$\displaystyle\displaystyle\leqslant\frac{a(|N_{1}(\undertilde{v})|+1)}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-1)a}\hskip
28.45274pt\mbox{ (as by Claim }\ref{cl_1},a_{i}(\hat{S}_{k_{1}+1})\geqslant
a)$ $\displaystyle\displaystyle=ar_{i}(\hat{S}_{k_{1}+1})$
$\displaystyle\displaystyle\leqslant\tau\hskip 28.45274pt\mbox{ (as
}a_{i}(\hat{S}_{k_{1}+1})r_{i}(\hat{S}_{k_{1}+1})\leqslant\tau\text{ and
}a_{i}(\hat{S}_{k_{1}+1})\geqslant a).$

Hence, agent $\displaystyle j$ will not get infected at $\displaystyle
k_{1}+2$. This concludes that $\displaystyle
I(S_{k_{2}+1})=N_{1}(\undertilde{v})\cup\\{1\\}$. Now using similar logic as
in Case 1, we may show that no agent would get infected after this and
$\displaystyle I(S_{\infty})=N_{1}(\undertilde{v})\cup\\{1\\}$.

Case 2.2: $\displaystyle|N_{1}(\undertilde{v})|+1\geqslant\bar{\alpha}$.
First assume that $\displaystyle\undertilde{v}_{k_{1}+1}=i$ where
$\displaystyle i\in N_{1}(\undertilde{v})\cup\\{1\\}$. We show that
$\displaystyle I(S_{\infty})=N$. Note that as $\displaystyle i\in
N_{1}(\undertilde{v})\cup\\{1\\}$, $\displaystyle a_{i}(\hat{S}_{k_{1}+1})=1$.
Thus, for any $\displaystyle j\notin N_{1}(\undertilde{v})\cup\\{1\\}$,

$\displaystyle
r_{j}(\hat{S}_{k_{1}+1})=\frac{|N_{1}(\undertilde{v})|+1}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-1)a}$

and hence,

$\displaystyle\displaystyle a_{j}(\hat{S}_{k_{1}+1})r_{j}(\hat{S}_{k_{1}+1})$
$\displaystyle\displaystyle=\frac{a(|N_{1}(\undertilde{v})|+1)}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-1)a}$
$\displaystyle\displaystyle=\frac{a(|N_{1}(\undertilde{v})|+1)}{(|N_{1}(\undertilde{v})|+1)(1-a)+(n-1)a}$
$\displaystyle\displaystyle\geqslant\frac{a\bar{\alpha}}{\bar{\alpha}(1-a)+(n-1)a}\hskip
28.45274pt\mbox{ (as }|N_{1}(\undertilde{v})|+1\geqslant\bar{\alpha})$
$\displaystyle\displaystyle>\tau.\hskip 99.58464pt\mbox{ (as
}\bar{\alpha}>\frac{(n-1)a\tau}{a-\tau(1-a)})$

Therefore, $\displaystyle I(S_{k_{1}+2})=N$ and $\displaystyle
I(S_{\infty})=N$.

Now assume that $\displaystyle\undertilde{v}_{k_{1}+1}=i$ where $\displaystyle
i\notin N_{1}(\undertilde{v})\cup\\{1\\}$. We show that $\displaystyle
I(S_{k_{1}+2})=N\setminus i$. Since $\displaystyle i\notin I(S_{k_{1}+1})$ and
$\displaystyle\undertilde{v}_{k_{1}+1}=i$, agent $\displaystyle i$ will not
get infected at $\displaystyle k_{1}+2$ (Observation 2.5). Consider
$\displaystyle j\notin N_{1}(\undertilde{v})\cup 1$ with $\displaystyle j\neq
i$. We first prove a claim.

######  Claim C.4.

$\displaystyle\tau<a_{i}(\hat{S}_{k_{1}+1})<a$.

Proof of the claim: We show that
$\displaystyle\tau<\frac{\tau}{r_{i}(S_{k_{1}+1})}<a$. This together with
$\displaystyle a<1$ proves the claim. As $\displaystyle
r_{i}(S_{k_{1}+1})=\frac{|N_{1}(\undertilde{v})|+1}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-1)a}<1$,
we have $\displaystyle\tau<\frac{\tau}{r_{i}(S_{k_{1}+1})}$. To see
$\displaystyle\frac{\tau}{r_{i}(S_{k_{1}+1})}<a$, recall that by (C.8)

$\displaystyle a_{i}(\hat{S}_{k_{1}+1})=\tau(1-a)+\frac{\tau
a(n-1)}{|N_{1}(\undertilde{v})|+1}.$

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

$\displaystyle\displaystyle a_{i}(\hat{S}_{k_{1}+1})$
$\displaystyle\displaystyle<\tau(1-a)+\frac{[\tau
a(n-1)](a-\tau(1-a))}{(n-1)a\tau}.$ $\displaystyle\displaystyle=a.$

This completes the proof of the claim. $\displaystyle\square$

For any $\displaystyle j\in I(S_{k_{1}+1})\setminus\\{i\\}$, $\displaystyle
a_{j}(\hat{S}_{k_{1}+1})=a$ and

$\displaystyle
r_{j}(\hat{S}_{k_{1}+1})=\frac{|N_{1}(\undertilde{v})|+1}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-2)a+a_{i}(\hat{S}_{k_{1}+1})}.$

Thus,

$\displaystyle\displaystyle a_{j}(\hat{S}_{k_{1}+1})r_{j}(\hat{S}_{k_{1}+1})$
$\displaystyle\displaystyle=\frac{a(|N_{1}(\undertilde{v})|+1)}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-2)a+a_{i}(\hat{S}_{k_{1}+1})}$
$\displaystyle\displaystyle>\frac{a(|N_{1}(\undertilde{v})|+1)}{|N_{1}(\undertilde{v})|+1+(n-1-|N_{1}(\undertilde{v})|-1)a}\hskip
28.45274pt\mbox{ (as by Claim }\ref{cl_2},a_{i}(\hat{S}_{k_{1}+1})<a)$
$\displaystyle\displaystyle=\frac{a(|N_{1}(\undertilde{v})|+1)}{(|N_{1}(\undertilde{v})|+1)(1-a)+(n-1)a}$
$\displaystyle\displaystyle\geqslant\frac{a\bar{\alpha}}{\bar{\alpha}(1-a)+(n-1)a}\hskip
28.45274pt\mbox{ (as }|N_{1}(\undertilde{v})|+1\geqslant\bar{\alpha})$
$\displaystyle\displaystyle>\tau.\hskip 28.45274pt\mbox{ (as
}\bar{\alpha}>\frac{(n-1)a\tau}{a-\tau(1-a)})$

Hence, agent $\displaystyle j$ will get infected at $\displaystyle k_{1}+2$.
This concludes that $\displaystyle I(S_{k_{1}+2})=N\setminus\\{i\\}$.

To determine the final infected set, we now distinguish two cases based on
whether $\displaystyle\undertilde{v}_{k_{1}+2}=i$ or not.

Case 2.2.1: $\displaystyle\undertilde{v}_{k_{1}+2}=i$
We show that the final infected set will be $\displaystyle N\setminus i$.
Since by our assumption $\displaystyle\undertilde{v}_{k_{1}+2}=i$ and
$\displaystyle i\notin I(S_{k_{1}+2})$, by Observation 2.5, $\displaystyle
i\notin I(S_{k_{1}+3})$. Hence, $\displaystyle
I(S_{k_{1}+3})=N\setminus\\{i\\}$. We now show that $\displaystyle i$ will not
get infected after this. At time point $\displaystyle k_{1}+2$,

$\displaystyle
r_{i}(\hat{S}_{k_{1}+2})=\frac{(|N_{1}(\undertilde{v})|+1)+a(n-1-|N_{1}(\undertilde{v})|-1)}{(|N_{1}(\undertilde{v})|+1)+(n-1-|N_{1}(\undertilde{v})|-1)a}=1.$

Therefore, $\displaystyle a_{i}(\hat{S}_{k_{1}+2})=\tau$. At $\displaystyle
k_{1}+3$, if $\displaystyle\undertilde{v}_{k_{1}+3}=i$, then agent
$\displaystyle i$ would not get infected at $\displaystyle k_{1}+4$
(Observation 2.5). On the other hand, if
$\displaystyle\undertilde{v}_{k_{1}+3}\neq i$ then as $\displaystyle
a_{i}(\hat{S}_{k_{1}+3})=a_{i}(\hat{S}_{k_{1}+2})=\tau$, agent $\displaystyle
i$ would remain uninfected at $\displaystyle k_{1}+4$. Continuing in this
manner, we may show that $\displaystyle i$ will not get infected after this.
Thus, $\displaystyle I(S_{\infty})=N\setminus\\{i\\}$.

Case 2.2.2: $\displaystyle\undertilde{v}_{k_{1}+2}\neq i$
We show that the final infected set will be $\displaystyle N$. Since
$\displaystyle I(S_{k_{1}+2})=N\setminus\\{i\\}$, $\displaystyle
r_{i}(\hat{S}_{k_{1}+2})=1$. Moreover as $\displaystyle
a_{i}(S_{k_{1}+2})=a_{i}(\hat{S}_{k_{1}+1})>\tau$ (by Claim C.4) and
$\displaystyle\undertilde{v}_{k_{1}+2}\neq i$, it follows that $\displaystyle
a_{i}(\hat{S}_{k_{1}+2})>\tau$. Combining this two we have $\displaystyle
a_{i}(\hat{S}_{k_{1}+2})r_{i}(\hat{S}_{k_{1}+2})>\tau$. Thus, agent
$\displaystyle i$ will get infected at $\displaystyle k_{1}+3$. Hence,
$\displaystyle I(S_{k_{1}+3})=N$ and $\displaystyle I(S_{\infty})=N$.

Step 2: We now find the distribution of $\displaystyle I(S_{\infty})$. Note
that by the definition, $\displaystyle\tilde{\alpha}\leqslant n-1$ and
$\displaystyle\bar{\alpha}\geqslant 2$. First assume that
$\displaystyle\tilde{\alpha}+1\leqslant\bar{\alpha}$. Therefore, by the above
cases we have

  * •

$\displaystyle I(S_{\infty})=\\{1\\}$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{0,\tilde{\alpha},\tilde{\alpha}+1,\ldots,n-1\\}$,

  * •

$\displaystyle I(S_{\infty})=N_{1}(\undertilde{v})\cup\\{1\\}$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{1,2,\ldots,\tilde{\alpha}-1\\}$.

Moreover, as $\displaystyle\mathbb{P}$ is uniform, any two subsets of
$\displaystyle N$ with same cardinality have the same probability. These
observations together with Lemma 2.8 yield

$\displaystyle\mathbb{P}(I(S_{\infty})=J)=\begin{cases}1-\frac{\tilde{\alpha}-1}{n}&\text{if
}J=\\{1\\},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{if }1\in J\text{
and }|J|=m\text{ where }m\in[2,\tilde{\alpha}],\\\ \\\ 0&\text{ otherwise.
}\end{cases}$

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

  1. (i)

$\displaystyle I(S_{\infty})=\\{1\\}$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{0,\tilde{\alpha},\tilde{\alpha}+1,\ldots,n-1\\},$

  2. (ii)

$\displaystyle|I(S_{\infty})|=|N_{1}(\undertilde{v})|+1$ with $\displaystyle
1\in I(S_{\infty})$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{1,2,\ldots,\bar{\alpha}-2\\},$

  3. (iii)

$\displaystyle|I(S_{\infty})|=n$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\\}$
and there is no $\displaystyle i\in N$ such that $\displaystyle k_{i}=k_{1}+1$
and $\displaystyle\undertilde{v}_{k_{1}+2}=i$, and

  4. (iv)

$\displaystyle|I(S_{\infty})|=n-1$ with $\displaystyle 1\in I(S_{\infty})$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\\}$
and there is $\displaystyle i\in N$ such that $\displaystyle k_{i}=k_{1}+1$
and $\displaystyle\undertilde{v}_{k_{1}+2}=i$.

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

$\displaystyle\mathbb{P}(I(S_{\infty})=J)=\begin{cases}1-\frac{\tilde{\alpha}-1}{n}&\text{if
}J=\\{1\\},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{if }1\in J\text{
and }|J|=m\text{ where }m\in[2,\bar{\alpha}-1].\end{cases}$

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

$\displaystyle\displaystyle\mathbb{P}(\undertilde{v}\mid|N_{1}(\undertilde{v})|\in\\{\bar{\alpha}-1,\ldots,\tilde{\alpha}\\}\text{
and }\exists i\neq 1\text{ such that }k_{i}=k_{1}+1\text{ and
}\undertilde{v}_{k_{1}+2}=i)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}P(\undertilde{v}\mid|N_{1}(\undertilde{v})|=w\text{
and }\exists i\neq 1\text{ such that }k_{i}=k_{1}+1\text{ and
}\undertilde{v}_{k_{1}+2}=i)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\sum_{t=w+1}^{\infty}P(\undertilde{v}\mid|N_{1}(\undertilde{v})|=w\text{
and }k_{1}=t\text{ and }\exists i\neq 1\text{ such that }k_{i}=t+1\text{ and
}\undertilde{v}_{t+2}=i)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\sum_{t=w+1}^{\infty}\prescript{n-1\mkern-0.5mu}{}{C}_{w}\times\left(\frac{w!\genfrac{\\{}{\\}}{0.0pt}{}{t-1}{w}}{n^{t-1}}\right)\times\frac{1}{n}\times\prescript{(n-w-1)\mkern-0.5mu}{}{C}_{1}\times\frac{1}{n^{2}}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\frac{(n-1)!}{n^{3}}\sum_{w=\bar{\alpha}-1}^{\tilde{\alpha}-1}\frac{1}{(n-w-2)!}\sum_{t=w+1}^{\infty}\left(\frac{\genfrac{\\{}{\\}}{0.0pt}{}{t-1}{w}}{n^{t-1}}\right)$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\eta(\tilde{\alpha},\bar{\alpha},n).$

Note that by (i)-(iv),

$\displaystyle\sum_{m=1}^{\bar{\alpha}-1}P(|I(S_{\infty})|=m)+P(|I(S_{\infty})|=n-1)+P(|I(S_{\infty})|=n)=1.$

Therefore,

$\displaystyle\displaystyle P(|I(S_{\infty})|=n)$
$\displaystyle\displaystyle=1-\sum_{m=1}^{\bar{\alpha}-1}P(|I(S_{\infty})|=m)-P(|I(S_{\infty})|=n-1)$
$\displaystyle\displaystyle=\frac{\tilde{\alpha}-(\bar{\alpha}-1)}{n}-\eta(\tilde{\alpha},\bar{\alpha},n).$

Since any two subsets of $\displaystyle N$ with the same cardinality have the
same probability, combining all the above observations, we have the following
distribution of the infected set.

$\displaystyle\mathbb{P}(I(S_{\infty})=J)=\begin{cases}1-\frac{\tilde{\alpha}-1}{n}&\text{if
}J=\\{1\\},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{if }1\in J\text{
and }|J|=m\text{ where }m\in[2,\bar{\alpha}-1],\\\ \\\
\frac{\eta(\tilde{\alpha},\bar{\alpha},n)}{n-1}&\text{if }1\in J\text{ and
}|J|=n-1,\\\ \\\
\frac{\tilde{\alpha}-(\bar{\alpha}-1)}{n}-\eta(\tilde{\alpha},\bar{\alpha},n)&\text{if
}|J|=n,\text{ i.e., }J=N,\\\ \\\ 0&\text{ otherwise.}\end{cases}$

This completes the proof of the theorem. ∎

###  C.5. Proof of Theorem 3.11

###### Proof.

We follow the same structure that we used in the proof of Theorem 3.2.

Step 1. Fix an agent sequence $\displaystyle\undertilde{v}\in N_{\infty}$ and
let $\displaystyle S$ be the DVSP induced by $\displaystyle\undertilde{v}$. To
shorten notation, for all $\displaystyle i\in N$, let us denote $\displaystyle
t_{i}(\undertilde{v})$ by $\displaystyle k_{i}$. The following claim
demonstrates how an agent $\displaystyle i$ with $\displaystyle k_{i}<k_{1}$
will update her action. Recall the set $\displaystyle N_{1}(\undertilde{v})$.
We distinguish two cases based on the value of
$\displaystyle|N_{1}(\undertilde{v})|$.

Case 1: $\displaystyle|N_{1}(\undertilde{v})|=0$.
We show that, for $\displaystyle\tau\leqslant\frac{a}{n-1}$, all the agents
will get infected under this assumption, i.e., $\displaystyle
I(S_{\infty})=N$. Note that by the assumption of the case,
$\displaystyle\undertilde{v}_{0}=1$. Recall that $\displaystyle\hat{S}_{0}$
denotes the intermediate state where the only change from $\displaystyle
S_{0}$ is that agent $\displaystyle\undertilde{v}_{0}$ has updated her action
to $\displaystyle b_{\undertilde{v}_{0}}(S_{0})$. Since
$\displaystyle\undertilde{v}_{0}=1$, we have $\displaystyle
a_{i}(S_{0})=a_{i}(\hat{S}_{0})=a$ for all $\displaystyle i\neq 1$. Moreover,
by Remark 2.1 and the definition of the process, $\displaystyle
a_{1}(\hat{S}_{0})=1$. Consider the time point $\displaystyle 1$. By the
definition of the process, an agent $\displaystyle i\neq 1$ will be in
$\displaystyle I(S_{1})$ if $\displaystyle
a_{i}(\hat{S}_{0})r_{i}(\hat{S}_{0})>\tau$. Since $\displaystyle
I(S_{0})=\\{1\\}$, $\displaystyle a_{i}(\hat{S}_{0})=a$ for all $\displaystyle
i\in N$, it follows that for all $\displaystyle i\in N\setminus\\{1\\}$

$\displaystyle\displaystyle ar_{i}(\hat{S}_{0})$
$\displaystyle\displaystyle=\frac{a}{(n-2)a+1}>\frac{a}{(n-1)}.$

Because $\displaystyle\tau\leqslant\frac{a}{n-1}$, this implies that all the
agents in $\displaystyle N\setminus\\{1\\}$ gets infected at the time point
$\displaystyle 1$. Hence, $\displaystyle I(S_{1})=N$. Therefore, by the
definition of the process $\displaystyle I(S_{\infty})=N$.

Case 2: $\displaystyle|N_{1}(\undertilde{v})|\geqslant 1$.
This means $\displaystyle\undertilde{v}_{0}\neq 1$. Let
$\displaystyle\undertilde{v}_{0}=i\in N\setminus 1$. Hence, by the definition
of the process, agent $\displaystyle i$ will choose her action as
$\displaystyle b_{i}(S_{0})$ at the intermediate state
$\displaystyle\hat{S}_{0}$. As $\displaystyle a_{j}(S_{0})=a>0$ for all
$\displaystyle j\in N$ and $\displaystyle I(S_{0})=\\{1\\}$, it follows that
$\displaystyle r_{i}(S_{0})\neq 0$. Therefore,

$b_{i}(S_{0})=\min\left\\{1,\frac{\tau}{r_{i}(S_{0})}\right\\}=\min\left\\{1,(n-1)\tau\right\\}=(n-1)\tau.$
(C.9)

Since by our assumption $\displaystyle\undertilde{v}_{0}=i$ and $\displaystyle
i\notin I(S_{0})$, by Observation 2.5, $\displaystyle i\notin I(S_{1})$. For
any other uninfected agent $\displaystyle j$,

$\displaystyle
r_{j}(\hat{S}_{0})=\frac{a}{(n-2)a+b_{i}(S_{0})}=\frac{a}{(n-2)a+(n-1)\tau}.$

This together with the fact that $\displaystyle a_{j}(\hat{S}_{0})=a$ implies

  1. (i)

if $\displaystyle\tau=\frac{a}{n-1}$ then $\displaystyle
a_{j}(\hat{S}_{0})r_{j}(\hat{S}_{0})=\frac{a}{n-1}=\tau$, and

  2. (ii)

if $\displaystyle\tau<\frac{a}{n-1}$ then $\displaystyle
a_{j}(\hat{S}_{0})r_{j}(\hat{S}_{0})>\frac{a}{n-1}>\tau$.

Combining the above observations, we may write if
$\displaystyle\tau=\frac{1}{n-1}$ then agent $\displaystyle j$ will not get
infected at time point $\displaystyle 1$ and if
$\displaystyle\tau<\frac{a}{n-1}$ then agent $\displaystyle j$ will get
infected at time point $\displaystyle 1$. Hence, we have

$\displaystyle\tau=\frac{a}{n-1}\implies I(S_{1})=\\{1\\}\mbox{ and
}\tau<\frac{a}{n-1}\implies I(S_{1})=N\setminus\\{i\\}.$

To decide the final outcome, we first assume
$\displaystyle\tau=\frac{a}{n-1}$. Note that by (C.9), $\displaystyle
b_{i}(S_{0})=a$. This means $\displaystyle a_{i}(S_{1})=a$. Moreover, as
$\displaystyle\undertilde{v}_{0}=i$, we have $\displaystyle a_{j}(S_{1})=a$
for all $\displaystyle j\neq i$. Using similar arguments, we can show that
$\displaystyle a_{k}(S_{k_{1}})=a$ for all $\displaystyle k\in N$ and
$\displaystyle I(S_{k_{1}})=\\{1\\}$. By the definition of the process,
$\displaystyle a_{1}(\hat{S}_{k_{1}})=1$ and $\displaystyle
a_{k}(\hat{S}_{k_{1}})=a$ for all $\displaystyle k\neq 1$. Therefore, for any
$\displaystyle k\neq 1$

$\displaystyle\displaystyle a_{k}(\hat{S}_{k_{1}})r_{k}(\hat{S}_{k_{1}})$
$\displaystyle\displaystyle=\frac{a}{(n-2)a+1}>\frac{a}{(n-1)}=\tau.$

Thus, all the agents other than agent 1 will get infected at $\displaystyle
k_{1}+1$. Hence, $\displaystyle I(S_{\infty})=N$.

Now assume $\displaystyle\tau<\frac{a}{n-1}$. We distinguish two subcases.

Case 2.1. $\displaystyle\undertilde{v}_{1}=i$.
We show that the final infected set will be $\displaystyle N\setminus i$.
Since by our assumption $\displaystyle\undertilde{v}_{1}=i$ and $\displaystyle
i\notin I(S_{1})$, by Observation 2.5, $\displaystyle i\notin I(S_{2})$.
Hence, $\displaystyle I(S_{2})=N\setminus\\{i\\}$. We now show that
$\displaystyle i$ will not get infected after this. At time point
$\displaystyle 2$,

$\displaystyle r_{i}(\hat{S}_{2})=\frac{(n-1)}{(n-1)}=1.$

Therefore, $\displaystyle a_{i}(\hat{S}_{2})=\tau$ (see Observation 2.5). At
time point $\displaystyle 3$, if $\displaystyle\undertilde{v}_{3}=i$, then
agent $\displaystyle i$ would not get infected at time point $\displaystyle 4$
(Observation 2.5). On the other hand, if $\displaystyle\undertilde{v}_{3}\neq
i$ then as $\displaystyle a_{i}(\hat{S}_{3})=a_{i}(\hat{S}_{2})=\tau$, it
follows that $\displaystyle
a_{i}(\hat{S}_{3})r_{i}(\hat{S}_{3})\leqslant\tau$. Hence, agent
$\displaystyle i$ would remain uninfected at time point $\displaystyle 4$.
Continuing in this manner, we may show that $\displaystyle i$ will not get
infected after this. Thus, $\displaystyle I(S_{\infty})=N\setminus\\{i\\}$.

Case 2.2.: $\displaystyle\undertilde{v}_{1}\neq i$
We show that the final infected set will be $\displaystyle N$. Since
$\displaystyle I(S_{1})=N\setminus\\{i\\}$, $\displaystyle
r_{i}(\hat{S}_{1})=1$. Moreover, as $\displaystyle
a_{i}(S_{1})=a_{i}(\hat{S}_{0})=b_{i}(S_{0})=(n-1)\tau>\tau$ (see C.9) and
$\displaystyle\undertilde{v}_{1}\neq i$, it follows that $\displaystyle
a_{i}(\hat{S}_{1})>\tau$. Combining this two we have $\displaystyle
a_{i}(\hat{S}_{1})r_{i}(\hat{S}_{1})>\tau$. Thus, agent $\displaystyle i$ will
get infected at time point $\displaystyle 2$. Hence, $\displaystyle
I(S_{2})=N$ and $\displaystyle I(S_{\infty})=N$.

Step 2. First assume $\displaystyle\tau=\frac{a}{n-1}$. Therefore, in view
Case 1 and Case 2 of the current proof, we have $\displaystyle
I(S_{\infty})=N$.

Now assume $\displaystyle\tau<\frac{a}{n-1}$. By Case 1 and Case 2 above, we
have

  1. (i)

$\displaystyle I(S_{\infty})=N\setminus i$ with $\displaystyle 1\in
I(S_{\infty})$ if $\displaystyle|N_{1}(\undertilde{v})|\geqslant 1$ and there
is $\displaystyle i\in N\setminus\\{1\\}$ such that $\displaystyle k_{i}=0$
and $\displaystyle\undertilde{v}_{1}=i$, and

  2. (ii)

$\displaystyle I(S_{\infty})=N$ if either
$\displaystyle|N_{1}(\undertilde{v})|=0$ or
$\displaystyle|N_{1}(\undertilde{v})|\geqslant 1$ and there is no
$\displaystyle i\in N\setminus\\{1\\}$ such that $\displaystyle k_{i}=0$ and
$\displaystyle\undertilde{v}_{1}=i$.

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

$\displaystyle\displaystyle
P(\undertilde{v}\mid|N_{1}(\undertilde{v})|\geqslant 1\text{ and }\exists
i\neq 1\text{ such that }k_{i}=0\text{ and }\undertilde{v}_{1}=i)$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle
P(\undertilde{v}\mid\exists i\neq 1\text{ such that }k_{i}=0\text{ and
}\undertilde{v}_{1}=i)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\prescript{n-1\mkern-0.5mu}{}{C}_{1}\times\frac{1}{n^{2}}$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle\frac{n-1}{n^{2}}.$

Note that by (i) and (ii),

$\displaystyle P(|I(S_{\infty})|=n-1)+P(I(S_{\infty})=N)=1.$

Therefore,

$\displaystyle\displaystyle P(I(S_{\infty})=N)$
$\displaystyle\displaystyle=1-P(|I(S_{\infty})|=n-1)$
$\displaystyle\displaystyle=1-\frac{n-1}{n^{2}}.$

Since any two subsets of $\displaystyle N$ with the cardinality $\displaystyle
n-1$ have the same probability, combining all the above observations, we have
the following distribution of the infected set.

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

This completes the proof of the theorem. ∎

##  Appendix D A few important lemmas

######  Lemma D.1.

Let $\displaystyle\undertilde{v}\in N_{\infty}$ and let $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Suppose $\displaystyle
t_{0}$ is such that $\displaystyle I(S_{t_{0}})=I(S_{t})$ for all
$\displaystyle t\geqslant t_{0}$ and $\displaystyle a_{k}(S_{t_{0}})=1$ for
all $\displaystyle k\in I(S_{t_{0}})$. Then for $\displaystyle i\notin
I(S_{t_{0}})$ and $\displaystyle\bar{t}>t_{0}$ with
$\displaystyle\undertilde{v}_{\bar{t}}=i$ implies

$\displaystyle a_{i}(\hat{S}_{\bar{t}})\geqslant
a_{j}(\hat{S}_{\bar{t}})\text{ for all }j\notin I(S_{t_{0}})\mbox{ with
}\undertilde{v}_{t}=j\text{ for some }t\in(t_{0},\bar{t}].$

###### Proof.

We use induction on $\displaystyle\bar{t}$ to prove the lemma. Note that for
the base case, that is, for $\displaystyle\bar{t}=t_{0}+1$, the lemma holds
vacuously. Next we introduce an introduction hypothesis.
Induction Hypothesis: Given $\displaystyle\bar{t}\in\mathbb{N}_{0}$ with
$\displaystyle\bar{t}>t_{0}+1$, the lemma holds for all $\displaystyle t$ with
$\displaystyle t_{0}+1\leqslant t<\bar{t}$.

We show that the lemma holds for $\displaystyle\bar{t}$. Suppose
$\displaystyle\undertilde{v}_{\bar{t}}=i$ where $\displaystyle i\notin
I(S_{t_{0}})$. If there is no $\displaystyle t\in(t_{0},\bar{t})$ such that
$\displaystyle\undertilde{v}_{t}\notin I(S_{t_{0}})$, the lemma holds
vacuously. So, assume that $\displaystyle\hat{t}$ is the last time point
before $\displaystyle\bar{t}$ such that
$\displaystyle\undertilde{v}_{\hat{t}}=j$ for some $\displaystyle j\notin
I(S_{t_{0}})$. This, together with the induction hypothesis, implies
$\displaystyle a_{j}(\hat{S}_{\hat{t}})\geqslant a_{k}(\hat{S}_{\hat{t}})$ for
all $\displaystyle k\notin I(S_{0})$ with $\displaystyle\undertilde{v}_{t}=k$
for some $\displaystyle t\in(t_{0},\hat{t})$. Also, by the definition of the
process, $\displaystyle a_{l}(\hat{S}_{\hat{t}})=a_{l}(\hat{S}_{\bar{t}})$ for
all $\displaystyle l\notin I(S_{0})\setminus i$. Therefore, to prove the lemma
it is enough to show that $\displaystyle a_{i}(\hat{S}_{\bar{t}})\geqslant
a_{j}(\hat{S}_{\bar{t}})$. Additionally, as $\displaystyle
a_{j}(\hat{S}_{\bar{t}})\leqslant 1$, we may assume that $\displaystyle
a_{i}(\hat{S}_{\bar{t}})=\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})}$. Moreover, as
$\displaystyle j\notin I(S_{0})$, $\displaystyle
a_{j}(\hat{S}_{\hat{t}})\leqslant\frac{\tau}{r_{j}(\hat{S}_{\hat{t}})}$. Now

$\displaystyle\displaystyle\frac{\tau}{r_{j}(\hat{S}_{\hat{t}})}$
$\displaystyle\displaystyle=\frac{\tau}{\frac{|I(S_{t_{0}})|}{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus j}a_{k}(\hat{S}_{\hat{t}})}}$
$\displaystyle\displaystyle=\frac{\tau[{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus j}a_{k}(\hat{S}_{\hat{t}})}]}{|I(S_{t_{0}})|}\text{ (as
}I(S_{t_{0}})=I(S_{\hat{t}})\text{ and for }k\in
I(S_{t_{0}}),\;a_{k}(S_{\hat{t}})=1)$
$\displaystyle\displaystyle=\frac{\tau[{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus\\{i,j\\}}a_{k}(\hat{S}_{\hat{t}})+a_{i}(\hat{S}_{\hat{t}})}]}{|I(S_{t_{0}})|}$
$\displaystyle\displaystyle\leqslant\frac{\tau[{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus\\{i,j\\}}a_{k}(\hat{S}_{\hat{t}})+a_{j}(\hat{S}_{\hat{t}})}]}{|I(S_{t_{0}})|}\text{
(as }a_{j}(\hat{S}_{\hat{t}})\geqslant a_{i}(\hat{S}_{\hat{t}}))$
$\displaystyle\displaystyle=\frac{\tau[{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus\\{i,j\\}}a_{k}(\hat{S}_{\bar{t}})+a_{j}(\hat{S}_{\bar{t}})}]}{|I(S_{t_{0}})|}\text{
(as }a_{k}(\hat{S}_{\hat{t}})=a_{k}(\hat{S}_{\bar{t}})\text{ for all }k\notin
I(S_{0})\setminus i)$
$\displaystyle\displaystyle=\frac{\tau[{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus i}a_{k}(\hat{S}_{\bar{t}})}]}{|I(S_{t_{0}})|}$
$\displaystyle\displaystyle=\frac{\tau}{\frac{|I(S_{t_{0}})|}{|I(S_{t_{0}})|+\sum_{k\notin
I(S_{t_{0}})\setminus i}a_{k}(\hat{S}_{\bar{t}})}}\text{ (as
}I(S_{t_{0}})=I(S_{\bar{t}})\text{ and for }k\in
I(S_{t_{0}}),\;a_{k}(S_{\bar{t}})=1)$
$\displaystyle\displaystyle=\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})}.$ (D.1)

(D.2) together with $\displaystyle
a_{i}(\hat{S}_{\bar{t}})=\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})}$ and
$\displaystyle
a_{j}(\hat{S}_{\hat{t}})\leqslant\frac{\tau}{r_{j}(\hat{S}_{\hat{t}})}$
implies $\displaystyle a_{i}(\hat{S}_{\bar{t}})\geqslant
a_{j}(\hat{S}_{\hat{t}})$. Hence, $\displaystyle
a_{i}(\hat{S}_{\bar{t}})\geqslant a_{j}(\hat{S}_{\bar{t}})$. This completes
the proof of the lemma. ∎

The following lemma provides an important property of the final action limit
for both infected and uninfected agents. It shows that an infected agent will
have the action limit 1 whereas any two uninfected agents will have the same
action limit, that is, for $\displaystyle i,j\notin I(S_{\infty})$,
$\displaystyle a_{i}(S_{\infty})=a_{j}(S_{\infty})$.

######  Lemma D.2.

Let $\displaystyle\undertilde{v}\in N_{\infty}$ and let $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Then, for

$\displaystyle[k\in I(S_{\infty})]\implies[a_{k}(S_{\infty})=1]$

and

$\displaystyle[i,j\notin
I(S_{\infty})]\implies[a_{i}(S_{\infty})=a_{j}(S_{\infty})].$

###### Proof.

Let $\displaystyle\undertilde{v}\in N_{\infty}$ and let $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Consider $\displaystyle
k\in I(S_{\infty})$. As $\displaystyle\undertilde{v}\in N_{\infty}$, agent
$\displaystyle k$ appears infinitely many times in
$\displaystyle\undertilde{v}$. And, after getting infected whenever she
updates her action, she will choose her action as 1. Thus, $\displaystyle
a_{k}(S_{\infty})=1$. Now consider $\displaystyle i,j\notin I(S_{\infty})$.
Let $\displaystyle b=a_{i}(S_{\infty})$ and consider
$\displaystyle\epsilon>0$. This means there exists $\displaystyle t_{0}$ such
that $\displaystyle a_{i}(S_{t})\geqslant b-\epsilon$ for all $\displaystyle
t\geqslant t_{0}$. Note that as $\displaystyle N$ is a finite set and
$\displaystyle I(S_{\infty})$ exists, there exists
$\displaystyle\tilde{t}\in\mathbb{N}_{0}$ such that $\displaystyle
I(S_{\tilde{t}})=I(S_{\infty})$. In view of this, we may assume that
$\displaystyle I(S_{t_{0}})=I(S_{\infty})$. Consider a time point
$\displaystyle\hat{t}$ such that

  1. (i)

$\displaystyle\hat{t}>t_{0}$ and $\displaystyle\undertilde{v}_{\hat{t}}=j$ and

  2. (ii)

there exists $\displaystyle\bar{t}\in[t_{o},\hat{t}]$ such that
$\displaystyle\undertilde{v}_{\bar{t}}=i$.

Such a time point $\displaystyle\hat{t}$ exists as
$\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by Lemma D.1,
$\displaystyle a_{j}(S_{\hat{t}})\geqslant a_{i}(S_{\hat{t}})$. As
$\displaystyle\hat{t}>t_{0}$, this means $\displaystyle
a_{j}(S_{\hat{t}})\geqslant b-\epsilon$. Further, as $\displaystyle
I(S_{t_{0}})=I(S_{\infty})$ and $\displaystyle\hat{t}>t_{0}$, by Claim A.1 in
Lemma 2.6, $\displaystyle a_{j}(S_{t})\geqslant a_{j}(S_{\hat{t}})$ for all
$\displaystyle t\geqslant\hat{t}$. Thus, $\displaystyle a_{j}(S_{t})\geqslant
b-\epsilon$ for all $\displaystyle t\geqslant\hat{t}$. Since
$\displaystyle\epsilon$ is arbitrary, this gives $\displaystyle
a_{j}(S_{\infty})\geqslant b$. Similarly, we can show that $\displaystyle
a_{i}(S_{\infty})\geqslant a_{j}(S_{\infty})$. Hence, $\displaystyle
a_{i}(S_{\infty})=a_{j}(S_{\infty})$. ∎

The next lemma determines the common action limit of the uninfected agents.

######  Lemma D.3.

Let $\displaystyle\undertilde{v}\in N_{\infty}$ and let $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Further, let
$\displaystyle\gamma$ be the common action limit of the uninfected agents.
Then,

$\displaystyle[(n-1)\tau<|I(S_{\infty})|]\implies\left[\gamma=\frac{\tau|I(S_{\infty})|}{(1+\tau)|I(S_{\infty})|-\tau(n-1)}<1\right],$

and

$\displaystyle[(n-1)\tau\geqslant|I(S_{\infty})|]\implies[\gamma=1].$

###### Proof.

Let $\displaystyle t_{0}\in\mathbb{N}_{0}$ be such that $\displaystyle
I(S_{t_{0}})=I(S_{\infty})$ and $\displaystyle a_{k}(S_{t_{0}})=1$ for all
$\displaystyle k\in I(S_{t_{0}})$. First assume that
$\displaystyle(n-1)\tau<|I(S_{\infty})|$. This implies
$\displaystyle\frac{\tau}{|I(S_{\infty})|}<\frac{1}{n-1}$. We first show that
for any time point $\displaystyle\bar{t}\geqslant t_{0}$, if
$\displaystyle\undertilde{v}_{\bar{t}}\notin I(S_{\infty})$ then
$\displaystyle a_{\undertilde{v}_{\bar{t}}}(\hat{S}_{\bar{t}})<1$. Let
$\displaystyle\undertilde{v}_{\bar{t}}=i$. Since $\displaystyle
a_{i}(\hat{S}_{\bar{t}})=\min\\{\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})},1\\}$,
it is enough to show that
$\displaystyle\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})}<1$.

$\displaystyle\displaystyle\frac{\tau}{r_{i}(\hat{S}_{\bar{t}})}$
$\displaystyle\displaystyle=\frac{\tau}{|I(\hat{S}_{\bar{t}})|}\left(|I(\hat{S}_{\bar{t}})|+\sum_{j\notin
I(\hat{S}_{\bar{t}})\cup\\{i\\}}a_{j}(\hat{S}_{\bar{t}})\right)$
$\displaystyle\displaystyle=\frac{\tau}{|I(S_{\infty})|}\left(|I(S_{\infty})|+\sum_{j\notin
I(S_{\infty})\cup\\{i\\}}a_{j}(\hat{S}_{\bar{t}})\right)\text{ (as
}I(S_{t_{0}})=I(S_{\infty})\text{ and }\bar{t}\geqslant t_{0})$
$\displaystyle\displaystyle<\frac{1}{n-1}\left(|I(S_{\infty})|+\sum_{j\notin
I(S_{\infty})\cup\\{i\\}}a_{j}(\hat{S}_{\bar{t}})\right)\text{ (as
}\frac{\tau}{|I(S_{\infty})|}<\frac{1}{n-1})$ $\displaystyle\displaystyle\leq
1\text{ (as }a_{j}(\hat{S}_{\bar{t}})\leqslant 1.\text{ for all }j\notin
I(S_{\infty}\cup\\{i\\}).$

Since $\displaystyle\bar{t}$ is arbitrary, it follows that $\displaystyle
a_{i}(\hat{S}_{t})=\frac{\tau}{r_{i}(\hat{S}_{t})}$ for all $\displaystyle
t\geqslant t_{0}$ with $\displaystyle\undertilde{v}_{t}=i$. Hence,

$\displaystyle\displaystyle a_{i}(\hat{S}_{t})$
$\displaystyle\displaystyle=\frac{\tau}{r_{i}(\hat{S}_{t})}$
$\displaystyle\displaystyle=\frac{\tau}{|I(\hat{S}_{t})|}\left(|I(\hat{S}_{t})|+\sum_{j\notin
I(\hat{S}_{t})\cup\\{i\\}}a_{j}(\hat{S}_{t})\right)$
$\displaystyle\displaystyle=\frac{\tau}{|I(S_{\infty})|}\left(|I(S_{\infty})|+\sum_{j\notin
I(S_{\infty})\cup\\{i\\}}a_{j}(\hat{S}_{t})\right).$ (D.2)

Taking limit on both the sides of D.2, we have

$\displaystyle\displaystyle\gamma=\frac{\tau}{|I(S_{\infty})|}\left(|I(S_{\infty})|+\sum_{j\notin
I(S_{\infty})\cup\\{i\\}}\gamma\right)$ $\displaystyle\displaystyle\implies$
$\displaystyle\displaystyle\gamma=\frac{\tau|I(S_{\infty})|}{(1+\tau)|I(S_{\infty})|-\tau(n-1)}$
$\displaystyle\displaystyle\implies$
$\displaystyle\displaystyle\gamma<\frac{\tau|I(S_{\infty})|}{(1+\tau)|I(S_{\infty})|-|I(S_{\infty})|}=1.$

Now assume $\displaystyle(n-1)\tau\geqslant|I(S_{\infty})|$. We have to show
that $\displaystyle\gamma=1$. Assume $\displaystyle\gamma<1$. Consider
$\displaystyle i\notin I(S_{\infty})$. Since by Claim A.1 in Lemma 2.6,
$\displaystyle a_{i}(S_{t})$ is an increasing sequence for $\displaystyle
t>t_{0}$, $\displaystyle\gamma<1$ implies $\displaystyle a_{i}(S_{t})<1$ for
all $\displaystyle t>t_{0}$. This means $\displaystyle
a_{i}(\hat{S}_{t})=\frac{\tau}{r_{i}(\hat{S}_{t})}$ for $\displaystyle
t>t_{0}$ with $\displaystyle\undertilde{v}_{t}=i$. Therefore, using similar
arguments as before we have

$\displaystyle\displaystyle\gamma=\frac{\tau|I(S_{\infty})|}{(1+\tau)|I(S_{\infty})|-\tau(n-1)}$
$\displaystyle\displaystyle\implies$
$\displaystyle\displaystyle\gamma\geqslant\frac{\tau|I(S_{\infty})|}{(1+\tau)|I(S_{\infty})|-|I(S_{\infty})|}=1.$

But this is a contradiction to $\displaystyle\gamma<1$. Therefore,
$\displaystyle\gamma=1$. ∎

##  Appendix E Proof of Theorem 3.3, Theorem3.5, Theorem 3.7, and Theorem 3.10

###  E.1. Proof of Theorem 3.3

###### Proof.

First assume $\displaystyle\tau\geqslant\frac{1}{n-1}$. We first explore the
limiting actions for a fixed agent sequence, and then we use this to find the
limiting probability distribution. Let $\displaystyle\undertilde{v}$ be an
agent sequence and $\displaystyle S$ be the DVSP induced by
$\displaystyle\undertilde{v}$. Note that by Remark 2.7, it is enough to assume
$\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by Lemma D.2, all the
agents outside $\displaystyle I(S_{\infty})$ have the same action limit, and
all the agents in $\displaystyle I(S_{\infty})$ have the action limit 1. Let
us denote the common limit by $\displaystyle\gamma$. We distinguish two cases
based on the value of $\displaystyle N_{1}(\undertilde{v})$ (as in the proof
of Theorem 3.2) to find $\displaystyle\gamma$. Note that by the assumption of
the theorem $\displaystyle\alpha\leqslant n-1$.
Case 1: $\displaystyle|N_{1}(\undertilde{v})|\geqslant\alpha$.
Recall that for this case the final infected set is $\displaystyle\\{1\\}$.
Moreover, by the assumption of the theorem, $\displaystyle\tau(n-1)\geqslant
1$. Therefore, by Lemma B.1, $\displaystyle\gamma=1$. Hence, $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Case 2: $\displaystyle|N_{1}(\undertilde{v})|\leqslant\alpha-1$.
Recall that for this case, the final infected set is $\displaystyle
N_{1}(\undertilde{v})\cup\\{1\\}$. Therefore, by Lemma B.1, if
$\displaystyle(n-1)\tau\geqslant|N_{1}(\undertilde{v})|+1$ then $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$, and if
$\displaystyle(n-1)\tau<|N_{1}(\undertilde{v})|+1$ then

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

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

  1. (i)

$\displaystyle|N_{1}(\undertilde{v})|+1\in[\beta,\alpha]$ implies

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

  2. (ii)

$\displaystyle|N_{1}(\undertilde{v})|+1\in[1,\beta-1]\cup[\alpha+1,n]$ implies
$\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

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

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1-\frac{\alpha-\beta+1}{n}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{ if
}\underaccent{\tilde}{x}\in A_{m}\text{ for some }m\in[\beta,\alpha],\\\ \\\
0&\text{ otherwise. }\end{cases}$

Now assume $\displaystyle\tau<\frac{1}{n-1}$. 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 $\displaystyle\undertilde{v}$ be an agent sequence and
$\displaystyle S$ be the DVSP induced by $\displaystyle\undertilde{v}$. Note
that by Remark 2.7, it is enough to assume $\displaystyle\undertilde{v}\in
N_{\infty}$. Therefore, by Lemma D.2, all the agents outside $\displaystyle
I(S_{\infty})$ have the same action limit, and all the agents in
$\displaystyle I(S_{\infty})$ have the action limit 1. Let us denote the
common limit by $\displaystyle\gamma$. As by the assumption of the theorem,
$\displaystyle(n-1)\tau<1$, we have $\displaystyle\alpha=n$, and hence,
$\displaystyle|N_{1}(\undertilde{v})|\leqslant\alpha-1$. Moreover, for
$\displaystyle|N_{1}(\undertilde{v})|\leqslant\alpha-1$ (shown in the proof of
Theorem 3.2), the final infected set is $\displaystyle
N_{1}(\undertilde{v})\cup\\{1\\}$. Thus,
$\displaystyle|I(S_{\infty})|>(n-1)\tau$. Hence, by Lemma B.1 if
$\displaystyle|N_{1}(\undertilde{v})|+1<n$

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

and if $\displaystyle|N_{1}(\undertilde{v})|+1=n$, $\displaystyle
a_{N}(S_{\infty})=1$. Recall the notation $\displaystyle A_{m}$. By the above
arguments, we have $\displaystyle a_{N}(S_{\infty})\in
A_{[|N_{1}(\undertilde{v})|+1]}$ when
$\displaystyle|N_{1}(\undertilde{v})|+1<n$. Moreover, as
$\displaystyle\mathbb{P}$ is uniform, any two vectors in $\displaystyle
A_{m}$, for $\displaystyle m\in[1,(n-1)]$, have the same probability. Thus, by
Theorem 3.2, we have the following distribution

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}\frac{1}{n}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{ if
}\underaccent{\tilde}{x}\in A_{m}\text{ for some }m\in[1,n-1],\\\ \\\ 0&\text{
otherwise.}\par\end{cases}$

∎

###  E.2. Proof of Theorem 3.5

###### Proof.

We first explore the limiting actions for a fixed agent sequence, and then we
use this to find the limiting probability distribution. Let
$\displaystyle\undertilde{v}$ be an agent sequence and $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Note that by Remark 2.7, it
is enough to assume $\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by
Lemma D.2, all the agents outside $\displaystyle I(S_{\infty})$ have the same
action limit, and all the agents in $\displaystyle I(S_{\infty})$ have the
action limit 1. Let us denote the common limit by $\displaystyle\gamma$. First
assume $\displaystyle\tau\geqslant\frac{1}{n-1}$. By Theorem 3.4,
$\displaystyle I(S_{\infty})=\\{1\\}$. Therefore,
$\displaystyle(n-1)\tau\geqslant|I(S_{\infty})|$, and hence, by Lemma D.3,
$\displaystyle\gamma=1$. Thus, $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Now assume that $\displaystyle\tau<\frac{1}{n-1}$. We distinguish two cases
based on the value of $\displaystyle N_{1}(\undertilde{v})$ (as in the proof
of Theorem 3.4) to find $\displaystyle\gamma$.
Case 1: $\displaystyle|N_{1}(\undertilde{v})|=0$.
Recall that for this case the final infected set is $\displaystyle N$. Hence,
$\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Case 2: $\displaystyle|N_{1}(\undertilde{v})|\geqslant 1$.
Recall that for this case, the final infected set has cardinality either
$\displaystyle n$ or $\displaystyle n-1$. If cardinality is $\displaystyle n$
then $\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$. If cardinality
is $\displaystyle n-1$, then as $\displaystyle(n-1)\tau<1$, by Lemma D.3,
$\displaystyle\gamma=\tau$. Hence,

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\ \tau&\text{if }i\notin I(S_{\infty}).\end{cases}$

Note that this implies $\displaystyle a_{N}(S_{\infty})\in A_{n-1}$. Also, as
$\displaystyle\mathbb{P}$ is uniform, any two vectors in $\displaystyle
A_{n-1}$ have the same probability. Thus, by Theorem 3.4, we have the
following distribution

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1-\frac{n-1}{n^{2}}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n^{2}}&\text{ if }\underaccent{\tilde}{x}\in A_{n-1},\\\ \\\ 0&\text{
otherwise. }\par\end{cases}$

∎

###  E.3. Proof of Theorem 3.7

###### Proof.

We first explore the limiting actions for a fixed agent sequence, and then we
use this to find the limiting probability distribution. Let
$\displaystyle\undertilde{v}$ be an agent sequence and $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Note that by Remark 2.7, it
is enough to assume $\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by
Lemma D.2, all the agents outside $\displaystyle I(S_{\infty})$ have the same
action limit, and all the agents in $\displaystyle I(S_{\infty})$ have the
action limit 1. Let us denote the common limit by $\displaystyle\gamma$. We
distinguish two cases based on the value of $\displaystyle
N_{1}(\undertilde{v})$ (as in the proof of Theorem 3.6) to find
$\displaystyle\gamma$.
Case 1: $\displaystyle|N_{1}(\undertilde{v})|\geqslant\hat{\alpha}$.
Recall that for this case the final infected set is $\displaystyle\\{1\\}$.
Moreover, by the assumption of the theorem, $\displaystyle(n-1)\tau\geqslant
1$. Therefore, by Lemma B.1, $\displaystyle\gamma=1$. Hence, $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Case 2: $\displaystyle|N_{1}(\undertilde{v})|\leqslant\hat{\alpha}-1$.
Recall that for this case, the final infected set is $\displaystyle
N_{1}(\undertilde{v})\cup\\{1\\}$. Note that as
$\displaystyle\hat{\alpha}\leqslant n-1$, $\displaystyle
N_{1}(\undertilde{v})\cup\\{1\\}\leqslant n-1$. Therefore, by Lemma B.1, if
$\displaystyle(n-1)\tau\geqslant|N_{1}(\undertilde{v})|+1$ then $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$, and if
$\displaystyle(n-1)\tau<|N_{1}(\undertilde{v})|+1$ then

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

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

  1. (i)

$\displaystyle|N_{1}(\undertilde{v})|+1\in[\hat{\beta},\hat{\alpha}]$ implies

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

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

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1-\frac{\hat{\alpha}-\hat{\beta}+1}{n}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{ if
}\underaccent{\tilde}{x}\in A_{m}\text{ for some
}m\in[\hat{\beta},\hat{\alpha}],\\\ \\\ 0&\text{ otherwise. }\end{cases}$

∎

###  E.4. Proof of Theorem 3.10

###### Proof.

We first explore the limiting actions for a fixed agent sequence, and then we
use this to find the limiting probability distribution. Let
$\displaystyle\undertilde{v}$ be an agent sequence and $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Note that by Remark 2.7, it
is enough to assume $\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by
Lemma D.2, all the agents outside $\displaystyle I(S_{\infty})$ have the same
action limit, and all the agents in $\displaystyle I(S_{\infty})$ have the
action limit 1. Let us denote the common limit by $\displaystyle\gamma$. First
assume $\displaystyle\tilde{\alpha}+1<\bar{\alpha}$. As shown in the proof of
Theorem 3.9, the infected set is either $\displaystyle\\{1\\}$ or
$\displaystyle|N_{1}(\undertilde{v})|+1$ where $\displaystyle
N_{1}(\undertilde{v})\in[1,\tilde{\alpha}]$. Since by the assumption of the
theorem, $\displaystyle(n-1)\tau\geqslant 1$, we have
$\displaystyle(n-1)\tau\geqslant I(S_{\infty})$ when the infected set is
$\displaystyle\\{1\\}$. Therefore, by Lemma D.3, $\displaystyle\gamma=1$ and
hence, $\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$. On the other
hand, if the final infected set is $\displaystyle
N_{1}(\undertilde{v})\cup\\{1\\}$, the limiting action depends on
$\displaystyle|N_{1}(\undertilde{v})|$. By Lemma B.1, if
$\displaystyle(n-1)\tau\geqslant|N_{1}(\undertilde{v})|+1$ then $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$, and if
$\displaystyle(n-1)\tau<|N_{1}(\undertilde{v})|+1$ then

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

Recall that the following was shown in the proof of Theorem 3.9 when
$\displaystyle\tilde{\alpha}+1\leqslant\bar{\alpha}$:

  * •

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

  * •

$\displaystyle|I(S_{\infty})|=|N_{1}(\undertilde{v})|+1$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{1,2,\ldots,\tilde{\alpha}-1\\}$.

Recall that
$\displaystyle\tilde{\beta}=\min\\{\lfloor(n-1)\tau\rfloor+1,\tilde{\alpha}+1\\}$.
Therefore, we have the following:

  1. (i)

$\displaystyle|N_{1}(\undertilde{v})|+1\in[\tilde{\beta},\tilde{\alpha}]$
implies

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|N_{1}(\undertilde{v})|+1)}{(1+\tau)(|N_{1}(\undertilde{v})|+1)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

  2. (ii)

$\displaystyle|N_{1}(\undertilde{v})|+1\in[1,\tilde{\beta}-1]\cup[\tilde{\alpha}+1,n]$
implies $\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

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

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1-\frac{\tilde{\alpha}-\tilde{\beta}+1}{n}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{ if
}\underaccent{\tilde}{x}\in A_{m}\text{ for some
}m\in[\bar{\beta},\bar{\alpha}],\\\ \\\ 0&\text{ otherwise. }\end{cases}$

Now assume $\displaystyle 2\leqslant\bar{\alpha}<\tilde{\alpha}+1$. Recall
that the following was shown in the proof of Theorem 3.9 when $\displaystyle
2\leqslant\bar{\alpha}<\tilde{\alpha}+1$:

  1. (i)

$\displaystyle|I(S_{\infty})|=1$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{0,\tilde{\alpha},\tilde{\alpha}+1,\ldots,n-1\\}$,

  2. (ii)

$\displaystyle|I(S_{\infty})|=|N_{1}(\undertilde{v})|+1$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{1,2,\ldots,\bar{\alpha}-2\\}$,

  3. (iii)

$\displaystyle|I(S_{\infty})|=n$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\\}$
and there is no $\displaystyle i\in N$ such that $\displaystyle k_{i}=k_{1}+1$
and $\displaystyle\undertilde{v}_{k_{1}+2}=i$, and

  4. (iv)

$\displaystyle|I(S_{\infty})|=n-1$ if
$\displaystyle|N_{1}(\undertilde{v})|\in\\{\bar{\alpha}-1,\ldots,\tilde{\alpha}-1\\}$
and there is $\displaystyle i\in N$ such that $\displaystyle k_{i}=k_{1}+1$
and $\displaystyle\undertilde{v}_{k_{1}+2}=i$.

By the assumption of the theorem, $\displaystyle(n-1)\tau\geqslant 1$ and
$\displaystyle\tau<1$. Thus, if $\displaystyle|I(S_{\infty})|=1$ we have
$\displaystyle(n-1)\tau\geqslant|I(S_{\infty})|$, and if
$\displaystyle|I(S_{\infty})|=(n-1)$ we have
$\displaystyle(n-1)\tau<|I(S_{\infty})|$. Recall that
$\displaystyle\bar{\beta}=\min\\{\lfloor(n-1)\tau\rfloor+1,\bar{\alpha}\\}$.
Combining all these observations, we may write the following

  1. (i)

$\displaystyle|I(S_{\infty})|\in[\bar{\beta},\bar{\alpha}-1]\cup\\{n-1\\}$
implies

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\
\frac{\tau(|I(S_{\infty})|)}{(1+\tau)(|I(S_{\infty})|)-\tau(n-1)}&\text{if
}i\notin I(S_{\infty}).\end{cases}$

  2. (ii)

$\displaystyle|I(S_{\infty})|\in[1,\bar{\beta}-1]\cup\\{n\\}$ implies
$\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Note that (i) implies $\displaystyle a_{N}(S_{\infty})\in
A_{(|I(S_{\infty})|)}$ when
$\displaystyle|I(S_{\infty})|\in[\bar{\beta},\bar{\alpha}-1]\cup\\{n-1\\}$.
Also, as $\displaystyle\mathbb{P}$ is uniform, any two vectors in
$\displaystyle A_{m}$, for $\displaystyle
m\in[\bar{\beta},\bar{\alpha}-1]\cup\\{n-1\\}$, have the same probability.
Therefore, using Theorem 3.9, we have the following distribution

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1+\frac{\bar{\beta}-\bar{\alpha}}{n}-\eta(\tilde{\alpha},\bar{\alpha},n)&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n\times\prescript{n-1\mkern-0.5mu}{}{C}_{m-1}}&\text{ if
}\underaccent{\tilde}{x}\in A_{m}\text{ for some
}m\in[\bar{\beta},\bar{\alpha}-1],\\\ \\\
\frac{\eta(\tilde{\alpha},\bar{\alpha},n)}{n-1}&\text{if
}\underaccent{\tilde}{x}\in A_{n-1},\\\ \\\ 0&\text{ otherwise. }\end{cases}$

∎

###  E.5. Proof of Theorem 3.12

###### Proof.

We first explore the limiting actions for a fixed agent sequence, and then we
use this to find the limiting probability distribution. Let
$\displaystyle\undertilde{v}$ be an agent sequence and $\displaystyle S$ be
the DVSP induced by $\displaystyle\undertilde{v}$. Note that by Remark 2.7, it
is enough to assume $\displaystyle\undertilde{v}\in N_{\infty}$. Therefore, by
Lemma D.2, all the agents outside $\displaystyle I(S_{\infty})$ have the same
action limit, and all the agents in $\displaystyle I(S_{\infty})$ have the
action limit 1. Let us denote the common limit by $\displaystyle\gamma$. First
assume $\displaystyle\tau=\frac{a}{n-1}$. By Theorem 3.11, $\displaystyle
I(S_{\infty})=N$. Therefore, by Lemma D.2, $\displaystyle
a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Now assume that $\displaystyle\tau<\frac{a}{n-1}$. We distinguish two cases
based on the value of $\displaystyle N_{1}(\undertilde{v})$ (as in the proof
of Theorem 3.11) to find $\displaystyle\gamma$.
Case 1: $\displaystyle|N_{1}(\undertilde{v})|=0$.
Recall that for this case the final infected set is $\displaystyle N$. Hence,
$\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$.

Case 2: $\displaystyle|N_{1}(\undertilde{v})|\geqslant 1$.
Recall that for this case, the final infected set has cardinality either
$\displaystyle n$ or $\displaystyle n-1$. If cardinality is $\displaystyle n$
then $\displaystyle a_{N}(S_{\infty})=\underaccent{\tilde}{1}$. If cardinality
is $\displaystyle n-1$, then as $\displaystyle(n-1)\tau<1$, by Lemma D.3,
$\displaystyle\gamma=\tau$. Hence,

$\displaystyle a_{i}(S_{\infty})=\begin{cases}1&\text{if }i\in
I(S_{\infty}),\\\ \tau&\text{if }i\notin I(S_{\infty}).\end{cases}$

Note that this implies $\displaystyle a_{N}(S_{\infty})\in A_{n-1}$. Also, as
$\displaystyle\mathbb{P}$ is uniform, any two vectors in $\displaystyle
A_{n-1}$ have the same probability. Thus, by Theorem 3.11, we have the
following distribution

$\displaystyle\mathbb{P}(a_{N}(S_{\infty})=\underaccent{\tilde}{x})=\begin{cases}1-\frac{n-1}{n^{2}}&\text{
if }\underaccent{\tilde}{x}=\underaccent{\tilde}{1},\\\ \\\
\frac{1}{n^{2}}&\text{ if }\underaccent{\tilde}{x}\in A_{n-1},\\\ \\\ 0&\text{
otherwise. }\par\end{cases}$

∎