Proof of Theorem 1. For the system with N nodes $\{0,\dots,N-1\}$ , $T_{N}$ is the sum of $N-1$ independent random variables, $T_{N} = \sum_{n=1}^{N-1} W_{n,N}$ , where we define $W_{n,N}$ as the time between the nth and ( $n+1$ )th nodes joining the informed group. By definition, $W_{0,N} = 0$ , since the initial condition is one node at level 0.

The transition from n informed nodes to $n+1$ occurs when one of the $N-n$ uninformed nodes interacts with one of the n informed nodes in any information level. This transition occurs at a rate $n\cdot (N-n)$ . The transition time $W_{n,N}$ is then exponentially distributed with rate $n(N-n)$ , mean $(n(N-n))^{-1}$ , and variance $(n(N-n))^{-2}$ . Hence,

\begin{align*} \mathbb{E}[T_{N}] & = \sum_{n=1}^{N-1} \mathbb{E}[W_{n,N}] = \sum_{n=1}^{N-1} \frac{1}{n(N-n)} = \frac{1}{N}\sum_{n=1}^{N-1}\bigg(\frac{1}{n} + \frac{1}{N-n}\bigg) = \frac{2}{N}\sum_{n=1}^{N-1} \frac{1}{n} = \frac{2H_{N-1}}{N}, \\[5pt] \text{Var}[T_{N}] & = \sum_{n=1}^{N-1} \text{Var}[W_{n,N}] = \sum_{n=1}^{N-1} \frac{1}{n^2(N-n)^2} = \frac{1}{N^2}\sum_{n=1}^{N-1} \bigg(\frac{1}{n} + \frac{1}{N-n}\bigg)^2 \\[5pt] & = \frac{2}{N^2}\Bigg(\sum_{n=1}^{N-1}\frac{1}{n^2}\Bigg) + \frac{4}{N^3}\Bigg(\sum_{k=1}^{N-1} \frac{1}{n}\Bigg) = \frac{2}{N^2}\bigg(\frac{\pi^2}{6} - \psi' (N)\bigg) + \frac{4}{N^3}H_{N-1}, \end{align*}

where $H_n = \sum_{k=1}^n ({1}/{k})$ is the nth harmonic number, and $\psi'(N+1)$ is the first derivative of the digamma function.

We will use the Borel–Cantelli lemma to prove almost sure convergence. Observe that $T_N > 0$ and $T_N > \varepsilon \Rightarrow T_N^2 > \varepsilon^2$ . Thus, for all $\varepsilon > 0$ ,

\begin{align*} \mathbb{P}(T_N > \varepsilon) = \mathbb{P}(T_N^2 > \varepsilon^2) & \leq \frac{\mathbb{E}[T_N^2]}{\varepsilon^2} \\[5pt] & \leq \frac{1}{\varepsilon^2}\bigg[\frac{2}{N^2}\bigg(\frac{\pi^2}{6} - \psi' (N)\bigg) + \frac{4}{N^3}H_{N-1} + \frac{4H_{N-1}^2}{N^2}\bigg] \end{align*}

by Chebyshev’s inequality, where we used $\mathbb{E}[T_N^2] = \text{Var}[T_N] + \mathbb{E}[T_N]^2$ .

Fix $\varepsilon > 0$ , and consider the set of events $A_N^\varepsilon = \{T_N > \varepsilon\}$ . It follows that

\begin{equation*} \sum_{N = 1}^\infty \mathbb{P}(A_N^\varepsilon) = \sum_{N = 1}^\infty \mathbb{P}\big(T_N > \varepsilon) \leq \sum_{N=1}^\infty \frac{1}{\varepsilon^2}\bigg[\frac{2}{N^2}\bigg(\frac{\pi^2}{6} - \psi' (N)\bigg) + \frac{4}{N^3}H_{N-1} + \frac{4H_{N-1}^2}{N^2}\bigg]. \end{equation*}

It is easy to see that the sum converges. By the Borel–Cantelli lemma, it follows that $\mathbb{P}(\limsup_{N\to\infty}A_N^\varepsilon) = \mathbb{P}(\limsup_{N\to\infty}\{T_N > \varepsilon\}) = 0$ .

To complete the proof, let $\varepsilon_k$ , $k\in\mathbb{N}$ , be a sequence of decreasing times such that $\varepsilon_{k} < \varepsilon_{k-1}$ , $\varepsilon_k > 0$ , and $\lim_{k\to\infty} \varepsilon_k = 0$ . Hence, $A_N^{\varepsilon_k} \subset A_N^{\varepsilon_{k+1}}$ , and

\begin{equation*} \mathbb{P}\Big(\cup_{\varepsilon>0}\limsup_{N\to\infty}A_N^\varepsilon\Big) \leq \sum_{k=1}^\infty\mathbb{P}\Big(\limsup_{N\to\infty}A_N^{\varepsilon_k}\Big) = 0. \end{equation*}

Consequently, $\mathbb{P}(\lim_{N\to\infty} T_N = 0) = 1$ , and the result is proved.

Proof of Theorem 2. Similarly to the previous case, $\sup_{i< N}\nu_i^N(t) = 1$ if and only if all nodes have interacted with the root, i.e. the graph is a star. Thus, the time until the star is formed can be written as the sum of the transition times between each node connecting to the root. That is, the time $T^*_{N}$ until the star forms is again the sum of $N-1$ independent random variables, $T^*_{N} = \sum_{n=1}^{N-1} V_{n,N}$ , where we define $V_{n,N}$ as the time between the nth node and the ( $n+1$ )th connecting to the root. Again, $V_{0,N} = 0$ , as there is one node at level 0 from the beginning.

The transition from n to $n+1$ nodes occurs when one of the remaining $N-n$ nodes not connected to the root interacts with the root. This transition has rate $1\cdot (N-n)$ . The transition time $V_{n,N}$ is then an exponentially distributed random variable with rate $(N-n)$ , mean $(N-n)^{-1}$ , and variance $(N-n)^{-2}$ . Consider now the set of random variables $\{\mathcal{V}_n\}_{n=1}^\infty$ such that $\mathcal{V}_n = ({1}/{n})X_n$ , where $X_n \sim \text{Exp}(1)$ are independent and identically distributed (i.i.d.). Observe that $\mathcal{V}_n\sim \text{Exp}(n)$ . Instead of writing $T_N^*$ as a sum of random variables that depend on N, we can use the set $\{\mathcal{V}_n\}_{n=1}^\infty$ to write $T_N^* = \sum_{n=1}^{N-1} \mathcal{V}_n = \sum_{n=1}^{N-1} ({1}/{n}) X_n$ .

By the monotone convergence theorem, $T_N^* \to T^*\in \mathbb{R}\cup \{\infty\}$ .

We will show that $T^*=\infty$ by contradiction. First, consider the random variable

(1) \begin{align} S_M & = \frac{1}{M-1}\sum_{N=2}^M T_N^* = \frac{1}{M-1}\sum_{N=2}^{M}\sum_{n=1}^{N-1}\frac{1}{n}X_n \nonumber \\[5pt] & = \frac{1}{M-1}\sum_{n=1}^{M-1}(M-n)\frac{X_n}{n} \nonumber \\[5pt] & = \bigg(1+\frac{1}{M-1}\bigg)T_M^* - \frac{1}{M-1}\sum_{n=1}^{M-1} X_n. \end{align}

Assume $T^*$ is finite. Then, $\lim_{M\to\infty} {T_M^*}/({M-1}) = 0$ almost surely (a.s.). Since $\lim_{M\to\infty}T_M^* \xrightarrow{\textrm{a.s.}} T^*$ , it follows that $\lim_{M\to\infty}S_M \xrightarrow{\textrm{a.s.}} T^*$ . Additionally, by the strong law of large numbers, $\lim_{M\to\infty}({1}/({M-1}))\sum_{n=1}^{M-1} X_n \xrightarrow{\textrm{a.s.}}1$ . Taking the limit $M\to\infty$ in (1) leads to $T^* = T^* - 1$ , contradicting our assumption that $T^*$ is finite. Thus, $T^*=\infty$ and $T_N^*\to \infty$ almost surely.

Proof of Theorem 3. Taking the system with nodes $\{0,\dots,N-1\}$ , consider first the probability that the maximum information level in the population has depth greater than two. This is equivalent to the probability that at least one node has depth greater than two.

For $i > 0$ , we define the variables $\tau^1_{ij}$ as the time when the first meeting between nodes i and j occurs. Note that the $\tau^1_{ij}$ are i.i.d. Exp(1) random variables. Now, fix $t > 0$ and consider the events $A_{ij}^t = \{\tau_{j0} < \tau^1_{ij} < t\}$ , i.e. the event that node j exchanged information with the root before interacting with node i for the first time before time t. Using the fact that the random variables $\tau^1_{ij}$ and $\tau_{j0}$ are independent, it follows that

(2) \begin{equation} \mathbb{P}(A_{ij}^t) = \int_0^t\int_0^s{\textrm{e}}^{-\tau}\times{\textrm{e}}^{-s}\,\text{d}\tau\,\text{d}s = \frac{1}{2}{\textrm{e}}^{-2t}({\textrm{e}}^t - 1)^2 = \varepsilon_t > 0. \end{equation}

The events $A_{ij}^t$ are mutually independent for fixed i and varying j, since meetings between nodes are independent. Furthermore, if event $A_{ij}^t$ occurred, then $\nu_i^N(t) \leq 2$ . Consequently,

\begin{equation*} \mathbb{P}(\nu_i^N(t) > 2) \leq \prod_{\substack{j\neq 0,\, j\neq i}}^{N-1}\mathbb{P}((A_{ij}^t)^\textrm{c}) = (1-\varepsilon_t)^{N-2}. \end{equation*}

Let $F_N^t = \{\max_{i<N}{\nu_i^N(t)} >2\}$ . Thus,

\begin{equation*} \mathbb{P}(F_N^t) = \mathbb{P}\Bigg(\bigcup_{i=1}^{N-1}\{\nu_i^N(t) > 2\}\Bigg) \leq \sum_{i=1}^{N-1}(1-\varepsilon_t)^{N-2} = (N-1)(1-\varepsilon_t)^{N-2}. \end{equation*}

Therefore, $\sum_{N=1}^\infty\mathbb{P}(F_N^t) = \sum_{N=1}^\infty(N-1)(1-\varepsilon_t)^{N-2} = \varepsilon_t^{-2} < \infty$ .

By the Borel–Cantelli lemma, $\mathbb{P}\big(\bigcap_{k=1}^\infty\bigcup_{N=k}^\infty\{\max_{i \leq N}\{\nu_i^N(t)\} > 2\}\big) = 0$ . Thus, $\mathbb{P}(\lim_{N\to \infty} \max_{i\leq N}\{\nu_i^N(t)\} \leq 2) = 1$ , and Theorem 3 is proved.

Proof. Consider the system with infinite nodes. As before, we define the variables $\tau_{ij}$ as the time when the first meeting between nodes i and j occurs, but this time only for $0 \leq i < j$ .

For $i > 0$ , we again define the random variables $\mu_i(t) = 1+\inf_{j > i}\{\mu_j(\tau_{ij})\colon\tau_{ij} < t\}$ , $\mu_i(0) = \infty$ , and $\mu_0(t) = \nu_0(t) = 0$ ; in this case, $\mu_i(t)$ is the quality of information of node i when we allow the nodes to meet and exchange information with nodes j, $j > i$ , only once. Considering again the events $A_{ij}^t = \{\tau_{j0} < \tau_{ij} < t\}$ , i.e. the event that node j exchanged information with the root before interacting with node i before time t. Recall that the events $A_{ij}^t$ are mutually independent for different j. Moreover, for all $t > 0$ , $\sum_{j = i+1}^\infty \mathbb{P}(A_{ij}^t) = \sum_{j = i+1}^\infty \varepsilon_t = \infty$ , where $\varepsilon_t$ is the same as in (2). Thus, by the converse Borel–Cantelli lemma, $\mathbb{P}\big(\bigcup_{j > i} A_{ij}^t\big) = 1$ for any $t > 0$ . Therefore, at least one of the events $A_{ij}^t$ occurred (in fact, infinitely many of them). Consequently, $\mathbb{P}(\nu_i(t) \leq 2) \geq \mathbb{P}(\mu_i(t) \leq 2) = 1$ . Since $\nu_i(t) \leq 2 \Rightarrow \nu_i(t) \in \{1,2\}$ , and $\mathbb{P}(\nu_i(t)=1) = 1-{\textrm{e}}^{-t}$ , the probability that node i interacted with the root, it follows that $\mathbb{P}(\nu_i(t) = 2) = {\textrm{e}}^{-t}$ . As a final step, note that $\nu_i(t) = 1 \iff \{\tau_{i0} < t\}$ , by definition. Ergo, $1 + \mathbf{1}_{\tau_{i0} > t}$ and $\nu_i(t)$ are defined in the same probability space $(\Omega,\mathcal{F},\mathbb{P})$ , and

\begin{equation*} \mathbb{P}(\{\omega \in \Omega\colon \nu_i(t)[\omega] = 1 + \mathbf{1}_{\tau_{i0} > t}[\omega]\}) = 1\Longrightarrow \mathbb{P}(\nu_i(t) = 1 + \mathbf{1}_{\tau_{i0} > t}) = 1. \end{equation*}

Proof of Theorem 4. To complete the proof of Theorem 4, note that $\lim_{N\to\infty} \nu_i^N(t) = \nu_i(t)$ by Definition 1. Thus, from Lemma 1, $\mathbb{P}\big(\lim_{N\to \infty} \nu_i^N(t) = 1 + \mathbf{1}_{\tau_{i0} > t}\big) = 1$ , and since $\mathbb{P}(\tau_{i0} > 0) = 1$ given $\tau_{i0}\sim \textrm{Exp}(1)$ , it follows that $\mathbb{P}\left(\lim_{t\to 0^+} \nu_i(t) = 2\right) = 1$ , and the theorem is proved.

Proof of Theorem 5. Define the random variables $X_n = \inf_{i\neq 0, i \leq n}\{\nu_i(n^{-\alpha})\}$ and $Y_n = \inf_{i\neq 0}\{\nu_i(n^{-\alpha})\}$ for fixed $\alpha \in (0,1)$ , i.e. $Y_n$ is the infimum information level when considering all nodes in the infinite tree at time $n^{-\alpha}$ , while $X_n$ is the infimum information level of the subset of nodes $\{1,\dots,n\}$ of the infinite tree.

Note that, almost surely, $Y_n \leq X_n$ for all n, and $1 \leq Y_n$ . Hence,

\begin{equation*} 1\leq \lim_{t\to 0^+} \inf_{i\neq0}\{\nu_i(t)\} = \lim_{n\to \infty} Y_n \leq \lim_{n\to \infty} X_n. \end{equation*}

We want to show that $X_n \xrightarrow{\textrm{a.s.}} 1$ as $n\to \infty$ . By definition, $\inf_{i\neq 0, i\leq n} \{\nu_i(t)\} > 1 \Leftrightarrow \text{ for all } i\neq 0$ , $i\leq n$ , $\nu_i(t) > 1$ , i.e. not a single node has interacted with the root before time t. As we know, the probability that a given node did not interact with the root before time t is ${\textrm{e}}^{-t}$ . Thus, $\mathbb{P}(X_n > 1) = {\textrm{e}}^{-n^{1-\alpha}}$ , and $\sum_{n=1}^\infty\mathbb{P}(X_n > 1) = \sum_{n=1}^\infty{\textrm{e}}^{-n^{1-\alpha}} < \infty$ .

By the Borel–Cantelli lemma,

\begin{equation*} \mathbb{P}\Big(\lim_{n\to\infty} X_n > 1\Big) = 0 \Longrightarrow \mathbb{P}\Big(\lim_{n\to\infty} X_n = 1\Big) = 1, \end{equation*}

because $\{ X_n > 1\}^\textrm{c} = \{X_n = 1\}$ . Therefore,

\begin{equation*} \mathbb{P}\Big(\lim_{n\to\infty} Y_n \leq 1\Big) = \mathbb{P}\Big(\lim_{n\to\infty} Y_n = 1\Big) = \mathbb{P}\Big(\lim_{t\to 0^+} \inf_{i\neq 0}\{\nu_i(t)\} = 1\Big) = 1. \end{equation*}

Proof of Theorem 6. In our model, the tree is frequently reshaped by reattachment events that move nodes closer to the root. Moreover, the rate and consequences of these events depends on the configuration of the tree. It will turn out, however, that long branches form before any reattachment event has occurred.

Recall that $\tau_{ij}$ denotes the set of times that nodes i and j are in contact, and the information level of node i at time t in the finite-N model is defined as

\begin{equation*} \nu_i^N(t) = 1 + \inf\{\nu_j^N(s) \colon j\neq i, s < t, j\leq N, s \in \tau_{ij}\}. \end{equation*}

Introduce the restricted set of times $\widetilde\tau_{ij}=\tau_{ij}\setminus(\inf\{t\colon\nu_i^N(t)<\infty\},\infty)$ in which all contact events after the initial attachment of i are discarded. These times define a coupled instance of the model in which reattachment is ignored via

\begin{equation*} \widetilde\nu_i^N(t) = 1 + \inf\{\widetilde\nu_j^N(s) \colon j\neq i, s < t, j\leq N, s \in \widetilde\tau_{ij}\}. \end{equation*}

Since contact events can only lower the information level of a node, we immediately have the bound

(3) \begin{equation} \nu_i^N(t)\leq\widetilde\nu_i^N(t). \end{equation}

Additionally, the two processes are identical up to the random time $t_N^\star$ of the first reattachment event. To control the maximal informational level in the original model, we consider the structure of the coupled process at time $t_N^\star$ and in its end state.

The network constructed by the process ignoring reattachment is an instance of a random recursive tree [Reference Mahmoud, Smythe and Szymański18, Reference Pittel20]. Write $\widetilde D_M$ for the length of the longest branch of this tree when it has M nodes. [Reference Pittel20] proved that $\widetilde D_M/\log M \xrightarrow{\textrm{a.s.}} {\textrm{e}}$ as $M\to\infty$ . This result translates immediately to an almost sure upper bound on the original model via (3). Specifically, $D_N<\widetilde{D}_N$ implies

\begin{equation*} \mathbb{P}\bigg(\lim_{N\to\infty}\frac{D_N}{\log N} \leq {\textrm{e}}\bigg) = 1, \end{equation*}

which in turn implies the convergence in probability upper bound in the statement of Theorem 6.

For the lower bound on $D_N$ , we compute the size $M_N^\star$ of the random recursive tree at time $t_N^\star$ . It turns out that $M_N^\star$ is of order $\sqrt{N}$ .

First, note that the rate of attachment events when the informed tree has size n is $n(N-n)$ , while the rate of reattachment events is smaller than $n(n-1)$ . Therefore, the probability $q_n$ that an attachment event occurs before a reattachment event when the tree has size n is bounded from below by

\begin{equation*} q_n \geq \frac{n(N-n)}{n(N-n) + n(n-1)} = \frac{N-n}{N-1}. \end{equation*}

It is in fact neater to study $M^\star_{N+1}$ , since we may compute

\begin{equation*} \mathbb{P}(M^\star_{N+1} \geq m) \geq \prod_{n=1}^m q_n = \frac{N!}{(N-m)!N^m}. \end{equation*}

Using the well-known factorial bounds

\begin{equation*} \sqrt{2\pi N}\bigg(\frac{N}{{\textrm{e}}}\bigg)^N{\textrm{e}}^{{1}/({12N+1})} < N! < \sqrt{2\pi N}\bigg(\frac{N}{{\textrm{e}}}\bigg)^N{\textrm{e}}^{{1}/{12N}}, \end{equation*}

it follows that

\begin{align*} \mathbb{P}(M^\star_{N+1} \geq m) & > \frac{\sqrt{N}N^N\exp\{-N+({1}/({12N+1}))\}}{\sqrt{N-m}(N-m)^{(N-m)}\exp\{-N+m+({1}/({12(N-m)}))\}N^m}, \\[5pt] & = \bigg(\frac{N}{N-m}\bigg)^{N-m+{1}/{2}}\exp\{-m+({1}/({12N+1}))-({1}/({12(N-m)}))\}, \\[5pt] & \geq \exp\bigg\{{-}\frac{m^2}{N} + \frac{m}{2N} + \frac{1}{12N+1} - \frac{1}{12(N-m)}\bigg\}, \end{align*}

where we have used the inequality $\log(N-m) \leq \log N - ({m}/{N})$ .

To understand the scaling law of $M_N$ we consider $m = N^\alpha$ , where $0 < \alpha \leq 1$ . From the above, we obtain

\begin{equation*} \mathbb{P}(M^\star_{N+1} \geq N^\alpha) \geq \exp\bigg\{{-}N^{2\alpha-1} + \frac{1}{2}N^{\alpha-1} + \frac{1}{12N+1} - \frac{1}{12N(1-N^{\alpha-1})}\bigg\}. \end{equation*}

Thus, for all $\alpha<\frac12$ , $\lim_{N\to\infty}\mathbb{P}(M_{N+1}^* \geq N^\alpha) = 1$ , and in particular

\begin{equation*} \lim_{N\to\infty}\mathbb{P}(2\log M_{N+1}^* \geq \log N)= 1. \end{equation*}

To end the proof of Theorem 6, we note that $D_N\geq \widetilde D_{M^\star_N}$ , where it is known that almost surely $\widetilde D_{M^\star_N}/\log M^\star_N \rightarrow {\textrm{e}}$ as $M^\star_N\to\infty$ . So, for any $\epsilon>0$ we may choose N large enough that

\begin{align*} \mathbb{P}(\{2D_{N} > (1-\epsilon){\textrm{e}}(2\log M^\star_N)\} \cap \{2\log M_N^\star \geq \log N\} )>1-\epsilon. \end{align*}

Sending $\epsilon\to0$ and passing to large N, we obtain

\begin{equation*} \lim_{N\to\infty}\mathbb{P}\bigg(\frac{D_N}{\log N} \geq \frac{{\textrm{e}}}{2}\bigg) = 1. \end{equation*}