2.1. Diffuse non-conflicting finite positive measures

Let D be a Polish space. We write $ \mathcal{B}$ for the Borel $ \sigma$ -field of D. A finite positive measure $ \eta$ on D is called diffuse if for every Borel set $ B \in \mathcal{B}$ with $ \eta(B) > 0$ there is a Borel subset $ C \subset B$ with $ 0 < \eta(C) < \eta(B)$ . One can define a Poisson process on D based on such a diffuse positive measure $ \eta$ on D; see e.g. [Reference Daley and Vere-Jones8, Chapter 9]. A point process on D is a random counting measure on D. The Poisson process on D with intensity measure $ \eta$ , where $ \eta$ is any finite positive diffuse measure on D, is the point process obtained by first selecting the total number of points according to the Poisson distribution on $ \mathbb N$ with mean $ \eta(D)$ , and then sampling independently the location of the points, if any, according to the measure $ \eta(.)/\eta(D)$ on $ \mathcal{B}$ .

Recall that d denotes the fixed metric on D generating its topology. The open Voronoi cells of any point process $ \Phi$ with respect to d can be defined in the usual way. Namely, for x in the support of the point process $ \Phi$ , the open Voronoi cell $ W_{\Phi}(x)$ consists of those $ z \in D$ such that $ d(x,z) < d(x^{\prime},z)$ for all points $ x^{\prime} \neq x$ in the support of $ \Phi$ . For our purposes, we need to impose on $ \eta$ the condition that the union of the open Voronoi cells of the Poisson process on D with intensity measure $ \eta$ has full measure $ \eta(D)$ with probability 1, conditioned on there being at least one point in this Poisson process. If $ \eta$ satisfies this condition, we call it non-conflicting. A sufficient condition for this to hold is given in Appendix A.

The importance of imposing a non-conflicting condition can be understood by considering the following example.

Throughout the paper, $ \eta$ is a finite positive diffuse non-conflicting measure on D which will be referred to as the base measure. It is straightforward to check that if $ \nu$ is any finite positive measure on D that is absolutely continuous with respect to $ \eta$ , then $ \nu$ is also diffuse and non-conflicting. We write $ \mathcal{M}(D)$ for the set of nonnegative finite measures on D.

2.2. Radon–Nikodym derivatives

In the Cox process Hotelling games considered in this paper, we think of each individual player as choosing a positive measure on D of fixed total mass that is absolutely continuous with respect to the base measure $ \eta$ . The measure chosen by the player then serves as the intensity measure for a Poisson process of points on D, which we think of as the points belonging to that player. By the Radon–Nikodym theorem, we may identify the measure chosen by the player with its likelihood function with respect to the base measure.

Figure 2.

A metric space comprising two intervals, each of length 1. Every point in the interval on the left is at distance 2 from every point in the interval on the right. The metric on each interval is the usual one.

With this viewpoint in mind, for any $ \rho > 0$ , let

(1) \begin{equation}\mathcal{C}(\rho) \,{:}\,{\raise-1.5pt{=}}\, \bigg\{f\,:\,D \to \mathbb{R}_+ \mbox{ s.t. } \int_D f(x) \eta(dx) = \rho \bigg\}.\end{equation}

The set $ \mathcal{C}(\rho)$ can be thought of as a subset of $ L^1(\eta)$ , i.e. the Lebesgue space of $ \mathcal{B}$ -measurable functions on D that are absolutely integrable with respect to the base measure $ \eta$ , and also as a set of measures on D by identifying $ f \in \mathcal{C}(\rho)$ with the measure $ f \eta$ . With the latter viewpoint in mind, we think of $ \mathcal{C}(\rho)$ as endowed with the topology of weak convergence of measures [Reference Parthasarathy26], which it inherits as a subset of $ \mathcal{M}(D)$ . Note that, for all $ \rho > 0$ and all $ f \in \mathcal{C}(\rho)$ , the measure $ f \eta$ is diffuse and non-conflicting. Also note that, for all $ \rho > 0$ , the set $ \mathcal{C}(\rho)$ is a convex subset of $ \mathcal{M}(D)$ . However, since we have endowed $ \mathcal{C}(\rho)$ with the topology of weak convergence of measures on D, this set is not closed.

For any $ \rho > 0$ , the subset of $ \mathcal{M}(D)$ comprising measures of total mass $ \rho$ , with the topology of weak convergence, can be metrized so as to make it a complete separable metric space [Reference Parthasarathy26, Theorem 6.2 and Theorem 6.5]. As a metric space it is first-countable, so it suffices to discuss convergence of sequences rather than of nets [Reference Kelley19, Theorem 8]. To discuss randomized strategies of the individual players we need to be able to discuss probability distributions on subsets of the type $ \mathcal{C}(\rho)$ of $ \mathcal{M}(D)$ . This is made possible by the following result.

2.3. Two-player games

In the two-player version of the game, a pure action of Alice consists of choosing an intensity measure which is absolutely continuous with respect to the measure $ \eta$ . Namely, Alice chooses a nonnegative measurable function $ f_A\,:\,D\to \mathbb{R}_+$ as the Radon–Nikodym derivative with respect to $ \eta$ of the intensity measure of its Poisson point process. We denote the set of functions from which this choice must be made by $ \mathcal{C}_A$ , i.e.,

(2) \begin{equation}\mathcal{C}_A \,{:}\,{\raise-1.5pt{=}}\, \mathcal{C}(\rho_A) = \bigg\{f_A\,:\,D \to \mathbb{R}_+ \mbox{ s.t. } \int_D f_A(x) \eta(dx) = \rho_A \bigg\}.\end{equation}

Thus $ \rho_A$ can be thought of as a constraint on the total intensity that Alice can deploy for her point process.

Similarly, Bob chooses a nonnegative measurable function $ f_B\,:\,D\to \mathbb{R}_+$ within the class of functions $ \mathcal{C}_B$ , where

(3) \begin{equation}\mathcal{C}_B \,{:}\,{\raise-1.5pt{=}}\, \mathcal{C}(\rho_B) = \bigg\{f_B\,:\,D \to \mathbb{R}_+ \mbox{ s.t. } \int_D f_B(x) \eta(dx) = \rho_B \bigg\}.\end{equation}

Here $ \rho_A$ and $ \rho_B$ are fixed positive constants. Note that $ \mathcal{C}_A$ and $ \mathcal{C}_B$ are convex subsets of $ \mathcal{M}(D)$ , but neither of them is closed in the topology of weak convergence on $ \mathcal{M}(D)$ .

We denote such a two-player Cox process Hotelling game by $ (D,\eta, \rho_A, \rho_B)$ .

Assume that Alice plays $ f_A$ and Bob plays $ f_B$ . The resulting value for Alice, denoted by $ V_A\big(\,f_A,f_B\big)$ , is defined as follows. Let $ \Phi_A$ and $ \Phi_B$ be independent Poisson point processes on D with intensity measures $ f_A \eta$ and $ f_B \eta$ , respectively. We think of the points of $ \Phi_A$ as Alice’s points, since these result from the choice of $ f_A$ by Alice. Similarly we think of the points of $ \Phi_B$ as Bob’s points. Let $ \Phi \,{:}\,{\raise-1.5pt{=}}\,\Phi_A+\Phi_B$ . For all $ x \in \Phi$ , let $ W_\Phi(x)$ denote the open Voronoi cell of x with respect to $ \Phi$ . Then

\[V_A\big(\,f_A,f_B\big) \,{:}\,{\raise-1.5pt{=}}\, \mathbb{E} \left[\sum_{x\in \Phi_A} \eta(W_\Phi(x)) \right],\]

where the expectation is with respect to the joint law of $ \Phi_A$ and $ \Phi_B$ , which are assumed to be independent. Here, by definition, a sum over an empty set is 0. The value for Bob resulting from this pair of actions, denoted by $ V_B\big(\,f_A,f_B\big)$ , is determined similarly; i.e., it is

\[V_B\big(\,f_A,f_B\big) \,{:}\,{\raise-1.5pt{=}}\, \mathbb{E} \left[\sum_{x\in \Phi_B} \eta(W_\Phi(x)) \right].\]

In words, the value of Alice is the mean value of the sum of the $ \eta$ -measures of the open $ \Phi$ -Voronoi cells centered at her points if she has points, and is 0 otherwise, and likewise for the value of Bob. Intuitively, one thinks of the points of Alice and those of Bob as competing to capture the ambient space, with a point in the ambient space D belonging to Alice if the closest point to it is one of Alice, and to Bob otherwise. Under our non-conflicting condition on the base measure $ \eta$ , there is no need to worry about how to break ties.

Note that

(4) \begin{equation}V_A\big(\,f_A,f_B\big) + V_B\big(\,f_A,f_B\big) = \eta(D) \big(1 - e^{- \rho }\big)\end{equation}

for all $ \big(\,f_A,f_B\big) \in \mathcal{C}_A \times \mathcal{C}_B$ , where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ . This is because, by virtue of our assumption that $ \eta$ is non-conflicting, the sum of the values of Alice and Bob is $ \eta(D)$ , except on the event where neither $ \Phi_A$ nor $ \Phi_B$ has any points, which is an event of probability $ 1 - e^{- \rho }$ . The formula in Equation (4) is formally established in Lemma 2 below.

2.4. Nash equilibrium in two-player games

In view of Lemma 1, any mixed-strategy pair in the two-player game $ (D, \eta, \rho_A, \rho_B)$ between Alice and Bob can be written as $ (\,f_A(M_A), f_B(M_B))$ , where $ M_A \in \mathcal{M}_A$ and $ M_B \in \mathcal{M}_B$ are independent random variables representing the randomizations used by Alice and Bob respectively in implementing randomized strategies, and $ f_A(m_A) \in \mathcal{C}_A$ (respectively, $ f_B(m_B) \in \mathcal{C}_B$ ) is the choice of Alice (respectively, Bob) in case the realization of her random variable $ M_A$ is $ m_A$ (respectively, the realization of his random variable $ M_B$ is $ m_B$ ). The value of Alice in such a mixed strategy is $ \mathbb{E}\big[ V_A\big(\,f_A\big(M_A\big), f_B\big(M_B\big)\big)\big]$ , while that of Bob is $ \mathbb{E}\big[ V_B\big(\,f_A\big(M_A\big), f_B\big(M_B\big)\big)\big]$ , the expectations being taken with respect to the joint distribution of $ M_A$ and $ M_B$ , which are independent. As a direct consequence of Equation (4), we have

(5) \begin{equation}\mathbb{E}\big[ V_A\big(\,f_A\big(M_A\big), f_B\big(M_B\big)\big)\big] + \mathbb{E}\big[ V_B\big(\,f_A\big(M_A\big), f_B\big(M_B\big)\big)\big]= \eta(D)\big( 1 - e^{-\rho}\big),\end{equation}

where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ .

Definition 1. The pair of independently randomized strategies $ (\,f_A(M_A), f_B(M_B)) \in \mathcal{C}_A \times \mathcal{C}_B$ is called a Nash equilibrium of the game between Alice and Bob if, for all $ g_A \in \mathcal{C}_A$ and $ g_B \in \mathcal{C}_B$ , we have

(6) \begin{equation}\mathbb{E}\big[V_A\big(\,f_A\big(M_A\big),f_B\big(M_B\big)\big)\big] \ge \mathbb{E}\big[V_A\big(g_A, f_B\big(M_B\big)\big)\big]\end{equation}

and

(7) \begin{equation}\mathbb{E}\big[V_B \big(\,f_A\big(M_A\big), f_B\big(M_B\big)\big)\big] \ge \mathbb{E}\big[V_B\big(\,f_A\big(M_A\big), g_B\big)\big].\end{equation}

In words, Alice sees no advantage in playing the strategy $ g_A$ instead of her randomized strategy $ f_A(M_A)$ , given that Bob is playing his randomized strategy $ f_B(M_B)$ , and similarly for Bob.

3.1. A formula for the value

Consider a two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ between Alice and Bob, and suppose that the players play the action pair $ (\,f_A, f_B)$ . Let $ \Phi_A$ and $ \Phi_B$ denote the Poisson processes of points of Alice and Bob respectively on D. Recall that these are independent point processes. Let $ \mathbb{P}_A^x$ $ \big($ respectively, $ \mathbb{P}_B^x\big)$ denote the Palm probability [Reference Kallenberg18, Chapter 6] with respect to $ \Phi_A$ (respectively, $ \Phi_B$ ) at x. By Slivnyak’s theorem [Reference Kallenberg18, Lemma 6.14], under the Palm probability with respect to $ \Phi_A$ at x, $ \Phi_A$ consists of a point at x and a Poisson process on D of intensity $ f_A \eta$ , and $ \Phi_B$ consists of an independent Poisson process on D of intensity $ f_B \eta$ . The Palm probability with respect to $ \Phi_B$ at x has a symmetrical description.

From Campbell’s formula [Reference Kallenberg18, Section 6.1] we have

\[V_A\big(\,f_A,f_B\big) = \int_D f_A(x) \mathbb{E}_A^x \big[\eta(W_\Phi(x))\big] \eta(dx).\]

Here the expectation is with respect to the law of $ \Phi$ under the Palm distribution $ \mathbb{P}_A^x$ . From the description of this Palm distribution, we have

\begin{align*} \mathbb{E}_A^x \big[\eta(W_\Phi(x))\big] & = \mathbb{E}_A^x \bigg[\int_{y\in D} 1_{\{y\in W_\Phi(x)\}} \eta(dy)\bigg] \\&\stackrel{(a)} = \int_{y\in D} \mathbb{P}_A^x\big( y\in W_\Phi(x)\big) \eta(dy) \\&\stackrel{(b)} = \int_{y\in D} e^{-\int_{B(y\to x)} (\,f_A(u)+f_B(u)) \eta(du)} \eta(dy),\end{align*}

where

\[B(y\to x) \,{:}\,{\raise-1.5pt{=}}\, \big\{z\in D \mbox{ s.t. } d(z,y)\le d(x,y) \big\}.\]

The notation $ B(y \rightarrow x)$ is supposed to bring to mind a closed ball centered at y having x at its boundary.

In the chain of equations above, Step (a) comes from an application of Fubini’s theorem and Step (b) from Slivnyak’s theorem. Hence

(8) \begin{equation}V_A(\,f_A, f_B) =\int_{x\in D} f_A(x)\int_{y\in D} e^{-\int_{B(y\to x)} (\,f_A(u)+f_B(u)) \eta(du)} \eta(dy) \eta(dx).\end{equation}

3.2. N-player games

The point process analysis above and the resulting formula in Equation (8) in the two-player case also leads to a formula in an N-player Cox process Hotelling game for the value $ V_i(\,f_1, \ldots, f_N)$ seen by player i when the pure strategies deployed by the individual players are $ f_1,\ldots, f_N$ respectively. For this, let $ f_j\,:\, D \to \mathbb{R}_+$ be a nonnegative measurable function on D belonging to

(9) \begin{equation}\mathcal{C}_j \,{:}\,{\raise-1.5pt{=}}\, \mathcal{C}(\rho_j) = \bigg\{ f_j\,:\,D \to \mathbb{R}_+ \mbox{ s.t. } \int_D f_j(x) \eta(dx) = \rho_j \bigg\},\end{equation}

where the $ \rho_j > 0$ are fixed for $ 1 \le j \le N$ . Note that each $ \mathcal{C}_j$ is a convex subset of $ \mathcal{M}(D)$ , but is not closed in the topology of weak convergence. Assume that the individual players play $ f_i \in \mathcal{C}_i$ for $ 1 \le i \le N$ . Let $ \Phi_i$ be a Poisson point process on D with intensity measure $ f_i \eta$ , with these processes being mutually independent for $ 1 \le i \le N$ , and let $ \Phi \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \Phi_i$ . We think of the points of $ \Phi_i$ as the points of player i, since these result from the choice of $ f_i$ , which was made by that player. Then the value of player i, denoted by $ V_i(\,f_1, \ldots, f_N)$ , is given by

\[V_i(\,f_1, \ldots, f_N) \,{:}\,{\raise-1.5pt{=}}\, \mathbb{E} \left[\sum_{x\in \Phi_i} \eta(W_\Phi(x))\right].\]

The expectation is with respect to the joint law of $ (\Phi_j, 1 \le j \le n)$ , which are independent. Here, by definition, a sum over an empty set is 0.

We denote such an N-player Cox process Hotelling game by $ (D, \eta, \rho_1, \ldots, \rho_N)$ .

We can write a formula for $ V_i(\,f_1, \ldots, f_N)$ , based on Equation (8), by thinking of the points of player i as competing for space in D with the union of the points of the other players. Thus we have

(10) \begin{equation}V_i(\,f_1, \ldots, f_N) = \int_{x \in D} f_i (x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx),\end{equation}

where $ f \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N f_i$ .

Note that we must have

(11) \begin{equation}\sum_{i=1}^N V_i(\,f_1, \ldots, f_N) = \eta(D) \big(1 - e^{- \rho }\big),\end{equation}

where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{j=1}^N \rho_j$ . This is because we have $ \int_{x \in D} f(x) \eta(dx) = \rho$ , so the law of the total number of points is Poisson with mean $ \rho$ , and thus, by virtue of our assumption that $ \eta$ is non-conflicting, the total value of all the players is $ \eta(D)$ except on the event that there are no points in $ \Phi$ , in which case the total value is 0. The formula in Equation (11) is established in Lemma 2 below.

3.3. Nash equilibrium in N-player games

In view of Lemma 1, any mixed-strategy pair in an N-player Cox process Hotelling game $ \big(D, \eta, \rho_1, \ldots, \rho_N\big)$ can be written as $ \big(\,f_1(M_1), \ldots, f_N(M_N)\big)$ , where $ \big(M_j \in \mathcal{M}_j, 1 \le j \le N\big)$ are independent random variables representing the randomizations used by the individual players in implementing their randomized strategies, and $ f_j(m_j) \in \mathcal{C}_j$ is the choice of action of player j in case the realization of her random variable $ M_j$ is $ m_j$ . The value of player i in such a mixed strategy is $ \mathbb{E}\big[ V_i\big(\,f_1(M_1), \ldots, f_N(M_N)\big)\big]$ , the expectation being taken with respect to the joint distribution of $ \big(M_j, 1 \le j \le N\big)$ , which are independent. As a direct consequence of Equation (11), we have

(12) \begin{equation}\sum_{i=1}^N \mathbb{E}[ V_i(\,f_1(M_1), \ldots, f_N(M_N))]= \eta(D)|\big( 1 - e^{-\rho}\big),\end{equation}

where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \rho_i$ .

Definition 2. The vector of independently randomized strategies

\[(\,f_1(M_1), \ldots, f_N(M_N)) \in \mathcal{C}_1 \times \ldots \times \mathcal{C}_N\]

is called a Nash equilibrium of the N-player game if, for all $ g_j \in \mathcal{C}_j$ , $ 1 \le j \le N$ , we have, for all $ 1 \le i \le N$ ,

(13) \begin{equation}\mathbb{E}\big[V_i(\,f_1(M_1), \ldots, f_N(M_N))\big] \ge\mathbb{E}\big[V_i\big(g_i, \big(\,f_j\big(M_j\big), j \neq i\big)\big)\big].\end{equation}

In words, the player i perceives no advantage in playing the strategy $ g_i$ instead of the randomized strategy $ f_i(M_i)$ , given that the other players, i.e. the players $ j \neq i$ , are playing the individually randomized strategies $ \big(\,f_j\big(M_j\big), j \neq i\big)$ .

3.4. A conservation law

To establish the conservation law in Equation (11) and the special case of it in Equation (4), it suffices to demonstrate that for all $ \rho > 0$ and all $ f \in \mathcal{C}(\rho)$ , we have Equation (14) below. This is because we would then immediately obtain the desired conservation laws from Equation (10) and the special case in Equation (8), respectively, by summing these formulas over the individual players. We establish (14) formally in the following lemma.

Lemma 2. For any $ \rho > 0$ and $ f \in \mathcal{C}(\rho)$ , we have

(14) \begin{equation}\int_{x\in D} f(x)\int_{y\in D} e^{-\int_{B(y\to x)} f(u) \eta(du)} \eta(dy) \eta(dx)= \eta(D) \!\left(1-e^{-\rho }\right).\end{equation}

Proof. We give two proofs of this formula. The first one is probabilistic. Let $ \Phi$ denote a Poisson process on D with intensity measure $ f \eta$ . By Campbell’s formula and Slivnyak’s theorem [Reference Kallenberg18], the integral on the left-hand side of (14) is just

\[ \mathbb{E} \left[\sum_{x \in \Phi} \eta(W_\Phi(x))\right],\]

where $ W_\Phi(x)$ denotes the open Voronoi cell of $ x \in \Phi$ with respect to the Poisson process $ \Phi$ . If $ \Phi(D)=0$ , which happens with probability $ e^{-\rho}$ , the sum is 0. On the complementary event the sum is $ \rho(D)$ , since, by virtue of the assumption that $ \eta$ is non-conflicting, the collection of sets $ (W_\Phi(x), x \in \Phi)$ form a partition of D up to a set of $ \eta$ -measure 0.

The second proof is analytical. We have, for each $ y \in D$ ,

\[\int_{x \in D} f(x) e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dx) =1 - e^{- \rho },\]

since, when $ \eta$ is non-conflicting, the integral on the left-hand side of the preceding equation is just the probability that a nonhomogeneous Poisson process with intensity function f(x) with respect to $ \eta$ has at least one point in D (to see this, think of moving out from y by balls of increasing radius till one covers all of D). Thus we have

\[\int_{y \in D} \int_{x \in D} f(x)e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx) =\eta(D) \big(1 - e^{- \rho }\big).\]

But the left-hand side of the preceding equation equals the left-hand side of Equation (14), by an application of Fubini’s theorem.

3.5. Constant strategies

Consider a two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ . Write $ \bar{\rho}_A$ for the strategy of Alice where she chooses $ f_A$ to be the constant $ \frac{\rho_A}{\eta(D)}$ , and similarly write $ \bar{\rho}_B$ for the strategy of Bob where he chooses $ f_B$ to be the constant $ \frac{\rho_B}{\eta(D)}$ . Note that $ \bar{\rho}_A \in \mathcal{C}_A$ and $ \bar{\rho}_B \in \mathcal{C}_B$ .

The following useful lemma gives an explicit formula for the value obtained by Alice and that obtained by Bob in the two-player game $ (D, \eta, \rho_A, \rho_B)$ when the players play the constant strategies $ \bar{\rho}_A$ and $ \bar{\rho}_B$ respectively. Note that, while the strategies are constant, the resulting intensities are $ \frac{\rho_A}{\eta(D)} \eta$ and $ \frac{\rho_B}{\eta(d)} \eta$ for Alice and Bob respectively.

Proof. We have

\begin{align*}V_A\big(\bar{\rho}_A, \bar{\rho}_B\big) &\stackrel{(a)}{=} \int_{x \in D} \frac{\rho_A}{\eta(D)} \int_{y \in D}e^{ - \frac{\rho}{\eta(D)} \eta(B(y \rightarrow x))} \eta(dy) \eta(dx)\\[3pt]& = \frac{\rho_A}{\rho}\int_{x \in D} \int_{y \in D}\frac{\rho}{\eta(D)} e^{ - \frac{\rho}{\eta(D)} \eta(B(y \rightarrow x))} \eta(dy) \eta(dx)\\[3pt]&\stackrel{(b)}{=} \frac{\rho_A}{\rho}\int_{y \in D} \int_{x \in D}\frac{\rho}{\eta(D)} e^{ - \frac{\rho}{\eta(D)} \eta(B(y \rightarrow x))} \eta(dx) \eta(dy)\\[3pt]&\stackrel{(c)}{=} \frac{\rho_A}{\rho} \eta(D) (1 - e^{- \rho}).\end{align*}

Here, Step (a) is from Equation (8), Step (b) is an application of Fubini’s theorem, and Step (c) comes from Lemma 2. The formula for $ V_B\big(\bar{\rho}_A, \bar{\rho}_B\big)$ results from interchanging the roles of Alice and Bob in this calculation.

In an N-player Cox process Hotelling game $ (D, \eta, \rho_1, \ldots, \rho_M)$ , write $ \bar{\rho}_i$ for the strategy of player i, $ 1 \le i \le N$ , where she chooses $ f_i$ to be the constant $ \frac{\rho_i}{\eta(D)}$ . We then have the following analogue of Lemma 3.

Lemma 4. Consider an N-player Cox process Hotelling game $ (D, \eta, \rho_1, \ldots, \rho_N)$ . Then, for all $ 1 \le i \le N$ , we have

\[V_i\big(\bar{\rho}_1, \ldots, \bar{\rho}_N\big) = \frac{\rho_i}{\rho} \eta(D) (1 - e^{- \rho}),\]

where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \rho_i$ .

3.6. Concavity of the value

To close this section, we record a convexity property that will play a key role in establishing the main claims of this paper.

Lemma 5. Consider a two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ . Fix $ f_B \in \mathcal{C}_B$ and $ x \in D$ . Then the mapping

\[f_A \mapsto \int_{y \in D}e^{ - \int_{B(y \rightarrow x)} (\,f_A(u) + f_B(u)) \eta(du) } \eta(dy)\]

is strictly convex on $ \mathcal{C}_A$ , which, we recall, is a convex set.

As a consequence, $ f_A \mapsto V_B(\,f_A, f_B)$ , for fixed $ f_B \in \mathcal{C}_B$ , is strictly convex on $ \mathcal{C}_A$ , and hence $ f_A \mapsto V_A(\,f_A, f_B)$ , for fixed $ f_B \in \mathcal{C}_B$ , is strictly concave on $ \mathcal{C}_A$ .

Proof. Let $ \,f_A, f^{\prime}_A \in \mathcal{C}_A$ and $ \theta \in [0,1]$ . Then we have

\begin{eqnarray*}&& \int_{y\in D} e^{-\int_{B(y\to x)} (\theta f_A(u) + (1-\theta) f^{\prime}_A(u) + f_B(u)) \eta(du)} \eta(dy)\\&& \le\theta\int_{y\in D} e^{{-}\int_{B(y\to x)} (\,f_A(u)+ f_B(u)) \eta(du)} \eta(dy)\\&& +\,(1-\theta)\int_{y\in D} e^{{-}\int_{B(y\to x)} \big(\,f^{\prime}_A(u)+ f_B(u)\big) \eta(du)} \eta(dy),\end{eqnarray*}

from the convexity of the exponential function, and this inequality is strict if $ \theta \notin \{0,1\}$ and $ f_A \neq f^{\prime}_A$ . This proves the first claim of the lemma.

From Equation (8), we have

(15) \begin{equation}V_B(\,f_A, f_B) = \int_{x \in D} f_B(x) \int_{y \in D}e^{ - \int_{B(y \rightarrow x)} (\,f_A(u) + f_B(u)) \eta(du) } \eta(dy) \eta(dx).\end{equation}

Since, for fixed $ f_B \in \mathcal{C}_B$ , the inner integral is strictly convex on $ \mathcal{C}_A$ for each $ x \in D$ , the overall integral is also strictly convex on $ \mathcal{C}_A$ , which establishes the second claim of the lemma.

Finally, from the conservation rule in Equation (4), we have $ V_A(\,f_A, f_B) = \eta(D)\big(1 - e^{-\rho}\big) - V_B\big(\,f_A,f_B\big)$ , where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ . From the second claim of the lemma, the third claim now follows immediately.

The main claim of Lemma 5 is the third one about the strict concavity of the value function. This also holds for N-player Cox process Hotelling games. We state this claim formally in the following lemma.

4.1. Exploiting concavity of the value

The following lemma, which it suffices to state in the two-player case, is the key technical result driving the game-theoretic results in this section.

Lemma 7. Consider a two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ where D is compact and admits a transitive group of metric-preserving automorphisms, and $ \eta$ is invariant under this group of automorphisms. Then, for any strategy $ f_A \in \mathcal{C}_A$ of Alice, we have

\[V_A \big(\,f_A, \bar{\rho}_B\big) \le V_A \big(\bar{\rho}_A, \bar{\rho}_B\big) = \frac{\rho_A}{\rho} \eta(D) \big(1 - e^{- \rho }\big),\]

where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ . Further, we have the strict inequality

\[V_A \big(\,f_A, \bar{\rho}_B\big) < V_A \big(\bar{\rho}_A, \bar{\rho}_B\big),\]

except in the case $ f_A = \bar{\rho}_A$ .

Proof. From Lemma 3, we have $ V_A\big(\bar{\rho}_A, \bar{\rho}_B\big) = \frac{\rho_A}{\rho} \eta(D) (1 - e^{- \rho})$ , which is one of the claims of this lemma. From Lemma 5 for the choice $ f_B = \bar{\rho}_B$ , we conclude that $ V_A\big(\,f_A, \bar{\rho}_B\big)$ is strictly concave on $ \mathcal{C}_A$ . We now use this to conclude that $ V_A\big(\,f_A, \bar{\rho}_B\big)$ is uniquely maximized over $ \mathcal{C}_A$ by the choice $ f_A = \bar{\rho}_A$ . Indeed, if $ f_A \in \mathcal{C}_A$ , $ f_A \neq \bar{\rho}_A$ , then we can find a translate $ f^{\prime}_A$ of $ f_A$ such that $ f_A \neq f^{\prime}_A$ . We have $ V_A\big(\,f_A, \bar{\rho}_B\big) = V_A\big(\,f^{\prime}_A,\bar{\rho}_B\big)$ , because $ f_A$ and $ f^{\prime}_A$ are translates of each other. However, since $ f_A \neq f^{\prime}_A$ , we have

\[V_A\bigg(\frac{1}{2}\big(\,f_A + f^{\prime}_A\big), \bar{\rho}_B\bigg) > \frac{1}{2} V_A\big(\,f_A, \bar{\rho}_B\big) + \frac{1}{2} V_A\big(\,f^{\prime}_A, \bar{\rho}_B\big) = V_A\big(\,f_A, \bar{\rho}_B\big).\]

Hence, $ f_A$ cannot be the maximizer of $ V_A\big(\,f_A, \bar{\rho}_B\big)$ over $ \mathcal{C}_A$ unless it is translation-invariant, i.e. unless it equals $ \bar{\rho}_A$ . This concludes the proof of the lemma.

4.2. Nash equilibrium structure for two-player games

We are now in a position to determine the Nash equilibrium structure of two-player Cox process Hotelling games $ (D, \eta, \rho_A, \rho_B)$ in the context of this section.

Proof. We need to show that if $ (\,f_A(M_A), f_B(M_B))$ is a Nash equilibrium for the game, then $ f_A(M_A) = \bar{\rho}_A$ and $ f_B(M_B) = \bar{\rho}_B$ with probability 1.

Suppose first that $ f_B(M_B) = \bar{\rho}_B$ with probability 1. If $ \mathbb{P}\big(\,f_A(M_A) \neq \bar{\rho}_A\big) > 0$ , then by Lemma 7 we have

\[\mathbb{E}\big[V_A\big(\,f_A(M_A),f_B(M_B)\big)\big] = \mathbb{E}\big[V_A\big(\,f_A(M_A), \bar{\rho}_B\big)\big]< V_A\big(\bar{\rho}_A, \bar{\rho}_B\big).\]

On the other hand, since $ f_A(M_A)$ is a best response by Alice to the strategy $ f_B(M_B)$ of Bob, we have

\[\mathbb{E}\big[V_A\big(\,f_A(M_A),f_B(M_B)\big)\big] \ge \mathbb{E}\big[V_A\big(\bar{\rho}_A,f_B(M_B)\big)\big]= V_A\big(\bar{\rho}_A, \bar{\rho}_B\big).\]

This contradiction establishes that if $ f_B(M_B) = \bar{\rho}_B$ with probability 1, then we must have $ f_A(M_A) = \bar{\rho}_A$ with probability 1. A similar argument works to show that if $ f_A(M_A) = \bar{\rho}_A$ with probability 1 then we must have $ f_B(M_B) = \bar{\rho}_B$ with probability 1.

Thus, it remains to handle the case where we have both $ \mathbb{P}\big(\,f_A(M_A) \neq \bar{\rho}_A\big) > 0$ and $ \mathbb{P}\big(\,f_B(M_B) \neq \bar{\rho}_B\big) > 0$ . In this case, since $ f_A(M_A)$ is a best response by Alice to the strategy $ f_B(M_B)$ of Bob, we have

(16) \begin{equation}\mathbb{E}\big[V_A\big(\,f_A(M_A),f_B(M_B)\big)\big] \ge \mathbb{E}\big[V_A\big(\bar{\rho}_A,f_B(M_B)\big)\big],\end{equation}

and, since $ f_B(M_B)$ is a best response by Bob to the strategy $ f_A(M_A)$ of Alice, we have

(17) \begin{equation}\mathbb{E}\big[V_B\big(\,f_A(M_A),f_B(M_B)\big)\big] \ge \mathbb{E}\big[V_B\big(\,f_A(M_A), \bar{\rho}_B\big)\big].\end{equation}

Now, since $ \mathbb{P}\big(\,f_B(M_B) \neq \bar{\rho}_B\big) > 0$ , by Lemma 7 we have

\[\mathbb{E}\big[V_B\big(\bar{\rho}_A,f_B(M_B)\big)\big] < V_B\big(\bar{\rho}_A, \bar{\rho}_B\big),\]

so that, by (5), we have

\[\mathbb{E}\big[V_A\big(\bar{\rho}_A,f_B(M_B)\big)\big] > V_A\big(\bar{\rho}_A, \bar{\rho}_B\big).\]

Combining this with (16), we get

\[\mathbb{E}\big[V_A\big(\,f_A(M_A),f_B(M_B)\big)\big] > V_A\big(\bar{\rho}_A,\bar{\rho}_B\big).\]

Similar reasoning, based on (17), gives

\[\mathbb{E}\big[V_B\big(\,f_A(M_A),f_B(M_B)\big)\big] > V_B\big(\bar{\rho}_A,\bar{\rho}_B\big).\]

However, putting these inequalities together contradicts the conservation law in (5). This completes the proof of the theorem.

4.3. Nash equilibrium structure for N-player games

Theorem 1 is actually a special case of a uniqueness theorem for Nash equilibria in the general N-player case. The proof in the N-player case also depends on a peculiar feature of Cox process Hotelling games, which is that a player faced with the strategies of the other players, i.e., their individual choices of likelihood functions with respect to the underlying measure $ \eta$ which result in their individual intensities, receives the same value as she would in a two-player game where she is faced with a single player playing a likelihood with respect to the underlying measure $ \eta$ that results in an intensity equal to the sum of the intensities corresponding to the strategies of the other players. With this in mind, we turn now to the N-player case.

Proof. We need to show that if $ (\,f_1(M_1), \ldots, f_N(M_N))$ is a Nash equilibrium, then $ \mathbb{P}\big(\,f_j\big(M_j\big) = \bar{\rho}_j\big) = 1$ for all $ 1 \le j \le N$ . To do this, suppose first, after reindexing if needed, that we have $ \mathbb{P}\big(\,f_1(M_1) \neq \bar{\rho}_1\big) > 0$ and

\[\mathbb{P} \!\left(\sum_{j=2}^N f_j\big(M_j\big) = \sum_{j=2}^N \bar{\rho}_j \right) = 1.\]

Then, because $ f_1(M_1)$ is a best reaction of player 1 to the individually randomized strategies $ \big(\,f_j\big(M_j\big), 2 \le j \le N\big)$ of the other players, we have

(18) \begin{equation}\mathbb{E}\big[V_1\big(\,f_1(M_1), f_2(M_2), \ldots, f_N(M_N)\big)\big] \ge\mathbb{E}\big[V_1\big(\bar{\rho}_1, f_2(M_2), \ldots, f_N(M_N)\big)\big].\end{equation}

On the other hand, by Lemma 7, we have

(19) \begin{equation}\mathbb{E}\big[V_1\big(\,f_1(M_1), f_2(M_2), \ldots, f_N(M_N)\big)\big] <\mathbb{E}\big[V_1\big(\bar{\rho}_1, f_2(M_2), \ldots, f_N(M_N)\big)\big].\end{equation}

To see this, observe that $ V_1(\,f_1(M_1), f_2(M_2), \ldots, f_N(M_N))$ is the same as the value of player 1 in the two-player game in which she plays the randomized strategy $ f_1(M_1)$ against a single opponent playing the strategy $ \sum_{j=2}^N f_j\big(M_j\big)$ , which we have assumed equals the constant $ \sum_{j=2}^N \bar{\rho}_j$ with probability 1, and also $ V_1(\bar{\rho}_1, f_2(M_2), \ldots, f_N(M_N))$ is the same as the value of player 1 in the two-player game in which she plays the constant strategy $ \bar{\rho}_1$ against a single opponent playing the strategy $ \sum_{j=2}^N f_j\big(M_j\big)$ , which we have assumed equals the constant $ \sum_{j=2}^N \bar{\rho}_j$ with probability 1, and so the scenario of Lemma 7 applies to allow us to compare these two values. Note that the inequalities (18) and (19) contradict each other. Thus, we can conclude that if $ (\,f_1(M_1), \ldots, f_N(M_N))$ is a Nash equilibrium, then for every player $ 1 \le i \le N$ for which $ \mathbb{P}\big(\,f_i(M_i) \neq \bar{\rho}_i\big) > 0$ , we must also have $ \mathbb{P} \!\left(\sum_{j \neq i} f_j\big(M_j\big) \neq \sum_{j \neq i} \bar{\rho}_j \right) > 0$ .

Suppose now, after reindexing if necessary, that $ \mathbb{P}\big(\,f_1(M_1) \neq \bar{\rho}_1\big) > 0$ . We have established that we must also have

\[\mathbb{P} \!\left(\sum_{j=2}^N f_j\big(M_j\big) \neq \sum_{j=2}^N \bar{\rho}_j \right) > 0.\]

Since $ f_1(M_1)$ is a best reaction of player 1 to the individually randomized strategies $ \big(\,f_j\big(M_j\big), 2 \le j \le N\big)$ of the other players, the inequality in (18) holds. Because $ \mathbb{P} \!\big(\sum_{j=2}^N f_j\big(M_j\big) \neq $ $ \sum_{j=2}^N \bar{\rho}_j \big) > 0$ , from Lemma 7 we also have

(20) \begin{equation}\sum_{j=2}^N \mathbb{E}\big[ V_j\big(\bar{\rho}_1, f_2(M_2), \ldots, f_N(M_N)\big)\big]< \sum_{j=2}^N V_j\big(\bar{\rho}_1, \bar{\rho}_2, \ldots, \bar{\rho}_N\big).\end{equation}

To see this, note that the sum of the values of the other players $ 2 \le j \le N$ , when player 1 plays the constant strategy $ \bar{\rho}_1$ , is the same as the value of a single player playing the strategy $ \sum_{j=2}^N f_j\big(M_j\big)$ against player 1 playing the constant strategy $ \bar{\rho}_1$ in a two-player game between player 1 and this single player, where the overall intensity constraint of player 1 continues to be $ \rho_1$ and that of this single player is $ \sum_{j=2}^N \rho_j$ . Since $ \mathbb{P} \!\left(\sum_{j=2}^N f_j\big(M_j\big) \neq \sum_{j=2}^N \bar{\rho}_j \right) > 0$ , Lemma 7 allows us to conclude that this value is strictly less than the value this single player would get by playing the constant strategy $ \sum_{j=2}^N \bar{\rho}_j$ against player 1, who is playing the constant strategy $ \bar{\rho}_1$ . But this is equal to the sum of the values of the individual players $ 2 \le j \le N$ in the given N-player game when they individually play the constant strategies $ \big(\bar{\rho}_j, 2 \le j \le N\big)$ respectively, and player 1 is playing the constant strategy $ \bar{\rho}_1$ .

Now, in view of the conservation law in Equation (12), we can conclude from the inequality (20) that

\[\mathbb{E}\big[V_1\big(\rho_1, f_2(M_2), \ldots, f_N(M_N)\big)\big]> V_1\big(\bar{\rho}_1, \bar{\rho}_2, \ldots, \bar{\rho}_N\big).\]

Thus, so far, what we have concluded is that if $ (\,f_1(M_1), \ldots, f_N(M_N))$ is a Nash equilibrium, then, for every $ 1 \le i \le N$ such that $ \mathbb{P}\big(\,f_i(M_i) \neq \bar{\rho}_i\big) > 0$ , we must have

(21) \begin{equation}\mathbb{E}\big[V_i\big(\bar{\rho}_i, \big(\,f_j\big(M_j\big), j \neq i\big)\big)\big]> V_i\big(\bar{\rho}_1, \bar{\rho}_2, \ldots, \bar{\rho}_N\big).\end{equation}

Finally, suppose that $ \mathbb{P}(\,f_i(M_i) = \bar{\rho}_i) = 1$ for some $ 1 \le i \le N$ . Then we must have

(22) \begin{equation}\mathbb{E}\big[V_i\big(\bar{\rho}_i, \big(\,f_j\big(M_j\big), j \neq i\big)\big)\big]\ge V_i\big(\bar{\rho}_1, \bar{\rho}_2, \ldots, \bar{\rho}_N\big).\end{equation}

To see this, note that the quantity $ \sum_{k \neq i} \mathbb{E}\big[V_k\big(\bar{\rho}_i, \big(\,f_j\big(M_j\big), j \neq i\big)\big)\big]$ is the same as the value of a single player who plays the strategy $ \sum_{j \neq i} f_j\big(M_j\big)$ in the two-player game against player i playing the constant strategy $ \bar{\rho}_i$ . By Lemma 7, this is no bigger that the value this single player would get if she played the constant strategy $ \sum_{j \neq i} \bar{\rho}_j$ , but this value is the same as $ \sum_{j \neq i} V_j\big(\bar{\rho}_1, \bar{\rho}_2, \ldots, \bar{\rho}_N\big)$ . To deduce (22) from this logic, we apply the conservation rule in Equation (12).

The inequalities in (21) and (22) together result in a contradiction of Equation (12) unless we have $ \mathbb{P}(\,f_i(M_i) = \bar{\rho}_i) = 1$ for all $ 1 \le i \le N$ . This concludes the proof of the theorem.

5.1. Structure of the Nash equilibria, assuming one exists

Leaving aside for the moment the question of existence of Nash equilibria, the strict concavity of the value function of a player for fixed choices of the pure actions of the other players, which was established in Lemmas 5 and 6, ensures that any Nash equilibrium that exists must be pure. We discuss this first for the two-player case.

Theorem 3. Consider a two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ between the players Alice and Bob. Suppose $ \big(\,f_A(M_A), f_B(M_B)\big) \in \mathcal{C}_A \times \mathcal{C}_B$ is a Nash equilibrium of the game, as defined in Definition 1. Then there exist $ g_A \in \mathcal{C}_A$ and $ g_B \in \mathcal{C}_B$ such that

\[\mathbb{P}((\,f_A(M_A), f_B(M_B)= (g_A, g_B)) = 1.\]

Proof. We write

\begin{align*}\mathbb{E}\big[V_A\big(\,f_A(M_A),f_B(M_B)\big)\big] & = \mathbb{E}\big[ \mathbb{E}\big[ V_A(\,f_A(M_A),f_B(M_B))|M_B\big]\big]\\&\stackrel{(a)}{\le} \mathbb{E}\big[ \mathbb{E}\big[V_A\big( \mathbb{E}\big[f_A(M_A)|M_B\big], f_B(M_B)\big)|M_B\big]\big]\\&\stackrel{(b)}{=} \mathbb{E}\big[ \mathbb{E}\big[ V_A\big(\mathbb{E}\big[f_A(M_A)\big], f_B(M_B)\big)|M_B\big]\big]\\& = \mathbb{E}\big[ V_A\big( \mathbb{E}\big[f_A(M_A)\big], f_B(M_B)\big].\end{align*}

Here, Step (a) comes from the concavity property of the value function established in Lemma 5, and Step (b) comes from independence of $ M_A$ and $ M_B$ . Since $ (\,f_A(M_A), f_B(M_B))$ is a Nash equilibrium pair, we see from Definition 1 that the inequality in the chain of equations above must be an equality. But then, by the strict concavity property of the value function established in Lemma 5, it follows that $ \mathbb{P}\big(\,f_A(M_A) = \mathbb{E}\big[f_A(M_A)\big]\big) = 1$ . A similar argument interchanging the roles of Alice and Bob completes the proof, with $ g_A$ being $ \mathbb{E}\big[f_A(M_A)\big]$ and $ g_B$ being $ \mathbb{E}\big[f_B(M_B)\big]$ in the notation of the statement of the lemma.

The analogue of Theorem 3 also holds in the N-player case.

Theorem 4. Consider an N-player Cox process Hotelling game $ (D, \eta, \rho_1, \ldots, \rho_N)$ . Suppose $ (\,f_1(M_1), \ldots, f_N(M_N)) \in \mathcal{C}_1 \times\ldots \times \mathcal{C}_N$ is a Nash equilibrium of the game, as defined in Definition 2. Then there exist $ g_i \in \mathcal{C}_i$ , $ 1 \le i \le N$ , such that

\[\mathbb{P}((\,f_1(M_1), \ldots, f_N(M_N) = (g_1, \ldots, g_N)) = 1.\]

In fact, the strict concavity property of the value function of a player, as established in Lemmas 5 and 6, together with the constant-sum nature of the game, ensures that if a Nash equilibrium exists, then it is not only pure, but also unique. We state this first in the two-player case.

Proof. We will first show that

(23) \begin{equation}V_A(g_A,g_B) = V_A\big(\,f_A,f_B\big).\end{equation}

The procedure to do this is standard in the theory of constant-sum games, but is reproduced here for convenience. To verify Equation (23), note that $ V_A(g_A, f_B) \le V_A(\,f_A, f_B)$ , because $ (\,f_A, f_B)$ is a Nash equilibrium; see (6). But then, by the constant-sum nature of the game (see (5)), we have $ V_B(g_A,f_B) \ge V_B\big(\,f_A,f_B\big)$ . However, since $ (g_A, g_B)$ is a Nash equilibrium, we have $ V_B(g_A,g_B) \ge V_B(g_A,f_B)$ , so we conclude that $ V_B(g_A,g_B) \ge V_B\big(\,f_A,f_B\big)$ . Interchanging the roles of $ (\,f_A, f_B)$ and $ (g_A, g_B)$ then gives $ V_B\big(\,f_A,f_B\big) \ge V_B(g_A,g_B)$ , which establishes (23).

We next show that

(24) \begin{equation}V_A(g_A, f_B) = V_A(\,f_A, f_B).\end{equation}

We have $ V_B(g_A,g_B) \ge V_B(g_A,f_B)$ because $ (g_A, g_B)$ is a Nash equilibrium. Hence, by the constant-sum nature of the game (see Equation (5)), we have $ V_A(g_A,g_B) \le V_A(g_A,f_B)$ . In view of Equation (23), this gives $ V_A\big(\,f_A,f_B\big) \le V_A(g_A,f_B)$ , but since $ (\,f_A, f_B)$ is a Nash equilibrium, this can only hold with equality; i.e. Equation (24) holds.

Now, since $ (\,f_A, f_B)$ is a Nash equilibrium, we know that $ f_A$ is a best response of Alice to the pure strategy $ f_B$ of Bob. Thus, Equation (24) tells us that $ g_A$ is also a best response of Alice in response to $ f_B$ . The strict concavity property of the value function of Alice, proved in Lemma 5, shows that this is possible only if $ g_A = f_A$ . Interchanging the roles of Alice and Bob, we conclude that we must also have $ g_B = f_B$ . This concludes the proof of the lemma.

Assume Nash equilibria exist. Beyond there being a unique pure Nash equilibrium, the special structure of the game allows us to say more about the form of this unique Nash equilibrium. We state the result first in the two-player case.

Proof. Suppose Alice reacts to $ f_B$ by playing $ \frac{\rho_A}{\rho_B} f_B$ . Let $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ . We have

\begin{align*}V_A\bigg(\frac{\rho_A}{\rho_B} f_B, f_B\bigg) &\stackrel{(a)}{=} \frac{\rho_A}{\rho_B} \int_{x\in D} f_B(x)\int_{y\in D} e^{-\int_{B(y\to x)} \frac{\rho}{\rho_B} f_B(u) \eta(du)} \eta(dy) \eta(dx)\\& = \frac{\rho_A}{\rho}\int_{x\in D} \frac{\rho}{\rho_B} f_B(x)\int_{y\in D} e^{-\int_{B(y\to x)} \frac{\rho}{\rho_B} f_B(u) \eta(du)} \eta(dy) \eta(dx)\\&\stackrel{(b)}{=} \frac{\rho_A}{\rho} \eta(D) \big(1 - e^{-\rho}\big),\end{align*}

where Step (a) is from Equation (8) and Step (b) is from Equation (14). Since $ (\,f_A, f_B)$ is a Nash equilibrium, it follows that

\[V_A\big(\,f_A,f_B\big) \ge \frac{\rho_A}{\rho} \eta(D) \big(1 - e^{-\rho}\big).\]

Interchanging the roles of Alice and Bob gives

\[V_B\big(\,f_A,f_B\big) \ge \frac{\rho_B}{\rho} \eta(D) \big(1 - e^{-\rho}\big).\]

In view of the constant-sum property in Equation (5), it then follows that we have equality in both these inequalities, i.e.

\[V_A\big(\,f_A,f_B\big) = \frac{\rho_A}{\rho} \eta(D) \big(1 - e^{-\rho}\big),\]

and

\[V_B\big(\,f_A,f_B\big) = \frac{\rho_B}{\rho} \eta(D) \big(1 - e^{-\rho}\big).\]

But then we have $ V_A\big(\,f_A,f_B\big) = V_A\big(\frac{\rho_A}{\rho_B} f_B, f_B\big)$ , so the strict concavity property of the value of Alice proved in Lemma 5, together with the assumption that $ (\,f_A, f_B)$ is a Nash equilibrium, implies that $ f_A = \frac{\rho_A}{\rho_B} f_B$ , which suffices to establish the claim. (One can also go through this argument again interchanging the roles of Alice and Bob, but it is not necessary to do so.)

The analogue of Theorem 5 and the stronger statement in Theorem 6 also hold in the N-player case. We state these claims together.

Proof. We prove the stronger statement. The proof is similar to that of Theorem 6. Suppose player i reacts to the strategy profile $ (\,f_j, j \neq i)$ by playing

\[\frac{\rho_i}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j.\]

Let $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \rho_i$ . We have

\begin{align*}& V_i\Bigg(\frac{\rho_i}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j,\big(\,f_j, j \neq i\big)\Bigg)\\& \stackrel{(a)}{=}\frac{\rho_i}{\sum_{j \neq i} \rho_j}\int_{x \in D} \sum_{j \neq i} f_j(x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)}\frac{\rho}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j(u) \eta(du)} \eta(dy) \eta(dx)\\& = \frac{\rho_i}{\rho}\int_{x \in D} \frac{\rho}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j(x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)}\frac{\rho}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j(u) \eta(du)} \eta(dy) \eta(dx)\\&\stackrel{(b)}{=} \frac{\rho_i}{\rho} \eta(D) \big(1 - e^{-\rho}\big),\end{align*}

where Step (a) is from Equation (10) and Step (b) is from Equation (14). Since $ (\,f_1, \ldots, f_N)$ is a Nash equilibrium, it follows that

\[V_i(\,f_1, \ldots, f_N) \ge \frac{\rho_i}{\rho} \eta(D) \big(1 - e^{-\rho}\big).\]

Since this holds for all $ 1 \le i \le N$ , it follows from Equation (11) that this inequality must hold with equality for all $ 1 \le i \le N$ , i.e. that we have

\[V_i(\,f_1, \ldots, f_N) = \frac{\rho_i}{\rho} \eta(D) \big(1 - e^{-\rho}\big)\]

for all $ 1 \le i \le N$ . But then, for each $ 1 \le i \le N$ we have

\[V_i\Bigg(\,f_i = \frac{\rho_i}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j,(\,f_j, j \neq i)\Bigg) = V_i(\,f_1, \ldots, f_N),\]

so from the strict concavity property of the value function of player i proved in Lemma 6 and the fact that $ f_i$ is a best response of player i to the strategy profile $ (\,f_j, j \neq i)$ of the other players, we must have

\[f_i = \frac{\rho_i}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j,\]

which is the same as $ f_i = \frac{\rho_i}{\rho} f$ , where $ f \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N f_i$ . This completes the proof of the theorem.

The properties of Nash equilibria of Cox process Hotelling games established so far, assuming a Nash equilibrium exists, can be gathered into the following statement.

Theorem 8. Consider an N-player Cox process Hotelling game $ (D, \eta, \rho_1, \ldots, \rho_N)$ . Let $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \rho_i$ . The game admits a Nash equilibrium if and only if there is a function $ f \in \mathcal{C}(\rho)$ such that

(25) \begin{align}& \int_{x \in D} f(x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx)\nonumber \\& \ge\int_{x \in D} g (x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx)\end{align}

for all $ g \in \mathcal{C}(\rho)$ . If such a function exists, the game has a unique Nash equilibrium, given by the pure strategy profile $ \big(\frac{\rho_1}{\rho} f, \ldots, \frac{\rho_N}{\rho} f\big)$ .

Proof. Suppose first that the game admits a Nash equilibrium. By Theorem 7, we know that this Nash equilibrium is unique and is a pure Nash equilibrium of the form $ \big(\frac{\rho_1}{\rho} f, \ldots, \frac{\rho_N}{\rho} f\big)$ , for some $ f \in \mathcal{C}(\rho)$ . Since $ \frac{\rho_i}{\rho} f$ is a best response of player i to the strategy profile $ \big(\,f_j = \frac{\rho_j}{\rho} f, j \neq i\big)$ of the other players, by the definition of Nash equilibrium (see (13)), we must have

\begin{align*}& \int_{x \in D} \frac{\rho_i}{\rho} f \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx)\\& \ge\int_{x \in D} g_i (x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx)\end{align*}

for all $ g_i \in \mathcal{C}(\rho_i)$ , which is the same as the condition in (25).

Conversely, suppose the condition in (25) holds for some function $ f \in \mathcal{C}(\rho)$ . Then, by the definition of Nash equilibrium in (13), we see that the strategy profile $ \big(\frac{\rho_1}{\rho} f, \ldots, \frac{\rho_N}{\rho} f\big)$ is a Nash equilibrium of the game. The existence of such a Nash equilibrium then guarantees, by Theorem 7, that it is the unique Nash equilibrium of the game.

5.2. Relationship with ordinal potential games

Recall that an n-player game, with player i having action set $ \mathcal{Y}_i$ , is called an ordinal potential game [Reference Monderer and Shapley23] if there is a function $ P\,:\, \prod_{i=1}^n \mathcal{Y}_i \to \mathbb{R}$ such that, for all $ 1 \le i \le n$ , $ y_i, z_i \in Y_i$ , and $ \big(y_j, j \neq i\big) \in \prod_{j \neq i} Y_j$ , we have $ V_i\big(y_i, \big(y_j, j \neq i\big)\big) > V_i\big(z_i, \big(y_j, j \neq i\big)\big)$ if and only if $$P({y_i},({y_j},j \ne i)) > P({z_i},({y_j},j \ne i)).$$

We have established that when a Cox process Hotelling game admits a Nash equilibrium it admits a unique pure Nash equilibrium. Since ordinal potential games are a well-known class of games that admit pure-strategy Nash equilibria, this leads naturally to the question of whether Cox process Hotelling games are ordinal potential games. However, it is possible to argue that in general this is not true.

To see this, consider the two-player Cox process Hotelling game between Alice and Bob on D, taken to be a circle of radius 1 centered at the origin in $ \mathbb{R}^2$ , the base measure $ \eta$ being the Lebesgue measure on D. Thus $ \eta(D) = 2 \pi$ . Suppose that $ \rho_A = \rho_B = \frac{\rho}{2}$ , where $ \rho$ should be thought of as being sufficiently large in a sense that we will make precise shortly. Let $ \epsilon > 0$ be sufficiently small $ \big($ to be precise, we require that $ \epsilon < \frac{2 \pi}{9}\big)$ .

We consider four pure strategies, i.e. elements of $ \mathcal{C}\big(\frac{\rho}{2}\big)$ , denoted by $ \sigma^R$ , $ \beta^R$ , $ \sigma^L$ , and $ \beta^L$ respectively, defined as follows:

• • The strategy $ \sigma^R$ is constant over the arc of the circle of length $ \epsilon$ centered at (1, 0) and is zero elsewhere.

• • Consider the arc of the circle of length $ \frac{\epsilon}{2}$ centered at $ \Big(\frac{\sqrt{3}}{2}, \frac{1}{2}\Big)$ and the arc of the circle of length $ \frac{\epsilon}{2}$ centered at $ \Big(\frac{\sqrt{3}}{2}, - \frac{1}{2}\Big)$ . The strategy $ \beta^R$ is constant over the union of these two arcs and is zero elsewhere.

• • The strategy $ \sigma^L$ is constant over the arc of the circle of length $ \epsilon$ centered at $ ({-}1,0)$ and is zero elsewhere. It can be considered to be the ‘left’ version of $ \sigma^R$ , which is the ‘right’ version of it.

• • The strategy $ \beta^L$ is the ‘left’ version of $ \beta^R$ , which is the ‘right’ version of it. Namely, $ \beta^L$ is uniform over the union of the two arcs of the circle of length $ \frac{\epsilon}{2}$ centered at $ \Big(\frac{- \sqrt{3}}{2},\frac{1}{2}\Big)$ and $ \Big(\frac{- \sqrt{3}}{2}, - \frac{1}{2}\Big)$ respectively, and is zero elsewhere.

To be consistent with the convention in the rest of the document, we will use a subscript to indicate the identity of the player playing the strategy. Thus, for instance, the strategy pair $ \big(\sigma^L_A, \sigma^R_B\big)$ indicates that Alice is playing the strategy $ \sigma^L$ and Bob is playing the strategy $ \sigma^R$ .

It is straightforward to check that

\[\lim_{\rho \to \infty} V_A\big(\sigma^R_A, \beta^R_B\big) = \frac{\pi}{6} + \frac{\epsilon}{4}\]

and

\[\lim_{\rho \to \infty} V_A\big(\sigma^L_A, \beta^R_B\big) = \frac{5 \pi}{6} + \frac{\epsilon}{4}.\]

We will use these facts and their obvious consequences in the following argument.

Suppose there were an ordinal potential function $ P\,:\, \mathcal{C}\big(\frac{\rho}{2}\big) \times\mathcal{C}\big(\frac{\rho}{2}\big) \to \mathbb{R}$ for this two-player Cox process Hotelling game.

For sufficiently large $ \rho$ , we have $ V_A\big(\sigma^R_A, \beta^R_B\big) < V_A\big(\sigma^L_A, \beta^R_B\big)$ , and so $ P\big(\sigma^R, \beta^R\big) < P\big(\sigma^L, \beta^R\big)$ .

For sufficiently large $ \rho$ we also have $ V_B\big(\sigma^L_A, \beta^R_B\big) < V_B\big(\sigma^L_A, \beta^L_B\big)$ , and so $ P\big(\sigma^L, \beta^R\big) < P\big(\sigma^L, \beta^L\big)$ .

But for sufficiently large $ \rho$ we also have $ V_A\big(\sigma^L_A, \beta^L_B\big) < V_A\big(\sigma^R_A, \beta^L_B\big)$ , and so $ P\big(\sigma^L, \beta^L\big) < P\big(\sigma^R, \beta^L\big)$ .

Finally, for sufficiently large $ \rho$ we also have $ V_B\big(\sigma^R_A, \beta^L_B\big) < V_B\big(\sigma^R_A, \beta^R_B\big)$ , and so $ P\big(\sigma^R, \beta^L\big) < P\big(\sigma^R, \beta^R\big)$ .

Putting these together leads to a contradiction. Hence this two-player Cox process Hotelling game is not an ordinal potential game.

5.3. Nash equilibria may not exist

Consider a two-player Cox process Hotelling game $ (D,\eta,\rho_A,\rho_B)$ . The results of Theorems 3, 5, and 6 are consistent with that of Theorem 1 in the case where D is compact and admits a transitive group of metric-preserving automorphisms, and where $ \eta$ is an invariant measure under the action of this group. This might lead one to expect that the constant intensity pair $ \big(\bar{\rho}_A, \bar{\rho}_B\big)$ is a Nash equilibrium for a general two-player Cox process Hotelling game $ (D, \eta, \rho_A, \rho_B)$ . The following simple example shows that this is not the case.

To see this, first note that $ \eta(D) = 1$ , so $ \bar{\rho}_A = \rho_A$ and $ \bar{\rho}_B = \rho_B$ . Recall that $ \rho \,{:}\,{\raise-1.5pt{=}}\, \rho_A + \rho_B$ . From Theorem 8, it suffices to find $ g \in \mathcal{C}(\rho)$ , i.e., $ g\,: \big[{-} \frac{1}{2}, \frac{1}{2}\big] \to \mathbb{R}_+$ with $ \int_{u = -\frac{1}{2}}^{\frac{1}{2}} g(u) du = \rho$ , such that

(26) \begin{equation}\int_{x = -\frac{1}{2}}^{\frac{1}{2}} g(x) \int_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \rho \eta(B(y \rightarrow x))} dy dx>\int_{x = -\frac{1}{2}}^{\frac{1}{2}}\rho \int_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \rho \eta(B(y \rightarrow x))} dy dx,\end{equation}

where we have written dx and dy in the integrals, instead of $ \eta(dx)$ and $ \eta(dy)$ , respectively, because $ \eta$ is the Lebesgue measure. By Equation (14), the integral on the right-hand side of (26) is $ 1 - e^{-\rho}$ . For the integral on the left-hand side of (26), let us first replace g(x) dx by $ \rho \delta_0(dx)$ , where $ \delta_0$ is the measure on $ \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ giving mass 1 to the point at the origin; i.e. let us consider the integral

\begin{align*} \rho \int_{x} & = -\frac{1}{2}^{\frac{1}{2}} \int_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \rho \eta(B(y \rightarrow x))} dy \delta_0(dx)\\& = \rho \int_{- \frac{1}{2}}^{\frac{1}{2}} e^{ - \rho \eta(B(y \rightarrow 0))} dy\\& = 2 \rho \int_0^{\frac{1}{4}} e^{-2 \rho y} dy+ 2 \rho \int_{\frac{1}{4}}^{\frac{1}{2}} e^{-\frac{\rho}{2}} dy\\& = 1 - e^{-\frac{\rho}{2}} +\frac{\rho}{2} e^{-\frac{\rho}{2}}.\end{align*}

For all $ \rho > 0$ this integral is strictly bigger than $ 1 - e^{ - \rho}$ . It follows that we can find $ g \in \mathcal{C}(\rho)$ to get the strict inequality in (26), as desired.

In the two-player game considered in Example 2, one can in fact conclude that there is no Nash equilibrium when the sum of the intensities of the two players is sufficiently small. We state this formally.

Proof. Theorem 8 tells us that, to prove that a Nash equilibrium does not exist for this two-player game, it suffices to show that for every $ f \in \mathcal{C}(\rho)$ there is some $ g \in \mathcal{C}(\rho)$ such that

(27) \begin{equation}\int\limits_{x = -\frac{1}{2}}^{\frac{1}{2}} g(x) \int\limits_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \int_{ u \in B(y \rightarrow x)} f(u) du} dy dx>\int\limits_{x = -\frac{1}{2}}^{\frac{1}{2}} f(x) \int\limits_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \int_{ u \in B(y \rightarrow x)} f(u) du} dy dx.\end{equation}

Let $ f \in \mathcal{C}(\rho)$ . For $ x \in \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ , define $ \psi_x \,:\, \big[{-} \frac{1}{2}, \frac{1}{2}\big] \to \mathbb{R}_+$ via

\[\psi_x(y) \,{:}\,{\raise-1.5pt{=}}\, e^{ - \int_{ u \in B(y \rightarrow x)} f(u) du}.\]

Suppose $ \int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_x(y) dy$ is not constant in x. Let $ x^* \in \mbox{arg max}_x \int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_x(y) dy$ , which exists because $ \int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_x(y) dy$ is continuous in x over $ \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ . Since

\[\int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_x(y) dy\]

is not constant in x over $ x \in \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ , and since $ \int_{x = - \frac{1}{2}}^{\frac{1}{2}} f(x) dx = \rho$ , this implies that

\[\rho \int_{y = - \frac{1}{2}}^{\frac{1}{2}} e^{ - \int_{ u \in B(y \rightarrow x^*)} f(u) du}dy> \int_{x = -\frac{1}{2}}^{\frac{1}{2}} f(x) \int_{y = - \frac{1}{2}}^{\frac{1}{2}}e^{ - \int_{ u \in B(y \rightarrow x)} f(u) du} dy dx,\]

from which we can conclude the existence of $ g \in \mathcal{C}(\rho)$ satisfying the strict inequality in (27).

From Theorem 8, we also know that if a Nash equilibrium exists it must be pure and of the form $ \big(\frac{\rho_A}{\rho} f, \frac{\rho_B}{\rho} f\big)$ for some $ f \in \mathcal{C}(\rho)$ . We claim that it must further be the case that $ f(u) = f({-}u)$ for all $ u \in \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ , i.e. that f is an even function. This is because, by symmetry, if $ \left(\frac{\rho_A}{\rho} f, \frac{\rho_B}{\rho} f \right)$ is a Nash equilibrium, then so is $ \left(\frac{\rho_A}{\rho} \tilde{f}, \frac{\rho_B}{\rho} \tilde{f}\right)$ , where $ \tilde{f}(u) \,{:}\,{\raise-1.5pt{=}}\, f({-}u)$ (so we also have $ \tilde{f} \in \mathcal{C}(\rho)$ ), and then, because the Nash equilibrium is unique, it must be the case that $ \tilde{f} = f$ .

Thus it suffices to show that when $ \rho < \log_e 4$ it is impossible to find an even function $ f \in \mathcal{C}(\rho)$ such that $ \int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_x(y) dy$ is constant in x. We will do this by establishing that for every even function $ f \in \mathcal{C}(\rho)$ we have

(28) \begin{equation}\int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_{-\frac{1}{2}}(y) dy<\int_{y = - \frac{1}{2}}^{\frac{1}{2}} \psi_0(y) dy.\end{equation}

Observe that $ \psi_0(y)$ is an even function of $ y \in \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ . Further, we have $ \psi_0(y) \ge e^{- \frac{\rho}{2}}$ for all $ y \in \big[{-} \frac{1}{2}, \frac{1}{2}\big]$ , and we have $ \psi_0(y) = e^{- \frac{\rho}{2}}$ for $ - \frac{1}{2} \le y \le - \frac{1}{4}$ .

Observe also that $ \psi_{-\frac{1}{2}}(y)$ is nonincreasing over $ y \in [{-} \frac{1}{2}, \frac{1}{2}]$ , with $ \psi_{-\frac{1}{2}}\big({-} \frac{1}{2}\big) = 1$ , $ \psi_{-\frac{1}{2}}\big({-} \frac{1}{4}\big) = e^{- \frac{\rho}{2}}$ , and $ \psi_{-\frac{1}{2}}(y) = e^{-\rho}$ for $ 0 \le y \le \frac{1}{2}$ .

From these two sets of observations, we have $ \psi_{-\frac{1}{2}}(y) \ge \psi_0(y)$ for $ y \in \big[{-} \frac{1}{2}, -\frac{1}{4}\big]$ and $ \psi_{-\frac{1}{2}}(y) \le \psi_0(y)$ for $ y \in \big[{-} \frac{1}{4}, \frac{1}{2}\big]$ . Further, we have

\[ \int_{y = - \frac{1}{2}}^{- \frac{1}{4}} \!\left( \psi_{-\frac{1}{2}}(y) - \psi_0(y) \right) dy \le \frac{1}{4} \Big(1 - e^{-\frac{\rho}{2}}\Big) \]

and

\[ \int_{y = 0}^{\frac{1}{2}} \!\left( \psi_0(y) -\psi_{-\frac{1}{2}}(y) \right) dy \ge \frac{1}{2} \Big( e^{-\frac{\rho}{2}} - e^{-\rho}\Big), \]

while we also have

\[ \int_{y = - \frac{1}{4}}^0 \!\left( \psi_0(y) -\psi_{-\frac{1}{2}}(y) \right) dy \ge 0. \]

From this we conclude that

\[ \int_{y = - \frac{1}{2}}^{\frac{1}{2}} \!\left( \psi_0(y) -\psi_{-\frac{1}{2}}(y) \right) dy \ge \frac{1}{2} \Big( e^{-\frac{\rho}{2}} - e^{-\rho}\Big) - \frac{1}{4} \Big(1 - e^{-\frac{\rho}{2}}\Big) > 0 \]

if $ 0 < \rho < \log_e 4$ , which establishes the strict inequality in (28) and completes the proof. A more careful analysis will increase the range of $ \rho$ for which one can prove that the game does not admit a Nash equilibrium.

5.4. Restricted Cox process Hotelling games

A more insightful characterization than the one in Theorem 8 of when Nash equilibria exist in Cox process Hotelling games remains an interesting open problem. The main technical difficulty in proving the existence of Nash equilibria in such games is that the action spaces of the individual players, which are of the form $ \mathcal{C}(\rho_i)$ , where $ \rho_i > 0$ is the intensity budget of player i, are not compact when endowed with the topology of weak convergence. Indeed, we have seen in Section 5.3 that Nash equilibria may not exist in some cases.

To better understand the question of when Nash equilibria exist in such games, we therefore propose to study a family of restricted Cox process Hotelling games, where the action space of each individual player is now a compact set. The restrictions can be imposed in such a way that a unique Nash equilibrium will be guaranteed to exist in each such restricted game, it will be pure, and, if the restrictions imposed on the individual players are proportional in a sense made precise below, this unique pure Nash equilibrium will be of proportional form. Furthermore, by varying the restriction, we can vary the compact action space of each player in such a way that the union over all such choices is the full space of allowed actions for that player, namely $ \mathcal{C}(\rho_i)$ for player i having a total intensity budget of $ \rho_i$ . If a Nash equilibrium did exist for the original Cox process Hotelling game, then this guarantees that it would be discovered as the Nash equilibrium for some profile of restrictions on the action spaces of the individual players, i.e., in one of the restricted Cox process Hotelling games that we consider. Pursuing this direction, which we leave as a topic for future research, may give more insight into what the characterization in Theorem 8 is actually saying.

To carry out this program, we first exhibit, for each $ \rho > 0$ , a family of compact subsets of $ \mathcal{C}(\rho)$ whose union is $ \mathcal{C}(\rho)$ . Recall that $ \mathcal{C}(\rho)$ is a subset of $ L^1(\eta)$ , where $ f \in \mathcal{C}(\rho)$ is identified with the measure $ f \eta$ on D, and $ \mathcal{C}(\rho)$ endowed with the topology of weak convergence of measures which it inherits as a subset of $ \mathcal{M}(D)$ . We will now consider $ L^1(\eta)$ with its weak topology defined by considering it to be a Banach space with Banach dual $ L^\infty(\eta)$ ; see [Reference Rudin29] or [Reference Diestel9, p. 44]. To avoid confusion, recall that the topology of weak convergence on $ \mathcal{C}(\rho)$ is called the narrow topology in the theory of Banach spaces (see e.g. [Reference Bogachev3, Vol. 1, p. 176]), and is weaker than the weak topology.

From the theorem of Dunford and Pettis ([Reference Diestel9, Theorem 3], [Reference Dunford and Pettis10]), the closure in the weak topology of a subset of $ L^1(\eta)$ is compact in the weak topology if and only if it is uniformly integrable. Here we recall (see e.g. [Reference Bogachev3, Vol. 1, Definition 4.5.1], [Reference Diestel9, p. 41]) that a subset $ S \subseteq L^1(\eta)$ is called uniformly integrable if

\[\lim_{c \to \infty} \sup_{f \in S}\int_D |f(x)| 1(|f(x)| \ge c) \eta(dx) = 0.\]

Furthermore, by the theorem of de la Vallée Poussin ([Reference Chafai5], [Reference Diestel9, Theorem 2]), for $ S \subseteq L^1(\eta)$ to be uniformly integrable it is necessary and sufficient that there be a nondecreasing convex function $ \Theta\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ , with $ \Theta(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta(x)}{x} = \infty$ , such that

\[\sup_{f \in S} \int_D \Theta(|f(x)|) \eta(dx) < \infty.\]

For $ \rho > 0$ , $ \Theta\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ a nondecreasing convex function with $ \Theta(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta(x)}{x} = \infty$ , and $ 0 < K < \infty$ , we propose to consider the subset of $ \mathcal{C}(\rho)$ defined by

(29) \begin{equation}\mathcal{C}(\rho,\Theta,K) \,:\!= \bigg\{f\,:\,D \to \mathbb{R}_+ \mbox{ s.t. } \int_D f(x) \eta(dx) = \rho, \int_D \Theta(\,f(x)) \eta(dx) \le K \bigg\}.\end{equation}

It can be checked that $ \mathcal{C}(\rho,\Theta,K)$ is a closed subset of $ L^1(\eta)$ in the weak topology. By the theorem of de la Vallée Poussin, $ \mathcal{C}(\rho,\Theta,K)$ is uniformly integrable, so by the Dunford–Pettis theorem it is a compact subset of $ L^1(\eta)$ in the weak topology. Since the weak topology on $ L^1(\eta)$ is stronger than the topology of weak convergence (i.e., the narrow topology) on $ L^1(\eta)$ , $ \mathcal{C}(\rho,\Theta,K)$ is compact in the topology of weak convergence on $ L^1(\eta)$ .

For every $ f \in L^1(\eta)$ it can be checked that there is some nondecreasing convex function $ \Theta\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ , with $ \Theta(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta(x)}{x} = \infty$ , such that $ \int_D \Theta(|f(x)|) \eta(dx) < \infty$ . It follows that the union of $ \mathcal{C}(\rho,\Theta,K)$ over all choices of $ \Theta$ and K equals $ \mathcal{C}(\rho)$ . We thus have a family of compact subsets of $ \mathcal{C}(\rho)$ whose union is $ \mathcal{C}(\rho)$ .

From the convexity of $ \Theta$ it is also straightforward to show that each $ \mathcal{C}(\rho,\Theta,K)$ is a convex subset of $ L^1(\eta)$ . As a closed subset of $ \mathcal{M}(D)$ in the topology of weak convergence, $ \mathcal{C}(\rho,\Theta,K)$ is a Borel subset of $ \mathcal{M}(D)$ , so we are able discuss probability measures on $ \mathcal{C}(\rho,\Theta,K)$ .

By an N-player restricted Cox process Hotelling game we mean a game with player i, for $ 1 \le i \le N$ , having the intensity budget $ \rho_i > 0$ and the space of pure actions some $ \mathcal{C}(\rho_i, \Theta_i, K_i)$ , which, as we have seen, is a compact subset of $ \mathcal{C}(\rho_i)$ in the topology of weak convergence. Here, for each $ 1 \le i \le N$ , $ \Theta_i\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ is a nondecreasing convex function with $ \Theta_i(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta_i(x)}{x} = \infty$ , and $ 0 < K_i < \infty$ . Suppose the individual players play the pure actions $ f_i \in \mathcal{C}(\rho_i, \Theta_i, K_i)$ for $ 1 \le i \le N$ . Let $ \Phi_i$ be a Poisson point process on D with intensity measure $ f_i \eta$ , with these processes being mutually independent for $ 1 \le i \le N$ , and let $ \Phi \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \Phi_i$ . We think of the points of $ \Phi_i$ as the points of player i, since these result from the choice of $ f_i$ , which was made by that player. Then, as before, the value of player i, denoted by $ V_i(\,f_1, \ldots, f_N)$ , is given by

\[V_i(\,f_1, \ldots, f_N) \,{:}\,{\raise-1.5pt{=}}\, \mathbb{E} \left[\sum_{x\in \Phi_i} \eta(W_\Phi(x)) \right].\]

The expectation is with respect to the joint law of $ (\Phi_j, 1 \le j \le n)$ , which are independent. Here, by definition, a sum over an empty set is 0.

Consider an N-player restricted Cox process Hotelling game where player i has the space of actions $ \mathcal{C}(\rho_i, \Theta_i, K_i)$ . Any mixed-strategy N-tuple in this game can be written as

\[(\,f_1(M_1), \ldots, f_N(M_N)),\]

where $ \big(M_i \in \mathcal{M}_j, 1 \le i \le N\big)$ are independent random variables representing the randomizations used by the individual players in implementing their randomized strategies, and $ f_i(m_i) \in \mathcal{C}\big(\rho_i, \Theta_i, K_i\big)$ is the choice of action of player i in case the realization of her random variable $ M_i$ is $ m_i$ . The value of player i in such a mixed strategy is $ \mathbb{E}\big[ V_i\big(\,f_1(M_1), \ldots, f_N(M_N)\big)\big]$ , the expectation being taken with respect to the joint distribution of $ \big(M_j, 1 \le j \le N\big)$ , which are independent. The vector of independently randomized strategies

\[\big(\,f_1(M_1), \ldots, f_N(M_N)\big) \in\mathcal{C}(\rho_1, \Theta_1, K_1) \times \ldots \times \mathcal{C}(\rho_N, \Theta_N, K_N)\]

is called a Nash equilibrium of the game if, for all $ g_j \in \mathcal{C}\big(\rho_j, \Theta_j, K_j\big)$ , $ 1 \le j \le N$ , we have, for all $ 1 \le i \le N$ ,

(30) \begin{equation}\mathbb{E}\big[V_i\big(\,f_1(M_1), \ldots, f_N(M_N)\big)\big] \ge\mathbb{E}\big[V_i\big(g_i, \big(\,f_j\big(M_j\big), j \neq i\big)\big)\big].\end{equation}

Since each $ \mathcal{C}\big(\rho_i, \Theta_i, K_i\big)$ is compact and each $ V_i\,:\, \prod_{j=1}^N \mathcal{C}\big(\rho_j, \Theta_j, K_j\big) \to \mathbb{R}_+$ is continuous, the existence of a mixed-strategy Nash equilibrium for every N-player restricted Cox process Hotelling game is guaranteed [Reference Glicksberg15].

We now formally state and prove that every restricted Cox process Hotelling game has unique Nash equilibrium, and that this consists of pure strategies. The following theorem is a combination of the analogue of Theorem 4 and the N-player version of Theorem 5 for restricted Cox process Hotelling games.

Theorem 10. Consider an N-player restricted Cox process Hotelling game on the Polish space D with base measure $ \eta$ where player i, for $ 1 \le i \le N$ , has the intensity budget $ \rho_i > 0$ and the space of pure actions $ \mathcal{C}(\rho_i, \Theta_i, K_i)$ , where $ \Theta_i\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ is a nondecreasing convex function with $ \Theta_i(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta_i(x)}{x} = \infty$ , and $ 0 < K_i < \infty$ . Suppose $ (\,f_1(M_1), \ldots, f_N(M_N))$ is a Nash equilibrium of the game, which we know exists. Then there exist $ g_i \in \mathcal{C}_i$ , $ 1 \le i \le N$ , such that

\[\mathbb{P}\big(\big(\,f_1(M_1), \ldots, f_N(M_N) = (g_1, \ldots, g_N)\big) = 1;\]

i.e., the Nash equilibrium is a pure-strategy Nash equilibrium. Furthermore, if $ (g_1, \ldots, g_N)$ and $ (h_1, \ldots, h_N)$ are two pure-strategy Nash equilibria for the game, then $ g_i = h_i$ for all $ 1 \le i \le N$ ; i.e. the Nash equilibrium is unique.

When the restrictions on the individual players in an N-player restricted Cox process Hotelling game are proportional to their allowed intensities, we can characterize the Nash equilibrium of the game, which we known by Theorem 10 is unique and consists of pure strategies, in a manner analogous to what was done in Theorem 7. To define what we mean by proportional restrictions, given $ \alpha > 0$ and a nondecreasing convex function $ \Theta\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ with $ \Theta(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta(x)}{x} = \infty$ , we define the function $ \Theta^{(\alpha)}\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ by

\[\Theta^{(\alpha)}(x) \,{:}\,{\raise-1.5pt{=}}\, \Theta\Big(\frac{x}{\alpha}\Big), x \in \mathbb{R}_+.\]

Note that $ \Theta^{(\alpha)}\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ is a nondecreasing convex function with $ \Theta^{(\alpha)}(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta^{(\alpha)}(x)}{x} = \infty$ . We can then make the following simple observation.

The following result characterizes the Nash equilibria of N-player restricted Cox process Hotelling games when the restrictions on the individual players are in proportion to their allowed intensities.

Proof. The proof is similar to that of Theorem 7, with the obvious modifications. The only thing that needs to be observed is that for any choice of $ f_i \in \mathcal{C}\big(\rho_i, \Theta^{(\rho_i)}, K\big)$ for $ 1 \le i \le N$ , we also have the following:

1. (i)

   \[\frac{\rho_i}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j \in \mathcal{C}\big(\rho_i, \Theta^{(\rho_i)}, K\big) \quad \text{for all } 1 \le i \le N;\]

2. (ii)

   \[\sum_{i=1}^N f_i \,=\!:\, f\in \mathcal{C}\big(\rho, \Theta^{(\rho)}, K\big);\]

3. (iii)

   \[\frac{\rho_i}{\rho} f\in \mathcal{C}\big(\rho_i, \Theta^{(\rho_i)}, K\big) \quad \text{for all } 1 \le i \le N.\]

To prove (i), by Lemma 8, what we need to show, for all $ 1 \le i \le N$ , is that

\[\frac{1}{\sum_{j \neq i} \rho_j} \sum_{j \neq i} f_j \in \mathcal{C}(1, \Theta, K).\]

By Lemma 8 again, we have $ \frac{f_j}{\rho_j} \in \mathcal{C}(1, \Theta, K)$ for all $ j \neq i$ . The desired claim follows from the convexity of $ \mathcal{C}(1, \Theta, K)$ .

To prove (ii), by Lemma 8, what we need to show is that $ \frac{f}{\rho}\in \mathcal{C}(1, \Theta, K)$ . By Lemma 8 we have $ \frac{f_i}{\rho_i} \in \mathcal{C}(1, \Theta, K)$ for all $ 1 \le i \le N$ . The desired claim follows from the convexity of $ \mathcal{C}(1, \Theta, K)$ .

As for (iii), it is an immediate consequence of Lemma 8 since we have established the claim in (ii).

Finally, with a view to using restricted Cox process Hotelling games as a vehicle for better understanding the characterization in Theorem 8, we can prove the following analogue of that result for restricted Cox process Hotelling games.

Theorem 12. Consider an N-player restricted Cox process Hotelling game on the Polish space D with base measure $ \eta$ where player i, for $ 1 \le i \le N$ , has the intensity budget $ \rho_i > 0$ and the space of pure actions $ \mathcal{C}\big(\rho_i, \Theta^{(\rho_i)}, K\big)$ , where $ \Theta\,:\, \mathbb{R}_+ \to \mathbb{R}_+$ is a nondecreasing convex function with $ \Theta(0) = 0$ and $ \lim_{x \to \infty} \frac{\Theta(x)}{x} = \infty$ , and $ 0 < K < \infty$ . The game admits a unique Nash equilibrium, consisting of pure strategies, defined in terms of a function $ f \in \mathcal{C}\big(\rho, \Theta^{(\rho)}, K\big)$ , where $ \rho \,{:}\,{\raise-1.5pt{=}}\, \sum_{i=1}^N \rho_i$ , such that

(31) \begin{align}& \int_{x \in D} f(x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx) \nonumber \\& \ge\int_{x \in D} g (x) \int_{y \in D}e^{ - \int_{u \in B(y \rightarrow x)} f(u) \eta(du)} \eta(dy) \eta(dx)\end{align}

for all $ g \in \mathcal{C}\big(\rho, \Theta^{(\rho)}, K\big)$ . The corresponding unique Nash equilibrium of the game is given by the pure strategy profile $ \big(\frac{\rho_1}{\rho} f, \ldots, \frac{\rho_N}{\rho} f\big)$ .