## Related questions with answers

Find the probability density for the distance from an event to its nearest neighbor for a Poisson process in three-dimensional space.

Solution

VerifiedRemember that in a Poisson process, the number of events that occur in a particular area (any subset of a 3D space, i.e. a subset of $\mathbb R^3$) is a Poisson random variable $X$ with the parameter $\lambda\cdot V(S),$ where $\lambda >0$ is a constant and $V(S)$ is the volume of that area $S.$

So,

$P(X=k \text{ in } S)=\cfrac{{\left(\lambda\cdot V(S)\right)}^k}{k!}\cdot e^{-\lambda\cdot V(S)}, \, \, \, \, k \in \mathbb N_0.$

Now, let $D$ be a random variable which "measures" the distance from an event to its nearest neighbour. So, we need to find the probability density function of the random variable $D$ (notice that $D$ is a continuous random variable).

Actually, let's first find the cumulative distribution function. First, let's examine what event $\{D>x\}$ is, for some $x \ge 0.$ This event means that the distance from an event to its nearest neighbour is $\textbf{greater}$ than $x.$ In fact, this means that there are no events (so, 0 events) in a sphere (around some point) of radius $x.$ Let's denote that sphere with $S,$ so that $V(S)=\frac{4}{3}x^3\pi.$ Now we can analyze:

$\begin{align*} F_D(x) & = P(D \le x) = 1-P(D > x) = 1- P(\{\text{There are 0 events in $S$}\}) \\\\ & = 1-P(X=0 \text{ in } S) = 1- \cfrac{{\left(\lambda\cdot V(S)\right)}^0}{0!}\cdot e^{-\lambda\cdot V(S)} \\\\ & = 1-e^{-\lambda\cdot \frac43 x^3\pi}. \end{align*}$

Finally, remember that a probability density function is simply a derivative of the cumulative distribution function, which means that

$\begin{align*} \boxed{f_D(x)} & = F_D'(x)=(1-e^{-\lambda\cdot \frac43 x^3\pi})' \\\\ & = e^{-\lambda\cdot \frac43 x^3\pi}\cdot \left(-\lambda\cdot \frac43 x^3\pi\right)' = e^{-\lambda\cdot \frac43 x^3\pi}\cdot 3\lambda\cdot\frac{4}{3}x^2\pi \\\\ & = \boxed{4\lambda x^2\pi\cdot e^{-\lambda\cdot \frac43 x^3\pi}}. \end{align*}$

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