Smallest distance to origin for points from Poisson process in $\mathbb{R}^d$

Question

Consider a Poisson process in $\mathbb{R}^d$ with uniform rate $\lambda$. What is the distribution of the smallest $\ell_2$-norm of the points (i.e., distance of the nearest point to the origin)? More generally, is there a formula for the inhomogeneous case with intensity function $\lambda(A), A\subseteq \mathbb{R}^d$?

Answer 1

Let $D$ denote the distance of the closest point to the origin and $N_{\epsilon}$ denote the number of points which have distance smaller or equal to $\epsilon$ to the origin. In the homogeneous Poisson process, $N_{\epsilon} \sim \text{Poisson}\left(\frac{\pi^{0.5d}\epsilon^{d}}{\Gamma(0.5d+1)}\lambda\right)$. Furthermore, \begin{align*} \mathbb{P}(D > \epsilon) &= \mathbb{P}(N_{\epsilon}=0) \\ &= \exp\left(-\frac{\pi^{0.5d}\epsilon^d}{\Gamma(0.5d+1)}\lambda\right) \\ F_{D}(\epsilon) &= 1-\exp\left(-\frac{\pi^{0.5d}\epsilon^d}{\Gamma(0.5d+1)}\lambda\right) \\ f_{D}(\epsilon) &= \frac{\pi^{0.5d}d\epsilon^{d-1}}{\Gamma(0.5d+1)}\lambda \cdot \exp\left(-\frac{\pi^{0.5d}\epsilon^d}{\Gamma(0.5d+1)}\lambda \right) \end{align*} When $d=1$, $D \sim \text{Exponencial}(2\lambda)$.

In the non-homogeneous case, the same idea applies. If you let $S_{\epsilon}=\{x \in \mathbb{R}^d: \|x\| \leq \epsilon\}$, then $N_{\epsilon} \sim \text{Poisson}(\lambda(S_\epsilon))$ and \begin{align*} \mathbb{P}(D > \epsilon) &= \mathbb{P}(N_{\epsilon}=0) \\ &= \exp(-\lambda(S_\epsilon)) \\ \mathbb{P}(D \leq \epsilon) &= 1-\exp(-\lambda(S_\epsilon)) \end{align*} You won't be able to simplify further without an explicit expression for $\lambda$.