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 $[0,R]$?

  • $\begingroup$ (1) Yes. (2) Given $n$, the $n$ points inside $\mathcal{B}$ will be uniformly distributed in the disc but this doesn't mean that $d_i$ has uniform distribution. A circle circumference is proportional to its radius, so $P(d_i=r)$ is proportional to $r$. $\endgroup$
    – Mick A
    Jul 6 '15 at 9:51
  • $\begingroup$ is it something like $\Pr [d_i = r] = \frac{2n}{r}$? $\endgroup$
    – Jeff
    Jul 6 '15 at 10:53
  • $\begingroup$ If $f_D(d)$ is the density of $d_i,\; i=1,\ldots,n,\;$ (i.e. they all have that pdf), then since we know $f_D(d)=ar$ for some $a$, then solve for $a$: $1=\int_0^R ar\;dr=[ar^2/2]_0^R=aR^2/2$ so $a=2/R^2$ and we have $f_D(d)=2d/R^2$. $\endgroup$
    – Mick A
    Jul 6 '15 at 11:49
  • $\begingroup$ @MickA: That seems more like an answer to me than a comment. $\endgroup$
    – joriki
    Oct 9 '15 at 8:37
  • 1
    $\begingroup$ You're right @joriki - I've written it up as an answer. Thanks for prompting me. :) $\endgroup$
    – Mick A
    Oct 9 '15 at 13:22

(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 $f_D(d) = \dfrac{2d}{R^2}.$


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