This is a misunderstanding probably caused by your unorthodox use of language; you seem to be referring merely to a subset of volume $V$ of $D$-dimensional Euclidean space. While such a subset can itself be regarded as a space, in that topological context one would expect you to explicitly specify the space you're talking about; since you seem to be assuming a Euclidean space, it's natural to parse your reference to "space" in a more elementary context, where it's usually not used to refer to subsets. – 2011-07-27
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0@joriki, right, my mistake, I'm just referring to a subset of the D-dimensional space's volume. – 2011-07-27
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0One more remark: Your nomenclature for the spheres is also non-standard; "sphere" usually only refers to the surface, whereas the interior (which you presumably mean, else the answer would simply be $p=0$) is called a $D$-ball. For example, a circle is a $1$-sphere, and its interior is a $2$-ball; an ordinary sphere is a $2$-sphere, and its interior is a $3$-ball. – 2011-07-27
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0@B.M.: "a subset of the D-dimensional space's volume" also doesn't make sense; you mean a subset of the $D$-dimensional space. Volume is a measure for the extent of a set; the entire $D$-dimensional space has no (finite) volume, and a volume is a number and doesn't have subsets. My advice would be to take more care with terminology; as you can see from Srivatsan's reaction, imprecise use of terminology can easily lead to misunderstandings. – 2011-07-27
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1@joriki, thanks, I'll be a lot more careful next time! – 2011-07-27
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I'm assuming that when you say you have a $D$-dimensional space of volume $V$, you mean a subset of volume $V$ of a $D$-dimensional Euclidean space, or at least of a space with a Euclidean metric.
The positioning and number of the spheres is irrelevant; they could be positioned randomly or fixed at arbitrary positions, and there could be any number of them. What matters is only the ratio $\rho$ of the volume of a sphere to the volume $V$ of the subset. The probability for finding $r$ of the $P$ points in a given sphere is then given by the binomial distribution: $\binom Pr\rho^r(1-\rho)^{P-r}$.
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0joriki, Why is it that the number of spheres doesn't matter? If we sample sufficiently many spheres, then we can will even (nearly) cover the whole space, no? In that case, your $\rho$ should really be the volume of the union of the $N$ spheres, normalized by the total volume $V$. Am I missing something here? – 2011-07-27
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0@Srivatsan: Yes, you seem to be missing the "given" at the end of the question, though, since the question seems to require some interpretation, it might be that B.M. actually meant "one of the spheres", as you seem to be interpreting it. – 2011-07-27
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0Oh, a trick question :-). Cool, thanks! – 2011-07-27
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0@Srivatsan, sorry about that... – 2011-07-27