3.1. Results when $a=0$

Here we consider the situation in which the (common) initial action a equals 0. Theorem 1 provides the limiting distribution of the infected set for arbitrary values of $\tau$ . Let

(3) \begin{align}\alpha=\min\left\{\left \lceil\frac{1}{\tau}\right \rceil,n\right\}.\end{align}

Theorem 1. Suppose $a=0$ . Then the limiting distribution of the infected set is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\alpha-1}{n} & if\ J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ 1\in J \ and\ |J|=m \ where\ m\in[2,\alpha], \\[3pt] 0 & otherwise, \end{cases} \]

where $\alpha$ is as defined in (3).

As $|N_1(\underline{v})| \leqslant n-1$ , the assumption of the case implies $\alpha=\left \lceil\frac{1}{\tau}\right \rceil$ . Hence, $\alpha\tau\geqslant 1$ . By Claim 1, $a_i(S_{k_1})=1$ for all $i\in N_1(\underline{v})$ . Also, by the definition of the process, $a_i(S_{k_1})=0$ for all $i\notin N_1(\underline{v})\cup \{1\}$ , as these agents have not updated their actions till the time point $k_1$ . Recall that $\hat{S}_{k_1}$ denotes the intermediate state whose only difference from $S_{k_1}$ is that agent $\underline{v}_{k_1}$ has updated their action to $b_{\underline{v}_{k_1}}(S_{k_1})$ . Since $\underline{v}_{k_1}=1$ , we have $a_i(S_{k_1})=a_i(\hat{S}_{k_1})$ for all $i\neq 1$ . Thus, $a_i(\hat{S}_{k_1})=1$ for all $i\in N_1(\underline{v})$ and $a_i(\hat{S}_{k_1})=0$ for all $i\notin N_1(\underline{v})\cup \{1\}$ .

By Remark 1 and the definition of the process, $a_1(\hat{S}_{k_1})=1$ . Consider the time point $k_1+1$ . By the definition of the process, an agent $i\neq 1$ will be in $I(S_{k_1+1})$ if $a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})>\tau$ . Since $I(S_{k_1})=\{1\}$ , $a_i(\hat{S}_{k_1})=1$ for all $i\in N_1(\underline{v}) \cup \{1\}$ , $a_i(\hat{S}_{k_1})=0$ for all $i\notin N_1(\underline{v})\cup \{1\}$ , and $g_{ij}=c$ for all $i\neq j$ , it follows that $r_i(\hat{S}_{k_1}) \leqslant \frac{1}{\alpha}$ for all $i\in N_1(\underline{v})$ . Because $\alpha\tau\geqslant 1$ , this implies that no agent in $N_1(\underline{v})$ gets infected at the time point $k_1+1$ . Moreover, since $a_i(\hat{S}_{k_1})=0$ for each agent $i\notin N_1(\underline{v})\cup \{1\}$ , we have $a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})=0 \leqslant \tau$ . Thus, no new agent gets infected at the time point $k_1+1$ , and hence $I(S_{k_1+1})=\{1\}$ .

We show that no new agent will get infected after this. We first show that $I(S_{k_1+2})=\{1\}$ . Let $\underline{v}_{k_1+1}=i$ . If $i\notin I(S_{k_1+1})$ , then as $I(S_{k_1})=I(S_{k_1+1})$ by Lemma 6, we have $I(S_{k_1+1})=I(S_{k_1+2})$ . If $i\in I(S_{k_1+1})$ then $i=1$ . Moreover, $a_1(S_{k_1+1})=a_1(\hat{S}_{k_1})=1$ . Hence, by Lemma 6, $I(S_{k_1+1})=I(S_{k_1+2})$ . Therefore, $I(S_{k_1+2})=\{1\}$ . Using the same arguments repeatedly, it follows that $I(S_t)=\{1\}$ for all $t\geqslant k_1+2$ . Thus, $I(S_\infty)=\{1\}$ .

Using similar arguments as in Case 1, we have $a_i(\hat{S}_{k_1})=1$ for all $i\in N_1(\underline{v})$ and $a_i(\hat{S}_{k_1})=0$ for all $i\notin N_1(\underline{v})\cup \{1\}$ . This, together with $g_{ij}=c$ for all $i\neq j$ , implies $r_i(\hat{S}_{k_1})\geqslant\frac{1}{\alpha-1}$ for all $i\in N_1(\underline{v})$ . As $\alpha=\min\left\{\left \lceil\frac{1}{\tau}\right \rceil,n\right\}$ , we have $(\alpha-1)\tau \lt1$ . Hence, all agents in $N_1(\underline{v})$ will get infected at time point $k_1+1$ . Moreover, as $a_i(\hat{S}_{k_1})=0$ for all $i\notin N_1(\underline{v})\cup \{1\}$ , the agents outside $N_1(\underline{v}) \cup \{1\}$ will not get infected at time point $k_1+1$ . Thus, we have $I(S_{k_1+1})=N_1(\underline{v})\cup \{1\}$ . Because $a_i(S_{k_1+1})=a_i(\hat{S}_{k_1})=1$ for all $i\in I(S_{k_1+1})$ and $a_i(S_{k_1+1})=0 \leqslant \tau$ for all $i\notin I(S_{k_1+1})$ , by Lemma 8 it follows that $I(S_{k_1+1})=I(S_\infty)$ . Hence, $I(S_\infty)=N_1(\underline{v})\cup \{1\}$ .

Step 2. Consider the probability space $(N_\infty, \mathcal{F}, \mathbb{P})$ and random variables S and $t_1,\ldots, t_n$ . Let $m \in \{2,\ldots, n\}$ be such that $m \leqslant \alpha$ . In view of Case 1 and Case 2 in the current proof, we have (i) $|I(S_\infty)| \leqslant \alpha$ , and (ii) $|I(S_\infty)| = m$ with $1\in I(S_\infty)$ if and only if $| \{i \in N \mid t_i \lt t_1\}| = m-1$ . Also, $I(S_\infty) = \{1\}$ if and only if $ |\{i \in N \mid t_i < t_1\}|\geqslant\alpha$ . Moreover, as $\mathbb{P}$ is uniform, any two subsets of N with the same cardinality have the same probability. These observations together yield

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\alpha-1}{n} & \text{if } J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & \text{if } 1\in J \text{ and } |J|=m \text{ where } m\in[2,\alpha], \\[3pt] 0 & \text{ otherwise }. \end{cases} \]

This completes the proof of the theorem.

Next we explore the limiting distribution of the action profile, and this is found to be dependent on the value of $\tau$ . Accordingly, the statement of Theorem 2 is split into two parts on the basis of whether $\tau$ exceeds $(n-1)^{-1}$ or not. We introduce the quantity

(4) \begin{align}\beta=\min \{\lfloor(n-1)\tau \rfloor+1,\alpha+1\}.\end{align}

Theorem 2. Suppose $a=0$ . For $\tau\geqslant \frac{1}{n-1}$ , the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{\alpha-\beta+1}{n} & i\ \vec{x}\in A_n, \ i.e., \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ \vec{x}\in A_m \ for\ some\ m\in [\beta,\alpha],\\[3pt] 0 & otherwise, \end{cases}\]

whereas for $\tau \lt \frac{1}{n-1}$ , the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} \frac{1}{n \times ^{n-1}C_{m-1}} & if\ \vec{x}\in A_m\ for\ some\ m\in [1,n],\\[3pt] 0 & otherwise, \end{cases}\]

with $\alpha$ and $\beta$ as defined in (3) and (4) respectively.

A brief discussion is in order regarding some of the startling findings that may be deduced from the two theorems of Section 3.1. Theorem 1 reveals that if we consider any two different subsets $J_{1}$ and $J_{2}$ of N, the probabilities $\mathbb{P}(I(S_{\infty}) = J_{1})$ and $\mathbb{P}(I(S_{\infty}) = J_{2})$ are the same as long as $J_{1}$ and $J_{2}$ have the same cardinality and either both of them contain agent 1 or neither contains agent 1. We note that the number of subsets J of N with $1 \in J$ and $|J| = m$ is given by $^{n-1}C_{m-1}$ , so that summing $\mathbb{P}(I(S_{\infty}) = J)$ over all such J yields $\mathbb{P}\left(\left|I(S_{\infty})\right| = m\right) = n^{-1}$ for each $m \in [2,\alpha]$ . These observations suggest a rather close resemblance between the limiting distribution of the infected set, as well as the limiting distribution of its cardinality, and certain suitably defined discrete uniform distributions. In fact, for $\tau \leqslant n^{-1}$ , we have $\alpha = n$ , reducing the distribution of $|I(S_{\infty})|$ to precisely the discrete uniform distribution on $\{ 1,2,\ldots, n \}$ . This uniform structure is somewhat marred when $\tau \gt n^{-1}$ . For example, when $n=5$ , $a=0$ , and $\tau= 0.25$ , we have

(5) \begin{align} &\mathbb{P}(|I(S_{\infty})|=1)=\frac{2}{5},\qquad \mathbb{P}(|I(S_{\infty})|=2)=\frac{1}{5},\qquad \mathbb{P}(|I(S_{\infty})|=3)=\frac{1}{5},\nonumber\\[3pt]& \mathbb{P}(|I(S_{\infty})|=4)=\frac{1}{5},\qquad \mathbb{P}(|I(S_{\infty})|=5)=0,\end{align}

whereas if $\tau=0.4$ , the probability distribution changes to

(6) \begin{align}&\mathbb{P}(|I(S_{\infty})|=1)=\frac{3}{5},\qquad \mathbb{P}(|I(S_{\infty})|=2)=\frac{1}{5},\qquad \mathbb{P}(|I(S_{\infty})|=3)=\frac{1}{5},\nonumber\\[3pt]& \mathbb{P}(|I(S_{\infty})|=4)=0,\qquad \mathbb{P}(|I(S_{\infty})|=5)=0.\end{align}

An intuitive explanation for this phenomenon is that with higher immunity, i.e. higher values of $\tau$ , the disease is less likely to spread to the entire community, instead having a higher probability of remaining confined to the initial infected set.

Conclusions of a similar flavor can be drawn as a consequence of Theorem 2. For any two different ordered tuples $\vec{x}$ and $\vec{y}$ that belong to the same $A_m$ , the probabilities $\mathbb{P}(a_{N}(S_{\infty}) = \vec{x})$ and $\mathbb{P}(a_{N}(S_{\infty}) = \vec{y})$ are equal, for both the cases $\tau \geqslant (n-1)^{-1}$ and $\tau < (n-1)^{-1}$ . Moreover, since $|A_m| = ^{n-1}C_{m-1}$ , we obtain $\mathbb{P}(a_{N}(S_{\infty}) \in A_m) = n^{-1}$ for every $m \in [\beta,\alpha]$ when $\tau \geqslant (n-1)^{-1}$ and for every $m \in [1,n]$ when $\tau \lt (n-1)^{-1}$ . These are, once again, reminiscent of suitably defined discrete uniform distributions.

A connection may be established between Theorem 1 and Theorem 2, for the case where $\tau \geqslant (n-1)^{-1}$ , via the following fact, whose justification has been included in the proof of Theorem 2: for any DVSP $S(\underline{v})$ , if the limiting infected set has cardinality $m\in [\beta,\alpha]$ (note that $\beta\geqslant 2$ ), the limiting action profile will be a tuple in $A_m$ , with all infected agents choosing action 1 and all uninfected agents choosing action $\tau m[(1+\tau)m-\tau(n-1)]^{-1}$ . On the other hand, if the limiting infected set for the DVSP $S(\underline{v})$ has cardinality strictly less than $\beta$ , the final action profile becomes $\vec{1}$ , signifying that all agents choose action 1 in the long run.

Recall that for this case the final infected set is $\{1\}$ . Moreover, by the assumption of the theorem, $\tau(n-1)\geqslant 1$ . Therefore, by Lemma 11, $\gamma=1$ . Hence, $a_{N}(S_\infty)=\vec{1}$ .

Recall that for this case, the final infected set is $N_1(\underline{v})\cup \{1\}$ . Therefore, by Lemma 11, if $(n-1)\tau\geqslant |N_1(\underline{v})|+1$ then $a_{N}(S_\infty)=\vec{1}$ , and if $(n-1)\tau \lt|N_1(\underline{v})|+1$ then

\[ a_i(S_\infty)= \begin{cases} 1 & \text{if } i \in I(S_{\infty}),\\[3pt] \frac{\tau(|N_1(\underline{v})|+1)}{(1+\tau)(|N_1(\underline{v})|+1)-\tau(n-1)} & \text{if } i \notin I(S_{\infty}). \end{cases} \]

Recall that $\beta=\min \{\lfloor(n-1)\tau \rfloor+1, \alpha+1\}$ . Hence, combining Cases 1 and 2, we have the following:

1. (i) $|N_1(\underline{v})|+1\in [\beta,\alpha]$ implies

\[ a_i(S_\infty)= \begin{cases} 1 & \text{if } i \in I(S_{\infty}),\\[3pt] \frac{\tau(|N_1(\underline{v})|+1)}{(1+\tau)(|N_1(\underline{v})|+1)-\tau(n-1)} & \text{if } i \notin I(S_{\infty}); \end{cases} \]

1. (ii) $|N_1(\underline{v})|+1\in [1,\beta-1]\cup [\alpha+1,n]$ implies $a_{N}(S_\infty)=\vec{1}$ .

Note that (i) implies $a_{N}(S_\infty)\in A_{(|N_1(\underline{v})|+1)}$ when $|N_1(\underline{v})|+1\in [\beta,\alpha]$ . Also, as $\mathbb{P}$ is uniform, any two vectors in $A_m$ , for $m\in [\beta,\alpha]$ , have the same probability. Thus, we have the following distribution:

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{\alpha-\beta+1}{n}& \text{ if } \vec{x}\in A_n, \text{ i.e., } \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} &\text{ if } \vec{x}\in A_m \text{ for some } m\in [\beta,\alpha], \\[3pt] 0 & \text{ otherwise. } \end{cases} \]

Now assume $\tau \lt \frac{1}{n-1}$ . We follow the same structure as in the previous case; that is, we first explore the limiting actions for a fixed agent sequence, and then we use this to find the limiting probability distribution. Let $\underline{v}$ be an agent sequence and S the DVSP induced by $\underline{v}$ . Note that by Remark 1, it is enough to assume $\underline{v}\in N_\infty$ . Therefore, by Lemma 10, all the agents outside $I(S_\infty)$ have the same action limit, and all the agents in $I(S_\infty)$ have the action limit 1. Let us denote the common limit by $\gamma$ . As $(n-1)\tau \lt1$ by the assumption of the theorem, we have $\alpha=n$ , and hence $|N_1(\underline{v})| \leqslant \alpha-1$ . Moreover, for $|N_1(\underline{v})| \leqslant \alpha-1$ (shown in the proof of Theorem 1), the final infected set is $N_1(\underline{v})\cup \{1\}$ . Thus, $|I(S_\infty)|>(n-1)\tau$ . Hence, by Lemma 11, if $|N_1(\underline{v})|+1<n$ , then

\[ a_i(S_\infty)= \begin{cases} 1 & \text{if } i \in I(S_{\infty}),\\[3pt] \frac{\tau(|N_1(\underline{v})|+1)}{(1+\tau)(|N_1(\underline{v})|+1)-\tau(n-1)} & \text{if } i \notin I(S_{\infty}), \end{cases} \]

and if $|N_1(\underline{v})|+1=n$ , then $a_{N}(S_\infty)=\vec{1}$ . Recall the notation $A_m$ . By the above arguments, we have $a_{N}(S_\infty)\in A_{[|N_1(\underline{v})|+1]}$ when $|N_1(\underline{v})|+1<n$ . Moreover, as $\mathbb{P}$ is uniform, any two vectors in $A_m$ , for $m\in [1,(n-1)]$ , have the same probability. Thus, by Theorem 1, we have the following distribution:

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} \frac{1}{n}& \text{ if } \vec{x}\in A_n, \text{ i.e., } \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} &\text{ if } \vec{x}\in A_m \text{ for some } m\in [1,n-1], \\[3pt] 0 & \text{ otherwise.} \end{cases} \]

3.2. Results when $ \boldsymbol{a}=1$

Here we consider the situation where the (common) initial action a equals 1. The following theorem provides the limiting distribution of the set of infected agents.

Theorem 3. Suppose $a=1$ . If $\tau\geqslant \frac{1}{n-1}$ , the limiting distribution of the infected set is given by

\begin{align*}\mathbb{P}(I(S_\infty)=\{1\})= 1 ,\end{align*}

whereas if $\tau \lt \frac{1}{n-1}$ , the limiting distribution is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} \frac{1}{n^2} & \text{if } 1\in J \ and\ |J|=n-1, \\[3pt] 1-\frac{n-1}{n^2} & if\ |J|=n, \ i.e.,\ J=N, \\[3pt] 0 & otherwise. \end{cases}\]

The proof of this theorem can be found in Appendix B (Subsection B.1). Note that since there are $n-1$ sets J such that $1\in J \text{ and } |J|=n-1$ , the above display exhibits a valid probability distribution.

Theorem 4. Suppose $a=1$ . If $\tau\geqslant \frac{1}{n-1}$ , the limiting distribution of the action profile is given by

\begin{align*}\mathbb{P}(a_{N}(S_\infty)=\vec{1})= 1 ,\end{align*}

whereas if $\tau \lt \frac{1}{n-1}$ , the limiting distribution is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{n-1}{n^2} & if\ \vec{x}\in A_n, \ i.e.,\ \vec{x}=\vec{1}, \\[3pt] \frac{1}{n^2}& if\ \vec{x}\in A_{n-1}, \\[3pt] 0 & otherwise. \end{cases}\]

The proof of this theorem can be found in Appendix D (Subsection D.1). We draw the reader’s attention to the fact that the results of Section 3.2 differ quite a bit in appearance from those of Section 3.1. While the limiting distribution of the infected set, for $a=0$ , is supported on all subsets of N that contain 1 and that have sizes bounded above by $\alpha$ (Theorem 1), the infected set, for $a = 1$ , converges to the singleton $\{1\}$ when $\tau \geqslant (n-1)^{-1}$ , and its limiting distribution is supported on only those subsets of N that contain 1 and have cardinality at least $n-1$ when $\tau \lt (n-1)^{-1}$ (Theorem 3). In some sense, for $a=0$ , the limiting distribution is ‘spread out’ over wider support, while for $a=1$ it is more ‘concentrated’.

Likewise, for $a = 0$ , the limiting distribution of the action profile is supported on all $A_m$ with $m \in \{n\} \cup [\beta,\alpha]$ when $\tau \geqslant (n-1)^{-1}$ , and it is supported on all $A_m$ with $m \in \{1, \ldots, n\}$ when $\tau \lt (n-1)^{-1}$ (Theorem 2). In contrast, for $a=1$ , the action profile converges to $\vec{1}$ when $\tau \geqslant (n-1)^{-1}$ , and the limiting distribution of the action profile is supported on just $A_{n} \cup A_{n-1}$ when $\tau \lt (n-1)^{-1}$ (Theorem 4).

3.3. Results when $0 \lt \boldsymbol{a} \boldsymbol{\leqslant} \tau$ and $\boldsymbol{a}\neq 1$

In this subsection, we consider the case where the (common) initial action a lies strictly between 0 and 1, and is bounded above by $\tau$ . Let

(7) \begin{align}\hat{\alpha}= \max\left\{1,\left\lceil \frac{\frac{1}{\tau}-(n-1)a}{1-a}\right\rceil\right\}.\end{align}

Theorem 5. Suppose $0 \lt a \leqslant \tau$ and $a\neq 1$ . Furthermore, suppose $\tau\geqslant \frac{1}{n-1}$ . Then the limiting distribution of the infected set is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\hat{\alpha}-1}{n} & if\ J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ 1\in J \ and\ |J|=m \ where\ m\in[2, \hat{\alpha}], \\[3pt] 0 & otherwise. \end{cases} \]

The proof of this theorem can be found in Appendix B (Subsection B.2). Next, we describe the limiting distribution of the action profile. We introduce the following notation in order to state our next result:

(8) \begin{align}\hat{\beta}=\min \{\lfloor(n-1)\tau \rfloor+1,\hat{\alpha}+1\}.\end{align}

Theorem 6. Suppose $0 \lt a \leqslant \tau$ and $a\neq 1$ . Furthermore, suppose $\tau\geqslant \frac{1}{n-1}$ . Then the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{\hat{\alpha}-\hat{\beta}+1}{n} & if\ \vec{x}\in A_n,\ i.e.,\ \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ \vec{x}\in A_m \ for\ some\ m\in [\hat{\beta},\hat{\alpha}],\\[3pt] 0 & otherwise. \end{cases}\]

The proof of the theorem can be found in Appendix D (Subsection D.2).

Discussions of a flavor similar to those in Section 3.1 can be included here as well. Even if $J_{1}$ and $J_{2}$ are two different subsets of N, Theorem 5 shows that the probabilities $\mathbb{P}(I(S_{\infty})=J_{1})$ and $\mathbb{P}(I(S_{\infty})=J_{2})$ are the same as long as $J_{1}$ and $J_{2}$ have the same cardinality and either both contain 1 or neither does. Summing over all subsets of N that contain 1 and are of cardinality m, we obtain $\mathbb{P}\left(\left|I(S_{\infty})\right|=m\right) = n^{-1}$ for each $2 \leqslant m \leqslant \hat{\alpha}$ . Likewise, for any two different ordered tuples $\vec{x}$ and $\vec{y}$ , Theorem 6 shows that the probabilities $\mathbb{P}(a_{N}(S_\infty)=\vec{x})$ and $\mathbb{P}(a_{N}(S_\infty)=\vec{y})$ are the same as long as both $\vec{x}$ and $\vec{y}$ belong to the same $A_m$ . Summing over all members of an $A_m$ yields $\mathbb{P}(a_{N}(S_\infty)\in A_m) = n^{-1}$ for every $\hat{\beta} \leqslant m \leqslant \hat{\alpha}$ .

3.4. Results when $\tau \lt \boldsymbol{a} \lt 1$

In this subsection, we consider the scenario where the (common) initial action a is strictly greater than $\tau$ . We introduce the following notation to facilitate the presentation of the subsequent results. Let

(9) \begin{align}\tilde{\alpha}=\max\left\{1, \left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil\right\} \quad\mbox{ and }\quad \bar{\alpha} =\left\lfloor \frac{(n-1)a\tau}{a-\tau(1-a)}\right\rfloor+1.\end{align}

Note that, in this regime, $\tau \lt a \lt a/(1-a)$ and thus $a-\tau(1-a)>0$ . Now since, $(n-1){}a\tau/(a-\tau(1-a))$ is increasing in $\tau$ , we have, for $\tau \geq 1/(n-1)$ ,

\begin{align*}\frac{(n-1)a\tau}{a-\tau(1-a)} \geq \frac{(n-1)a\left(\frac{1}{n-1}\right)}{a-\frac{1-a}{n-1}}=\frac{a}{a- \frac{1-a}{n-1}} >1. \end{align*}

This yields $\bar{\alpha} \geqslant 2$ .

Theorem 7. Suppose $\frac{1}{n-1} \leqslant \tau \lt a \lt 1$ . If $\tilde{\alpha}+1 \leqslant \bar{\alpha}$ , the limiting distribution of the infected set is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\tilde{\alpha}-1}{n} & if\ J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ 1\in J\ and\ |J|=m \ where\ m\in [2,\tilde{\alpha}], \\[3pt] 0 & otherwise, \end{cases} \]

whereas if $2 \leqslant \bar{\alpha} \lt \tilde{\alpha}+1$ , the limiting distribution is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\tilde{\alpha}-1}{n} & if\ J=\{1\}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ 1\in J\ and\ |J|=m \ where\ m\in[2, \bar{\alpha}-1], \\[3pt] \frac{\eta(\tilde{\alpha},\bar{\alpha},n)}{n-1} & if\ 1\in J \ and\ |J|=n-1,\\[3pt] \frac{\tilde{\alpha}-(\bar{\alpha}-1)}{n}-\eta(\tilde{\alpha},\bar{\alpha},n) & if\ |J|=n, \ i.e.,\ J=N,\\[3pt] 0 & otherwise, \end{cases} \]

where

\begin{align*}\eta(\tilde{\alpha},\bar{\alpha},n)=\frac{(n-1)!}{n^3}\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\frac{1}{(n-w-2)!}\sum_{t= w+1}^{\infty} \left(\frac{\left\{\begin{smallmatrix}{t-1}\\ {w}\end{smallmatrix}\right\}}{n^{t-1}}\right), \end{align*}

and $\left\{\begin{smallmatrix} p \\ q \end{smallmatrix} \right\}$ is the Stirling number of the second kind with parameters p and q.

Proof. Note that if $\hat{t}=0$ then there is nothing to show. So assume $\hat{t}\geqslant 1$ . We use induction to prove the statement. As the base case, we show that $I(S_1)=\{1\}$ . Let $\underline{v}_0=i$ . Since $\hat{t}\geqslant 1$ , $i\neq 1$ . Moreover, as $g_{ij}=c$ for all $i\neq j$ , $r_i(S_0)=\frac{1}{(n-1)}$ . Hence,

\begin{align*}b_i(S_0)=\min\left\{1,\frac{\tau}{\frac{1}{(n-1)}}\right\}=\min\{1,(n-1)\tau\}=1, \end{align*}

as by our assumption $\tau\geqslant \frac{1}{(n-1)}$ . This means agent i will not get infected. For any $j\notin \{1,i\}$ , $a_j(\hat{S}_0)=a_j(S_0)=a$ and $r_j(\hat{S}_0)=\frac{a}{(n-2)a+1} \leqslant \frac{1}{(n-1)}$ . Thus,

\begin{align*}a_j(\hat{S}_0)r_j(\hat{S}_0)=a\frac{a}{(n-2)a+1} \leqslant \frac{a}{(n-1)} \leqslant \frac{1}{(n-1)} \leqslant \tau. \end{align*}

So agent j will also not get infected at $t=1$ . Thus, $I(S_1)=\{1\}$ .

Next, we introduce the following induction hypothesis: Given $\bar{t}\in \mathbb{N}_0$ with $\hat{t}\geqslant \bar{t}>1$ , we have $I(S_1)=\cdots=I(S_{\bar{t}-1})=\{1\}$ .

We now show that $I(S_{\bar{t}})=\{1\}$ . Let $\underline{v}_{\bar{t}-1}=i$ . Since $\hat{t}\geqslant \bar{t}$ , this means $i\neq 1$ . Hence, $i\notin I(S_{\bar{t}-1})$ . As $\bar{t} \gt 1$ , we have $I(S_{\bar{t}-2})=I(S_{\bar{t}-1})$ . This, together with Lemma 6, implies $I(S_{\bar{t}})=I(S_{\bar{t}-1})=\{1\}$ . Thus, by induction, we have $I(S_{\hat{t}})=\{1\}$ . This completes the proof of the lemma.

We complete the proof of the theorem in two steps. In Step 1, we explore how the infection spreads when agents update their actions according to a fixed agent sequence, and in Step 2 we use this to explore how infection spreads when agents update their actions randomly.

Step 1: Fix an agent sequence $\underline{v}\in N_\infty$ and let S be the DVSP induced by $\underline{v}$ . To shorten notation, for all $i \in N$ , let us denote $t_i(\underline{v})$ by $k_i$ .

Proof of the claim. Let $\underline{v}_0=i$ . As $k_1>0$ , $i\neq 1$ . Since $a_j(S_0)=a>0$ for all $j\in N$ , $I(S_0)=\{1\}$ , and $g_{ij}=c$ for all $i\neq j$ , we have $r_i(S_0)=\frac{1}{(n-1)}$ . This means

\begin{align*}b_i(S_0)=\min\left\{1,\frac{\tau}{\frac{1}{(n-1)}}\right\}=\min\{1,(n-1)\tau\}=1, \end{align*}

as by our assumption $\tau \geqslant \frac{1}{(n-1)}$ . Thus, $a_i(S_1)=a_i(\hat{S}_0)=1$ .

Next, we introduce the following induction hypothesis: Given $\bar{t}\in \mathbb{N}_0$ with $\bar{t} \lt k_1$ , we have for all $t<\bar{t}$ , $a_j(S_{t+1})=1$ where $\underline{v}_t=j$ .

Now, let $\underline{v}_{\bar{t}}=i^{\prime}$ ; we show that $a_{i^{\prime}}(S_{\bar{t}+1})=1$ . Note that by Lemma 4, $I(S_{\bar{t}})=\{1\}$ . Moreover, by the induction hypothesis, $a_j(S_{\bar{t}})\geqslant a$ for all $j\in N\setminus \{1\}$ . Also, as $\bar{t} \lt k_1$ , we have $a_1(S_{\bar{t}})=a$ . Combining all these observations, we have

(10) \begin{align} \frac{1}{(n-1)}\geqslant r_{i^{\prime}}(S_{\bar{t}})\geqslant \frac{a}{(n-1)}. \end{align}

Since $r_{i^{\prime}}(S_{\bar{t}})>0$ ,

\begin{align*}b_{i^{\prime}}(S_{\bar{t}})=\min\left\{1,\frac{\tau}{r_{i^{\prime}}(S_{\bar{t}})}\right\}; \end{align*}

see Lemma 1. Therefore, using (10) and the fact that $\tau \geqslant \frac{1}{(n-1)}$ , we have $b_{i^{\prime}}(S_{\bar{t}})=1$ . Thus, $a_{i^{\prime}}(S_{\bar{t}+1})=a_{i^{\prime}}(\hat{S}_{\bar{t}})=1$ .

We distinguish some cases based on $|N_1(\underline{v})|$ .

We show that no new agent will get infected and $I(S_\infty)=\{1\}$ . By Claim 1, $a_i(S_{k_1})=1$ for all $i\in N_1(\underline{v})$ . By the definition of the process, $a_i(S_{k_1})=a$ for all $i\notin N_1(\underline{v})\cup \{1\}$ , as these agents have not updated their actions till the time point $k_1$ . Recall that $\hat{S}_{k_1}$ denotes the intermediate state whose only difference from $S_{k_1}$ is that agent $\underline{v}_{k_1}$ has updated their action to $b_{\underline{v}_{k_1}}(S_{k_1})$ . Since $\underline{v}_{k_1}=1$ , we have $a_i(S_{k_1})=a_i(\hat{S}_{k_1})$ for all $i\neq 1$ . Thus, $a_i(\hat{S}_{k_1})=a$ for all $i\in |N_1(\underline{v})|$ and $a_i(\hat{S}_{k_1})=1$ for all $i\notin N_1(\underline{v})\cup \{1\}$ .

By Lemma 1 and the definition of the process, $a_1(\hat{S}_{k_1})=1$ . Consider the time point $k_1+1$ . By the definition of the process, an agent $i\neq 1$ will be in $I(S_{k_1+1})$ if $a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})>\tau$ . Since $I(S_{k_1})=\{1\}$ , $a_i(\hat{S}_{k_1})=1$ for all $i\in N_1(\underline{v}) \cup \{1\}$ , $a_i(\hat{S}_{k_1})=a$ for all $i\notin N_1(\underline{v})\cup \{1\}$ , and $g_{ij}=c$ for all $i\neq j$ , it follows that for all $i\in N_1(\underline{v})$ ,

\begin{align*}r_i(\hat{S}_{k_1})= \frac{1}{|N_1(\underline{v})|+(n-1-|N_1(\underline{v})|)a} \end{align*}

and

(11) \begin{align} a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})&= \frac{1}{|N_1(\underline{v})|+(n-1-|N_1(\underline{v})|)a}\nonumber\\[3pt] &=\frac{1}{|N_1(\underline{v})|(1-a)+(n-1)a}. \end{align}

Recall that by the assumption of the case, $|N_1(\underline{v})|\geqslant \tilde{\alpha}$ . This together with the fact that

\begin{align*}\tilde{\alpha}=\max\left\{1, \left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil\right\} \end{align*}

implies

(12) \begin{align} |N_1(\underline{v})|\geqslant \frac{1-(n-1)a\tau}{\tau(1-a)} \implies \tau \geqslant \frac{1}{|N_1(\underline{v})|(1-a)+(n-1)a}. \end{align}

Combining (11) and (12), we may conclude that agent i will not be infected at the time point $t+1$ . Similar arguments show that any agent $j\notin N_1(\underline{v})\cup \{1\}$ will not be infected at the time point $t+1$ . Hence, $I(S_{k_1+1})=\{1\}$ .

We show that no new agent will get infected after this. We first show that $I(S_{k_1+2})=\{1\}$ . Let $\underline{v}_{k_1+1}=i$ . If $i\notin I(S_{k_1+1})$ , then as $I(S_{k_1})=I(S_{k_1+1})$ by Lemma 6, we have $I(S_{k_1+1})=I(S_{k_1+2})$ . If $i\in I(S_{k_1+1})$ then $i=1$ . Moreover, $a_1(S_{k_1+1})=a_1(\hat{S}_{k_1})=1$ . Hence, by Lemma 6, $I(S_{k_1+1})=I(S_{k_1+2})$ . Therefore, $I(S_{k_1+2})=\{1\}$ . Using the same arguments repeatedly, it follows that $I(S_t)=\{1\}$ for all $t\geqslant k_1+2$ . Thus, $I(S_\infty)=\{1\}$ .

In the following claim, we show that at time point $k_1+1$ , the infected set is $N_1(\underline{v})\cup \{1\}$ .

Proof of the claim. Recall that $\tilde{\alpha}=\max\left\{1, \left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil\right\}$ . First assume $\tilde{\alpha}\neq \left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil$ , i.e., $\tilde{\alpha}=1$ and $1-(n-1)a\tau \leqslant 0$ . This, together with the assumption of the case, implies $|N_1(\underline{v})|=0$ . Therefore, $k_1=1$ . Hence, to prove the claim, it is enough to show that $I(S_1)=\{1\}$ . Note that by the definition of the process, $a_1(\hat{S}_0)=1$ , $a_i(\hat{S}_{0})=a$ for all $i\neq 1$ , and $g_{ij}=c$ for all $i\neq j$ . Thus,

\begin{align*} r_i(\hat{S}_{0})&=\frac{1}{1+(n-2)a} \\[3pt] & \leqslant\frac{1}{(n-1)a} \\[3pt] & \leqslant \tau \hspace{20mm}\text{ (since } 1-(n-1)a\tau \leqslant 0).\end{align*}

This implies $I(S_{1})=\{1\}$ . Now assume $\tilde{\alpha}=\left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil$ . Consider an agent $i\in N_1(\underline{v})$ . Using similar arguments as in (11), we may show that

\begin{align*}a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})=\frac{1}{|N_1(\underline{v})|(1-a)+(n-1)a}. \end{align*}

This, together with the fact that $\tilde{\alpha}=\left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil$ and $|N_1(\underline{v})| \leqslant \tilde{\alpha}-1$ , implies $a_i(\hat{S}_{k_1})r_i(\hat{S}_{k_1})>\tau$ , and hence agent i will get infected at the time point $t+1$ . For any $j\notin N_1(\underline{v})\cup \{1\}$ ,

\begin{align*}r_j(\hat{S}_{k_1})= \frac{1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a} \end{align*}

and

\begin{align} a_j(\hat{S}_{k_1})r_j(\hat{S}_{k_1})&= \frac{a}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a}\nonumber\\[3pt] &=\frac{a}{|N_1(\underline{v})|(1-a)+(n-2)a+1}. \nonumber \end{align}

Hence agent j gets infected at $t+1$ if

\begin{align} \frac{a}{|N_1(\underline{v})|(1-a)+(n-2)a+1}>\tau \implies \frac{a-\tau(n-1)a}{\tau(1-a)}>|N_1(\underline{v})|. \nonumber \end{align}

But this does not hold, as $\frac{a-\tau(n-1)a}{\tau(1-a)} \leqslant 0$ and $|N_1(\underline{v})|\geqslant 0$ . So agent j does not get infected at $t+1$ . Thus, $I(S_{k_1+1})=N_1(\underline{v})\cup 1$ . This completes the proof of the claim.

We now determine the final infected set. To do so we consider two sub-cases based on the value of $|N_1(\underline{v})|$ .

We show that no new agent will get infected after $k_1+1$ . We first show that $I(S_{k_1+2})=N_1(\underline{v})\cup \{1\}$ . Let $\underline{v}_{k_1+1}=i$ . If $i\in I(S_{k_1+1})$ then $i\in N_1(\underline{v})\cup \{1\}$ . Moreover, $a_i(S_{k_1+1})= a_i(\hat{S}_{k_1})=1$ . Hence, by Lemma 6, $I(S_{k_1+1})=I(S_{k_1+2})$ . If $i\notin I(S_{k_1+1})$ then since

\begin{align*}r_i(S_{k_1+1})=\frac{|N_1(\underline{v})|+1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a}\neq 0, \end{align*}

agent i will choose $\min\left\{1,\frac{\tau}{r_i(S_{k_1+1})}\right\}$ as their action $a_i(\hat{S}_{k_1+1})$ at $\hat{S}_{k_1+1}$ . This means $a_i(\hat{S}_{k_1+1})r_i(\hat{S}_{k_1+1}) \leqslant \tau$ and agent i will not get infected at $k_1+2$ . To show that any agent $j\in I(S_{k_1+1})\setminus \{i\}$ will not get infected at $k_1+2$ , we first prove a claim.

Proof of the claim. Note that if $a_i(\hat{S}_{k_1+1})=1$ , then the claim holds, as $a \leqslant 1$ . If $a_i(\hat{S}_{k_1+1})=\frac{\tau}{r_i(S_{k_1+1})}$ , then

(13) \begin{align} a_i(\hat{S}_{k_1+1})&=\frac{\tau}{r_i(S_{k_1+1})} \nonumber\\[3pt] &=\tau(1-a)+\frac{\tau a(n-1)}{|N_1(\underline{v})|+1}. \end{align}

Moreover, by the assumption of the case $|N_1(\underline{v})|+1<\bar{\alpha}$ . This together with $\bar{\alpha} =\left\lfloor \frac{(n-1)a\tau}{a-\tau(1-a)}\right\rfloor +1$ and (13) implies

\begin{align} a_i(\hat{S}_{k_1+1})&\geqslant \tau(1-a)+\frac{[\tau a(n-1)](a-\tau(1-a))}{(n-1)a\tau} \nonumber \\[3pt] &=a. \nonumber \end{align}

This completes the proof of the claim.

For any $j\notin I(S_{k_1+1})\setminus \{i\}$ , $a_j(\hat{S}_{k_1+1})=a$ and

\begin{align*}r_j(\hat{S}_{k_1+1})=\frac{|N_1(\underline{v})|+1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-2)a+a_i(\hat{S}_{k_1+1})} \quad \text{ (as } g_{ij}=c \text{ for all } i\neq j). \end{align*}

Thus,

Hence agent j will not get infected at $k_1+2$ . This allows us to conclude that $I(S_{k_1+2})=N_1(\underline{v})\cup \{1\}$ . Now, using similar logic as in Case 1, we can show that no agent will get infected after this and $I(S_\infty)=N_1(\underline{v})\cup \{1\}$ .

First assume that $\underline{v}_{k_1+1}=i$ where $i\in N_1(\underline{v})\cup \{1\}$ . We show that $I(S_{\infty})=N$ . Note that as $i\in N_1(\underline{v})\cup \{1\}$ , $a_i(\hat{S}_{k_1+1})=1$ . Thus, for any $j\notin N_1(\underline{v})\cup \{1\}$ ,

\begin{align*}r_j(\hat{S}_{k_1+1})=\frac{|N_1(\underline{v})|+1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a} \quad \text{ (as } g_{ij}=c \text{ for all } i\neq j), \end{align*}

and hence

\begin{align} a_j(\hat{S}_{k_1+1})r_j(\hat{S}_{k_1+1})&= \frac{a(|N_1(\underline{v})|+1)}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a} \nonumber \\[3pt] &= \frac{a(|N_1(\underline{v})|+1)}{(|N_1(\underline{v})|+1)(1-a)+(n-1)a}\nonumber \\[3pt] &\geqslant \frac{a\bar{\alpha}}{\bar{\alpha}(1-a)+(n-1)a} \hspace{10mm} \mbox{ (as } |N_1(\underline{v})|+1\geqslant \bar{\alpha}) \nonumber \\[3pt] &> \tau \hspace{35mm} \mbox{ (as } \bar{\alpha}>\frac{(n-1)a\tau}{a-\tau(1-a)}). \nonumber \end{align}

Therefore, $I(S_{k_1+2})=N$ and $I(S_\infty)=N$ .

Now assume that $\underline{v}_{k_1+1}=i$ where $i\notin N_1(\underline{v})\cup \{1\}$ . We show that $I(S_{k_1+2})=N\setminus i$ . Since $i\notin I(S_{k_1+1})$ and $\underline{v}_{k_1+1}=i$ , agent i will not get infected at $k_1+2$ (Observation 4). Consider $j\notin N_1(\underline{v})\cup 1$ with $j\neq i$ . We first prove a claim.

Proof of the claim. We show that $\tau \lt \frac{\tau}{r_i(S_{k_1+1})} \lt a$ . This together with $a<1$ proves the claim. As

\begin{align*}r_i(S_{k_1+1})=\frac{|N_1(\underline{v})|+1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-1)a}<1, \end{align*}

we have $\tau \lt \frac{\tau}{r_i(S_{k_1+1})}$ . To see that $\frac{\tau}{r_i(S_{k_1+1})} \lt a$ , recall that by (13),

\begin{align*}a_i(\hat{S}_{k_1+1})=\tau(1-a)+\frac{\tau a(n-1)}{|N_1(\underline{v})|+1}. \end{align*}

Moreover, by the assumption of the case, $|N_1(\underline{v})|+1\geqslant \bar{\alpha}$ . This, together with $\bar{\alpha} > \frac{(n-1)a\tau}{a-\tau(1-a)}$ , implies

\begin{align} a_i(\hat{S}_{k_1+1})&\lt \tau(1-a)+\frac{[\tau a(n-1)](a-\tau(1-a))}{(n-1)a\tau} \nonumber \\[3pt] &=a. \nonumber \end{align}

This completes the proof of the claim.

For any $j\notin I(S_{k_1+1})\setminus \{i\}$ , $a_j(\hat{S}_{k_1+1})=a$ and

\begin{align*}r_j(\hat{S}_{k_1+1})=\frac{|N_1(\underline{v})|+1}{|N_1(\underline{v})|+1+(n-1-|N_1(\underline{v})|-2)a+a_i(\hat{S}_{k_1+1})} \quad \text{ (as } g_{ij}=c \text{ for all } i\neq j). \end{align*}

Thus,

Hence agent j will get infected at $k_1+2$ . This allows us to conclude that $I(S_{k_1+2})=N\setminus \{i\}$ .

To determine the final infected set, we now distinguish two cases based on whether $\underline{v}_{k_1+2}=i$ or not.

We show that the final infected set will be $N\setminus i$ . Since by our assumption $\underline{v}_{k_1+2}=i$ and $i\notin I(S_{k_1+2})$ , by Observation 4, $i\notin I(S_{k_1+3})$ . Hence, $I(S_{k_1+3})=N\setminus \{i\}$ . We now show that agent i will not get infected after this. At time point $k_1+2$ ,

\begin{align*}r_i(\hat{S}_{k_1+2})=\frac{(|N_1(\underline{v})|+1)+a(n-1-|N_1(\underline{v})|-1)}{(|N_1(\underline{v})|+1)+(n-1-|N_1(\underline{v})|-1)a}=1. \end{align*}

Therefore, $a_i(\hat{S}_{k_1+2})=\tau$ . At $k_1+3$ , if $\underline{v}_{k_1+3}=i$ , then agent i will not get infected at $k_1+4$ (Observation 4). On the other hand, if $\underline{v}_{k_1+3}\neq i$ , then as $a_i(\hat{S}_{k_1+3})=a_i(\hat{S}_{k_1+2})=\tau$ , agent i will remain uninfected at $k_1+4$ . Continuing in this manner, we can show that agent i will not get infected after this. Thus, $I(S_\infty)=N\setminus \{i\}$ .

We show that the final infected set will be N. Since $I(S_{k_1+2})=N\setminus \{i\}$ , $r_i(\hat{S}_{k_1+2})=1$ . Moreover, as $a_i(S_{k_1+2})=a_i(\hat{S}_{k_1+1})>\tau$ (by Claim 4) and $\underline{v}_{k_1+2}\neq i$ , it follows that $a_i(\hat{S}_{k_1+2})>\tau$ . Combining these two facts, we have $a_i(\hat{S}_{k_1+2})r_i(\hat{S}_{k_1+2})>\tau$ . Thus, agent i will get infected at $k_1+3$ . Hence, $I(S_{k_1+3})=N$ and $I(S_\infty)=N$ .

Step 2: To begin with, we claim $\tilde{\alpha} \leqslant n-1$ . To see this, observe that

\begin{align*}\left\lceil \frac{1-(n-1)a\tau}{\tau(1-a)}\right\rceil \leqslant \left\lceil \frac{1-a}{\tau(1-a)}\right\rceil \leqslant n-1, \end{align*}

as $(n-1)\tau\geqslant 1$ . Hence, $\tilde{\alpha} \leqslant n-1$ . Moreover, recall that $\bar{\alpha}\geqslant 2$ . We now find the distribution of $I(S_\infty)$ . First, assume that $\tilde{\alpha}+1 \leqslant \bar{\alpha}$ . Therefore, by the above cases we have the following:

• • $I(S_\infty)=\{1\}$ if $|N_1(\underline{v})|\in \{0,\tilde{\alpha},\tilde{\alpha}+1,\ldots,n-1\}$ , and

• • $I(S_\infty)=N_1(\underline{v})\cup \{1\}$ if $|N_1(\underline{v})|\in \{1,2,\ldots,\tilde{\alpha}-1\}$ .

Moreover, as $\mathbb{P}$ is uniform, any two subsets of N with the same cardinality have the same probability. These observations together with Lemma 3 yield

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\tilde{\alpha}-1}{n} & \text{if } J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & \text{if } 1\in J \text{ and } |J|=m \text{ where } m\in [2,\tilde{\alpha}], \\[3pt] 0 & \text{ otherwise. } \end{cases} \]

Now assume that $\tilde{\alpha}+1> \bar{\alpha}\geqslant 2$ . By Case 1 and Case 2, we have the following:

1. (i) $I(S_\infty)=\{1\}$ if $|N_1(\underline{v})|\in \{0,\tilde{\alpha},\tilde{\alpha}+1,\ldots,n-1\},$

2. (ii) $|I(S_\infty)|=|N_1(\underline{v})|+1$ with $1\in I(S_\infty)$ if $|N_1(\underline{v})|\in \{1,2,\ldots,\bar{\alpha}-2\},$

3. (iii) $|I(S_\infty)|=n$ if $|N_1(\underline{v})|\in \{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\}$ and there is no $i\in N$ such that $k_i=k_1+1$ and $\underline{v}_{k_1+2}=i$ , and

4. (iv) $|I(S_\infty)|=n-1$ with $1\in I(S_\infty)$ if $|N_1(\underline{v})|\in \{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\}$ and there is $i\in N$ such that $k_i=k_1+1$ and $\underline{v}_{k_1+2}=i$ .

Since $|N_1|$ follows a uniform distribution on $\{0,1,\ldots,n-1\}$ and any two subsets of N with the same cardinality have the same probability, by (i) and (ii) we have

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\tilde{\alpha}-1}{n} & \text{if } J=\{1\},\\[3pt]\frac{1}{n \times ^{n-1}C_{m-1}} & \text{if } 1\in J \text{ and } |J|=m \text{ where } m\in[2,\bar{\alpha}-1]. \end{cases} \]

We calculate the probability of $|I(S_\infty)|=n-1$ . By (iv) we have

\begin{align*} &\mathbb{P}(\underline{v}\mid |N_1(\underline{v})|\in \{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\} \text{ and } \exists i\neq 1 \text{ such that } k_i=k_1+1 \text{ and } \underline{v}_{k_1+2}=i)\\[3pt] =&\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}P(\underline{v}\mid |N_1(\underline{v})|=w \text{ and } \exists i\neq 1 \text{ such that } k_i=k_1+1 \text{ and } \underline{v}_{k_1+2}=i)\\[3pt] =&\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\sum_{t= w+1}^{\infty}P(\underline{v}\mid |N_1(\underline{v})|=w \text{ and } k_1=t \text{ and } \exists i\neq 1 \text{ such that } k_i=t+1 \text{ and } \underline{v}_{t+2}=i)\\[3pt] =&\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\sum_{t= w+1}^{\infty} {^{n-1}C_{w}}\times \left(\frac{w!\{\begin{smallmatrix}{t-1}\\[3pt]{w} \end{smallmatrix}\}}{n^{t-1}}\right)\times \frac{1}{n} \times ^{(n-w-1)}C_{1}\times \frac{1}{n^2}\\[3pt] =&\,\frac{(n-1)!}{n^3}\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\frac{1}{(n-w-2)!}\sum_{t= w+1}^{\infty} \left( \frac{ \{\begin{smallmatrix}{t-1}\\[3pt]{w} \end{smallmatrix}\}}{n^{t-1}} \right)\\[3pt] =&\,\eta(\tilde{\alpha},\bar{\alpha},n). \end{align*}

Note that by (i)–(iv),

\begin{align*}\sum_{m=1}^{\bar{\alpha}-1}P(|I(S_\infty)|=m)+P(|I(S_\infty)|=n-1)+P(|I(S_\infty)|=n)=1. \end{align*}

Therefore,

\begin{align*} P(|I(S_\infty)|=n)&=1-\sum_{m=1}^{\bar{\alpha}-1}P(|I(S_\infty)|=m)-P(|I(S_\infty)|=n-1)\\[3pt] &=\frac{\tilde{\alpha}-(\bar{\alpha}-1)}{n}-\eta(\tilde{\alpha},\bar{\alpha},n). \end{align*}

Since any two subsets of N with the same cardinality have the same probability, combining all of the above observations yields the following distribution of the infected set:

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} 1-\frac{\tilde{\alpha}-1}{n} & \text{if } J=\{1\} , \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & \text{if } 1\in J \text{ and } |J|=m \text{ where } m\in[2, \bar{\alpha}-1], \\[3pt] \frac{\eta(\tilde{\alpha},\bar{\alpha},n)}{n-1} & \text{if } 1\in J \text{ and } |J|=n-1,\\[3pt] \frac{\tilde{\alpha}-(\bar{\alpha}-1)}{n}-\eta(\tilde{\alpha},\bar{\alpha},n) & \text{if } |J|=n, \text{ i.e., } J=N,\\[3pt] 0 & \text{ otherwise.} \end{cases} \]

This completes the proof of the theorem.

As in the previous subsections, we now proceed to explore the limiting distribution of the action profile. The following notation will help us present the results:

(14) \begin{align}\tilde{\beta}=\min \{\lfloor(n-1)\tau \rfloor+1,\tilde{\alpha}+1\}, \qquad \bar{\beta}=\min \{\lfloor(n-1)\tau \rfloor+1,\bar{\alpha}\}.\end{align}

Theorem 8. Suppose $\frac{1}{n-1} \leqslant \tau \lt a \lt 1$ . If $\tilde{\alpha}+1 \leqslant \bar{\alpha}$ , the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{\tilde{\alpha}-\tilde{\beta}+1}{n} & if\ \vec{x}\in A_n, \ i.e.,\ \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ \vec{x}\in A_m \ for\ some\ m\in [\tilde{\beta},\tilde{\alpha}],\\[3pt] 0 & otherwise, \end{cases}\]

whereas if $2 \leqslant \bar{\alpha} \lt \tilde{\alpha}+1$ , the limiting distribution is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1+\frac{\bar{\beta}-\bar{\alpha}}{n}-\eta(\tilde{\alpha},\bar{\alpha},n) & if\ \vec{x}\in A_n, \ i.e.,\ \vec{x}=\vec{1}, \\[3pt] \frac{1}{n \times ^{n-1}C_{m-1}} & if\ \vec{x}\in A_m \ for\ some\ m\in [\bar{\beta},\bar{\alpha}-1],\\[3pt] \frac{\eta(\tilde{\alpha},\bar{\alpha},n)}{n-1}& if\ \vec{x}\in A_{n-1}, \\[3pt] 0 & otherwise, \end{cases}\]

where

\begin{align*}\eta(\tilde{\alpha},\bar{\alpha},n)=\frac{(n-1)!}{n^3}\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\frac{1}{(n-w-2)!}\sum_{t= w+1}^{\infty} \left(\frac{\{\begin{smallmatrix}{t-1}\\[3pt]{w} \end{smallmatrix}\}}{n^{t-1}}\right), \end{align*}

and $\left\{\begin{smallmatrix} p \\[3pt]q \end{smallmatrix} \right\}$ is the Stirling number of the second kind with parameters p and q.

The proof of this theorem can be found in Appendix D (Subsection D.3). Once again, observations similar to those made in Sections 3.1 and 3.3 could be made here for Theorems 7 and 8, but to avoid unnecessary repetition, we do not elaborate upon them.

The next two theorems deal with the situation when $\tau$ is less than or equal to $\frac{a}{(n-1)}$ . Theorem 9 characterizes the limiting distribution of the infected set and Theorem 10 characterizes the limiting distribution of the action profile.

Theorem 9. Suppose $\tau \lt a \lt 1$ . If $\tau= \frac{a}{n-1}$ , the limiting distribution of the infected set is given by

\begin{align*}\mathbb{P}(I(S_\infty)=N)= 1, \end{align*}

whereas if $\tau \lt \frac{a}{n-1}$ , the limiting distribution is given by

\[ \mathbb{P}(I(S_\infty)=J)= \begin{cases} \frac{1}{n^2} & if\ 1\in J \ and\ |J|=n-1, \\[3pt] 1-\frac{n-1}{n^2} & if\ |J|=n, \ i.e.,\ J=N, \\[3pt] 0 & otherwise. \end{cases}\]

The proof of this theorem can be found in Appendix B (Subsection B.3).

Theorem 10. Suppose $\tau \lt a \lt 1$ . If $\tau= \frac{a}{n-1}$ , the limiting distribution of the action profile is given by

\begin{align*}\mathbb{P}(a_{N}(S_\infty)=\vec{1})= 1, \end{align*}

whereas if $\tau \lt \frac{a}{n-1}$ , the limiting distribution is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{x})= \begin{cases} 1-\frac{n-1}{n^2} & if\ \vec{x}\in A_n, \ i.e.,\ \vec{x}=\vec{1}, \\[3pt] \frac{1}{n^2}& if\ \vec{x}\in A_{n-1}, \\[3pt] 0 & otherwise. \end{cases}\]

The proof of this theorem can be found in Appendix D (Subsection D.4). The results in this final set of theorems are also consistent with our intuition, which says that if $\tau$ is small enough, the eventually infected set is either of size $n-1$ or of size n with probability 1.

4.1. Results when $\boldsymbol{a}=0$

In this subsection, we consider the situation where the (common) initial action a equals 0. The next theorem describes the limiting distribution of the infected set of agents and their action profiles. As we stated earlier, it shows that all the agents will recover in the long run, and the action profile will have a degenerate distribution at $\vec{1}$ .

Theorem 11. Suppose $a=0$ . Then the limiting distribution of the infected set is given by

\[ \mathbb{P}(I(S_\infty)=\emptyset)=1, \]

and the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{1})=1.\]

The assumption of the claim implies that agent 1 will recover before getting a chance to update their action. As their initial action is 0 and they are the only agent infected at the beginning, this means that for any other agent i, $r_i(\hat{S}_t)=0$ for all t with $0\leq t\leq \kappa$ . Therefore, no new agent will be infected till $\kappa$ . At $\kappa$ , agent 1 will recover as per the process, and no one further will get infected.

Note that the assumption implies that agent 1 gets a chance to update their action before they recover. Thus, as in Theorem 1, we consider two sub-cases. Recall the definition of $\alpha$ as defined in (3).

As in Case 1 of Theorem 1, we can show that $I(S_1)=\cdots=I(S_{k_1+1})=\{1\}$ and $a_{1}(\hat{S}_{k_1})=1$ . Hence, by Lemma 5, $I(S_\infty)=\emptyset$ .

As $k_1< \kappa$ , using similar arguments as in Case 2 of Theorem 1, we can show that $N_1(\underline{v}) \subseteq I(S_{k_1+1})\subseteq N_1(\underline{v})\cup \{1\}$ with $a_{i}(S_{k_1+1})=1$ for all $i\in I(S_{k_1+1})$ and $a_{i}(S_{k_1+1})=0$ for all $i\notin I(S_{k_1+1})$ . At epoch $k_1+1$ , if $\underline{v}_{k_1+1}\in I(S_{k_1+1})$ then they will choose the the same action 1, and we will have $I(S_{k_1+2})\subseteq I(S_{k_1+1})$ . If $\underline{v}_{k_1+1}\notin I(S_{k_1+1})$ , they will choose their action $b_{\underline{v}_{k_1+1}}(S_{k_1+1})=\tau$ , as all the uninfected agents have action 0 and all the infected agents have action 1. Therefore, no new agent will get infected at $k_1+2$ , and we have $I(S_{k_1+2})\subseteq I(S_{k_1+1})$ . As $a_i(\hat{S}_{k_1+1})=1$ for all $i\in I(S_{k_1+1})$ , by Lemma 5, it follows that $I(S_\infty)=\emptyset$ .

As $\underline{v}$ is an arbitrary agent sequence in $N_\infty$ and the cases are exhaustive, we have $\mathbb{P}(I(S_\infty)=\emptyset)=1$ .

For the second part of the theorem, take any agent sequence $\underline{v} \in N_\infty$ . Then we have $I(S_\infty)=\emptyset$ . This means there exists $\hat{t}$ such that $I(S_{\hat{t}})=I(S_{\hat{t}+1})=\cdots=\emptyset$ . As there is no infected agent after epoch $\hat{t}$ , any agent who updates their action after epoch $\hat{t}$ will choose their best response as 1. This, together with the fact that at $\underline{v}$ every agent appears infinitely many times, lets us conclude that $a_N(S_\infty)=\vec{1}$ . Therefore, $ \mathbb{P}(a_{N}(S_\infty)=\vec{1})=1$ .

4.2. Results when $\boldsymbol{a}=1$

This subsection considers the situation where the (common) initial action a equals 1. As was the case for $a=0$ , here too we show that in the long run the epidemic vanishes, and people roam freely.

Theorem 12. Suppose $a=1$ . Then the limiting distribution of the infected set is given by

\[ \mathbb{P}(I(S_\infty)=\emptyset)=1, \]

and the limiting distribution of the action profile is given by

\[ \mathbb{P}(a_{N}(S_\infty)=\vec{1})=1.\]

Proof. We prove the theorem by showing that for any agent sequence $\underline{v}$ in $N_\infty$ , $I(S_\infty)=\emptyset$ , where S is the DVSP induced by $\underline{v}$ . Consider an agent sequence $\underline{v}$ and the DVSP S induced by $\underline{v}$ . First, assume that $\kappa\geq 2$ . Because of this assumption, using similar arguments as in the proof of Theorem 3, we can show the following:

1. (i) If $\tau\geq \frac{1}{n-1}$ , then $I(S_1)=\{1\}$ .

2. (ii) If $\tau \lt \frac{1}{n-1}$ , then one of the following holds:

   1. (a) $I(S_1)=N$ and $a_j(S_1)=1$ for all $j\in N$ ,

   or

3. (b) there exists $i\in N\setminus 1$ with $a_j(S_1)=a_j(S_2)=1$ for all $j\in N\setminus i$ such that either $I(S_1)=I(S_2)=N\setminus i$ or $I(S_1)=N\setminus i$ and $I(S_2)=N$ .

It is easy to see that if either (i) or (ii)-(a) holds, by Lemma 5, we will have $I(S_\infty)=\emptyset$ . Suppose (ii)-(b) holds with $I(S_1)=I(S_2)=N\setminus i$ , $a_j(S_1)=a_j(S_2)=1$ for all $j\in N\setminus i$ . This means $I(S_1)=I(S_2)=N\setminus i$ and $a_j(\hat{S}_1)=1$ for all $j\in N\setminus i$ . Thus, by Lemma 5, $I(S_\infty)=\emptyset$ . Now, suppose (ii)-(b) holds with $I(S_1)=N\setminus i$ , $I(S_2)=N$ with $a_j(S_1)=a_j(S_2)=1$ for all $j\in N\setminus i$ . At epoch $\kappa\geq 2$ , agent 1 recovers, i.e., $I(S_\kappa)=N\setminus 1$ . This means $r_i(S_\kappa)=1$ . So, if agent 1 gets a chance to update their action, they will choose $a_1(\hat{S}_\kappa)=\tau$ as their best response. Otherwise, it would be $a_1(\hat{S}_\kappa)=0$ . Thus, agent 1 will not get infected at $\kappa+1$ . Hence $I(S_{\kappa+1})=i$ , as all the agents except i will recover at $\kappa+1$ . At epoch $\kappa+1$ , all the uninfected agents have action less than or equal to $\tau$ ; thus, whichever agent updates their action at $\kappa+1$ , no one will get infected at $\kappa+2$ . Furthermore, at epoch $\kappa+2$ agent i will get infected. Therefore, $I(S_{\kappa+2})=\emptyset$ . No one will get infected after this; hence $I(S_\infty)=\emptyset$ . This shows that if $\kappa\geq 2$ , $I(S_\infty)=\emptyset$ .

Now, assume that $\kappa=1$ . This means agent 1 will recover at epoch 1 and $a_1(S_1)=0$ . Suppose $\tau\geq \frac{1}{n-1}$ . Using similar arguments as in the proof of Theorem 3, we can show that at epoch 1, no new agent will get infected, implying $I(S_1)=\emptyset$ . Therefore, $I(S_\infty)=\emptyset$ . Suppose $\tau \lt \frac{1}{n-1}$ . If $\underline{v}_0=1$ , then, using similar arguments as in the proof of Theorem 3, it can be shown that at epoch 1, all the agents other than 1 will get infected. We claim that $I(S_2)=\emptyset$ , yielding $I(S_\infty)=\emptyset$ . To see this, observe that if agent 1 updates their action at epoch 2, they will change it to $\tau$ , as all the other agents are infected and have non-zero actions. Otherwise, their action will remain the same, i.e., $a_1(\hat{S}_1)=0$ . Thus, agent 1 will not get further infected at epoch 2. Also, as $\kappa=1$ , all the other agents will recover at epoch 2. Hence, $I(S_2)=\emptyset$ .

If $\underline{v}_0=i\;(\neq 1)$ , then at epoch 1, all the agents other than agents i and 1 will get infected (using similar arguments as in Case 2 of Theorem 3). Moreover, agent i will update their action to $(n-1)\tau$ , i.e., $a_i(S_1)=(n-1)\tau$ . We consider different possibilities for $\underline{v}_1$ . First, consider $\underline{v}_1=i$ . This means that at epoch 2, agent i will not get infected, and as $a_1(S_1)=0$ , agent 1 will not get infected either. Furthermore, all the other agents recover at epoch 2. Combining these facts, we have $I(S_2)=\emptyset$ . Now consider $\underline{v}_1=j\; (\notin \{1,i\})$ . As $j\in I(S_1)$ and $a_j(S_1)=1$ , this means $a_j(\hat{S}_1)=1$ . Thus, at epoch 2, agent i will get infected (see Case 2.2 in the proof of Theorem 3). Also, as $\kappa=1$ , all agents except 1 and i will recover at this epoch. This means $a_k(S_2)=0$ for all $k\neq i$ (recall that $a_(S_1)=0$ and $\underline{v}_1\neq 1$ ). Therefore, none of these agents will get infected at the next epoch, i.e., at epoch 3. In addition, agent i will be recovered, implying $I(S_3)=\emptyset$ .

Finally, consider $\underline{v}_1=1$ . As $I(S_1)=N\setminus \{1,i\}$ with $a_k(S_1)=1$ for all $k\in I(S_1)$ and $a_i(S_1)=(n-1)\tau$ , agent 1 will choose their action as

\begin{align*}a_1(\hat{S}_1)=\tau\frac{(n-2)+(n-1)\tau}{(n-2)}. \end{align*}

Also, at epoch 2, agent i will get infected (see Case 2.2 in the proof of Theorem 3), and all the agents in $N\setminus \{1,i\}$ will be recovered. Thus,

\begin{align*}a_1(S_2)=\tau\frac{(n-2)+(n-1)\tau}{(n-2)} \end{align*}

and $a_k(S_2)=0$ for all $k\in N\setminus \{1,i\}$ . If $\underline{v}_2=1$ , then as $I(S_2)=i$ , $a_i(S_2)>0$ , and $a_k(S_2)=0$ for all $k\in N\setminus \{1,i\}$ , agent 1 will update their action to $\tau$ and will not get infected in the next epoch. Furthermore, agent i will be recovered at the next epoch, implying that $I(S_3)=\emptyset$ . If $\underline{v}_2=i$ , as $I(S_2)=i$ , agent i will update their action to 1, and as a result agent 1 will get infected again at epoch 3. However, as $\kappa=1$ , agent i will recover at epoch 3. This means all the uninfected agents at epoch 3 have action 0. Hence, at epoch 4, no new agent will get infected and agent 1 will recover, implying $I(S_4)=\emptyset$ . If $\underline{v}_2=r\; (\notin \{1,i\})$ , agent r will update their action to

\begin{align*}\frac{\tau}{b_r(S_2)}=\frac{\tau ((n-1)\tau+a_1(S_2))}{(n-1)\tau}. \end{align*}

We claim $I(S_3)=\emptyset$ . To see this, note that agent i will be recovered at epoch 3. Among the other agents, only agents 1 and r have positive actions. So it is enough to show that agents 1 and r will not get infected. As agent r updates their action to $a_r(\hat{S}_2)=\frac{\tau}{b_r(S_2)}$ , they will not get infected. For agent 1, we show that they will not get infected by showing that $a_r(\hat{S}_2)>a_1(\hat{S}_2)$ . Note that $a_1(\hat{S}_2)=a_1(\hat{S}_1)$ and

\begin{align*}a_r(\hat{S}_2)=\frac{\tau ((n-1)\tau+a_1(S_2))}{(n-1)\tau}. \end{align*}

Thus, $a_r(\hat{S}_2)>a_1(\hat{S}_2)$ holds if and only if

\begin{align*} &\tau\bigg(\frac{(n-1)\tau+\tau\frac{(n-2)+(n-1)\tau}{(n-2)}}{(n-1)\tau}\bigg)>\tau \frac{(n-2)+(n-1)\tau}{(n-2)} \\[3pt] \iff & 1+\frac{(n-2)+(n-1)\tau}{(n-2)(n-1)}>\frac{(n-2)+(n-1)\tau}{(n-2)}\\[3pt] \iff& \frac{(n-2)+(n-1)\tau}{(n-2)(n-1)}>\frac{(n-1)\tau}{(n-2)}\\[3pt] \iff& \frac{1}{(n-1)}>\tau.\end{align*}

Therefore, $I(S_3)=\emptyset$ . This shows that $I(S_\infty)=\emptyset$ for $\kappa=1$ and completes the first part of the theorem. The second part of the theorem follows from the same arguments used in the proof of Theorem 11.

4.3. Evidence from simulation

Although we furnish a thorough simulation study for the model without recovery in Section 6, here we briefly mention that we also ran a simulation study for this general model with recovery, and the following are our findings:

• • For any $0 \lt \tau \lt1$ and $\kappa>0$ , for both $a=0$ and $a=1$ , the population does indeed become completely uninfected.

• • Consistent with our intuition, the time it takes for the population to become disease-free is monotonic with $\kappa$ ; that is, with higher recovery time, it takes longer for the disease to vanish.

• • Similar results indicating a disease-free population seem to be true for the $0 \lt a<1$ cases as well; however, since we leave the proofs for these scenarios of general a to future work, we do not comment here on the time it takes for the population to become disease-free, which depends on the value of a.

5.1. Nash equilibria of the game at the initial state

We now investigate the Nash equilibria of the game induced at the initial state. Since we treat it as a static game (in particular, the state is not allowed to change), we consider a general setup where an arbitrary set of agents S, $S \subseteq N$ , is infected at the initial state. Following the formulation in Section 2 (and simplifying certain expressions for the present case), the game at the initial state of the infection is defined as $G=\langle N, (A_i)_{i \in N}, (v_i)_{i \in N} \rangle $ , where

• • $N=\{1,2,\ldots,n\}$ is the set of players;

• • $A_i =[0,1]$ is the set of actions of each player $i \in N$ ;

• • for each action profile $\vec{a}:=(a_1,\ldots,a_n)\in [0,1]^n,$

  • • for $i \in S$ , $v_i(\vec{a})=f(a_i)$ , and

  • • for $i \in N\setminus S$ ,

    (16) \begin{align} v_i(\vec{a}) = \begin{cases} 1+f(a_i) & \text{if } a_i r_i(\vec{a}) \leqslant \tau, \\[3pt] f(a_i) & \text{if } a_i r_i(\vec{a})> \tau, \end{cases}\end{align}
    where $f\,:\,[0,1]\to [0,1]$ is a strictly increasing function, and
    \[ r_i(\vec{a}) = \begin{cases} \left(\frac{\sum_{j\in S}a_j}{\sum_{j\in N\setminus \{i\}}a_j}\right) & \text{if } \sum_{j\in N\setminus \{i\}}a_j \neq 0, \\[3pt] 0 & \text{if } \sum_{j\in N\setminus \{i\}}a_j = 0. \end{cases}\]

We now define the notion of (pure) Nash equilibrium for static games.

Definition 1. An action profile $a_N=(a_1,\ldots,a_n)$ is a (pure) Nash equilibrium of a game $G = \langle N, (A_i)_{i \in N}, (v_i)_{i \in N} \rangle $ if, for all $i\in N$ and all $a_i^{\prime}\in A_i$ ,

\begin{align*}v_i(a_N)\geq v_i(a_i^{\prime},a_{-i}). \end{align*}

The following theorem characterizes all Nash equilibria of the game G.

Proof. We first show that in any Nash equilibrium of the game G, an uninfected agent will remain uninfected. Assume for the sake of contradiction that there is a Nash equilibrium $\vec{a}=(a_1,\ldots,a_n)$ where the uninfected agent i becomes infected. Since $\vec{a}$ is a Nash equilibrium, we have

(17) \begin{align} u_i(\vec{a})\geq u_i(a_i^{\prime},a_{-i}) \text{ for all } a_i^{\prime}\in [0,1].\end{align}

As, by our assumption, agent i is infected at $\vec{a}$ , we have $u_i(\vec{a})=f(a_i)$ . Consider $a_i^{\prime}=\tau$ . This means agent i will remain uninfected at $(a_i^{\prime},a_{-i})$ , and hence $u_i(a_i^{\prime},a_{-i})=1+f(a_i^{\prime})$ . But this contradicts (17), as $f\,:\,[0,1]\to [0,1]$ is a strictly increasing function. Therefore, in any Nash equilibrium of the game G, all the agents other than agents in S will remain uninfected. Also, as agents in S have the utility function f, they will always choose the action 1 in a Nash equilibrium.

Next we show that in any Nash equilibrium of the game G, any two uninfected agents will have the same action. Assume for the sake of contradiction that there is a Nash equilibrium of the game $\vec{a}=(a_1,\ldots,a_n)$ where two uninfected agents i and j have different actions, i.e., $a_i\neq a_j$ . Without loss of generality we may assume that $a_i>a_j$ . We show that $u_j(\vec{a})<u_j(a_j^{\prime},a_{-j})$ where $a_j^{\prime}=a_i$ , a contradiction to the fact that $\vec{a}$ is a Nash equilibrium. Note that as $\vec{a}$ is a Nash equilibrium, agent j will remain uninfected, and hence $u_j(\vec{a})=1+f(a_j)$ . For the profile $(a_j^{\prime},a_{-j})$ ,

\begin{align*} a_j^{\prime}r_j((a_j^{\prime},a_{-j}))=\frac{a_j^{\prime} |S|}{|S|+\sum_{k\notin S\cup j}a_k} &=\frac{a_j^{\prime}|S|}{1+\sum_{k\notin S\cup \{i,j\}}a_k+a_i} \\[3pt] & <\frac{a_j^{\prime}}{1+\sum_{k\notin S\cup \{i,j\}}a_k+a_j} \hspace{20mm} (\text{ as } a_i>a_j)\\[3pt] & =\frac{a_i}{1+\sum_{k\notin S\cup \{i,j\}}a_k+a_j} \hspace{20mm} (\text{ as } a_j^{\prime}=a_i)\\[3pt] &\leq \tau \hspace{30mm} (\text{ as agent } i \text{ is uninfected at } \vec{a}).\end{align*}

This means agent j will get infected at $(a_j^{\prime},a_{-j})$ . Hence, $u_j(a_j^{\prime},a_{-j})=1+f(a_j^{\prime})>1+f(a_j)=u_j(\vec{a})$ . This shows that in any Nash equilibrium of the game G, all the uninfected agents will have the same action.

We are now ready to complete the proof of the lemma. Let $\vec{a}$ be a Nash equilibrium of the game G. First, assume that $\tau \lt \frac{|S|}{n-1}$ . As discussed before, all the agents in S will choose their action as 1. Therefore, $a_i=1$ for all $i\in S$ . Consider $i\notin S$ . As agent i will remain uninfected at $\vec{a}$ and $a_i$ is their best action given the actions of the others, it must be that

\begin{align*}a_ir_i(\vec{a})=a_i\frac{|S|}{|S|+\sum_{k\notin S\cup i}a_k}=\tau. \end{align*}

This together with the fact that $a_j=a_l$ for all $j,l \in N\setminus S$ implies

\begin{align*} a_i\frac{|S|}{|S|+\sum_{k\notin S\cup i}a_k}=\tau &\implies x\frac{|S|}{|S|+\sum_{k\notin S\cup i}x}=\tau \hspace{15mm} (\text{where } a_j=x \text{ for all } j\neq 1)\\[3pt] &\implies x=\frac{\tau |S|}{|S|-(n-|S|-1)\tau}.\end{align*}

Note that as $\tau \lt \frac{|S|}{n-1}$ , the following holds:

\begin{align*} x=\frac{\tau |S|}{|S|-(n-|S|-1)\tau} \leq \frac{\tau |S|}{|S|-(n-|S|-1)\frac{|S|}{n-1}}=\frac{\tau (n-1)}{|S|}\leq 1,\end{align*}

\begin{align*}\text{ and } |S|-(n-|S|-1)\tau \geq |S|-(n-|S|-1)\frac{|S|}{n-1}= \frac{|S|^2}{n-1}\geq 0.\end{align*}

Thus, $0\leq x\leq 1$ . Now assume that $\tau \geq \frac{|S|}{n-1}$ . As for the action profile $\vec{a}=\vec{1}$ , we have

\begin{align*}a_ir_i(\vec{a})=\frac{|S|}{n-1}\leq \tau \end{align*}

for all $i\in N\setminus S$ , so no agent in $N\setminus S$ will get infected if they choose the action 1. As 1 is the maximum possible action, the unique Nash equilibrium will be $\vec{1}$ .

6.1. Empirical distribution of the number of infected agents up to a few epochs

In our paper so far, we have provided the asymptotic distribution of the cardinality of the final infected set, but it remains to understand how fast this distribution or a close approximation of it is reached. In this section, we provide a very thorough exploration of the cardinality of the infected set up to a few epochs of time. Note that, for the distribution of the total number of infected people in the population, we required an exact enumeration of all possible sequences; thus it becomes difficult to compute this beyond 10 or 11 epochs. So the number we will report for this case is an exact probability enumeration based on 10 epochs. We set $n=5$ and consider the various values of a and $\tau$ listed in Table 1.

Table 1. Exact enumeration of empirical distribution after 10 epochs. Here $(p_1,\ldots,p_5)$ denotes $(\mathbb{P}(|I(S_{\infty})|=1),\ldots,\mathbb{P}(|I(S_{\infty})|=5)) $ , and $*$ denotes regions that are not covered by our theoretical results.

Table 2. Theoretical distributions of action profiles and their sum. Here vectors stand for the class where the last four entries can be permuted. For example, (1,1,1,0.33,0.33) stands for the class containing the six vectors (1,1,1,0.33,0.33), (1,1,0.33,0.33,1), (1,0.33,0.33,1,1), (1,1,0.33,1,0.33), (1,0.33,1,0.33,1), and (1,0.33,1, 1, 0.33).

In particular, we see that after only 10 epochs for $n=5$ , we can reach very good approximations of the final distributions. Moreover, the number of epochs needed to get a close approximation is much smaller. One can see that except for the very first case of $a=0$ , $\tau=0.12$ , we have fairly good convergence to the actual distribution in 10 steps. Generally speaking, for smaller a and $\tau$ it takes longer to get close to the asymptotic distribution.

6.2. Empirical distribution of the action profile

A very pertinent question about some simulation evidence for the asymptotic distribution of the final action set was asked by one reviewer. First we must acknowledge that, even for $n=5$ , exactly tabulating all possible empirical distributions of a 5-dimensional vector is a challenging problem. Moreover, while we did run the simulation, we saw that for most cases the empirical distribution (with complete enumeration) after 10 or 11 epochs was somewhat far from the theoretical ones. For this reason, we decided to extend our empirical probability calculation to a larger number of epochs. As complete enumeration was computationally challenging, even for $n=5$ , we decided to randomly sample 50000 sequences of epoch lengths 50, 200, and 400. After a lot of deliberation on how to concisely report the average performance of the 5-dimensional vector after these epoch lengths, we decided to report the following:

• • the theoretical distributions in Table 2, and

• • the empirical distribution of the sum of the action profiles after 50, 200, and 400 epochs.

We choose the following cases:

• • $a=0$ , with $\tau=0.1$ , $0.3$ , $0.6$ , and $0.9$ ;

• • $a=0.4$ , with $\tau=0.05$ , $0.3$ , $0.4$ , and $0.9$ ;

• • $a=0.7$ , with $\tau=0.1$ , $0.3$ , $0.6$ , and $0.9$ ;

• • $a=1$ , with $\tau=0.1$ , $0.5$ , $0.6$ , and $0.9$ .

Figure 1. Sum of action profiles for $a=0$ , for $n=5$ .

Figure 2. Sum of action profiles for $a=0.4$ , for $n=5$ .

Figure 3. Sum of action profiles for $a=0.7$ , for $n=5$ .

Solely for presentation purposes, we have omitted a few cases from Table 2 and presented them instead in Figures 1, 2, 3, and 4. One can see in Figure 1 that for very small $\tau$ the uniform distribution in the five possible classes was reached after 200 epochs, whereas for larger $\tau$ the theoretical distributions were reached quite quickly. For large a, however, say $a=1$ , as shown in Figure 4, the convergence is faster. This is consistent with our findings in Table 1, which shows the results of a complete enumeration.

Figure 4. Sum of action profiles for $a=1$ , for $n=5$ .

Apart from matching the sum of actions in each case, which we present visually, we would like to draw the attention of the reader to the following cases:

• • $a=0.4$ , $\tau=0.3$ , and

• • $a=0.7$ , $\tau=0.3$

In each of these cases, the theoretical distribution from Table 2 includes a class with a negligible theoretical probability (of 0.006 in the first case and 0.01 in the second). One can see that even though these small classes were not prominent after 50 epochs, they became so after 200 epochs. Such strong empirical evidence bolsters the likelihood that our theoretical findings are accurate. Moreover, it says that the time needed for the action profile to eventually reach the limiting distribution depends on the values of a and $\tau$ .

6.3. Large sample size

Lastly we want to understand how well our results hold empirically for a large population, and, more importantly, how fast we converge to the theoretical distribution. For this as well, we have to rely on evaluating a random sample of sequences, as complete enumeration for even moderately large n, around 20 or 50, is nearly impossible to achieve. That said, we were able to achieve the theoretical distributions by evaluating a large number (50000) of permutations of the same length as the corresponding epoch. We choose the scenario $a=0$ , $\tau <1/(n-1)$ to exhibit this dynamics. Our theory in Theorem 1 says that the number of infected agents asymptotically converges to Uniform( $1, \cdots, n$ ). In Figure 5, we choose $n=5,20,50$ and obtain the dynamics after 20, 50, and 100 epochs.