Proof. We adapt the idea of the proof in [Reference Penrose and Yukich38] to our setup. We only show the claim for the total edge length of $NG\!\left(\Xi_{\Gamma_{\alpha}}\right)$ since $NG^{\prime}\left(\Xi_{\Gamma_{\alpha}}\right)$ can be handled with the same idea. The score function in this case is

\begin{equation*}{\eta\!\left(x,\mathscr{X},\Gamma_\alpha\right)\,:\!=\,\frac{1}{2}\sum_{y\in \mathscr{X}_{\Gamma_\alpha}}\|y-x\|\textbf{1}_{\big\{(x,y)\in NG\big(\mathscr{X}_{\Gamma_\alpha}\big)\big\}},}\end{equation*}

which is clearly translation-invariant.

To apply Theorem 2.2, we need to check the moment condition (2.4), the non-singularity in Assumption 2.4, and the stabilisation condition in Definition 2.2. For simplicity, we show these conditions in two-dimensional cases; the argument can easily be extended to $\mathbb{R}^d$ with $d\ge 3$ (resp. $d=1$ ) by using cones (resp. intervals) instead of triangles in the proof.

We start with the exponential stabilisation, and fix $\alpha>0$ and $x\in \Gamma_\alpha$ . Referring to Figure 3, for each $t>0$ , we construct six disjoint sectors of the same size $T_j(t)$ , $1\le j\le 6$ , with x as the centre and angle $\frac{\pi}{3}$ . In consideration of edge effects near the boundary of $\Gamma_\alpha$ , the sectors are rotated around x so that all straight edges of the sectors have angles at least $\pi/12$ with respect to the edges of $\Gamma_\alpha$ . It is clear that $T_j(t)\subset T_j(u)$ for all $0<t<u$ . Set $T_{j}(\infty)=\cup_{t>0}T_j(t)$ for $1\le j\le 6$ ; then from the properties of the Poisson point process, there are infinitely many points in $\Xi\cap T_{j}(\infty)$ for all j, a.s. Let $|A|$ denote the cardinality of the set A; define

(3.1) \begin{equation}t_{x,\alpha}=\inf\big\{t\,:\, |T_j(t)\cap \Gamma_\alpha\cap \Xi|\ge k+1\mbox { or }T_j(t)\cap \Gamma_\alpha=T_j(\infty)\cap \Gamma_\alpha,\ 1\le j\le 6\big\} \end{equation}

and $\bar{R}\!\left(x,\alpha\right)=3t_{x,\alpha}$ . We show that $\bar{R}$ is a radius of stabilisation, and its tail distribution can be bounded by an exponentially decaying function independent of $\alpha$ and x.

Figure 3. k-nearest: stabilisation.

For the radius of stabilisation, there are two cases to consider. The first case is that none of $T_j(t_{x,\alpha})\cap \Gamma_\alpha\cap\Xi$ , $1\le j\le 6$ , contains at least $k+1$ points; then $B\big(x, \bar{R}\big(x,\alpha\big)\big)\supset \Gamma_\alpha$ , and (2.1) is obvious. The second case is that at least one of $T_j(t_{x,\alpha})\cap \Gamma_\alpha\cap\Xi$ , $1\le j\le 6$ , contains at least $k+1$ points, which means that the k nearest neighbours $\{x_1,\dots,x_k\}$ of x are in $B\!\left(x,t_{x,\alpha}\right)$ . If a point $y\in \Gamma_\alpha\backslash B(x,t_{x,\alpha})$ , then $y\in \Gamma_\alpha\cap (T_j(\infty)\backslash T_j(t_{x,\alpha}))$ for some j. Since $\Gamma_\alpha\cap (T_j(\infty)\backslash T_j(t_{x,\alpha}))$ is non-empty, $T_j(t_{x,\alpha})$ contains at least $k+1$ points $\{y_1,\dots, y_{k+1}\}$ and $d(x,y)>d(y_i,y)$ for all $i\le k+1$ ; hence y cannot have x as one of its k nearest neighbours. This ensures that all points having x as one of their k nearest neighbours are in $B(x, t_{x,\alpha})$ . Noting that the diameter of $B\!\left(x,t_{x,\alpha}\right)$ is $2t_{x,\alpha}$ and there are at least $k+1$ points in $B\!\left(x,t_{x,\alpha}\right)$ , we can see that whether a point y in $B\!\left(x,t_{x,\alpha}\right)$ has x as one of its k nearest neighbours is entirely determined by $\Xi\cap B(y,2t_{x,\alpha})\subset \Xi\cap B(x,3t_{x,\alpha})$ . This guarantees that $\eta\!\left(x,\mathscr{X},\Gamma_\alpha\right)$ is $\sigma\!\left(\Xi_{B(x,3t_{x,\alpha})}\right)$ -measurable, and $\bar{R}\!\left(x,\alpha\right)$ is a radius of stabilisation. For the tail distribution of $\bar{R}$ , referring to Figure 4, we consider the number of points of $\Xi$ falling into a triangle $A_t$ as a result of a sector being chopped off by the edge of $\Gamma_\alpha$ . This is the worst situation for capturing the number of points by one sector intersecting $\Gamma_{\alpha}$ in the sense of volume. A routine trigonometry calculation gives that the area of $A_t$ is at least $0.116t^2$ . Define $\tau\,:\!=\,\inf\{t\,:\,|\Xi\cap A_t|\ge k+1\}$ ; then

(3.2) \begin{equation}\mathbb{P}\!\left(\bar{R}\!\left(x,\alpha\right)>t\right)\le 6\mathbb{P}\!\left(\tau>t/3\right)\le 6e^{-0.116\lambda (t/3)^2}\sum_{i=0}^k\frac{\left(0.116\lambda (t/3)^2\right)^i}{i!},\ t>0,\end{equation}

which implies the exponential stabilisation.

Figure 4. k-nearest: $A_t$ .

The non-singularity in Assumption 2.4 can be proved through the corresponding unrestricted score function $\bar{\eta}(x,\mathscr{X}\,)=\frac{1}{2}\sum_{y\in \mathscr{X}}\|y-x\|\textbf{1}_{\{(x,y)\in NG(\mathscr{X}\,)\}}.$ Referring to Figure 5, we take $N_0\,:\!=\,B(0,{0.5})$ , observe that $\partial B(0,6)$ can be covered by finitely many B(x, 3) with $\|x\|=5$ , and write the centres of these balls as $x_1$ , $\dots$ , $x_n$ (in the two-dimensional case, $n=5$ ). Let E be the event that $|B(x_i,1)\cap \Xi|\ge k+1$ for all $1\le i\le n$ , $|\left(B(0,{1})\backslash B(0,{0.5})\right)\cap\Xi|=k$ , and $\Xi\cap \left(B(0,6)\backslash \left(\cup_{i\le n}B(x_i,1)\cup B(0,{1})\right)\right)$ is empty; then E is $\sigma\Big(\Xi_{N_0^c}\Big)$ -measurable and $\mathbb{P}(E)>0$ . Conditional on E, we can see that $E_1\,:\!=\,\{|\Xi\cap B(0,{0.5})|=1\}$ satisfies $\mathbb{P}(E_1|E)>0$ , and on $E_1$ , all of the summands in (2.6) that are random are those involving the point of $\Xi_{B(0,{0.5})}$ ; thus we now establish that these random score functions are entirely determined by $\Xi_{B(0,{1})}$ . As a matter of fact, any point in $\Xi\cap \left(\cup_{1\le i\le n}B(x_i,1)\right)$ has k nearest points with distances no larger than 2, so points in $\Xi\cap B(0,{1})$ cannot be among the k nearest points to points in $\Xi\cap B(x_i,1)$ . For any point $y\in \Xi\cap B(0,6)^c$ , the line between 0 and y intersects $\partial B(0,6)$ at y ^′, which is in $B(x_{i_0},3)$ for some $1\le i_0\le n$ , so the distances between y and points in $\Xi\cap B(x_{i_0},3)$ are at most $\|y-y^{\prime}\|+4=\|y\|-2$ . The distances between y and points in $\Xi\cap B(0,{1})$ are at least $\|y\|-1$ , which ensures that points of $\Xi\cap B(0,6)^c$ cannot have points in $\Xi\cap B(0,{1})$ as their k nearest neighbours. On the other hand, on $E\cap E_1$ , there are $k+1$ points in $\Xi\cap B(0,1)$ . For any point $x\in B(0,1)$ , since the distances between x and other points in B (0, 1) are less than 2, the points outside $B(x,2)\subset B(0,3)$ will not be among the k nearest points to x:

\begin{equation*}\bar{\eta}(x,\Xi)=\frac{1}{2}\sum_{y\in \left(\Xi\cap B(0,1)\right)\backslash\{x\}}\|x-y\|.\end{equation*}

Hence, given E, all random score functions contributing to the sum of (2.6) are those completely determined by $\Xi_{B(0,2)}$ , giving

where X is $\sigma\Big(\Xi_{N_0^c}\Big)$ -measurable. Since this is a continuous function of $y\in \Xi\cap B(0,{0.5})$ , the non-singularity in Assumption 2.4 follows.

Figure 5. k-nearest: non-singularity.

For the moment condition (2.4), recalling the definition of $t_{x,\alpha}$ in (3.1), we replace x with $x_3$ to get $t_{x_3,\alpha}$ . We now establish that

(3.3) \begin{equation} \eta\!\left(x_3,\Xi_{\Gamma_{\alpha}}+a_1\delta_{x_1}+a_2\delta_{x_2}+\delta_{x_3}\right)\le 3.5kt_{x_3,\alpha}.\end{equation}

In fact, the k nearest neighbours to $x_3$ have the contribution of the total edge length $\le \frac12 kt_{x_3,\alpha}$ . On the other hand, for $1\le j\le 6$ , each point in $\left(\Xi_{\Gamma_{\alpha}}+a_1\delta_{x_1}+a_2\delta_{x_2}\right)\cap T_{j}\!\left(t_{x_3,\alpha}\right)$ may take $x_3$ as its k nearest neighbour, with the contribution to the total edge length $\le \frac12 t_{x_3,\alpha}$ . As there are six sectors $\left(\Xi_{\Gamma_{\alpha}}+a_1\delta_{x_1}+a_2\delta_{x_2}\right)\cap T_{j}\!\left(t_{x_3,\alpha}\right)$ , $1\le j\le 6$ , and each sector has no more than k points, with $x_3$ as one of their k nearest neighbours, the contribution to the total edge length from this part is bounded by $3kt_{x_3,\alpha}$ . By adding these two bounds, we obtain (3.3). Finally, we combine (3.3) and (3.2) to get

\begin{align*} &\mathbb{E}\!\left\{\eta\!\left(x_3,\Xi_{\Gamma_{\alpha}}+a_1\delta_{x_1}+a_2\delta_{x_2}+\delta_{x_3}\right)^3\right\} \le 42.875k^3\mathbb{E}\!\left\{t_{x_3,\alpha}^3\right\}\le C <\infty,\end{align*}

and the proof is completed by applying Theorem 2.2.

Proof. The stabilising part of the proof is inspired by [Reference Penrose and Yukich38], and the idea can also be used to show the stabilisation property for Laguerre tessellations, which are a generalisation of Voronoi tessellations, as in [Reference Flimmel, Pawlas and Yukich23].

Before going into details, we observe that

\begin{align*} \mathscr{V}\left(\Xi_{\Gamma_{\alpha}},\alpha\right)=&\sum_{\{x,y\}\subset \Xi_{\Gamma_{\alpha}},x\neq y}l\!\left(\partial C\big(x, \Xi_{\Gamma_{\alpha}}\big)\cap \partial C\big(y, \Xi_{\Gamma_{\alpha}}\big)\right)\\ &+l(\partial \Gamma_\alpha),\end{align*}

where $l({\cdot})$ is the volume of a set in dimension $d-1$ . We restrict our attention to Voronoi tessellations of random point sets in $\mathbb{R}^2$ ; with greater notational complexity, the approach also works in $\mathbb{R}^d$ with $d\ge3$ , using cones instead of triangles. Because $l(\partial \Gamma_\alpha)=4\alpha^{\frac{1}{2}}$ is a constant, by removing this constant, we have $\mathscr{V}^{\,\,\prime}\!\left(\Xi_{\Gamma_{\alpha}},\alpha\right)\,:\!=\,\mathscr{V}\left(\Xi_{\Gamma_{\alpha}},\alpha\right)-4\alpha^{\frac{1}{2}}$ and $d_{TV}\!\left(\mathscr{V}\left(\Xi_{\Gamma_{\alpha}}, \alpha\right),{\bar{Z}}_\alpha\right)=d_{TV}\!\left(\mathscr{V}^{\,\,\prime}\!\left(\Xi_{\Gamma_{\alpha}},\alpha\right),{\bar{Z}}^{\prime}_\alpha\right)$ , where ${\bar{Z}}^{\prime}_\alpha$ is a normal random variable having the same mean and variance as $\mathscr{V}^{\,\,\prime}\!\left(\Xi_{\Gamma_{\alpha}},\alpha\right)$ . We can set the score function corresponding to $\mathscr{V}^{\,\,\prime}$ as

\begin{equation*}{\eta(x,\mathscr{X},\Gamma_\alpha)=\frac{1}{2}\sum_{y\in \mathscr{X},\,y\neq x}l\!\left(\partial C(x,\mathscr{X}\,)\cap \partial C(y, \mathscr{X}\,)\right)=\frac{1}{2}l\!\left(\partial (C(x, \mathscr{X}\,)\cap \Gamma_\alpha)\backslash (\partial \Gamma_\alpha)\right)}\end{equation*}

for all $x\in \mathscr{X}\subset \Gamma_\alpha$ , i.e., $\eta(x,\mathscr{X},\Gamma_\alpha)$ is half of the total length of edges of $C(x, \mathscr{X}\,)\cap \Gamma_\alpha$ , excluding the boundary of $\Gamma_\alpha$ . The score function $\eta$ is clearly translation-invariant. Thus, to apply Theorem 2.2, we need to verify the stabilisation in Definition 2.2, the moment condition (2.4), and the non-singularity in Assumption 2.4.

Figure 7. Voronoi: stabilisation.

We start by showing that the score function is exponentially stabilising. Referring to Figure 7, similarly to Section 3.1, we construct six disjoint equilateral triangles $T_{xj}(t)$ , $1\le j\le 6$ , such that x is a vertex of these triangles, and the triangles are rotated so that all edges with x as a vertex have angles at least $\pi/12$ relative to the edges of $\Gamma_\alpha$ . Let $T_{xj}(\infty)=\cup_{t\ge 0} T_{xj}(t)$ , $1\le j\le 6$ ; then $\cup_{1\le j\le 6}T_{xj}(\infty)=\mathbb{R}^2$ . Define

\begin{equation*}R_{xj}\,:\!=\,R_{xj}\!\left(x,\alpha,\Xi_{\Gamma_\alpha}\right)\,:\!=\,\inf\{t\,:\,T_{xj}(t)\cap \Xi_{\Gamma_\alpha}\ne \emptyset\mbox{ or }T_{xj}(t)\cap\Gamma_\alpha=T_{xj}(\infty)\cap\Gamma_\alpha\}\end{equation*}

and

\begin{equation*}R_{x0}\,:\!=\,R_{x0}\!\left(x,\alpha,\Xi_{\Gamma_\alpha}\right)\,:\!=\,\max_{1\le j\le 6}R_{xj}\!\left(x,\alpha,\Xi_{\Gamma_\alpha}\right).\end{equation*}

We note that there is a minor issue with the counterpart of $R_{x0}$ defined in [Reference McGivney and Yukich34] when x is close to the corners of $\Gamma_\alpha$ . We now show that $\bar{R}(x,\alpha)\,:\!=\,3R_{x0}\big(x,\alpha,\Xi_{\Gamma_\alpha}\big)$ is a radius of stabilisation. In fact, for any point x ^′ in $\Gamma_\alpha\backslash \overline{\left(\cup_{1\le j\le 6}T_{xj}(R_{x0})\right)}$ , x ^′ is contained in a triangle $T_{xj_0}(\infty)\backslash\overline{T_{xj_0}(R_{x0})}$ . This implies that $T_{xj_0}(R_{x0})\cap \Xi_{\Gamma_\alpha}\ne\emptyset$ , i.e., we can find a point $y\in T_{xj_0}(R_{x0})\cap \Xi_{\Gamma_\alpha}$ and the point y satisfies $d(x^{\prime},y)\le d(x,x^{\prime})$ ; hence $x^{\prime}\notin C(x,\Xi_{\Gamma_\alpha})\cap \Gamma_\alpha$ , which ensures that $C(x,\Xi_{\Gamma_\alpha})\subset \overline{\left(\cup_{1\le j\le 6}T_{xj}(R_0)\right)}$ . Consequently, if a point y in $\Xi_{\Gamma_\alpha}$ generates an edge of $C(X)\cap\Gamma_\alpha$ , then $d(x,y)\le 2R_0$ , and $\bar{R}(x,\alpha)$ satisfies Definition 2.2. As in Section 3.1, we use $A_t$ in Figure 4 again to define $\tau\,:\!=\,\inf\{t\,:\,|\Xi\cap A_t|\ge 1\}$ ; then

(3.4) \begin{equation} \mathbb{P}\!\left(\bar{R}\!\left(x\right)>t\right)\le 6\mathbb{P}\!\left(\tau>t/3\right)\le 6e^{-0.116\lambda (t/3)^2},\ t>0.\end{equation}

This completes the proof of the exponential stabilisation of $\eta$ .

The non-singularity in Assumption 2.4 can be examined by using a non-restricted counterpart $\bar{\eta}$ of $\eta$ , taking $N_0 =B(0,1)$ , and filling the moat $B(0,4)\backslash B(0,3)$ with sufficiently dense points of $\Xi$ such that, when $\Xi_{N_0^c}$ is fixed, the random score functions contributing to the sum of (2.6) are purely determined by a point in $\Xi\cap N_0$ . More precisely, we cover the circle $\partial B(0,3)$ with disjoint squares having side length $\sqrt{2}/4$ , and enumerate the squares as $S_i,\ 1\le i\le k$ . Note that all the squares are contained in $B(0,4)\backslash B(0,2)$ . Let $E=\cap_{1\le i\le k}\{|\Xi\cap(S_i)|\ge 1\}$ , $E_1=\{|\Xi\cap N_0|=1\}$ ; then E is $\sigma\Big(\Xi_{N_0^c}\Big)$ -measurable, $\mathbb{P}(E)>0$ , and $\mathbb{P}(E_1|E)>0$ . Since the points in $\Xi\cap \left(\cup_{i=1}^kS_i\right)$ have neighbours within distance 1, for any $x\in N_0$ , $T_{xj}(6)$ contains at least one point from $\Xi\cap \left(B(0,4)\backslash B(0,2)\right)$ . As argued in the stabilisation, points in $\Xi\cap \left(B(0,12)^c\right)$ do not affect the cell centred at $x\in N_0$ , and by symmetry, $x\in N_0$ does not affect the Voronoi cells centred at points in $\Xi\cap \left(B(0,12)^c\right)$ . This ensures that, conditional on E, all random score functions contributing to the sum of (2.6) are those completely determined by $\Xi_{N_0}$ , giving

where X is $\sigma\Big(\Xi_{N_0^c}\Big)$ -measurable. As $\textbf{1}_{E_1}\bar{\eta}\!\left(x,\Xi\right)$ is an almost surely (a.s.) (in terms of the volume measure in $\mathbb{R}^2$ ) continuous function of $x\in \Xi\cap N_0$ , the proof of the non-singularity in Assumption 2.4 is completed.

It remains to show the moment condition (2.4). In fact, as shown in the stabilisation property, we can see that $\bar{R}(x,\alpha)$ will not increase when points are added, so $C\big(x_3,\Xi_{\Gamma_\alpha}+a_1\delta_{x_1}+a_2\delta_{x_2}+\delta_{x_3}\big)\cap \Gamma_\alpha\subset B(x, \bar{R}(x,\alpha))$ ; then the number of edges of $C\big(x_3,\Xi_{\Gamma_\alpha}+a_1\delta_{x_1}+a_2\delta_{x_2}+\delta_{x_3}\big)\cap \Gamma_\alpha$ excluding those in the edge of $ \Gamma_\alpha$ is less than or equal to $(\Xi_{\Gamma_\alpha}+a_1\delta_{x_1}+a_2\delta_{x_2})\left(B\!\left(x,\bar{R}(x{,\alpha})\right)\right)\le\Xi_{\Gamma_\alpha}\left(B\!\left(x,\bar{R}(x{,\alpha})\right)\right)+2$ , and each of them has length less than $2\bar{R}(x,\alpha)$ . We observe that $\Xi$ restricted to outside of $\cup_{j=1}^6T_{xj}(R_{xj})$ is independent of $\bar{R}(x,\alpha)$ ; hence

where

stands for ‘stochastically less than or equal to’, and $\Xi^{\prime}$ is an independent copy of $\Xi$ . Therefore, using (3.4), we obtain

\begin{align*} &\mathbb{E}\!\left(\eta\!\left(x_3,\Xi_{\Gamma_\alpha}+a_1\delta_{x_1}+a_2\delta_{x_2}+\delta_{x_3}\right)^3\right)\\ \le&\mathbb{E}\!\left(\left(\Xi^{\prime}\left(B\!\left({x_3},\bar{R}(x{,\alpha})\right)\right)+8\right)^3\left(2\bar{R}({x_3}{,\alpha})\right)^3\right)\\ \le &\int_0^\infty \sum_{i=0}^\infty (i+8)^3(2r)^3\frac{e^{-\lambda \pi r^2} \big(\lambda \pi r^2\big)^i}{i!}6e^{-0.116\lambda (t/3)^2}*\left(0.116\lambda/9\right)2rdr\\ \le& C<\infty,\end{align*}

which ensures (2.4). The proof of Theorem 3.2 is completed by using Theorem 2.2.

Proof. Before going into details, we first construct a new marked Poisson point process $\Xi^{\prime}\,:\!=\,\sum_{x\in \bar{\Xi}}\delta_{(x, (M_x, {\epsilon}_x))}$ with i.i.d. marks $(M_x,{\epsilon}_x)\in T\times \mathbb{R}$ independent of the ground process $\bar{\Xi}^{\prime}=\bar{\Xi}$ , and incorporate ${\epsilon}_x$ into a new score function on $\Xi^{\prime}$ as

\begin{align*} \eta^{\prime}((x,{(m,{\epsilon})}),\Xi^{\prime},\Gamma_\alpha)\,:\!=\,&\eta^{\prime}((x,(m,{\epsilon})),\Xi^{\prime}_{\Gamma_\alpha},\Gamma_\alpha)\\ \,:\!=\,&\left[\left(\eta(\left(x,m\right)\!, \Xi_{\Gamma_\alpha},\Gamma_\alpha)+{\epsilon}\right)\vee 0\right]\textbf{1}_{(x,(m,{\epsilon}))\in \Xi^{\prime}_{\Gamma_\alpha}}.\end{align*}

We can see that

\begin{equation*}{\bar{W}}_\alpha=\sum_{x\in \bar{\Xi}^{\prime}_{\Gamma_\alpha}}\eta^{\prime}\Big((x,(m_x,{\epsilon}_x)),\Xi^{\prime}_{\Gamma_\alpha},\Gamma_\alpha\Big).\end{equation*}

The score function $\eta^{\prime}$ is clearly translation-invariant. Thus, to apply Theorem 2.2, it is sufficient to verify that $\eta^{\prime}$ is range-bounded as in Definition 2.2 and satisfies the moment condition (2.4) and the non-singularity in Assumption 2.4.

The range-bounded property of the score function $\eta^{\prime}$ is inherited from the range-bounded property of $\eta$ with the same radius of stabilisation $\bar{R}(x,\alpha)\,:\!=\,r$ . The moment condition (2.4) is a direct consequence of the boundedness of $\eta$ , the finite third moment of ${\epsilon}_x$ , and the Minkowski inequality. Hence it remains to show the non-singularity. To this end, we observe that the corresponding unrestricted counterpart $\bar{\eta}$ of $\eta^{\prime}$ is defined by

\begin{align*} \bar{\eta}\big((x,(m_x,{\epsilon}_x)),\mathscr{X}^{\,\prime}\big)&= \lim_{\alpha\rightarrow \infty}\eta^{\prime}\big((x,(m_x,{\epsilon}_x)),\mathscr{X}^{\,\prime},\Gamma_{\alpha}\big)\\ &=\eta^{\prime}\big((x,(m_x,{\epsilon}_x)),\mathscr{X}^{\,\prime},\Gamma_{\alpha_x}\big)= \left(\eta\big((x,m_x),\mathscr{X},\Gamma_{\alpha_x}\big)+{\epsilon}_x\right)\vee 0,\end{align*}

where $\alpha_x=4(\|x\|+r)^2$ , and $\mathscr{X}$ is the projection of $\mathscr{X}^{\,\prime}$ on $\mathbb{R}^2\times T$ . Let $N_0=B(0,1)$ , and let

\begin{equation*}E=\left\{\left|\Xi^{\prime}_{B(N_0,r)\backslash N_0}\right|=0\right\}, \qquad E_1=\left\{\left|\Xi^{\prime}_{N_0}\right|=1\right\};\end{equation*}

then E is $\sigma\!\left(\Xi^{\prime}_{N_0^c}\right)$ -measurable, $\mathbb{P}(E)>0$ , and $\mathbb{P}(E_1|E)>0$ . Writing ${\epsilon}_x^-=-\min\!({\epsilon}_x,0)$ , $\bar{\Xi}^{\prime}_{N_0}=\{x_0\}$ in $E_1$ , given E we have

\begin{align*} &\quad \textbf{1}_{E_1}\sum_{x\in\bar{\Xi^{\prime}}}\bar{\eta}\!\left((x,(m_x,{\epsilon}_x)),\Xi^{\prime}\right)\textbf{1}_{d(x, N_0)<r}\\ &=\textbf{1}_{E_1} \!\left(\eta\big(\big(x_0,m_{x_0}\big),\delta_{\big({x_0},m_{x_0}\big)},\Gamma_{4(r+1)^2}\big)+{{\epsilon}_{x_0}}\right)\vee0 \\ &= \textbf{1}_{E_1}\!\left[\textbf{1}_{{{\epsilon}_{x_0}}>0}\left(\eta\big(\big({x_0},m_{x_0}\big),\delta_{\big({x_0},m_{x_0}\big)},\Gamma_{4(r+1)^2}\big)+{\epsilon}_{x_0}^+\right)\right.\\ & \quad +\left.\textbf{1}_{{\epsilon}_{x_0}\le 0}\left(\eta\big(\big({x_0},m_{x_0}\big),\delta_{\big({x_0},m_{x_0}\big)},\Gamma_{4(r+1)^2}\big)-{\epsilon}_{x_0}^-\right)\right]\vee 0.\end{align*}

On $\{{\epsilon}_{x_0}>0\}$ , ${\epsilon}_{x_0}^+$ is independent of $\eta\Big(\big({x_0},m_{x_0}\big),\delta_{\big({x_0},m_{x_0}\big)},\Gamma_{4(r+1)^2}\Big)$ and has a positive non-singular component; hence $\eta\Big(\big({x_0},m_{x_0}\big),\delta_{\big({x_0},m_{x_0}\big)},\Gamma_{4(r+1)^2}\Big)+{\epsilon}_{x_0}^+$ is also non-singular. Together with the fact that $\{{\epsilon}_{x_0}>0\}$ and $\{{\epsilon}_{x_0}\le 0\}$ are disjoint, this implies the non-singularity.

Proof of Theorem 3.4. As the score function $d({\cdot},P)$ is not translation-invariant in the sense of Definition 2.4, we first turn the problem into that of a marked Poisson point process with independent marks. The idea is to project the points of $\Xi_{\Gamma_{\alpha,r}}$ on their first $d-1$ coordinates to obtain the ground Poisson point process, and send the last coordinate to marks with $T=[0,r]$ . To this end, define a mapping $h^{\prime}\,:\,\Gamma_{\infty,r}\,:\!=\,\cup_{\alpha>0}\Gamma_{\alpha,r}\rightarrow[0,\infty)^{d-1}\times [0,r]$ such that

\begin{equation*}h^{\prime}(x_1,\dots,x_d)=(x_1,\dots,x_d)+\left(0,\dots,0,\sum_{i=1}^{d-1}x_i\cot\!(\theta_i)\right)\end{equation*}

and $h(\mathscr{X}\,)\,:\!=\,\{h^{\prime}(x)\,:\,x\in \mathscr{X}\}$ . Then h ^′ is a one-to-one mapping, and $\Xi^{\prime}\,:\!=\,h\!\left(\Xi_{\Gamma_{\infty,r}}\right)$ can be regarded as a marked Poisson point process on $[0,\infty)^{d-1}\times [0,r]$ with rate $r\lambda$ and independent marks following the uniform distribution on [0, r]. Write the mark of $x\in \Xi^{\prime}$ as $m_x$ ; then

(3.5) \begin{equation} \bar{W}_{k,r,\alpha}=C(\theta_1,\dots,\theta_{d-1})\sum_{x\in h(\Xi_{k,r,\alpha})}(r-m_x),\end{equation}

where $C(\theta_1,\dots,\theta_{d-1})$ is a constant determined by $\theta_1,\dots,\theta_{d-1}$ . Let $\Gamma_\alpha^{\prime}\,:\!=\,\big[0,\alpha^{\frac{1}{d-1}}\big]^{d-1}$ ; then $h\big(\Xi_{\Gamma_{\alpha,r}}\big)=\Xi^{\prime}_{\Gamma^{\prime}_\alpha}$ . For a point $(x,m)\in [0,\infty)^{d-1}\times[0,r]$ , we write

\begin{equation*}A_{x,m,r,\alpha}^{\prime} =h^{\prime}\big(\big((h^{\prime})^{-1}(x,m)+[0,\infty)^d\big)\cap \Gamma_{\alpha,r}\big)\end{equation*}

(see Figure 8(b)) and

(3.6) \begin{equation}\ Xi^{\prime}_{k,r,\alpha}\,:\!=\,h(\Xi_{k,r,\alpha})=\sum_{x\in \bar{\Xi}^{\prime}_{\Gamma^{\prime}_\alpha}}\delta_{(x,m_x)}\textbf{1}_{A^{\prime}_{x,m_x,r,\alpha}\cap\left(\Xi^{\prime}_{\Gamma^{\prime}_\alpha}\backslash\cup_{i=1}^{k-1}\Xi^{\prime}_{i,r,\alpha}\right)=\{(x,m_x)\}}.\end{equation}

Combining (3.5) and (3.6), we can represent ${\bar{W}}_{k,r,\alpha}$ as the sum of values of the score function

\begin{equation*}\eta\big((x,m_x),\Xi^{\prime},\Gamma^{\prime}_\alpha\big)\,:\!=\,C(\theta_1,\dots,\theta_{d-1})(r-m_x)\textbf{1}_{(x,m_x)\in \Xi^{\prime}_{k,\alpha}}\end{equation*}

over the range $\Gamma^{\prime}_\alpha$ . To apply Theorem 2.2, we need to check that $\eta$ is range-bounded as in Definition 2.2, and that it satisfies the moment condition (2.4) and the non-singularity in Assumption 2.4.

For simplicity, we only show the claim in the two-dimensional case; the argument for $d>2$ is the same except for notational complexity. When $d=2$ , $\Gamma_{\alpha,r}$ is a parallelogram with angle $\theta_1$ as in Figure 8(a), and P and $C(\theta_1,\dots,\theta_{d-1})$ reduce to an edge in $\mathbb{R}^2$ and $\sin\!(\theta_1)$ , respectively. Since $\sin\!(\theta_1)(r-m_x)$ is given by the mark of x, to show $\eta$ is range-bounded, it is sufficient to show that $\textbf{1}_{\big\{(x,m_x)\in\Xi^{\prime}_{k,r,\alpha}\big\}}$ is completely determined by $\Xi^{\prime}\cap A^{\prime}_{x,m_x,r,\alpha}$ . In fact, we can accomplish this by observing that $(x,m_x)\in \Xi^{\prime}_{k,r,\alpha}$ if and only if there is a sequence $\{(x_j,m_{x_j}),\ 1\le j\le k\}\subset \Xi^{\prime}\cap A^{\prime}_{x,m_x,r,\alpha}$ $\big($ which ensures $A^{\prime}_{x_j,m_{x_j},r,\alpha}\subset A^{\prime}_{x,m_x,r,\alpha}\big)$ such that $(x_k,m_{x_k})=(x,m)$ and

\begin{equation*}A^{\prime}_{x_j,m_{x_j},r,\alpha}\cap\Big(\Xi^{\prime}\backslash \cup_{i=1}^{j-1}\Xi^{\prime}_{i,r,\alpha}\Big)=\big\{\big(x_j,m_{x_j}\big)\big\}\end{equation*}

for $1\le j\le k$ . Since

\begin{equation*}\Xi^{\prime}\cap A^{\prime}_{x,m_x,\alpha}\subset \Xi^{\prime}_{[x,x+r\tan\!(\theta_1)]},\end{equation*}

we can see that $\eta$ is range-bounded as in Definition 2.2 with $\bar{R}(x,\alpha)\,:\!=\,r\tan\!(\theta_1){+1}$ . The moment condition follows from the fact that $\eta$ is bounded above by r. For the non-singularity, we extend $\Xi^{\prime}$ to $\mathbb{R}^{d-1}\times[0,r]$ , write

\begin{equation*}\Gamma_{\infty,r}^e\,:\!=\,\{x\in\mathbb{R}^d\,:\, \mbox{there exists } y\in P\ \mbox{such that }y-r(0,\dots,0,1)\le x\le y\},\end{equation*}

and let $(\Xi_{\Gamma_{\infty,r}^e})_j$ be the jth maximal layer of $\Xi_{\Gamma_{\infty,r}^e}$ and $\Xi^{\prime}_j=h((\Xi_{\Gamma_{\infty,r}^e})_j)$ . We can see that the corresponding unrestricted score function is $\bar{\eta}(x,\Xi^{\prime})=\sin\!(\theta_1)(r-m_x)\textbf{1}_{(x,m_x)\in \Xi^{\prime}_k}$ with the stabilising radius $R(x)=r\tan\!(\theta_1){+1}$ . Referring to Figure 8(b), we set $N_0\,:\!=\,\left(0,\frac{r\tan\!(\theta_1)}{2}\right)$ , $B_0=\{(x,m);\,x\in N_0,0\le m\le x\cot\!(\theta_1)\}$ , $B_i$ as the triangle region with vertices

\begin{align*}&\left(r\tan\!(\theta_1)\left(\frac12+\frac{i-1}{4(k-1)}\right), r\!\left(\frac12+\frac{2i-1}{4(k-1)}\right)\right),\\&\left(r\tan\!(\theta_1)\left(\frac12+\frac{i}{4(k-1)}\right),r\!\left(\frac12+\frac{2i-1}{4(k-1)}\right)\right),\\&\left(r\tan\!(\theta_1)\left(\frac12+\frac{i}{4(k-1)}\right), r\!\left(\frac12+\frac{2i}{4(k-1)}\right)\right)\end{align*}

for $1\le i\le k-1$ , and

\begin{equation*}{D}=\left({\left(\left[-r\tan\!(\theta_1){-1},\frac{3r\tan\!(\theta_1)}{2}{+1}\right]\backslash N_0\right)}\times[0,r]\right)\backslash\left(\cup_{i=1}^{k-1}B_i\right).\end{equation*}

Define $E\,:\!=\,\left\{\Xi^{\prime}\cap D=\emptyset, \left|\Xi^{\prime}\cap B_i\right|=1,1\le i\le k-1\right\}$ and $E_0\,:\!=\,\big\{\big|\Xi^{\prime}_{N_0}\big|=\left|\Xi^{\prime}\cap B_0\right|=1\big\}$ . Then $E\in\sigma\!\left(\Xi^{\prime}_{N_0^c}\right)$ , $\mathbb{P}(E)>0$ , and $\mathbb{P}(E_0|E)>0$ . We can see that given E, the point in $\Xi^{\prime}\cap B_i$ is in $\Xi^{\prime}_{k-i}$ for all $1\le i\le k-1$ . Moreover, on $E\cap E_0$ , the point $({x_0},m_{x_0})$ in $\Xi^{\prime}_{N_0}$ is in $\Xi^{\prime}_k.$ Hence, given E,

\begin{equation*}\textbf{1}_{E_0}\sum_{x\in\bar{\Xi^{\prime}}}\bar{\eta}\!\left(x,\Xi^{\prime}\right)\textbf{1}_{d(x, N_0)<R(x)}=\textbf{1}_{E_0}\sin\!(\theta_1)(r-m_{x_0})\end{equation*}

is non-singular.