# Nearest-neighbour Markov point processes on graphs with Euclidean edges

We define nearest-neighbour point processes on graphs with Euclidean edges and linear networks. They can be seen as the analogues of renewal processes on the real line. We show that the Delaunay neighbourhood relation on a tree satisfies the Baddeley--Møller consistency conditions and provide a characterisation of Markov functions with respect to this relation. We show that a modified relation defined in terms of the local geometry of the graph satisfies the consistency conditions for all graphs with Euclidean edges.

## Authors

• 3 publications
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## 1 Introduction

In recent years, a theory of point processes on linear networks has been emerging so as to be able to analyse, for example, the prevalence of accidents on motorways, the occurrence of street crimes and other data described in the first chapter of the pioneering monograph by Okabe and Sugihara [10]. Although there exists a mature theoretical framework for point processes on Euclidean spaces [6], the development of a similar theory on linear networks is complicated by the geometry inherent in the network. In particular, it is not possible to define strictly stationary models, as the network may not be closed under translations. For this reason, most attention has focussed on the development of second order summary statistics [11].

Little attention has been paid to model building with a few notable exceptions. The first serious work in this direction seems to be that by Baddeley et al. [4], who construct certain types of Cox processes as well as a Switzer-type and a cell process. The authors conclude that familiar procedures for constructing models tend not to produce processes on a linear network that are pseudostationary with respect to the shortest path distance, except when the network is a tree – an unrealistic assumption for a road network. Another important contribution is the work by Anderes et al. [1] who expand the modelling framework in various directions. They relax the assumption of [4, 10, 11] that a linear network consists of a finite union of straight line segments that intersect only in vertices, in the sense that the segments are replaced by parametrised rectifiable curves that may or may not overlap. The parametrisations have the additional advantage of naturally defining a weighted shortest path distance. In the motivating example where the linear network represents a road network, such a generalisation allows for bridges or tunnels and for distance to be measured in travel time where appropriate. Additionally, Anderes et al. [1] construct log-Gaussian Cox processes in terms of a Gaussian process on the network specified by an isotropic covariance function.

All models discussed so far are clustered in nature, that is, exhibit a positive association between the points. In this paper, our aim is to develop appropriate analogues of renewal processes, exploiting the one-dimensional nature of a linear network. Recall that renewal processes exhibit the property that the probability (in an infinitesimal sense) of an event at a given location conditional on the realisation of the process elsewhere depends only on the two nearest points, regardless of how far away they may be. Such models are known as nearest-neighbour Markov point processes

[2] and may incorporate both inhibition and clustering [9]. In contrast to Cox models, the second order summary statistics may not be available in closed form, but the conditional intensities and likelihood are. The latter can be expressed as a product of interaction functions, which may be chosen to be isotropic.

The plan of this paper is as follows. In Section 2, we recall the definitions of Anderes et al. [1] regarding graphs with Euclidean edges, the weighted shortest path metric thereon and Poisson process defined on them. In Section 3, we extend the notion of a Markov point process with respect to the Delaunay nearest-neighbour relation [2] to graphs with Euclidean edges and state our main results. More specifically, we show that the Delaunay relation on a tree satisfies the Baddeley–Møller consistency conditions and we provide a characterisation of the Markov functions with respect to this relation. We then use the graph structure to define a modified Delaunay relation and show that it satisfies the consistency conditions on any graph with Euclidean edges. The proofs are given in Sections 45.

## 2 Preliminaries

### 2.1 Graphs with Euclidean edges

In their pioneering monograph on the subject, Okabe and Sugihara [10, page 31] define a network as a finite union

 L=n⋃i=1li,n∈N,

of straight line segments in or that intersect only at their endpoints in such a way that is connected. The representation is not unique since a line segment may arbitrarily be split in two pieces without affecting the union .

A more general definition is given by Anderes, Møller and Rasmussen [1]. They replace the straight line segments by curves that are parametrised by bijections. In order to define a weighted shortest path distance on the graph, we impose the further condition that these parametrisations are diffeomorphisms and follow [1] to define the weight of an edge as the ‘length’ induced by the parametrisation.

###### Definition 1

A graph with Euclidean edges in is a triple such that

• is a finite, simple connected graph, i.e. has neither loops nor multiple edges;

• every edge , , is parametrised by the inverse of a bijection , that is, for a non-empty open interval with endpoints , .

In other words, an edge is associated with a set that does not contain the endpoints.

A graph with Euclidean edges gives rise to the space

 L=({0}×V)∪n(E)⋃i=1({i}×ϕ−1i(Ji),

where is the cardinality of . The labels serve to identify the edges and will prevent paths from ‘jumping from one edge to another’ in case their interiors overlap. For instance, if represents a road network, overlap is typically present due to tunnels and bridges [1].

As an aside, if there is no overlap between the edge interiors, one may drop the labels and simply consider the disjoint union

 V∪n(E)⋃i=1ϕ−1i(Ji),

which in turn reduces to the classic linear networks if all edges are straight line segments. From now on, we will work with the general space including the labels.

### 2.2 Weighted shortest path metric

The family of parametrisations that is part of the definition of a graph with Euclidean edges can be used to define concepts of length and distance [1].

###### Definition 2

Let be a graph with Euclidean edges. For every , define the -induced length measure on the -algebra on generated by the functions as follows. The induced length of the set is equal to the Euclidean length of in the closed interval . In particular, the edge has length equal to the Euclidean length of .

If we use the arc length parametrisation, the induced length of Definition 2 corresponds to the usual Euclidean length. A different set of parametrisations may induce a different length measure.

###### Definition 3

Let be a graph with Euclidean edges. A walk between two elements and of travels alternatingly from to along a finite number of nodes and edges such that the nodes are endpoints of the edges. If , the first bit of the walk travels from along ; similarly for , the last bit of the walk is along to . In particular, a walk between two points along the same curve does not reverse its tracks.

###### Definition 4

Let be a graph with Euclidean edges. A path between two different elements and of is a walk in which all edge segments and all vertices are different. The weight of the path is the sum of the lengths of the edge segments in it, and the shortest path distance between two elements and of is the smallest weight carried by any path between them.

Note that it is possible that the shortest weighted path between two vertices joined by an edge is not along that edge! The assumption that the parametrisations are diffeomorphisms ensures that the edge lengths are finite. Moreover, the weighted shortest path distance on a graph with Euclidean edges defines a metric. For further details, see [1].

### 2.3 Poisson process on graphs with Euclidean edges

Our goal in this paper is to define point process on graphs with Euclidean edges in terms of a density with respect to a unit rate Poisson process [9]. To do so, we recall Anderes et al.’s definition of a Poisson process on a graph with Euclidean edges [1].

###### Definition 5

Let be a graph with Euclidean edges and the corresponding network. The Lebesgue–Stieltjes measure is defined as follows. For every and every set in the -algebra generated by , set

 λiG(Bi)=∫Ji1{ϕ−1i(t)∈Bi}∣∣∣ddtϕ−1i(t)∣∣∣dt,

where denotes the norm of the gradient. Then

 λG(∪n(E)i=1({i}×Bi))=n(E)∑i=1λiG(Bi)

is a measure on equipped with the finite disjoint union -algebra in which a set is measurable if and only if it can be written as with in the -algebra generated by for and the power set of for . By default, the set has Lebesgue–Stieltjes measure zero.

Note that Definition 5 does not depend on the parametrisations.

###### Definition 6

Let be a graph with Euclidean edges and the corresponding network. The unit rate Poisson process on is defined as follows. For every and every set in the -algebra generated by ,

• the number of points in

is Poisson distributed with expectation

;

• given that the number of points falling in is , these

points are independent and identically distributed according to the probability density function

.

In words, the unit rate Poisson process scatters a Poisson number of points independently and uniformly on every edge, and the average number of such points is equal to the arc length of the edge. The definition does not depend on the parametrisation.

The integral of a measurable function with respect to the Lebesgue–Stieltjes measure is defined as the sum of the line integrals of along the rectifiable curves parametrised by the diffeomorphism :

 ∫LfdλG=n(E)∑i=1∫Jif(ϕ−1i(t))∣∣∣ddtϕ−1i(t)∣∣∣dt.

Higher order integrals are defined analogously. Note that the definition does not depend on the parametrisation.

###### Example 1

As an example, consider the functions defined on by

 fiG(x)=∣∣∣(ddtϕ−1i)(ϕi(x))∣∣∣−1.

Then, for in the -algebra generated by ,

 ∫L1{x∈Bi}fiG(x)dλG(x) = ∫Ji1{ϕ−1i(t)∈Bi}∣∣∣(ddtϕ−1i)(ϕi(ϕ−1i(t))∣∣∣−1∣∣∣ddtϕ−1i(t)∣∣∣dt = |ϕi(Bi)|,

the Euclidean length of . In other words, the functions , , define the weighted shortest path distance on .

A point process on is said to have probability density with respect to the unit rate Poisson process if

 (1) P(X∈F)=∞∑n=0e−λG(L)n!∫L⋯∫L1{{x1,…,xn}∈F}p({x1,…,xn})dλG(x1)⋯dλG(xn)

for all in the usual -algebra on locally finite point configurations in generated by the counts with in the finite disjoint union -algebra on [6]. For convenience’s sake, we will use the notation for configurations of finitely many distinct points in . For further details on graphs with Euclidean edges, the reader may consult Anderes et al. [1].

## 3 Nearest-neighbour Markov point processes

The purpose of this section is to define appropriate analogues of the well-known class of renewal processes on the real line. Intuitively speaking, we are looking for a class of point processes in which the conditional behaviour at a given point depends on the remainder of the configuration only through its ‘nearest neighbours’. We shall show that in the case that the network is a tree, the shortest path distance may be used to define which points are each other’s nearest neighbours. In the general case, we will use the geometry of the network to define a local neighbourhood relation.

### 3.1 The Delaunay relation

In this section, we adapt Baddeley and Møller’s definition of configuration dependent neighbourhoods and cliques in Euclidean spaces [2] to our context.

###### Definition 7

Let be a graph with Euclidean edges and the corresponding network. Let be a finite configuration of distinct points and define the Delaunay relation as the symmetric, reflexive relation on given by

 x1∼xx2⇔C(x1|x)∩C(x2|x)≠∅

where, for , ,

 C(xj|x)={(i,y)∈L:dG((i,y),xj)≤dG((i,y),x) % for all x∈x}

is the Voronoi cell of in . The –neighbourhood of a subset is defined as

 N(z|x)={x∈x:x∼xz for some z∈z}.

The configuration is an -clique if for each , . By convention, the empty set and singletons are cliques too.

A few remarks are in order. Although the space can be seen as one dimensional, the Voronoi cells are not necessarily line segments and the number of neighbours a point can have is not restricted to two. Some further elementary properties of the relation are collected in the next lemma.

###### Lemma 1

Let be a graph with Euclidean edges and the corresponding network. The Delaunay relation on satisfies the following properties.

• If then also for all .

• If then also for all .

Here if is an -clique and zero otherwise.

The proof can be found in Section 4.

### 3.2 The Delaunay relation on trees

Recall that is said to be a tree if it has no cycles, that is, there is no closed path , , of positive length (). It is well-known that a graph is a tree if and only if there is exactly one path between any two vertices [5].

A graph with Euclidean vertices is said to be a tree if is.

###### Lemma 2

A graph with Euclidean edges is a tree if and only if there is exactly one path between any two points and in .

We are particularly interested in the geometrical arrangement of the paths between three points.

###### Lemma 3

Let be a graph with Euclidean edges that is a tree and its associated network. Consider a triple . Then there exist unique paths between the elements of that

• either form a wheel with three stokes of strictly positive length emanating from a vertex ,

• or combine into a single path.

The proofs of Lemma 2 and 3 will be given in Section 4.

A configuration is said to be in general position if no three points lie on the boundary of the same -ball. In other words, by Lemma 3, a configuration in general position cannot form a wheel with three or more equally long spokes. Such a configuration would lead to Delaunay cliques with more than two elements. Restriction to configurations in general position ensures that the size of cliques is at most two. Clearly, the class of all configurations in general position is hereditary.

###### Lemma 4

Let be a graph with Euclidean edges that is a tree and its associated network. Then the clique sizes with respect to the Delaunay relation are at most two on the class of configurations in general position. Moreover, for all with in general position, if and only if the midpoint of with respect to the weighted shortest path metric along the unique path between and lies in .

Since the clique structure depends on the configuration, consistency conditions must be imposed on the family of neighbourhood relations [2].

###### Definition 8

Let be a graph with Euclidean edges and its associated network. Consider finite point configurations and points such that . Then the Baddeley–Møller consistency conditions read as follows:

• implies ;

• if for , then

 χ(y|z∪{u})+χ(y|z∪{v})=χ(y|z)+χ(y|x)

where is the clique indicator function, i.e.  is an -clique.

In general, the consistency conditions of Definition 8 do not hold, as illustrated by the following counterexample.

###### Example 2

Take and connect all three vertices by straight line segments to form the complete graph and the linear network . Let be a configuration containing a single point on each edge. Then . However, any additional point would split up two of the points in . Hence , although the third point of is no Delaunay neighbour of in . Consequently, (C1) does not hold. Upon the addition of a second point that is not a Delaunay neighbour of in the resulting five point configuration, although , in violation with (C2).

The main theorem of this section is the following. Its proof will be given in Section 5.

###### Theorem 1

Let be a graph with Euclidean edges that is a tree and its associated network. Then the Delaunay relation satisfies (C1)–(C2) on the family of configurations in general position.

### 3.3 Markov functions

We are now ready to define Markov functions on graphs with Euclidean edges, in analogy with the spatial models of [2].

###### Definition 9

Let be a graph with Euclidean edges and the corresponding network. Let be a family of reflexive, symmetric relations on finite point configurations in . Then a function from the set of finite point configurations into is a Markov function with respect to if for all such that ,

(a)

for all ;

(b)

for all , depends only on , on and on the relations restricted to .

The next theorem provides a Hammersley–Clifford factorisation. Similar results for spatial point processes in Euclidean spaces can be found in [2], [3], [8] or [12]. Recall that a function from the space of finite point configurations into is an interaction function [2] if the following properties hold. If then also (a) for all and (b) if additionally then .

###### Theorem 2

Let be a graph with Euclidean edges such that is a tree and the corresponding network. Let a measurable function from the set of finite point configurations into . Then is a Markov function on the family of configurations in general position with respect to the Delaunay relation if and only if

 p(x)∝1{γ(x)>0}∏xi∈xγ({xi})∏i

for some measurable, non-negative interaction function .

If can be normalised into a probability density, e.g. by assuming that is bounded and , is a Markov density and we can define a nearest-neighbour Markov point process by (1).

The characterisation allows some flexibility in the choice of .

###### Example 3

Set , , and suppose that the interaction function for pairs depends only on the weighted shortest path distance between its elements:

 γ({xi,xj})=g(dG(xi,xj))

for some function . For configurations with , set

 γ(x)=1{g(dG(xi,xj))>0 for all xi,xj∈x}.

Clearly if and only if for all .

The function thus defined is an interaction function provided that if the conditions

• for all and

• for all such that

hold, then also for all . An alternative condition is [2, (G) in §5.2] in combination with the interaction function for configurations with . In either case, a sufficient condition is that the function takes strictly positive values. Then the Papangelou conditional intensity reads

in complete analogy with the conditional intensity of a renewal process on the real line.

### 3.4 The local Delaunay relation

As shown in Example 2, the Delaunay relation does not satisfy the consistency relations (C1)–(C2) if the graph is not a tree. However, we may employ the neighbourhood relation implicit in the graph to define a local Delaunay relation. Such a procedure is similar to one employed in image analysis for edge detection and texture analysis [7].

###### Definition 10

Let be a graph with Euclidean edges and its associated network. Define a symmetric and reflexive relation on as follows:

 (i,x)∼E(j,y)⇔⎧⎪ ⎪⎨⎪ ⎪⎩ϕ−1i(∂Ji)∩ϕ−1j(∂Jj)≠∅,i,j≠0ϕ−1i(∂Ji)∩{y}≠∅,i≠0,j=0{x,y}∈E,i=j=0.

Write, for , for the Delaunay relation restricted to

 L∩(({i,0}×ϕ−1i(¯Ji))∪({j,0}×ϕ−1j(¯Jj))),

the restriction of to at most two edges and their incident vertices, and define

 (i,x)∼Ez(j,y)⇔⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(i,x)∼i,jz∩({i,0}×ϕ−1i(¯Ji))∪({j,0}×ϕ−1j(¯Jj))(j,y);(i,x)∼E(j,y),i,j≠0(i,x)∼i,iz∩({i,0}×ϕ−1i(¯Ji))(0,y);(i,x)∼E(0,y),i≠0,j=0(0,x)∼k,kz∩({k,0}×ϕ−1k(¯Jk))(0,y);{x,y}=ek∈E,i=j=0.

In words, a vertex is a -neighbour of the edges it is incident with and edges are neighbours if they share a common vertex. Thus, the -relation does not depend on the configuration. It does, however, crucially depend on the geometry of the graph – splitting an edge will result in a different relation. After combination with the Delaunay relation, the resulting relation is configuration dependent and depends on the geometry of the graph.

The main result of this section is the following. Its proof will be given in Section 5.

###### Theorem 3

Let be a graph with Euclidean edges and its associated network. Then the local Delaunay relation satisfies (C1)–(C2).

## 4 Proofs of Lemmata

Proof of Lemma 1:
We claim that implies that for any and . To see this, note that

 χ({y1,y2}|x)=1⇔∃ξ∈L:dG(ξ,x)≥dG(ξ,yi)=dG(ξ,y2) for all x∈x.

A fortiori, for this , for all in the smaller set , so that , that is, .

By convention, the clique indicator function takes the value one for singletons and the empty set regardless of the configuration. Hence we may focus on configurations of cardinality at least two.

For the first statement, suppose that . Pick any . Then and, by the above claim, . Since and are chosen arbitrarily, .

For the second statement, if , there exist such that . By the claim, also , hence .

Proof of Lemma 2:
Suppose that is a tree. For vertices, since is a tree, there is a unique path between each pair of vertices. Hence we may restrict ourselves to a pair of points and of which at least one belongs to the set .

If , one path runs along the edge. Since a walk does not reverse its tracks by definition, any other walk from to must visit at least one of the two end vertices of before returning, hence use twice.

If and , note there is a unique path from to in the graph and hence a corresponding one in the labelled space . If this path includes edge , by deleting the -labelled curve segment from to , one obtains a path from to , otherwise such a path is found by extending the path to with this segment. Moreover, any other path would have to pass at least one of the vertices in , that is, coincide with the construction up to there and hence entirely.

Finally, if , as seen in the previous case, there is a unique path from to . If this path includes edge , this yields the path from to . Otherwise, precede by the segment along from to . Any other path would have to pass at least one of the vertices in , that is, coincide with the construction from there and hence entirely.

Reversely, let be such that there is a unique path between any pair of distinct points and in (which exists since a graph with Euclidean edges is connected by definition). In particular, there is a unique path between any pair of -labelled vertices, and therefore is a tree.

Proof of Lemma 3:
Since is a tree, by Lemma 2, there are unique paths from to and from to , say with consecutive labelled vertices

 y1,(0,v1),…,(0,vp),y2,(0,vp+1),…,(0,vp+q),y3,

and no vertex replication in nor in .

Suppose that for some and . Then the paths and both connect and in , and must therefore coincide. Extending if possible, we may assume that . If some earlier vertex , would be identical to for , a cycle would be created from via to . Hence the sequences and do not intersect and the paths , and are connected at and therefore form three stokes connected at a single hub provided the lengths are positive. It therefore remains to consider the cases , or .

If , may lie on the path from to ; if it does not, the curve segment from to forms a stoke of positive length. Similarly, if , may lie on the path from to ; if it does not it, the curve segment from to forms a stoke of positive length. Also, if , may lie along the path between and ; if it does not, the curve segment from to forms a stoke of positive length.

Finally, if there is no such that for some , the unique path from to runs via .

Proof of Lemma 4:
Suppose that for some . By Lemma 1, hence one needs only consider .

By Lemma 3, the elements of either form a wheel or a path. First consider the case that is a wheel with three stokes emanating from a hub . Ordering the stokes according to their length, without loss of generality suppose that where , and . Since by assumption is in general position, . Therefore, as is a tree, and .

It remains to consider the case that all three elements of lie along a path, without loss of generality from via to . Then , again using the tree property to ensure that there are no paths to connect and other than via . Therefore . In conclusion, there cannot be a clique of cardinality three or larger.

To prove the second assertion, let and write for the midpoint with respect to along the path between and . Since is assumed to be a tree, by Lemma 2, the path is unique. Since all are bijections, also the midpoint is unique. Let be some configuration in general position.

If , clearly . Reversely, suppose that is a clique in . Then there exists some such that . By Lemma 3, the triple either forms a path or a wheel. In the first case, the property of equidistance to and implies that and the proof is complete.

Next consider the case that and form three stokes of strictly positive length connected at some hub . Since , by the uniqueness of the paths from to (cf. Lemma 2), . Hence , now using the uniqueness of the path between and . It remains to show that no other point of lies closer to than the , . Since is a tree, any such is connected either to the hub or to exactly one of the tree stokes. Any connected to the end points of the stokes lies further from than and .

Any connected to the stoke of satisfies, since is assumed to lie in ,

 dG(η,x)≥dG(η,yj)=dG(η,ξ)+dG(ξ,yj),j=1,2,

again using the uniqueness of paths for the last equality. If the connection is at the endpoint , then hence, by the above equation, If the connection is at a vertex , ,

 dG(ξ,x) = dG(ξ,(0,w))+dG(x,(0,w)) = dG(ξ,(0,w))+dG(η,x)−dG(η,(0,w)) ≥ dG(ξ,(0,w))+dG(η,yj)−dG(η,(0,w)) = dG(ξ,(0,w))+dG(yj,(0,w))≥dG(ξ,yj).

Finally, any connected to the stoke of a at or some other vertex satisfies, since is assumed to lie in ,

 dG(η,x)≥dG(η,yj)=dG(η,ξ)+dG(ξ,yj)

so that and the proof is complete.

## 5 Proof of main theorems

Proof of Theorem 1:
First, note that for of size zero or one, (C1) and (C2) are automatically satisfied. The assumption that is a tree and the restriction to configurations in general position imply, by Lemma 4, that for all when the cardinality of is three or more. Hence (C1) and (C2) hold for with as well and it suffices to consider pairs .

#### Condition (C1)

First, consider (C1). Take , with , and suppose that differs from . By Lemma 1, if , also , so it suffices to consider the case that but .

Let be the point lying halfway between and along the unique path between them (cf. Lemma 2) and let be the halfway point between and . These points exist since the parametrisations are bijections. By Lemma 4, since , , so that for all , , . Also by construction We shall show that, for , and therefore . To do so, we must demonstrate that

 (2) dG(~yj,z)≥dG(~yj,u)=dG(~yj,yj)

for .

By Lemma 3, since is a tree, the paths between the three points form either a wheel with three stokes joined at some hub with , or the points lie on a path.

First, consider the case that , and lie on a path. If were an extremity of this path, say is on the path between and , then

 dG(ξ,u)=dG(ξ,y1)+dG(y1,u)>dG(ξ,y1).

By the assumption and Lemma 4, for any , also so that , thus violating the assumption that . We conclude that has to lie on the path from to . Suppose that (2) does not hold, i.e. that for some and some , the distance . Then

 dG(ξ,z)≤dG(ξ,~yj)+dG(~yj,z)<