Distances from a Fixed Location
In a spatial Poisson model, a basic question is how far one must search from a fixed location to find a point. The answer depends on the volume of the search region: in a plane, for example, a search radius \(r\) covers area \(\pi r^2\), while in three dimensions it covers volume \(\tfrac43\pi r^3\). Counting points in that region turns the distance question into a Poisson probability.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(\Phi\) be a homogeneous Poisson point process on the state space \(\mathbb{R}^d\), where \(d\in\{1,2,3\}\). This means that for every Borel set \(A\subseteq\mathbb{R}^d\) of finite volume, the number \(\Phi(A)\) of points in \(A\) has a Poisson distribution with mean \(\lambda |A|\), where \(\lambda>0\) is the intensity per unit \(d\)-dimensional volume; counts in disjoint sets are independent. Write \(\|x\|\) for Euclidean distance from \(x\) to the origin. The distance \(R_k\) to the kth-nearest point is the kth smallest distance among the points of \(\Phi\), for a positive integer \(k\). A point at the origin has probability zero because a singleton has volume zero.
Let \(B_r=\{x:\|x\|\le r\}\), and let \(v_d\) be the volume of the unit ball in dimension \(d\). Thus \(|B_r|=v_dr^d\), with \(v_1=2\), \(v_2=\pi\), and \(v_3=\tfrac43\pi\). The count \(\Phi(B_r)\) is Poisson with mean \(c r^d\), where \(c=\lambda v_d\). For \(r\ge0\), the event \(R_k>r\) is exactly the event that fewer than \(k\) points lie in \(B_r\).
Proof. For any \(r\ge0\), the number of points in \(B_r\) has a Poisson distribution with mean \(cr^d\). Therefore \(\mathbb{P}(R_k>r)=\mathbb{P}(\Phi(B_r)\le k-1)=e^{-cr^d}\sum_{j=0}^{k-1}(cr^d)^j/j!\). Taking the complement gives the stated distribution function. As \(r\) increases to infinity, \(cr^d\) increases to infinity, and each term \(e^{-cr^d}(cr^d)^j/j!\) tends to zero. Hence \(\mathbb{P}(R_k=\infty)=\lim_{r\to\infty}\mathbb{P}(R_k>r)=0\). To obtain the density, set \(x=cr^d\). Differentiating the finite sum gives \(\frac{d}{dx}\bigl(1-e^{-x}\sum_{j=0}^{k-1}x^j/j!\bigr)=e^{-x}x^{k-1}/(k-1)!\): the derivative of \(e^{-x}\sum_{j=0}^{k-1}x^j/j!\) is \(-e^{-x}x^{k-1}/(k-1)!\), since all lower-power terms cancel. Also \(dx/dr=cd r^{d-1}\). The chain rule yields the stated density. \(\square\)
For \(k=1\), the sum in the distribution function has only its \(j=0\) term. The nearest-point distance therefore has survival function \(e^{-cr^d}\) and density \(cd r^{d-1}e^{-cr^d}\). For larger \(k\), the additional polynomial factor accounts for the possibility that one or more points are already within radius \(r\) but fewer than \(k\) are there.
Expected Distance and Dimension
Proof. Substitute \(x=cr^d\) into the density from the kth-neighbor law. Then \(r=(x/c)^{1/d}\), and the density times \(dr\) becomes \(e^{-x}x^{k-1}/(k-1)!\,dx\). Consequently, \(\mathbb{E}[R_k]=c^{-1/d}\frac{1}{(k-1)!}\int_0^\infty x^{k+1/d-1}e^{-x}\,dx =c^{-1/d}\frac{\Gamma(k+1/d)}{(k-1)!}\). Integration by parts in the definition of \(\Gamma\) gives \(\Gamma(k)=(k-1)!\) for a positive integer \(k\); the boundary terms vanish because \(x^m e^{-x}\) tends to zero at infinity for each nonnegative integer \(m\), and the base case is \(\Gamma(1)=1\). Substituting \(c=\lambda v_d\) proves the formula. \(\square\)
The mean formula gives exact dimension-specific results. Here the density is given for \(r>0\), and the mean is for any positive integer \(k\).
| Dimension | Density of \(R_k\) | Mean of \(R_k\) | Nearest-point mean |
|---|---|---|---|
| One | \(\frac{2\lambda}{(k-1)!}r^{k-1}(2\lambda)^{k-1}e^{-2\lambda r}\) | \(\frac{\Gamma(k+1)}{\Gamma(k)}(2\lambda)^{-1}=\frac{k}{2\lambda}\) | \(\frac{1}{2\lambda}\) |
| Two | \(\frac{2\pi\lambda}{(k-1)!}r(\pi\lambda r^2)^{k-1}e^{-\pi\lambda r^2}\) | \((\pi\lambda)^{-1/2}\frac{\Gamma(k+1/2)}{\Gamma(k)}\) | \(\frac{1}{2\sqrt{\lambda}}\) |
| Three | \(\frac{4\pi\lambda}{(k-1)!}r^2(\frac{4\pi}{3}\lambda r^3)^{k-1}e^{-(4\pi/3)\lambda r^3}\) | \((\frac{4\pi}{3}\lambda)^{-1/3}\frac{\Gamma(k+1/3)}{\Gamma(k)}\) | \((\frac{4\pi}{3}\lambda)^{-1/3}\Gamma(\frac43)\) |
In one dimension, \(v_1=2\) because a radius-\(r\) ball is the interval \([-r,r]\). In two dimensions, \(v_2=\pi\), and in three dimensions \(v_3=\tfrac43\pi\). In each case the intensity enters through the scale \((\lambda v_d)^{-1/d}\): multiplying the intensity by a factor of \(a>0\) multiplies the mean distance by \(a^{-1/d}\). The mean for the kth neighbor also increases with \(k\), as the gamma-function ratio shows.
Worked Calculations in One, Two, and Three Dimensions
Worked Example: The Second-Nearest Point on a Line
Consider a homogeneous Poisson point process on \(\mathbb{R}\) with intensity \(\lambda=\tfrac14\). Find the density and mean distance to the second-nearest point, and compute the probability that this distance is at most \(2\).
In one dimension, \(c=\lambda v_1=2\lambda=2(\tfrac14)=\tfrac12\). With \(k=2\), the density formula gives \(f_{R_2}(r)=\frac{(1/2)}{1!}r^{1}(1/2)^{1}e^{-r/2}=\tfrac{r}{4}e^{-r/2}\), for \(r>0\).
Its mean is \(\frac{k}{2\lambda}=\frac{2}{2(\tfrac14)}=\frac{2}{1/2}=4\). At \(r=2\), the Poisson mean in the interval \([-2,2]\) is \(cr^d=(\tfrac12)(2)=1\). Thus \(\mathbb{P}(R_2\le2)=1-e^{-1}(1+1)=1-\tfrac2e\).
Check: Differentiating \(1-e^{-r/2}(1+r/2)\) gives \(\tfrac{r}{4}e^{-r/2}\), the density above. Its value at \(r=2\) is \(\tfrac12e^{-1}\), also obtained from \(\tfrac{2}{4}e^{-2/2}\). The mean follows independently from the dimension-one formula \(k/(2\lambda)=4\).
Worked Example: The Nearest Point in a Plane
Take a homogeneous Poisson point process on \(\mathbb{R}^2\) with intensity \(\lambda=\tfrac1\pi\). Find the nearest-point density and mean, then calculate the probability that the nearest point is within distance \(1\).
Here \(c=\lambda v_2=(\tfrac1\pi)\pi=1\). For \(k=1\), the density is \(2cr e^{-cr^2}=2r e^{-r^2}\), for \(r>0\). The mean is \((\pi\lambda)^{-1/2}\Gamma(\tfrac32)=1\cdot\Gamma(\tfrac32)=\tfrac{\sqrt{\pi}}2\), using \(\Gamma(\tfrac32)=\tfrac12\Gamma(\tfrac12)=\tfrac{\sqrt{\pi}}2\).
The radius-one disk has area \(\pi\), so its Poisson mean is \(\lambda\pi=(\tfrac1\pi)\pi=1\). The nearest-point probability is therefore \(\mathbb{P}(R_1\le1)=1-e^{-1}\).
Check: The distribution function is \(1-e^{-r^2}\); differentiating it gives \(2r e^{-r^2}\). At \(r=1\) the density equals \(2e^{-1}\), which agrees with substituting \(r=1\) into the density. The density is nonnegative and its integral is \(1\), since \(\int_0^\infty 2r e^{-r^2}\,dr=1\) by the substitution \(x=r^2\).
Worked Example: The Second-Nearest Point in Three Dimensions
Let \(\Phi\) be a homogeneous Poisson point process on \(\mathbb{R}^3\) with intensity \(\lambda=\tfrac{3}{4\pi}\). Find the second-neighbor density and mean, and compute the probability that the second-nearest point is within distance \(1\).
The constant is \(c=\lambda v_3=(\tfrac{3}{4\pi})(\tfrac43\pi)=1\). For \(k=2\), the density becomes \(f_{R_2}(r)=\frac{3}{1!}r^2(r^3)e^{-r^3}=3r^5e^{-r^3}\), for \(r>0\).
The mean is \(c^{-1/3}\Gamma(\tfrac73)/\Gamma(2)=\Gamma(\tfrac73)\), since \(c=1\) and \(\Gamma(2)=1\). Applying the recurrence \(\Gamma(a+1)=a\Gamma(a)\) gives \(\Gamma(\tfrac73)=(\tfrac43)(\tfrac13)\Gamma(\tfrac13)=\tfrac49\Gamma(\tfrac13)\). At radius \(1\), the mean count in the ball is \(cr^3=1\), so \(\mathbb{P}(R_2\le1)=1-e^{-1}(1+1)=1-\tfrac2e\).
Check: Differentiating \(1-e^{-r^3}(1+r^3)\) gives \(3r^5e^{-r^3}\). At \(r=1\), this derivative is \(3e^{-1}\), equal to the density \(3(1)^5e^{-(1)^3}\). The mean integral also matches the gamma expression: substituting \(x=r^3\) into \(\int_0^\infty r\,3r^5e^{-r^3}\,dr\) gives \(\int_0^\infty x^{4/3}e^{-x}\,dx=\Gamma(\tfrac73)\).
Common Pitfall: Using the Nearest-Point Formula for Every Neighbor
The probability of at least one point in a ball is \(1-e^{-cr^d}\), but this is the distribution function of \(R_1\), not \(R_k\) for \(k>1\). For the kth neighbor, the event \(R_k\le r\) requires at least \(k\) points in the ball. In the three-dimensional example above, at radius \(1\) the Poisson mean is \(1\). The incorrect nearest-point calculation would give \(1-e^{-1}\) for the second-neighbor probability. The correct calculation is \(1-\mathbb{P}(\Phi(B_1)=0)-\mathbb{P}(\Phi(B_1)=1)=1-e^{-1}(1+1)=1-\tfrac2e\). The missing term \(e^{-1}\) is the probability of exactly one point, which does not suffice to reach the second neighbor.
The same count-based reasoning applies at any radius: \(R_k\le r\) exactly when \(\Phi(B_r)\ge k\). This makes the Poisson count formula the safest starting point. After obtaining a density or mean, check it against the relevant distribution function and make sure the spatial volume is \(v_dr^d\), rather than a lower-dimensional expression.
Check Your Understanding
Use the Poisson count in a radius-\(r\) ball to answer each question. Take \(k\) to be a positive integer and \(\lambda>0\).
- In one dimension with intensity \(\lambda\), write the distribution function of \(R_3\) for \(r\ge0\).
- In two dimensions, how does the mean nearest-point distance change if the intensity is multiplied by \(4\)?
- For a three-dimensional process with intensity \(\lambda=\tfrac{3}{4\pi}\), find \(\mathbb{P}(R_1\le1)\).
- Explain why the event \(R_2\le r\) involves the probabilities of zero and one point in \(B_r\).
- Use the mean formula to write \(\mathbb{E}[R_2]\) for a planar process of intensity \(\lambda\).