Consider n random poison points which are distributed according to parameter $\lambda$. Two random points are chosen. Is it true to say CDF of the distance between two points equals to: $$F_{d}(x)=\frac{\sum\limits_{k=1}^{k=n}\left(1- \sum\limits_{l=0}^{k-1} \frac{\left(\lambda x \right) ^{k}}{k!}\right) e^{-\lambda x}} {n}$$
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Statement is confusing. "$\lambda$ per unit area" "uniformly in the disk". What is the distribution? – herb steinberg Nov 28 '18 at 22:39
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@herbsteinberg: presumably the number of points in any part of the disk with area $a$ has a Poisson distribution with expectation $\lambda a$ and the numbers of points in two disjoint parts are independent – Henry Nov 28 '18 at 23:36
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I think there is a positive probability $(1+\lambda \pi R^2)\exp(-\lambda \pi R^2)$ that there are not two points in your disk – Henry Nov 28 '18 at 23:38
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But if you consider all points in the disk, then this is a different question – Henry Nov 28 '18 at 23:44