Proof of Theorem 2.1. As in the proofs of [Reference Lodewijks and Ortgiese12, Theorem 2.9] and [Reference Devroye and Lu6, Theorem 1], we first prove the convergence holds in probability, and then discuss how to improve it to almost sure convergence.

We take $\eta\in(0,1-\log(\theta_m)/(\theta_m-1))$ and write

(4.1) \begin{align} \mathbb{P}_W\!\left(\Big|\frac{\log I_n}{\log n}-\mu_m\Big|\geq \varepsilon\right) \leq {}&\mathbb{P}_W\!\left(\big\{I_n\leq n^{\mu_m-\varepsilon}\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\big\}\right)\nonumber\\[5pt] &+\mathbb{P}_W\!\left(\{I_n\geq n^{\mu_m+\varepsilon}\big\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\big\}\right)\nonumber\\[5pt] &+\mathbb{P}_W\!\left(\max_{i\in[n]}\mathcal{Z}_n(i) < (1-\eta)\log_{\theta_m}\!n\right). \end{align}

We start by dealing with the first two terms on the right-hand side, and then use Theorem 3.1 to deal with the final term. The first two probabilities can be bounded from above by

(4.2) \begin{equation} \mathbb{P}_W\!\left( \max_{i\in[n^{\mu_m-\varepsilon}]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\right)+\mathbb{P}_W\!\left(\max_{n^{\mu_m+\varepsilon}\leq i\leq n}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\right). \end{equation}

The aim is thus to show that vertices with a label ‘far away’ from $n^{\mu_m}$ are unlikely to have a high degree. With $I_n^-\;:\!=\;n^{\mu_m-\varepsilon}$ , $I_n^+\;:\!=\;n^{\mu_m+\varepsilon}$ , we first apply a union bound to obtain the upper bound

\begin{equation*} \sum_{i\in[n]\backslash [I_n^-,I_n^+]}\mathbb{P}_W\!\left(\mathcal{Z}_n(i)\geq (1-\eta)\log_{\theta_m}\!n\right). \end{equation*}

With the same approach that leads to the upper bound in (3.2), that is, using a Chernoff bound with $t=\log((1-\eta)\log_{\theta_m}\!n)-\log\!\big(mW_i\sum_{j=i}^{n-1}1/S_j\big)$ , we arrive at the upper bound

(4.3) \begin{equation}\begin{aligned} \sum_{i\in[n]\backslash [I_n^-,I_n^+]}\!\!\!\!\!\!\!{\textrm{e}}^{-t(1-\eta)\log_{\theta_m}\!\!n}\prod_{j=i}^{n-1}\bigg(1+\big({\textrm{e}}^t-1\big)\frac{W_i}{S_j}\bigg)^m\leq\sum_{i\in[n]\backslash [I_n^-,I_n^+]}\!\!\!\!\!\!{\textrm{e}}^{-(1-\eta)\log_{\theta_m}\!\!n(u_i-1-\log u_i)}, \end{aligned} \end{equation}

where

\begin{equation*} u_i\;:\!=\;\frac{mW_i}{(1-\eta)\log_{\theta_m}\!n}\sum_{j=i}^{n-1} \frac{1}{S_j}. \end{equation*}

We now set

\begin{equation*} \begin{aligned} \delta\;:\!=\;\min\bigg\{{}&\frac{1-\eta}{2\log\theta_m}\bigg(\frac{\log\theta_m}{(\theta_m-1)(1-\eta)}-1-\log\!\bigg(\frac{\log\theta_m}{(\theta_m-1)(1-\eta)}\bigg)\!\bigg),\\[5pt] &-\frac{(\theta_m-1)(1-\eta)}{2\log \theta_m}W_0\big(\!-\!\theta_m^{-1/(1-\eta)}{\textrm{e}}^{-1}\big)\bigg\}, \end{aligned} \end{equation*}

with $W_0$ the (main branch of the) W Lambert function, the inverse of $f\;:\;[\!-\!1,\infty)\to [\!-\!1/{\textrm{e}},\infty)$ , $f(x)\;:\!=\;x{\textrm{e}}^x$ . Note that, when $\varepsilon$ is sufficiently small, $\delta\in(0,\min\{\mu_m-\varepsilon,1-\mu_m-\varepsilon\})$ . We use this $\delta$ to split the union bound in (4.3) into three parts,

(4.4) \begin{align} R_1&\;:\!=\;\sum_{i=1}^{\lfloor n^{\delta}\rfloor} {\textrm{e}}^{-(1-\eta)\log_{\theta_m}\!\!n(u_i-1-\log u_i)},\nonumber\\[5pt] R_2&\;:\!=\;\sum_{i=\lceil n^{1-\delta}\rceil}^n {\textrm{e}}^{-(1-\eta)\log_{\theta_m}\!\!n(u_i-1-\log u_i)},\nonumber\\[5pt] R_3&\;:\!=\;\sum_{i\in [n^\delta, n^{1-\delta}]\backslash [I_n^-,I_n^+]}{\textrm{e}}^{-(1-\eta)\log_{\theta_m}\!\!n(u_i-1-\log u_i)}, \end{align}

and we aim to show that each of these terms converges to zero with n almost surely. For $R_1$ we use that, uniformly in $i\leq n^\delta$ , almost surely

(4.5) \begin{equation} u_i\leq \frac{m}{(1-\eta)\log_{\theta_m}\!n}\sum_{j=1}^{n-1}\frac{1}{S_j}=\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}(1+o(1)), \end{equation}

where the final step follows from Lemma 3.3. Using the fact that the upper bound is at most 1 by the choice of $\eta$ and that $x\mapsto x-1-\log x$ is decreasing on (0,1), and applying this in $R_1$ in (4.4), we bound $R_1$ from above by

(4.6) \begin{equation} \begin{aligned} \sum_{i=1}^{\lfloor n^\delta\rfloor}{}&\exp\!\bigg(\!-\!\frac{(1-\eta)\log n}{\log\theta_m}\bigg( \frac{\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg)\\[5pt] ={}&\exp\!\bigg(\!\log n\bigg(\delta-\frac{1-\eta}{\log\theta_m}\bigg( \frac{\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg), \end{aligned} \end{equation}

which converges to zero by the choice of $\delta$ . In a similar way, uniformly in $n^{1-\delta}\leq i\leq n$ , almost surely

(4.7) \begin{equation} u_i\leq \frac{m}{(1-\eta)\log_{\theta_m}\!n}\sum_{j=\lceil n^{1-\delta}\rceil }^{n-1}\frac{1}{S_j}=\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}(1+o(1)), \end{equation}

so that we can bound $R_2$ from above by

(4.8) \begin{equation}\begin{aligned} \sum_{i= \lceil n^{1-\delta}\rceil}^n\!\!\!\!\!\!{}&\exp\!\bigg(\!-\!(1-\eta)\log_{\theta_m}\! n\bigg( \frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg)\\[5pt] ={}&\exp\!\bigg(\!\log n\bigg(1-\frac{1-\eta}{\log\theta_m}\bigg( \frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg). \end{aligned}\end{equation}

Again, by the choice of $\delta$ , the exponent is strictly negative, so that the upper bound converges to zero with n. It remains to bound $R_3$ . We aim to approximate the sum by an integral, using the same approach as in the proof of [Reference Lodewijks and Ortgiese12, Theorem 2.9, bounded case]. We first bound $u_i\leq m(H_n-H_i)/((1-\eta)\log_{\theta_m}\! n)\;=\!:\;\widetilde u_i$ almost surely for any $i\in[n]$ , where $H_n\;:\!=\;\sum_{j=1}^{n-1}1/S_j$ . Then we define $u\;:\;(0,\infty)\to \mathbb{R}$ and $\phi\;:\;(0,\infty)\to\mathbb{R}$ by

\begin{equation*} u(x)\;:\!=\;\bigg(1-\frac{\log x}{\log n}\bigg)\frac{\log\theta_m}{(1-\eta)(\theta_m-1)} \quad \text{and}\quad \phi(x)\;:\!=\;x-1-\log x, \qquad x>0. \end{equation*}

For i in $[n^\delta, n^{1-\delta}]\backslash [I_n^-,I_n^+]$ such that $i=n^{\beta+o(1)}$ for some $\beta\in[\delta,1-\delta]$ (where the o(1) is independent of $\beta$ ) and $x\in[i,i+1)$ ,

(4.9) \begin{align} |\phi(\widetilde u_i)-\phi(u(x))|\leq{}& |\widetilde u_i-u(x)|+|\log(\widetilde u_i/u(x))|\nonumber\\[5pt] ={}&\Bigg|\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg(1-\frac{\log x}{\log n}\bigg)-\frac{\log\theta_m}{(1-\eta)(\theta_m-1)\log n}\sum_{j=i}^{n-1}\frac{1}{S_j}\Bigg|\nonumber\\[5pt] &+\Bigg|\log\!\Bigg(\frac{\mathbb{E}\left[W\right]}{\log n-\log x}\sum_{j=i}^{n-1}\frac{1}{S_j}\Bigg)\Bigg|. \end{align}

By (3.4) and since i diverges with n, we have $\sum_{j=i}^{n-1}1/S_j-\log(n/i)/\mathbb{E}\left[W\right]=o(1)$ almost surely as $n\to\infty$ . Applying this to the right-hand side of (4.9) yields

\begin{equation*} |\phi(\widetilde u_i)-\phi(u(x))|\leq \frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg|\frac{\log x-\log i}{\log n}\bigg|+\bigg|\log\!\bigg(1+\frac{\log x-\log i+o(1)}{\log n-\log x}\bigg)\bigg|. \end{equation*}

Since $x\geq i\geq n^\delta$ and $|x-i|\leq 1$ , we thus obtain that, uniformly in $[n^\delta, n^{1-\delta}]\backslash [I_n^-,I_n^+]$ and $x\in[i,i+1)$ , we have $|\phi(\widetilde u_i)-\phi(u(x))|=o(1/(n^\varepsilon\log n))$ almost surely as $n\to\infty$ . Applying this to $R_3$ in (4.4) yields the upper bound

(4.10) \begin{align} \sum_{i\in [n^\delta, n^{1-\delta}]\backslash [I_n^-,I_n^+]}\!\!\!\!\!\!\!\!\!\!{}&{\textrm{e}}^{-(1-\eta)\phi(\widetilde u_i)\log_{\theta_m}\! n}\nonumber\\[5pt] \leq{}& \sum_{i\in [n^\delta, n^{1-\delta}]\backslash [I_n^-,I_n^+]}\int_i^{i+1}{\textrm{e}}^{-(1-\eta)\log_{\theta_m}\! n( \phi(u(x))+|\phi(\widetilde u_i)-\phi(u(x))|)}\,\textrm{d} x\nonumber\\[5pt] \leq {}&(1+o(1))\int_{[n^\delta,n^{1-\delta}]\backslash [I_n^-I_n^+]}{\textrm{e}}^{-(1-\eta)\phi(u(x))\log_{\theta_m}\! n}\,\textrm{d} x. \end{align}

Using the variable transformation $w=\log x/\log n$ and setting $U\;:\!=\;[\delta,1-\delta]\backslash[\mu_m-\varepsilon,\mu_m+\varepsilon]$ yields

(4.11) \begin{align} (1{}&+o(1))\int_{U}\exp\!\bigg(\!-\!\log n\frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log \theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)n^w\log n \, \textrm{d} w\nonumber\\[5pt] &=(1+o(1))\int_{U}\exp\!\bigg(\!-\!\log n\bigg(\frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log \theta_m}{(1-\eta)(\theta_m-1)}\bigg)-w\bigg)+\log\log n\bigg) \, \textrm{d} w. \end{align}

We now observe that the mapping

\begin{equation*} w\mapsto \frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg) \end{equation*}

has two fixed points, namely

(4.12) \begin{align} w^{(1)}&\;:\!=\;1+\frac{(1-\eta)(\theta_m-1)}{\theta_m\log\theta_m}W_0\big(\!-\!\theta_m^{-\eta/(1-\eta)}{\textrm{e}}^{-1}\big), \nonumber\\[5pt] w^{(2)}&\;:\!=\;1+\frac{(1-\eta)(\theta_m-1)}{\theta_m\log\theta_m}W_{-1}\big(\!-\!\theta_m^{-\eta/(1-\eta)}{\textrm{e}}^{-1}\big), \end{align}

where we recall that $W_0$ is the inverse of $f\;:\;[\!-\!1,\infty)\to [\!-\!1/{\textrm{e}},\infty)$ , $f(x)=x{\textrm{e}}^x$ , also known as the main branch of the Lambert W function, and where $W_{-1}$ is the inverse of $g\;:\;(\!-\!\infty,-1]\to (\!-\!\infty,-1/{\textrm{e}}]$ , $g(x)=x{\textrm{e}}^x$ , also known as the negative branch of the Lambert W function. Moreover, we also have the inequalities

(4.13) \begin{equation} \begin{aligned} w&<\frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg), \qquad w\in\big(0,w^{(2)}\big),\ w\in(w^{(1)},1), \\[5pt] w&>\frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg), \qquad w\in\big(w^{(2)},w^{(1)}\big), \end{aligned} \end{equation}

and we claim that the following statements hold:

(4.14) \begin{equation} \forall\ \eta>0\text{ sufficiently small},\ w^{(2)}<\mu_m<w^{(1)},\quad\text{ and }\quad\; \lim_{\eta\downarrow 0}w^{(1)}=\lim_{\eta\downarrow 0}w^{(2)}=\mu_m. \end{equation}

We defer the proof of these inequalities and claims to the end. For now, let us use these properties and set $\eta$ sufficiently small so that $\mu_m-\varepsilon<w^{(2)}<\mu_m<w^{(1)}<\mu_m+\varepsilon$ , so that $U\subset [\delta,w^{(2)})\cup (w^{(1)},1-\delta]$ . If we define

\begin{equation*} \phi'_{\!\!U}\;:\!=\;\inf_{w\in U}\bigg[\frac{1-\eta}{\log \theta_m}\phi\bigg(\frac{(1-w)\log \theta_m}{(1-\eta)(\theta_m-1)}\bigg)-w\bigg], \end{equation*}

then it follows from the choice of $\eta$ , from (4.13), and from the definition of U that $\phi'_{\!\!U}>0$ , so that the integral in (4.11) can be bounded from above by

(4.15) \begin{equation} (1+o(1))\exp\!\Big(\!-\!\phi'_{\!\!U}\log n+\log\log n\Big), \end{equation}

which converges to zero with n. We have thus established that $R_1, R_2, R_3$ converge to zero almost surely as n tends to infinity. Combined, this yields that the upper bound in (4.3) converges to zero almost surely, so that, using (4.2), we find that

(4.16) \begin{align} \mathbb P_W{}&\Big(\big\{I_n\leq n^{\mu_m-\varepsilon}\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\big\}\Big)\nonumber\\[5pt] &+\mathbb{P}_W\!\left(\{I_n\geq n^{\mu_m+\varepsilon}\big\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\big\}\right)\buildrel {a.s.}\over{\longrightarrow} 0, \end{align}

We now return to (4.1). Taking the mean yields

\begin{equation*}\begin{aligned} \limsup_{n\to\infty}\mathbb{P}\!\left(\Big|\frac{\log I_n}{\log n}-\mu_m\Big|\geq \varepsilon\!\right)\leq{}& \!\limsup_{n\to\infty}\mathbb E\bigg[\mathbb{P}_W\!\left(\!\big\{I_n\leq n^{\mu_m-\varepsilon}\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\!\log_{\theta_m}\!n\big\}\!\right)\\[5pt] {}&+\mathbb{P}_W\!\left(\{I_n\geq n^{\mu_m+\varepsilon}\big\}\cap \{\max_{i\in[n]}\mathcal{Z}_n(i) \geq (1-\eta)\log_{\theta_m}\!n\big\}\right)\!\bigg]\\[5pt] &+\limsup_{n\to\infty}\mathbb{P}\!\left(\max_{i\in[n]}\mathcal{Z}_n(i)<(1-\eta)\log_{\theta_m}\! n\right). \end{aligned}\end{equation*}

Using the uniform integrability of the conditional probability (this is clearly the case as the conditional probability is bounded from above by one) combined with (4.16) implies that the first limsup on the right-hand side equals zero. The second limsup also equals zero by Theorem 3.1. Since $\varepsilon>0$ is arbitrary, this proves that $\log I_n/\log n\buildrel {\mathbb{P}}\over{\longrightarrow} \mu_m$ .

Now that we have obtained the convergence in probability of $\log I_n/\log n$ to $\mu_m$ , we strengthen it to almost sure convergence. We obtain this by constructing the following inequalities: first, for any $\varepsilon\in(0,\mu_m)$ , using the monotonicity of $\max_{i\in[n^{\mu_m-\varepsilon}]}\mathcal{Z}_n(i)$ and $\log_{\theta_m}\!n$ ,

\begin{equation*} \begin{aligned} \sup_{2^N\leq n}\!\frac{\max_{i\in[ n^{\mu_m-\varepsilon}]}\mathcal{Z}_n(i)}{\log_{\theta_m}\!n}&=\sup_{k\in\mathbb{N}}\sup_{2^{N+(k-1)}\leq n<2^{N+k}}\!\!\!\frac{\max_{i\in[ n^{\mu_m-\varepsilon}]}\mathcal{Z}_n(i)}{\log_{\theta_m}\!n}\\[5pt] &\leq \sup_{N\leq n}\frac{\max_{i\in[2^{(n+1)(\mu_m-\varepsilon)}]}\mathcal{Z}_{2^{n+1}}(i)}{n\log_{\theta_m}\!2}. \end{aligned} \end{equation*}

With only a minor modification, we can obtain a similar result for $\max_{n^{\mu_m+\varepsilon}\leq i\leq n}\mathcal{Z}_n(i)$ , where now $\varepsilon\in(0,1-\mu_m)$ . Here, we can no longer use that this maximum is monotone. Rather, we write

\begin{equation*} \begin{aligned} \sup_{2^N\leq n}\!\frac{\max_{ n^{\mu_m+\varepsilon}\leq i\leq n}\mathcal{Z}_n(i)}{\log_{\theta_m}\!n}&=\sup_{k\in\mathbb{N}}\sup_{2^{N+(k-1)}\leq n<2^{N+k}}\!\!\!\frac{\max_{n^{\mu_m+\varepsilon}\leq i\leq n}\mathcal{Z}_n(i)}{\log_{\theta_m}\!n}\\[5pt] &\leq \sup_{k\in\mathbb{N}}\frac{\max_{2^{(N+(k-1))(\mu_m+\varepsilon)}\leq i\leq 2^{N+k}}\mathcal{Z}_{2^{N+k}}(i)}{(N+(k-1))\log_{\theta_m}\!2}\\[5pt] &=\sup_{N\leq n}\frac{\max_{2^{n(\mu_m+\varepsilon)}\leq i\leq 2^{n+1}}\mathcal{Z}_{2^{n+1}}(i)}{n\log_{\theta_m}\!2}. \end{aligned} \end{equation*}

It thus follows that for any $\eta>0$ , the inequalities

(4.17) \begin{equation} \limsup_{n\to\infty} \frac{\max_{i\in[n^{\mu_m-\varepsilon}]}\mathcal{Z}_n(i)}{(1-\eta)\log_{\theta_m}\!n}\leq 1,\quad\limsup_{n\to\infty}\frac{\max_{n^{\mu_m+\varepsilon}\leq i\leq n}\mathcal{Z}_n(i)}{(1-\eta)\log_{\theta_m}\!n}\leq 1, \quad \mathbb P_W\text{-a.s.,} \end{equation}

are implied by

(4.18) \begin{align} \limsup_{n\to\infty}\frac{\max_{i\in[2^{(n+1)(\mu_m-\varepsilon)}]}\mathcal{Z}_{2^{n+1}}(i)}{(1-\eta) n\log_{\theta_m}\!2}&\leq 1,\quad &\mathbb P_W-\text{a.s.},&\nonumber\\[5pt] \limsup_{n\to\infty} \frac{\max_{2^{n(\mu_m+\varepsilon)}\leq i\leq 2^{n+1}}\mathcal{Z}_{2^{n+1}}(i)}{(1-\eta) n\log_{\theta_m}\!2}&\leq 1,\quad &\mathbb P_W-\text{a.s.},& \end{align}

respectively. We start by proving the first inequality in (4.18). Define

\begin{equation*} \begin{aligned} \mathcal E_n^1&\;:\!=\;\bigg\{\max_{i\in[2^{(n+1)(\mu_m-\varepsilon)}]}\mathcal{Z}_{2^{n+1}}(i)> (1-\eta)n\log_{\theta_m}\!2\bigg\}, \\[5pt] \mathcal E_n^2&\;:\!=\;\bigg\{\max_{2^{n(\mu_m+\varepsilon)}\leq i\leq 2^{n+1}}\mathcal{Z}_{2^{n+1}}(i)>(1-\eta)n\log_{\theta_m}\! 2\bigg\}. \end{aligned}\end{equation*}

Let us abuse notation to write $I_n^-=2^{(n+1)(\mu_m-\varepsilon)}$ , $I_n^+=2^{n(\mu_m+\varepsilon)}$ . By a union bound, we again find

(4.19) \begin{align} \mathbb{P}\left(\mathcal E_n^1\cup \mathcal E_n^2\right)\leq{}& \sum_{i=1}^{\lfloor 2^{(n+1)\delta}\rfloor }\mathbb{P}\left(\mathcal{Z}_{2^{n+1}}(i)>(1-\eta)n\log_{\theta_m}\!2\right)\nonumber \\[5pt] &+\sum_{i=\lceil 2^{(n+1)(1-\delta)}\rceil }^{2^{n+1}}\mathbb{P}\left(\mathcal{Z}_{2^{n+1}}(i)>(1-\eta)n\log_{\theta_m}\!2\right)\nonumber \\[5pt] &+\sum_{i\in[ 2^{(n+1)\delta},2^{(n+1)(1-\delta)}]\backslash [I_n^-,I_n^+]}\mathbb{P}\left(\mathcal{Z}_{2^{n+1}}(i)>(1-\eta)n \log_{\theta_m}\! 2\right), \end{align}

and these three sums are the equivalents of $R_1,R_2,R_3$ in (4.4). We again take $\eta$ small enough so that $\mu_m-\varepsilon<w^{(2)}<\mu_m<w^{(1)}<\mu_m+\varepsilon$ , where we recall $w^{(1)}, w^{(2)}$ from (4.12). With the same steps as in (4.3), (4.5), and (4.6), we obtain that we can almost surely bound the first sum on the right-hand side from above by

\begin{equation*}\begin{aligned} \sum_{i=1}^{\lfloor 2^{(n+1)\delta}\rfloor}{}&\exp\!\bigg(\!-\!\frac{(1-\eta)n\log 2}{\log\theta_m}\bigg( \frac{\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg)\\[5pt] ={}&\exp\!\bigg(n\log 2\bigg(\delta-\frac{1-\eta}{\log\theta_m}\bigg( \frac{\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\!\bigg), \end{aligned} \end{equation*}

which is summable by the choice of $\delta$ . Similarly, using the same steps as in (4.7) and (4.8), we can almost surely bound the second sum on the right-hand side of (4.19) from above by

\begin{equation*}\begin{aligned} \sum_{i= \lceil 2^{(n+1)(1-\delta)}\rceil}^{2^n}\!\!\!\!\!\!{}&\exp\!\bigg(\!-\!\frac{(1-\eta)n\log 2}{\log \theta_m}\bigg( \frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\bigg)\\[5pt] ={}&\exp\!\bigg(\!n\log 2\bigg(\!1-\frac{1-\eta}{\log\theta_m}\bigg(\!\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}-1-\log\!\bigg(\!\frac{\delta\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)(1+o(1))\!\bigg), \end{aligned}\end{equation*}

which again is summable by the choice of $\delta$ . Finally, the last sum on the right-hand side of (4.19) can be approximated by an integral, as in (4.10). By the choice of $\eta$ , we can then use the same steps as in (4.11)–(4.15) to obtain the almost sure upper bound

\begin{equation*} (1+o(1))\exp\!\Big(\!-\!n\phi'_{\!\!U}\log 2(1+o(1))+\log n+\mathcal O(1)\Big), \end{equation*}

which again is summable. As a result, $\mathbb P_W$ -almost surely, $\mathcal E_n^1\cup \mathcal E_n^2$ occurs only finitely often by the Borel–Cantelli lemma. This implies that both bounds in (4.18) hold, and these imply the bounds in (4.17). Defining the events

\begin{equation*} \begin{aligned} \mathcal C^1_n&\;:\!=\;\{|\log I_n/\log n-\mu_m|\geq \varepsilon\}, \quad \mathcal C^2_n\;:\!=\;\big\{I_n\leq n^{\mu_m-\varepsilon}\big\}, \quad \mathcal C^3_n\;:\!=\;\big\{I_n\geq n^{\mu_m+\varepsilon}\big\},\\[5pt] \mathcal C^4_n&\;:\!=\;\big\{\max_{i\in[n]}\mathcal{Z}_n(i)> (1-\eta)\log_{\theta_m}\!n\big\}, \end{aligned}\end{equation*}

we can use the same approach as in (4.1) to bound

\begin{equation*} \sum_{n=1}^\infty \mathbb{1}_{\mathcal C^1_n}\leq \sum_{n=1}^\infty \mathbb{1}_{\mathcal C^2_n\cap \mathcal C^4_n}+\mathbb{1}_{\mathcal C^3_n\cap\mathcal C^4_n}+\mathbb{1}_{(\mathcal C^4_n)^c}. \end{equation*}

By the proof of Theorem 3.1 given in [Reference Lodewijks and Ortgiese12], $(\mathcal C^4_n)^c$ occurs for finitely many n $\mathbb P_W$ -almost surely (not just $\mathbb P$ -almost surely, as follows directly from Theorem 3.1). The bounds in (4.17) imply that, $\mathbb P_W$ -almost surely, the events $\mathcal C^2_n\cap\mathcal C^4_n$ and $\mathcal C^3_n\cap \mathcal C^4_n$ occur for only finitely many n, via similar reasoning as in (4.2). Combining these statements, we obtain that $\mathcal C^1_n$ occurs only finitely many times $\mathbb P_W$ -almost surely. As a final step we write

\begin{equation*} \begin{aligned} \mathbb P({}&\forall\, \varepsilon>0\ \exists\, N\in \mathbb{N}\;:\; \forall\, n\geq N\ |\log I_n/\log n-\mu_m|<\varepsilon)\\[5pt] &=\mathbb{E}\left[\mathbb{P}_W\!\left(\forall\, \varepsilon>0\, \exists\, N\in \mathbb{N}\;:\; \forall\, n\geq N |\log I_n/\log n-\mu_m|<\varepsilon\right)\right]=1, \end{aligned} \end{equation*}

so that $\log I_n/\log n\xrightarrow{\mathbb P\text{-a.s.}} \mu_m$ .

It remains to prove the inequalities in (4.13) and the claims in (4.14). Let us start with the inequalities in (4.13). We compute

\begin{equation*} \frac{\textrm{d} }{\textrm{d} w}\bigg(w-\frac{1-\eta}{\log\theta_m}\phi\bigg(\frac{(1-w)\log\theta_m}{(1-\eta)(\theta_m-1)}\bigg)\!\bigg)=1+\frac{1}{\theta_m-1}-\frac{1-\eta}{\log \theta_m}\frac{1}{1-w}, \end{equation*}

which equals zero when $w=w^*\;:\!=\;1-(1-\eta)(\theta_m-1)/(\theta_m\log\theta_m)$ , is positive when $w\in(0,w^*)$ , and is negative when $w\in(w^*,1)$ . Moreover, as $W_0(x)\geq -1$ for all $x\in[\!-\!1/{\textrm{e}},\infty)$ and $ W_{-1}(x)\leq -1$ for all $x\in[\!-\!1/{\textrm{e}},0)$ , it follows from the definition of $w^{(1)}$ and $w^{(2)}$ in (4.12) that $w^{(2)}<w^*<w^{(1)}$ for any choice of $\eta>0$ . This implies both inequalities in (4.13).

We now prove the claims in (4.14). Again using that $W_0(x)\geq -1$ for all $x\in[\!-\!1/{\textrm{e}},\infty)$ directly yields $w^{(1)}>\mu_m$ . The inequality $w^{(2)}<\mu_m$ is implied by

\begin{equation*} W_{-1}\big(\!-\!\theta_m^{-\eta/(1-\eta)}{\textrm{e}}^{-1}\big)<-\frac{1}{1-\eta}, \end{equation*}

or, equivalently,

\begin{equation*} -\theta_m^{-\eta/(1-\eta)}{\textrm{e}}^{-1}>-\frac{1}{1-\eta}{\textrm{e}}^{-1/(1-\eta)}. \end{equation*}

Setting $\beta\;:\!=\;1/(1-\eta)$ yields

\begin{equation*} \frac{\theta_m}{{\textrm{e}}}<\beta \bigg(\frac{\theta_m}{{\textrm{e}}}\bigg)^{\beta}. \end{equation*}

This inequality is then satisfied when $\beta\in(1,W_{-1}(\!\log(\theta_m/{\textrm{e}})\theta_m/{\textrm{e}})/\log(\theta_m/{\textrm{e}}))$ , or, equivalently, when $\eta\in(0,1-\log(\theta_m/{\textrm{e}})/W_{-1}(\!\log(\theta_m/{\textrm{e}})\theta_m/{\textrm{e}}))$ , as required. By the definition of $w^{(1)}$ and $w^{(2)}$ in (4.12) and since $\mu_m\;:\!=\;1-(\theta_m-1)/(\theta_m\log \theta_m)$ , the second claim in (4.14) directly follows from the continuity of $W_0$ and $W_{-1}$ and since $W_0(\!-\!1/{\textrm{e}})=W_{-1}(\!-\!1/{\textrm{e}})=-1$ , which concludes the proof.

Proof of Lemma 5.1 subject to Proposition 5.1. Fix $\varepsilon\in(0\vee (c(1-\theta^{-1})-(1-\mu)),\mu)$ . We note that such an $\varepsilon$ exists, since $c<\theta/(\theta-1)$ . We start with the first implication. By Theorem 2.1 and a union bound we have

(6.1)

where $v_1$ is a vertex selected uniformly at random from [n]. We now apply Proposition 5.1 with $k=1$ , $d_1=d_n$ , $\ell_1=n^{\mu-\varepsilon}$ (we observe that, since $\varepsilon<\mu$ and by the bound on $d_n$ , the conditions in Proposition 5.1 for $\ell_1$ and $d_1$ are satisfied) to obtain the upper bound

\begin{align*} &\mathbb{P}\left(\max_{i\in[n]}\mathcal{Z}_n(i)\geq d_n\right)\leq n\mathbb{E}\left[\Bigg(\frac{W}{\theta-1+W}\Bigg)^{d_n} \mathbb{P}_W\!\left(X\leq \Bigg(1+\frac{W}{\theta-1}\Bigg)\log (n^{1-\mu+\varepsilon})\right)\right]\\[5pt] & (1+o(1))+o(1), \end{align*}

where $X\sim\text{Gamma}(d+1,1)$ . We can simply bound the conditional probability from above by one, so that the assumption yields the desired implication.

For the second implication, we use the Chung–Erdös inequality. If we let $v_1,v_2$ be two vertices selected uniformly at random without replacement from [n] and set $A_{i,n}\;:\!=\;\{\mathcal{Z}_n(i)\geq d_n\}$ , then

(6.2) \begin{equation} \mathbb{P}\left({\max_{i\in[n]}\mathcal{Z}_n(i)\geq d_n}\right)=\mathbb{P}\left({\cup_{i=1}^n A_{i,n}}\right)\geq \mathbb{P}\left({\cup_{i=\lceil n^{\mu-\varepsilon}\rceil}^n A_{i,n}}\right)\geq \frac{\big(\sum_{i=\lceil n^{\mu-\varepsilon}\rceil}^n \mathbb{P}\left(A_{i,n}\right)\big)^2}{\sum_{i,j= \lceil n^{\mu-\varepsilon}\rceil}^n \mathbb{P}\left(A_{i,n}\cap A_{j,n}\right)}. \end{equation}

As in (6.1), we can write the numerator as $(n\mathbb{P}(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}))^2$ . The denominator can be written as

\begin{equation*} \begin{aligned} \sum_{\substack{i,j= \lceil n^{\mu-\varepsilon}\rceil\\ i\neq j}}^n \mathbb{P}\left(A_{i,n}\cap A_{j,n}\right)+\sum_{i=\lceil n^{\mu-\varepsilon}\rceil}^n \mathbb{P}(A_{i,n})={}&n(n-1)\mathbb{P}\left(\mathcal{Z}_n(v_i)\geq d_n, v_i\geq n^{\mu-\varepsilon}, i\in\{1,2\}\right)\\[5pt] &+n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right). \end{aligned} \end{equation*}

By applying Proposition 5.1 to the right-hand side, we find that it equals

\begin{equation*} \left(n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right)\right)^2(1+o(1))+n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right). \end{equation*}

It follows that the right-hand side of (6.2) equals

\begin{equation*} \frac{n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right)}{n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right)(1+o(1))+1}. \end{equation*}

It thus suffices to prove that the implication

(6.3) \begin{equation} \lim_{n\to\infty}n\mathbb{E}\left[\Bigg(\frac{W}{\theta-1+W}\Bigg)^{d_n}\right]=\infty\quad \Rightarrow \quad \lim_{n\to\infty}n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right)=\infty \end{equation}

holds to conclude the proof. Again using Proposition 5.1, we have that

\begin{align*} &\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right) \\[5pt] & \geq \mathbb{E}\left[\Bigg(\frac{W}{\theta-1+W}\Bigg)^{d_n}\mathbb{P}_W\!\left(\widetilde X\leq \Bigg(1+\frac{W}{\theta-1}\Bigg)\log (n^{1-\mu+\varepsilon})\right)\right](1+o(1)), \end{align*}

where $\widetilde X\sim\text{Gamma}(d+\lfloor d^{1/4}\rfloor+1,1)$ . Hence, it follows from Lemma A.2 in the appendix and the choice of $\varepsilon$ that

\begin{equation*} n\mathbb{P}\left(\mathcal{Z}_n(v_1)\geq d_n, v_1\geq n^{\mu-\varepsilon}\right)\geq n\mathbb{E}\left[\Bigg(\frac{W}{\theta-1+W}\Bigg)^{d_n}\right](1-o(1)), \end{equation*}

which implies (6.3) as desired and concludes the proof.

Proof of Proposition 5.2 subject to Proposition 5.1. Recall that $c\in(0,\theta/(\theta-1))$ ; that $\mu=1-(\theta-1)/(\theta\log \theta)$ , $\sigma^2=1-(\theta-1)^2/(\theta^2\log \theta)$ ; and that we have a non-decreasing integer sequence $(j_k)_{k\in[K]}$ with $K'=\min\{k\;:\; j_{k+1}=j_K\}$ such that $j_1+\log_\theta n =\omega(1)$ , $j_K+\log_\theta n<c\log n$ , as well as a sequence $(B_k)_{k\in[K]}$ such that $B_k\in \mathcal {B}(\mathbb{R})$ and $B_k\cap B_\ell=\varnothing$ when $j_k=j_\ell$ and $k\neq \ell$ . Then let $(c_k)_{k\in[K]}\in \mathbb{N}_0^K$ and set $M\;:\!=\;\sum_{k=1}^K c_k$ and $M'\;:\!=\;\sum_{k=1}^{K'}c_k$ .

We define $\bar d=(d_i)_{i\in[M]}\in \mathbb{Z}^M$ and $\bar A=(A_i)_{i\in[M]}\subset \mathcal {B}(\mathbb{R})^M$ as follows. For each $i\in[M]$ , find the unique $k\in[K]$ such that $\sum_{\ell=1}^{k-1}c_\ell<i\leq \sum_{\ell=1}^k c_\ell$ , and set $d_i\;:\!=\;\lfloor \log_\theta n\rfloor +j_k, A_i\;:\!=\;B_k$ . We note that this construction implies that the first $c_1$ -many $d_i$ and $A_i$ equal $\lfloor \log_\theta n\rfloor +j_1$ and $B_1$ , respectively, that the next $c_2$ -many $d_i$ and $A_i$ equal $\lfloor \log_\theta n\rfloor +j_2$ and $B_2$ , respectively, etc. Moreover, we let $(v_i)_{i\in[M]}$ be M vertices selected uniformly at random without replacement from [n]. We then define the events

\begin{equation*} \begin{aligned} \mathcal {L}_{\bar A,\bar d}&\;:\!=\;\Bigg\{ \frac{\log v_i-(\!\log n-(1-\theta^{-1})d_i)}{\sqrt{(1-\theta^{-1})^2 d_i}}\in A_i, i\in[M]\bigg\},\\[5pt] \mathcal {D}_{\bar d}(M',M)&\;:\!=\;\{\mathcal{Z}_n(v_i)=d_i, i\in[M'], \mathcal{Z}_n(v_j)\geq d_j, M'<j\leq M\},\\[5pt] {\mathcal{E}}_{\bar d}(S)&\;:\!=\;\{\mathcal{Z}_n(v_i)\geq d_i+\mathbb{1}_{\{i\in S\}}, i\in [M]\}. \end{aligned} \end{equation*}

We know from [Reference Addario-Berry and Eslava1, Lemma 5.1] that by the inclusion–exclusion principle,

\begin{equation*} \mathbb{P}\left(\mathcal {D}_{\bar d}(M',M)\right)=\sum_{j=0}^{M'}\sum_{\substack{ S\subseteq [M']:\\ |S|=j}}(\!-\!1)^j\mathbb{P}\left({\mathcal{E}}_{\bar d}(S)\right), \end{equation*}

so that intersecting the event $\mathcal {L}_{\bar A.\bar d}$ in the probabilities on both sides yields

(6.4) \begin{equation} \mathbb{P}\left(\mathcal {D}_{\bar d}(M',M)\cap \mathcal {L}_{\bar A,\bar d}\right)=\sum_{j=0}^{M'}\sum_{\substack{ S\subseteq [M']:\\ |S|=j}}(\!-\!1)^j\mathbb{P}\left({\mathcal{E}}_{\bar d}(S)\cap \mathcal {L}_{\bar A,\bar d}\right). \end{equation}

We define $\ell_d\;:\;\mathbb{R}\to (0,\infty)$ by $\ell_d(x)\;:\!=\; \exp\!\big(\!\log n-(1-\theta^{-1})d+x\sqrt{(1-\theta^{-1})^2d}\big)$ , $x\in \mathbb{R}$ , abuse this notation to also write $\ell_d( A)\;:\!=\;\{\ell_d(x)\;:\; x\in A\}$ for $A\subseteq \mathbb{R}$ , and note that $\mathcal {L}_{\bar A,\bar d}=\{v_i\in \ell_{d_i}(A_i), i\in[M]\}$ . We also observe, since $d_i$ diverges with n for all $i\in[M]$ , that $\ell_{d_i+\mathbb{1}_{\{i\in S\}}}(x)=\ell_{d_i}(x(1+o(1)))$ for any $i\in[M]$ and $x\in\mathbb{R}$ . This can be extended to the sets $(A_i)_{i\in[M]}$ rather than $x\in\mathbb{R}$ as well. As a result, we can use Corollary A.1 in the appendix (with the observations made in Remark A.1) to then obtain

\begin{align*} \mathbb{P}\!\left({\mathcal{E}}_{\bar d}(S)\cap \mathcal {L}_{\bar A,\bar d}(S)\right)&=(1+o(1))\prod_{i=1}^M q_0\theta^{-(d_i+\mathbb{1}_{\{i\in S\}})}\Phi(A_i)\\[5pt] &=(1+o(1))q_0^M\theta^{-|S|-\sum_{i=1}^M d_i}\prod_{i=1}^M \Phi(A_i). \end{align*}

Using this in (6.4) we arrive at

(6.5) \begin{equation}\begin{aligned} \mathbb{P}\left(\mathcal {D}_{\bar d}(M',M)\cap \mathcal {L}_{\bar A,\bar d}\right)&=(1+o(1))q_0^M \theta^{-\sum_{i=1}^M d_i}\prod_{i=1}^M\Phi(A_i)\sum_{j=0}^{M'}\sum_{\substack{ S\subseteq [M']:\\ |S|=j}}(\!-\!1)^j \theta^{-j}\\[5pt] &=(1+o(1))q_0^M \theta^{-\sum_{i=1}^M d_i}(1-\theta^{-1})^{M'}\prod_{i=1}^M\Phi(A_i), \end{aligned} \end{equation}

where the $1+o(1)$ and the product on the right-hand side are independent of S and j and can therefore be taken out of the double sum. Now, recall the definition of the variables $X_j^{(n)}(B)$ , $X_{\geq j}^{(n)}(B)$ as in (5.2). Combining (6.4) and (6.5), we arrive at

(6.6) \begin{equation}\begin{aligned} \mathbb E\Bigg[\prod_{k=1}^{K'}\Big(X_{j_k}^{(n)}(B_k)\Big)_{c_k}\prod_{k=K'+1}^K \Big(X_{\geq j_k}^{(n)}(B_k)\Big)_{c_k}\Bigg] &=(n)_M \mathbb{P}\left( \mathcal {D}_{\bar d}(M',M)\cap \mathcal {L}_{\bar A.\bar d} \right)\\[5pt] &\sim q_0^M \theta^{M\log_\theta n-\sum_{i=1}^M d_i}(1-\theta^{-1})^{M'}\prod_{i=1}^M\Phi(A_i), \end{aligned}\end{equation}

since $(n)_M\;:\!=\;n(n-1)\cdots (n-(M-1))=(1+o(1))n^M$ , and where we recall that $a_n\sim b_n$ denotes $\lim_{n\to\infty}a_n/b_n=1$ . We now recall that there are exactly $c_k$ -many $d_i$ and $A_i$ that equal $\lfloor \log_2n\rfloor +j_k$ and $B_k$ , respectively, for each $k\in[K]$ , and that $j_{K'+1}=\ldots =j_K$ , so that

\begin{equation*} \begin{aligned} \prod_{i=1}^M \Phi(A_i)&=\prod_{k=1}^K \Phi(B_k)^{c_k},\\[5pt] M\log_\theta n-M'-\sum_{i=1}^M d_i &= -\sum_{k=1}^{K'}(j_k+1-\varepsilon_n)c_k-\sum_{k=K'+1}^K (j_K-\varepsilon_n)c_k, \end{aligned} \end{equation*}

which, combined with (6.6), finally yields

\begin{equation*} \begin{aligned} \mathbb E\Bigg[\!\prod_{k=1}^{K'}\!\bigg(\!X_{j_k}^{(n)}(B_k)\!\bigg)_{c_k}\prod_{k=K'+1}^K\!\! \bigg(\!X_{\geq j_k}^{(n)}(B_k)\!\bigg)_{c_k}\!\Bigg]\!={}&(1+o(1))\prod_{k=1}^{K'} \big(q_0(1-\theta^{-1})\theta^{-j_k+\varepsilon_n}\Phi(B_k)\big)^{c_k}\\[5pt] &\times \prod_{k=K'+1}^K \big(q_0\theta^{-j_K+\varepsilon_n}\Phi(B_k)\big)^{c_k}. \end{aligned} \end{equation*}

To prove the second result, we observe that for $j_1,\ldots, j_K=o(\sqrt{\log n})$ ,

\begin{equation*} \frac{\log v_i-(\!\log n-(1-\theta^{-1})d_i)}{\sqrt{(1-\theta^{-1})^2 d_i}}=\frac{\log v_i-\mu\log n}{\sqrt{(1-\sigma^2)\log n}}(1+o(1))+o(1). \end{equation*}

Hence, the same steps as above can be applied to the random variables $\widetilde X^{(n)}_j(B)$ , $\widetilde X^{(n)}_{\geq j}(B)$ to obtain the desired result.

Proof of Proposition 5.1, Equation (5.4), upper bound. We assume without loss of generality that $\ell_1, \ldots, \ell_k$ are integer-valued. (If they are not, we can use $\lceil \ell_1\rceil, \ldots, \lceil \ell_k\rceil$ , which yields the same result.) By first conditioning on the value of $v_1, \ldots, v_k$ , we obtain

\begin{equation*} \mathbb{P}\left(\mathcal{Z}_n(v_i)= d_i, v_i>\ell_i, i\in[k]\right)=\frac{1}{(n)_k}\sum_{j_1=\ell_1+1}^n \sum_{\substack{j_2=\ell_2+1\\ j_2\neq j_1}}^n \cdots \sum_{\substack{j_k=\ell_k+1\\ j_k\neq j_{k-1}, \ldots, j_1}}^n \mathbb{P}\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right). \end{equation*}

If we let $\mathcal {P}_k$ be the set of all permutations on [k], we can write the sums on the right-hand side as

(6.7) \begin{equation} \frac{1}{(n)_k}\sum_{\pi\in\mathcal {P}_k}\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n \sum_{j_{\pi(2)}=(\ell_{\pi(2)}\vee j_{\pi(1)})+1}^n \cdots \sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n \mathbb{P}\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right). \end{equation}

To prove an upper bound for this expression, we first consider the identity permutation, i.e. $\pi(i)=i$ for all $i\in[k]$ , and take

(6.8) \begin{equation} \frac{1}{(n)_k}\sum_{j_1=\ell_1}^n \sum_{j_2=(\ell_2\vee j_1)+1}^n \cdots \sum_{j_k=(\ell_k\vee j_{k-1})+1}^n \mathbb{P}\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right). \end{equation}

One can think of this as all realisations $v_i=j_i$ , $i\in[k]$ , where $j_1<j_2<\ldots <j_k$ and $j_i>\ell_i$ for all $i\in[k]$ . (Later on we will discuss what changes when one uses other choices of $\pi\in\mathcal {P}_k$ in (6.7).) Let us introduce the event

(6.9) \begin{equation} E_n^{(1)}\;:\!=\;\Bigg\{ \sum_{\ell=1}^j W_\ell \in ((1-\zeta_n)\mathbb{E}\left[W\right]j,(1+\zeta_{n})\mathbb{E}\left[W\right]j),\ \forall \ n^\eta\leq j\leq n\Bigg\}, \end{equation}

where $\zeta_n=n^{-\delta\eta}/\mathbb{E}\left[W\right]$ for some $\delta\in(0,1/2)$ and where we recall that $n^\eta$ is a lower bound for all $\ell_i$ , $i\in[k]$ , with $\eta\in(0,1)$ . It follows from Lemma 3.1 that $\mathbb P((E_n^{(1)})^c)=o(n^{-\gamma})$ for any $\gamma>0$ . We can hence bound (6.8) from above, for any $\gamma>0$ , by

(6.10) \begin{equation} \begin{aligned} \frac{1}{(n)_k}\sum_{j_1=\ell_1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\!\!\!\!\!\!\!\!\mathbb E[\mathbb{P}_W\!\left(\mathcal{Z}_n(j_\ell)=m_\ell,\ell\in[k]\right)\mathbb{1}_{E_n^{(1)}}]+o(n^{-\gamma}). \end{aligned} \end{equation}

Now, to express the first term in (6.10) we introduce the ordered indices $j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n$ , $i\in[k]$ , which denote the steps at which vertex $j_i$ increases its degree by one. Note that for every $i\in[k]$ these indices are distinct by definition, but we also require that $m_{s,i}\neq m_{t,h}$ for any distinct $i,h\in[k]$ , $s\in[d_i]$ , $t\in[d_h]$ (equality is allowed only when $i=h$ and $s=t$ ). We denote this constraint by adding an asterisk $*$ to the summation symbol. If we also define $j_{k+1}\;:\!=\;n$ , we can write the first term in (6.10) as

\begin{equation*}\begin{aligned} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{(n)_k}{}&\sum_{j_1=\ell_1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\mathbb E\!\left[\prod_{t=1}^k\prod_{s=1}^{d_t}\frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell}\phantom{\times \prod_{u=1}^k\!\!\prod_{\substack{s=j_u+1\\ s\neq m_{i,t},t\in[d_i],i\in[k]}}^{j_{u+1}}\!\!\!\!\bigg(1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{\sum_{\ell=1}^{s-1}W_{\ell}}\bigg)\mathbb{1}_{E_n^{(1)}}} \right.\\[5pt] & \left. \times \prod_{u=1}^k\!\!\prod_{\substack{s=j_u+1\\ s\neq m_{i,t},t\in[d_i],i\in[k]}}^{j_{u+1}}\!\!\!\!\Bigg(1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{\sum_{\ell=1}^{s-1}W_{\ell}}\Bigg)\mathbb{1}_{E_n^{(1)}}\right]\!. \end{aligned} \end{equation*}

We then include the terms where $s=m_{i,t}$ for $i\in[d_t]$ , $t\in[k]$ in the second double product. To do this, we need to change the first double product to

\begin{equation*} \prod_{t=1}^k \prod_{s=1}^{d_t}\frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell-\sum_{\ell=1}^k W_{j_\ell}\mathbb{1}_{\{m_{s,t}>j_\ell\}}}\leq\prod_{t=1}^k \prod_{s=1}^{d_t} \frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell-k}; \end{equation*}

that is, we subtract the vertex-weight $W_{j_\ell}$ in the numerator when the vertex $j_\ell$ has already been introduced by step $m_{s,t}$ . In the upper bound we use that the weights are bounded from above by one. We thus arrive at the upper bound

(6.11) \begin{equation}\begin{aligned} \frac{1}{(n)_k}\sum_{j_1=\ell_1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\mathbb E\Bigg[{}&\prod_{t=1}^k\prod_{s=1}^{d_t}\frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell-k}\\[5pt] &\times \prod_{u=1}^k\prod_{s=j_u+1}^{j_{u+1}}\Bigg(1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{\sum_{\ell=1}^{s-1}W_\ell}\Bigg)\mathbb{1}_{E_n^{(1)}}\Bigg]. \end{aligned} \end{equation}

For ease of writing, for now we consider only the inner sum, until we actually intend to sum over the indices $j_1,\ldots,j_k$ later on in (6.16). We use the bounds from the event $E_n^{(1)}$ defined in (6.9) to bound

\begin{equation*} \sum_{\ell=1}^{m_{s,t}-1}W_\ell\geq (m_{s,t}-1)\mathbb{E}\left[W\right](1-\zeta_n),\qquad \sum_{\ell=1}^{s-1}W_\ell \leq s\mathbb{E}\left[W\right] (1+\zeta_n). \end{equation*}

For n sufficiently large, we observe that $(m_{s,t}-1)\mathbb{E}\left[W\right](1-\zeta_n)-k\geq m_{s,t}\mathbb{E}\left[W\right](1-2\zeta_n)$ , which yields the upper bound

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\!\mathbb E\Bigg[\prod_{t=1}^k\prod_{s=1}^{d_t}\frac{W_{j_t}}{m_{s,t}\mathbb{E}\left[W\right] (1-2\zeta_n)} \prod_{u=1}^k\prod_{s=j_u+1}^{j_{u+1}}\!\!\bigg(1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{s\mathbb{E}\left[W\right](1+\zeta_n)}\bigg)\mathbb{1}_{E_n^{(1)}}\Bigg]. \end{aligned} \end{equation*}

We can now bound the indicator from above by one. Moreover, relabelling the vertex-weights $W_{j_t}$ to $W_t$ for $t\in[k]$ does not change the distribution of the terms within the expected value, so that the expected value remains unchanged. We thus arrive at the upper bound

(6.12)

We bound the final product from above by

(6.13) \begin{equation} \begin{aligned} \prod_{s=j_u+1}^{j_{u+1}}\bigg(1-\frac{\sum_{\ell=1}^u W_\ell}{s\mathbb{E}\left[W\right](1+\zeta_n)}\bigg)&\leq \exp\!\bigg(\!-\!\frac{1}{\mathbb{E}\left[W\right](1+\zeta_n)} \sum_{s=j_u+1}^{j_{u+1}}\frac{\sum_{\ell=1}^u W_\ell}{s}\bigg)\\[5pt] &\leq \exp\!\bigg(\!-\!\frac{1}{\mathbb{E}\left[W\right](1+\zeta_n)} \sum_{\ell=1}^u W_\ell \log\!\bigg(\frac{j_{u+1}}{j_u+1}\bigg)\!\bigg)\\[5pt] &= \bigg(\frac{j_{u+1}}{j_u+1}\bigg)^{-\sum_{\ell=1}^u W_\ell/(\mathbb{E}\left[W\right](1+\zeta_n))}. \end{aligned}\end{equation}

As the weights are almost surely bounded by one, we thus find

\begin{equation*}\begin{aligned} \prod_{s=j_u+1}^{j_{u+1}}\bigg(1-\frac{\sum_{\ell=1}^u W_\ell}{s\mathbb{E}\left[W\right](1+\zeta_n)}\bigg)&\leq \bigg(\frac{j_{u+1}}{j_u}\bigg)^{-\sum_{\ell=1}^u W_\ell/(\mathbb{E}\left[W\right](1+\zeta_n))}\bigg(1+\frac{1}{j_u}\bigg)^{k/(\mathbb{E}\left[W\right](1+\zeta_n))}\\[5pt] &=\bigg(\frac{j_{u+1}}{j_u}\bigg)^{-\sum_{\ell=1}^u W_\ell/(\mathbb{E}\left[W\right](1+\zeta_n))}(1+o(1)). \end{aligned} \end{equation*}

Using this upper bound in (6.12) and setting

\begin{equation*} a'_{\!\!t}\;:\!=\;\frac{W_t}{\mathbb{E}\left[W\right](1+\zeta_n)},\qquad t\in[k], \end{equation*}

we obtain

(6.14)

where in the last step we recall that $j_{k+1}=n$ . Since $d_t\leq c\log n$ for all $t\in[k]$ , $j_t>\ell_t>n^\eta$ for all $t\in[k]$ , and $\zeta_n=n^{-\delta\eta}/\mathbb{E}\left[W\right]$ , it readily follows that

(6.15)

We can thus omit the first term from (6.14) as well as use $a_t\;:\!=\;W_t/\mathbb{E}\left[W\right]$ instead of $a'_{\!\!t}$ at the cost of an additional $1+o(1)$ term. So we obtain

\begin{equation*} \frac{1}{(n)_k}\ \,\sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\mathbb E\Bigg[\prod_{t=1}^k\Bigg( a_t^{d_t} (j_t/n)^{a_t}\prod_{s=1}^{d_t}\frac{1}{m_{s,t}}\Bigg) \Bigg](1+o(1)). \end{equation*}

We then bound this from above even further by no longer constraining the indices $m_{s,t}$ to be distinct (so that the $*$ in the sum is omitted). That is, for different $t_1,t_2\in[k]$ , we allow $m_{s_1,t_1}=m_{s_2,t_2}$ to hold for any $s_1\in[d_{t_1}]$ , $s_2\in[d_{t_2}]$ . This also allows us to interchange the sum and the first product. We bound the sums from above by multiple integrals, which yields

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}\mathbb{E}\left[ \prod_{t=1}^k a_t^{d_t}(j_t/n)^{a_t} \int_{j_t}^n \int _{x_{1,t}}^n\cdots \int_{x_{d_t-1,t}}^n \prod_{s=1}^{d_t}x_{s,t}^{-1}\,\textrm{d} x_{d_t,t}\ldots\textrm{d} x_{1,t}\right](1+o(1)). \end{aligned} \end{equation*}

Applying Lemma 3.2 with $a=j_t$ , $b=n$ , we then obtain

\begin{equation*} \frac{1}{(n)_k}\mathbb{E}\left[ \prod_{t=1}^k (n/j_t)^{-a_t} \frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right](1+o(1)). \end{equation*}

Reintroducing the sums over the indices $j_1,\ldots, j_k$ (which were omitted after (6.11)), we arrive at

(6.16) \begin{equation} \frac{1}{(n)_k} \sum_{j_1=\ell_1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\mathbb{E}\left[ \prod_{t=1}^k(n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right](1+o(1)). \end{equation}

We observe that switching the order of the indices $j_1,\ldots,j_k$ (and their respective bounds $\ell_1,\ldots, \ell_k$ ) achieves the same result as permuting the $d_1,\ldots,d_k$ and $a_1,\ldots, a_k$ . Hence, if we take any $\pi\in\mathcal {P}_k$ , then as in (6.7) and (6.10),

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}{}&\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n \sum_{j_{\pi(2)}=(\ell_{\pi(2)}\vee j_{\pi(1)})+1}^n\!\!\! \cdots\!\!\! \sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n \!\!\!\!\!\!\mathbb{E}\left[\mathbb{P}_W\!\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right)\mathbb{1}_{E_n^{(1)}}\right]\\[5pt] &\leq \frac{1}{(n)_k}\mathbb{E}\left[\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n\!\!\!\cdots \!\!\!\!\sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n\prod_{t=1}^k(n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right]. \end{aligned} \end{equation*}

As a result, reintroducing the sum over all $\pi\in\mathcal {P}_k$ , we arrive at

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}{}&\sum_{\pi\in\mathcal {P}_k}\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n \sum_{j_{\pi(2)}=(\ell_{\pi(2)}\vee j_{\pi(1)})+1}^n\!\!\! \cdots\!\!\! \sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n\!\!\!\!\!\! \mathbb{E}\left[\mathbb{P}_W\!\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right)\mathbb{1}_{E_n^{(1)}}\right]\\[5pt] &\leq\frac{1}{(n)_k}\mathbb{E}\left[ \sum_{\pi\in\mathcal {P}_k}\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n\!\!\!\cdots\!\!\! \!\!\!\!\sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n\prod_{t=1}^k(n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right](1+o(1))\\[5pt] &=\frac{1}{(n)_k}\mathbb{E}\left[ \sum_{j_1=\ell_1+1}^n\sum_{\substack{j_2=\ell_2+1\\ j_2\neq j_1}}\ldots \!\!\!\!\sum_{\substack{j_k=\ell_k+1\\ j_k\neq j_{k-1,\ldots, j_1}}}^n\prod_{t=1}^k(n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right](1+o(1)). \end{aligned} \end{equation*}

We now bound these sums from above by allowing each index $j_i$ to take any value in $\{\ell_i+1,\ldots, n\}$ for all $i\in[k]$ , independent of the values of the other indices. Moreover, since the weights $W_1,\ldots, W_k$ , and hence $a_1, \ldots, a_k,$ are independent, this yields the upper bound

(6.17) \begin{equation} \prod_{t=1}^k\mathbb{E}\left[ \frac1n\sum_{j_t=\ell_t+1}^n(n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\right](1+o(1)), \end{equation}

so that we can now deal with each sum independently instead of k sums at the same time. First, we note that $(n/j_t)^{a_t}(\!\log(n/j_t))^{d_t}$ is increasing on $(0,n\exp\!(\!-\!d_t/a_t))$ , maximised at $n\exp\!(\!-\!d_t/a_t)$ , and decreasing on $(n\exp\!(\!-\!d_t/a_t),n]$ for all $t\in[k]$ . To provide the optimal bound, we want to know whether this maximum is attained in $[\ell_t+1,n]$ or not—that is, whether $n\exp\!(\!-\!d_t/a_t)\in[\ell_t+1,n]$ or not. To this end, we let

\begin{equation*} c_t\;:\!=\;\limsup_{n\to\infty}\frac{d_t}{\log n}, \qquad t\in[k], \end{equation*}

and consider two cases:

1. (1) $c_t\in[0,1/(\theta-1)]$ , $t\in[k]$ .

2. (2) $c_t\in(1/(\theta-1),c)$ , $t\in[k]$ .

Clearly, when $c\leq 1/(\theta-1)$ the second case can be omitted, so that without loss of generality we can assume $c>1/(\theta-1)$ . In the second case, it directly follows that the maximum is almost surely attained at

\begin{equation*} n\exp\!(\!-\!d_t/a_t)\leq n\exp\!(\!-\!c_t\log n (\theta-1)(1+o(1)))=n^{1-c_t(\theta-1)(1+o(1))}=o(1), \end{equation*}

so that the summand $(n/j_t)^{-a_t}(a_t\log(n/j_t))^{d_t}$ is almost surely decreasing in $j_t$ when $\ell_t< j_t\leq n$ . In the first case, such a conclusion cannot be made in general and depends on the precise value of $W_t$ . Therefore, the first case requires a more involved approach. We first assume Case (1) holds and discuss what simplifications can be made when Case 2 holds afterwards. In Case (1), we use Lemma A.4 to bound each sum from above by

\begin{equation*} \begin{aligned} \frac{1}{n} \sum_{j_t=\ell_t+1}^n (n/j_t)^{-a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}\leq \frac{1}{n} \int_{\ell_t}^n (n/x_t)^{-a_t}\frac{(a_t\log(n/x_t))^{d_t}}{d_t!}\,\textrm{d} x_t+\frac{1}{n}. \end{aligned}\end{equation*}

Here, we use that the summand is at most one, since

(6.18) \begin{equation} \Big(\frac{j_t}{n}\Big)^{a_t}\frac{(a_t\log(n/j_t))^{d_t}}{d_t!}=\mathbb{P}_W\!\left(\text{Poi}(a_t,j_t)=d_t\right)\leq 1, \end{equation}

irrespective of $a_t\in(0,\infty)$ and $j_t\in\mathbb{N}$ , and where $\text{Poi}(a_t,j_t)$ is a Poisson random variable with rate $a_t\log(n/j_t)$ , conditionally on $W_t$ . In Case (2), the summand on the left-hand side is decreasing in $j_t$ , so that we arrive at an upper bound without the additional error term $1/n$ . Using a substitution $y_t\;:\!=\;\log(n/x_t)$ , we obtain

(6.19) \begin{equation} \begin{aligned} \frac{a_t^{d_t}}{(1+a_t)^{d_t+1}}{}&\int_0^{\log(n/\ell_t)}\frac{(1+a_t)^{d_t+1}}{d_t!}y_t^{d_t}{\textrm{e}}^{-(1+a_t)y_t}\,\textrm{d} y_t+\frac1n\\[5pt] &=\frac{a_t^{d_t}}{(1+a_t)^{d_t+1}}\mathbb{P}_W\!\left(Y_t<\log(n/\ell_t)\right)+\frac1n, \end{aligned} \end{equation}

where, conditionally on $W_t$ , $Y_t\sim \text{Gamma}(d_t+1,1+a_t)$ . We recall that we redefined $a_t\;:\!=\;W_t/\mathbb{E}\left[W\right]=W_t/(\theta-1)$ . Since $X_t\;:\!=\;(1+W_t/(\theta-1))Y_t\sim\text{Gamma}(d_t+1,1)$ , we obtain

(6.20) \begin{equation} \frac{\theta-1}{\theta-1+W_t}\bigg(\frac{W_t}{\theta-1+W_t}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W_t}{(\theta-1)}\bigg)\log(n/\ell_t)\right)+\frac1n. \end{equation}

Using this in (6.17), we arrive at an upper bound of the form

\begin{equation*} \prod_{t=1}^k \mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W}{(\theta-1)}\bigg)\log(n/\ell_t)\right)+\frac1n\right](1+o(1)), \end{equation*}

where we recall that in each term of the product, the additive term $1/n$ is present only when $d_t$ satisfies Case (1) and can be omitted when $d_t$ satisfies Case (2). Moreover, we have omitted the indices of the weights, as they are all i.i.d. By Lemma A.3 in the appendix, the term $1/n$ can be included in the o(1) in the square brackets when $d_t$ satisfies Case (1). Thus, we finally obtain

(6.21) \begin{equation} \prod_{t=1}^k \mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W}{(\theta-1)}\bigg)\log(n/\ell_t)\right)\right](1+o(1)), \end{equation}

as desired. This concludes the upper bound for the first term in (6.10). Since we can choose $\gamma$ arbitrarily large in the second term in (6.10), we can use the same argument as in Lemma A.3 ((A.33) through (A.36) in particular), but now using that $d_t\leq c\log n<\theta/(\theta-1)\log n$ , to obtain that the second term in (6.10) can be included in the o(1) term of the final expression of the upper bound, as well in both Case (1) and Case (2), which concludes the proof of the upper bound.

Proof of Proposition 5.1, Equation (5.4), lower bound. We define the event

\begin{equation*} E_n^{(2)}\;:\!=\;\bigg\{\sum_{\ell=k+1 }^j W_\ell\in(\mathbb{E}\left[W\right](1-\zeta_n)j,\mathbb{E}\left[W\right](1+\zeta_n)j),\ \forall\ n^\eta\leq j\leq n\bigg\}. \end{equation*}

We then again have (6.7) and start by considering the identity permutation, $\pi(i)=i$ for all $i\in[k]$ , as in (6.8), by omitting the second term in (6.10), and using the event $E_n^{(2)}$ instead of $E_n^{(1)}$ . This yields the lower bound

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}{}&\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\!\!\!\!\!\!\!\!\mathbb E[\mathbb{P}_W\!\left(\mathcal{Z}_n(j_\ell)=m_\ell,\ell\in[k]\right)\mathbb{1}_{E_n^{(2)}}]\\[5pt] \geq{}&\frac{1}{(n)_k}\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\mathbb E\!\left[\prod_{t=1}^k\prod_{s=1}^{d_t}\frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell}\phantom{\times \prod_{u=1}^k\!\!\prod_{\substack{s=j_u+1\\ s\neq m_{i,t},t\in[d_i],i\in[k]}}^{j_{u+1}}}\right.\\[5pt] &\left.\times \prod_{u=1}^k\!\!\prod_{\substack{s=j_u+1\\ s\neq m_{i,t},t\in[d_i],i\in[k]}}^{j_{u+1}}\!\!\!\!\bigg(1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{\sum_{\ell=1}^{s-1}W_{\ell}}\bigg)\mathbb{1}_{E_n^{(2)}}\right]. \end{aligned} \end{equation*}

We omit the constraint $s\neq m_{\ell,i}$ , $\ell\in[d_i]$ , $i\in[k]$ in the final product. As this introduces more multiplicative terms smaller than one, we obtain a lower bound. Then, in the two denominators, we bound the vertex-weights $W_{j_1},\ldots, W_{j_k}$ from above by one and below by zero, respectively, to obtain a lower bound

\begin{equation*}\begin{aligned} \frac{1}{(n)_k}\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\!\mathbb E\Bigg[{}&\prod_{t=1}^k\prod_{s=1}^{d_t}\frac{W_{j_t}}{\sum_{\ell=1}^{m_{s,t}-1}W_\ell\mathbb{1}_{\{\ell\neq j_t,t\in[k]\}}+k}\\[5pt] &\times \prod_{u=1}^k\prod_{s=j_u+1}^{j_{u+1}}\!\!\!\bigg(\!1-\frac{\sum_{\ell=1}^u W_{j_\ell}}{\sum_{\ell=1}^{s-1}W_\ell\mathbb{1}_{\{\ell\neq j_t,t\in[k]\}}}\bigg)\mathbb{1}_{E_n^{(2)}}\!\Bigg]. \end{aligned}\end{equation*}

As a result, we can now swap the labels of $W_{j_t}$ and $W_t$ for each $t\in[k]$ , which again does not change the expected value, but changes the value of the two denominators to $\sum_{\ell=k+1}^{m_{s,t}}W_\ell+k$ and $\sum_{\ell=k+1}^{m_{s,t}}W_\ell$ , respectively. After this we use the bounds in $E_n^{(2)}$ on these sums in the expected value to obtain a lower bound. Finally, we note that the (relabelled) weights $W_t$ , $t\in[k],$ are independent of $E_n^{(2)}$ , so that we can take the indicator out of the expected value. Combining all of the above steps, we arrive at the lower bound

(6.22) \begin{equation} \begin{aligned} \frac{1}{(n)_k}\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \,{}& \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\mathbb E\Bigg[\prod_{t=1}^k \Big(\frac{W_t}{\mathbb{E}\left[W\right]}\Big)^{d_t} \prod_{s=1}^{d_t}\frac{1}{m_{s,t}(1+2\zeta_n)}\\[5pt] &\times\prod_{u=1}^k\prod_{s=j_u+1}^{j_{u+1}}\bigg(1-\frac{\sum_{\ell=1}^u W_\ell}{(s-1)\mathbb{E}\left[W\right](1-\zeta_n)}\bigg)\Bigg]\mathbb P(E_n^{(2)}).\\[5pt] \end{aligned} \end{equation}

The $1+2\zeta_n$ in the fraction on the first line arises from the fact that, for n sufficiently large, $(m_{s,t}-1)(1+\zeta_n)+k\leq m_{s,t}(1+2\zeta_n)$ . It follows from Lemma 3.1 that $\mathbb P(E_n^{(2)})=1-o(n^{-\gamma})$ for any $\gamma>0$ . Similarly to the calculations in (6.13) and using $\log(1-x)\geq -x-x^2$ for x small, we obtain an almost sure lower bound for the final product, for n sufficiently large, of the form

\begin{equation*} \prod_{s=j_u+1}^{j_{u+1}}\bigg(1-\frac{\sum_{\ell=1}^u W_\ell}{(s-1)\mathbb{E}\left[W\right](1-\zeta_n)}\bigg)\geq \bigg(\frac{j_{u+1}}{j_u}\bigg)^{-\sum_{\ell=1}^u W_\ell/(\mathbb{E}\left[W\right](1-\zeta_n))}(1-o(1)). \end{equation*}

Using this in (6.22) yields the lower bound

\begin{equation*} \frac{1}{(n)_k}\!\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\!\!\!\!\!\!\!\!\!\!(1-o(1))\bigg(\frac{1-\zeta_n}{1+2\zeta_n}\bigg)^{\sum_{t=1}^k d_t} \mathbb E\Bigg[\prod_{t=1}^k \widetilde a_t^{d_t} \Big(\frac{j_t}{n}\Big)^{\widetilde a_t}\prod_{s=1}^{d_t}\frac{1}{m_{s,t}} \Bigg], \end{equation*}

where $\widetilde a_t\;:\!=\;W_t/(\mathbb{E}\left[W\right](1-\zeta_n))$ . Since $d_t\leq c\log n$ and $j_t\geq \ell_t\geq n^\eta$ for all $t\in[k]$ , and $\zeta_n=n^{-\eta\delta}/\mathbb{E}\left[W\right]$ for some $\delta\in(0,1/2)$ , we have as in (6.15) that

\begin{equation*} \bigg(\frac{1-\zeta_n}{1+2\zeta_n}\bigg)^{\sum_{t=1}^k d_t}=1-o(1), \quad \text{and}\quad \widetilde a_t^{d_t}\Big(\frac{j_t}{n}\Big)^{\widetilde a_t}=a_t^{d_t}\Big(\frac{j_t}{n}\Big)^{a_t}(1-o(1)), \end{equation*}

where $a_t\;:\!=\;W_t/\mathbb{E}\left[W\right]$ . This yields

\begin{equation*} \frac{1}{(n)_k}\!\sum_{j_1=\ell_1+1}^n\ldots \!\!\!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^n\ \, \sideset{}{^*}\sum_{\substack{j_i<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\!\! \mathbb E\Bigg[\prod_{t=1}^k a_t^{d_t} \Big(\frac{j_t}{n}\Big)^{ a_t}\prod_{s=1}^{d_t}\frac{1}{m_{s,t}} \Bigg](1-o(1)). \end{equation*}

We now bound the sum over the indices $m_{s,i}$ from below. We note that the expression in the expected value is decreasing in $m_{s,i}$ , and we restrict the range of the indices to $j_i+\sum_{t=1}^k d_t<m_{1,i}<\ldots< i_{d_i,i}\leq n$ for all $i \in[k]$ , but no longer constrain the indices to be distinct (so that we can drop the $*$ in the sum). In the distinct sums and the suggested lower bound, the numbers of values the $m_{s,i}$ take on equal

\begin{equation*} \prod_{i=1}^k \binom{n-(j_i-1)-\sum_{t=1}^{i-1}d_t}{d_i} \quad \text{and} \quad \prod_{i=1}^k \binom{n-(j_i-1)-\sum_{t=1}^k d_t}{d_i}, \end{equation*}

respectively. It is straightforward to see that the former allows for more possibilities than the latter, as $\binom{b}{c}> \binom{a}{c}$ when $b> a\geq c$ . As we omit the largest values of the expected value (since it decreases in $m_{s,t}$ and we omit the smallest values of $m_{s,t}$ ), we thus arrive at the lower bound

\begin{equation*} \begin{aligned} \frac{1}{(n)_k} \!\!\!\sum_{j_1=\ell_1+1}^{n-\sum_{t=1}^k d_t}\!\!\!\!\ldots \!\!\!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^{n-\sum_{t=1}^k d_t}\ \, \sideset{}{^*}\sum_{\substack{j_i+\sum_{t=1}^k d_t<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\!\!\!\!\mathbb E\Bigg[\prod_{t=1}^k a_t^{d_t} \Big(\frac{j_t}{n}\Big)^{ a_t}\prod_{s=1}^{d_t}\frac{1}{m_{s,t}} \bigg](1-o(1)), \end{aligned} \end{equation*}

where we also restrict the upper range of the indices of the outer sums, as otherwise there would be a contribution of zero from these values of $j_1,\ldots,j_k$ . We now use techniques similar to those used for the proof of the upper bound to switch from summation to integration. However, because of the altered bounds on the range of the indices over which we sum and the fact that we require lower bounds rather than upper bounds, we face some more technicalities.

For now, we omit the expected value and focus on the terms

(6.23) \begin{equation} \frac{1}{(n)_k} \!\!\!\sum_{j_1=\ell_1+1}^{n-\sum_{t=1}^k d_t}\!\!\!\!\ldots \!\!\!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^{n-\sum_{t=1}^k d_t} \sum_{\substack{j_i+\sum_{t=1}^k d_t<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\prod_{t=1}^k a_t^{d_t}\Big(\frac{j_t}{n}\Big)^{ a_t}\prod_{s=1}^{d_t}\frac{1}{m_{s,t}}. \end{equation}

We start by restricting the upper bound on the k outer sums to $n-2\sum_{i=1}^k d_i$ . This will prove useful later. We set $h_k\;:\!=\;\sum_{t=1}^k d_t$ and bound the inner sum over the indices $m_{s,t}$ from below by

\begin{equation*}\begin{aligned} {}&\sum_{\substack{j_i+h_k<m_{1,i}<\ldots<m_{d_i,i}\leq n,\\ i\in[k]}}\prod_{t=1}^k \prod_{s=1}^{d_t}\frac{1}{m_{s,t}} \geq \prod_{t=1}^k \int_{j_t+1+h_k}^{n}\int_{x_{1,t}+1}^{n}\cdots \int_{x_{d_t-1,t}+1}^{n}\prod_{s=1}^{d_t}x_{s,t}^{-1}\,\textrm{d} x_{d_t,t}\ldots \textrm{d} x_{1,t}. \end{aligned} \end{equation*}

Applying Lemma 3.2 with $a=j_t+1+h_k$ and $b=n$ , and using that $j_t\leq n-2h_k $ (recall that we restricted the upper bound on the outer sums in (6.23) to $n-2h_k$ ), yields the lower bound

\begin{equation*} \prod_{t=1}^k \frac{ a_t^{d_t}}{d_t!}\bigg(\!\log\!\bigg(\frac{n}{j_t +2\sum_{i=1}^k d_i}\bigg)\!\bigg)^{d_t}. \end{equation*}

Substituting this in (6.23) with the restriction on the outer sum discussed after (6.23) yields

(6.24) \begin{equation} \frac{1}{(n)_k} \!\!\!\sum_{j_1=\ell_1+1}^{n-2\sum_{i=1}^k d_i}\!\!\!\!\ldots \!\!\!\!\!\!\sum_{j_k=(\ell_k\vee j_{k-1})+1}^{n-2\sum_{i=1}^k d_i}\prod_{t=1}^k \bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ a_t^{d_t}}{d_t!}\bigg(\!\log\!\bigg(\frac{n}{j_t +2\sum_{i=1}^k d_i}\bigg)\!\bigg)^{d_t} . \end{equation}

To simplify the summation over $j_1,\ldots,j_k$ , we write the summand as

\begin{equation*} \prod_{t=1}^k\bigg(\bigg(j_t+2\sum_{i=1}^k d_i\bigg)/n\bigg)^{ a_t}\frac{ a_t^{d_t}}{d_t!}\bigg(\!\log\!\bigg(\frac{n}{j_t +2\sum_{i=1}^k d_i}\bigg)\!\bigg)^{d_t}\bigg(1-\frac{2\sum_{i=1}^k d_i}{j_t +2\sum_{i=1}^k d_i}\bigg)^{ a_t}. \end{equation*}

Using that $d_t\leq c \log n$ , $j_t\geq \ell_t\geq n^\eta$ , and $x^{ a_t}\geq x^{1/\mathbb{E}\left[W\right]}$ for $x\in(0,1)$ almost surely, we can write the last term as $(1-o(1))$ almost surely. We then shift the bounds on the range of the sums in (6.24) by $2\sum_{i=1}^k d_i$ and let $\widetilde \ell_i\;:\!=\;\ell_i+2\sum_{t=1}^k d_t$ for all $ i\in[k]$ , to obtain the lower bound

\begin{equation*} \frac{1}{(n)_k} \sum_{j_1=\widetilde \ell_1+1}^n\sum_{j_2=(\widetilde \ell_2\vee j_1)+1}^n\!\!\!\ldots \!\!\!\sum_{j_k=(\widetilde \ell_k\vee j_{k-1})+1}^n\!\!\!\!\!\!\!\!\!\!\!\!(1-o(1))\prod_{t=1}^k \bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}. \end{equation*}

We recall that this lower bound is achieved for the permutation $\pi$ such that $\pi(i)=i$ for all $i\in[k]$ . As the product is invariant to permuting the indices $t\in[k]$ , we can use this in (6.7) to obtain

(6.25) \begin{equation}\begin{aligned} \frac{1}{(n)_k}{}&\sum_{\pi\in\mathcal {P}_k}\sum_{j_{\pi(1)}=\ell_{\pi(1)}}^n \sum_{j_{\pi(2)}=(\ell_{\pi(2)}\vee j_{\pi(1)})+1}^n \cdots \sum_{j_{\pi(k)}=(\ell_{\pi(k)}\vee j_{\pi(k-1)})+1}^n \mathbb{P}\left(\mathcal{Z}_n(j_i)=d_i, i\in[k]\right)\\[5pt] \geq{}&\frac{1}{(n)_k}\sum_{j_1=\widetilde \ell_1+1}^n \sum_{\substack{j_2=\widetilde \ell_2+1\\ j_2\neq j_1}}^n \cdots \sum_{\substack{j_k=\widetilde \ell_k+1\\ j_k\neq j_1,\ldots, j_{k-1}}}^n (1-o(1))\prod_{t=1}^k \mathbb{E}\left[\Big(\frac{j_t}{n}\Big)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\right]. \end{aligned}\end{equation}

We now want to allow for the indices $j_1,\ldots, j_k$ to have the same value. This way, we can more easily evaluate the sums. To do this, we distinguish between two cases in terms of the sizes of $d_1,\ldots, d_k$ , namely Case (1) and Case (2). In Case (1), we subtract all terms where two or more indices have the same value to avoid creating an upper bound. That is, we write the multiple sums as

\begin{equation*} \begin{aligned} \frac{1}{(n)_k}{}&\sum_{j_1=\widetilde \ell_1+1}^n \sum_{j_2=\widetilde \ell_2+1}^n \cdots \sum_{j_k=\widetilde \ell_k+1}^n (1-o(1))\prod_{t=1}^k \mathbb{E}\left[\Big(\frac{j_t}{n}\Big)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\right]\\[5pt] -\frac{1}{(n)_k}{}&\sum_{m=2}^k \sum_{\substack{S\subseteq [k]\\ |S|=m}}\sum_{i\in S}\sum_{j_i=\widetilde \ell_i+1}^n\ \sideset{}{^*}\sum_{\substack{j_s=\widetilde \ell_s+1\\ s\in [k]\backslash S}}^n\!\! \mathbb{E}\left[ \prod_{u\in S}\Big(\frac{j_i}{n}\Big)^{a_u}\frac{(a_u\log(n/j_i))^{d_u}}{d_u!} \!\!\prod_{s\in [k]\backslash S}\!\!\!\Big(\frac{j_s}{n}\Big)^{d_s}\frac{(a_s\log(n/j_s))^{d_s}}{d_s!}\!\right]. \end{aligned} \end{equation*}

Here, the $*$ in the final sum on the second line indicates that the indices $j_s$ with $s\in [k]\backslash S$ are not allowed to have the same value, nor to be equal to $j_i$ for any $i\in S$ . The error term on the second line can be bounded from below by bounding the multiple sums from above, following an approach equivalent to that used in the proof of the upper bound. By (6.18) we can omit all terms $u\neq i$ in the product over $u\in S$ , as they can be bounded from above by one. Furthermore, we can omit the $*$ in the final sum to obtain an upper bound, so that all indices $j_i$ and $j_s$ , $s\in[k]\backslash S$ , can be equal in value. Finally, let us write $S_i\;:\!=\;S\backslash \{i\}$ . It then follows from (6.17)–(6.21) that the error term is at least

\begin{equation*} \begin{aligned} -{}&\sum_{m=2}^k \!\frac{1+o(1)}{n^{m-1}}\! \sum_{\substack{S\subseteq [k]\\ |S|=m}}\sum_{i\in S} \prod_{t\in[k]\backslash S_i}\!\!\!\!\mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\widetilde \ell_t)\right)\right]\\[5pt] &\geq -C\sum_{m=2}^k \frac{1}{n^{m-1}}\!\!\!\!\!\!\!\!\!\! \sum_{\substack{S'\subseteq [k]\\ |S|=k-(m-1)}}\!\!\!\!\!\!\!\prod_{t\in S'}\!\mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\widetilde \ell_t)\right)\right], \end{aligned} \end{equation*}

for some large constant $C>0$ . It remains to take care of the main term,

(6.26) \begin{equation} \begin{aligned} \frac{1}{(n)_k}{}&\sum_{j_1=\widetilde \ell_1+1}^n \sum_{j_2=\widetilde \ell_2+1}^n \cdots \sum_{j_k=\widetilde \ell_k+1}^n (1-o(1))\prod_{t=1}^k \mathbb{E}\left[\bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\right]\\[5pt] &\geq \prod_{t=1}^k\mathbb{E}\left[\frac1n \sum_{j_t=\widetilde \ell_t+1}^n\bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\right](1-o(1)). \end{aligned} \end{equation}

We bound each sum from below by an integral, similarly to the proof of the upper bound. We again consider the two cases used in the upper bound, Case (1) and Case (2). In Case (2), the summand is decreasing in $j_t$ and hence we can replace the sum by an integral from $\ell_t$ to n. In Case (1), we use Lemma A.4 and (6.18) to obtain the lower bound

\begin{equation*} \frac1n\sum_{j_t=\widetilde \ell_t+1}^n\bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\geq \int_{\widetilde \ell_t}^n \bigg(\frac{x_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/x_t))^{d_t}\,\textrm{d} x_t -\frac1n. \end{equation*}

The same steps as in (6.19) and (6.20) yield that this equals

\begin{equation*} \frac{\theta-1}{\theta-1+W_t}\bigg(\frac{W_t}{\theta-1+W_t}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W_t}{(\theta-1)}\bigg)\log(n/\widetilde \ell_t)\right)-\frac1n. \end{equation*}

Using this in (6.26) and combining it with the bound for the error term, we arrive at the final lower bound

\begin{equation*} \begin{aligned} \prod_{t=1}^k{}& \mathbb{E}\left[\frac{\theta-1}{\theta-1+W_t}\bigg(\frac{W_t}{\theta-1+W_t}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W_t}{(\theta-1)}\bigg)\log(n/\widetilde \ell_t)\right)-\frac1n\right](1-o(1))\\[5pt] &-C\sum_{m=2}^k \frac{1}{n^{m-1}}\!\!\!\!\!\!\!\!\!\! \sum_{\substack{S'\subseteq [k]\\ |S|=k-(m-1)}}\!\!\!\!\!\!\!\prod_{t\in S'}\!\mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\widetilde \ell_t)\!\right)\!\right]\!. \end{aligned} \end{equation*}

We can replace $\widetilde \ell_t$ with $\ell_t$ at the cost of a $1-o(1)$ term, since $\log(n/\widetilde \ell_t)=\log(n/\ell_t)-o(1)$ . It then follows from Lemma A.3 that both the $1/n$ term on the first line and that of the second line can be incorporated into the $1-o(1)$ term.

In Case (2), we know that the summand in (6.25) is decreasing in $j_t$ for all $t\in[k]$ . Hence, we can omit the smallest values of $j_1, \ldots, j_k$ to obtain a lower bound. This yields

\begin{equation*} \frac{1}{(n)_k}{}\sum_{j_1=\widetilde \ell_1+1}^n \sum_{j_2=\widetilde \ell_2+2}^n \cdots \sum_{j_k=\widetilde \ell_k+k}^n (1-o(1))\prod_{t=1}^k \mathbb{E}\left[\bigg(\frac{j_t}{n}\bigg)^{ a_t}\frac{ 1}{d_t!}(a_t\log(n/j_t))^{d_t}\right], \end{equation*}

which can be evaluated in the same manner as in Case (1) to yield the lower bound

\begin{equation*} \prod_{t=1}^k \mathbb{E}\left[\frac{\theta-1}{\theta-1+W_t}\bigg(\frac{W_t}{\theta-1+W_t}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W_t}{(\theta-1)}\bigg)\log(n/(\widetilde \ell_t+(t-1)))\right)\right](1-o(1)). \end{equation*}

Again, since $\log(n/(\widetilde \ell_t+(t-1)))=\log(n/\ell_t)-o(1)$ for each $t\in[k]$ , we can replace $\widetilde \ell_t+(t-1)$ with $\ell_t$ for each $t\in[k]$ at the cost of a $1-o(1)$ term. We thus conclude that

\begin{equation*} \begin{aligned} \mathbb P(\mathcal{Z}_n{}&(v_i)= d_i, v_i>\ell_i,i\in[k])\\[5pt] \geq{}& (1-o(1))\prod_{t=1}^k \bigg[\mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{d_t}\mathbb{P}_W\!\left(X_t<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_t)\right)\right], \end{aligned} \end{equation*}

which concludes the proof of the lower bound.

Proof of Proposition 5.1, Equation (5.5). We prove the two bounds in (5.5) by using (5.4). We assume that $d_i$ diverges with n, and we note that if

\begin{equation*} d_i\leq c\log n\quad \text{and}\quad \ell_i\leq n\exp\!(\!-\!(1-\xi)(1-\theta^{-1})(d_i+1)) \end{equation*}

for any $\xi\in(0,1)$ and for all sufficiently large n, then for any $j\in[\lfloor d_i^{1/4}\rfloor]$ , it also holds that

\begin{equation*} d_i +j \leq c'\log n,\quad \text{and}\quad \ell_i\leq n\exp\!(\!-\!(1-\xi)(1-\theta^{-1})(d_i+j+1)), \end{equation*}

for any $\xi\in(0,1)$ and for all sufficiently large n as well, where we can choose $c'\in(c,\theta/(\theta-1))$ arbitrarily close to c. As a result, we can write

\begin{equation*} \begin{aligned} \mathbb P(\mathcal{Z}_n{}&(v_i)\geq d_i, v_i> \ell_i, i\in[k])\\[5pt] \leq & \sum_{j_1=d_1}^{d_1+\lfloor d_1^{1/4}\rfloor}\cdots \sum_{j_k=d_k}^{d_k+\lfloor d_k^{1/4}\rfloor}\mathbb{P}\left(\mathcal{Z}_n(v_i)=j_i, v_i> \ell_i, i\in[k]\right)\\[5pt] &+\sum_{t=1}^k \mathbb{P}\left(\mathcal{Z}_n(v_t)\geq d_t+\lceil d_t^{1/4}\rceil ,\mathcal{Z}_n(v_i)\geq d_i, i\neq t, v_i>\ell_i, i\in[k]\right). \end{aligned}\end{equation*}

We first provide an upper bound for the multiple sums on the first line. By (5.4), this equals

\begin{equation*} \sum_{j_1=d_1}^{d_1+\lfloor d_1^{1/4}\rfloor}\!\!\!\cdots\!\!\! \sum_{j_k=d_k}^{d_k+\lfloor d_k^{1/4}\rfloor}\!\!\!\!(1+o(1))\prod_{i=1}^k \mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{j_i}\mathbb{P}_W\!\left(X_{j_i}<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\!\right)\!\right]\!, \end{equation*}

where we write $X_{j_i}\sim\text{Gamma}(j_i+1,1)$ instead of $X_i$ to explicitly state the dependence on $j_i$ . If $X_{j_i}\sim\text{Gamma}(j_i+1,1)$ and $X_{j'_{\!\!i}}\sim\text{Gamma}(j'_{\!\!i}+1,1)$ , then $X_{j_i}$ stochastically dominates $X_{j'_{\!\!i}}$ when $j_i>j'_{\!\!i}$ . Hence, we obtain the upper bound

(6.27) \begin{equation}\begin{aligned} \sum_{j_1=d_1}^\infty\!\!\!{}&\ldots\!\!\! \sum_{j_k=d_k}^\infty\!(1+o(1))\prod_{i=1}^k \mathbb E\bigg[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{j_i}\!\mathbb P_W\!\bigg(\!X_{d_i}\!<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\!\bigg)\bigg]\\[5pt] &=(1+o(1))\prod_{i=1}^k \mathbb{E}\left[\bigg(\frac{W}{\theta-1+W}\bigg)^{d_i}\mathbb{P}_W\!\left(X_i<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\right)\right], \end{aligned} \end{equation}

where we note that $X_i\equiv X_{d_i}$ by the definition of $X_i$ and $X_{d_i}$ . It thus remains to show that

(6.28) \begin{equation} \sum_{t=1}^k \mathbb{P}\left(\mathcal{Z}_n(v_t)\geq d_t+\lceil d_t^{1/4}\rceil ,\mathcal{Z}_n(v_i)\geq d_i, i\neq t, v_i>\ell_i, i\in[k]\right) \end{equation}

is negligible compared to (6.27). We show this holds for each term in the sum, and since all $d_i$ , $i\in[k]$ , diverge, it suffices to show this holds for $t=1$ . The in-degrees in the WRT model are negative quadrant dependent under the conditional probability measure $\mathbb P_W$ . That is, by [Reference Lodewijks and Ortgiese12, Lemma 7.1], for any indices $r_1, \ldots, r_k\in[n]$ , $r_i\neq r_j$ when $i\neq j$ ,

\begin{equation*} \mathbb{P}_W\!\left(\mathcal{Z}_n(r_i)\geq d_i, i\in[k]\right)\leq \prod_{i=1}^k \mathbb{P}_W\!\left(\mathcal{Z}_n(r_i)\geq d_i\right). \end{equation*}

We can thus bound the term with $t=1$ in (6.28) from above by

\begin{equation*} \begin{aligned} \sum_{j_1=\ell_1+1}^n{}&\sum_{\substack{j_2=\ell_2+1\\ j_2\neq j_1}}^n \cdots \sum_{\substack{j_k=\ell_k+1\\ j_k\neq j_{k-1},\ldots, j_1}}^n \mathbb{E}\left[\mathbb{P}_W\!\left(\mathcal{Z}_n(j_1)\geq d_1+\lceil d_1^{1/4}\rceil\right)\prod_{i=2}^k \mathbb{P}_W\!\left(\mathcal{Z}_n(j_i)\geq d_i\right)\right] \\[5pt] &\leq \mathbb{E}\left[\mathbb{P}_W\!\left(\mathcal{Z}_n(v_1)\geq d_1+\lceil d_1^{1/4}\rceil, v_1>\ell_1\right)\prod_{i=2}^k \mathbb{P}_W\!\left(\mathcal{Z}_n(v_i)\geq d_i, v_i>\ell_i\right)\right], \end{aligned} \end{equation*}

where the last step follows by allowing the indices $j_i$ to take on any value between $\ell_i+1$ and n, $i\in[k]$ . We can now deal with each of these probabilities individually instead of with all the events at the same time, which makes obtaining an explicit bound for the probability of the event $\{\mathcal{Z}_n(v_i)\geq d_i, v_i>\ell_i\}$ easier. We claim that, with a very similar approach to that in the proof of the upper bound for (5.4) (see also the steps (5.47)–(5.51) in the proof of [Reference Eslava, Lodewijks and Ortgiese7, Lemma 5.11] for the case $\ell_1=\ldots \ell_k=n^{1-\varepsilon}$ for some $\varepsilon\in(0,1)$ ), it can be shown that this expected value is bounded from above by

\begin{equation*} \begin{aligned} (1+o(1)){}&\mathbb{E}\left[\bigg(\frac{W}{\theta-1+W}\bigg)^{d_1+\lceil d_1^{1/4}\rceil}\mathbb{P}_W\!\left(X_1\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_1)\right)\right]\\[5pt] &\times \prod_{i=2}^k \mathbb{E}\left[\bigg(\frac{W}{\theta-1+W}\bigg)^{d_i}\mathbb{P}_W\!\left(X_i\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\right)\right]\\[5pt] \leq{}& (1+o(1))\theta^{-\lceil d_1^{1/4}\rceil}\prod_{i=1}^k \mathbb{E}\left[\bigg(\frac{W}{\theta-1+W}\bigg)^{d_i}\mathbb{P}_W\!\left(X_i\leq \bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\right)\right]. \end{aligned}\end{equation*}

This upper bound can be achieved for each term in (6.28) (with $\lceil d_1^{1/4}\rceil$ changed accordingly), so that (6.28) is indeed negligible compared to (6.27) and hence can be included in the o(1) term in (6.27). This proves the upper bound in (5.5).

For a lower bound we directly obtain

\begin{equation*} \mathbb{P}\left(\mathcal{Z}_n(v_i)\geq d_i, v_i> \ell_i, i\in[k]\right)\geq \sum_{j_1=d_1}^{d_1+\lfloor d_1^{1/4}\rfloor}\cdots \sum_{j_k=d_k}^{d_k+\lfloor d_k^{1/4}\rfloor}\mathbb{P}\left(\mathcal{Z}_n(v_i)=j_i, v_i> \ell_i, i\in[k]\right). \end{equation*}

With an approach similar to that used for the upper bound, we can use (5.4) and now bound the probability from below by replacing $X_{j_i}$ with $\widetilde X_i\equiv X_{d_i+\lfloor d_i^{1/4}\rfloor}$ instead of $X_{d_i}$ , to arrive at the lower bound

\begin{equation*} \begin{aligned} \sum_{j_1=d_1}^{d_1+\lfloor d_1^{1/4}\rfloor}\!\!\!{}&\cdots\!\!\! \sum_{j_k=d_k}^{d_k+\lfloor d_k^{1/4}\rfloor}\!\!\!\!(1+o(1))\prod_{i=1}^k \mathbb{E}\left[\frac{\theta-1}{\theta-1+W}\bigg(\frac{W}{\theta-1+W}\bigg)^{j_i}\mathbb{P}_W\!\left( X_{j_i}<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\!\right)\!\right]\\[5pt] \geq (1+{}&o(1))\prod_{i=1}^k \mathbb{E}\!\left[\!\bigg(\frac{W}{\theta-1+W}\bigg)^{d_i}\bigg(1-\bigg(\frac{W}{\theta-1+W}\bigg)^{\lfloor d_i^{1/4}\rfloor}\bigg)\mathbb{P}_W\!\left(\widetilde X_i<\bigg(1+\frac{W}{\theta-1}\!\bigg)\log(n/\ell_i)\!\right)\!\right]\\[5pt] \geq(1+{}&o(1))\prod_{i=1}^k\mathbb{E}\left[\bigg(\frac{W}{\theta-1+W}\bigg)^{d_i}\mathbb{P}_W\!\left(\widetilde X_i<\bigg(1+\frac{W}{\theta-1}\bigg)\log(n/\ell_i)\right)\right], \end{aligned} \end{equation*}

where in the last step we use that $1-(W/(\theta-1+W))^{\lfloor d_i^{1/4}\rfloor}\geq 1-\theta^{-\lfloor d_i^{1/4}\rfloor}=1-o(1)$ almost surely, since $d_i$ diverges for any $i\in[k]$ . This concludes the proof of the lower bound in (5.5) and hence of Proposition 5.1.