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Assume a homogeneous Poisson point process in a plane (2D) with density $\lambda$. Let $n$, the number of points, be random according to the homogeneous Poisson point process. Let $\{r_1, r_2, \ldots, r_n\}$ be the set of radial distances of the points from the origin.

How can I find distribution of $r_i$ where $i \in \{1,2,...,n\}$ ? The expression for the distribution should include $\lambda$ if possible.

In distance distribution in Poisson point process, a similar distance distribution was calculated but it was assumed that the number of points are known. With this assumption we no longer have a Poisson point process.

There are also two distribution functions related to what I am looking for: Contact distribution function (https://en.wikipedia.org/wiki/Spherical_contact_distribution_function) and Nearest neighbour distribution (https://en.wikipedia.org/wiki/Nearest_neighbour_distribution). Contact distribution function is defined based on first contact with a point in a point process. Nearest neighbour distribution is defined for a point already in the point process (not any given point or origin). Therefore, these two are different from the distribution I am looking for because in my question the distance needs to be considered from origin to $\bf{any}$ point of the point process, not the first contact or not the nearest point.

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Assume that $y$ is the radius of a circle centered at the origin so that $r_n \leq y$. The cumulative distribution function (CDF) of $r_n$ is then:

$$ \begin{align} \Pr(r_n \leq y) &= \sum_{j=n}^{\infty} \frac{(\lambda \pi y^2)^j}{j!}\exp(-\lambda \pi y^2) \\ &= 1 - \sum_{j=0}^{n-1} \frac{(\lambda \pi y^2)^j}{j!}\exp(-\lambda \pi y^2) \end{align} $$

By differentiation, we find that the sought distribution $f_n(r_n)$ is

$$ f_n(r_n)dr_n = 2 \frac{(\lambda \pi)^n}{(n-1)!}r_n^{2n-1}\exp(-\lambda \pi r_n^2)dr_n $$

for $0 \leq r_n < \infty$.

The reference and proof can be found in the book "An Introduction to Geometrical Probability" by A. M. Mathai, pages 224-5.

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