Consider a homogeneous Poisson point process in 2D space with density $\lambda$ per unit area. Let $\mathcal{B}(o,R)$ denote a disk centered at origin with radius $R$. Let $n$ be the number of points inside the disk $\mathcal{B}$. Given $n \geq 1$, let $\{d_1, d_2, \ldots, d_n\}$ be the set of radial distances of points inside the disk respect to the origin.
(1) What is the distribution of $n$? Is it a Poisson random variable with density $\lambda \pi R^2$?
(2) What is the distribution of $d_i$, given $n$. Given $n$, are $d_i$s i.i.d random variables with uniform distribution in $[0,R]$?
Best Answer
(1) Yes, the Poisson point process is homogeneous (constant intensity $\lambda$ throughout $\mathcal{B}$) so the distribution of $n$ is Poisson with parameter: $\lambda\times(\text{Area of $\mathcal{B}(0,R)$}) = \lambda\pi R^2.$
(2) Given $n,\;$ the $n$ points are uniformly distributed throughout $\mathcal{B}$. Since a circle's circumference is proportional to its radius, the common density $f_D\;$ of $d_i$ for all $i$ is also proportional to its radius. That is, $f_D(d) = ad$, for some constant $a$. We find $a$ as follows:
\begin{align} 1 &= \int_{r=0}^R ar\;dr \\ &= \left[ \dfrac{ar^2}{2}\right]_{r=0}^R \\ &= \dfrac{aR^2}{2}. \\ \therefore\quad a &= \dfrac{2}{R^2} \end{align}
So $f_D(d) = \dfrac{2d}{R^2}.$