# Supercritical Spatial SIR Epidemics: Spreading Speed and Herd Immunity

Xinghua Zheng, Qingsan Zhu Department of ISOM, Hong Kong University of Science
and Technology, Clear Water Bay, Kowloon, Hong Kong. Email:
<EMAIL_ADDRESS>Jockey Club Institute for Advanced Study, Hong Kong
University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong.
Email<EMAIL_ADDRESS>
###### Abstract

We study supercritical spatial SIR epidemics on
$\mathbb{Z}^{2}\times\\{1,2,\ldots,N\\}$, where each site in $\mathbb{Z}^{2}$
represents a village and $N$ stands for the village size. We establish several
key asymptotic results as $N\to\infty$. In particular, we derive the
probability that the epidemic will last forever if the epidemic is started by
one infected individual. Moreover, conditional on that the epidemic lasts
forever, we show that the epidemic spreads out linearly in all directions and
derive an explicit formula for the spreading speed. Furthermore, we prove that
the ultimate proportion of infection converges to a number that is constant
over space and find its explicit value. An important message is that if there
is no vaccination, then the ultimate proportion of population who will be
infected can be _much higher_ than the vaccination proportion that is needed
in order to prevent sustained spread of the infection.

##  1 Introduction

###  1.1 SIR Model

The Susceptible, Infected and Recovered (SIR) epidemic model is a fundamental
model in epidemiology. In the usual SIR model, there is a fixed population,
and at any time, individuals in the population fall in one of the three
categories: susceptible, infected and recovered ([Kermack and
McKendrick(1927)]). Infected individuals remain infected for one unit of time,
then recover and gain immunity. The disease is spread from an infected
individual to a susceptible individual with a fixed probability.

In the afore-mentioned model, spatial information is not considered. This is
not suitable for various applications because many epidemics, including
COVID-19, only transmit _locally_.

In this paper, we focus on a spatial version of SIR model.

###  1.2 Spatial SIR Model

We consider the spatial SIR epidemic model introduced in [Lalley(2009)]. The
epidemic takes place on ${\mathbb{Z}^{2}}\times\\{1,2,\ldots,N\\}$. The space
dimension can be more general, but in this paper we focus on the two
dimensional case because of its practical relevance. In the model, each site
$\mathbf{x}\in\mathbb{Z}^{2}$ represents a village, which hosts $N$
individuals, which are labeled as
$(\mathbf{x},1),(\mathbf{x},2),\ldots,(\mathbf{x},N)$. The general rule is the
same as the usual SIR model, except that the disease can only be spread from
an infected individual to a susceptible individual who is either at the same
site or in a nearest neighbor site. We assume that the infection probability
is the same for all such pairs of infected and susceptible individuals, and
the infection probability is given by

$P_{N}^{\theta}=\frac{1+\theta}{5N}.$ (1)

From (1), we see that the basic reproduction number is $R_{0}=(5N-1)\cdot
P_{N}^{\theta}$, which is approximately $1+\theta$ when the village size $N$
is large. The epidemic is hence supercritical, critical, or subcritical
according to whether $\theta>0$, $\theta=0$, or $\theta<0$.

The critical case when $\theta=0$, or more generally, the near-critical case
when $\theta=\theta_{0}/N^{1/2}$ for some fixed constant $\theta_{0}$, has
been studied in [Lalley and Zheng(2010)] and [Lalley, Perkins, and
Zheng(2014)]. In the first paper, the authors prove that the process, suitably
scaled, converges to a measure-valued super-process. In [Lalley, Perkins, and
Zheng(2014)], the authors further establish a survival-extinction phase
transition for the limiting process.

COVID-19, as well as many other epidemics, are supercritical. This is the case
we focus on in this paper. Henceforth, we assume that $\theta$ is a fixed
positive constant.

###  1.3 Main Results

We answer the following three fundamental questions:

  1. Q1.

How big is the probability that the epidemic will last forever?

  2. Q2.

How fast does the epidemic spread out?

  3. Q3.

What is the ultimate proportion of individuals who will be infected?

We start with Q1. Note that because the total population size is infinity,
there _is_ a positive probability that the epidemic will last forever.

We first define related random variables:

$\left\\{\begin{aligned}
I_{t}(\mathbf{x})&=I_{t}^{N}(\mathbf{x})=\\#\mbox{infected individuals at site
}\mathbf{x}\mbox{ at time }t,\\\
R_{t}(\mathbf{x})&=R_{t}^{N}(\mathbf{x})=\\#\mbox{recovered individuals at
site }\mathbf{x}\mbox{ at time }t,\mbox{and}\\\
S_{t}(\mathbf{x})&=S_{t}^{N}(\mathbf{x})=\\#\mbox{susceptible individuals at
site }\mathbf{x}\mbox{ at time }t\\\
&=N-I_{t}(\mathbf{x})-R_{t}(\mathbf{x}).\\\ \end{aligned}\right.$

The evolution of the SIR process can be described as the following:

$\displaystyle I_{t+1}(\mathbf{x})|(I_{t}(\mathbf{x}),R_{t}(\mathbf{x}))$
$\displaystyle\stackrel{{\scriptstyle\mathrm{d}}}{{=}}\mathrm{Bin}\left(S_{t}(\mathbf{x}),1-\left(1-P_{N}^{\theta}\right)^{\widetilde{I}_{t}(\mathbf{x})}\right),$
(2) $\displaystyle\mbox{ where }\widetilde{I}_{t}(\mathbf{x})$
$\displaystyle=\sum_{||\mathbf{y}-\mathbf{x}||_{1}\leq
1}I_{t}(\mathbf{y}),\quad\mbox{ and}$ $\displaystyle\mbox{
}R_{t+1}(\mathbf{x})$
$\displaystyle=I_{t}(\mathbf{x})+R_{t}(\mathbf{x}),\quad\mbox{ for all }t\geq
0.$

Here, $\stackrel{{\scriptstyle\mathrm{d}}}{{=}}$ means “equal in
distribution”, $\mathrm{Bin}(n,p)$ represents the binomial distribution with
parameters $n$ and $p$, and $||\cdot||_{1}$ denotes the $\ell_{1}$ norm,
namely, $||(a,b)||_{1}=|a|+|b|$ for any $(a,b)\in\mathbb{R}^{2}.$ The way we
define the function $\widetilde{I}$ will be used throughout the rest of the
paper, namely, for any function $\phi(\cdot)$ on $\mathbb{Z}^{2}$,
$\widetilde{\phi}(\cdot)$ is a function defined as follows:

$\widetilde{\phi}(\mathbf{x})=\sum_{||\mathbf{y}-\mathbf{x}||_{1}\leq
1}\phi(\mathbf{y}),\quad\mbox{for all }\mathbf{x}\in\mathbb{Z}^{2}.$

We focus on the following two initial conditions:

  * •

IC1:

$I_{0}^{N}(\mathbf{0})=1,\mbox{ and }I_{0}^{N}(\mathbf{x})=0\mbox{ for all
}\mathbf{x}\neq\mathbf{0};$ (3)

  * •

IC2: for some fixed $\gamma\in(0,1]$,

$I_{0}^{N}(\mathbf{0})\sim\gamma N,\mbox{ and }I_{0}^{N}(\mathbf{x})=0\mbox{
for all }\mathbf{x}\neq\mathbf{0},$ (4)

where for any two sequences $(a_{n})$ and $(b_{n})$, we write $a_{n}\sim
b_{n}$ if $a_{n}/b_{n}\to 1$.

The default setting is that $R_{0}^{N}\equiv 0$, unless otherwise stated.

IC1 corresponds to the situation where the epidemic starts with one infected
individual, so-called “patient zero”. IC2 can be considered as the situation
where there is a virus outbreak, which causes an instant infection of a
significant proportion of population at the outbreak point.

We now state the result about the probability that the epidemic lasts forever.
Let us recall that for a Galton-Watson process started by one particle and
with offspring distribution Poisson$(1+\theta)$, its survival probability,
denoted by $\iota$, is the unique solution to the following equation:

$1-\iota=\exp(-(1+\theta)\iota),\quad\iota\in(0,1).$ (5)

More general results can be found in, e.g., [Athreya and Ney(1972)].

######  Theorem 1.

Let $q_{N}$ be the survival probability of the SIR process with village size
$N$, namely, $q_{N}=P(I_{t}^{N}(\cdot)\not\equiv 0\mbox{ for all }t>0)$. Then,
we have

$\lim_{N\rightarrow\infty}q_{N}=\left\\{\begin{aligned}
\iota,&\quad\mbox{under the initial condition IC1};\\\ 1,&\quad\mbox{under the
initial condition IC2}.\end{aligned}\right.$

Theorem 1 states that as the village size gets larger, the survival
probability of the SIR process under the initial condition IC1 approaches the
survival probability of a supercritical Galton-Watson process. The intuition
is the following. During the initial period, the population of people who are
immune to the disease is small, hence if the village size $N$ is large, the
epidemic evolves like a supercritical Galton-Watson process, causing the
number of infected individuals to blow up quickly. Afterwards, due to the
fixed village size and the accumulation of people who are immune to the
disease, the SIR process starts to evolve differently than the Galton-Watson
process. However, because the number of infected individuals is already large,
with a high probability, the epidemic will last forever. Whether the epidemic
lasts forever or not hence depends mainly the initial period, during which
time the process behaves similarly to a Galton-Watson process.

Next, we turn to Q2, the spreading speed of the epidemic. Because the disease
can only be spread between neighboring sites, the speed, under the $\ell_{1}$
norm, is at most one. Intuitively, the bigger the $\theta$, the more likely
the disease spreads out, hence the higher the speed is. Our second main result
states that there is a double phase transition, according to how $\theta$
compares with $1.5$ or $4$, and the speed can be strictly smaller than one in
all directions, or smaller than one in some but not all directions, or equal
to one in all directions.

To give the explicit formula of the spreading speed, we need to define several
functions. Let

$h(t)=t\log t+(1-t)\log(1-t),\quad t\in[0,1],$
$a(\phi)=\frac{\min(|\sin(\phi)|,|\cos(\phi)|)}{|\sin(\phi)|+|\cos(\phi)|},\quad\phi\in[0,2\pi),$

and

$\displaystyle G(v,\phi)$
$\displaystyle=\inf_{t\in[v,1]}h(t)+t\left(h\left(\frac{1}{2}-\frac{v}{2t}\right)+h\left(\frac{1}{2}-\frac{(1-2a(\phi))v}{2t}\right)\right),$
(6) $\displaystyle\quad\mbox{ for }v\in(0,1]\mbox{ and }\phi\in[0,2\pi).$

The function $G$ is strictly decreasing in $v\in(0,1]$ with

$\lim_{v\rightarrow 0}G(v,\phi)=\log\frac{1}{5},\quad\mbox{and}\quad
G(1,\phi)=h(a(\phi)).$

It follows that for any $\theta\in(0,1.5)$, there exits a unique
$\upsilon=\upsilon(\theta,\phi)\in(0,1)$ such that

$G(\upsilon,\phi)=\log\left(\frac{1+\theta}{5}\right).$ (7)

When $\theta\geq 1.5,$ it may occur that $h(a(\phi))\leq\log((1+\theta)/5)$,
in which case we define

$\upsilon(\theta,\phi)=1.$ (8)

Finally, we define for any $\mathbf{0}\neq(x,y)\in\mathbb{R}^{2}$, a function
$\arg(x,y)\in[0,2\pi)$ to be the angle from the positive real axis to the
vector representing the complex number $x+\mathbf{i}y$.

######  Theorem 2.

For any $\theta\in(0,\infty)$, either under the initial condition IC1 and
conditional on that the epidemic lasts forever, or under the initial condition
IC2, the following results hold:

  1. (i)

for any $\varepsilon>0$, any village size $N$, along any sequence
$(i_{k},j_{k})_{k}\subset\mathbb{Z}^{2}$ satisfying
$|i_{k}|+|j_{k}|\rightarrow\infty$ and $\arg(i_{k},j_{k})\rightarrow\phi$ for
some $\phi\in[0,2\pi)$, we have

$\lim_{k\to\infty}P(R_{\lfloor({|i_{k}|+|j_{k}|})(\upsilon(\theta,\phi)^{-1}-\varepsilon)\rfloor}(i_{k},j_{k})>0)=0;$

  2. (ii)

For any $\varepsilon>0$, when the village size $N$ is sufficiently large,
along any sequence $(i_{k},j_{k})_{k}\subset\mathbb{Z}^{2}$ satisfying
$|i_{k}|+|j_{k}|\rightarrow\infty$ and $\arg(i_{k},j_{k})\rightarrow\phi$ for
some $\phi\in[0,2\pi)$, we have

$\limsup_{k\to\infty}P\left({\left|\frac{R_{\lfloor({|i_{k}|+|j_{k}|})(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor}(i_{k},j_{k})}{N}-\iota\right|>\varepsilon}\right)<\varepsilon.$

Results like Theorem 2 are known as “Shape Theorem”. A number of such theorems
have been proved for other models; see, e.g., [Richardson(1973), Durrett and
Liggett(1981), Cox and Durrett(1988), Zhang(1993), Lalley(2003)]. However, in
all these situations, the exact form of the limiting shape is unknown. To the
best of our knowledge, Theorem 2 is the first such result for which the exact
shape is known.

Let us explain Theorem 2 in words. For any direction $\phi\in[0,2\pi)$, any
site $(i,j)$ that is far away from the origin and on this direction, namely,
$(i,j)\approx(|i|+|j|)(\cos\phi,\sin\phi)$, before generation
$\lfloor(|i|+|j|)(\upsilon(\theta,\phi)^{-1}-\varepsilon)\rfloor$, with a high
probability, there is no recovered individual, in other words, the epidemic
has not reached the site. However, after a short period of time, at generation
$\lfloor(|i|+|j|)(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor$, conditional
on that the epidemic lasts forever, with a high probability, there is roughly
a constant proportion of recovered individuals. The constant proportion, as we
shall see from Theorems 3 and 4 below, turns out to be the ultimate proportion
of recovered individuals over the whole period. To sum up, the epidemic
spreads at the speed of $\upsilon(\theta,\phi)$ along direction $\phi$, and
most infections occur during a short period of time.

Let us visualize the statement about speed. Theorem 2 asserts that the
“frontier” of the epidemic, after scaled by the generation, converges to the
curve $(\upsilon(\theta,\phi):\phi\in[0,2\pi))$. The following pictures plot
the curve for three different values of $\theta:$ $\theta=1,2$ and $5$.

Figure 1: Limiting shape of the frontier of the epidemic when $\theta=1,2$ and
$5$.

We note that there is an interesting phase transition:

  * •

when $\theta<1.5$, the speed is smaller than one in all directions;

  * •

when $\theta\in[1.5,4)$, there are four cones along the $x$ and $y$ axes,
inside of which the speed is smaller than one while the speed equals one
outside;

  * •

when $\theta\geq 4$, the speed equals one in all directions.

To see how well the curve $(\upsilon(\theta,\phi):\phi\in[0,2\pi))$ describes
the frontier, we run simulations for $\theta=1,2$ and $5$ with village size
$N=1,000$ and up to generation $T=1,000$ under the initial condition IC1. The
following pictures show the population of infected/recovered individuals.

Figure 2: Population of infected/recovered individuals from random simulations
when $\theta=1,2$ and $5$ with village size $N=1,000$ and at generation
$T=1,000$ under the initial condition IC1. Above: infected population; bottom:
recovered population. The pink curve is
$(T\upsilon(\theta,\phi):\phi\in[0,2\pi))$.

We see that the curves provide remarkably accurate description of the frontier
in all three cases.

Now we turn to Q3, the ultimate proportion of individuals who will be
infected. In fact, the three pictures in the bottom panel of Figure 2 already
shed some light on the proportion. We see that the proportion is almost
constant behind the frontier. Our next result confirms this observation and
gives the explicit value for the constant. Recall that $\iota$ is defined in
(5). Note that for any fixed $N$ and $\mathbf{x}$, $R^{N}_{n}(\mathbf{x})$ is
increasing in $n$, so we can define
$R^{N}_{\infty}(\mathbf{x})=\lim_{n\to\infty}R^{N}_{n}(\mathbf{x})$.

######  Theorem 3.

For any $\varepsilon>0$, there exists $N_{0}\in\mathbb{N}$ such that for all
$N\geq N_{0}$, under the initial condition IC1, conditional on that the
epidemic lasts forever, we have

$\sup_{\mathbf{x}\in\mathbb{Z}^{2}}P\left(\left|\frac{R^{N}_{\infty}(\mathbf{x})}{N}-\iota\right|>\varepsilon\right)<\varepsilon.$
(9)

Under the initial condition IC2, the ultimate proportion of infection is not
constant over the space, rather, it is specified by a difference equation.
More precisely, let
$\mathcal{R}_{\infty}(\mathbf{x}):\mathbb{Z}^{2}\rightarrow[0,1]$ be the
solution to the following difference equation:

$\displaystyle f(\mathbf{x})$
$\displaystyle=1-\exp(-\frac{1+\theta}{5}\widetilde{f}(\mathbf{x})),\,\mbox{
for all }\mathbf{x}\in\mathbb{Z}^{2}\setminus\\{\mathbf{0}\\},\mbox{ and }$
(10) $\displaystyle f(\mathbf{0})$
$\displaystyle=\gamma+(1-\gamma)\left(1-\exp(-\frac{1+\theta}{5}\widetilde{f}(\mathbf{0}))\right).$

We will show in Proposition 30 that there is a unique solution to the above
equation. Moreover, the function is symmetric and satisfies that
$\mathcal{R}_{\infty}(\mathbf{x})\searrow\iota$ in the sense that
$\mathcal{R}_{\infty}(\mathbf{x})\geq\max\\{\mathcal{R}_{\infty}(\mathbf{x}+(1,0)),\mathcal{R}_{\infty}(\mathbf{x}+(0,1))\\}$
for all $\mathbf{x}\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$, and
$\mathcal{R}_{\infty}(\mathbf{x})\to\iota$ as
$||\mathbf{x}||_{1}\rightarrow\infty$.

######  Theorem 4.

Under the initial condition IC2, we have

$\frac{R^{N}_{\infty}(\mathbf{x})}{N}\stackrel{{\scriptstyle\mathbb{P}}}{{\longrightarrow}}\mathcal{R}_{\infty}(\mathbf{x}),\mbox{
for all }\mathbf{x}\in\mathbb{Z}^{2}\mathrm{~{}as~{}}N\rightarrow\infty.$

Moreover, for any $\varepsilon>0$, when $N$ is sufficiently large, we have

$\limsup_{\mathbf{x}\rightarrow\infty}P\left(\left|\frac{R^{N}_{\infty}(\mathbf{x})}{N}-\iota\right|>\varepsilon\right)<\varepsilon.$

Theorems 3 and 4 have a significant implication on “herd immunity”. In
general, for an epidemic with a basic reproduction number $R_{0}$, the minimum
proportion of the population that needs to be vaccinated to prevent sustained
spread of the infection is $1-1/R_{0}$. In our case, $R_{0}\sim 1+\theta$,
hence the required vaccination proportion is $\sim 1-1/(1+\theta)$. Theorem 3,
however, states that under the initial condition IC1, if there is no
vaccination (hence the only way for an individual to gain immunity is via
recovery from infection), then on the event that the the epidemic lasts
forever, the proportion of individuals who will be infected throughout the
whole process is around $\iota$. Under the initial condition IC2, the
proportion of individuals who will be infected on site far away from the
origin is also around $\iota$. A natural question is: How do the two
proportions compare with each other? The comparison is illustrated in Figure
3.

Figure 3: Comparison between the ultimate proportion of infection and the
minimum required vaccination proportion for different values of $\theta$.

Figure 3 reveals a sharp difference between “natural immunization”, namely,
recovery from infection, and vaccination. Via natural immunization, the
ultimate proportion of population who will be infected can be _much higher_
than the required vaccination proportion to prevent sustained spread of the
infection. For an epidemic with a basic reproduction number of 3, the two
proportions are 0.94 and $2/3$, respectively. In other words, without
vaccination, ultimately, there will be $94\%$ of population who will be
infected! This is substantially higher than the required vaccination level of
$2/3.$ The difference between the two proportions highlights the huge benefit
of vaccination.

Finally, we give a theorem to describe how infection evolves inside the “speed
one” cone. Recall that by Theorem 2, in order to have such a cone, $\theta$
needs to be larger than $1.5$. When $\theta\in(1.5,4)$, let $\kappa$ be the
unique solution in $(0,0.5]$ to

$\kappa^{\kappa}(1-\kappa)^{1-\kappa}=\frac{1+\theta}{5}.$ (11)

When $\theta\geq 4$, let $\kappa=0$. Also note that $\upsilon(\theta,\phi)=1$
if and only if $\kappa\leq|\cos(\phi)|/(|\cos(\phi)|+|\sin(\phi)|)\leq
1-\kappa$.

######  Remark 1.

The value $\kappa$ equals to the velocity of the leftmost position of a
supercritical branching random walk. Specifically, consider a branching random
walk started by one particle at the origin. At every generation, each particle
produces a random number of offspring according to
$\mathrm{Bin}(2N,(1+\theta)/(5N))$ and dies out; each offspring independently
moves to the upper neighbour or the right neighbour with equal probability. It
is easy to see that such defined process stochastically dominates
$(I_{n}^{N}(m,n-m))_{m\in\mathbb{Z}}$. Write $(m(n),n-m(n))$ as the leftmost
position of the branching random walk in the $n$-th generation. Then, the
classical results in branching random walk (see, e.g., [Hammersley(1974)],
[Kingman(1975)], [Biggins(1976)] and [Bramson(1978)]) state that, conditional
on the event that the branching random walk survives forever, one has

$\frac{m(n)}{n}\rightarrow\kappa.$

Before giving our next theorem, we define a sequence
$\\{\ell^{(k)}\\}_{k\geq-1}$ recursively as follows:
$\ell^{(-1)}=\ell^{(0)}=0,$ and for $i\geq 1$, $\ell^{(i)}$ is the unique
solution in $(0,1)$ to

$\ell^{(i)}=\left(1-\sum_{1\leq
j<i}\ell^{(j)}\right)\left(1-\exp\left(-\frac{1+\theta}{5}\cdot(2\ell^{(i)}+\ell^{(i-1)}+2\ell^{(i-2)})\right)\right).$
(12)

We will show in Lemma 31 that $\iota=\sum_{i=1}^{\infty}\ell^{(i)}$ when
$\theta>1.5$. Further define a sequence of random variables
$\\{\mathcal{K}(n)=\mathcal{K}^{N}(n)\\}$ as

$\mathcal{K}(n)=\mathcal{K}^{N}(n)=\inf\left\\{k:\sum_{(m,l)\in\mathbb{Z}^{2}:m+l=n}I^{N}_{n+k}(m,l)>0\right\\},$

with the convention that $\inf\emptyset=\infty$. In words, $\mathcal{K}(n)$
measures after how many generations the epidemic reaches the line
$\\{(m,l)|m+l=n\\}$. It is easy to see that $\mathcal{K}^{N}(n)$ is
increasing, and we can define

$\mathcal{K}^{N}=\lim_{n\rightarrow\infty}\mathcal{K}^{N}(n).$

######  Theorem 5.

Suppose that $\theta>1.5$, then the following results hold:

  * •

Under the initial condition IC2, for any $\varepsilon>0$ and $i\geq 0$, when
$N$ is sufficiently large, we have

$\limsup_{n\rightarrow\infty}\sup_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\left|\frac{I_{n+i}^{N}(m,n-m)}{N}-\ell^{(i+1)}\right|>\varepsilon\right)<\varepsilon.$

  * •

Under the initial condition IC1, we have

    1. (i)

For any $i\geq 0$,

$\lim_{N\rightarrow\infty}P(\mathcal{K}^{N}=i)=\ell^{(i+1)}.$

    2. (ii)

For any $i,j\geq 0$, $\varepsilon>0$, when $N$ is sufficiently large, we have

$\limsup_{n\rightarrow\infty}\sup_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}\\\
P\left(\left|\frac{I^{N}_{n+j+i}(m,n-m)}{N}-\ell^{(i+1)}\right|>\varepsilon\
|\ {\mathcal{K}^{N}}=j\right)<\varepsilon.$ (13)

######  Remark 2.

Theorem 5 is stated for the first quadrant. Analogous results hold for the
three other quadrants.

Let us give some explanations to the above theorem. Under the initial
condition IC2, when $N$ is sufficiently large, inside the “speed one” cone,
Theorem 5 asserts that the epidemic spreads out exactly at speed one in the
sense that as the generation $n$ increases, with a high probability, the
epidemic would reach the diamond $\\{(x,y):|x|+|y|=n\\}$, and the infection
proportion at sites on the diamond inside the cone is approximately
$\ell^{(1)}$. Moreover, for any $i\geq 1$, at $i$ generations afterwards, the
infection proportion would be approximately $\ell^{(i+1)}$. Under the initial
condition IC1, similar conclusions hold except that the frontier of the
epidemic in the first quadrant is $\mathcal{K}(n)$ layers behind the line
$\\{(x,y):x+y=n\\}$. Analogous results hold for other three quadrants.

Figure 4 visualizes Theorem 5. We take $\theta=2$ with village size $N=1,000$
and run the process up to generation $T=1,000$ under IC1. The left picture in
Figure 4 plots the distribution of infected population at time $T=1,000$. We
then compute the infection proportions on sites along the direction $y=x$ near
the frontier. The results are shown as the black curve in the right picture in
Figure 4. The red curve, on the other hand, plots the values of
$\\{\ell^{(i)}\\}$. We can see that $\\{\ell^{(i)}\\}$ approximates the
(random) infection proportions well.

Figure 4: Comparison between $\\{\ell^{(i)}\\}$ and the infection proportions
near the frontier from a random simulation when $\theta=2$ with village size
$N=1,000$ and at generation $T=1,000$ under the initial condition IC1. The
random variable $\mathcal{K}^{N}(998)=2$ for this simulation.

###  1.4 Connection to percolation

Let us recall a connection between our SIR process and a percolation model
([Lalley(2009)]). We will call the percolation the $N$-percolation, which is
defined as follows. Consider the graph with vertex set
$\mathbb{Z}^{2}\times[N]$, where we write $[N]$ for $\\{1,2\dots,N\\}$, and
edge set
$\\{(\mathbf{x},i)-(\mathbf{y},j):||\mathbf{x}-\mathbf{y}||_{1}\in\\{0,1\\}\\}$.
Each edge is open with probability $P_{N}^{\theta}$. For any initial condition
$(I^{N}_{0}(\cdot),R^{N}_{0}(\cdot)\equiv 0)$, let $A$ be any subset of
$\mathbb{Z}^{2}\times[N]$ such that
$\\#(A\cap\mathbf{x}\times[N])=I^{N}_{0}(\mathbf{x})$ for any
$\mathbf{x}\in\mathbb{Z}^{2}$. Then, $I^{N}_{n}(\mathbf{x})$
($R^{N}_{n}(\mathbf{x})$, respectively) equals the number of vertices at site
$\mathbf{x}$ that are of graph distance $n$ to $A$ (of graph distance $\leq
n-1$, respectively). Moreover,
$R^{N}_{\infty}(\mathbf{x})=\lim_{n\rightarrow\infty}R^{N}_{n}(\mathbf{x})$ is
just the number of vertices at site $\mathbf{x}$ that are connected to $A$.
From this, we see that $R^{N}_{t}$(but not $I^{N}_{t}$) is increasing in the
initial condition $I_{0}^{N}$. Based on such a connection, Theorems 1-5 can be
interpreted correspondingly for the percolation model.

###  1.5 Generalization to other dimensions and distributions

Our methods of proof are robust enough to extend to other dimensions and
infection distributions. We state the assumptions and results below without
giving detailed proofs.

Suppose that the dimension $d\geq 2$, and
$p=\\{p(\mathbf{x})\\}_{\mathbf{x}\in\mathbb{Z}^{d}}$ is a probability measure
on $\mathbb{Z}^{d}$ that satisfies the following regularity assumptions:

  1. (i)

$p$ has a finite support, namely,
$\mathrm{Supp}(p):=\\{\mathbf{x}:p(\mathbf{x})>0\\}$ is finite;

  2. (ii)

$p$ is symmetric, i.e., $p(\mathbf{x})=p(-\mathbf{x})$ for all $\mathbf{x}$;

  3. (iii)

$p$ is irreducible, i.e., $\mathrm{Supp}(p)$ is not contained in any strict
subgroup of $\mathbb{Z}^{d}$;

  4. (iv)

$p$ is aperiodic, i.e., there exist an odd $n$ and
$\mathbf{x}_{1},\dots,\mathbf{x}_{n}\in\mathbb{Z}^{d}$ such that
$\sum_{i=1}^{n}\mathbf{x}_{i}=\mathbf{0}$ and
$\prod_{i=1}^{n}p(\mathbf{x}_{i})\neq 0$.

The general SIR process on $\mathbb{Z}^{d}\times[N]$ evolves as follows:

$\displaystyle I_{t+1}(\mathbf{x})|(I_{t}(\mathbf{x}),R_{t}(\mathbf{x}))$
$\displaystyle\stackrel{{\scriptstyle\mathrm{d}}}{{=}}\mathrm{Bin}\left(S_{t}(\mathbf{x}),1-\prod_{\mathbf{y}\in\mathrm{Supp}(p)}\left(1-(1+\theta)\frac{p(\mathbf{y})}{N}\right)^{I_{t}(\mathbf{x}+\mathbf{y})}\right),$
$\displaystyle\mbox{ }R_{t+1}(\mathbf{x})$
$\displaystyle=I_{t}(\mathbf{x})+R_{t}(\mathbf{x}),\quad\mbox{ for all }t\geq
0.$

The initial conditions are still IC1 and IC2.

Under the above setting, Theorems 1 and 3 still hold. Theorem 4 hold as well
except that $\mathcal{R}_{\infty}(\cdot)$ is defined via the following
difference equation:

$\displaystyle f(\mathbf{x})$
$\displaystyle=1-\exp(-(1+\theta)\mathcal{A}f(\mathbf{x})),\,\mbox{ for all
}\mathbf{x}\in\mathbb{Z}^{d}\setminus\\{\mathbf{0}\\},\mbox{ and }$
$\displaystyle f(\mathbf{0})$
$\displaystyle=\gamma+(1-\gamma)\left(1-\exp(-(1+\theta)\mathcal{A}f(\mathbf{0}))\right),$

where $\mathcal{A}$ stands for the Markov operator with respect to $p$,
namely,
$\mathcal{A}f(\mathbf{x})=\sum_{\mathbf{y}\in\mathrm{Supp}(p)}p(\mathbf{y})f(\mathbf{x}+\mathbf{y})$.

As to the spreading speed, we need to define several functions. Write
$\mathcal{R}^{1}$ and $\mathcal{Q}^{1}$ for the unit real and rational
spheres:

$\mathcal{R}^{1}:=\\{\mathbf{x}\in\mathbb{R}^{d}|\,||\mathbf{x}||_{1}=1\\},\quad\mathcal{Q}^{1}:=\mathcal{R}^{1}\cap\\{(x_{1},\dots,x_{d})\,|\,x_{i}\in\mathbb{Q},\
i=1,\ldots,d\\},$

where we write $\mathbb{Q}$ for the set of rational numbers. For any
$\mathbf{x}\in\mathbb{Z}^{d}\setminus\\{\mathbf{0}\\},u\in(0,+\infty)\cap\mathbb{Q}$,
for all $n\geq 1$ with $nu\in\mathbb{Z}$, define

$b_{n}=b_{n}(\mathbf{x},u)=P(S_{n||\mathbf{x}||_{1}}=nu\mathbf{x}),$

where $(S_{m})$ is a random walk starting at $\mathbf{0}$ with jump
distribution $p$. Clearly, $b_{n+m}\geq b_{n}b_{m}$, hence the sequence
$(-\log b_{n})$ is subadditive. Therefore, by Fekete’s Lemma,
$b_{n}\approx\varrho^{n}$ for some $\varrho=\varrho(\mathbf{x},u)\in[0,1]$.
Note that $\varrho(\mathbf{x},u)=\varrho(q\mathbf{x},u)$ for any
$q\in\mathbb{Q}\setminus\\{0\\}$ and for fixed $\mathbf{x}$,
$\varrho(\mathbf{x},u)$ is strictly decreasing in $u$ for
$u\in\mathbb{Q}\cap(0,\sup\\{u:b(\mathbf{x},u)>0\\})$. Therefore, $\varrho$
can be extended to be a function in
$\mathcal{Q}^{1}\times((0,\infty)\cap\mathbb{Q})$. For any fixed $\theta>0$
and $\mathbf{x}\in\mathcal{Q}^{1}$, define

$\upsilon(\mathbf{x})=\upsilon(\theta,\mathbf{x}):=\inf\left\\{u:\varrho(\mathbf{x},u)\leq\frac{1}{1+\theta}\right\\}.$

It can be easily verified that $\upsilon(\cdot)$ is a concave function on
$\mathcal{Q}^{1}$ and therefore can be extended to be a continuous concave
function in $\mathcal{R}^{1}$. Moreover, it can be shown that
$\upsilon(\theta,\mathbf{x})>0$ for all $\theta>0$ and
$\mathbf{x}\in\mathcal{R}^{1}$.

We now state the generalization of Theorem 2.

######  Theorem 6.

For any $\theta\in(0,\infty)$, either under the initial condition IC1
conditional on that the epidemic lasts forever, or under the initial condition
IC2, the following results hold:

  1. (i)

for any $\varepsilon>0$, any village size $N$, along any sequence
$\mathbf{x}_{k}\subset\mathbb{Z}^{d}$ satisfying
$||\mathbf{x}_{k}||_{1}\rightarrow\infty$ and
$\mathbf{x}_{k}/||\mathbf{x}_{k}||_{1}\rightarrow\mathbf{y}$ for some
$\mathbf{y}\in\mathcal{R}^{1}$, we have

$\lim_{k\to\infty}P(R_{\lfloor||\mathbf{x}_{k}||_{1}(\upsilon(\theta,\mathbf{y})^{-1}-\varepsilon)\rfloor}(\mathbf{x}_{k})>0)=0;$

  2. (ii)

For any $\varepsilon>0$, when the village size $N$ is sufficiently large,
along any sequence $\mathbf{x}_{k}\subset\mathbb{Z}^{d}$ satisfying
$||\mathbf{x}_{k}||_{1}\rightarrow\infty$ and
$\mathbf{x}_{k}/||\mathbf{x}_{k}||_{1}\rightarrow\mathbf{y}$ for some
$\mathbf{y}\in\mathcal{R}^{1}$, we have

$\limsup_{k\to\infty}P\left({\left|\frac{R_{\lfloor||\mathbf{x}_{k}||_{1}(\upsilon(\theta,\mathbf{y})^{-1}+\varepsilon)\rfloor}(\mathbf{x}_{k})}{N}-\iota\right|>\varepsilon}\right)<\varepsilon.$

###  1.6 Plan of the paper

The remainder of the paper is organized as follows. In Section 2, we introduce
a deterministic process, which arises as the limiting process of our SIR
process as the village size $N\to\infty$. In Section 3, we focus on the
deterministic process and prove the analogous version of Theorem 5 for this
deterministic process. In Section 4, we prove Theorem 5. In Section 5, we
study the spreading speed for both the deterministic process and the SIR
process, and prove Theorem 2(i) and a weak version of Theorem 2(ii), see
Theorem 11. In Section 6, we consider the ultimate proportion of recovered
particles and prove Theorems 1, 3, 4 and Theorem 2(ii).

##  2 The Deterministic Large $N$ Limit Process

It is well known that the mean-field supercritical SIR model admits a
deterministic limit process as the population size gets to infinity.
Similarly, our spatial SIR model admits a deterministic large $N$ limit
process. It will play a fundamental role in the analysis of the original SIR
process.

The deterministic process
$(\mathcal{S}_{t}(\mathbf{x}),\mathcal{I}_{t}(\mathbf{x}),\mathcal{R}_{t}(\mathbf{x}))_{\\{t\in\mathbb{Z}_{\geq
0},\mathbf{x}\in\mathbb{Z}^{2}\\}}$ is defined as follows. For any initial
condition $\mathcal{I}_{0},\mathcal{R}_{0}\in[0,1]^{\mathbb{Z}^{2}}$ with
$\mathcal{S}_{0}(\mathbf{x})=1-\mathcal{I}_{0}(\mathbf{x})-\mathcal{R}_{0}(\mathbf{x})\geq~{}0$
for all $\mathbf{x}\in\mathbb{Z}^{2}$, the process evolves as

$\displaystyle\mathcal{I}_{t+1}(\mathbf{x})$
$\displaystyle=\mathcal{S}_{t}(\mathbf{x})\left(1-\exp\left(-\frac{1+\theta}{5}\widetilde{\mathcal{I}}_{t}(\mathbf{x})\right)\right),\mbox{
where }$ (14) $\displaystyle\mathcal{S}_{t}(\mathbf{x})$
$\displaystyle=1-\mathcal{I}_{t}(\mathbf{x})-\mathcal{R}_{t}(\mathbf{x})\quad\mbox{
and}$ $\displaystyle\mathcal{R}_{t+1}(\mathbf{x})$
$\displaystyle=\mathcal{I}_{t}(\mathbf{x})+\mathcal{R}_{t}(\mathbf{x}),\quad\mbox{
for all }t\geq 0\mbox{ and }\mathbf{x}\in\mathbb{Z}^{2}.$

The following proposition states that
$(\mathcal{S}_{t}(\mathbf{x}),\mathcal{I}_{t}(\mathbf{x}),\mathcal{R}_{t}(\mathbf{x}))$
is the scaling limit of the original SIR process as $N\to\infty$. Note that
since $\mathcal{S}+\mathcal{I}+\mathcal{R}\equiv 1$, we can drop $\mathcal{S}$
and only consider $(\mathcal{I},\mathcal{R})$.

######  Proposition 3.

Suppose that the initial conditions
$(I_{0}^{N}(\mathbf{x}),R_{0}^{N}(\mathbf{x}))$ and
$(\mathcal{I}_{0}(\mathbf{x}),\mathcal{R}_{0}(\mathbf{x}))$ satisfy

$\frac{I_{0}^{N}(\mathbf{x})}{N}\rightarrow\mathcal{I}_{0}(\mathbf{x}),\frac{R_{0}^{N}(\mathbf{x})}{N}\rightarrow\mathcal{R}_{0}(\mathbf{x}),\mbox{
for all }\mathbf{x}\in\mathbb{Z}^{2}\quad\mbox{as }N\rightarrow\infty.$ (15)

Then the process $(I_{t}^{N}(\mathbf{x})/N,R_{t}^{N}(\mathbf{x})/N)$
converges, under finite dimensional distribution, to the process
$(\mathcal{I}_{t}(\mathbf{x}),\mathcal{R}_{t}(\mathbf{x}))$.

###### Proof.

By the Markovian property, to prove finite dimensional convergence, it
suffices to prove the convergence for $t=1$, which follows from (2) and the
law of large numbers. ∎

######  Remark 4.

The initial condition IC1, after scaling, converges to the trivial initial
condition $\mathcal{I}_{0}\equiv 0$. For this reason, it requires additional
techniques to study the asymptotics under the initial condition IC1.

##  3 Distribution Near the Frontier: the Deterministic Case

In this section, we derive the distribution on the frontier for the
deterministic limit process. We only focus on the first quadrant; identical
arguments apply to the other three quadrants.

We assume that that $\mathcal{R}_{0}\equiv 0$ and $\mathcal{I}_{0}$ is
supported by the line $x+y=0$. Hence, the frontier at time $n$ is on the lines
$x+y=\pm n$. We remark that the frontier process is monotone, i.e., when the
initial condition $\\{\mathcal{I}_{0}(-m,m)\\}_{\\{m\in\mathbb{Z}\\}}$ gets
larger, the frontier $\\{\mathcal{I}_{n}(n-m,m)\\}_{\\{m\in\mathbb{Z}\\}}$
also becomes larger. This motivates us to consider the following two extreme
initial conditions, which serve as the upper and lower bounds: for some
$\gamma\in(0,1]$,

$\mathcal{I}_{0}(m,-m)=\gamma\quad\mbox{for all }m\in\mathbb{Z},\mbox{ and
}\mathcal{I}_{0}(\mathbf{x})=0\mbox{ otherwise, }$ (16)

or

$\mathcal{I}_{0}(0,0)=\gamma,\mbox{ and }\mathcal{I}_{0}(\mathbf{x})=0\mbox{
otherwise. }$ (17)

For notational ease, we write

$\alpha=\exp((1+\theta)/5),$ (18)

and

$\psi(x)=1-\alpha^{-x},\quad\mbox{for }x\in[0,1].$ (19)

This function plays a fundamental role in the analysis below. Note that it is
increasing and concave . Moreover, when $\alpha\leq e$, namely, when
$\theta\leq 4$, $\psi(x)<x$ for $x\in(0,1]$ and it has no fixed point in
$(0,1]$. On the other hand, when $\alpha>e$, namely, when $\theta>4$,
$\psi(\cdot)$ has a unique fixed point in $(0,1)$, denoted by $\gamma_{1}$, to
the left (right, respectively) of which $\psi(x)>x$ ($\psi(x)<x$,
respectively).

###  3.1 Convergence on the frontier

######  Lemma 5.

Under the initial condition (16), for all $n$, $\mathcal{I}_{n}(n-k,k)$ is
constant in $k$. Moreover, as $n\to\infty$, the constant converges to 0 when
$\theta\leq 1.5$ and to the unique positive solution
$\ell=\ell(\theta)=\ell^{(1)}$ to

$\ell=1-\exp\left(-\frac{2\ell(1+\theta)}{5}\right)$ (20)

when $\theta>1.5$.

###### Proof.

By symmetry, $\mathcal{I}_{n}(n-k,k)$ is constant in $k$. Denote the constant
value by $y_{n}$. By (14), we have

$y_{0}=\gamma,\quad y_{n+1}=1-\alpha^{-2y_{n}}.$ (21)

When $\theta\leq 1.5$, it is easy to see that $y_{n}$ is decreasing in $n$,
and the limit will solve (20). The only nonnegative solution to (20) is zero,
so $\lim_{n\to\infty}y_{n}=0.$

On the other hand, when $\theta>1.5$, $y_{n}$ is increasing when
$\gamma\leq\ell(\theta)$ and decreasing otherwise, so the limit again exists
and solves (20). The conclusion follows. ∎

Now we consider the initial condition (17). In this case, there are two phase
transitions: one at $\theta=1.5$ and the other at $\theta=4$. Precisely, we
have

######  Theorem 7.

Under the initial condition (17), the process $\\{\mathcal{I}_{n}(m,n-m)\\}$
behaves differently depending on the value of $\theta$.

  1. (i)

When $\theta\leq 1.5$, the process converges to zero uniformly:

$\lim_{n\rightarrow\infty}\sup_{m\in\\{0,\dots,n\\}}\mathcal{I}_{n}(m,n-m)=0;$

  2. (ii)

When $\theta\in(1.5,4]$, there is a cone along the diagonal $y=x$ such that
inside the cone the frontier converges to $\ell(\theta)$ while outside the
cone the process converges to zero. Moreover, as $\theta$ increases, the cone
grows continuously from the diagonal line to the first quadrant.

  3. (iii)

when $\theta\in(4,\infty)$, the process restricted to the first quadrant
converges to $\ell(\theta)$ uniformly:

$\lim_{n,k\rightarrow\infty:k\leq n/2}\sup_{m:k\leq m\leq
n-k}|\mathcal{I}_{n}(m,n-m)-\ell(\theta)|=0.$

Note that the result for the case when $\theta\in(0,1.5]$ is implied by Lemma
5 and the monotonicity of $\\{\mathcal{I}_{n}(m,n-m)\\}$ with respect to the
initial condition.

We first address the case when $\theta\in(4,\infty)$, which is relatively
easier. We will show a little bit more than the content of the above theorem.
Recall that $\gamma_{1}$ represents the unique fixed point of $\psi(\cdot)$
defined in (19).

######  Proposition 6.

Let $y_{n}(k)=\mathcal{I}_{n}(n-k,k)$. When $\theta>4$, under the initial
condition (17), if $\gamma\leq\gamma_{1}$, then for any $k\in\mathbb{Z}_{\geq
0}$, we have

$\lim_{n\rightarrow\infty}y_{n}(k)=\ell_{k},$ (22)

where $\ell_{k}$ is determined recursively by

$\ell_{0}=\gamma_{1},\quad\ell_{k+1}=1-\alpha^{-(\ell_{k}+\ell_{k+1})}.$ (23)

Moreover, we have

$\lim_{\min\\{m,n\\}\rightarrow\infty}\mathcal{I}_{m+n}(m,n)=\ell(\theta).$
(24)

######  Remark 7.

Note that when $\theta>4$ and $a\geq 0$, the equation $x=1-\alpha^{-(x+a)}$
has a unique solution in $(0,1)$.

###### Proof.

Write

$Y_{n}=(y_{n}(0),y_{n}(1),\ldots).$

Then $Y_{0}=({\gamma},0,\dots,)$ and

$y_{n+1}(k)=1-\alpha^{-(y_{n}(k)+y_{n}(k-1))}.$ (25)

Because $\gamma\leq\gamma_{1}$, we have $y_{1}(0)\geq y_{0}(0)$. Hence,
$Y_{1}\geq Y_{0}$. By monotonicity and induction, we get

$Y_{n+1}\geq Y_{n}.$

Hence, we can define

$Y_{\infty}:=\lim_{n\rightarrow\infty}Y_{n}=:(\ell_{0},\ell_{1},\ldots).$

Letting $n\rightarrow\infty$ in (25), we get that $\ell_{n}$ satisfies the
recursive equation (23). Note that $\ell_{n}$ increases to $\ell(\theta)$.

On the other hand, by induction, one can show that $\mathcal{I}_{n}(i,n-i)$ is
increasing for $i\in[0,\dots,\lfloor n/2\rfloor]$. Combining this with
$\ell_{n}\nearrow\ell(\theta)$, we get the last assertion. ∎

######  Remark 8.

By monotonicity, conclusion (24) holds for all initial conditions
$\mathcal{I}_{0}=\gamma\bm{\delta}_{(0,0)}$ with $\gamma\in(\gamma_{1},1]$.

Now we address the case when $\theta\in(1.5,4]$. Recall that $\kappa$ is
defined in (11).

######  Proposition 9.

Assume $\theta\in(1.5,4]$. Under the initial condition (17), for any
$\varepsilon>0$, we have

$\lim_{n\rightarrow\infty}\sup_{m:\,m<(\kappa-\varepsilon)n\,\mathrm{or}\,m>(1-\kappa+\varepsilon)n}\mathcal{I}_{n}(m,n-m)=0;$
(26)

and

$\lim_{n\rightarrow\infty}\sup_{m:\,(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}|\mathcal{I}_{n}(m,n-m)-\ell(\theta)|=0.$
(27)

###### Proof.

Write $Y:\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}\rightarrow[0,1]$ be the
value when the frontier arrives, i.e.,

$Y(m,n)=\mathcal{I}_{m+n}(m,n).$ (28)

The key observation is the following, which can be easily proved by induction.

Claim 1: For any $\mathbf{a}\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$
with

$Y(\mathbf{a})>Y(0,0)=\gamma,$

then, for any $\mathbf{x}\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$,

$Y(\mathbf{x}+\mathbf{a})>Y(\mathbf{x}).$

Inspired by this, we define

$g(\mathbf{x})=\left.\frac{\partial
Y}{\partial\gamma}\right|_{\gamma=0}(\mathbf{x}).$

Since $Y(m,n)=1-\alpha^{-(Y(m-1,n)+Y(m,n-1))}$, a simple application of the
chain rule yields

$g(m,n)=(\log\alpha)(g(m-1,n)+g(m,n-1)).$

Now, by induction, one can easily show

Claim 2:

$g(m,n)=(\log\alpha)^{m+n}\binom{m+n}{m}.$

By Stirling’s formula, we get

$g(m,n)={\sqrt{\frac{m+n+1}{2\pi(m+1)(n+1)}}}\left(\frac{\log\alpha}{s^{s}(1-s)^{1-s}}\right)^{m+n}\left(1+O\left(\frac{1}{m+1}+\frac{1}{n+1}\right)\right),$
(29)

where $s=m/(m+n)$.

Now we have finished the preparation and are ready to show (26) and (27).

For (26), note that $1-\alpha^{-x}\leq x\log\alpha$. We can get (by induction)

$Y(\mathbf{x})\leq g(\mathbf{x})\gamma.$

When $\mathbf{x}=(m,n)$ is outside the cone (meaning the condition inside the
supreme in (26)), $s^{s}(1-s)^{1-s}>\kappa^{\kappa}(1-\kappa)^{1-\kappa}$
($s=m/m+n$ as before). Therefore,
$\log\alpha/s^{s}(1-s)^{1-s}<1-\varepsilon_{0}$, for some $\varepsilon_{0}$
depending on $\varepsilon$. We see that $g(\mathbf{x})$ converges to zero
uniformly. Hence, $Y(\mathbf{x})\leq g(\mathbf{x})\gamma\rightarrow 0$
uniformly. The proof of (26) is complete.

For (27), we need more work. Since for the upper extreme case
($\mathcal{I}_{0}(n,-n)\equiv 1$), (27) holds, we just need to show the lower
extreme case, i.e., we can assume that $\gamma$ is small enough. For any
$\varepsilon>0$, by (29), we can find some large
$\mathbf{b},\mathbf{b}_{1},\mathbf{b}_{2}\in\mathbb{Z}_{\geq
0}\times\mathbb{Z}_{\geq 0}$ in the cone
$\\{(b_{1},b_{2}):{\kappa+0.5\varepsilon<b_{2}/(b_{1}+b_{2})<\kappa+\varepsilon}\\}$
such that $\mathbf{b}=\mathbf{b}_{1}+\mathbf{b}_{2}$,
$\min\\{g(\mathbf{b}),g(\mathbf{b}_{1}),g(\mathbf{b}_{2})\\}>1$, and
$\\{\mathbf{b}_{1},\mathbf{b}_{2}\\}$ generates $\mathbb{Z}^{2}$ in the sense
that any $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ can be written as
$m\mathbf{b}_{1}+n\mathbf{b}_{2}$ for some $m,n\in\mathbb{Z}$. Letting
$\gamma$ be small enough, we can assume
$\min\\{Y(\mathbf{b}),Y(\mathbf{b}_{1}),Y(\mathbf{b}_{2})\\}>\gamma$. We will
show that

$\lim_{k\rightarrow\infty}Y((x,y)+k\mathbf{b})=\ell(\theta),\;\mbox{ for all
}x\in\\{0,\ldots,b_{1}\\},y\in\\{0,\ldots,b_{2}\\}.$ (30)

Note that once we get (30), we have (27). The reason is as follows. By
induction, one can easily show that for any fixed $n$, $Y_{n}(n-m,m)$ is
increasing in $m\in\\{0,\dots,\lfloor n/2\rfloor\\}$. Hence, for any $(m,n-m)$
with $(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n$, we can find some
$(x,y)$ in the rectangle in (30),$k\in\mathbb{Z}_{\geq 0}$, such that
$Y(n-m,m)\geq Y((x,y)+k\mathbf{b})$. If the latter term goes to
$\ell(\theta)$, then so does the former term (note that all $Y$’s are less
than $\ell(\theta)$).

We are remaining to prove (30). By Claim 1, we know that for any
$(x,y)\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$, $Y((x,y)+k\mathbf{b})$
is increasing in $k$ and hence that the limit does exist. We denote the limit
by $h(x,y)$. Obviously,

$h(x,y)=h((x,y)+\mathbf{b}).$

By Claim 1, we have $Y((x,y)+\mathbf{b}_{i})>Y(x,y)$. Hence,

$h(x,y)\leq h((x,y)+\mathbf{b}_{i})\leq
h((x,y)+\mathbf{b}_{1}+\mathbf{b}_{2})=h((x,y)+\mathbf{b})=h(x,y).$

Therefore, $h(x,y)=h((x,y)+\mathbf{b}_{i})$. Since
$\\{\mathbf{b}_{1},\mathbf{b}_{2}\\}$ generates $\mathbb{Z}\times\mathbb{Z}$,
we obtain that all $h(x,y)$’s are equal. Moreover since
$Y(m,n)=1-\alpha^{-(Y(m-1,n)+Y(m,n-1))}$, by taking the limit, we get that
$h(x,y)$ satisfies the same equation of $\ell(\theta)$. Finally, since
$h(x,y)\geq\gamma>0$, we exclude the case $h(x,y)=0$ and reach the conclusion
that $h(x,y)=\ell(\theta)$. ∎

###  3.2 Convergence on all layers near the frontier

We consider the proportion of infection in all layers near the frontier. As in
the previous subsection, we focus on the initial conditions supported by the
line $x+y=0$. By the $k$-th layer at time $n$ near the frontier, we mean
$\\{\mathcal{I}_{n}(m,n+1-k-m)\\}_{m\in\mathbb{Z}}$. By symmetry, we only
consider the frontier on the positive direction.

We start with the initial condition (16). Denote

$y_{n}^{(i)}=\mathcal{I}_{n}(n+1-i,0)(=\mathcal{I}_{n}(n+1-i-k,k),\,\mbox{for
all }k),i=1,2,\ldots,n={i-1,i,\ldots}.$

When $i=1$, $(y_{n}^{(1)})$ satisfies the recursive equation (21). For $i=2$,
by (14) we have

$\displaystyle y_{1}^{(2)}$
$\displaystyle=(1-y^{(1)}_{0})(1-\alpha^{-y^{(1)}_{0}}),$ (31) $\displaystyle
y_{n+1}^{(2)}$
$\displaystyle=(1-y_{n}^{(1)})(1-\alpha^{-(2y_{n}^{(2)}+y_{n}^{(1)})}),\mbox{
for all }n\geq 1.$

In general, for $i\geq 3$, we have

$\displaystyle y_{i-1}^{(i)}=$ $\displaystyle\left(1-\sum_{1\leq
j<i}y_{j-1}^{(j)}\right)(1-\alpha^{-({2y_{i-2}^{(i-2)}+y_{i-2}^{(i-1)}+2y_{i-2}^{(i-2)}})});$
(32) $\displaystyle y_{n+1}^{(i)}=$ $\displaystyle\left(1-\sum_{1\leq
j<i}y_{n+1-i+j}^{(j)}\right)(1-\alpha^{-(2y_{n}^{(i)}+y_{n}^{(i-1)}+2y_{n}^{(i-2)})}),\,\mbox{
for all }n\geq i-1.$

Let us now define the limiting constants. Recall the sequence
$\ell^{(i)}=\ell^{(i)}(\theta)$ defined in (12). It is easy to verify that
when $\theta\leq 1.5$, (12) has no solution in $(0,1]$, in which case we
define $\ell^{(i)}\equiv 0$. On the other hand, when $\theta>1.5$, (12) has a
unique solution in $(0,1)$, which is defined as $\ell^{(i)}$.

######  Proposition 10.

Under the initial condition (16), for any $i\geq 1$, $y_{n}^{(i)}$ converges
to $\ell^{(i)}=\ell^{(i)}(\theta)$.

###### Proof.

The case $i=1$ has been proved in Lemma 5. We shall only prove for the case
when $i=2$; the case when $i\geq 3$ can be proved similarly.

We first consider the case when $\theta\neq 1.5$. Define

$\displaystyle f(x,y)$ $\displaystyle=(1-y)(1-\alpha^{-(2x+y)}),\quad
x,y\in[0,1].$

We have $y_{n+1}^{(2)}=f(y_{n}^{(2)},y_{n}^{(1)})$, and
$\ell^{(2)}=f(\ell^{(2)},\ell^{(1)})$. Elementary calculus yields that

$\displaystyle|f(x_{1},y_{1})-$ $\displaystyle f(x_{2},y_{2})|\leq$ (33)
$\displaystyle|y_{1}-y_{2}|+(1-\min\\{y_{1},y_{2}\\})\log(\alpha)(2|x_{1}-x_{2}|+|y_{1}-y_{2}|).$

Therefore,

$\displaystyle|y_{n+1}^{(2)}$ $\displaystyle-\ell^{(2)}|\leq$
$\displaystyle|y_{n}^{(1)}-\ell^{(1)}|+2(1-\min\\{y_{n}^{(1)},\ell^{(1)}\\})\log(\alpha)\cdot(\frac{1}{2}|y_{n}^{(1)}-\ell^{(1)}|+|y_{n}^{(2)}-\ell^{(2)}|).$

When $\theta\neq 1.5$, it is easy to show using equation (20) that
$2(1-\ell^{(1)})\log(\alpha)<1$. It follows that

$\displaystyle|y_{n+1}^{(2)}-\ell^{(2)}|\leq$
$\displaystyle(1.5-\frac{1}{2}\delta+\log(\alpha)|y_{n}^{(1)}-\ell^{(1)}|)|y_{n}^{(1)}-\ell^{(1)}|$
(34)
$\displaystyle+(1-\delta+2\log(\alpha)|y_{n}^{(1)}-\ell^{(1)}|)|y_{n}^{(2)}-\ell^{(2)}|,$

where $\delta>0$ is a fixed constant. Because $y_{n}^{(1)}\to\ell^{(1)}$,
taking limsup on both sides of (34) yields that $y_{n}^{(2)}\to\ell^{(2)}$.

We now turn to the case when $\theta=1.5$. In this case, $\alpha=\sqrt{e}$ and

$f(x,y)=(1-y)(1-e^{-(2x+y)/2}),\quad x,y\in[0,1].$

It is easy to show that for any $\varepsilon\in(0,1)$, there exist
$\varepsilon_{0}\in(0,\varepsilon)$ and $\delta\in(0,1)$ such that for any
$y\in[0,\delta]$, we have,

$f(x,y)<\left\\{\begin{array}[]{cc}\varepsilon-\varepsilon_{0},&\mathrm{~{}when~{}}x\in[0,\varepsilon];\\\
f(\varepsilon,y)+\exp(-\varepsilon)(x-\varepsilon)<x-\varepsilon_{0},&\mathrm{~{}when~{}}x\in(\varepsilon,1].\\\
\end{array}\right.$

Using such a fact and a similar argument to above, we get that $y_{n}^{(2)}\to
0$. ∎

Simple modifications of the proof of Proposition 10 yield the following
generalization.

######  Corollary 11.

For any initial condition $\\{\mathcal{I}_{0}(m,-m)\\}_{m\in\mathbb{Z}}$ such
that

$\lim_{n\rightarrow\infty}\sup_{m\in\mathbb{Z}}|\mathcal{I}_{n}(m,n-m)-\ell(\theta)|=0,$
(35)

we have,

$\lim_{n\rightarrow\infty}\sup_{m\in\mathbb{Z}}|\mathcal{I}_{n}(m,n+1-k-m)-\ell^{(k)}(\theta)|=0,\quad\quad\mbox{for
all }k\geq 1.$

######  Remark 12.

Thanks to the results in Section 3.1, we can get some sufficient conditions to
guarantee (35). For example, if there exists $\varepsilon>0$ such that
$\mathcal{I}_{0}(m,-m)>\varepsilon$ for all $m\in\mathbb{Z}$, then (35) holds.
The condition can be further weakened to be the following: If there exist
$\varepsilon>0$ and $k\in\mathbb{N}$ such that for any $m\in\mathbb{Z}$, there
exists $n\in\mathbb{Z}$ such that $|n-m|<k$ and
$\mathcal{I}_{0}(n,-n)>\varepsilon$, then the frontier at time $2k$ is
uniformly bounded from below and hence (35) holds.

Next result describes some asymptotics about the sequence $(\ell^{(i)})$.

######  Lemma 13.

If $\theta>1.5$, then,

$\sum_{i=1}^{\infty}\ell^{(i)}>\frac{\theta}{1+\theta}.$ (36)

Moreover, there exist $A>0$ and $b\in(0,1)$ depending on $\theta$ such that

$\ell^{(n)}\leq Ab^{n},\quad\mbox{for all }n\geq 1.$ (37)

###### Proof.

Write $s_{n}=\sum_{i=1}^{n}\ell^{(i)}$. By (12), $\ell^{(i)}<1-s_{i-1}$.
Therefore, $s_{n}\leq 1$ and $\ell^{(i)}\rightarrow 0$. Note that
$1-\alpha^{-t}\sim t\log\alpha$ as $t\to 0$. Hence as $n\to\infty$,

$\ell^{(n+1)}\sim(1-s_{n})(2\ell^{(n+1)}+\ell^{(n)}+2\ell^{(n-1)})\log\alpha.$
(38)

We will prove (36) by contradiction. Suppose that
$s_{\infty}=\sum_{i=1}^{\infty}\ell^{(i)}\leq\frac{\theta}{1+\theta}$.

If $\sum_{i=1}^{\infty}\ell^{(i)}<\frac{\theta}{1+\theta}$, then by (38), we
can find some $\delta>0$ and $n_{0}$ such that when $n\geq n_{0}$, we have

$\ell^{(n+1)}\geq\left(\frac{1}{5}+\delta\right)(2\ell^{(n+1)}+\ell^{(n)}+2\ell^{(n-1)}).$

Then, one can use induction to show that
$\ell^{(n)}\geq\min\\{\ell^{(n_{0})},\ell^{(n_{0}-1)}\\}$. This is
contradictory to that $\ell^{(n)}\rightarrow 0$.

Now assume $\sum_{i=1}^{\infty}\ell^{(i)}=\frac{\theta}{1+\theta}$. By (38),
for any $\delta>0$, we can find some $n_{0}$ such that when $n\geq n_{0}$, we
have

$\ell^{(n+2)}\geq(1-\delta)\frac{\ell^{(n+1)}+2\ell^{(n)}}{3}.$

By induction, one can show that when $n\geq n_{0}$ and $k\geq 2$,

$\ell^{(n+k)}\geq(1-\delta)^{k-1}\frac{3\ell^{(n+1)}+\ell^{(n)}}{9}.$

By summation, we get

$\sum_{k\geq
2}\ell^{(n+k)}\geq\frac{1-\delta}{\delta}\frac{3\ell^{(n+1)}+\ell^{(n)}}{9}.$
(39)

On the other hand, by Taylor expansion we have

$1-\alpha^{-t}\geq
t\log\alpha-\frac{t^{2}(\log\alpha)^{2}}{2},\mathrm{~{}for~{}}t\in[0,5].$ (40)

Using the above estimates, we get that with
$A_{n}=2\ell^{(n+1)}+\ell^{(n)}+2\ell^{(n-1)}$,

$\displaystyle\ell^{(n+1)}$
$\displaystyle=(1-s_{n})(1-\alpha^{-A_{n}})\geq(1-s_{n})A_{n}\log\alpha-\frac{1}{2}A_{n}^{2}(\log\alpha)^{2}$
(41) $\displaystyle=(1-s_{\infty})A_{n}\log\alpha+A_{n}\log\alpha\
\left(\sum_{k\geq 1}\ell^{(n+k)}-\frac{1}{2}A_{n}\log\alpha\right).$

Using (39) twice, we get

$\sum_{k\geq
1}\ell^{(n+k)}\geq\frac{1-\delta}{2\delta}\frac{3\ell^{(n+1)}+\ell^{(n)}+3\ell^{(n)}+\ell^{(n-1)}}{9}.$

It follows that by choosing $\delta$ to be small enough, we have $\sum_{k\geq
1}\ell^{(n+k)}>{1}/{2}\cdot A_{n}\log\alpha$ for all $n\geq n_{0}(\delta)$.
Therefore, by (41), we get that when $n\geq n_{0}$,

$\ell^{(n+1)}\geq(1-s_{\infty})A_{n}\log\alpha=\frac{2\ell^{(n+1)}+\ell^{(n)}+2\ell^{(n-1)}}{5}.$

Therefore,

$\inf_{n\geq n_{0}}\ell^{(n)}\geq\min\\{\ell^{(n_{0})},\ell^{(n_{0}+1)}\\},$

which again contradicts to that $\ell^{(n)}\rightarrow 0$.

We now prove (37). By (38) and (36), there exists $c>0$ such that when $n$ is
large enough, we have

$\ell^{(n+1)}\leq\left(\frac{1}{5}-c\right)(2\ell^{(n+1)}+\ell^{(n)}+2\ell^{(n-1)}).$

It follows that for a $c^{\prime}>0$, when $n$ is large enough, say, $n\geq
n_{1}$,

$\ell^{(n+1)}\leq(1-c^{\prime})\frac{\ell^{(n)}+2\ell^{(n-1)}}{3}.$

Let $b$ be the unique solution in $(0,1)$ to $x^{2}=(1-c^{\prime})(x+2)/3$.
Then one can show that if (37) holds for $n=k$ and $k+1$, then it holds for
$n=k+2$ as well. Pick an $A$ large enough such that (37) holds for all $n\leq
n_{1}+1$. The conclusion follows. ∎

##  4 Distribution Near the Frontier: the Stochastic Case

In this section, we prove analogous results for the SIR process. Again, we
focus on the first quadrant.

###  4.1 When $\theta\leq 1.5$

This case is relatively easy because the expectation is decreasing.

######  Proposition 14.

Suppose that $\theta\leq 1.5$ and $I^{N}_{0}(\mathbf{x})=0$ for all
$\mathbf{x}\not\in\\{(m,-m):m\in\mathbb{Z}\\}$.

  1. (i)

If $\theta<1.5$, then for all $N$,

$\sum_{n=0}^{\infty}\sup_{m\in\mathbb{Z}}E(I^{N}_{n}(n-m,m))<\infty.$

Consequently, for any sequence $(m_{n})\subseteq\mathbb{Z}$,

$I^{N}_{n}(n-m_{n},m_{n})\to 0\mbox{ a.s.}.$

  2. (ii)

If $\theta=1.5$, then for any sequence $(m_{n})\subseteq\mathbb{Z}$,

$I^{N}_{n}(n-m_{n},m_{n})\stackrel{{\scriptstyle\mathbb{P}}}{{\longrightarrow}}0\mbox{
}.$

  3. (iii)

Suppose further that $I^{N}_{0}(m,-m)$ is constant and $\theta<1.5$, then for
all $k\in\mathbb{Z}_{\geq 0}$,

$\sum_{n=0}^{\infty}E(I^{N}_{n}(0,n-k))<\infty.$

Consequently, for any sequence $(m_{n})\subseteq\mathbb{Z}$,

$I^{N}_{n}(n-m_{n}-k,m_{n})\to 0\mbox{ a.s.}.$

###### Proof.

We first prove Part (i). For any $n$ and $m$, it is easy to see that
$I^{N}_{n}(n-m,m)$ is increasing in the initial condition as long as the
initial condition is supported by the diagonal line
$\\{(m,-m):m\in\mathbb{Z}\\}$. Therefore, for Part (i) and (ii), it suffices
to prove the conclusions under the initial condition that

$I^{N}_{0}(m,-m)=N,\mbox{ and }I^{N}_{0}(\mathbf{x})=0\mbox{ for all
}\mathbf{x}\not\in\\{(m,-m):m\in\mathbb{Z}\\}.$ (42)

Under such an initial condition, $E(I^{N}_{n}(n-m,m))$ is constant in $m$.
Write $E(I^{N}_{n}(n-m,m))=y^{N}_{n}$ for any $m$. By (2), we have

$y^{N}_{n+1}=N\cdot
E\left(1-\left(1-\frac{1+\theta}{5N}\right)^{\tilde{I}_{n}^{N}(n+1,0)}\right)=:NE(g_{N}(\tilde{I}_{n}^{N}(n+1,0))),$

where $g_{N}(x)=1-(1-(1+\theta)/(5N))^{x}$ and
$\tilde{I}_{n}^{N}(n+1,0)=I_{n}^{N}(n,0)+I_{n}^{N}(n+1,-1)$. Elementary
calculus yields

$g_{N}(x)\leq\frac{1+\theta}{5N}x,\quad\mbox{ for all }x\in\mathbb{N}.$ (43)

Hence, $y^{N}_{n+1}\leq 2(1+\theta)/5\cdot y^{N}_{n}$. When $\theta<1.5$, we
see that $y^{N}_{n}$ decays to zero exponentially and the conclusion follows.

We turn to Part (ii). When $\theta=1.5$, we only have $y^{N}_{n+1}\leq
y^{N}_{n}$. This guarantees that $\delta:=\lim_{t\to\infty}y^{N}_{t}$ exists.
We want to show that $\delta=0$. By induction, it is easy to show that

$g_{N}(x)\leq\frac{1}{2N}x-\frac{1}{4N^{2}}(x-1),\quad\mbox{ for all
}x\in\mathbb{N}.$

It follows that

$y^{N}_{n+1}\leq y^{N}_{n}-\frac{1}{4N}P(I_{n}^{N}(n,0)\geq 2).$

Therefore, $P(I_{n}^{N}(n,0)\geq 2)\rightarrow 0$. On the other hand,

$\displaystyle P(I_{n}^{N}(n,0)\geq 2)$ $\displaystyle\geq$ $\displaystyle
P(I_{n-1}^{N}(n-1,0)\geq 1)\cdot P(I_{n}^{N}(n,0)\geq
2|I_{n-1}^{N}(n-1,0)+I_{n-1}^{N}(n,-1)=1).$

The latter probability is constant in $n$, and so $P(I_{n-1}^{N}(n-1,0)\geq
1)\to 0$.

Finally, we prove Part (iii). For $k\geq 1$, $E(I^{N}_{n}(n-k,0))$ is no
longer increasing in the initial condition. However, if $I^{N}_{0}(m,-m)$ is
constant in $m$, then $E(I^{N}_{n}(n-m-k,m))$ is constant in $m$ as well.
Write $E(I^{N}_{n}(n-m-k,m))=y^{N,k}_{n}$ for any $m$. By (2) and (43),

$y^{N,k}_{n+1}\leq\zeta(y^{N,k}_{n}+1/2\cdot y^{N,k-1}_{n}+y^{N,k-2}_{n}),$

where $\zeta=2(1+\theta)/{5}<1$. Therefore,

$(1-\zeta)\sum_{n\geq 0}y^{N,k}_{n+1}\leq
y^{N,k}_{0}+1/2\sum_{n}y^{N,k-1}_{n}+\sum_{n}y^{N,k-2}_{n}.$

It follows by induction that $\sum_{n\geq 0}y^{N,k}_{n}<\infty$ for all $k$. ∎

###  4.2 When $\theta>1.5$

####  4.2.1 Convergence on the frontier

######  Theorem 8.

Under the initial condition IC2, the process $\\{I^{N}_{n}(m,n-m)\\}$ behaves
differently depending on the value of $\theta$:

  1. (i)

If $\theta\in(1.5,4]$, then for any $\varepsilon>0$, for all $N$ sufficiently
large,

$\lim_{n\rightarrow\infty}P\left(\sum_{m:m<(\kappa-\varepsilon)n\,\mathrm{or}\,m>(1-\kappa+\varepsilon)n}I_{n}^{N}(m,n-m)>0\right)=0,$
(44)

and

$\limsup_{n\rightarrow\infty}\max_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\left|\frac{I_{n}^{N}(m,n-m)}{N}-\ell(\theta)\right|>\varepsilon\right)<\varepsilon,$
(45)

where $\ell(\theta)$ and $\kappa=\kappa(\theta)$ are defined in (20) and (11),
respectively.

  2. (ii)

If $\theta\in(4,\infty)$, then for any $\varepsilon>0$, for all $N$
sufficiently large, we have

$\limsup_{n\rightarrow\infty}\max_{m:\varepsilon
n<m<(1-\varepsilon)n}P\left(\left|\frac{I_{n}^{N}(m,n-m)}{N}-\ell(\theta)\right|>\varepsilon\right)<\varepsilon.$
(46)

The proof of Theorem 8 is divided into several parts. To simplify the
notation, we write

$W(m,l)=W^{N}(m,l)=I_{m+l}^{N}(m,l).$

Note that similarly to the limiting process, $W$ is monotone in the initial
condition as long as the initial condition is supported by the line
$\\{(x,y)|x+y=0\\}$, namely, if $\\{I^{\prime}_{0}(m,-m)\\}_{m\in\mathbb{Z}}$
stochastically dominates $\\{I^{\prime\prime}_{0}(m,-m)\\}_{m\in\mathbb{Z}}$,
then the corresponding $\\{W^{\prime}(n+m,-m)\\}$ stochastically dominates
$\\{W^{\prime\prime}(n+m,-m)\\}$.

As we mentioned in Section 1.4, the SIR process has an equivalent description
in terms of the $N$-percolation. Similarly, the “frontier” process $W$ can be
described in terms of an oriented percolation defined as follows, which we
call the oriented $N$-percolation. Consider the graph with vertex set
$\\{(x,y)\in\mathbb{Z}^{2}:x+y\geq 0\\}\times[N]$. The edge set consists of
all oriented edges pointing up or right, i.e., of the form
$\\{(\mathbf{x},i)\rightarrow(\mathbf{y},j):\mathbf{y}\in\\{\mathbf{x}+(0,1),\mathbf{x}+(1,0)\\}\\}$.
Each directed edge is open with probability $P_{N}^{\theta}$. For any initial
condition $I^{N}_{0}$ supported by the line $x+y=0$, let $A$ be any subset of
$\mathbb{Z}^{2}\times[N]$ such that
$\\#(A\cap\mathbf{x}\times[N])=I^{N}_{0}(\mathbf{x})$ for any
$\mathbf{x}\in\mathbb{Z}^{2}$. Then, for any site $\mathbf{x}=(x,y)$
satisfying $x+y\geq 0$, $W(\mathbf{x})$ equals the number of vertices at site
$\mathbf{x}$ that can be reached via open directed edges from $A$.

The conclusion in (44) follows from Remark 1. To show the other assertions, we
first give two lemmas, which are implied by the corresponding results of the
limiting process.

######  Lemma 15.

Let the initial condition be $I^{N}_{0}(m,-m)=N$ for all $m\in\mathbb{Z}$.
Then, for any $\varepsilon>0$, for all $N$ sufficiently large,

$\lim_{n\rightarrow\infty}P\left(\frac{W^{N}(n,0)}{N}>\ell(\theta)+\varepsilon\right)<\varepsilon.$
(47)

###### Proof.

By Lemma 5, for the limiting process of $W^{N}/N$, denoted by
$\mathcal{I}^{\prime}$, we can find $n_{0}$ such that
$\mathcal{I}^{\prime}(n_{0},0)<\ell(\theta)+\varepsilon/2$. Because
$W^{N}(\mathbf{x})/N\stackrel{{\scriptstyle\mathbb{P}}}{{\longrightarrow}}\mathcal{I}^{\prime}(\mathbf{x})$
for any $\mathbf{x}\in\mathbb{Z}^{2}$, we can find an $N_{0}$ such that
$P(W^{N}(n_{0},0)/N>\ell(\theta)+\varepsilon)<\varepsilon$ for $N>N_{0}$. By
the monotonicity of $W$ with regard to the initial condition, we have that
$W(n,0)$ is stochastically bounded by $W(n_{0},0)$ for $n\geq n_{0}$. The
conclusion follows. ∎

Similarly, one can show

######  Lemma 16.

For any $\varepsilon>0$ and $a\in(0,1]$, there exist $n_{0}$ and $N_{0}$ such
that when $N>N_{0}$ and the initial condition $I_{0}^{N}(0,0)\geq aN$, we
have, for any $m\in[(\kappa+\varepsilon)n_{0},(1-\kappa-\varepsilon)n_{0}]$,

$P\left(\frac{W(m,n_{0}-m)}{N}<\ell(\theta)-\varepsilon\right)<\varepsilon.$
(48)

The key step to prove Theorem 8 is to show that the survival probability for
the oriented $N$-percolation is positive when $\theta>1.5$. Our idea is to
introduce constructions which will allow us to reduce questions about our
oriented $N$-percolation to corresponding questions about the oriented
$1$-percolation, a situation in which much is known.

###### Proof of (46).

The upper bound is easy: by (47), when $N$ is large enough,

$\limsup_{n}\sup_{m}P\left(\frac{W(m,n-m)}{N}>\ell(\theta)+\varepsilon\right)<\varepsilon.$

We need to show the lower bound: when $N$ is large enough,

$\limsup_{n}\sup_{m:\varepsilon n<m<(1-\varepsilon
n)}P\left(\frac{W(m,n-m)}{N}<\ell(\theta)-\varepsilon\right)<\varepsilon.$
(49)

We will define an oriented bond percolation process and describe its
relationship to our original SIR process. Without loss of generality, assume
that $\gamma<\ell(\theta)$. For any oriented edge
$\vec{e}=(m,n)\rightarrow(m,n+1)\,\mathrm{or}\,(m,n)\rightarrow(m+1,n)$,
define a $\\{0,1\\}$-valued random variable $\eta(\vec{e})$ as follows.
Initially at the origin, there are $\lceil\gamma N\rceil$ infected particles,
which we call red particles. In general, suppose at time $t$ and at site
$(m,n)$, there are $\lceil\gamma N\rceil$ red particles. If at time $t+1$,
there are at least $\lceil\gamma N\rceil$ in the endpoint of $\vec{e}$
infected by those red particles, then we declare $\eta(\vec{e})=1$, otherwise
$\eta(\vec{e})=0$. For a site $(m,n)\neq(0,0)$, if there is an edge $\vec{e}$
pointing to $(m,n)$ with $\eta(\vec{e})=1$, then there are at least
$\lceil\gamma N\rceil$ particles infected by the red particles. In this case,
we randomly choose $\lceil\gamma N\rceil$ infected particles and declare them
red. On the other hand, if $\eta(\vec{e})=0$ for both edges pointing to
$(m,n)$, then we randomly choose $\lceil\gamma N\rceil$ particles on the site
$(m,n)$ and call them red particles.

Note that when $\theta>4$,
$\psi(\gamma)=1-\exp(-\frac{1+\theta}{5}\gamma)>\gamma$ for
$\gamma<\ell(\theta)$. It follows from (2) and Hoeffding’s inequality that as
$N$ gets large, $P(\eta(\vec{e})=1)$ can be arbitrarily close to $1$.
Therefore, using the standard results on $2$-dimensional oriented percolation
(see, e.g., ([Durrett(1984)])), we conclude that for any fixed
$\varepsilon>0$, when $N$ is sufficiently large, the probability that a
distant point with its argument in $(\varepsilon,\pi/2-\varepsilon)$ is
connected to the origin is close to $1$. In other words, for any
$\varepsilon>0$, when $N$ is sufficiently large, we have

$\limsup_{n}\sup_{m:\varepsilon n<m<(1-\varepsilon
n)}P\left(\frac{W(m,n-m)}{N}<\gamma\right)<\varepsilon.$ (50)

We are ready to prove (49). By Lemma 16, letting $a=\gamma$, if we can find
some $n_{0}$ such that

$W\left(m-\lfloor\frac{n_{0}}{2}\rfloor,l-n_{0}+\lfloor\frac{n_{0}}{2}\rfloor\right)\geq\gamma
N,$ (51)

then we have

$P\left(\frac{W(m,l)}{N}<\ell(\theta)-\varepsilon\right)<\varepsilon.$

The event in (51) holds with a high probability by (50), therefore (49)
follows. ∎

###### Proof of (45).

In order to make the description more inline with the usual oriented
percolation, we will rotate the first quadrant counterclockwise by $\pi/4$ and
work with the grid

$\mathcal{L}=\\{(m,n)\in\mathbb{Z}^{2}\,:\,m+n\textrm{ is even and }\,n\geq
0\\}.$

After rotation, the cone
$\\{(m,l)\in\mathbb{Z}^{2}:\kappa(m+l)<m<(1-\kappa)(m+l)\\}$ in the first
quadrant becomes $\\{(x,y)\in\mathcal{L}:|x|<(1-2\kappa)y\\}$, and the edges
in the oriented $N$-percolation point to either upper-right or upper-left .
Write $\chi=(1-2\kappa)$.

We will define an oriented site percolation process as follows. For each
$\mathbf{b}\in\mathcal{L}$, pick a $\beta<\chi$, a small $\delta<\beta/10$, a
large $L\in\mathbb{N}$ satisfying $\beta L\in\mathbb{Z}_{\geq 0}$, and define
for each $\mathbf{b}=(b_{1},b_{2})\in\mathcal{L}$,

$C(\mathbf{b})=(\beta Lb_{1},Lb_{2}){\in\mathcal{L}},\quad
A(\mathbf{b})=C(\mathbf{b})+(-(\beta+\delta)L,(\beta+\delta)L)\times[0,L].$

Assume that there are $\lceil\gamma N\rceil$ red particles at $C(\mathbf{b})$.
We set $\eta(\mathbf{b})=1$ if these red particles are connected to at least
$\lceil\gamma N\rceil$ particles on both sites $C(b_{1}+1,b_{2}+1)$ and
$C(b_{1}-1,b_{2}+1)$ via only the oriented edges inside $A(\mathbf{b})$, and
$\eta(\mathbf{b})=0$ otherwise.

From this construction, it is easy to see that if percolation occurs in the
$\eta$-system, then percolation also occurs in the oriented $N$-percolation
percolation. To see when the oriented site percolation occurs, we need the
following lemma whose proof will be given later.

######  Lemma 17.

For any positive and rational $\beta<\chi$, $\delta<\beta/10$, we can find
some small $\gamma$ and large $L$ such that

$\lim_{N\rightarrow\infty}P(\eta(\mathbf{b})=1)=1.$

Note that the rectangle $A(\mathbf{b})$ does not intersect with
$A(\mathbf{b}+(0,2))$ or $A(\mathbf{b}+(4,0))$. This guarantees that the
$\eta$-system is $4$-dependent, that is, when the (graph) distance between
$B_{1}$ and $B_{2}$ is bigger than $4$, then $\\{\eta(\mathbf{b}):b\in
B_{1}\\}$ and $\\{\eta(\mathbf{b}):b\in B_{2}\\}$ are independent. It follows
from Lemma 17 that the $\eta$-system percolates with a high probability for
all $N$ sufficiently large; see, e.g., see [Liggett, Schonmann, and
Stacey(1997)]. Moreover, by standard results in the two dimensional oriented
percolation ([Durrett(1984)]), for any $\varepsilon>0$,

$\lim_{N\rightarrow\infty}\max_{(x,y)\in\mathcal{L}:|x|<(1-\varepsilon)y}P_{\eta}((0,0)\rightarrow(x,y))=1,$

where $P_{\eta}$ is the law of the $\eta$-system and here
$\mathbf{x}\rightarrow\mathbf{y}$ means the event that $\mathbf{y}$ is reached
from $\mathbf{x}$ by open oriented edges.

Returning to the oriented $N$-percolation, we get

$\lim_{N\rightarrow\infty}\max_{(x,y)\in\mathcal{L}:|x|<(1-\varepsilon)y}P\left(\frac{W(\beta
Lx,Ly)}{N}\geq\gamma\right)=1.$ (52)

Similarly to the argument that (47), (48) and (50) imply (46), one can deduce
(45) from (47), (48) and (52). ∎

We now prove Lemma 17.

###### Proof of Lemma 17.

We will deduce the lemma from its limiting process. Consider the following
deterministic system on $\mathcal{L}$. Write
$A=(-(\beta+\delta)L,(\beta+\delta)L)\times[0,L]$. Let $Y(0,0)=\gamma$ and
define $Y:\mathcal{L}\rightarrow\mathbb{R}$ recursively:

$\displaystyle Y(m,n)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{cc}1-\exp(-\frac{1+\theta}{5}(Y(m-1,n-1)+Y(m+1,n-1))),&\mathrm{if}\,(m,n)\in
A;\\\ 0,&\mbox{otherwise}.\end{array}\right.$

As in the proof of Proposition 9, we will analyze

$g(m,n)=\left.\frac{\partial Y}{\partial\gamma}\right|_{\gamma=0}(m,n).$

As before, one can get $g(0,0)=1$ and for $(m,n)\in A$,

$g(m,n)=\frac{1+\theta}{5}(g(m-1,n-1)+g(m+1,n-1)).$

By induction, one can show that

Claim 1:
$g(m,n)=(\frac{1+\theta}{5})^{n}\times\\#\mathrm{oriented~{}paths~{}from~{}}(0,0)\rightarrow(m,n)\mathrm{~{}in~{}}A$.

Moreover, by the reflection principle, we can do exact counting and get

Claim 2: The number of oriented paths from $(0,0)$ to $(m,n)$ inside
$(-k,k)\times[0,n]$ is

$\binom{n}{\frac{n+m}{2}}-\binom{n}{\frac{n+2k+m}{2}}-\binom{n}{\frac{n+2k-m}{2}}+\binom{n}{\frac{n+4k+m}{2}}+\binom{n}{\frac{n+4k-m}{2}}.$

Using Stirling’s formula, one can show that the last four corresponding terms
for $g(\beta L,L)$ are negligible compared with the first one, in other words,
the restriction that the path is inside $A$ can be ignored. Therefore, we have
that

$g(\beta
L,L)\sim\sqrt{\frac{1}{2L\pi}}\left(\frac{1+\theta}{5}\right)^{L}h\left(\frac{1+\beta}{2}\right),$

where

$h(t)=\left(\frac{1}{t^{t}(1-t)^{(1-t)}}\right)^{L}\frac{1}{\sqrt{t(1-t)}}.$

By assumption, $\beta<\chi=1-2\kappa$, hence $(1+\beta)/{2}<1-\kappa$.
Therefore, $g(\beta L,L)\rightarrow\infty$ exponentially. In particular, we
can find a large $L$ such that $g(\beta L,L)>1$. For such an $L$, we can find
some small $\gamma$ satisfying $Y(\beta L,L)>\gamma$. Because $Y$ is the limit
of $W/N$ as $N\rightarrow\infty$, the conclusion follows. ∎

####  4.2.2 Convergence on all layers near the frontier

We have shown in Theorem 8 that when $\theta>1.5$, the infection proportion on
the first layer inside the cone is roughly $\ell(\theta)=\ell^{(1)}(\theta)$.
In this subsection, we show that for the $i$-th layer in the cone, the
infection proportion is roughly $\ell^{(i)}=\ell^{(i)}(\theta)$ defined in
(12).

######  Theorem 9.

Suppose that $\theta>1.5$. Consider the SIR process with initial condition
IC2. For any $\varepsilon>0$ and $i\in\mathbb{N}$, when $N$ is sufficiently
large, we have

$\limsup_{n\rightarrow\infty}\max_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\left|\frac{I_{n+i-1}^{N}(m,n-m)}{N}-\ell^{(i)}\right|>\varepsilon\right)<\varepsilon.$

######  Remark 18.

For future use, we note that the above conclusion also holds under the initial
condition

$I_{0}^{N}(x,y)=\lfloor\gamma N\rfloor\bm{\delta}_{(0,0)},\mbox{ and
}R_{0}^{N}(x,y)=(N-I^{N}_{0})\bm{\delta}_{x+y\leq 0}.$ (53)

###### Proof.

We shall only prove for the case when $i=2$; the case when $i\geq 3$ can be
proved similarly. The idea is similar to the one used in the limiting process.
Note that conditionally on the process $(I_{n}^{N}(m,n-m))_{n\geq
0,m\in\mathbb{Z}}$, the process $(I_{n+1}^{N}(m,n-m))_{n\geq
0,m\in\mathbb{Z}}$ is a Markov chain.

For notational ease, we write

$\displaystyle W(m,l)$
$\displaystyle=W^{N}(m,l)=\frac{I^{N}_{m+l}(m,l)}{N},\mbox{ and}$
$\displaystyle Z(m,l)$
$\displaystyle=Z^{N}(m,l)=\frac{I^{N}_{m+l+1}(m,l)}{N},$

which represent the infection proportion on the first and second layers,
respectively. For any $(m,l)\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$
and $j\leq k\in\mathbb{N}$, denote $T_{k}(m,l)$ and $L_{k}^{j}(m,l)$ by

$\displaystyle T_{k}(m,l)$ $\displaystyle=\\{(x,y)\in\mathbb{Z}^{2}:x\leq
m,\,y\leq l,\,x+y\geq m+l-k\\};$ $\displaystyle L_{k}^{j}(m,l)$
$\displaystyle=\\{(x,y)\in T_{k}(m,l):x+y=m+l-k+j\\}.$

Note that the distribution of $Z(m,l)$ is determined by $\\{W(x,y):(x,y)\in
T_{k}(m,l)\\}$ and $\\{Z(x,y):(x,y)\in L_{k}^{0}(m,l)\\}$. Note that because
the “frontier” processes $W$ under the initial conditions IC2 and (53) are the
same, Theorem 8 also holds for the initial condition (53). Therefore, to prove
the theorem, it suffices to prove the following lemma.

######  Lemma 19.

For any $\varepsilon>0$, there exist $k\in\mathbb{N}$ and $\delta>0$ such that
as long as

$|W(\mathbf{y})-\ell(\theta)|<\delta,\quad\mbox{for all }\mathbf{y}\in
T_{k}(\mathbf{x}),$

we have, when $N$ is sufficiently large,

$P(|Z(\mathbf{x})-\ell^{(2)}|>\varepsilon)<\varepsilon.$

∎

###### Proof of Lemma 19.

Write $M_{i}=\sup_{\mathbf{y}\in
L_{k}^{i}(\mathbf{x})}|Z(\mathbf{y})-\ell^{(2)}|$. Obviously $M_{0}\leq 1$. We
will use induction to show that with high probability, $M_{i}\leq
4\delta+(1-\varepsilon_{0})M_{i-1}$, for some $\varepsilon_{0}$ depending only
on $\theta$. Then we can get that
$|Z(\mathbf{x})-\ell^{(2)}|=M_{k}<4k\delta+(1-\varepsilon_{0})^{k}$ holds with
a high probability, and the lemma follows.

By (2), we have that, conditionally on
$(W(\mathbf{y}),Z(\mathbf{y}-(0,1)),Z(\mathbf{y}-(1,0)))$,

$Z(\mathbf{y})\stackrel{{\scriptstyle\mathrm{d}}}{{=}}\frac{1}{N}\mathrm{Bin}\left(N(1-W(\mathbf{y})),g_{N}\left(Z(\mathbf{y}-(0,1))+Z(\mathbf{y}-(1,0))+W(\mathbf{y}\right))\right),$

where $g_{N}(t)=1-(1-\frac{1+\theta}{5N})^{Nt}$. Note that $\ell(\theta)<1$,
hence by letting $\delta$ be small enough, we have
$1-W(\mathbf{y})\geq\varepsilon_{1}$. Recall the Chernoff bound for the
binomial distribution: for
$X\stackrel{{\scriptstyle\mathrm{d}}}{{=}}\mbox{Bin}(n,p)$, we have
$P(|X/n-p|\geq\delta)\leq\exp(-n\delta/(3p))\leq\exp(-n\delta/3)$. It follows
that

$P(|Z(\mathbf{y})-E(Z(\mathbf{y}))|>\delta_{1})<a^{N},$ (54)

where $a\in(0,1)$ depends only on $\varepsilon_{1}$ and $\delta_{1}$.

We need to estimate $E(Z(\mathbf{y}))$(conditioned on
$(W(\mathbf{y}),Z(\mathbf{y}-(0,1)),Z(\mathbf{y}-(1,0)))$). Similarly to (34),
we can get that when $\delta$ is small enough, $N$ is sufficiently large, and
$|W(\mathbf{y})-\ell^{(1)}|<\delta$,

$|E(Z(\mathbf{y}))-\ell^{(2)}|<3\delta+(1-\varepsilon_{0})\left|\frac{Z(\mathbf{y}-(0,1))+Z(\mathbf{y}-(1,0))}{2}-\ell^{(2)}\right|,$

where $\varepsilon_{0}$ depends only on $\theta$. When $\mathbf{y}\in
L_{k}^{i}(\mathbf{x})$, we get

$|E(Z(\mathbf{y}))-\ell^{(2)}|<3\delta+(1-\varepsilon_{0})M_{i-1}.$

By induction and using (54), we get that, with probability at least
$1-k^{2}a^{N}$, $M_{k}<4k\delta+(1-\varepsilon_{0})^{k}$. Letting $k$ be large
enough first and then $\delta$ be small enough, we finish the proof of the
lemma. ∎

Similarly, one can show

######  Proposition 20.

Suppose that $\theta>1.5$. Consider the SIR process with initial condition
$\lfloor\gamma N\rfloor\bm{\delta}_{x+y=0}$ for some fixed $\gamma\in(0,1]$.
For any $\varepsilon>0$ and $i\in\mathbb{N}$, when $N$ is sufficiently large,
we have

$\limsup_{n\rightarrow\infty}P\left(\left|\frac{I_{n+i-1}^{N}(0,n)}{N}-\ell^{(i)}\right|>\varepsilon\right)<\varepsilon.$

###  4.3 The one-infected-particle initial condition

In this subsection, we always assume that $\theta>1.5$ and the initial
condition is IC1. Recall that

$\mathcal{K}(n)=\mathcal{K}^{N}(n)=\inf\left\\{k:\sum_{(x,y)\in\mathbb{Z}^{2}:x+y=n}I^{N}_{n+k}(x,y)>0\right\\}\mbox{
and }\mathcal{K}^{N}=\lim_{n\rightarrow\infty}\mathcal{K}^{N}(n).$

Note that the event $\mathcal{K}^{N}\leq k$ is equal to
$\cap_{n}\\{k:(\mathbf{0},1)\stackrel{{\scriptstyle
n+k}}{{\longleftrightarrow}}\mathcal{L}(n)\\}$, where
$\mathcal{L}(n)=\mathcal{L}^{N}(n)=\\{(x,y):x+y=n\\}\times[N]$, and
$(\mathbf{0},1)\stackrel{{\scriptstyle
n+k}}{{\longleftrightarrow}}\mathcal{L}(n)$ denotes the event that
$(\mathbf{0},1)$ is connected to $\mathcal{L}(n)$ via $\leq n+k$ open edges in
the $N$-percolation.

######  Theorem 10.

  1. (i)

For any $i\in\mathbb{Z}_{\geq 0}$,

$\lim_{N\rightarrow\infty}P(\mathcal{K}^{N}=i)=\ell^{(i+1)}.$ (55)

  2. (ii)

For any $k,j\in\mathbb{Z}_{\geq 0}$, $\varepsilon>0$, when $N$ is sufficiently
large, we have

$\displaystyle\limsup_{n\rightarrow\infty}\sup_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\left|\frac{I^{N}_{n+k+j}(m,n-m)}{N}-\ell^{(k+1)}\right|>\varepsilon|\mathcal{K}^{N}=j\right)$
(56) $\displaystyle<$ $\displaystyle\varepsilon.$

###### Proof.

We start with (55). It suffices to prove that for any $k\in\mathbb{N}$,

$\lim_{N\rightarrow\infty}P(\mathcal{K}^{N}\leq
k-1)=\sum_{i=1}^{k}\ell^{(i)}.$ (57)

We will work mainly with $\sum_{i=0}^{k}I^{N}_{n+i}(\cdot,n-\cdot)$ instead of
$I^{N}_{k}(\cdot,n-\cdot)$, because the former is increasing in the initial
condition while the latter is not. We first consider the upper bound. By
Proposition 10, for the deterministic limiting process
$(\mathcal{I}_{n}(\cdot))$, under the initial condition
$\mathcal{I}_{0}(x,y)=\bm{\delta}_{x+y=0}$, for any $\varepsilon>0$, we can
find some $n\in\mathbb{N}$ such that

$\sum_{i=0}^{k-1}\mathcal{I}_{n+i}(n,0)\leq\sum_{i=1}^{k}\ell^{(i)}+\varepsilon/2.$

It follows from Proposition 3 that when $N$ is sufficiently large, under the
initial condition $I_{0}(x,y)=N\bm{\delta}_{x+y=0}$,

$P\left(\sum_{i=0}^{k-1}I^{N}_{n+i}(n,0)/N\leq\sum_{i=1}^{k}\ell^{(i)}+\varepsilon\right){>}1-\varepsilon.$

By symmetry, we get that

$P\left(\frac{\sum_{i=1}^{N}\mathbf{1}_{(\mathbf{0},i)\stackrel{{\scriptstyle
n+k-1}}{{\longleftrightarrow}}\mathcal{L}(n)}}{N}\leq\sum_{i=1}^{k}\ell^{(i)}+\varepsilon\right)>1-\varepsilon.$

Therefore,

$E\left(\frac{\sum_{i=1}^{N}\mathbf{1}_{(\mathbf{0},i)\stackrel{{\scriptstyle
n+k-1}}{{\longleftrightarrow}}\mathcal{L}(n)}}{N}\right)\leq\sum_{i=1}^{k}\ell^{(i)}+2\varepsilon.$

In other words, $P((\mathbf{0},{1})\stackrel{{\scriptstyle
n+k-1}}{{\longleftrightarrow}}\mathcal{L}(n))\leq\sum_{i=1}^{k}\ell^{(i)}+2\varepsilon$,
and we get the upper bound.

Now we turn to the lower bound. By Theorem 5 for IC2, under the initial
condition $I_{0}^{N}=N\bm{\delta}_{(0,0)}$, for any $\varepsilon>0$, when $N$
is sufficiently large, we have

$\limsup_{n\rightarrow\infty}\max_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\sum_{i=1}^{k}\frac{I_{n+i-1}^{N}(m,n-m)}{N}\leq\sum_{i=1}^{k}\ell^{(i)}-\varepsilon\right)<\varepsilon.$

Hence,

$\liminf_{n\rightarrow\infty}\min_{m:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}E\left(\sum_{i=1}^{k}\frac{I_{n+i-1}^{N}(m,n-m)}{N}\right)\geq(1-\varepsilon)\left(\sum_{i=1}^{k}\ell^{(i)}-\varepsilon\right).$

By monotonicity, we get that under the initial condition
$I_{0}^{N}=N\bm{\delta}_{x+y=0}$, we also have

$\liminf_{n\rightarrow\infty}E\left(\sum_{i=1}^{k}\frac{I_{n+i-1}^{N}(n,0)}{N}\right)\geq(1-\varepsilon)\left(\sum_{i=1}^{k}\ell^{(i)}-\varepsilon\right).$

In other words,

$\liminf_{n\rightarrow\infty}P((\mathbf{0},1)\stackrel{{\scriptstyle
n+k-1}}{{\longleftrightarrow}}\mathcal{L}(n))\geq(1-\varepsilon)\left(\sum_{i=1}^{k}\ell^{(i)}-\varepsilon\right).$

This is the lower bound we need.

We now prove (56). It suffices to show the following upper and lower bounds:
For any $k,j\in\mathbb{Z}_{\geq 0},\varepsilon>0$, and $N$ sufficiently large,

$\displaystyle\limsup_{n,m\rightarrow\infty:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\frac{\sum_{i=0}^{k-1}I^{N}_{n+i+j}(m,n-m)}{N}>\sum_{i=1}^{k}\ell^{(i)}+\varepsilon,\mathcal{K}=j\right)$
(58) $\displaystyle<$ $\displaystyle\varepsilon;$

and

$\displaystyle\limsup_{n,m\rightarrow\infty:(\kappa+\varepsilon)n<m<(1-\kappa-\varepsilon)n}P\left(\frac{\sum_{i=0}^{k-1}I^{N}_{n+i+j}(m,n-m)}{N}<\sum_{i=1}^{k}\ell^{(i)}-\varepsilon,\mathcal{K}=j\right)$
(59) $\displaystyle<$ $\displaystyle\varepsilon.$

With Theorem 5 for IC2 in mind, we aim to reduce our case to IC2. Introduce
the following events: fix a $\gamma_{0}\in(0,1]$, say, $0.1$, and define

$\displaystyle B_{n}$ $\displaystyle:=\\{\mathcal{K}^{N}(n+1)\geq j,\mbox{ and
there exists }m\in\mathbb{Z}\mbox{ such that
}I^{N}_{n+j}(m,n-m)\geq\gamma_{0}N\\},$ $\displaystyle A_{n}$
$\displaystyle:=\cup_{k=n}^{2n-1}B_{k}.$

We will prove (58) and (59) based on the following lemma, whose proof will be
given later.

######  Lemma 21.

For any $N\in\mathbb{N}$ and $j\in\mathbb{Z}_{\geq 0}$, we have

$\lim_{n\rightarrow\infty}P(A^{\mathrm{c}}_{n},\mathcal{K}^{N}=j)=0.$

Now we prove (58) and (59). For notational ease, we write $\Box^{+}$
($\Box^{-}$, respectively) for the event
$\\{\sum_{i=0}^{k-1}I^{N}_{n+i+j}(m,n-m)/N-\sum_{i=1}^{k}\ell^{(i)}>\varepsilon\\}$
($<-\varepsilon$, respectively). By Lemma 21, we can find $n_{0}$ large enough
such that

$\displaystyle P(\Box^{+},\mathcal{K}^{N}=j)$ $\displaystyle\leq
P(\Box^{+},\mathcal{K}^{N}=j,A_{n_{0}})+P(\Box^{+},\mathcal{K}^{N}=j,A^{\mathrm{c}}_{n_{0}})$
$\displaystyle\leq
P(\Box^{+},A_{n_{0}})+P(\mathcal{K}^{N}=j,A^{\mathrm{c}}_{n_{0}})$
$\displaystyle\leq P(\Box^{+}|A_{n_{0}})+\varepsilon/2.$

By monotonicity, we have

$P(\Box^{+}|A_{n_{0}})\leq P(\Box^{+}|B^{\prime}(2n_{0}-1)),$

where $B^{\prime}(n)=\\{\mathcal{K}^{N}(n+1)\geq
j,\,I^{N}_{n+j}(m,n-m)=N\mbox{ for all }m\in\mathbb{Z}\\}$. By Proposition 20,
we see that $P(\Box^{+}|B^{\prime}(2n_{0}-1))$ is small when both $N$ and
$n-2n_{0}$ are large. The upper bound (58) follows.

For the lower bound (59), by Lemma 21 again, we can find $n_{0}$ large enough
such that

$\displaystyle P(\Box^{-},\mathcal{K}^{N}=j)$ $\displaystyle\leq
P(\Box^{-},\mathcal{K}=j,A_{n_{0}})+P(\Box^{-},\mathcal{K}^{N}=j,A^{\mathrm{c}}_{n_{0}})$
$\displaystyle\leq
P(\Box^{-},A_{n_{0}})+P(\mathcal{K}^{N}=j,A^{\mathrm{c}}_{n_{0}})$
$\displaystyle\leq P(\Box^{-}|A_{n_{0}})+\varepsilon/2.$

By monotonicity, we have

$P(\Box^{-}|A_{n_{0}})\leq\max_{\mathbf{x}\in
T_{n_{0}}}P(\Box^{-}|B(\mathbf{x})),$

where $T_{n_{0}}=\\{(x,y)\in\mathbb{Z}^{2}:x,y\geq-j\mbox{ and }n_{0}\leq
x+y\leq 2n_{0}-1\\}$, and $B(m,l)$ stands for that event that
$I_{t_{0}}=\lceil\gamma_{0}N\rceil\bm{\delta}_{(m,l)}$ and
$R_{t_{0}}=(N-I_{t_{0}})\bm{\delta}_{x+y\leq m+l}$ with $t_{0}=m+l+j$. By
Remark 18, we see that $\max_{\mathbf{x}\in
T_{n_{0}}}P(\Box^{-}|B(\mathbf{x}))$ is sufficiently small when both $N$ and
$n-2n_{0}$ are large. This finishes the proof of (59). ∎

###### Proof of Lemma 21.

Note that $\mathcal{K}^{N}(n)\nearrow\mathcal{K}^{N}$. It follows that

$\lim_{n\rightarrow\infty}P(\mathcal{K}^{N}=j,\mathcal{K}^{N}(n)<j)=0.$ (60)

Write

$K_{n}=\\{\mathcal{K}^{N}(n+1)\geq j,\mbox{ and there exists
}m\in\mathbb{Z}\mbox{ such that }I^{N}_{n+j}(m,n-m)\geq 1\\}.$

It is easy to see that there exists $a=a(N,\theta)>0$ such that

$\inf_{\vec{b}\in{\\{0,\ldots,N\\}}^{\mathbb{Z}},\vec{b}\not\equiv\vec{0}}P(B_{n+1}|K_{n}\cap\\{I^{N}_{n+j}(m,n-m)=\vec{b}(m)\mbox{
for all }m\\})\\\
=P(B_{n+1}|K_{n}\cap\\{I^{N}_{n+j}(m,n-m)=\mathbf{1}_{m=0}\mbox{ for all
}m\\})\geq a.$ (61)

Note that conditionally on the event $K_{n}$, the laws of
$(I^{N}_{{n+1+j}}(m,n+1-m))_{m\in\mathbb{Z}},\mathbf{1}_{K_{n+1}}$ and
$\mathbf{1}_{B_{n+1}}$ depend only on
$(I^{N}_{{n+j}}(m,n-m))_{m\in\mathbb{Z}}$. Therefore,

$\displaystyle P$
$\displaystyle(\mathcal{K}^{N}=j,\mathcal{K}^{N}(n)=j,A_{n}^{\mathrm{c}})\leq
P(\cap_{k=n}^{2n-1}K_{k},\cap_{k=n}^{2n-1}B_{k}^{\mathrm{c}})$
$\displaystyle\leq\prod_{k=n}^{2n-2}\max_{\vec{b}\in{\\{0,\ldots,N\\}}^{\mathbb{Z}},\vec{b}\not\equiv\vec{0}}P(K_{k+1}\cap
B_{k+1}^{\mathrm{c}}|K_{k}\cap\\{I^{N}_{{k}+j}(m,{k}-m)=\vec{b}(m),\forall
m\\})$ $\displaystyle\leq\prod_{k=n}^{2n-2}(1-a)=(1-a)^{n-1},$

where in the last inequality we used (61). Combining the result above with
(60), we prove the lemma.

∎

##  5 Spreading Speed

Recall the limiting (deterministic) system
$(\mathcal{I}_{n}(\mathbf{x}),\mathcal{R}_{n}(\mathbf{x}))_{n\in\mathbb{Z}_{\geq
0},\mathbf{x}\in\mathbb{Z}^{2}}$ from Section 2. As before, we always assume
that $\mathcal{R}_{0}\equiv 0$ and focus only on the first quadrant. In this
section, we assume further that $\mathcal{I}_{0}$ is supported on the origin.
To simplify notation, we define
$\mathcal{D}_{n}(\mathbf{x})=\mathcal{R}_{n+1}(\mathbf{x})=\mathcal{I}_{n}(\mathbf{x})+\mathcal{R}_{n}(\mathbf{x})$,
and
$D_{n}(\mathbf{x})=D^{N}_{n}(\mathbf{x})=R^{N}_{n+1}(\mathbf{x})=I^{N}_{n}(\mathbf{x})+R^{N}_{n}(\mathbf{x})$.
Note that $D_{n}$ represents the number of particles that are infected before
or at time $n$ and is increasing in the initial condition $I_{0}$. We show a
weaker version of Theorem 2 in this section.

######  Theorem 11.

For any fixed $\theta\in(0,\infty)$, the following results hold:

  1. (i)

Consider the deterministic system with the initial condition
$\mathcal{I}_{0}=\gamma\bm{\delta}_{\mathbf{0}}$ for some $\gamma\in(0,1]$.
For any $\varepsilon>0$, along any sequence
$(i_{k},j_{k})_{k}\subset\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$
satisfying $i_{k}+j_{k}\rightarrow\infty$ and
$\arg(i_{k},j_{k})\rightarrow\phi$ for some $\phi\in[0,\pi/2]$, we have

$\lim_{k\rightarrow\infty}\mathcal{D}_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}-\varepsilon)\rfloor}(i_{k},j_{k})=0;$
(62)

Moreover, there exists $\delta=\delta(\theta,\gamma,\varepsilon)>0$ such that

$\liminf_{k\rightarrow\infty}\mathcal{D}_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor}(i_{k},j_{k})>\delta.$
(63)

  2. (ii)

Consider the SIR process with the initial condition IC2 (4). For any
$\varepsilon>0$ and $N$, along any sequence
$(i_{k},j_{k})_{k}\subset\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq 0}$
satisfying $i_{k}+j_{k}\rightarrow\infty$ and
$\arg(i_{k},j_{k})\rightarrow\phi$ for some $\phi\in[0,\pi/2]$, we have,

$\lim_{k\rightarrow\infty}P(D^{N}_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}-\varepsilon)\rfloor}(i_{k},j_{k})>0)=0;$
(64)

Moreover, for any $\varepsilon>0$, there exists
$\delta=\delta(\theta,\varepsilon,\gamma)>0$ such that when $N$ is
sufficiently large,

$\liminf_{k\rightarrow\infty}P(D^{N}_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor}(i_{k},j_{k})/N\geq\delta)>1-\varepsilon.$
(65)

######  Remark 22.

We will show a little bit more than (65): we can replace $D^{N}(i,j)$ by the
corresponding number of particles that are infected via only those edges lying
in the region $\\{(x,y):\ 0<x+y<i+j\\}$ except possibly with one endpoint at
either $(0,0)$ or $(i,j)$. We need this stronger version when proving Theorem
2(ii). Note that Theorem 2(i) for IC2 is just (64), which, by the monotonicity
of the recovered process with respect to the initial condition, also implies
the conclusion for IC1 conditional on survival.

###### Proof of Theorem 11.

Similarly to what we did in Section 4.2, we will analyze the asymptotics of

$g_{n}(\mathbf{x})=\left.\frac{\partial\mathcal{I}_{n}}{\partial\gamma}\right|_{\gamma=0}(\mathbf{x}).$

From the recursive equation, a simple application of chain rule yields

$g_{n+1}(m,l)=\frac{1+\theta}{5}\widetilde{g}_{n}(m,l).$

Similarly as before, by induction, one can show

Claim 1:

$g_{n}(m,l)=\left(\frac{1+\theta}{5}\right)^{n}\times\\#\mathrm{~{}LRW~{}paths~{}from~{}}(0,0)\rightarrow(m,l)\mathrm{~{}with~{}}n\mathrm{~{}steps}.$

Here “LRW” stands for lazy random walk, namely, at each time, the walker
either stays at the same position or moves to one of the four neighbouring
site with equal probability $1/5$.

As before, we need to analyze the asymptotic behavior of $g_{n}(m,l)$. Recall
that Simple Random Walk (SRW) is such that at each time, the walker moves to
one of the four neighbouring site with equal probability $1/4$. Write
$\sharp_{\mathrm{S}}(m,l;n)$($\sharp_{\mathrm{L}}(m,l;n)$, respectively) for
the number of SRW (LRW, respectively) paths from $(0,0)$ to $(m,l)$ with $n$
steps. We have

$\sharp_{\mathrm{S}}(m,l;n)=\binom{n}{\frac{n-(m+l)}{2}}\binom{n}{\frac{n-(m-l)}{2}}.$
(66)

By considering the number of times that the walker does not move before time
$n$, we get

$\sharp_{\mathrm{L}}(m,l;n)=\sum_{i:m+l\leq i\leq
n,2|i-m-l}\binom{n}{n-i}\sharp_{\mathrm{S}}(m,l;i).$ (67)

By Stirling’s formula, we have

$\binom{n}{k}\approx\left(\frac{1}{u^{u}(1-u)^{1-u}}\right)^{n},\mathrm{~{}where~{}}u=\frac{k}{n},$

where for any two sequences $(a_{n})$ and $(b_{n})$, $a_{n}\approx b_{n}$
means that

$\lim_{n\rightarrow\infty}\log a_{n}/\log b_{n}=1.$

It follows that when both $(m+l)/n$ and $m/(m+l)$ converge, we have

$\displaystyle\sharp_{\mathrm{L}}(m,l;n)\approx$
$\displaystyle\max_{i}\left(\frac{1}{t^{t}(1-t)^{1-t}}\right)^{n}\left(\frac{1}{r^{r}(1-r)^{1-r}}\right)^{i}\left(\frac{1}{s^{s}(1-s)^{1-s}}\right)^{i}$
(68) $\displaystyle\approx$
$\displaystyle\left(\sup_{t\in(\frac{m+l}{n},1]}\frac{1}{t^{t}(1-t)^{1-t}r^{tr}(1-r)^{t(1-r)}s^{ts}(1-s)^{t(1-s)}}\right)^{n}$
$\displaystyle\approx$
$\displaystyle\left(\frac{1}{\exp(G(\frac{m+l}{n},\frac{m}{m+l}))}\right)^{n},$

where $t=i/n,r={(t-(m+l)/n)}/{(2t)},s=(t-(m-l))/n/(2t)$,

$\displaystyle h(t)$ $\displaystyle=t\log t+(1-t)\log(1-t),$ $\displaystyle
g(t,v,a)$
$\displaystyle=h(t)+t\left(h\left(\frac{1}{2}-\frac{v}{2t}\right)+h\left(\frac{1}{2}-\frac{(1-2a)v}{2t}\right)\right),$

and

$G(v,a)=\inf_{t\in[v,1]}g(t,v,a).$ (69)

Note that $g(t,v,a)$ is negative, continuous on all variables, strictly
increasing in $v\in(0,1]$ and decreasing in $a\in[0,1/2]$. Hence, $G(v,a)$ has
the same property and

$G(1,a)=h(a),\quad\lim_{v\rightarrow
0}G(v,a)=G(0,a)=\min_{t\in(0,1)}h(t)+2t\,h({1}/{2})=\log(1/5).$

Combining the estimates above yields

$\sharp_{\mathrm{L}}(m,l;n)\approx\left(\frac{1}{\exp(G(\frac{m+l}{n},\frac{m}{m+l}))}\right)^{n},\,g_{n}(m,l)\approx\left(\frac{(1+\theta)/5}{\exp(G(\frac{m+l}{n},\frac{m}{m+l}))}\right)^{n}.$

Using the continuity of $g(t,v,a)$, one gets

######  Lemma 23.

There exists sequence $c(n)\rightarrow 0$ such that for any integer-valued
functions $m=m(n),l=l(n)$ with $|m|+|l|\leq n$, we have

$\sharp_{\mathrm{L}}(m,l;n)\in\left[\left(\frac{1}{\exp(G(\frac{|m|+|l|}{n},\frac{|m|}{|m|+|l|}))}-c(n)\right)^{n},\left(\frac{1}{\exp(G(\frac{|m|+|l|}{n},\frac{|m|}{|m|+|l|}))}+c(n)\right)^{n}\right].$

When $h(a)\geq\log((1+\theta)/5)$, we can find a unique
$\upsilon=\upsilon(\theta,a)\in(0,1]$ such that

$G(\upsilon,a)=\log((1+\theta)/5).$ (70)

When $h(a)<\log((1+\theta)/5)$, we define $\upsilon(\theta,\phi)=1.$

We now continue the proof of the theorem. We will mainly work with
$a=a(\phi)=|\cos(\phi)|/(|\cos(\phi)|+|\sin(\phi)|)$ instead of $\phi$. Note
that $\arg(i_{k},j_{k})\rightarrow\phi$ implies that
$|i_{k}|/(|i_{k}|+|j_{k}|)\rightarrow a(\phi)$. Without loss of generality, we
assume $a\in[0,1/2]$.

We first show the upper bounds, (62) and (64), which are relatively easier.
Note that when $\upsilon(\theta,\phi)=1$, (62) and (64) hold trivially because
$R_{n}(m,l)=R^{N}_{n}(m,l)=0$ for $n\leq m+l$. Now assume that
$\upsilon(\theta,\phi)<1$. By Claim 1, the lemma above and the definition of
$\upsilon$, when $m+l$ is large enough and $|m/(m+l)-a|$ is small enough, we
have

$g_{k}(m,l)\leq b^{k},\mathrm{~{}when~{}}m+l\leq
k\leq(m+l)(\upsilon(\theta,a)^{-1}-\varepsilon),$

where $b<1$ depends only on $\theta,a(\phi)$ and $\varepsilon$. It is easy to
use induction to show that $\mathcal{I}_{n}(\mathbf{x})\leq g_{n}(\mathbf{x})$
and $E(I^{N}_{n}(\mathbf{x})/N)\leq g_{n}(\mathbf{x})$. Therefore,

$\displaystyle\mathcal{D}_{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}(m,l)=\sum_{k=m+l}^{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}\mathcal{I}_{k}(m,l)$
$\displaystyle\,\leq\sum_{k=m+l}^{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}g_{k}(m,l)\leq\sum_{k=m+l}^{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}b^{k}\leq
b^{m+l}/(1-b)\stackrel{{\scriptstyle\mathrm{as~{}}m+l\rightarrow\infty}}{{\longrightarrow}}0.$

The proof of (62) is complete. Similarly,

$\displaystyle E(D^{N}_{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}$
$\displaystyle(m,l)/N)=\sum_{k=m+l}^{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}E\left(I_{k}(m,l)/N\right)$
$\displaystyle\leq\sum_{k=m+l}^{\lfloor(m+l)(\upsilon^{-1}-\varepsilon)\rfloor}g_{k}(m,l)\leq
b^{m+l}/(1-b)\stackrel{{\scriptstyle\mathrm{as~{}}m+l\rightarrow\infty}}{{\longrightarrow}}0.$

The proof of (64) is complete.

Next, we prove (63) and (65). Our method is similar to the proof of (45). As
in the proof of (45), it would be more convenient to rotate the space
counterclockwise by $\pi/4$ and work with

$\mathcal{L}=\\{(m,n)\in\mathbb{Z}^{2}\,:\,m+n\textrm{ is even }\,n\geq 0\\}.$

Note that the cone $\\{(m,l)\in\mathbb{Z}_{\geq 0}\times\mathbb{Z}_{\geq
0}:a(m+l)<m<(1-a)(m+l)\\}$ in the first quadrant corresponds to
$\\{(x,y)\in\mathcal{L}:|x|<\lambda y\\}$ in $\mathcal{L}$, with
$\lambda=(1-2a)$.

We will define an oriented site percolation process, called $\eta_{t}$-system,
as follows. For $t>0$, $\beta_{1},\beta_{2}>0$, $L\in\mathbb{Z}_{\geq 0}$,
define for each $\mathbf{b}=(b_{1},b_{2})\in\mathcal{L}$,

$\displaystyle C(\mathbf{b})$
$\displaystyle=(b_{1}\lfloor(\lambda+\beta_{1})L\rfloor,b_{2}L)\in\mathcal{L},$
$\displaystyle A(\mathbf{b})$
$\displaystyle=C(\mathbf{b})+(-\lfloor(\lambda+\beta_{1}+\beta_{2})L\rfloor,\lfloor(\lambda+\beta_{1}+\beta_{2})L\rfloor)\times[0,L].$

For some small $\delta_{0}\in(0,\gamma)$ to be chosen below, assume that there
are $\lceil\delta_{0}N\rceil$ infected particles at $C(\mathbf{b})$ at time
$0$. Consider the SIR process restricted to $A(\mathbf{b})$, namely, a
particle inside $A(\mathbf{b})$ cannot infect particles outside. For
notational ease, we use the same notation for this restricted SIR process as
the original SIR. We set $\eta_{t}(\mathbf{b})=1$ if

$\min\\{D_{\lfloor tL\rfloor}(C(\mathbf{b}+(1,1)),D_{\lfloor
tL\rfloor}(C(\mathbf{b}+(-1,1)))\\}\geq\lceil{\delta_{0}N}\rceil,$ (71)

and $\eta_{t}(\mathbf{b})=0$ otherwise.

From this construction, it is easy to see that if $(b_{1},b_{2})$ is connected
to the origin in the $\eta_{t}$-system, then in the SIR process with initial
condition $I_{0}^{N}(0,0)=\lceil\delta_{0}N\rceil$, we have $D^{N}_{\lfloor
tL\rfloor
b_{2}}(b_{1}\lfloor(\lambda+\beta_{1})L\rfloor,Lb_{2})\geq\delta_{0}N$.
Similarly to Lemma 17, we have the following lemma whose proof will be given
later.

######  Lemma 24.

For any $\varepsilon>0$, there exist $\beta_{1},\beta_{2}>0$ sufficiently
small, $L\in\mathbb{N}$ sufficiently large and $\delta_{0}>0$ sufficiently
small such that

$\lim_{N\rightarrow\infty}\inf_{\lambda\in[0,1]}P(\eta_{\upsilon(\theta,(1-\lambda)/2)^{-1}+\varepsilon}(\mathbf{b})=1)=1.$

Note that the $\eta_{t}$-system with
$t=\upsilon(\theta,(1-\lambda)/2)^{-1}+\varepsilon$ is $k$-dependent with
$k=2\lceil 2(\beta_{1}+\beta_{2})/\beta_{1})\rceil~{}+~{}2$ (when $L$ is
large). It follows from Lemma 24 that the $\eta_{t}$-system percolates with
high probability for all $N$ sufficiently large. Moreover, similarly to the
proof of Lemma 17, from the standard result in two dimensional oriented
percolation ([Durrett(1984)]), we know that when $N$ is sufficiently large,
for any fixed $\varepsilon_{1}>0$, as long as
$|b_{1}|<(1-\varepsilon_{1})b_{2}$, $\mathbf{b}=(b_{1},b_{2})$ is connected to
the origin in the $\eta_{t}$-system for
$t={\upsilon(\theta,(1-\lambda)/2)^{-1}+\varepsilon}$ with high probability.
In other words, $D^{N}_{\lfloor(\upsilon^{-1}+\varepsilon)L\rfloor
b_{2}}(b_{1}\lfloor(\lambda+\beta_{1})L\rfloor,Lb_{2})\geq\delta_{0}N$ with
high probability. From this, we see that (65) holds for $\delta=\delta_{0}$
when $(i_{k},j_{k})_{k}\subset\bigcup\\{C(\mathbf{b})\\}$.

When $(i_{k},j_{k})_{k}\not\subset\bigcup\\{C(\mathbf{b})\\}$, we use an
argument similar to the one in the proof of (46). From Proposition 3, it is
easy to see that when $L$ is fixed, we can find some
$\delta_{1}\leq\delta_{0}$ such that for the SIR process restricted to
$\mathbb{Z}\times[0,2L]$ with initial condition
$\lceil\delta_{0}N\rceil\bm{\delta}_{\mathbf{0}}$, one has

$\lim_{N\rightarrow\infty}\sup_{\mathbf{x}\in[-2L,2L]\times[0,L]}P(D_{5L}(\mathbf{x})<\delta_{1}N)\rightarrow
0.$ (72)

Similarly to the argument that (47), (48) and (50) imply (46), we have that
(65) for
$\delta=\delta_{0},(i_{k},j_{k})_{k}\subset\bigcup\\{C(\mathbf{b})\\}$ and
(72) imply (65) for $\delta=\delta_{1}$ and
$(i_{k},j_{k})_{k}\not\subset\bigcup\\{C(\mathbf{b})\\}$.

Finally, we prove the stronger version of (65) stated in Remark 22. In order
to get the desired result, by a similar argument to (72), we have the
following:

$\lim_{N\rightarrow\infty}\sup_{\mathbf{x}\in[-2L,2L]\times[L,2L]}P(D^{-}_{5L}(\mathbf{x})<\delta_{1}N)\rightarrow
0,$ (73)

where $D_{n}^{-}(i,j)$ stands for the number of particles that are infected
before or at time $n$ via only those edges in the region
$\\{(x,y)\in\mathcal{L}:\ 0<y<j\\}$ except possibly with one endpoint at
either $(0,0)$ or $(i,j)$. To prove the stronger version of (65), for any
$(i_{k},j_{k})\in\mathcal{L}$, find an
$\mathbf{x}\in\bigcup\\{C(\mathbf{b})\\}$ such that
$(i_{k},j_{k})\in\mathbf{x}+[-2L,2L]\times[L,2L]$. By (73) and the proof above
for (65), and translating the result back to original orientation, we get the
stronger version of (65) with all edges lying in
$\\{(x,y)\in\mathbb{Z}^{2}:\,0\leq x+y<i_{k}+j_{k}\\}$ except possibly with
one endpoint at $(i,j)$. We need to further rule out those edges on the line
$\\{(x,y)\in\mathbb{Z}^{2}:\ x+y=0\\}$. We can do so by changing the starting
point to $(1,0)$. Indeed, with a high probability,
$I_{1}(1,0)/N\geq\gamma_{0}$ for some $\gamma_{0}$ depending only on $\gamma$.
Then regarding $(1,0)$ as the starting point and using the argument above, we
get the desired stronger version.

∎

###### Proof of Lemma 24.

The proof is similar to that of Lemma 17. Write
$A=A(\mathbf{0})=(-\lfloor(\lambda+\beta_{1}+\beta_{2})L\rfloor,\lfloor(\lambda+\beta_{1}+\beta_{2})L\rfloor)\times[0,L]$.
Let $Y_{0}(\mathbf{x})=\gamma\bm{\delta}_{(0,0)}(\mathbf{x})$ and define
$Y_{n}:\mathcal{L}\rightarrow\mathbb{R}$ recursively:

$\displaystyle Y_{n}(m,l)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{cc}(1-\sum_{0}^{n-1}Y_{i}(m,l))(1-\exp(-\frac{1+\theta}{5}(\widetilde{Y}_{n-1}(m,l))),&\mathrm{if}\,(m,l)\in
A;\\\ 0,&\mbox{otherwise}.\end{array}\right.$

Note that in $\mathcal{L}$,
$\widetilde{f}(m,l)=f(m+1,l+1)+f(m-1,l+1)+f(m+1,l-1)+f(m-1,l-1)$.

As before, we will analyze

$g^{\prime}_{n}(m,l):=\left.\frac{\partial
Y_{n}}{\partial\gamma}\right|_{\gamma=0}(m,l).$

We have $g{{}_{0}}^{\prime}(0,0)=1$, and for $(m,l)\in A$,

$g^{\prime}_{n}(m,l)=\frac{1+\theta}{5}\widetilde{g^{\prime}_{n-1}}(m,l).$

By induction, one gets

Claim 1:

$g^{\prime}_{n}(m,l)=\left(\frac{1+\theta}{5}\right)^{n}\times\sharp_{\mathrm{L}}(m,l;n;A),$

where $\sharp_{\mathrm{L}}(m,l;n;A)$ is the number of LRW paths from $(0,0)$
to $(m,l)$ with $n$ steps inside $A$. Write
$\sharp_{\mathrm{{\mathrm{S}}}}(m,l;n;A)$ as the corresponding number for SRW.

Similarly to the proof of Lemma 17, it suffices to show that when $L$ is
large, $n=\lfloor(\upsilon^{-1}+\varepsilon)L\rfloor$, for any
${\lambda}\in[0,1]$,

$g^{\prime}_{n}(m,L)>1,\,\mathrm{~{}with~{}}m=\lfloor(\lambda+\beta_{1})L\rfloor.$
(74)

We aim to show $\sharp_{\mathrm{L}}(m,L;n;A)\approx\sharp_{\mathrm{L}}(m,L;n)$
for some small and fixed $\beta_{1},\beta_{2}$. If so, then by the asymptotics
of $\sharp_{\mathrm{L}}(m,L;n)$, namely, Lemma 23, the above inequality holds
for large $L$ and we finish the proof. Obviously,
$\sharp_{\mathrm{L}}(m,l;n;A)\leq\sharp_{\mathrm{L}}(m,l;n)$. We need to show
the other direction. In principal, $\sharp_{\mathrm{L}}(m,l;n;A)$ can be
written as the sum of $\sharp_{\mathrm{L}}$’s without restriction by using the
reflection principle. However the process will be too involved. Here we
separate it into two steps. We will first consider the number of paths in
$A_{1}=\mathbb{Z}\times[0,L]$.

Using the reflection principle, we have

$\displaystyle\sharp_{\mathrm{{S}}}$
$\displaystyle(m,l;n;A_{1})=\sharp_{\mathrm{S}}(m,l;n)-2\sharp_{\mathrm{S}}(m,l+2;n)+\sharp_{\mathrm{S}}(m,l+4;n)+\sharp_{\mathrm{S}}(m,3l+4;n)$
$\displaystyle\geq\sharp_{\mathrm{S}}(m,l;n)-2\sharp_{\mathrm{S}}(m,l+2;n)+\sharp_{\mathrm{S}}(m,l+4;n).$

Note that for the lattice $\mathcal{L}$, (66) and the aysmptotics in Lemma 23
become (when $m,l\geq 0$)

$\sharp_{\mathrm{S}}(m,l;n)=\binom{n}{\frac{n-l}{2}}\binom{n}{\frac{n-m}{2}};$
$\sharp_{\mathrm{L}}(m,l;n)\approx\left(\exp^{-1}(G(\frac{m\vee
l}{n},\frac{1-\frac{m\wedge l}{m\vee l}}{2}))\right)^{n},$ (75)

where we write $m\vee l=\max\\{m,l\\},m\wedge l=\min\\{m,l\\}$.

Using this, we get

$\displaystyle\frac{\sharp_{\mathrm{S}}(m,l;n)-2\sharp_{\mathrm{S}}(m,l+2;n)+\sharp_{\mathrm{S}}(m,l+4;n)}{\sharp_{\mathrm{S}}(m,l;n)}$
$\displaystyle=1-2\frac{n-l}{n+l+2}+\frac{(n-l)(n-l-2)}{(n+l+2)(n+l+4)}\geq\left(\frac{l}{n}\right)^{2}-O({1}/{n}).$

Hence,

$\frac{\sharp_{\mathrm{S}}(m,l;n;A_{1})}{\sharp_{\mathrm{S}}(m,l;n)}\geq\left(\frac{l}{n}\right)^{2}-O({1}/{n}).$

Using the relation between the number of LRW and that of SRW, namely, (67) one
gets

$\frac{\sharp_{\mathrm{L}}(m,l;n;A_{1})}{\sharp_{\mathrm{L}}(m,l;n)}\geq\left(\frac{l}{n}\right)^{2}-O({1}/{{n}})=\left(\frac{l}{n}\right)^{2}-O({1}/{{l}}).$
(76)

Note that $\sharp_{\mathrm{L}}(m,l;n)$ is decreasing in $m$ in $[0,n]$.
Therefore, with
$k=\lfloor(\lambda+\beta_{1}+\beta_{2})L\rfloor,n=\lfloor(\upsilon^{-1}+\varepsilon)L\rfloor$,
when $m=\lfloor(\lambda+\beta_{1})L\rfloor$,

$\displaystyle\sharp_{\mathrm{L}}(m,L;n;A)=\sharp_{\mathrm{L}}(m,L;n;A_{1})-\sum_{i=\pm
m}\sharp_{\mathrm{L}}(2k+i,L;n;A_{1})$
$\displaystyle\quad\quad\quad\quad\quad\quad+\sum_{i=\pm
m}\sharp_{\mathrm{L}}(4k+i,L;n;A_{1})$
$\displaystyle\geq\left(\left(\frac{L}{n}\right)^{2}-O({1}/{L})\right)\sharp_{\mathrm{L}}(m,L;n)-2\sharp_{\mathrm{L}}(2k-m,L;n)$
$\displaystyle\geq\left(\left(\frac{L}{n}\right)^{2}-O({1}/{L})\right)\sharp_{\mathrm{L}}(\lfloor(\lambda+\beta_{1})L\rfloor,L;n)-2\sharp_{\mathrm{L}}(\lfloor(\lambda+\beta_{1}+2\beta_{2})L\rfloor-2,L;n)$
$\displaystyle\geq C\left(\exp\left(-G\left(\frac{(\lambda+\beta_{1})\vee
1}{\upsilon^{-1}+{\varepsilon}},\frac{1-\frac{(\lambda+\beta_{1})\wedge
1}{(\lambda+\beta_{1})\vee 1}}{2}\right)\right)-c(n)\right)^{n}$
$\displaystyle\quad\quad\quad-\left(\exp\left(-G\left(\frac{(\lambda+\beta_{1}+2\beta_{2})\vee
1}{\upsilon^{-1}+{\varepsilon}},\frac{1-\frac{(\lambda+\beta_{1}+2\beta_{2})\wedge
1}{(\lambda+\beta_{1}+2\beta_{2})\vee 1}}{2}\right)\right)+c(n)\right)^{n},$

where in the last step we use Lemma 23 and (75). Note that
$G(\upsilon,a)=G(\upsilon,(1-\lambda)/2)\leq\log((1+\theta)/5)$ and
$G(v,\cdot)$ is strictly decreasing in $v$. By the continuity of
$G(\cdot,\cdot)$ and $\upsilon(\theta,a)$, for any $\varepsilon>0$, we can
find $\beta_{1},\beta_{2}$ small enough such that for some $\varepsilon_{1}>0$
and when $L$ is large, the above term is bigger than
$\left({5}/{(1+\theta)}+\varepsilon_{1}\right)^{n}$ for all
${\lambda}\in[0,1]$. Therefore, by Claim 1, (74) holds. ∎

##  6 Proportion of Recovery

We prove Theorems 1, 2(ii), 3 and 4 in this section.

We consider the SIR process and the limiting process simultaneously. The
default setting is that their initial conditions are consistent, namely,

$\frac{I^{N}_{0}(\mathbf{x})}{N}\stackrel{{\scriptstyle}}{{\longrightarrow}}\mathcal{I}_{0}(\mathbf{x}),\quad\mbox{
for all }{\mathbf{x}}\in\mathbb{Z}^{2}\mbox{ as }N\rightarrow\infty.$ (77)

We also assume that the initial condition is nontrivial, i.e.,
$\mathcal{I}_{0}\not\equiv 0$. Recall that we write
$\mathcal{D}_{n}=\mathcal{R}_{n+1}$ and $D_{n}=R_{n+1}$,
$\mathcal{R}_{\infty}(\mathbf{x})=\lim_{n\rightarrow\infty}\mathcal{R}_{n}(\mathbf{x})$,
and $R_{\infty}(\mathbf{x})=\lim_{n\rightarrow\infty}R_{n}(\mathbf{x})$.

######  Theorem 12.

  1. (i)

Consider the deterministic system. For any nontrivial initial condition,
$\\{\mathcal{R}_{\infty}(\mathbf{x})\\}_{\mathbf{x}\in\mathbb{Z}^{2}}$
satisfies that

$\frac{\mathcal{R}_{\infty}(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x})}{1-\mathcal{I}_{0}(\mathbf{x})}>\iota,\mbox{
for all }\mathbf{x}\mbox{ such that }\mathcal{I}_{0}(\mathbf{x})<1;$ (78)

and

$\displaystyle\frac{\mathcal{R}_{\infty}(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x})}{1-\mathcal{I}_{0}(\mathbf{x})}$
(79) $\displaystyle=$ $\displaystyle
1-\exp\left(-\frac{1+\theta}{5}\widetilde{\mathcal{R}}_{\infty}(\mathbf{x})\right),\mbox{
for all }\mathbf{x}\mbox{ such that }\mathcal{I}_{0}(\mathbf{x})<1.$

Moreover, if the initial condition is either
$\mathcal{I}_{0}(m,l)=\gamma\cdot\bm{\delta}_{m+l=0}$ or
$\mathcal{I}_{0}(m,l)=\gamma\cdot\bm{\delta}_{(0,0)}$ for some
$\gamma\in(0,1]$, then we have

$\lim_{(m,l)\rightarrow\infty:m,l\geq 0}\mathcal{R}_{\infty}(m,l)=\iota.$ (80)

  2. (ii)

Consider the SIR system. For any initial condition satisfying (77) with
$\mathcal{I}_{0}\not\equiv 0$, we have

$\frac{R^{N}_{\infty}(\mathbf{x})}{N}\stackrel{{\scriptstyle\mathbb{P}}}{{\longrightarrow}}\mathcal{R}_{\infty}(\mathbf{x}),\quad\mbox{
for all }\mathbf{x}\in\mathbb{Z}^{2}\mbox{ as }N\rightarrow\infty.$ (81)

Moreover, if the initial condition is either $I^{N}_{0}(m,l)=\lfloor\gamma
N\rfloor\bm{\delta}_{m+l=0}$ or $I^{N}_{0}(m,l)=\lfloor\gamma
N\rfloor\bm{\delta}_{(0,0)}$ for some $\gamma\in(0,1]$, then for any
$\varepsilon>0$, there exists $N_{0}$ such that for all $N\geq N_{0}$,

$\limsup_{(m,l)\rightarrow\infty:m,l\geq
0}P\left(\left|\frac{R^{N}_{\infty}(m,l)}{N}-\iota\right|>\varepsilon\right)<\varepsilon.$
(82)

Note that Theorem 4 is contained in Theorem 12(ii). The proof of this theorem
is divided into several parts. The original recursive formula (14) is for
$\mathcal{I}_{n}$, which is not convenient when addressing $\mathcal{R}_{n}$
or $\mathcal{D}_{n}$. We start by giving a recursive formula for
$\mathcal{D}_{n}$.

######  Lemma 25.

The process $\\{\mathcal{D}_{n}(\mathbf{x})\\}_{n\in\mathbb{Z}_{\geq 0}}$
satisfies the following recursive equation:

$\displaystyle\frac{\mathcal{D}_{n+1}(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x})}{1-\mathcal{I}_{0}(\mathbf{x})}$
(83) $\displaystyle=$ $\displaystyle
1-\exp\left(-\frac{1+\theta}{5}\widetilde{\mathcal{D}}_{n}(\mathbf{x})\right),\mbox{
for all }\mathbf{x}\mbox{ satisfying }\mathcal{I}_{0}(\mathbf{x})<1.$

Letting $n$ go to infinity yields (79).

###### Proof.

One can show (83) by induction. Here we give a probabilistic proof, which is
more intuitive.

For any fixed $\mathbf{x}$ satisfying $\mathcal{I}_{0}(\mathbf{x})<1$,
consider the following adjusted percolation model. There are $N$ particles at
each site $\mathbf{y}\in\mathbb{Z}^{2}$ except that at $\mathbf{x}$, there are
$N+1$ particles. Other settings are the same as before: any two particles have
an edge if and only if they are in the same or in the neighbor site(s); each
edge is open with probability $P_{N}=(1+\theta)/(5N)$. It is easy to see that
this adjusted percolation model shares the same limiting process as the
original one, and Proposition 3 also holds for the adjusted model.

Note that the underlying graph of the original SIR model can be embedded into
the adjusted one, and we can couple the two models in a same probability
space. Write $V$ for the extra vertex at $\mathbf{x}$. Write $A_{N}$ for the
set of initial infected particles, $Y_{0},Y_{1},\dots,Y_{4}$ for the numbers
of particles at $\mathbf{x}$ and the four neighboring sites that are connected
to $A_{N}$ within $n$ edges, without passing through $V$. Conditionally on the
values of $\\{Y_{i}\\}$, we have

$P(V\mathrm{~{}is~{}connected~{}to~{}}A_{N}\mathrm{~{}within~{}}n+1\mathrm{~{}steps})=1-(1-P_{N})^{\sum_{i=0}^{4}Y_{i}}.$

Letting $N\rightarrow\infty$, by Proposition 3, we get the assertion. ∎

######  Lemma 26.

If the initial condition is $\mathcal{I}_{0}(m,l)=\bm{\delta}_{m+l=0}$, then

$\mathcal{R}_{\infty}(m,0):=\mathcal{R}_{\infty}(m)\searrow\iota.$

Moreover, if the initial condition is
$\mathcal{I}_{0}=\gamma\bm{\delta}_{\mathbf{0}}$ for some $\gamma\in{(0,1]}$,
then

$\lim_{\mathbf{x}\rightarrow\infty}\mathcal{R}_{\infty}(\mathbf{x})=\iota.$

###### Proof.

First assume that the initial condition is $\bm{\delta}_{m+l=0}$. Using (83),
by induction, one can see $\mathcal{D}_{n}(m,0)\geq\mathcal{D}_{n}(m+1,0)$.
Letting $n$ go to infinity, we see that $\mathcal{R}_{\infty}(m)$ is
decreasing in $m$. Hence, $\lim_{m\rightarrow\infty}\mathcal{R}_{\infty}(m)$
exists. We denote it by $s$. Letting $\mathbf{x}\rightarrow\infty$ in (79), we
see that $s$ satisfies the same equation as (5), the equation for $\iota$.
Moreover, by Theorem 11, $s\geq\delta$ for some positive $\delta$. Therefore,
$s=\iota$.

The second assertion can be proved in a similar way. Using the recursive
equation (83) and induction, we can show that
$\mathcal{D}_{n}(m,l)\geq\max\\{\mathcal{D}_{n}(m+1,l),\mathcal{D}_{n}(m,l+1)\\}$
when $m,l\geq 0$. Therefore,

$\mathcal{R}_{\infty}(m,l)\geq\max\\{\mathcal{R}_{\infty}(m+1,l),\mathcal{R}_{\infty}(m,l+1)\\},\mbox{
for all }m,l\geq 0.$

Hence, for any $(m,l)\in\mathbb{Z}^{2}$,
$\lim_{k\rightarrow\infty}\mathcal{R}_{\infty}(m+k,l+k)$ exists, denoted by
$s(m,l)$. Because
$\mathcal{R}_{\infty}(m+k,l+k)\geq\mathcal{R}_{\infty}(m+k+1,l+k)\geq\mathcal{R}_{\infty}(m+k+1,l+k+1)$,
we get that $s(m,l)=s(m+1,l)$. Similarly, one can get that $s(m,l)=s(m,l+1)$.
Therefore, all $s(m,l)$’s are equal, say, to $s$. By the recursive equation,
we see that $s$ satisfies the same equation as (5). Theorem 11 guarantees that
$s\neq 0$, hence $s$ must be $\iota$. Combining the conclusions above, we see
that $\mathcal{R}_{\infty}(m,l)\geq\iota$. On the other hand,
$\mathcal{R}_{\infty}(m,l)$ is not more than the corresponding
$\mathcal{R}_{\infty}(m,l)$ when the initial condition is
$\bm{\delta}_{x+y=0}$. By the first assertion, we get the upper bound and
finish the proof of the second assertion. ∎

We now prove (78).

###### Proof of (78).

For any nontrivial initial condition, by the previous lemma and monotonicity,
we have that $\mathcal{R}_{\infty}(\mathbf{x})\geq\iota$ for all
$\mathbf{x}\in\mathbb{Z}^{2}$. Plugging this into (79) yields

$\frac{\mathcal{R}_{\infty}(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x})}{1-\mathcal{I}_{0}(\mathbf{x})}\geq\iota,\mbox{
for all }\mathbf{x}\mbox{ such that }\mathcal{I}_{0}(\mathbf{x})<1.$ (84)

We need to further show that “=” cannot hold in (84). Assume otherwise that
for some $\mathbf{x}\in\\{\mathbf{x}:\mathcal{I}_{0}(\mathbf{x})\neq 1\\}$,
the above inequality becomes equality. Then by (79) and (84), we get that for
any $\mathbf{y}$ with $||\mathbf{y}-\mathbf{x}||_{1}=1$, we have

$\mathcal{R}_{\infty}(\mathbf{y})=\iota,\mbox{ and
}\mathcal{I}_{0}(\mathbf{y})=0.$

Repeating this argument yields that the above equalities hold for all
$\mathbf{z}\in\mathbb{Z}^{2}$, which contradicts to the assumption that
$\mathcal{I}_{0}$ is nontrivial. Therefore, (84) can not take “$=$” for any
$\mathbf{x}\in\mathbb{Z}^{2}$. ∎

######  Lemma 27.

If the initial condition satisfies that

$\inf_{{\mathbf{x}}\in\mathbb{Z}^{2}}\mathcal{R}_{\infty}(\mathbf{x})>\theta/(1+\theta),$
(85)

then the conclusion (81) holds.

######  Remark 28.

Note that $\iota>\theta/(1+\theta)$. One can see this by showing that
$t<1-\exp(-(1+\theta)t)$ for $t=\theta/(1+\theta)$. From this, the proof of
(81) is complete once we prove the lemma above.

###### Proof of Lemma 27.

It suffices to show that for any fixed $\mathbf{x}\in\mathbb{Z}^{2}$,
$\varepsilon>0$ and $n_{1}\in\mathbb{N}$, there is an $n>n_{1}$ such that when
$N$ is sufficiently large, we have

$E\left(\frac{R_{\infty}^{N}(\mathbf{x})-D^{N}_{n}(\mathbf{x})}{N}\right)<\varepsilon.$
(86)

For any $K\in\mathbb{N}$, write
$\mathcal{C}_{K}(\mathbf{x})=\\{\mathbf{z}\in\mathbb{Z}^{2}:||\mathbf{x}-\mathbf{z}||_{1}\leq
K\\}$ and
$\partial\mathcal{C}_{K}(\mathbf{x})=\\{\mathbf{z}\in\mathbb{Z}^{2}:||\mathbf{x}-\mathbf{z}||_{1}=K+1\\}$.
By (85), we can find a $\delta>0$ such that
$\inf_{{\mathbf{x}}\in\mathbb{Z}^{2}}\mathcal{R}_{\infty}(\mathbf{x})>(\theta+\delta)/(1+\theta)$.
For any $K$, we can find an $n_{0}$ such that

$\mathcal{D}_{n_{0}}(\mathbf{y})>\frac{\theta+\delta}{1+\theta},\mbox{ for all
}\mathbf{y}\in\mathcal{C}_{K}(\mathbf{x}).$

Therefore, as $N\to\infty$, with a high probability we have that

$\frac{D_{n_{0}}^{N}(\mathbf{y})}{N}>\frac{\theta+\delta}{1+\theta},\mbox{ for
all }\mathbf{y}\in\mathcal{C}_{K}(\mathbf{x}).$

By (2), when $k\geq n_{0}$ and $y\in\mathcal{C}_{K}{(\mathbf{x})}$,

$\displaystyle E\left(\frac{I^{N}_{k+1}(\mathbf{y})}{N}\right)$
$\displaystyle=\left(1-\frac{D^{N}_{k}(\mathbf{y})}{N}\right)E\left(1-\left(1-\frac{1+\theta}{5N}\right)^{\widetilde{I}^{N}_{k}(\mathbf{y})}\right)$
$\displaystyle\leq\frac{1-\delta}{1+\theta}\frac{1+\theta}{5}\frac{E\widetilde{I}_{k}^{N}(\mathbf{y})}{N}=\frac{1-\delta}{5}\frac{E\widetilde{I}_{k}^{N}(\mathbf{y})}{N}.$

By induction, we get that when $y\in\mathcal{C}_{K}{(\mathbf{x})}$,

$\displaystyle E\left(\frac{I^{N}_{n_{0}+m}(\mathbf{y})}{N}\right)$
$\displaystyle\leq\sum_{\mathbf{z}\in\mathcal{C}_{K}{(\mathbf{x}})}E\left(\frac{I^{N}_{n_{0}}(\mathbf{z})}{N}\right)\sharp_{\mathrm{L}}(\mathbf{y},\mathbf{z};m;\mathcal{C})\left(\frac{1-\delta}{5}\right)^{m}$
$\displaystyle+\sum_{\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})}\sum_{i=0}^{m}E\left(\frac{I^{N}_{n_{0}+i}(\mathbf{z})}{N}\right)\sharp_{\mathrm{L}}(\mathbf{y},\mathbf{z};m-i;\mathcal{C})\left(\frac{1-\delta}{5}\right)^{m-i},$

where we write $\sharp_{\mathrm{L}}(\mathbf{y},\mathbf{z};j;\mathcal{C})$ for
the number of LRW paths of length $j$ from $\mathbf{y}$ to $\mathbf{z}$ inside
$\mathcal{C}_{K}{(\mathbf{x})}$ (except the last step when
$\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})$).

Obviously, $\sharp_{\mathrm{L}}(\mathbf{y},\mathbf{z};m;\mathcal{C})\leq
5^{m}$ and for $i\leq K,\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})$,
$\sharp_{\mathrm{L}}(\mathbf{x},\mathbf{z};i;\mathcal{C})=0$. Therefore,

$\displaystyle E\left(\frac{I^{N}_{n_{0}+m}(\mathbf{x})}{N}\right)$
$\displaystyle\leq\sum_{\mathbf{z}\in\mathcal{C}_{K}{(\mathbf{x}})}(1-\delta)^{m}+\sum_{\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})}\sum_{i=0}^{m-K}E\left(\frac{I^{N}_{n_{0}+i}(\mathbf{z})}{N}\right)(1-\delta)^{m-i}$
$\displaystyle\leq(5K^{2})(1-\delta)^{m}+\sum_{\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})}\sum_{i=0}^{m-K}E\left(\frac{I^{N}_{n_{0}+i}(\mathbf{z})}{N}\right)(1-\delta)^{m-i}.$

It follows that for any $n_{2}>0,$

$\displaystyle E\left(\frac{\sum_{m\geq
n_{2}}I^{N}_{n_{0}+m}(\mathbf{x})}{N}\right)$ $\displaystyle\leq$
$\displaystyle 5K^{2}\sum_{m\geq
n_{2}}(1-\delta)^{m}+\sum_{\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})}\sum_{m\geq
n_{2}}\sum_{i=0}^{m-K}E\left(\frac{I^{N}_{n_{0}+i}(\mathbf{z})}{N}\right)(1-\delta)^{m-i}$
$\displaystyle\leq$ $\displaystyle
5K^{2}\frac{(1-\delta)^{n_{2}}}{\delta}+\sum_{\mathbf{z}\in\partial\mathcal{C}_{K}{(\mathbf{x}})}\sum_{i\geq
0}E\left(\frac{I^{N}_{n_{0}+i}(\mathbf{z})}{N}\right)\sum_{m\geq\max(n_{2},i+K)}(1-\delta)^{m-i}$
$\displaystyle\leq$ $\displaystyle
5K^{2}\frac{(1-\delta)^{n_{2}}}{\delta}+8K\frac{(1-\delta)^{K}}{\delta}.$

Letting $K$ be large enough first and then $n_{2}$ be large enough, we get
(86). ∎

###### Proof of (82).

For the upper bound, it is sufficient to consider the initial condition
$N\bm{\delta}_{x+y=0}$. By (80), we can find an $n_{0}$ such that
$\mathcal{R}_{\infty}(n_{0},0)<\iota+\varepsilon/2$. It follows from (81) that
there is an $N_{0}$ such that when $N\geq N_{0}$,

$P\left(\frac{R_{\infty}^{N}(n_{0},0)}{N}>\iota+\varepsilon\right)<\varepsilon.$

Note that for any $(m,l)$ with $m+l>n_{0}$, $R_{\infty}^{N}(m,l)$ is
stochastically dominated by $R_{\infty}^{N}(n_{0},0)$. Therefore, we have

$P\left(\frac{R_{\infty}^{N}(m,l)}{N}>\iota+\varepsilon\right)\leq
P\left(\frac{R_{\infty}^{N}(n_{0},0)}{N}>\iota+\varepsilon\right)<\varepsilon.$

Next, we prove the lower bound of (82). For this, it suffices to consider the
initial condition $\lfloor\gamma N\rfloor\bm{\delta}_{\mathbf{0}}$. We
consider the corresponding system in the half space
$\mathcal{H}=\\{(m,l)\in\mathbb{Z}^{2}:m+l\geq 0\\}$. Let
$\mathcal{Q}_{0}(\mathbf{x})=\gamma\bm{\delta}_{(0,0)}(\mathbf{x})$ and define
$\mathcal{Q}_{n}:\mathbb{Z}^{2}\rightarrow\mathbb{R}$ recursively as follows:

$\mathcal{Q}_{n}(m,l)=\left\\{\begin{array}[]{cc}1-\exp(-\frac{1+\theta}{5}\widetilde{\mathcal{Q}}_{n-1}(m,l)),&\mathrm{if}\,(m,l)\in\mathcal{H}\setminus\\{(0,0)\\};\\\
\gamma+(1-\gamma)(1-\exp(-\frac{1+\theta}{5}\widetilde{\mathcal{Q}}_{n-1}(m,l))),&\mathrm{if}\,(m,l)=(0,0);\\\
0,&\mbox{otherwise}.\end{array}\right.$ (87)

Let $\mathcal{Q}(m,l)=\lim_{n\rightarrow\infty}\mathcal{Q}_{n}(m,l)$.
Analogously to (27), we have

######  Proposition 29.

For any $b>1,\gamma\in(0,1]$,

$\lim_{n\rightarrow\infty}\sup_{m:\,-bn\leq m\leq
bn}|\mathcal{Q}(m,n-m)-\iota|=0.$ (88)

The proof of the above proposition is similar to the one of (27) and will be
given after the current proof.

Consider the $N$-percolation model. In the proof of (65), we have shown that
under the initial condition $I^{N}_{0}\sim\gamma N\bm{\delta}_{\mathbf{0}}$,
for any $\varepsilon>0$, there exist positive constants $\delta$ and $a$ such
that when $N$ is large,

$\limsup_{(m,l)\rightarrow\infty:m,l\geq
0}P(D^{N}_{a(m+l)}(m,l)/N<\delta)<\varepsilon.$ (89)

In fact, we have shown the above inequality holds when we replace
$D^{N}_{a(m+l)}(m,l)$ with $D^{N,-}_{a(m+l)}(m,l)$, the number of particles
that are connected to the $\lfloor\gamma N\rfloor$ initial infected particles
at the origin via no more than $a(m+l)$ edges inside the lower half space
$\\{(x,y):x+y<l+m\\}$.

On the other hand, by Proposition 29 and Proposition 3, for any
$\varepsilon>0$, there exist $n_{0}$ and $n_{1}$ such that when $N$ is large
enough,

$P\left(\frac{D^{N,+}_{n_{0}}({n_{1},0})}{N}<\iota-\varepsilon\right)<\varepsilon,$
(90)

where $D^{N,+}_{n_{0}}({n_{1},0})$ stands for the number of particles at
$(n_{1},0)$ that are connected to the $\lceil\delta N\rceil$ initial infected
particles at the origin via no more than $n_{0}$ edges in $\\{(x,y):x+y\geq
0\\}$.

Combining the results in the last two paragraphs, by connecting paths in the
$N$-percolation model, one can get the lower bound of (82). By symmetry, we
assume $m\geq l$. In fact, (89) for $D^{N,-}_{a{(m_{0}+l_{0})}}(m_{0},l_{0})$
says that, with high probability the number of particles at $(m_{0},l_{0})$
that are connected to the $\lfloor\gamma N\rfloor$ infected particles at the
origin via open edges inside the lower half space
$\\{(x,y):x+y<m_{0}+l_{0}\\}$ is at least $\lceil\delta N\rceil$. On the other
hand, it follows (90) that from those $\lceil\delta N\rceil$ infected
particles at $(m_{0},l_{0})$, with high probability, there are at least
$\lceil(\iota-\varepsilon)N\rceil$ particles at site $(m_{0}+n_{1},l_{0})$
that are are connected to the $\lceil\delta N\rceil$ infected particles at
$(m_{0},n_{0})$ via open edges inside the upper half space $\\{(x,y):x+y\geq
m_{0}+l_{0}\\}$. Hence, with high probability, the total number of the
infected particles at site $(m_{0}+n_{1},l_{0})$ is at least
$\lceil(\iota-\varepsilon)N\rceil$. Note that the two events are independent
because they depend on the edges that are lying on different sides of
$x+y=m_{0}+{l_{0}}$. ∎

###### Proof of Proposition 29.

The proof is very similar to the one of (27). Note that by (80), we only need
to show the lower bound. Moreover, by monotonicity, we can assume that
$\gamma$ is sufficiently small. Analogous to Claim 1 in the proof of
Proposition 9, we have

Claim: For any $\mathbf{a}\in\mathcal{H}$ and $n\in\mathbb{Z}_{\geq 0}$ with

$\mathcal{Q}_{n}(\mathbf{a})>\mathcal{Q}_{0}(0,0)=\gamma\mathrm{~{}and~{}}$
(91)
$1-\exp\left(-\frac{1+\theta}{5}\mathcal{Q}_{n}(\mathbf{a}-(1,0))\right)\geq\gamma,$
(92)

we have

$\mathcal{Q}_{n+k}(\mathbf{x}+\mathbf{a})>\mathcal{Q}_{k}(\mathbf{x}),\mbox{
for any }\mathbf{x}\in\mathcal{H}\mbox{ and }k\in\mathbb{Z}_{\geq 0}.$ (93)

As before, in order to find some $\mathbf{a}\in\mathcal{H}$ and
$n\in\mathbb{Z}_{\geq 0}$ satisfying the conditions in the claim, we need to
analyze the derivative respect to $\gamma$. Note that

$\left.\frac{\partial}{\partial\gamma}\right|_{\gamma=0}\left(1-\exp(-\frac{1+\theta}{5}\mathcal{Q}_{n}((i,m-i))\right)=\frac{1+\theta}{5}\left.\frac{\partial\mathcal{Q}_{n}(i,m-i)}{\partial\gamma}\right|_{\gamma=0}.$

Hence, it suffices to analyze $\partial\mathcal{Q}_{n}/\partial\gamma$. It
turns out that it has a similar recursive formula as $g^{\prime}_{n}$ in the
proof of Lemma 24:

$\left.\frac{\partial\mathcal{Q}_{n}}{\partial\gamma}\right|_{\gamma=0}(m,l)=\left\\{\begin{array}[]{cc}\frac{1+\theta}{5}\left.\frac{\partial\widetilde{\mathcal{Q}_{n-1}}}{\partial\gamma}\right|_{\gamma=0}(m,l),&\mathrm{if}\,(m,l)\in\mathcal{H}\setminus\\{(0,0)\\};\\\
1+\frac{1+\theta}{5}\left.\frac{\partial\widetilde{\mathcal{Q}_{n-1}}}{\partial\gamma}\right|_{\gamma=0}(m,l),&\mathrm{if}\,(m,l)=(0,0);\\\
0,&\mbox{otherwise}.\end{array}\right.$

Therefore, we get that

$\displaystyle\left.\frac{\partial}{\partial\gamma}\right|_{\gamma=0}$
$\displaystyle\mathcal{Q}_{n}(m,l)\geq\left(\frac{1+\theta}{5}\right)^{n}\sharp_{\mathrm{L}}(m,l;n;\mathcal{H})$
$\displaystyle=\left(\frac{1+\theta}{5}\right)^{n}(\sharp_{\mathrm{L}}(m,l;n)-\sharp_{\mathrm{L}}(m+1,l+1;n))$
$\displaystyle\geq
C(b)\left(\frac{1+\theta}{5}\right)^{n}\sharp_{\mathrm{L}}(m,l;n),$

where we write $\sharp_{\mathrm{L}}(m,l;n;\mathcal{H})$ for the number of LRW
paths from $(0,0)$ to $(m,l)$ with $n$ steps inside $\mathcal{H}$, in the
second line we use the reflection principle, and $C(b)$ is a constant that is
independent of $(m,l)$ as long as $|m|\leq b|m+l|$ and $1\leq n/(m+l)$ is
bounded from above. The last inequality can be proved in a very similar way to
(76).

By Lemma 23, we can find $n_{0}$ and $m_{0}$ large enough such that when
$n=n_{0}$ and $m\in[m_{0},3m_{0}]$,

$\inf_{i:|i|\leq
bm}\left.\frac{\partial\mathcal{Q}_{n}(i,m-i)}{\partial\gamma}\right|_{\gamma=0}\gg
1,\mathrm{~{}and~{}}\inf_{i:|i|\leq
bm}\left.\frac{\partial\mathcal{Q}_{n}(i-1,m-i)}{\partial\gamma}\right|_{\gamma=0}\gg
1.$

Therefore, there exists a positive $\gamma$ such that (91) and (92) hold for
all $\mathbf{a}\in\\{(i,j)||i|\leq b(i+j),\,m_{0}\leq i+j\leq 2m_{0}+1\\}:=A$
and $n=n_{0}$. By the Claim, (93) holds for such $\mathbf{a}$ and $n_{0}$.
Letting $k\rightarrow\infty$ in (93) we get

$\mathcal{Q}(\mathbf{x}+\mathbf{a})\geq\mathcal{Q}(\mathbf{x}),\,\mbox{ for
all }\mathbf{x}\in\mathcal{H},\mathbf{a}\in A.$ (94)

Hence, we can define, for any $\mathbf{x}\in\mathcal{H},\mathbf{a}\in A$,

$\lim_{n\rightarrow\infty}\mathcal{Q}(\mathbf{x}+n\mathbf{a}):=h(\mathbf{x},\mathbf{a}).$

For $\mathbf{a}_{1},\mathbf{a}_{2}\in A$ with
$\mathbf{a}:=\mathbf{a}_{1}+\mathbf{a}_{2}\in A$, (94) implies

$h(\mathbf{x},\mathbf{a})\leq h(\mathbf{x}+\mathbf{a}_{1},\mathbf{a})\leq
h(\mathbf{x}+\mathbf{a}_{1}+\mathbf{a}_{2},\mathbf{a})=h(\mathbf{x},\mathbf{a}).$

Hence,

$h(\mathbf{x},\mathbf{a})=h(\mathbf{x}+\mathbf{a}_{1},\mathbf{a}).$

Similarly we can get that

$h(\mathbf{x},\mathbf{a})=h(\mathbf{x}+\mathbf{a}_{2},\mathbf{a}).$

By choosing $\\{\mathbf{a}_{1},\mathbf{a}_{2}\\}$ so that they generate
$\mathbb{Z}^{2}$, we see that for $h(\mathbf{x},\mathbf{a})$ is independent of
$\mathbf{x}$. Letting $n$ in (87) go to infinity first and then
$\mathbf{x}=(m,l)$ go to infinity, we get that
$h(\mathbf{x},\mathbf{a})=\iota$. From this, one can get the lower bound, in a
similar way that (30) implies the lower bound of (27). ∎

We give the following proposition about the uniqueness of the solution to
equation (79).

######  Proposition 30.

For any nontrivial $\mathcal{I}_{0}:\mathbb{Z}^{2}\rightarrow[0,1]$, there
exists a unique solution $f:\mathbb{Z}^{2}\rightarrow[0,1]$ to the following
difference equation

$f(\mathbf{x})=\mathcal{I}_{0}(\mathbf{x})+(1-\mathcal{I}_{0}(\mathbf{x}))\left(1-\exp\left(-\frac{1+\theta}{5}\widetilde{f}_{n}(\mathbf{x})\right)\right),\mbox{for
all }\mathbf{x}\in\mathbb{Z}^{2}.$ (95)

Moreover, for such solution $f$, we have

$\frac{f(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x})}{1-\mathcal{I}_{0}(\mathbf{x})}>\iota,\,\mbox{for
all }\mathbf{x}\in\\{\mathbf{x}:\mathcal{I}_{0}(\mathbf{x})\neq 1\\},\mbox{
and
}\lim_{\mathbf{x}:\mathbf{d}(\mathbf{x},\mathrm{Supp}(\mathcal{I}_{0}))\rightarrow\infty}f(\mathbf{x})=\iota,$
(96)

where $\mathrm{Supp}(\mathcal{I}_{0})$ represents the support of
$\mathcal{I}_{0}$, namely, the set
$\\{\mathbf{x}:\mathcal{I}_{0}(\mathbf{x})\neq 0\\}$, and for any
$\mathbf{x}\in\mathbb{Z}^{2}$ and $A\subseteq\mathbb{Z}^{2},$
$\mathbf{d}(\mathbf{x},A):=\inf_{\mathbf{y}\in
A}||\mathbf{x}-\mathbf{y}||_{1}$.

###### Proof.

Define $u_{0}=\mathcal{I}_{0},v_{0}\equiv 1$ and recursively,

$\displaystyle u_{n+1}(\mathbf{x})$
$\displaystyle=\mathcal{I}_{0}(\mathbf{x})+(1-\mathcal{I}_{0}(\mathbf{x}))\left(1-\exp(-\frac{1+\theta}{5}\widetilde{u_{n}}(\mathbf{x}))\right);$
$\displaystyle v_{n+1}(\mathbf{x})$
$\displaystyle=\mathcal{I}_{0}(\mathbf{x})+(1-\mathcal{I}_{0}(\mathbf{x}))\left(1-\exp(-\frac{1+\theta}{5}\widetilde{v_{n}}(\mathbf{x}))\right).$

By induction, it is easy to show that $u_{n}$ increases to the minimal
solution to (95), denoted by $u$. Similarly, $v_{n}$ decreases to the maximal
solution to (95), denoted by $v$.

We first argue that

$(u(\mathbf{x})-\mathcal{I}_{0}(\mathbf{x}))/(1-\mathcal{I}_{0}(\mathbf{x}))>\iota.$

From (83), we see that $\mathcal{R}_{n}=u_{n}$ and hence
$\mathcal{R}_{\infty}=u$. By (78), we get the above assertion.

Note that when $a>b\in[\iota,1]$, we have

$(1-\exp(-(1+\theta)a))-(1-\exp(-(1+\theta)b))\leq(a-b).$

Therefore, we have

$\displaystyle v(\mathbf{x})-u(\mathbf{x})$
$\displaystyle\leq\frac{v(\mathbf{x})-\mathcal{I}_{0}(x)}{1-\mathcal{I}_{0}(\mathbf{x})}-\frac{u(\mathbf{x})-\mathcal{I}_{0}(x)}{1-\mathcal{I}_{0}(\mathbf{x})}$
$\displaystyle=\left(1-\exp(-\frac{1+\theta}{5}\widetilde{v}(\mathbf{x}))\right)-\left(1-\exp(-\frac{1+\theta}{5}\widetilde{u}(\mathbf{x}))\right)$
$\displaystyle\leq\frac{\widetilde{v}(x)}{5}-\frac{\widetilde{u}(x)}{5}.$

This implies that $v-u$ is a (discrete) subharmonic function. On the other
hand, $v-u$ is bounded. The classical result in subharmonic functions states
that a (two dimensional continuous) subharmonic function bounded from above
must be constant. This result also holds for discrete subharmonic functions in
$\mathbb{Z}^{2}$; see, e.g., [Rigoli, Salvatori, and Vignati(1997)]. Hence,
$v-u$ is constant, say $c$, and the above display must take equality. In order
for the above display to take equality, $c$ must be $0$. Now we finish the
proof of the proposition except for the last assertion in (96). We proceed to
prove the last assertion in (96). Write $w_{n}$ for the solution $f$ when
$\mathcal{I}_{0}(x,y)=\bm{\delta}_{|x|+|y|\geq n}$. By monotonicity, we get
that when $\mathbf{d}(\mathbf{x},\mathrm{Supp}(\mathcal{I}_{0}))\geq n$,
$f(\mathbf{x})\leq w_{n}(0)$. It suffices to show that

$\liminf_{n\rightarrow\infty}w_{n}(\mathbf{0})\leq\iota.$ (97)

By monotonicity, $w_{n}(\mathbf{x})$ decreases in $n$. Hence, we define

$w(\mathbf{x})=\lim_{n\rightarrow\infty}w_{n}(\mathbf{x}).$

Note that $w_{n}$ satisfies (95). Letting $n$ go to infinity, we get that $w$
satisfies (95) for all $\mathbf{x}\in\mathbb{Z}^{2}$ with
$\mathcal{I}_{0}\equiv 0$. Now for this trivial initial condition, we analyze
the maximal solution. As before, define $v_{0}\equiv 1$,

$v_{n+1}(\mathbf{x})=\mathcal{I}_{0}(\mathbf{x})+(1-\mathcal{I}_{0}(\mathbf{x}))\left(1-\exp(-\frac{1+\theta}{5}\widetilde{v_{n}}(\mathbf{x}))\right).$

One can see that $v_{n}$ is constant and so is the maximal solution
$v:=\lim_{n\rightarrow\infty}v_{n}$. The constant solution to (95) is either
$\iota$ or $0$. Hence we get that $v\equiv\iota$ and therefore (97). ∎

We now prove the result about $\sum_{i=1}^{\infty}\ell^{(i)}$ mentioned in the
Introduction.

######  Lemma 31.

When $\theta>1.5$, the $\ell^{(i)}$’s defined in (12) satisfy that

$\iota=\sum_{i=1}^{\infty}\ell^{(i)}.$

###### Proof.

Consider the deterministic system with initial condition
$\mathcal{I}_{0}=\bm{\delta}_{x+y=0}$. By Proposition 10, we have that
$\lim_{n\rightarrow\infty}\mathcal{D}_{k}(n,0)=\sum_{i=1}^{k}\ell^{(i)}$.
Letting $\mathbf{x}=(n,0)$ with $n\rightarrow\infty$ in (83), we get that
$s_{k}=\sum_{i=1}^{k}\ell^{(i)}$ satisfies

$s_{k+1}=1-\exp\left(-\frac{1+\theta}{5}(2s_{k+1}+s_{k}+2s_{k-1})\right).$

Letting $k\rightarrow\infty$, we get that
$s=\lim_{k\rightarrow\infty}s_{k}=\sum_{i=1}^{\infty}\ell^{(i)}$ satisfies

$s=1-\exp(-(1+\theta)s),$

which is the same equation as (5), the equation for $\iota$. Note that because
$\theta>1.5$, $s\geq\ell^{(1)}>0$. Hence, $s$ must equal $\iota$. ∎

We are ready to prove Theorems 1, 2(ii) and 3, starting with Theorem 1.

###### Proof of Theorem 1.

Note that Theorem 2(ii) guarantees that $q_{N}\rightarrow 1$ for IC2. Hence we
assume that the initial condition is IC1.

Clearly, $q_{N}$ is not more than the survival probability of the branching
process with offspring distribution $\mathrm{Bin}(5N,(1+\theta)/(5N))$. It is
easy to show that the last survival probability converges to $\iota$. This
gives the upper bound.

We now show the lower bound. Write
$\mathcal{L}(k)=\\{(m,l)\in\mathbb{Z}^{2}:m+l=k\\}\times[N]$. We first
consider the SIR process with initial condition $N\bm{\delta}_{m+l=0}$. For
any $m,l\in\mathbb{Z}$, we have

$E\left(\frac{R_{\infty}^{N}(m,l)}{N}\right)=P^{N}(((m,l),1)\mathrm{~{}is~{}connected~{}to~{}}\mathcal{L}(0)),$

where $((m,l),1)$ represents the first particle at site $(m,l)$ and we use
superscript $N$ to indicate that the village size is $N$. By (82), for any
$\varepsilon>0$, there is an $N_{0}$ such that when $N>N_{0}$, we have

$\limsup_{n\rightarrow\infty}P^{N}(((n,0),1)\mathrm{~{}is~{}connected~{}to~{}}\mathcal{L}(0))>\iota-\varepsilon.$

Note that the probability above is decreasing in $n$, hence we get

$\inf_{n\in\mathbb{Z}_{\geq
0}}P^{N}(((n,0),1)\mathrm{~{}is~{}connected~{}to~{}}\mathcal{L}(0))>\iota-\varepsilon.$

By symmetry, we get

$\inf_{n\in\mathbb{Z}_{\geq
0}}P^{N}(({(0,0)},1))\mathrm{~{}is~{}connected~{}to~{}}\mathcal{L}(n))>\iota-\varepsilon.$

Therefore, we have $q_{N}>\iota-\varepsilon$. ∎

Next, we show Theorem 2(ii).

###### Proof of Theorem 2(ii).

By symmetry, we assume all $(i_{k},j_{k})$ satisfy $i_{k}\geq j_{k}\geq 0$. We
first show the upper bound for IC1 conditional on survival and IC2:

$\limsup_{k\to\infty}P\left(\frac{R_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor}(i_{k},j_{k})}{N}>\iota+\varepsilon\right)<\varepsilon.$

Note that in order to show the inequality above for IC1 conditional on
survival, it suffices to show the inequality unconditionally, and by
monotonicity, it suffices to do so for IC2. This follows from (82) because
$R_{n}\leq R_{\infty}$ for any $n$.

Now we turn to the lower bound:

$\limsup_{k\to\infty}P\left(\frac{R_{\lfloor(i_{k}+j_{k})(\upsilon(\theta,\phi)^{-1}+\varepsilon)\rfloor}(i_{k},j_{k})}{N}<\iota-\varepsilon\right)<\varepsilon.$
(98)

We first consider the initial condition IC2. We will show a stronger version: