Proof. If $\underline{\lambda}$ is not fully supercritical (cf. Definition 2.8), then there exists a color $i \in [k]$ for which $\lambda^{\setminus i} \le 1$ ; let $i \in [k]$ be such a color. Clearly, $i \notin I_{\underline{\lambda}}$ , thus $\underline{t}^*(r) \ne \underline{1}$ since the lengths of these vectors are not equal. Now we need to show that $\big\{\big|\widetilde{\mathcal{C}}^*(r)\big| = \infty\big\}$ almost surely does not occur. By Lemma 3.1, the branching process $\big(\big|R_{\infty,d}^{\setminus i}\big|\big)$ almost surely dies out, which means that r is not i-avoiding connected to infinity (cf. Definition 2.5), thus the set of i-avoiding friends of r is $\mathcal{C} \big( r, G_{\infty}^{\setminus i} \big)$ , which is almost surely finite. Since each friend of r is also an i-avoiding friend of it, $\widetilde{\mathcal{C}}^* (r, G_{\infty}) \subseteq \mathcal{C} \big( r, G_{\infty}^{\setminus i} \big)$ holds, which implies that r has almost surely finitely many friends.

If $\underline{\lambda}$ is fully supercritical, then assume first that the event $\big\{\big|\widetilde{\mathcal{C}}^*(r)\big| = \infty\big\}$ occurs. We need to show that r is i-avoiding connected to infinity for all $i \in [k]$ . Let $i \in [k]$ be an arbitrary color. Since $|\widetilde{\mathcal{C}}^*(r)| = \infty$ , clearly $\Big\{v \stackrel{G_{\infty}^{\setminus i} }{\longleftrightarrow} r\Big\} \cup \Big(\Big\{v \stackrel{G_{\infty}^{\setminus i}}{\longleftrightarrow} \infty\Big\} \cap \Big\{r \stackrel{G_{\infty}^{\setminus i}}{\longleftrightarrow} \infty\Big\}\Big)$ holds for infinitely many vertices v. Now this event is the union of two events, and the second one contains the desired event $\Big\{r \stackrel{G_\infty^{\setminus i}}{\longleftrightarrow} \infty\Big\}$ , therefore if the second event occurs for any vertex v, then we are done. So assume that the first event holds for infinitely many vertices v. Noting that the degree of every vertex of $G_\infty^{\setminus i}$ is $\mathbb{P}$ -almost surely finite, we obtain, by König’s lemma [Reference König14], that there exists a path from the root r to infinity in $G_\infty^{\setminus i}$ , and we are done.

Now let us assume that $\underline{t}^*(r) = \underline{1}$ , and we need to show that this implies $\big\{\big|\widetilde{\mathcal{C}}^*(r, G_\infty)\big| = \infty\big\}$ . Let $V_{\infty}(\underline{1}) \,:\!=\, \{v \in V(G_\infty) \colon \underline{t}^*(v, G_\infty) = \underline{1}\}$ . By the definitions of friendship and type vectors (cf. Definitions 3.7 and 3.7, respectively), we only need to show that

(3.1) \begin{equation} \mathbb{P}\big(|V_{\infty}(\underline{1})| = +\infty \mid \underline{t}^*(r) = \underline{1}\big) = 1. \end{equation}

Let $i \in [k]$ be an arbitrary color. By Lemma 3.1, $\big(\big|R_{\infty,d}^{\setminus i}\big|\big)_{d \in \mathbb{N}}$ is a supercritical branching process and the Kesten–Stigum theorem (see, e.g., [Reference Lyons, Pemantle and Peres22, Theorem A]) implies that, conditional on $(\underline{t}^*(r))_i = 1$ , we have $\big|R_{\infty,d}^{\setminus i}\big| \stackrel{\mathbb{P}}{\longrightarrow} \infty$ as $d \to \infty$ . For any $d \in \mathbb{N}$ , conditional on $G_{\infty}\big[R^{\le}_d\big(r, G_{\infty}^{\setminus i}\big)\big]$ , the vertices of $R_{\infty,d}^{\setminus i}$ have type vectors $\underline{1}$ independently of each other with probability $p^*(\underline{1})$ , and thus the distribution of $\Big|R_{\infty,d}^{\setminus i} \cap V_{\infty}(\underline{1})\Big|$ is $\textrm{Bin}\Big(\big|R_{\infty,d}^{\setminus i}\big|, p^*(\underline{1})\Big)$ . By the Harris–FKG inequality [Reference Fortuin, Kasteleyn and Ginibre8, Reference Harris12] and by the full supercriticality of $\underline{\lambda}$ , we obtain

\begin{align*} p^*(\underline{1}) \ge \prod_{i \in [k]} \mathbb{P}((\underline{t}^*(r))_i = 1) = \prod_{i \in [k]} \theta^{\setminus i} > 0. \end{align*}

Therefore, conditional on the event $\{ \underline{t}^*(r) = \underline{1} \}$ , we have $\Big|R_{\infty,d}^{\setminus i} \cap V_{\infty}(\underline{1})\Big| \stackrel{\mathbb{P}}{\longrightarrow} \infty$ as $d \to \infty$ . Hence, (3.1) holds.

Proof of Proposition 2.1. As we saw in the proof of Lemma 3.2, if $\underline{\lambda}$ is not fully supercritical, then r has almost surely finitely many friends, i.e. $f^*_{\infty} = 0$ , and if $\underline{\lambda}$ is fully supercritical, then $f^*_{\infty} = p^*(\underline{1}) > 0$ .

Proof of Proposition 2.2. Let $G_{\infty} \,:\!=\, G_{\infty}(r) \sim \mathcal{G}_{\infty}(\underline{\lambda})$ .

We begin with the proof of (i). Let $\underline{\lambda} \in \mathbb{R}_+^k$ be fully critical-subcritical. We need to show that r has almost surely no friends other than itself (cf. Definition 2.5). Since $\underline{\lambda}$ is fully critical-subcritical, $\lambda^{\setminus i} \le 1$ holds for all $i \in [k]$ and thus, by Lemma 3.1, the root r is almost surely not i-avoiding connected to infinity. Let $v \in V(G_{\infty})$ , $v \ne r$ , be an arbitrary vertex, and let $i \in [k]$ be a color appearing on the unique path connecting r and v in $G_{\infty}$ . Then , and thus r and v are almost surely not i-avoiding friends, so they are not friends either. Therefore, $\mathbb{P}(\widetilde{\mathcal{C}}^*(r) = \{r\}) = 1$ and thus, by Claim 2.1, we are done.

Next, we prove (ii). Let $\underline{\lambda} \in \mathbb{R}_+^k$ be not fully critical-subcritical. By the definition of $I_{\underline{\lambda}}$ , we have $I_{\underline{\lambda}} \ne \emptyset$ ; let $i_1 \in I_{\underline{\lambda}}$ and $i_2\in[k]\setminus\{i_1\}$ , and let $\ell \in \mathbb{N}_+$ . Let F be the following rooted, edge-colored tree: the root of F is r, which has exactly $\ell$ neighbors $v_1, \ldots, v_{\ell}$ , each of which with one further neighbor $w_1, \ldots, w_{\ell}$ , respectively, and the color of the edges $rv_1, \ldots, r v_{\ell-1}$ is $i_1$ , and the color of the edges $r v_{\ell}$ and $v_1 w_1, \ldots, v_{\ell} w_{\ell}$ is $i_2$ . It is not difficult to see that if the event

\begin{align*} \{G_{\infty,2}(r) \simeq F\} \cap \bigg\{\textrm{for all } j \in [\ell] \colon w_j \stackrel{G_\infty^{\setminus i_1}}{\longleftrightarrow} \infty \bigg\} \end{align*}

occurs, then $\widetilde{\mathcal{C}}^*(r, G_{\infty}) = \{ r, v_1, \ldots, v_{\ell-1} \}$ (where we identified the vertices of $G_{\infty,2}(r)$ and F). All we are left to show is that the above event occurs with positive probability. Clearly, we have

\begin{equation*} \mathbb{P}\Big(\{G_{\infty,2}(r) \simeq F\} \cap \Big\{\textrm{for all } j \in [\ell] \colon w_j \stackrel{G_\infty^{\setminus i_1}}{\longleftrightarrow} \infty\Big\}\Big) = (\theta^{\setminus i_1})^{\ell} \cdot \mathbb{P}(G_{\infty,2}(r) \simeq F). \end{equation*}

Since $i_1 \in I_{\underline{\lambda}}$ , we have $\theta^{\setminus i_1} > 0$ , and it is not difficult to see that $\mathbb{P}(G_{\infty,2}(r) \simeq F) > 0$ , therefore we are done.

Proof. By the definition of $p_I$ , the independence of the branches, and the definition of $\widetilde{N}_{\underline{s}}(r)$ (cf. Definition 3.3), we obtain

where (i) denotes the color string consisting of the color i only. Since the number of i-colored children (i.e. the number of children that are connected to r by an edge of color i) has distribution $\textrm{Poi}(\lambda_i)$ , its generating function is $\exp({-}\lambda_i(1-z))$ for any $i \in [k]$ . Thus we obtain

\begin{equation*} 1 - p_I = \exp\Bigg({-}\sum_{i \in I}\lambda_i p_{I \setminus \{i\}}\Bigg) \cdot \exp\Bigg({-}p_I\sum_{i \in [k] \setminus I}\lambda_i\Bigg). \end{equation*}

Proof. Let $G_{\infty} \,:\!=\, G_{\infty}(r) \sim \mathcal{G}_{\infty}(\underline{\lambda})$ . First, assume that $\underline{\lambda}$ is fully supercritical. Then, for any nonempty $I \subseteq [k]$ ,

\begin{align*} p_I = \mathbb{P}\Bigg(\bigcup_{i \in I}\bigg\{r \stackrel{G_{\infty}^{\setminus i}}{\longleftrightarrow} \infty\bigg\}\Bigg) \ge \mathbb{P}\Big(r \stackrel{G_{\infty}^{\setminus k}}{\longleftrightarrow} \infty\Big) > 0, \end{align*}

and hence $(p_I)_{I \subseteq [k]}$ is indeed a relevant solution.

Now assume that $\underline{\lambda}$ is not fully supercritical, i.e. there exists a color $i \in [k]$ such that $\lambda^{\setminus i} \le 1$ . Then $p_{\{i\}} = \mathbb{P}\Big(r \stackrel{G_{\infty}^{\setminus i}}{\longleftrightarrow} \infty\Big) = 0$ , and thus $(p_I)_{I \subseteq [k]}$ is not a relevant solution.

Proof. By the definition of a relevant solution, it is enough to show that

(4.7) \begin{equation} x_I = 1 - \exp\Bigg({-}\sum_{j \in I}\lambda_j p_{I \setminus \{j\}} - x_{I}\sum_{j \in [k] \setminus I}\lambda_j\Bigg) \end{equation}

has a unique solution in (0, 1), namely $p_I$ , for any nonempty $I \subseteq [k]$ . We prove this by induction on $|I|$ .

First, assume that $|I| = 1$ , i.e. $I = \{i\}$ for some $i \in [k]$ . Then we need to show that

(4.8) \begin{equation} x_{\{i\}} = 1 - \exp\Bigg({-}x_{\{i\}}\sum_{j \in [k] \setminus \{i\}}\lambda_j\Bigg). \end{equation}

Now, the left-hand side of (4.8) is linear in the variable $x_{\{ i \}}$ , while its right-hand side is exponential in $x_{\{ i \}}$ , and it is well known that such equations have at most two solutions. As we saw earlier, $x_{\{ i \}} = 0$ and $x_{\{ i \}} = p_{\{ i \}}$ are both solutions, and by Lemma 4.2, $p_{\{ i \}} > 0$ . Thus, (4.8) indeed has a unique solution in (0, 1).

Now let $I \subseteq [k]$ with $|I| \ge 2$ and assume that (4.7) holds for any $I^{\prime} \subseteq [k]$ with $|I^{\prime}| < |I|$ . Consider first the case $I \ne [k]$ . Again, the left-hand side of (4.7) is linear, while its right-hand side is exponential in $x_I$ , so there are at most two solutions. Clearly, $p_{I}$ is a solution, and by Lemma 4.2, it is in (0, 1). In addition, in $x_{I} = 0$ the left-hand side of (4.7) is clearly 0, while its right-hand side is

\begin{align*}1 - \exp\Bigg({-}\sum_{i \in I}\lambda_i p_{I \setminus \{i\}}\Bigg) > 1 - \textrm{e}^0 = 0,\end{align*}

so by the continuity of the linear and exponential functions, the other solution of (4.7) is negative.

Now consider the case $I = [k]$ . Then (4.7) becomes

\begin{equation*} x_{[k]} = 1 - \exp\Bigg({-}\sum_{i \in [k]}\lambda_i p_{[k] \setminus \{i\}}\Bigg), \end{equation*}

which clearly has a unique solution. By Lemma 4.1, this solution is $p_I$ and it is in (0, 1).

Proof of Lemma 4.4. First, note that each color string of length $h+1$ can be uniquely written as the concatenation of a color string of length h and a color not appearing in this color string. Therefore, by the tower rule,

\begin{equation*} \Phi_{h+1}\big((z_{\underline{s}})_{\underline{s} \in S_{h+1}}\big) = \Phi_{h+1}\big((z_{\underline{s}i})_{\underline{s} \in S_h, i \in [k] \setminus \textrm{set}(\underline{s})}\big) = \mathbb{E}\left(\mathbb{E}\left( \prod_{\substack{\underline{s}\in S_h,\\i\in[k]\setminus\textrm{set}(\underline{s})}}z_{\underline{s}i}^{|\widetilde{R}_{\underline{s}i}(r)|} \,\Big|\, (\widetilde{R}_{\underline{s}}(r))_{\underline{s} \in S_h}\right)\right). \end{equation*}

By the independence of the different branches of an ECBP tree,

\begin{align*} \mathbb{E}\left(\prod_{\substack{\underline{s} \in S_h, \\ i \in [k] \setminus \textrm{set}(\underline{s})}} z_{\underline{s}i}^{|\widetilde{R}_{\underline{s}i}(r)|} \,\Big|\, \big(\widetilde{R}_{\underline{s}}(r)\big)_{\underline{s}\in S_h}\right) = \prod_{\substack{\underline{s} \in S_h, \\ i \in [k] \setminus \textrm{set}(\underline{s})}} \mathbb{E}\bigg(z_{\underline{s}i}^{|\widetilde{R}_{\underline{s}i}(r)|} \mid \widetilde{R}_{\underline{s}}(r)\bigg). \end{align*}

Let $\underline{s} \in S_h$ and $i \in [k] \setminus \textrm{set}(\underline{s})$ . Using that for any $v \in \widetilde{R}_{\underline{s}}(r)$ , the function $z \mapsto \exp(\lambda_i \cdot (F_{\underline{s}i}(z) - 1))$ is the generating function of the cardinality of the set of descendants of v belonging to $\widetilde{R}_{\underline{s}i}(r)$ , we obtain

\begin{align*} \mathbb{E}\bigg(z_{\underline{s}i}^{|\widetilde{R}_{\underline{s}i}(r)|} \mid \widetilde{R}_{\underline{s}}(r)\bigg) = (\!\exp(\lambda_i \cdot (F_{\underline{s}i}(z) - 1)))^{\big|\widetilde{R}_{\underline{s}}(r)\big|}. \end{align*}

Therefore, and by the definition of the function $\Phi_h$ , we obtain

\begin{align*} \Phi_{h+1}\big((z_{\underline{s}})_{\underline{s} \in S_{h+1}}\big) & = \mathbb{E}\left(\prod_{\substack{\underline{s} \in S_h, \\ i \in [k] \setminus \textrm{set}(\underline{s})}} (\!\exp(\lambda_i \cdot (F_{\underline{s}i}(z) - 1)))^{\big|\widetilde{R}_{\underline{s}}(r)\big|}\right) \\ & = \mathbb{E}\left(\prod_{\underline{s} \in S_h}\left(\prod_{i \in [k] \setminus \textrm{set}(\underline{s})} \exp(\lambda_i \cdot (F_{\underline{s}i}(z) - 1))\right)^{\big|\widetilde{R}_{\underline{s}}(r)\big|}\right) \\ & = \Phi_h\left(\left(\prod_{i \in [k] \setminus \textrm{set}(\underline{s})} \exp(\lambda_i \cdot (F_{\underline{s}i}(z_{\underline{s}i}) - 1))\right)_{\underline{s} \in S_h}\right). \end{align*}

Proof. If $v \in V(G_\infty)$ and $i \in [k]$ , let us denote by $\widetilde{N}^*_{(i)}(v)$ the set of children of v that are connected to v by an edge of color i. Let $\mathcal{F}$ denote the sigma-algebra

\begin{align*} \mathcal{F} \,:\!=\, \sigma\left(\textbf{1}[\{u_1, u_2\} \in E_i] \colon u_1 \in \bigcup_{j=0}^{k-2}\bigcup_{\underline{s} \in S_j}\widetilde{R}_{\underline{s}}(r),\; u_2 \in V(G_\infty), \; i \in [k]\right). \end{align*}

In words: $\mathcal{F}$ contains the information that we collect if we explore the set of vertices of $G_\infty$ that can be reached by a path from the root that uses at most $k-2$ colors, and all edges emanating from this set. Assumption 2.1 implies that this set is almost surely finite.

By Lemma 3.1 and using that each offspring of any vertex in a branching process tree spans a branching process tree with the same offspring distribution, and by the definition of $\Phi_{k-2}$ , for any $L \subseteq S_{k-1}$ , we obtain

(4.10) \begin{align} \mathbb{P}\left(\bigcap_{\underline{s}^{\prime} \in L}\{|\widetilde{R}_{\underline{s}^{\prime}}(r)| < +\infty\}\right) & = \mathbb{E}\left(\mathbb{E}\left(\prod_{\underline{s} \in S_{k-2}}\prod_{v \in \widetilde{R}_{\underline{s}}(r)} \prod_{\substack{i \in [k] \setminus \textrm{set}(\underline{s}): \\ \underline{s}i \in L}}\prod_{u \in \widetilde{N}^*_{(i)}(v)} \textbf{1}\big[\big|R\big(u,G_{\infty}^{\textrm{set}(\underline{s}i)}\big)\big| < +\infty\big] \,\Big|\, \mathcal{F}\right)\right) \nonumber \\ & = \mathbb{E}\left(\prod_{\underline{s} \in S_{k-2}}\prod_{v \in \widetilde{R}_{\underline{s}}(r)} \prod_{\substack{i \in [k] \setminus \textrm{set}(\underline{s}): \\ \underline{s}i \in L}}\prod_{u \in \widetilde{N}^*_{(i)}(v)} F_{\underline{s}i}(1_-)\right) \nonumber \\ & = \mathbb{E}\left(\prod_{\underline{s} \in S_{k-2}}\prod_{v \in \widetilde{R}_{\underline{s}}(r)} \prod_{\substack{i \in [k] \setminus \textrm{set}(\underline{s}): \\ \underline{s}i \in L}} \exp(\lambda_i(F_{\underline{s}i}(1_-) - 1))\right) \nonumber \\ & = \Phi_{k-2}\left(\left(\prod_{i \in [k] \setminus \textrm{set}(\underline{s})} q_{L,\underline{s}i}\right)_{\underline{s} \in S_{k-2}}\right). \end{align}

By the definition of $\widetilde{R}_{\underline{s}}(r)$ (cf. Definition 3.3) we have

\begin{align*} f^*_\infty = \mathbb{P}\bigg(\textrm{for all}\ j \in [k] \colon r \stackrel{G_{\infty}^{\setminus j} }{\longleftrightarrow} \infty\bigg) = \mathbb{P}\left(\bigcap_{j \in [k]}\left(\bigcap_{\underline{s}^{\prime} \in S_{k-1}^{\setminus j}} \big\{\big|\widetilde{R}_{\underline{s}^{\prime}}(r)\big| < +\infty\big\}\right)^\textrm{c}\,\right). \end{align*}

Then, by the inclusion–exclusion formula, the definition of $L_J$ , and (4.10), the previous equation implies

\begin{align*} f^*_\infty & = \sum_{J \subseteq [k]}({-}1)^{|J|} \cdot \mathbb{P}\left(\bigcap_{j \in J}\bigcap_{\underline{s}^{\prime} \in S_{k-1}^{\setminus j}} \big\{\big|\widetilde{R}_{\underline{s}^{\prime}}(r)\big| < +\infty\big\}\right) \\ & = \sum_{J \subseteq [k]}({-}1)^{|J|} \cdot \mathbb{P}\left(\bigcap_{\underline{s}^{\prime} \in L_J} \big\{\big|\widetilde{R}_{\underline{s}^{\prime}}(r)\big| < +\infty\big\}\right) \\ & = \sum_{J \subseteq [k]}({-}1)^{|J|} \cdot \Phi_{k-2}\left(\left(\prod_{i \in [k] \setminus \textrm{set}(\underline{s})} q_{L_J,\underline{s}i}\right)_{\underline{s} \in S_{k-2}}\right). \end{align*}

Proof of Theorem 2.1. Let $\varepsilon \ge 0$ and let $G_{\infty}^{\varepsilon} \,:\!=\, G_{\infty}^{\varepsilon}(r) \sim \mathcal{G}_{\infty}(\underline{\lambda}(\varepsilon))$ , where

\begin{align*} \underline{\lambda}(\varepsilon) = \bigg(\frac{1+\varepsilon}{k-1},\ldots,\frac{1+\varepsilon}{k-1}\bigg) \in \mathbb{R}_+^k. \end{align*}

Since $\underline{\lambda}(\varepsilon)$ is homogeneous, $\theta^{\setminus k}(\varepsilon) \,:\!=\, \mathbb{P}\Big(r \stackrel{(G_{\infty}^{\varepsilon})^{\setminus k}}\leftrightarrow \infty\Big) = \mathbb{P}\Big(r \stackrel{{(G_{\infty}^{\varepsilon})^{\setminus i}}}\leftrightarrow \infty\Big)$ holds for any $i \in [k]$ .

Note that if $\varepsilon < {1}/({k-2})$ then Assumption 2.1 holds, which implies that $\bigcup_{j=0}^{k-2} \bigcup_{\underline{s} \in S_j}\!\widetilde{R}_{\underline{s}}(r)$ is almost surely finite (cf. Definition 3.3), and thus every i-avoiding path from r to infinity must pass through $\widetilde{N}_{k-1}^{\setminus i} (r)$ (cf. Definition 3.3); we therefore have

\begin{align*} f^*_{\infty}(\varepsilon) = \mathbb{P}\bigg(\textrm{for all}\ i \in [k]\ \textrm{there exists}\ v \in \widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big) \colon v\stackrel{{(G_{\infty}^{\varepsilon})^{\setminus i}}}\leftrightarrow \infty\bigg). \end{align*}

So, by (4.3) and the disjointness of the sets $\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)$ for all $i \in [k]$ , the previous equation implies that

(4.11) \begin{equation} f^*_{\infty}(\varepsilon) = \mathbb{E}\left(\prod_{i \in [k]}\Big(1 - (1 - \theta^{\setminus k}(\varepsilon))^{|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)|}\Big)\right). \end{equation}

Clearly, $\lambda^{\setminus k}(\varepsilon) \,:\!=\, \sum_{i \in [k-1]}\lambda_i(\varepsilon) = 1 + \varepsilon$ holds, and thus by [Reference Van der Hofstad31, Corollary 3.19] we have $\lim_{\varepsilon \to 0_+}{\theta^{\setminus k}(\varepsilon)}/{\varepsilon} = 2$ . Hence,

(4.12) \begin{equation} \lim_{\varepsilon \to 0_+}\frac{f^*_{\infty}(\varepsilon)}{\varepsilon^k} = 2^k\lim_{\varepsilon \to 0_+}\frac{f^*_{\infty}(\varepsilon)}{(\theta^{\setminus k}(\varepsilon))^k}, \end{equation}

and now we show that

(4.13) \begin{equation} \lim_{\varepsilon \to 0_+}\frac{f^*_{\infty}(\varepsilon)}{(\theta^{\setminus k}(\varepsilon))^k} = \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Bigg). \end{equation}

By Assumption 2.1, there exists $\varepsilon_0 \ge 0$ such that

(4.14) \begin{equation} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon_0}\big)\big|\Bigg) < +\infty. \end{equation}

Then, by the monotone coupling of the random variables $\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|$ for all $\varepsilon \in [0, \varepsilon_0]$ , and by Bernoulli’s inequality, we obtain

(4.15) \begin{equation} \prod_{i \in [k]}\bigg(1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|}\bigg) \le \prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big| \le \prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon_0}\big)\big|. \end{equation}

By (4.14) and (4.15), we may apply the dominated convergence theorem (using the random variable that appears on the right-hand side of (4.15) as the dominating random variable) to conclude that the first equality below holds. Then we can calculate (using L’Hospital’s rule in the last equality):

\begin{align*} & \lim_{\varepsilon \to 0_+}\mathbb{E}\left(\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|}} {\theta^{\setminus k}(\varepsilon)}\right) \\ & = \mathbb{E}\left(\lim_{\varepsilon \to 0_+}\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|}} {\theta^{\setminus k}(\varepsilon)}\right)\end{align*}

\begin{align*} & = \mathbb{E}\left(\lim_{\varepsilon \to 0_+}\left( \textbf{1}\Big[\textrm{for all}\ i \in [k] \colon \big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big| = \big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Big]\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|}} {\theta^{\setminus k}(\varepsilon)} \right.\right.\\ & \left.\left. \quad + \textbf{1}\Big[\textrm{there exists}\ i \in [k] \colon \big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big| \ne \big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Big]\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^{\varepsilon}\big)\big|}} {\theta^{\setminus k}(\varepsilon)}\right)\right)\\ & = \mathbb{E}\left(\lim_{\varepsilon \to 0_+}\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|}} {\theta^{\setminus k}(\varepsilon)}\right) \\ & = \mathbb{E}\left(\lim_{\theta^{\setminus k}(\varepsilon) \to 0_+}\prod_{i \in [k]} \frac{1 - (1 - \theta^{\setminus k}(\varepsilon))^{\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|}} {\theta^{\setminus k}(\varepsilon)}\right) = \mathbb{E}\left(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\right). \end{align*}

Thus, by (4.11), we obtain that (4.13) holds. Therefore, by (4.12), we obtain

\begin{align*} \lim_{\varepsilon \to 0_+}\frac{f^*_{\infty}(\varepsilon)}{\varepsilon^k} = 2^k\mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Bigg) \,=\!:\, C(k). \end{align*}

Now we want to express this constant C(k) using the generating function $\Phi_{k-2}$ , where we take the underlying parameter $\underline{\lambda}$ to be

\begin{align*} \underline{\lambda} = \underline{\lambda}(0) = \bigg( \frac{1}{k-1}, \ldots, \frac{1}{k-1} \bigg). \end{align*}

Since the sets $\widetilde{N}_{\underline{s}}\big(r, G_{\infty}^0\big)$ for all $\underline{s} \in S_{k-1}$ are disjoint, it follows that

\begin{equation*} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Bigg) = \mathbb{E}\left(\prod_{i \in [k]}\sum_{\underline{s} \in S_{k-1}^{\setminus i}} \big|\widetilde{N}_{\underline{s}}\big(r, G_{\infty}^0\big)\big|\right) = \sum_{(\underline{s}_i \in S_{k-1}^{\setminus i})_{i \in [k]}} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{\underline{s}_i}\big(r, G_{\infty}^0\big)\big|\Bigg). \end{equation*}

Let $\underline{s}_i \in S_{k-1}^{\setminus i}$ for all $i \in [k]$ . Since the random variables $\big|\widetilde{N}_{\underline{s}_i}\big(r, G_{\infty}^0\big)\big|$ for all $i \in [k]$ are conditionally independent given $\big(\big|\widetilde{R}_{\underline{s}}\big(r, G_{\infty}^0\big)\big|\big)_{\underline{s} \in S_{k-2}}$ , and their conditional distribution is Poisson with parameter $\big|\widetilde{R}_{\underline{s}}\big(r, G_{\infty}^0\big)\big|/(k-1)$ , we obtain

\begin{align*} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{\underline{s}_i}\big(r, G_{\infty}^0\big)\big|\Bigg) = \frac{1}{(k-1)^k}\mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{R}_{\underline{s}_i^-}\big(r, G_{\infty}^0\big)\big|\Bigg). \end{align*}

Let us define the multivariate moment-generating function

\begin{align*} \varphi\big((x_{\underline{s}})_{\underline{s} \in S_{k-2}}\big) \,:\!=\, \Phi_{k-2}\big(\big(\textrm{e}^{x_{\underline{s}}}\big)_{\underline{s} \in S_{k-2}}\big) = \mathbb{E}\Bigg(\!\exp\Bigg(\sum_{\underline{s} \in S_{k-2}}\big|\widetilde{R}_{\underline{s}}\big(r, G_{\infty}^0\big)\big| \cdot x_{\underline{s}}\Bigg)\Bigg). \end{align*}

Since the function $\varphi$ is the multivariate moment-generating function of the random vector $\big(\big|\widetilde{R}_{\underline{s}}\big(r, G_{\infty}^0\big)\big|\big)_{\underline{s} \in S_{k-2}}$ , we obtain

\begin{align*} \frac{\partial^k}{\partial x_{{\underline{s}_1^-}} \cdots \partial x_{{\underline{s}_k^-}}} \varphi\big(\big(x_{\underline{s}} \big)_{\underline{s} \in S_{k-2}}\big)\bigg|_{(x_{\underline{s}})_{\underline{s} \in S_{k-2}} = \underline{0}} = \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{R}_{\underline{s}_i^-}\big(r, G_{\infty}^0\big)\big|\Bigg). \end{align*}

Hence,

\begin{align*} C(k) & = 2^k\mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{k-1}^{\setminus i}\big(r, G_{\infty}^0\big)\big|\Bigg) \\ & = 2^k\sum_{(\underline{s}_i \in S_{k-1}^{\setminus i})_{i \in [k]}} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{N}_{\underline{s}_i}\big(r, G_{\infty}^0\big)\big|\Bigg) \\ & = \frac{2^k}{(k-1)^k}\sum_{(\underline{s}_i \in S_{k-1}^{\setminus i})_{i \in [k]}} \mathbb{E}\Bigg(\prod_{i \in [k]}\big|\widetilde{R}_{\underline{s}_i^-}\big(r, G_{\infty}^0\big)\big|\Bigg) \\ & = \frac{2^k}{(k-1)^k}\sum_{\big(\underline{s}_i \in S_{k-1}^{\setminus i}\big)_{i \in [k]}} \frac{\partial^k}{\partial x_{{\underline{s}_1^-}} \cdots \partial x_{{\underline{s}_k^-}}} \Phi_{k-2}\big((\textrm{e}^{x_{\underline{s}}})_{\underline{s} \in S_{k-2}}\big) \bigg|_{(x_{\underline{s}})_{\underline{s} \in S_{k-2}} = \underline{0}} , \end{align*}

which can be expressed explicitly using an iterated composition of the Lambert W function and elementary functions since the same is true for $\Phi_{k-2}$ by Lemma 4.4.

In particular, when $k=2$ , then $\Phi_0(\textrm{e}^x) = \textrm{e}^x$ , and thus $C(2)=4$ . When $k=3$ , by Lemma 4.4,

\begin{align*} \Phi_1\big(\textrm{e}^{x_1}, \textrm{e}^{x_2}, \textrm{e}^{x_3}\big) = \prod_{i \in [3]}\exp\bigg(\frac{1+\varepsilon}{2}(F_{(i)}(\textrm{e}^{x_i}) - 1)\bigg), \end{align*}

where, by (4.9), $F_{(i)}(z) = {-W\big({-}\frac{1}{2}\cdot\textrm{e}^{-{1}/{2}}\cdot z\big)}/{\frac{1}{2}}$ for any $i \in [3]$ . Using the identities

\begin{alignat*}{2} \frac{\textrm{d}}{\textrm{d}z} W(z) & = \frac{W(z)}{z(1 + W(z))} \quad & & \textrm{if } z \notin \{0, -1/\textrm{e}\}, \\ W(z\textrm{e}^z) & = z & & \textrm{if } z \ge 0, \end{alignat*}

we obtain $C(3)=32$ . Similarly, but with significantly more work, we obtain the value $C(4)=624$ . Recall that we obtained the same values for C(k) when $k \in \{ 2, 3, 4 \}$ using a different approach at the end of Section 4.1.