0

This is a follow up question to (Distribution of distance from origin to any point of Poisson point process)

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 ordered in increasing distance. Let $\{(x_1,y_1), (x_2,y_2) \ldots, (x_n,y_n)\}$ be the coordinates of the points ordered by the same increasing radial distance.

In (Distribution of distance from origin to any point of Poisson point process) the distribution of $r_i$, $i \in \{1,2,...,n\}$ is found. I am interested in the distribution of $\theta_i=\arctan\frac{y_i}{x_i}$.

Intuitively I believe it should be uniform over $(-\pi/2,\pi/2)$ as I can subdivide the expanding disc used to find contact distances into equal spaced non-overlapping segments, and by the properties of a PPP the probability of points being inside each segment must be equal.

The uniform distribution also is confirmed via numerical simulation. I am hoping there is a formal way to show that it is a uniform random variable.

Pablitorun
  • 201

1 Answers1

2

Yes, it is uniform.

One algorithm (informally stated) for simulating a homogeneous Poisson process on the plane is the following:

  1. Divide the plane into a sequence of finite-area regions $S_1, S_2, \dots$.
  2. For each $k$, sample $N_k \sim \operatorname{Poisson}(\lambda |S_k|)$, where $|S_k|$ is the area of $S_k$.
  3. For each $k$, let $P_k$ be $N_k$ i.i.d. points sampled from $\operatorname{Unif}(S_k)$.
  4. The union $P_1 \cup P_2 \cup \dots$ is the desired Poisson process.

Picking rotationally symmetrical regions $S_i$, e.g. letting $S_k$ be all points with Euclidean distance from the origin in $[k-1, k)$, means that each point in $P$ is generated by choosing a uniformly random angle in step 3.

Ziv
  • 1,279