Proof. We choose the cones $S_i$ , $i \le I_2 $ , sufficiently thin and not axes-parallel so that for any $r>0 $ , the angle generated by starting from the origin, proceeding to any point in $S_i\cap B_r(0) $ , and ending at any point in $S_i\cap \partial B_r(0)$ exceeds $\gamma $ . Now, if $x \in \varphi$ is the closest point to 0 contained in $S_i$ , then $\angle 0xy > \gamma$ for every $y \in S_i$ with $|y| \ge |x|$ and $x\neq y$ . Thus, there cannot be an edge between the origin and y.

To construct $\theta$ , we first let $P_i$ denote the point closest to the origin in $\varphi \cap (S_i\setminus\{0\})$ , if the intersection is non-empty (resolving potential ties by choosing the lexicographic minimum). Then we put $\theta \;:\!=\; \{P_i \,:\, \varphi \cap S_i \ne\emptyset\}\cup\{0\}$ .

Proof of Lemma 2.

Part (i). In the setting of Lemma 2, assume that f intersects $\Delta_{M(e)}(e) $ and note that the nodes $f_1 $ and $f_2 $ have to be outside C(e) for e to exist. But since f intersects $\Delta_{M(e)}(e) $ , at least one of $e_1 $ and $e_2 $ is in $B_{|f|/2}(h_{|f|/2}(f)) $ . Therefore, f would not exist in the GG, and thus also not in the $\beta$ -skeleton. Hence, $\Delta_{M(e)}(e) $ cannot intersect f.

Part (ii). Repeating the above argument for the second choice of M(e) yields a rhombus with centroid $h_{|e|/2}(e) $ that cannot be intersected by other edges. However, since the side lengths of this rhombus are of order $|e|>a $ , there exists a constant $c_\textsf{disj} = c_\textsf{disj}(\beta)\in(0,1/2) $ such that any disk with center between $h_{a/2}(e) $ and $h_{|e|-a/2}(e)$ and radius $c_\textsf{disj}a $ also has distance of more than $c_\textsf{disj}a $ to the boundary of the rhombus (and similarly for e replaced by f). Since the rhombus linked to any edge cannot be intersected by another edge, it follows that the disk associated with e and the disk associated with f are disjoint.

Proof of Lemma 3. Since e is an edge in the $\beta$ -skeleton, the nodes $f_1 , f_2 $ lie outside the interior of C(e). We first consider the case where $f_1, f_2$ are contained in the boundary of C(e), and the segments $[M(f), f_1]$ , $[M(f), f_2]$ are tangent to C(e). Then $\Delta_{M(f)}(f)\cap C(e) $ yields a full circular segment of $B_{\beta|e|/2}(M(e))$ so that $y \in \Delta_{M(f)}(f)$ . We assert that if $[M(f), f_1]$ and $[M(f), f_2]$ are tangent to C(e), then $|f|=\tan(\tau)|e|$ . Since $y \in \Delta_{M(f)}(f)$ will remain true if we shorten $|f|$ , this will conclude the proof of the lemma.

To prove that $|f| = \tan(\tau)|e|$ , note that the tangency implies that $M(e)f_1M(f)$ is a right triangle. Thus,

\begin{align*} \frac{|f_1-M(f)|}{|f_1 - M(e)|}= \frac{|f_1-M(f)|}{\beta|e|/2} = \tan(\tau).\end{align*}

Next, $f_1M(f) h_{|f|/2}(f) $ also defines a right triangle so that ${|f|}/{(2|f_1-M(f)|)} = \cos(\tau) = \beta^{-1}.$ Finally, combining these two relations yields the asserted $|f| = \tan(\tau)|e|.$

Proof of Lemma 4. First, we note that $E_\textsf{REC}(\varphi,y, e)$ contains at most one edge that is shorter than $\tan(\tau)|e|$ . Indeed, suppose that $f \ne f^{\prime}$ are two such edges with $S_y(f) \subseteq S_y(f^{\prime})$ . Now, from Lemma 2(i), we know that f ^′ cannot intersect $\Delta_{M(f)}(f) $ and therefore also not $\Delta_y(e)\cap\Delta_y(f) $ .

This contradicts Lemma 3.

Hence, it suffices to bound the number of $f \in E_\textsf{REC}(\varphi,y, e)$ with $|f| \ge \tan(\tau)|e|$ . To achieve this goal, let $f^{(1)},\dots,f^{(c)}\in E_\textsf{REC}(\varphi,y,e)$ be disjoint edges, each of length at least $\tan(\tau)|e|$ . Note that none of these edges can intersect. Furthermore, for the edge e to exist, the edges $f^{(1)},\dots,f^{(c)} $ must also fully cross the disk segment C(e), as drawn in Figure 3.

Then, for all $i \le c$ , the edge $f^{(i)}$ crosses the cone $S_y(e)$ somewhere since $S_y(e)\subseteq S_y(f^{(i)})$ . In particular, $f^{(i)}$ has to cross the triangle $\Delta_y(e)$ . If that were not the case and $f^{(i)}$ were to cross $S_y(e)\setminus \Delta_y(e)$ , then $e_1,e_2\in \Delta_y(f^{(i)})\subseteq C(f)$ , which would contradict the existence of the edge $f^{(i)}$ . It is impossible for $f^{(i)}$ to cross both $S_y(e)\setminus \Delta_y(e)$ and $\Delta_y(e)$ , because then it would have to intersect e.

Then, by Lemma 2, each $f^{(i)}$ generates a disk with radius $c_\textsf{disj}\tan(\tau)|e| $ whose center has to be within distance $\tan(\tau)|e| $ of C(e), disjoint from the disks created by other edges longer than $\tan(\tau)|e| $ . Thus, the total number of long edges that can cross $\Delta_y(e) $ is bounded by

\begin{equation*} \frac{2|B_{2\tan(\tau)|e|+\beta|e|/2}(M(e))|}{\pi(c_\textsf{disj}\tan(\tau)|e|)^2}=\frac{2|B_{2\tan(\tau)+\beta/2}(0)|}{\pi c_\textsf{disj}^2\tan(\tau)^2} =: c_\textsf{edges}(\beta)-1, \end{equation*}

thereby concluding the proof.

Proof of the upper bound of Theorem 1. Let $a\in (0,(1-d/\alpha)/2)$ and $\varepsilon\in(0,ad/\alpha)$ . Then

(24) \begin{align}&\mathbb P\Bigg(\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x) > \mu_\alpha n^d + rn^d\Bigg) \nonumber \\[5pt] &\le \mathbb P\Bigg(\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x)\mathbb{1}\{\xi^{(\alpha)}(X-x) < n^a\} - \mu_\alpha n^d > \varepsilon rn^d\Bigg) \nonumber \\[5pt] &\quad + \mathbb P\Bigg(\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x)\mathbb{1}\{\xi^{(\alpha)}(X-x) \ge n^a\} \ge (1-\varepsilon)rn^d\Bigg) \nonumber \\[5pt] &\le \mathbb P\Bigg(\frac1{n^d}\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x)\mathbb{1}\{\xi^{(\alpha)}(X-x) < n^a\} > \mu_\alpha + \varepsilon r,\mathcal{R}_n(X) \le n\Bigg) \nonumber \\[5pt] & \quad + \mathbb P\Big(J_n^{(a)}(X) > n^{d^2/\alpha-\varepsilon}\Big) \nonumber \\[5pt] &\quad + 2\mathbb P(\mathcal{R}_{3n}(X) > n) + \mathbb P\!\left(\sum_{x\in\mathcal{J}_n^{(a)} (X)}\! \xi^{(\alpha)}(X-x) \ge (1-\varepsilon)rn^d, J_n^{(a)}(X)\le n^{d^2/\alpha-\varepsilon}, \mathcal{R}_{3n}(X) \le n\!\right)\!. \end{align}

From Lemmas 7 and 8 we know that with our choices of a and $\varepsilon$ , the first two summands after the last inequality of (24) do not play a role in large-volume asymptotics. Moreover, with the help of Markov’s inequality and Mecke’s formula [Reference Last and Penrose13, Theorem 4.4], we get that

(25) \begin{align}\begin{split}\mathbb P(\mathcal{R}_{3n}(X) > n) &= \mathbb P(\#\{x\in X\cap Q_{3n}\,:\, \mathcal{R}(X-x) > n\} \ge 1) \le \mathbb E[\#\{x\in X\cap Q_{3n}\,:\, \mathcal{R}(X-x) > n\}] \\[5pt] &= \mathbb E\Bigg[\sum_{x\in X\cap Q_{3n}} \mathbb{1}\{\mathcal{R}(X-x) > n\}\Bigg] = \int_{ Q_{3n}} \mathbb P(\mathcal{R}((X\cup\{x\})-x) \ge n)dx \\[5pt] &\le \int_{ Q_{3n}} \sum_{i\le I_d} \mathbb P(\mathcal{S}_i((X\cup\{x\})-x) \ge n)dx.\end{split}\end{align}

From here, owing to the characteristics of ( STA ), it is implied that for each $i\le I_d$ and $r>0$ it holds that

\begin{align*}|S_i\cap B_r(0)| \ge r^d \underbrace{\min_{j\le I_d} |S_j\cap B_1(0)|}_{ =: c_\textsf{cones}},\end{align*}

and by applying a Poisson concentration bound [Reference Penrose14, Lemma 1.2] for a large enough n, we can continue our computations for each $i\le I_d$ and $x\in Q_{3n}$ with

\begin{align*}\mathbb P(\mathcal{S}_i((X\cup\{x\})-x) \ge n) &\le \mathbb P\big(X(S_i\cap B_{n/c_\textsf{STA}}(0)) \le c_\textsf{STA}\big) \\[5pt] &\le \exp\!\bigg({-}c_\textsf{cones} \bigg(\frac{n}{c_\textsf{STA}}\bigg)^d + c_\textsf{STA} - c_\textsf{STA}\log\!\bigg(\frac{c_\textsf{STA}^{d+1}}{c_\textsf{cones} n^d}\bigg)\bigg),\end{align*}

where we recall that if X is interpreted as a Poisson random measure, we can denote the random number of points in a Borel set by $X(\!\cdot\!)$ . Therefore, continuing from (25), we arrive at

(26) \begin{equation}\frac1{n^{d^2/\alpha}}\log \mathbb P(\mathcal{R}_{3n}(X) > n) \le - \frac{c_\textsf{cones}}{c_\textsf{STA}^d} n^{d(1-d/\alpha)} + \frac{c_\textsf{STA}}{n^{d^2/\alpha}} - \frac1{n^{d^2/\alpha}}\log\!\bigg(\frac{c_\textsf{STA}^{d+1}}{c_\textsf{cones} n^d}\bigg) \overset{n\uparrow\infty}{\longrightarrow} -\infty.\end{equation}

Thus, it remains to consider the fourth summand after the last inequality of (24). Here, Lemma 9 yields

\begin{align*} &\limsup_{n \uparrow \infty}\frac1{n^{d^2/\alpha}} \log\mathbb P\!\left(\sum_{x\in\mathcal{J}_n^{(a)} (X)} \xi^{(\alpha)}(X-x) \ge (1-\varepsilon)rn^d, J_n^{(a)}(X)\le n^{d^2/\alpha-\varepsilon}, \mathcal{R}_{3n}(X) \le n\right) \\[5pt] &\le -((1-\varepsilon)r)^{d/\alpha} \inf_{(\varphi,\psi)\in B} |A(\varphi,\psi)|.\end{align*}

In brief, we arrive at

\begin{align*}\limsup_{n \uparrow \infty}\frac1{n^{d^2/\alpha}}\log\mathbb P(H_n > \mu_\alpha + r) \le -((1-\varepsilon)r)^{d/\alpha} \inf_{(\varphi,\psi)\in B} |A(\varphi,\psi)|.\end{align*}

Letting $\varepsilon\downarrow 0$ concludes the proof of the upper bound.

Proof of Lemma 7. We start by introducing some of the notation from [Reference Bachmann and Peccati3]. For the Poisson process X, we define the functional

(27) \begin{equation}F_n^{(\alpha)}(X) \;:\!=\; \sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X\cap Q_{3n}-x)\mathbb{1}\{\xi^{(\alpha)}(X\cap Q_{3n}-x) < n^a\}.\end{equation}

Before we can apply the concentration bound, we need to find a link between the typical value $n^d \mu_\alpha$ and the expectation of the functional defined in (27). We can find the connection using the fact that, because of ( STA ), under the event $\{\mathcal{R}_n(X) \leq n\}$ , this functional is equal to the one considered in Lemma 7:

\begin{align*}n^d\mu_\alpha &\ge \mathbb E\Bigg[\sum_{x\in X\cap Q_n}\xi^{(\alpha)}(X-x)\mathbb{1}_{\{\xi^{(\alpha)}(X-x) < n^a\}}\Bigg]\\ & \ge \mathbb E\Bigg[\sum_{x\in X\cap Q_n}\xi^{(\alpha)}(X-x)\mathbb{1}_{\{\xi^{(\alpha)}(X-x) < n^a\}}\mathbb{1}_{\{\mathcal{R}_n(X) \le n\}}\Bigg] \\[5pt] &= \mathbb E\Bigg[\sum_{x\in X\cap Q_n}\xi^{(\alpha)}(X\cap Q_{3n}-x)\mathbb{1}{\{\xi^{(\alpha)}(X\cap Q_{3n}-x) < n^a, \, \mathcal{R}_n(X) \le n\}}\Bigg] \\[5pt] &= \mathbb E\Bigg[\sum_{x\in X\cap Q_n}\xi^{(\alpha)}(X\cap Q_{3n}-x)\mathbb{1}{\{\xi^{(\alpha)}(X\cap Q_{3n}-x) < n^a\}}(1-\mathbb{1}{\{\mathcal{R}_n(X) > n\}})\Bigg] \\[5pt] &= \mathbb E[F_n^{(\alpha)}(X)] - \mathbb E\Bigg[\sum_{x\in X\cap Q_n}\xi^{(\alpha)}(X\cap Q_{3n}-x)\mathbb{1}{\{\xi^{(\alpha)}(X\cap Q_{3n}-x) < n^a,\,\mathcal{R}_n(X) > n\}}\Bigg] \\[5pt] &\ge \mathbb E[F_n^{(\alpha)}(X)] - \mathbb E[X(Q_n)n^a\mathbb{1}{\{\mathcal{R}_n(X) > n\}}].\end{align*}

Subsequently, the Cauchy–Schwarz inequality yields

\begin{align*}& n^d\mu_\alpha \ge \mathbb E[F_n^{(\alpha)}(X)] - \sqrt{\mathbb E[X(Q_n)^2 n^{2a}] \mathbb E[\mathbb{1}_{\{\mathcal{R}_n(X) > n\}}]} \\[5pt] & \quad = \mathbb E[F_n^{(\alpha)}(X)] - \sqrt{(n^{2d}+n^d)n^{2a} \mathbb P(\mathcal{R}_n(X)>n)}.\end{align*}

As argued in (26), the factor $\mathbb P(\mathcal{R}_n(X)>n)$ decays exponentially with n, and consequently we can assume that n is large enough to guarantee that $n^d\mu_\alpha \ge \mathbb E[F_n^{(\alpha)}(X)] - \varepsilon$ . Therefore,

(28) \begin{align}\begin{split}&\mathbb P\Bigg(\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x)\mathbb{1}_{\{\xi^{(\alpha)}(X-x) < n^a\}} > n^d\mu_\alpha + n^d\varepsilon r,\mathcal{R}_n(X) \leq n\Bigg)\\[5pt] & \le \mathbb P\big(F_n^{(\alpha)}(X) > n^d\mu_\alpha +n^d\varepsilon r\big) \\[5pt] &\le \mathbb P(F_n^{(\alpha)}(X) > \mathbb E[F_n^{(\alpha)}(X)] + n^d\varepsilon r - \varepsilon).\end{split}\end{align}

Furthermore, we need the difference operator $D_y$ , $y \in \mathbb R^d$ , defined by $D_yF_n^{(\alpha)}(X) \;:\!=\; F_n^{(\alpha)}(X\cup\{y\}) - F_n^{(\alpha)}(X).$ For $\beta\ge0$ , we now set

(29) \begin{equation}V_\beta^+(F_n^{(\alpha)}(X)) \;:\!=\; \int_{\mathbb R^d} \big(D_yF_n^{(\alpha)}(X)\mathbb{1}_{\{D_yF_n^{(\alpha)}(X) \le \beta\}}\big)^2 dy + \sum_{x\in X} \Big(D_xF_n^{(\alpha)}(X\setminus\{x\})\mathbb{1}_{\{D_xF_n^{(\alpha)}(X\setminus\{x\}) > \beta\}}\Big)^2.\end{equation}

To apply [Reference Bachmann and Peccati3, Corollary 3.3(i)], we need to find an almost sure upper bound for $V_\beta^+(F_n^{(\alpha)})$ . Points outside of $Q_{3n}$ do not affect the functional. Thus, choosing $y\in\mathbb R^d\setminus Q_{3n}$ in the difference operator has no effect and yields $D_yF_n^{(\alpha)}(X) = 0$ . Moreover, by ( FIN ), adding a point to any configuration can only affect the outgoing edges of $c_\textsf{max}$ nodes, and the degree of each node is bounded by $c_\textsf{max}$ as well. Hence, $\sup_{y \in Q_{3n}}|D_yF_n^{(\alpha)}(X)| \le(c_\textsf{max}+1)^2 n^a =: \beta$ , and by the same reasoning, $\sup_{x \in X}|D_xF_n^{(\alpha)}(X\setminus\{x\})| \le \beta$ . Thus, we bound (29) by

\begin{align*}V_\beta^+(F_n^{(\alpha)}(X)) \le \int_{Q_{3n}} (c_\textsf{max}+1)^4 n^{2a} dy = ((c_\textsf{max}+1)^2 n^a)^2(3n)^d.\end{align*}

Then, by applying [Reference Bachmann and Peccati3, Corollary 3.3(i)], we have

\begin{align*}& \mathbb P(F_n^{(\alpha)}(X) > \mathbb E [F_n^{(\alpha)}(X)] + n^d\varepsilon r - \varepsilon)\\[5pt] \quad & \le \exp\!\Bigg({-}\frac{n^d\varepsilon r - \varepsilon}{2|\beta|}\log\!\Bigg(1+\frac{|\beta|(n^d\varepsilon r - \varepsilon)}{((c_\textsf{max}+1)^2 n^a)^2(3n)^d}\Big)\Bigg) \\[5pt] & \quad = \exp\!\Bigg({-}\frac{n^d\varepsilon r - \varepsilon}{2(c_\textsf{max}+1)^2 n^a}\log\!\Bigg(1+\frac{(c_\textsf{max}+1)^2 n^a (n^d\varepsilon r - \varepsilon)}{(c_\textsf{max}+1)^4 n^{2a}(3n)^d}\Bigg)\Bigg) \\[5pt] & \quad = \exp\!\Bigg({-}\frac{n^{d-a}(\varepsilon r - \varepsilon/n^d)}{2(c_\textsf{max}+1)^2}\log\!\Bigg(1+\frac{n^{-a}(\varepsilon r - \varepsilon/n^d)}{(c_\textsf{max}+1)^2 3^d}\Bigg)\Bigg)\end{align*}

if $n^d\varepsilon r - \varepsilon \ge 0$ . Finally, with the help of (28), we obtain

\begin{align*}\limsup_{n\uparrow\infty} \frac1{n^{d^2/\alpha}} \log \mathbb P\Bigg(\sum_{x\in X\cap Q_n} \xi^{(\alpha)}(X-x)\mathbb{1}_{\{\xi^{(\alpha)}(X-x) < n^a\}} > n^d\mu_\alpha +n^d\varepsilon r, \mathcal{R}_n(X) \le n\Bigg) = - \infty\end{align*}

for $a\in(0,(1-d/\alpha)/2)$ .

Proof of Lemma 8. Let $a\in(0,1)$ . We divide $Q_n$ into a grid consisting of $\lfloor n^{1-a/\alpha}\rfloor^d$ smaller boxes with side length $l_n \;:\!=\; n/\lfloor n^{1-a/\alpha}\rfloor$ . The set of all of these cubes is

\begin{align*}\mathcal{Q} \;:\!=\; \big\{Q \,:\, Q = l_n z + [{-}n/2,-n/2+l_n]^d, z\in\{0,\dots,\lfloor n^{1-a/\alpha}\rfloor - 1\}^d\big\}.\end{align*}

Furthermore, we label each box in such a way that between two boxes of the same label there are always two boxes with a different label. For instance, we can label the boxes according to elements of the set $\mathcal{L} = \{0,1,2\}^d$ , thus using $\#\mathcal{L} = 3^d$ different labels; see Figure 4. For $m\in \mathcal{L}$ , we denote the set of label-m cubes by

\begin{align*}\mathcal{Q}^{(m)} & \;:\!=\; \big\{Q \,:\, Q = l_n z + [{-}n/2,-n/2+l_n]^d,\\[5pt] z & =(z_1,\dots,z_d)\in\{0,\dots,\lfloor n^{1-a/\alpha}\rfloor - 1\}^d \text{ with } z_i\bmod 3 = m_i\big\},\end{align*}

so that $\#\mathcal{Q}^{(m)} = \lfloor n^{1-a/\alpha}\rfloor^d/3^d + o( n^{1-a/\alpha}) =: K_n.$

Figure 4.

Labeling of the boxes in three dimensions, where 27 labels are sufficient, and two dimensions, where 9 are sufficient.

Setting $P_n\;:\!=\; (n^a/c_\textsf{max})^{1/\alpha}$ , we start bounding the probabilities under consideration using the bounded node degree:

(30) \begin{align}\begin{split}\mathbb P(J_n^{(a)}(X) > n^{d^2/\alpha-\varepsilon}) &\le \mathbb P\Big(\xi^{(\alpha)}(X-x) > n^a \text{ for all }x\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge n^{d^2/\alpha-\varepsilon}\Big) \\[5pt] &\le \mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ for all }x\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge n^{d^2/\alpha-\varepsilon}\bigg).\end{split}\end{align}

We now thin out the configuration consisting of all x as in the previous line, as follows. Starting with any point $x\in\varphi$ , we omit all points of $\varphi$ that are at distance at most $P_n$ to x. According to ( FIN2 ) with $M=P_n$ , this operation removes at most $c_\textsf{max}-1$ points. Repeating iteratively for the other points of $\varphi$ yields a configuration $\varphi$ that contains at least $N_n \;:\!=\; n^{d^2/\alpha-\varepsilon}/c_\textsf{max}$ nodes satisfying $\max_{y\in\mathcal{E}(X-x)} |y| > P_n$ and $|x-y| > P_n$ for all $x,y\in\varphi$ with $x\ne y$ . Thus, we can continue in (30) with

(31) \begin{align}\begin{split}&\mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ for all }x\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge n^{d^2/\alpha-\varepsilon}\bigg) \\[5pt] &\le \mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ and } |x-y| > P_n \text{ for all }x\ne y\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge N_n \bigg).\end{split}\end{align}

Next, we note that in a ball of radius $\sqrt dl_n$ , only a limited number of points can be placed so that all of their mutual distances are larger than $P_n$ . For large n, this number is bounded by the number of balls with radius $(n^a/c_\textsf{max})^{1/\alpha}/2$ that fit in a ball with radius $4\sqrt d n^{a/\alpha}$ in such a way that none of the smaller balls overlap. The ratio between the volume of $B_{4\sqrt d n^{a/\alpha}}(0)$ and the volume of $B_{(n^a/c_\textsf{max})^{1/\alpha}/2}(0)$ yields the bound $8^d c_\textsf{max}^{d/\alpha} d^{d/2}$ for n large. Thus, after setting $M_n \;:\!=\; N_n/(8^d c_\textsf{max}^{d/\alpha + 1} d^{d/2})$ , we can use this argument to proceed in (31) and estimate, for n sufficiently large,

(32) \begin{align}\begin{split}&\mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ and } |x-y| > P_n \text{ for all }x\ne y\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge N_n \bigg) \\[5pt] &\le \mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ and } |x-y| > \sqrt{d} l_n \text{ for all }x\ne y\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge M_n \bigg).\end{split}\end{align}

In the event on the right-hand side of (32), each hypercube $Q\in\mathcal{Q}$ contains at most one node that has an edge larger than $P_n$ . Furthermore, if $\max_{y\in\mathcal{E}(X-x)} |y| > P_n$ holds for an $x\in X$ , then ( STA ) gives that $\mathcal{R}(X-x) > P_n$ . Thus, by a union bound, we arrive at

(33) \begin{align}&\mathbb P\bigg(\max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ and } |x-y| > \sqrt{d} l_n \text{ for all }x\ne y\text{ in some }\varphi\subseteq X\cap Q_n\text{ with }\#\varphi\ge M_n \bigg)\nonumber \\[5pt] &\le \sum_{m\in \mathcal{L}} \mathbb P\bigg(\#\{Q\in \mathcal{Q}^{(m)} \,:\, \max_{y\in\mathcal{E}(X-x)} |y| > P_n \text{ for some } x\in Q\cap X\} \ge M_n/3^d\bigg) \\[5pt] &\le \sum_{m\in \mathcal{L}} \mathbb P\bigg(\#\{Q\in \mathcal{Q}^{(m)} \,:\, \mathcal{R}(X-x) \ge P_n \text{ for some } x\in Q\cap X\} \ge M_n/3^d\bigg).\nonumber \end{align}

Through a calculation performed in the same fashion as in (26), we get

\begin{align*}\mathbb P\bigg(\max_{x \in Q \cap X}\mathcal{R}(X-x) \ge P_n \bigg) \le \int_Q\mathbb P\big(\mathcal{R}(X\cup\{x\}-x) \ge (n^a/c_\textsf{max})^{1/\alpha}\big)dx \le l_n^ d e^{-cn^{ad/\alpha}}\end{align*}

for n large enough and a value $c>0$ . Next, note that $l_n\ge P_n$ . Thus, for a fixed $m\in\mathcal{L}$ , the events of finding a Poisson point with a stabilization radius exceeding $P_n$ in a box Q are independent for different choices of $Q \in \mathcal{Q}^{(m)}$ . Therefore, a binomial concentration bound [Reference Penrose14, Lemma 1.1] gives that for each $m\in \mathcal{L}$ ,

(34) \begin{align}\begin{split} &\mathbb P\big(\#\{Q\in \mathcal{Q}^{(m)} \,:\, \max_{x \in Q \cap X}\mathcal{R}(X-x) \ge P_n \} \ge M_n/3^d\big) \\[5pt] &\le \exp\!\bigg({-}\frac{M_n}{3^d 2} \log\!\bigg(\frac{M_n/3^d}{K_n l_n^d e^{-cn^{ad/\alpha}}}\bigg)\bigg),\end{split}\end{align}

assuming n is sufficiently large. Now, note that

(35) \begin{equation}\lim_{n\uparrow\infty} -\frac1{n^{d^2/\alpha}}\frac{M_n}{3^d 2}\log\!\bigg(\frac{M_n/3^d}{K_n l_n^ d e^{-cn^{ad/\alpha}}}\bigg) = -\infty\end{equation}

holds if $\varepsilon\in (0,ad/\alpha)$ . Finally, combining (30), (31), (32), (33), (34), and (35) yields the desired result.

Proof of Lemma 9. Let $m\ge1$ and $\tau>0$ , and let us assume that we are under the event that we would like to bound in Lemma 9. Note that by (7), under $\{\mathcal{R}_{3n}(X) \le n\}$ we have that $\mathcal{E}(X-x) = \mathcal{E}(X\cap Q_{5n}-x)$ for all $x\in X\cap Q_{3n}$ . Under the event $\{J_n^{(a)}(X) \le m\}$ , we choose $\varphi=\mathcal{J}_n^a(X)$ and $\psi^{\prime}=\cup_{x\in\varphi}\mathcal{E}_x(X)$ . From ( STA ), we obtain configurations $\theta_x$ , $x\in\varphi\cup\psi^{\prime}$ , with $\mathcal{E}(X-x)=\mathcal{E}(\theta_x-x)$ and $\#\theta_x\le I_d c_\textsf{STA}$ . The condition ( STA ) also implies that $\mathcal{E}(X-x)=\mathcal{E}(\psi-x)$ for every $x\in\varphi\cup\psi^{\prime}$ where $\psi \;:\!=\; \varphi\cup\psi^{\prime}\cup_{x\in\varphi\cup\psi^{\prime}} \theta_x$ . Note that, thanks to the bound on the stabilization radius, the set $\psi$ is entirely contained in $Q_{5n}$ . Moreover, the bounded node degree implies that

(36) \begin{equation}\#\psi \le \#\varphi + \#\psi^{\prime} + (\#\varphi + \#\psi^{\prime}) I_d c_\textsf{STA} \le (I_d c_\textsf{STA} + 1) (c_\textsf{max} + 1) m \le (I_d c_\textsf{max} + 1)^2 m.\end{equation}

Below, in the case that $\#\psi<c_\textsf{INF}$ , we add $c_\textsf{INF}-\#\psi$ points in $X\cap Q_{5n}$ to $\psi$ to be able to apply ( INF ). To justify that $X(Q_{5n}) \ge c_\textsf{INF}-\#\psi$ can be assumed here, we remark that under $\big\{\!\sum_{x\in\mathcal{J}_n^{(a)}(X)} \xi^{(\alpha)}(X-x) \ge \tau, \mathcal{R}_{3n}(X) \le n\big\}$ , it must hold that $X(Q_{5n})\ge c_\textsf{INF}$ . The reason for this is that the fact that $\xi^{(\alpha)}(X-x) > 0$ for some $x\in X\cap Q_n$ implies that $X\cap Q_n$ cannot be empty, and $\mathcal{R}_{3n}(X) \le n$ implies that there have to be at least $I_d (c_\textsf{max}-1)$ other Poisson points within distance n of any point in $Q_n$ . Thus, we even get that $X(Q_{5n})\ge I_d(c_\textsf{max}-1) + 1 \ge c_\textsf{INF}$ , which concludes this argument. Next, using (36) together with $\mathcal{E}(X-x)=\mathcal{E}(\psi-x)$ and ( INF ), we obtain that

\begin{align*}&\mathbb P\Bigg(\sum_{x\in\mathcal{J}_n^{(a)}(X)} \xi^{(\alpha)}(X-x) \ge \tau, J_n^{(a)}(X) \le m, \mathcal{R}_{3n}(X) \le n\Bigg) \\[5pt] &\le\mathbb P\Bigg(\sum_{x\in\varphi} \xi^{(\alpha)}(\psi-x) \ge \tau, \text{ for some }\varphi\subseteq X\cap Q_n, \#\varphi \le m \text{ and } \varphi\subseteq\psi\subseteq X\cap Q_{5n}, \\[5pt] &\qquad\quad c_\textsf{INF} \le \#\psi\le (I_d c_\textsf{max} + 1)^2 m, \mathcal{E}(\psi-x)=\mathcal{E}(X-x) \text{ for all } x\in\varphi \cup \cup_{z\in\varphi} \mathcal{E}_z(\psi)\Bigg) \\[5pt] &\le\mathbb P\Bigg(\sum_{x\in\varphi} \xi^{(\alpha)}(\psi-x) \ge \tau, \text{ for some }\varphi\subseteq X\cap Q_n, \#\varphi \le m \text{ and } \varphi\subseteq\psi\subseteq X\cap Q_{5n}, \\[5pt] &\qquad\quad c_\textsf{INF} \le \#\psi\le (I_d c_\textsf{max} + 1)^2 m, \mathcal{E}(\psi-x)\subseteq\mathcal{E}((\psi\cup\{y\}-x) \text{ for all } y\in X \text{ and } \\[5pt] &\qquad\quad x\in\varphi \cup \cup_{z\in\varphi}\mathcal{E}_z(\psi)\Bigg) \\[5pt] &=\mathbb P\Bigg(\sum_{x\in\varphi} \xi^{(\alpha)}(\psi-x) \ge \tau, \text{ for some }\varphi\subseteq X\cap Q_n, \#\varphi \le m \text{ and } \varphi\subseteq\psi\subseteq X\cap Q_{5n}, \\[5pt] &\qquad\quad c_\textsf{INF} \le \#\psi\le (I_d c_\textsf{max} + 1)^2 m, X\cap A(\varphi,\psi) = \emptyset\Bigg) =: (\star).\end{align*}

We remind the reader of the sets $D^{\prime}_{l}$ , $l\in\mathbb N$ , and $\mathcal{N}^{\prime}$ that were defined in Section 2, before Equation (9). Note that by the assumptions in ( CON ),

\begin{align*}0=|N_{l+1}| = \int_{\{(x_1,\dots,x_l)\in\mathbb R^{dl}:\text{ pw.~distinct}\}} |D(\{x_1,\dots,x_l\})| d(x_1,\dots,x_l),\end{align*}

which implies that $|D^{\prime}_{l}|=0$ , and thus $\mathcal{N}^{\prime}$ is a zero-set. In the following, let $\textbf{x}=(x_1,\dots,x_{l_1})$ and $\textbf{y}=(y_1,\dots,y_{l_2})$ represent $\varphi$ and $\psi\setminus\varphi$ , respectively. We will abuse notation and allow $\textbf{x}$ and $\textbf{y}$ to be treated as sets. A combination of the union bound, Markov’s inequality, and Mecke’s formula yields

\begin{align*}(\star) &\le \sum_{0\le l_1,l_2\le (I_d c_\textsf{max} + 1)^2 m} \int_{Q_{5n}^{l_2}}\int_{Q_{5n}^{l_1}} \mathbb P\bigg(\sum_{x\in\textbf{x}} \xi^{(\alpha)}(\textbf{x}\cup\textbf{y}-x) \ge \tau,\#\textbf{y} \ge c_\textsf{INF}, X\cap A(\textbf{x},\textbf{x}\cup\textbf{y})=\emptyset\bigg) d\textbf{x} d\textbf{y} \\[5pt] &= \sum_{0\le l_1,l_2\le (I_d c_\textsf{max} + 1)^2 m} \int_{Q_{5n}^{l_2}}\int_{Q_{5n}^{l_1}} \mathbb{1}_{\{\sum_{x\in\textbf{x}} \xi^{(\alpha)}(\textbf{x}\cup\textbf{y}-x) \ge \tau\}} \mathbb{1}_{\{\textbf{x}\cup\textbf{y}\not\in\mathcal{N}^{\prime},\,\#\textbf{y} \ge c_\textsf{INF}\}} \exp\!({-}|A(\textbf{x},\textbf{x}\cup\textbf{y})|) d\textbf{x} d\textbf{y} \\[5pt] &\le (I_d c_\textsf{max} + 1)^4 m^2 (5n)^{d 2 (I_d c_\textsf{max} + 1)^2 m}\exp\!\Big({-}\tau^{d/\alpha}\inf_{(\textbf{x},\textbf{x}\cup\textbf{y})\in B} |A(\textbf{x},\textbf{x}\cup\textbf{y})|\Big),\end{align*}

from which the assertion follows.

Proof of the lower bound of Theorem 1. First, fix two configurations $(\varphi, \psi) \in B$ such that $|A(\varphi, \psi)| \le \inf_{(\varphi^{\prime},\psi^{\prime})\in B} |A(\varphi^{\prime},\psi^{\prime})| + \varepsilon$ . Because $\inf_{(\varphi^{\prime},\psi^{\prime})\in B} |A(\varphi^{\prime},\psi^{\prime})|<\infty$ was assumed at the start of this section, $|A(\varphi,\psi)|<\infty$ also has to be satisfied. Now, let $\delta>0$ be such that (a) and (b) from Lemma 11 are satisfied. Under the event $\cap_{x\in\tau_n\varphi} \{X(B_1(x))=1\}$ , we can find $(z_x)_{x\in\tau_n\psi} \subseteq B_1(0)$ such that $\{x+z_x\} = X\cap B_1(x)$ for each $x\in\tau_n\psi$ . Furthermore, if n is so large that $\tau_n \delta \ge 1$ , then under $\{X(\tau_nA_{\delta}^-)=0\}$ it is guaranteed by ( INF ) that for each $x+z_x\in\{y+z_y \,:\, y\in\tau_n\varphi\}\cup\cup_{w\in\tau_n\varphi} (\mathcal{E}_{w+z_w}(\{y+z_y \,:\, y\in\tau_n\psi\}))$ ,

(40) \begin{equation}\mathcal{E}(X-(x+z_x)) \supseteq \mathcal{E}(\{y+z_y \,:\, y\in\tau_n\psi\}-(x+z_x)).\end{equation}

Then, also if $\tau_n \delta \ge 1$ , Lemma 11(b) and (40) give that

(41) \begin{equation}\sum_{x\in\tau_n\psi} \xi^{(\alpha)}(X-(x+z_x)) \ge \sum_{x\in\tau_n\varphi} \xi^{(\alpha)}(\{y+z_y \,:\, y\in\tau_n\psi\}-(x+z_x)) > \tau_n^\alpha/(1+\varepsilon) = (r+\varepsilon)n^d.\end{equation}

Note that the index set in the sum before the first inequality in (41) contains more points than the one after it. The reason for this is that when a point is added outside of the influence zone of $(\tau_n\varphi,\tau_n\psi)$ , our framework for graphs does not exclude new edges from being created between two already existing nodes in $\tau_n\psi$ . While it is admittedly hard to come up with an actual example of a graph for which the following is possible, it might potentially happen that when the points from $X\setminus(\tau_n\psi)$ are added to $\tau_n\psi$ , an additional edge arises from a point in $\tau_n\psi\setminus(\tau_n\varphi)$ to a point in $\tau_n\varphi$ . In an undirected graph this could have the effect that the power-weighted edge lengths of some outgoing edges from points in $\tau_n\varphi$ are only taken into account with the factor $1/2$ on the left-hand side of the first inequality in (41), while being considered with their full weight on the right-hand side of it. Summing over all points in $\tau_n\psi$ lets us avoid this issue.

As the remark after Lemma 12 suggets, we can assume that all of the occurring sets and configurations are contained in $U_n$ , from which point (41) implies that $G_{2,n}(\delta) \subseteq \{H_n(U_n,\mathbb R^d) > r + \varepsilon\}.$

We now define the set $W_n^- \;:\!=\; \big[(\log n)^2, b_n - (\log n)^2\big]$ and assert that under the event $G_{1, n}$ from (38), we have

(42) \begin{align}H_n(W_n\setminus W^-_n, W_n) < \varepsilon/2\end{align}

for all large n. Once (42) is established, we can conclude the proof of the lower bound of Theorem 1. Indeed, under $E_n^\textsf{good}$ each box in $W_n\setminus W_n^-$ contains at least $c_\textsf{max}$ Poisson points, and therefore, if n is chosen large, then each of the cones around an $x\in X\cap W_n^-$ has to contain $c_\textsf{max}$ Poisson points before the base of the cone leaves $W_n$ , which more formally means that $\cup_{i \le I_d} \big ((S_i+x)\cap B_{\mathcal{S}_i (X-x)} (x)\big) \subseteq W_n$ . Thus, under $E_n^\textsf{good}$ , again by ( STA ), we get that $\xi^{(\alpha)}(X\cap W_n-x) = \xi^{(\alpha)}(X-x)$ for all points $x\in X\cap W^-_n$ if n is sufficiently large. In other words, the layer of boxes containing points would not admit the score of points in $W_n^-$ being influenced by any points outside of $W_n$ . With (42) we get that under $G_{1,n}$ ,

\begin{align*}H_n\big(W_n,\mathbb R^d\big) \ge H_n\big(W^-_n,\mathbb R^d\big) = H_n\big(W^-_n, W_n\big) = H_n\big(W_n, W_n\big) - H_n\big(W_n\setminus W^-_n, W_n\big)> \mu_\alpha - \varepsilon.\end{align*}

In addition, $G_{1, n}$ and $G_{2,n}(\delta)$ are independent for large n. Next, shifting the coordinate system shows that $\mathbb P(H_n > \mu_\alpha + r) = \mathbb P\big(H_n([0,n]^d,\mathbb R^d) > \mu_\alpha + r\big).$ Hence,

\begin{align*}\mathbb P\big(H_n > \mu_\alpha + r\big) &\ge \mathbb P\big(H_n\big(U_n,\mathbb R^d\big) > r + \varepsilon, H_n\big(W_n,\mathbb R^d\big) > \mu_\alpha - \varepsilon\big)\\[5pt] &\ge \mathbb P\big(G_{2,n}(\delta), G_{1,n}\big)\\[5pt] &= \mathbb P\big(G_{2,n}(\delta)\big) \mathbb P\big(G_{1,n}\big).\end{align*}

Using Lemmas 10 and 13, it follows that

(43) \begin{equation}\liminf_{n\uparrow\infty} \frac1{n^{d^2/\alpha}} \log \mathbb P(H_n > \mu_\alpha + r) \ge -\bigg(\!\inf_{(\varphi,\psi)\in B} |A(\varphi,\psi)|+\varepsilon\bigg)(r+\varepsilon)^{d/\alpha},\end{equation}

and letting $\varepsilon\downarrow 0$ gives the asserted result.

It remains to prove (42) under the event $G_{1, n}$ . To that end, we recall that

\begin{align*}H_n(W_n\setminus W_n^-,W_n) = \frac1{n^d} \sum_{x\in X\cap (W_n\setminus W^-_n)} \xi^{(\alpha)}(X\cap W_n - x).\end{align*}

Henceforth, we bound the summands on the right-hand side separately in the cases where $\textrm{dist}(x,\partial W_n)\ge c\log n$ and where $\textrm{dist}(x,\partial W_n)< c\log n$ for a suitable $c > 0$ .

First, consider the case $\textrm{dist}(x,\partial W_n)\ge c\log n$ . If we cut off the cone $S_i + x$ at a distance $c\log n$ for large enough c, then it still contains one of the boxes $Q^{\prime}_j$ . By definition of the event $E_n^\textsf{good}$ , each of these boxes contains at least $c_\textsf{max}$ nodes. Therefore, $\mathcal{S}_i(X\cap W_n-x)$ is of order $\log n$ for all $1\le i\le I_d$ .

Now, consider the case that $\textrm{dist}(x,\partial W_n) < c\log n$ . If a cone that arises from x does not intersect $W_n$ anymore after a distance from the apex of order $\log n$ , then it contains Poisson points of $X\cap W_n$ only up until a distance of order $\log n$ . Since, by ( STA ), none of the lateral boundaries of any cone are parallel to an axis of the coordinate system, we obtain that otherwise the cone envelopes a whole box $Q^{\prime}_j$ after a distance from the apex of order $\log n$ . Then $\mathcal{S}_i(X\cap W_n-x)$ is of order $\log n$ as argued for above. Hence, after n is chosen sufficiently large, ( STA ) yields a finite configuration $\theta_x$ with $\theta_x\subseteq B_{(\log n)^2}(x)\cap W_n$ satisfying that $\mathcal{E}(X\cap W_n-x)=\mathcal{E}(\theta_x-x)$ . Together with the bounded node degree, we have for n large

\begin{align*} n^dH_n(W_n\setminus W_n^-,W_n) &\le \hspace{-.5cm} \sum_{\substack{x\in X\cap (W_n\setminus W^-_n)\\ y\in\mathcal{E}(\theta_x -x)}} \hspace{-.5cm}|y| \\[5pt] &\le X(W_n\setminus W^-_n) c_\textsf{max} (\log n)^{2\alpha} \\[5pt] &\le (\log n)^{2d}\left\lceil\frac{n^{d-1}}{(\log n)^{d-3}}\right\rceil c_\textsf{max} (\log n)^{2\alpha},\end{align*}

where the final inequality follows from (37), the upper bound on the number of Poisson points in each box $Q^{\prime}_{i}$ . Hence, we can choose n sufficiently large to ensure that $H_n(W_n\setminus W_n^-,W_n)<\varepsilon/2$ .

Proof of Lemma 10. We consider separately each of the two events whose intersection forms $G_{1,n}$ . For $E_n^{\textsf{good}}$ we use a Poisson bound from [Reference Penrose14, Lemma 1.2] and calculate, for n sufficiently large,

(44) \begin{align}\mathbb P\big(E_n^\textsf{good}\Big) &= 1-\mathbb P\big(X(Q^{\prime}_{i}) < c_\textsf{max} \text{ or } X(Q^{\prime}_{i}) \ge (\log n)^{2d} \text{ for some } i\big) \nonumber\\[5pt] &\ge 1-\left\lceil\frac{n^{d-1}}{(\log n)^{d-3}}\right\rceil \Bigg(e^{-(\log n)^d} \sum_{i \le c_\textsf{max}-1} \frac{(2\log n)^{di}}{i!} + e^{-\frac12(\log n)^{2d}\log\!\big(\frac{(\log n)^{2d}}{(2\log n)^d}\big)}\Bigg) \nonumber\\[5pt] &\ge 1-n^{d-1}\Big(e^{-(\log n)^d} (\log n)^{d c_\textsf{max}+2} + e^{-\frac12(\log n)^{2d}}\Big).\end{align}

Next, we deal with $\{H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\}$ . Under the condition that $X(W_n) = b_n^d$ , we can deduce

\begin{align*}\big(H_n(W_n,W_n)\ \big|\ X(W_n) = b_n^d\big) &\overset d= \frac{b_n^d}{n^d} \frac1{b_n^d} \sum_{i \le b_n^d} \xi^{(\alpha)}\big(\{X_1^{(n)},\dots,X_{b_n^d}^{(n)}\} - X_i^{(n)}\big) \\[2pt] &\overset d= \frac{b_n^d}{n^d} \frac1{b_n^d} \sum_{i \le b_n^d} \xi^{(\alpha)}\big(b_n\{\widehat X_1,\dots,\widehat X_{b_n^d}\} - b_n\widehat X_i\big)\end{align*}

for $X_1^{(n)}, X_2^{(n)},\dots$ being independent and identically distributed (i.i.d.) uniform random variables on $W_n$ and $\widehat X_1,\widehat X_2,\dots$ being i.i.d. uniform random variables on $[0,1]^d$ . In order to apply [Reference Penrose and Yukich16, Theorem 2.1], we need to check the moment condition, i.e., that for $p>2$ ,

(45) \begin{equation}\sup_{n\ge 1} \mathbb E\big[\xi^{(\alpha)}\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big)^p\big] <\infty.\end{equation}

We can use the bound on the node degree to get

(46) \begin{align}\begin{split}&\mathbb E\big[\xi^{(\alpha)}\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big)^p\big] = \int_0^\infty \mathbb P\big(\xi^{(\alpha)}\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big)^p > s\big)ds \\[2pt] &\le \int_0^\infty \mathbb P\big(|y| > \big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha} \text{ for some }y\in \mathcal{E}\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big)\big)ds.\end{split}\end{align}

Next, from ( STA ) we can deduce that, for every $s>0$ , if $b_n \widehat X_1$ has an out-neighbor among $b_n\big\{\widehat X_2,\dots, \widehat X_{b_n^d}\big\}$ that is farther away than $\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}$ , then one of the cones arising from $b_n \widehat X_1$ has to extend until at least a distance of $\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}$ from its apex before it contains $c_\textsf{STA}$ vertices. More precisely, there has to be an $i\le I_d$ such that $\mathcal{S}_i\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big) > \big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}$ . Additionally, the intersection of $W_n$ and $\big(S_i + b_n \widehat X_1\big)\setminus B_{\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}}\big(b_n \widehat X_1\big)$ cannot be empty, since the aforementioned out-neighbor has to be within $W_n$ . Therefore, under the event from the last line of (46), it is implied by the definition of $\mathcal{S}_i(\!\cdot\!)$ that for some $i \le I_d$ it holds that $b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}\cap\big(S_i+b_n\widehat X_1\big)\cap B_{\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}}\big(b_n\widehat X_1\big)$ contains at most $c_\textsf{max}$ points, while $W_n\cap \big(S_i+b_n\widehat X_1\big)\setminus B_{\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}}\big(b_n\widehat X_1\big)\neq\emptyset$ . With these arguments we arrive at

(47) \begin{align}\begin{split}&\int_0^\infty \mathbb P\big(|y| > \big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha} \text{ for some }y\in \mathcal{E}\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}- b_n\widehat X_1\big)\big)ds \\[2pt] &\le \sum_{i \le I_d} \int_0^\infty \mathbb P\big(\#\big(b_n\big\{\widehat X_1,\dots, \widehat X_{b_n^d}\big\}\cap\big(S_i+b_n\widehat X_1\big)\cap B_{\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}}\big(b_n\widehat X_1\big)\big) \le c_\textsf{max} \\[2pt] &\qquad\qquad\qquad\ \text{and } W_n\cap \big(S_i+b_n\widehat X_1\big)\setminus B_{\big(s^{1/p}/c_\textsf{max}\big)^{1/\alpha}}\big(b_n\widehat X_1\big) \ne\emptyset \big)ds.\end{split}\end{align}

Furthermore, since the cones do not have lateral boundaries parallel to any axes, under the event after the last inequality in (47), the volume of the set $W_n\cap (S_i+b_n\widehat X_1)\cap B_{(s^{1/p}/c_\textsf{max})^{1/\alpha}}(b_n\widehat X_1)$ is of order $s^{d/(p\alpha)}$ and therefore at least $c s^{d/(p\alpha)}$ for all i, where $c>0$ depends only on the layout of the cones, $\alpha$ , and $c_\textsf{max}$ . Thus, by the independence of $\widehat X_2,\dots,\widehat X_{b_n^d}$ , when conditioned on $\widehat X_1$ , we can bound the probability of the event after the last inequality in (47) by the probability of a binomial random variable consisting of $b_n^d-1$ trials with success probability

\begin{align*} & \mathbb P(b_n \widehat X_2 \in (S_i+b_n\widehat X_1)\cap B_{(s^{1/p}/c_\textsf{max})^{1/\alpha}}(b_n\widehat X_1) \mid\, W_n\cap (S_i+b_n\widehat X_1)\setminus B_{(s^{1/p}/c_\textsf{max})^{1/\alpha}}(b_n\widehat X_1) \ne\emptyset)\\[5pt] & \ge cs^{d/(p\alpha)}/b_n^d\end{align*}

realizing a value of at most $c_\textsf{max}-1$ . Thus, by a binomial concentration bound [Reference Penrose14, Lemma 1.1],

\begin{align*}&\mathbb E\big[\xi^{(\alpha)}(b_n\{\widehat X_1,\dots, \widehat X_{b_n^d}\}- b_n\widehat X_1)^p\big] \\[5pt] &\le I_d \int_0^\infty \mathbb P\big(\textsf{Bin}\big(b_n^d-1, c s^{d/(p\alpha)} / b_n^d\big) \le c_\textsf{max}-1\big)ds \\[5pt] & \le I_d \int_0^\infty \mathbb P\big(\textsf{Bin}(b_n^d, c s^{d/(p\alpha)} / b_n^d) \le c_\textsf{max}\big)ds \\[5pt] &\le I_d (c_\textsf{max}/c)^{p\alpha/d} + I_d \int_{(c_\textsf{max}/c)^{p\alpha/d}}^\infty \mathbb P\big(\text{Bin}(b_n^d, c s^{d/(p\alpha)} / b_n^d) \le c_\textsf{max}\big)ds \\[5pt] &\le I_d (c_\textsf{max}/c)^{p\alpha/d} +I_d \int_{(c_\textsf{max}/c)^{p\alpha/d}}^\infty \exp\!\big({-}c s^{d/(p\alpha)} (1- \tfrac{c_\textsf{max}}{c s^{d/(p\alpha)}} + \tfrac{c_\textsf{max}}{c s^{d/(p\alpha)}} \log(\tfrac{c_\textsf{max}}{c s^{d/(p\alpha)}})\big)ds < \infty.\end{align*}

In particular, the bound does not depend on n. Therefore, the moment condition (45) is satisfied. Now, [Reference Penrose and Yukich16, Theorem 2.1] gives

\begin{align*}\frac1{b_n^d} \sum_{i=1}^{b_n^d} \xi^{(\alpha)}\big(b_n\big\{\widehat X_1,\dots,\widehat X_{b_n^d}\big\} - b_n\widehat X_i\big) \overset{P}{\longrightarrow} \mu_\alpha,\end{align*}

and since $b_n^d/n^d \overset{n\uparrow\infty}{\longrightarrow} 1$ , it follows that

(48) \begin{equation}\lim_{n\uparrow\infty} \mathbb P\big(H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\, |\, X(W_n) = b_n^d\big) = 1.\end{equation}

Now we can use the union bound to arrive at

\begin{align*}\mathbb P(G_{1,n}) &\ge \mathbb P\big(H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\big) + \mathbb P\big(E_n^\textsf{good}\big) - 1 \\[5pt] &\ge \mathbb P\big(H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\, |\, X(W_n) = b_n^d\big) \mathbb P(X(W_n) = b_n^d) + \mathbb P\big(E_n^\textsf{good}\big) - 1\\[5pt] &\ge \mathbb P\big(H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\, |\, X(W_n) = b_n^d\big) \frac{\exp\left({-}\frac1{12 b_n^d}\right)}{\sqrt{2\pi b_n^d}} + \mathbb P\big(E_n^\textsf{good}\big) - 1,\end{align*}

where in the last line we used [Reference Penrose14, Lemma 1.3]. By (48) we can assume n large enough so that

\begin{align*}\mathbb P\big((H_n(W_n,W_n) > \mu_\alpha - \varepsilon/2\, |\, X(W_n) = b_n^d\big) \ge 1/2.\end{align*}

Hence, combining this with (44), we get that

\begin{align*}\mathbb P(G_{1,n}) \ge \frac12\frac{\exp\left({-}\frac1{12b_n^d}\right)}{\sqrt{2\pi b_n^d}} - n^{d-1}\big(e^{-(\log n)^d} (\log n)^{d c_\textsf{max}+2} + e^{-\frac12(\log n)^{2d}}\big),\end{align*}

as asserted.

Proof of Lemma 11. First, we show that

(49) \begin{equation}\bigcap_{\delta>0}\underbrace{\bigcup_{(z_y)_{y\in\psi}\subseteq B_\delta(0)} A(\{x+z_x \,:\, x\in\varphi\},\{x+z_x \,:\, x\in\psi\})}_{=: A_\delta(\varphi,\psi)} \subseteq A(\varphi,\psi) \cup D(\psi),\end{equation}

where we recall that $D(\psi) = \{y \in \mathbb R^d \,:\, \varphi \cup\{y\} \in N_{\#\varphi +1}\}$ . The subset relation in (49) holds because if we let $y\in\cap_{\delta>0} A_\delta(\varphi,\psi)\setminus D(\psi)$ , then for every $\delta\in(0,1)$ there exists a family $(z_w)_{w\in\psi}\subseteq B_\delta(0)$ such that $y\in A(\{w+z_w \,:\, w\in\varphi\},\{w+z_w \,:\, w\in\psi\})$ . Hence,

\begin{align*}\mathcal{E}_{x+z_x}(\{w+z_w \,:\, w\in\psi\}) \not\subseteq \mathcal{E}_{x+z_x}(\{w+z_w \,:\, w\in\psi\}\cup\{y\})\end{align*}

for some

\begin{align*}x+z_x\in\{w+z_w \,:\, w\in\varphi\}\cup\cup_{v+z_v\in\{w+z_w \,:\, w\in\varphi\}} (\mathcal{E}_{v+z_v}(\{w+z_w \,:\, w\in\psi\})).\end{align*}

Since $y\not\in D(\psi)$ , we can apply ( CON ) to both sides, choosing $\delta$ sufficiently small, which gives $\mathcal{E}_x(\psi) \not\subseteq \mathcal{E}_x(\psi\cup\{y\})$ for some $x\in\varphi\cup\cup_{v\in\varphi} (\mathcal{E}_v(\psi))$ and therefore $y\in A(\varphi,\psi)$ . Since $|D(\psi)|=0$ , we deduce from (49) that for $\delta$ sufficiently small we have $|A_\delta(\varphi,\psi)|\leq |A(\varphi,\psi)| + \varepsilon.$

To prove Part (b), note that since $\psi$ is finite, we can use ( CON ) and find $\delta\in(0,1)$ small enough so that for all choices of $(z_x)_{x\in\psi} \subseteq B_\delta(0)$ we have

(50) \begin{equation}\{w - z_w \,:\, w\in\mathcal{E}_{x+z_x}(\{y+z_y \,:\, y\in\psi\})\} = \mathcal{E}_x(\psi).\end{equation}

This means the graph looks the same despite some small noise of at most $\delta$ for every node. But the finiteness of the configuration combined with (50) guarantees that for $\delta$ sufficiently small,

\begin{align*}\inf_{(z_x)_{x\in\psi} \subseteq B_\delta(0)} \sum_{x\in\varphi} \xi^{(\alpha)}(\{y+z_y \,:\, y\in\psi\}-(x+z_x)) > 1/(1+\varepsilon).\end{align*}

Proof of Lemma 12. Let $(\varphi,\psi)\in B$ be such that $|A(\varphi,\psi)|<\infty$ . The key step is to construct a finite set of points $\theta \subseteq \mathbb R^d$ and a scalar $R > 0$ such that, for all $x\in\eta\;:\!=\;\varphi \cup\cup_{z\in\varphi} (\mathcal{E}_z (\psi))$ and $(z_y)_{y\in\psi}\subseteq B_\delta(0)$ , we have (i) $\mathcal{R}\big((\{y+z_y\,:\, y\in\psi\} \cup \theta) - (x + z_x)\big) \le R$ , and (ii)

(51) \begin{align} \mathcal{E}\big(\{y+z_y\,:\, y\in\psi\}-(x+z_x)\big)\subseteq\mathcal{E}\big((\{y+z_y\,:\, y\in\psi\}\cup\theta )-(x+z_x)\big).\end{align}

Once $\theta$ is constructed, we assert that $A_\delta(\varphi, \psi) \subseteq \bigcup_{x \in \eta} B_{R + 1}(x)$ . Indeed, for any $v \in \mathbb R^d\setminus \bigcup_{x \in \eta} B_{R + 1}(x)$ , the definition of the stabilization radius implies that

(52) \begin{equation} \mathcal{E}\big((\{y+z_y\,:\, y\in\psi\}\cup\theta)-(x+z_x)\big) = \mathcal{E}\big((\{y+z_y\,:\, y\in\psi\}\cup\theta\cup\{v\})-(x+z_x)\big).\end{equation}

Hence, combining (51), (52), and ( STA ) gives that

\begin{align*} \mathcal{E}\big(\{y+z_y\,:\, y\in\psi\}-(x+z_x)\big) \subseteq \mathcal{E}\big((\{y+z_y\,:\, y\in\psi\}\cup\{v\})-(x+z_x)\big),\end{align*}

thereby proving the assertion that $v \not\in A_\delta(\varphi, \psi)$ .

It remains to prove the existence of $R > 0$ and $\theta \subseteq \mathbb R^d$ . To that end, first note that $|T_i(x,\delta)| = \infty$ for all $i \le I_d$ and $x\in\mathbb R^d$ , where $T_i(x,\delta)\;:\!=\; \cap_{y\in B_\delta(x)} (S_i+y)$ . Since $|A_\delta(\varphi,\psi)|<\infty$ , by Lemma 11(a), if $\delta$ is chosen appropriately, it follows that for every $x\in \eta$ and $i\le I_d$ there are distinct $w_{x,i}^{(1)},\dots,w_{x,i}^{(c_\textsf{STA})}\in T_i(x,\delta)\setminus A_\delta(\varphi,\psi)$ . Then, defining $\theta\;:\!=\; \{w_{x,i}^{(j)} \,:\, x\in\eta, i \le I_d, j \le c_\textsf{STA}\}$ , we note that ( INF ) implies the property (51). Now, set

\begin{align*}R\;:\!=\; \max_{\substack{x\in\eta, \, z\in B_\delta(x)\\ i\le I_d, j\le c_\textsf{STA}}} c_\textsf{STA}|z-w_{x,i}^{(j)}| \le \max_{\substack{x\in\eta\\ i \le I_d, j\le c_\textsf{STA}}} c_\textsf{STA} |x-w_{x,i}^{(j)}|+c_\textsf{STA} \delta,\end{align*}

and note that the finiteness of the configurations in B implies that $R < \infty$ . Then the definition of the stabilization radius yields $\mathcal{R}\big((\{y+z_y\,:\, y\in\psi\} \cup \theta) - (x + z_x)\big) \le R$ , as asserted.

Proof of Lemma 13. Let $(\varphi,\psi)\in B$ and $\delta\in(0,1)$ be given according to the setting of Lemma 11. First note that $B_1(x) \subseteq B_{\tau_n \delta}(x)$ for sufficiently large $n\ge 1$ . Moreover, the events $\{X(B_1(x)) = 1 \text{ for all } x\in\tau_n\psi\}$ and $\{X(A_{\delta, n}^-) = 0\}$ are independent. Thus, we can examine them separately and start with the first one. We assume that n is chosen large enough so that all of the balls around points in $\tau_n\psi$ are disjoint. Therefore,

\begin{align*}\mathbb P\big(X(B_1 (x)) = 1 \text{ for all } x\in\tau_n\psi\big) = \prod_{x\in\tau_n\psi}\mathbb P\big(X(B_1 (x)) = 1\big) = \kappa_d ^{\#\psi} e^{-\#\psi \kappa_d}.\end{align*}

For the second event, Lemma 11(a) yields

\begin{align*}\mathbb P\big(X(A_{\delta, n}^-) = 0\big) \ge \mathbb P\big(X(\tau_nA_{\delta}(\varphi,\psi)) = 0\big) \ge \exp\!\big({-}\big( |A(\varphi,\psi)|+\varepsilon\big)\tau_n^d\big).\end{align*}

All of this combined shows that for large enough n,

\begin{align*} \frac1{n^{d^2/\alpha}} \log \mathbb P(G_{2,n}(\delta)) &\ge \frac1{n^{d^2/\alpha}} \log\!\Big( \kappa_d^{\#\psi} e^{-\#\psi \kappa_d}\Big) - \Big(\!\inf_{(\varphi,\psi)\in B} |A(\varphi,\psi)|+\varepsilon\Big)((r+\varepsilon)(1+\varepsilon))^{d/\alpha} \\[5pt] &\overset{n\uparrow\infty}{\longrightarrow} - \Big(\!\inf_{(\varphi,\psi)\in B} |A(\varphi,\psi)|+\varepsilon\Big)((r+\varepsilon)(1+\varepsilon))^{d/\alpha},\end{align*}

as asserted.