# Functional law of large numbers and PDEs for epidemic models
with infection-age dependent infectivity

Guodong Pang∗ ∗Department of Computational Applied Mathematics and Operations
Research, George R. Brown College of Engineering, Rice University, Houston, TX
77005<EMAIL_ADDRESS>and  Étienne Pardoux† †Aix–Marseille Univ, CNRS, I2M,
Marseille, France<EMAIL_ADDRESS>
###### Abstract.

We study epidemic models where the infectivity of each individual is a random
function of the infection age (the elapsed time of infection). To describe the
epidemic evolution dynamics, we use a stochastic process that tracks the
number of individuals at each time that have been infected for less than or
equal to a certain amount of time, together with the aggregate infectivity
process. We establish the functional law of large numbers (FLLN) for the
stochastic processes that describe the epidemic dynamics. The limits are
described by a set of deterministic Volterra-type integral equations, which
has a further characterization using PDEs under some regularity conditions.
The solutions are characterized with boundary conditions that are given by a
system of Volterra equations. We also characterize the equilibrium points for
the PDEs in the SIS model with infection-age dependent infectivity. To
establish the FLLNs, we employ a useful criterion for weak convergence for the
two-parameter processes together with useful representations for the relevant
processes via Poisson random measures.

###### Key words and phrases:

Functional law of large numbers, deterministic Volterra integral equations,
PDEs, non–Markovian epidemic models, infection-age dependent (varying)
infectivity, Poisson random measure, SIR, SIS, equilibrium in the SIS model

##  1\. Introduction

Kermack and McKendrick pioneered the introduction of PDE models to describe
the epidemic dynamics for models with infection-age dependent (variable)
infectivity in 1932 [19]. The underlying assumption of their model is that the
infectious periods have a general distribution with density which is modeled
through an infection–age dependent recovery rate, the infectious individuals
having an infection–age dependent infectivity, and the recovered ones a
recovery–age susceptibility. In the present paper, we do not consider possible
loss of immunity. We defer to a work in preparation the study of variable
susceptibility. In the present paper, we mainly consider the SIR model
(although we can allow for an exposed period, as will be explained below) and
the SIS model. This work is a continuation of our first work on non–Markov
epidemic models [25], and our work on varying infectivity models [10], see
also [24]. In those papers, we show that certain deterministic Volterra type
integral equations are Functional Law of Large Numbers (FLLN) limits of
adequate individual based stochastic models. An important feature of our
stochastic models is that they are non–Markov (since the infectious duration
need not have an exponential distribution), and as a result the limiting
deterministic models are equations with memory. Note that as early as in 1927,
Kermack and McKendrick introduced in their seminal paper [18] a SIR model with
both infection–age dependent infectivity and infection–age dependent recovery
rate, the latter allowing the infectious period to have an arbitrary
absolutely continuous distribution (the infection–age dependent recovery rate
is the hazard rate function of the infectious period). One part of that paper
is devoted to the simpler case of constant rates, and apparently most of the
later literature on epidemic models has concentrated on this special case,
which leads to simpler ODE models, the corresponding stochastic models being
Markov models, at the price of the models being less realistic. For example,
the recent studies in Covid-19 [11, 30] indicates that using the ODE models
can lead to an underestimation of the basic reproduction number $R_{0}$.

In this paper, we go back to the original model of Kermack and McKendrick
[18], with two new aspects. First, as in our previous publications, we want to
obtain the deterministic model as a law of large numbers limit of stochastic
models, and second, we distribute the various infected individuals at time $t$
according to their infection–age, and establish a PDE for the “density of
individuals” being infected at time $t$, with infection–age $x$.

In our stochastic epidemic model, each individual is associated with a random
infectivity, which varies as a function of the age of infection (elapsed time
since infection). The random infectivity functions, effective during the
infected period, are assumed to be i.i.d. for the various individuals, and
will also generate the infectious period. The infectivity function is assumed
to be càdlàg with a given number of discontinuities, and upper bounded by a
deterministic constant. In particular, the law of the infectious period can be
completely arbitrary. Our modeling approach allows the random infectivity
functions to have an initial period of time during which they take zero
values, corresponding to the exposed period. Thus our model generalizes both
the classical SIR and SEIR models. To describe the epidemic dynamics of the
model, we use a (two-parameter or measure-valued) stochastic process that
tracks at each time $t$ the number of individuals that have been infected for
a duration less than or equal to a certain amount of time $x$, and an
associated aggregate infectivity process which at each time $t$ sums up the
infectivities of all individuals who are infected. From these processes, we
can describe the cumulative infection process, the total number of infected
individuals as well as the number of recovered ones at each time. We use
similar processes to describe the epidemic dynamics for the SIS model with
infection-age dependent infectivity.

In the asymptotic regime of a large population (i.e., as the total population
size $N$ tends to infinity), we establish the FLLN for the epidemic dynamics.
The limits are characterized by a set of deterministic Volterra-type integral
equations (Theorem 2.1). Under certain regularity conditions, the density
function of the two-parameter (calendar time and infection age) limit process
can be described by a one-dimensional PDE (Proposition 3.1 in the case where
the distribution of the infectious period is absolutely continuous). Its
solution is characterized with a boundary condition satisfying a one-
dimensional Volterra-type integral equation. The aggregate infectivity limit
process can be described by an integral of the average infectivity function
with respect to the limiting two-parameter infectious process (Corollary 3.1,
see also Remark 3.4). For the classical SIR model, we recover the well-known
linear PDE first proposed by Kermack and McKendrick [19]. We further derive
the PDE model when the distribution of the infectious period need not be
absolutely continuous (Proposition 3.2 and see also Corollary 3.3 where the
infectious periods are deterministic). These PDE models are new to the
literature of epidemiology. For the SIS model, we also describe the limiting
epidemic dynamics and the PDE representations, and derive the equilibrium
quantities associated with the PDE and total count limit (assuming convergence
to the equilibria).

###  1.1. Literature review

Non–Markov stochastic epidemic models lead (via the FLLN) to deterministic
models, which are either low dimensional evolution equation with memory (i.e.,
Volterra type integral equations), or else coupled ODE/PDE models, where the
two variables are the time and the age of infection (time since infection).
The first paper of Kermack and McKendrick [18] adopts the first point of view,
and the two next [19, 20] the second one. In our recent previous work on this
topic [25, 10], we have adopted the first description. The goal of the present
paper is to show that in the limit of a large population, our stochastic
individual based model with age of infection dependent infectivity and
recovery rate converges as well to a limiting system of PDE/ODEs.

While the general model from [18] was largely neglected until rather recently,
most of the literature concentrating on the particular case of constant rates,
there has been since the 1970s some papers considering infection-age dependent
epidemic models, see in particular [14]. More recently, several papers have
introduced coupled PDE/ODE models for studying age of infection dependent both
infectivity and recovery rate, see in particular [28, 16, 29, 22, 6] and
Chapter 13 in [23]. In [8], the authors consider a stochastic epidemic model
with contract-tracing, tracking the infection duration since detection for
each individual, and use a measure-valued Markov process to describe the
epidemic dynamics. They prove a FLLN with a large population and establish a
PDE limit, and also prove a FCLT with a SPDE limit process. Since the
beginning of the Covid–19 pandemic, a huge number of papers have been
produced, with various models of the propagation of this disease. Most of them
use ODE models, but a few, notably [17, 13, 12, 9] consider age of infection
dependent infectivity, and possibly recovery rate. The last two derive the
ODE/PDE model as a law of large numbers limit of stochastic individual based
models. The article [12] considers a branching process approximation of the
early phase of an epidemic, and the way they model the dependence of the rate
of infection with respect to the age of infection is less general than in our
model. Recently, the authors in [9] study contact tracing in an individual-
based epidemic model via an “infection graph” of the population, and prove the
local convergence of the random graph to a Poisson marked tree and a Kermack
and McKendrick type of PDE limit for the dynamics by tracking the infection
age.

Note also that one way that many authors have chosen in order to improve the
realism of ODE models is by increasing the number of compartments. For
instance, dividing the infectious compartment into subcompartments, each one
corresponding to a different infection rate, is a way to introduce a
(piecewise constant) infection age dependent infectivity. In a way, this means
approaching a non-Markov process of a given dimension by a higher dimensional
Markov process, or approaching a system differential equations with memory by
a higher dimensional system of ODEs. In the present paper, we show that the
system of integral equations with memory introduced in our earlier work [25]
can be replaced by an ODE/PDE system, i.e., an infinite dimensional
differential equation. At the level of the stochastic finite population model,
this means replacing a non-Markov finite dimensional Markov process by a high
dimensional process (whose dimension is bounded by the total population size
$N$, which tends to infinity in our asymptotic). See Remarks 3.2 and 3.7
below.

We also like to mention the relevant work in queueing systems where the
elapsed service times are tracked using two-parameter or measure-valued
processes. The most relevant to us are the infinite-server (IS) queueing
models studied in [26, 27, 1], where FLLN and FCLT are established for two-
parameter processes to tracking elapsed and residual service times. However,
the proof techniques we employ in this paper are very different from those
papers. Here we exploit the representations with Poisson random measures and
use a new weak convergence criterion (Theorem 5.1). In addition, despite
similarities with the IS queueing models, the stochastic epidemic models have
an arrival (infection) process that depend on the state of the system. As a
consequence, the limits in the FLLNs result in PDEs while the IS queueing
models do not.

###  1.2. Organization of the paper

The paper is organized as follows. In Section 2, we describe the stochastic
epidemic model with infection-age dependent infectivity, and state the FLLN.
In Section 3, we present the PDE models from the FLLN limits, and also
characterize the solution properties of the PDEs. The limits and PDE for the
SIS model are presented in Section 4, which also considers the equilibrium
behavior. In Sections 5, we prove the FLLN. The Appendix gives the proof of
the convergence criterion in Theorem 5.1.

###  1.3. Notation

All random variables and processes are defined on a common complete
probability space $(\Omega,{\mathcal{F}},\mathbb{P})$. The notation
$\Rightarrow$ means convergence in distribution. We use
${\mathbf{1}}_{\\{\cdot\\}}$ for the indicator function, and occasionally use
${\mathbf{1}}\\{\cdot\\}$ for better readability. Throughout the paper,
${\mathbb{N}}$ denotes the set of natural numbers, and
${\mathbb{R}}^{k}({\mathbb{R}}^{k}_{+})$ denotes the space of $k$-dimensional
vectors with real (nonnegative) coordinates, with
${\mathbb{R}}({\mathbb{R}}_{+})$ for $k=1$. For $x,y\in{\mathbb{R}}$, we
denote $x\wedge y=\min\\{x,y\\}$ and $x\vee y=\max\\{x,y\\}$. Let
$D=D({\mathbb{R}}_{+};{\mathbb{R}})$ denote the space of ${\mathbb{R}}$–valued
càdlàg functions defined on ${\mathbb{R}}_{+}$. Throughout the paper,
convergence in $D$ means convergence in the Skorohod $J_{1}$ topology, see
Chapter 3 of [4]. Also, $D^{k}$ stands for the $k$-fold product equipped with
the product topology. Let $C$ be the subset of $D$ consisting of continuous
functions. Let $C^{1}$ consist of all differentiable functions whose
derivative is continuous. Let $D_{\uparrow}$ denote the set of increasing
functions in $D$. Let
$D_{D}=D({\mathbb{R}}_{+};D({\mathbb{R}}_{+};{\mathbb{R}}))$ be the $D$-valued
$D$ space, and the convergence in the space $D_{D}$ means that both $D$ spaces
are endowed with the Skorohod $J_{1}$ topology. The space $C_{C}$ is
equivalent to $C({\mathbb{R}}_{+}^{2};{\mathbb{R}}_{+})$. Let
$C_{\uparrow}(\mathbb{R}_{+}^{2};\mathbb{R}_{+})$ denote the space of
continuous functions from $\mathbb{R}_{+}^{2}$ into $\mathbb{R}_{+}$, which
are increasing as a function of their second variable. For any increasing
càdlàg function $F(\cdot):\mathbb{R}_{+}\to\mathbb{R}_{+}$, abusing notation,
we write $F(dx)$ by treating $F(\cdot)$ as the positive (finite) measure on
$\mathbb{R}_{+}$ whose distribution function is $F$. For any
${\mathbb{R}}$–valued càdlàg function $\phi(\cdot)$ on $\mathbb{R}_{+}$, the
integral $\int_{a}^{b}\phi(x)F(dx)$ represents $\int_{(a,b]}\phi(x)F(dx)$ for
$a<b$.

##  2\. Model and FLLN

###  2.1. Model description

We consider an epidemic model in which the infectivity rate depends on the age
of infection (that is, how long the individuals have been infected).
Specifically, each individual $i$ is associated with an infectivity process
$\lambda_{i}(\cdot)$, and we assume that these random functions are i.i.d..
Let $\eta_{i}=\inf\\{t>0:\lambda_{i}(r)=0,\,\forall r\geq t\\}$ be the
infected period corresponding to the individual that gets infected at time
$\tau^{N}_{i}$. The $\eta_{i}$’s are i.i.d., with a cumulative distribution
function (c.d.f.) $F$. Let $F^{c}=1-F$.

Individuals are grouped into susceptible, infected and recovered ones. Let the
population size be $N$ and $S^{N}(t)$, $I^{N}(t)$ and $R^{N}(t)$ denote the
numbers of the susceptible, infected and recovered individuals at time $t$. We
have the balance equation: $N=S^{N}(t)+I^{N}(t)+R^{N}(t)$, $t\geq 0.$ Assume
that $S^{N}(0)>0$, $I^{N}(0)>0$ and $R^{N}(0)=0$. Let $\mathfrak{I}^{N}(t,x)$
be the number of infected individuals at time $t$ that have been infected for
a duration less than or equal to $x$. Note that for each $t$,
$\mathfrak{I}^{N}(t,x)$ is nondecreasing in $x$, which is the distribution of
$I^{N}(t)$ over the infection-ages. Let $A^{N}(t)$ be the cumulative number of
newly infected individuals in $(0,t]$, with the infection times
$\\{\tau^{N}_{i}:i\in{\mathbb{N}}\\}$.

Let $\\{\tau_{j,0}^{N},j=1,\dots,I^{N}(0)\\}$ be the times at which the
initially infected individuals at time 0 became infected. Then
$\tilde{\tau}_{j,0}^{N}=-\tau_{j,0}^{N}$, $j=1,\dots,I^{N}(0)$, represent the
amount of time that an initially infected individual has been infected by time
$0$, that is, the age of infection at time $0$. WLOG, assume that
$0>\tau_{1,0}^{N}>\tau_{2,0}^{N}>\cdots>\tau_{I^{N}(0),0}^{N}$ (or
equivalently
$0<\tilde{\tau}_{1,0}^{N}<\tilde{\tau}_{2,0}^{N}<\cdots<\tilde{\tau}_{I^{N}(0),0}^{N}$).
Set $\tilde{\tau}_{0,0}^{N}=0$. We define $\mathfrak{I}^{N}(0,x)=\max\\{j\geq
0:\tilde{\tau}_{j,0}^{N}\leq x\\}$, the number of initially infected
individuals that have been infected for a duration less than or equal to $x$
at time $0$. Assume that there exists $0\leq\bar{x}<\infty$ such that
$I^{N}(0)=\mathfrak{I}^{N}(0,\bar{x})$ a.s.

Each initially infected individual $j=1,\dots,I^{N}(0)$, is associated with an
infectivity process $\lambda_{j}^{0}(\cdot)$, and we assume that they are also
i.i.d., with the same law as $\lambda_{i}(\cdot)$. This is reasonable since it
is for the same disease, and the infectivity for the initially and newly
infected individuals with the same infection age should have the same law. The
infectivity processes take effect at the epochs of infection. For each $j$,
let
$\eta^{0}_{j}=\inf\\{t>0:\lambda_{j}^{0}(\tilde{\tau}_{j,0}^{N}+r)=0,\,\forall
r\geq t\\}$ be the remaining infectious period, which depends on the elapsed
infection time $\tilde{\tau}_{j,0}^{N}$, but is independent of the elapsed
infection times of other initially infected individuals. In particular, the
conditional distribution of $\eta^{0}_{j}$ given that
$\tilde{\tau}_{j,0}^{N}=s>0$ is given by

$\displaystyle\mathbb{P}(\eta^{0}_{j}>t|\tilde{\tau}_{j,0}^{N}=s)=\frac{F^{c}(t+s)}{F^{c}(s)},\quad\mbox{for}\quad
t,s>0.$ (2.1)

Note that the $\eta^{0}_{j}$’s are independent but not identically
distributed.

For an initially infected individual $j=1,\dots,I^{N}(0)$, the infection age
is given by $\tilde{\tau}^{N}_{j,0}+t$ for $0\leq t\leq\eta^{0}_{j}$, during
the remaining infectious period. For a newly infected individual $i$, the
infection age is given by $t-\tau^{N}_{i}$, for $\tau^{N}_{i}\leq
t\leq\tau^{N}_{i}+\eta_{i}$ during the infectious period. Note that
$\lambda_{i}(\cdot)$ and $\lambda^{0}_{j}(\cdot)$ are equal to zero on
${\mathbb{R}}_{-}$.

The aggregate infectivity process at time $t$ is given by

$\displaystyle\mathcal{I}^{N}(t)=\sum_{j=1}^{I^{N}(0)}\lambda_{j}^{0}(\tilde{\tau}^{N}_{j,0}+t)+\sum_{i=1}^{A^{N}(t)}\lambda_{i}(t-\tau^{N}_{i}),\quad
t\geq 0.$ (2.2)

(Note that the notation $\mathfrak{I}^{N}$ was used for the infectivity
process in [10, 24].) The instantaneous infection rate at time $t$ can be
written as

$\displaystyle\Upsilon^{N}(t)$
$\displaystyle=\frac{S^{N}(t)}{N}\mathcal{I}^{N}(t),\quad t\geq 0.$ (2.3)

The counting process of newly infected individuals $A^{N}(t)$ can be written
as

$\displaystyle
A^{N}(t)=\int_{0}^{t}\int_{0}^{\infty}{\mathbf{1}}_{u\leq\Upsilon^{N}(s^{-})}Q(ds,du)\,,$
(2.4)

where $Q$ is a standard Poisson random measure on ${\mathbb{R}}^{2}_{+}$ (see,
e.g., [7, Chapter VI]).

Among the initially infected individuals, the number of individuals who have
been infected for a duration less than or equal to $x$ at time $t$ is equal to

$\displaystyle\mathfrak{I}^{N}_{0}(t,x)=\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\eta_{j}^{0}>t}{\mathbf{1}}_{\tilde{\tau}_{j,0}^{N}\leq(x-t)^{+}}=\sum_{j=1}^{\mathfrak{I}^{N}(0,(x-t)^{+})}{\mathbf{1}}_{\eta_{j}^{0}>t}\,,\quad
t,x\geq 0,$ (2.5)

Recall the age limit of the initially infected individuals $\bar{x}$ at time
zero. Thus, the number of the initially infected individuals that remain
infected at time $t$ can be written as

$I^{N}_{0}(t)\;=\;\mathfrak{I}^{N}_{0}(t,\bar{x}+t)\,,\quad t\geq 0\,.$ (2.6)

Among the newly infected individuals, the number of individuals who have been
infected for a duration less than or equal to $x$ at time $t$ is equal to

$\displaystyle\mathfrak{I}^{N}_{1}(t,x)$
$\displaystyle=\sum_{i=1}^{A^{N}(t)}{\mathbf{1}}_{(t-x)^{+}<\tau^{N}_{i}\leq
t}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}=\sum_{i=1}^{A^{N}(t)}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-\sum_{i=1}^{A^{N}((t-x)^{+})}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}$
$\displaystyle=\sum_{i=A^{N}((t-x)^{+})+1}^{A^{N}(t)}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}$
(2.7)

Thus, the number of newly infected individuals that remain infected at time
$t$ can be written as

$I^{N}_{1}(t)=\mathfrak{I}^{N}_{1}(t,t).$ (2.8)

We also have the total number of individuals infected at time $t$ that have
been infected for a duration which is less than or equal to $x$:

$\mathfrak{I}^{N}(t,x)=\mathfrak{I}^{N}_{0}(t,x)+\mathfrak{I}^{N}_{1}(t,x),\quad
t\geq 0,\,x\geq 0.$

Note that for each $t$, the support of the measure
$\mathfrak{I}^{N}_{0}(t,dx)$ is included in $[0,t+\bar{x}]$ and the support of
the measure $\mathfrak{I}^{N}_{1}(t,dx)$ is included in $[0,t]$. Thus

$I^{N}(t)=\mathfrak{I}^{N}_{0}(t,t+\bar{x})+\mathfrak{I}^{N}_{1}(t,t)=\mathfrak{I}^{N}(t,\infty),\quad
t\geq 0.$

Here we occasionally use $\infty$ in the second component for convenience with
the understanding that
$\mathfrak{I}^{N}_{0}(t,x)=\mathfrak{I}^{N}_{0}(t,t+\bar{x})$ for
$x>t+\bar{x}$ and $\mathfrak{I}^{N}_{1}(t,x)=\mathfrak{I}^{N}_{1}(t,t)$ for
$x>t$.

We also have for $t\geq 0$,

$\displaystyle S^{N}(t)$ $\displaystyle=S^{N}(0)-A^{N}(t),$ (2.9)
$\displaystyle R^{N}(t)$
$\displaystyle=\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\eta_{j}^{0}\leq
t}+\sum_{i=1}^{A^{N}(t)}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}\leq t}\,.$ (2.10)

We remark that the sample paths of $\mathfrak{I}^{N}(t,x)$ belong to the space
$D_{D}$, denoting $D({\mathbb{R}}_{+};D({\mathbb{R}}_{+};{\mathbb{R}}))$, the
$D$-valued $D$ space, but not in the space
$D({\mathbb{R}}_{+}^{2};{\mathbb{R}})$. We prove the weak convergence in the
space $D_{D}$ where both $D$ spaces are endowed with the Skorohod $J_{1}$
topology. Note that the space $D({\mathbb{R}}_{+}^{2};{\mathbb{R}})$ is a
strict subspace of $D_{D}$, although they are equivalent in the continuous
cases, that is, $C({\mathbb{R}}_{+}^{2};{\mathbb{R}})=C_{C}$. See more
discussions on these spaces in [26, 27, 2, 3].

######  Remark 2.1.

The SEIR model. Suppose that $\lambda_{i}(t)=0$ for $t\in[0,\xi_{i})$, where
$\xi_{i}<\eta_{i}$, and denote $I$ as the compartment of infected (not
necessarily infectious) individuals. An individual who gets infected at time
$\tau^{N}_{i}$ is first exposed during the time interval
$[\tau^{N}_{i},\tau^{N}_{i}+\zeta_{i})$, and then infectious during the time
interval $(\tau^{N}_{i}+\zeta_{i},\tau^{N}_{i}+\eta_{i})$. One may state that
the individual is infected during the time interval
$[\tau^{N}_{i},\tau^{N}_{i}+\eta_{i})$. At time $\tau^{N}_{i}+\eta_{i}$, he
recovers. All what follows covers perfectly this situation. In other words,
our model accomodates perfectly an exposed period before the infectious
period, which is important for many infectious diseases, including the
Covid–19. However, we distinguish only three compartments, $S$ for
susceptible, $I$ for infected (either exposed or infectious), $R$ for
recovered.

In the sequel, the time interval $[\tau^{N}_{i},\tau^{N}_{i}+\eta_{i})$ will
be called the infectious period, although it might rather be the period during
which the individual is infected (either exposed or infectious).

###  2.2. FLLN

Define the LLN-scaled processes $\bar{X}^{N}=N^{-1}X^{N}$ for any processes
$X^{N}$. We make the following assumptions on the initial quantities.

######  Assumption 2.1.

There exists a deterministic continuous nondecreasing function
$\bar{\mathfrak{I}}(0,x)$ for $x\geq 0$ with $\bar{\mathfrak{I}}(0,0)=0$ such
that $\bar{\mathfrak{I}}^{N}(0,\cdot)\to\bar{\mathfrak{I}}(0,\cdot)$ in $D$ in
probability as $N\to\infty$. Let $\bar{I}(0)=\bar{\mathfrak{I}}(0,\bar{x})$.
Then $(\bar{I}^{N}(0),\bar{S}^{N}(0))\to(\bar{I}(0),\bar{S}(0))\in(0,1)^{2}$
in probability as $N\to\infty$ where $\bar{S}(0)=1-\bar{I}(0)\in(0,1)$.

######  Remark 2.2.

Recall that $\bar{\mathfrak{I}}^{N}(0,\cdot)$ describes the distribution of
the initially infected individuals over the ages of infection. The assumption
means that there is a corresponding limiting continuous distribution as the
population size goes to infinity.

Suppose now that the r.v.’s $\\{\tau_{j,0}^{N}\\}_{1\leq j\leq N}$ are not
ordered, but rather i.i.d., with a common distribution function $G$ which we
assume to be continuous. It then follows from the law of large numbers that
Assumption 2.1 holds in this case.

We make the following assumption on the random function $\lambda$.

######  Assumption 2.2.

Let $\lambda(\cdot)$ be a process having the same law of
$\\{\lambda_{j}^{0}(\cdot)\\}_{j}$ and $\\{\lambda_{i}(\cdot)\\}_{i}$. Assume
that there exists a constant $\lambda^{*}$ such that for each $0<T<\infty$,
$\sup_{t\in[0,T]}\lambda(t)\leq\lambda^{*}$ almost surely. Assume that there
exist an integer $k$, a random sequence
$0=\zeta^{0}\leq\zeta^{1}\leq\cdots\leq\zeta^{k}$ and associated random
functions $\lambda^{\ell}\in C({\mathbb{R}}_{+};[0,\lambda^{\ast}])$,
$1\leq\ell\leq k$, such that

$\displaystyle\lambda(t)=\sum_{\ell=1}^{k}\lambda^{\ell}(t){\mathbf{1}}_{[\zeta^{\ell-1},\zeta^{\ell})}(t).$
(2.11)

In addition, we assume that there exists a deterministic nondecreasing
function $\varphi\in C({\mathbb{R}}_{+};{\mathbb{R}}_{+})$ with $\varphi(0)=0$
such that $|\lambda^{\ell}(t)-\lambda^{\ell}(s)|\leq\varphi(t-s)$ almost
surely for all $t,s\geq 0$ and for all $\ell\geq 1$. Let
$\bar{\lambda}(t)=\mathbb{E}[\lambda_{i}(t)]=\mathbb{E}[\lambda^{0}_{j}(t)]$
and $v(t)=\text{\rm
Var}(\lambda(t))=\mathbb{E}\big{[}\big{(}\lambda(t)-\bar{\lambda}(t)\big{)}^{2}\big{]}$
for $t\geq 0$.

######  Remark 2.3.

Recall that the basic reproduction number $R_{0}$ is the mean number of
susceptible individuals whom an infectious individual infects in a large
population otherwise fully susceptible. In the present model, clearly

$R_{0}=\int_{0}^{\infty}\bar{\lambda}(t)dt\,.$

Suppose that $\lambda_{i}(t)=\tilde{\lambda}(t){\mathbf{1}}_{t<\eta_{i}}$,
where $\tilde{\lambda}(t)$ is a deterministic function. Then

$R_{0}=\int_{0}^{\infty}\tilde{\lambda}(t)F^{c}(t)dt\,.$

In the standard SIR model with $\tilde{\lambda}(t)\equiv\lambda$ and
$\mathbb{E}[\eta]=\int_{0}^{\infty}F^{c}(t)dt$, the formula above reduces to
the well known $R_{0}=\lambda\mathbb{E}[\eta]$. See, e.g., [5]. We obtain the
same formula if the deterministic function $\tilde{\lambda}(t)$ is replaced by
a process $\lambda_{i}(t)$ independent of $\eta_{i}$, with mean
$\tilde{\lambda}(t)$. More precisely, in that case the sequence
$({\lambda}_{i}(t),\eta_{i})_{i\geq 1}$ is assumed to be i.i.d., and for each
$i$, ${\lambda}_{i}$ and $\eta_{i}$ are independent.

The proof of the following Theorem, which is the main result of this section,
will be given in section 5. For a function $u(t,x)\in D_{D_{\uparrow}}$, we
use the equivalent notations $d_{x}u(t,x)$ and $u_{x}(t,x)$ for the partial
derivative w.r.t. $x$, while $u(t,dx)$ denotes the measure whose distribution
function is $x\mapsto u(t,x)$, which coincides with $u_{x}(t,x)dx$ if that
last map is differentiable. In particular, $u_{x}(t,0)$ indicates the partial
derivative evaluated at $x=0$.

######  Theorem 2.1.

Under Assumptions 2.1 and 2.2, as $N\to\infty$,

$\displaystyle\big{(}\bar{S}^{N},\overline{{\mathcal{I}}}^{N},\bar{\mathfrak{I}}^{N},\bar{R}^{N}\big{)}\to\big{(}\bar{S},\overline{{\mathcal{I}}},\bar{\mathfrak{I}},\bar{R}\big{)}\
\text{in probability, locally uniformly in $t$ and $x$},$ (2.12)

where the limits are the unique continuous solution to the following set of
integral equations, for $t,x\geq 0$,

$\displaystyle\bar{S}(t)$
$\displaystyle=1-\bar{I}(0)-\int_{0}^{t}\bar{\Upsilon}(s)ds,$ (2.13)
$\displaystyle\overline{{\mathcal{I}}}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\Upsilon}(s)ds\,,$
(2.14) $\displaystyle\bar{\mathfrak{I}}(t,x)$
$\displaystyle=\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy)+\int_{(t-x)^{+}}^{t}F^{c}(t-s)\bar{\Upsilon}(s)ds,$
(2.15) $\displaystyle\bar{R}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\left(1-\frac{F^{c}(t+y)}{F^{c}(y)}\right)\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}F(t-s)\bar{\Upsilon}(s)ds,$
(2.16)

with

$\bar{\Upsilon}(t)=\bar{S}(t)\overline{{\mathcal{I}}}(t)=\bar{\mathfrak{I}}_{x}(t,0)\,.$
(2.17)

The function $\bar{\mathfrak{I}}(t,x)$ is nondecreasing in $x$ for each $t$.
As a consequence, $\bar{I}^{N}\to\bar{I}$ in $D$ in probability as
$N\to\infty$ where

$\displaystyle\bar{I}(t)=\bar{\mathfrak{I}}(t,t+\bar{x})=\int_{0}^{\bar{x}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}F^{c}(t-s)\bar{\Upsilon}(s)ds,\quad
t\geq 0.$ (2.18)

##  3\. PDE Models

One can regard $\bar{\mathfrak{I}}(t,x)$ as the ‘distribution function’ of
$\bar{I}(t)=\bar{\mathfrak{I}}(t,t+\bar{x})$ over the ‘ages’
$x\in[0,t+\bar{x})$ for each fixed $t$. If $x\mapsto\bar{\mathfrak{I}}(t,x)$
is absolutely continuous, we denote by
$\bar{\mathfrak{i}}(t,x)=\bar{\mathfrak{I}}_{x}(t,x)$ the density function of
$\bar{\mathfrak{I}}(t,x)$ with respect to $x$. Note that $\bar{S}(t)=0$ for
$t<0$ and $\bar{\mathfrak{i}}(t,x)=0$ both for $t<0$ and $x<0$.

###  3.1. The case $F$ absolutely continuous

In this subsection, we assume that $F$ is absolutely continuous,
$F(dx)=f(x)dx$, and we denote by $\mu(x)$ the hazard function of the r.v.
$\eta$, i.e., $\mu(x):=f(x)/F^{c}(x)$ for $x\geq 0$. If the density function
$\bar{\mathfrak{i}}(t,x)$ exists, we obtain the following PDE representation.

######  Proposition 3.1.

Suppose that $F$ is absolutely continuous, with the density $f$, and that
$\bar{\mathfrak{I}}(0,x)$ is differentiable with respect to $x$, with the
density function $\bar{\mathfrak{i}}(0,x)$. Then for $t>0$, the increasing
function $\bar{\mathfrak{I}}(t,\cdot)$ is absolutely continuous, and $(t,x)$
a.e. in $(0,+\infty)^{2}$,

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}$
$\displaystyle=-\mu(x)\bar{\mathfrak{i}}(t,x)\,,$ (3.1)

with the initial condition
$\bar{\mathfrak{i}}(0,x)=\bar{\mathfrak{I}}_{x}(0,x)$ for $x\in[0,\bar{x}]$,
and the boundary condition

$\bar{\mathfrak{i}}(t,0)=\bar{S}(t)\int_{0}^{t+\bar{x}}\frac{\bar{\lambda}(x)}{\frac{F^{c}(x)}{F^{c}(x-t)}}\bar{\mathfrak{i}}(t,x)dx\,,$
(3.2)

with the convention that $F^{c}=1$ on $\mathbb{R}_{-}$, and that the integrand
in (3.2) is zero when $F^{c}(x)=0$.

In addition,

$\bar{S}^{\prime}(t)=-\bar{\mathfrak{i}}(t,0),\quad\text{and }\
\bar{S}(0)=1-\bar{I}(0)\,.$ (3.3)

Moreover, the PDE (3.1) has a unique solution which is given as follows. For
$x\geq t$,

$\bar{\mathfrak{i}}(t,x)=\frac{F^{c}(x)}{F^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)\,,$
(3.4)

while for $t>x$,

$\bar{\mathfrak{i}}(t,x)=F^{c}(x)\bar{\mathfrak{i}}(t-x,0)\,,$ (3.5)

and the boundary function is the unique solution of the integral equation

$\bar{\mathfrak{i}}(t,0)=\left(\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\right)\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right)\,.$
(3.6)

######  Remark 3.1.

The PDE (3.1) can be considered as a linear equation, with a nonlinear
boundary condition which is the integral equation (3.6).

It follows from (3.4) and (3.5) that $F^{c}(x)=0$ implies that
$\bar{\mathfrak{i}}(t,x)=0$. This is why we can impose that the integrand in
the right hand side of (3.2) is zero whenever $F^{c}(x)=0$.

We remark that the PDE given in [19] resembles that given in (3.1), see
equations (28)–(29), see also equation (2.2) in [15]. In particular, the
function $\mu(x)$ is interpreted as the recovery rate at infection age $x$.
Equivalently, it is the hazard function of the infectious duration.

######  Remark 3.2.

In a sense, what we do in the present paper can be interpreted as follows: we
replace the two–dimensional system of equations with memory (2.13)–(2.14)
(with, see (2.17), $\bar{\Upsilon}(t)$ replaced by
$\bar{S}(t)\overline{{\mathcal{I}}}(t)$) by the infinite dimensional system of
ODE-PDE (2.13)–(3.1)-(3.2) (with, see again (2.17), $\bar{\Upsilon}(t)$
replaced by $\bar{\mathfrak{i}}(t,0)$).

At the level of our population of size $N$, we have a two–dimensional
non–Markov process $(S^{N}(t),\mathcal{I}^{N}(t))$. For any $t\geq 0$, let
$\bar{\mathfrak{i}}^{N}(t)$ denote the measure whose distribution function is
$x\mapsto\bar{\mathfrak{I}}^{N}(t,x)$. Theorem 2.1 implies that locally
uniformly in $t$, $\bar{\mathfrak{i}}^{N}(t)$ converges weakly to the measure
which has the density $\bar{\mathfrak{i}}(t,x)$ w.r.t. Lebesque’s measure.
$\bar{\mathfrak{i}}^{N}(t)$ is a point measure which assigns the mass $N^{-1}$
to any $x$ which is the infection age of one of the individuals infected at
time $t$. Clearly, from the knowledge of $\bar{\mathfrak{i}}^{N}(t)$, we can
deduce the values of both $I^{N}(0)$ and $A^{N}(t)$, hence of $S^{N}(0)$ and
of $S^{N}(t)$ (see (2.9)). Note that the points of the measure
$\bar{\mathfrak{i}}^{N}(t)$ which are larger than $t$ are the
$\\{\tilde{\tau}^{N}_{j,0}+t,1\leq j\leq I^{N}(0)\\}$, and those which are
less than $t$ are the $\\{t-\tau^{N}_{i},1\leq i\leq A^{N}(t)\\}$. Hence from
(2.2), $\mathcal{I}^{N}(t)$ is a function of both $\bar{\mathfrak{i}}^{N}(t)$
and the $\lambda_{i}$’s. The same is true for $\Upsilon^{N}(t)$. Conditionally
upon the $\lambda_{i}$’s, the process $\bar{\mathfrak{i}}^{N}(t)$ is a
measure-valued Markov process, which evolves as follows. Each point $x$ which
belongs to it increases at speed $1$, dies at rate $\mu(x)$, and new points
are added at rate $\Upsilon^{N}(t)$. $\bar{\mathfrak{i}}^{N}(t)$ is determined
by a sequence of at most $N$ positive numbers; it can be considered as an
element of $\cup_{k=1}^{N}\mathbb{R}^{k}$. We have “Markovianized” the
two–dimensional non–Markov process $(S^{N}(t),\mathcal{I}^{N}(t))$, at the
price of increasing dramatically the dimension.

Note that the pair composed of $\bar{\mathfrak{i}}^{N}(t)$ and the collection
$\\{\lambda^{0}_{j},1\leq j\leq I^{N}(0);\lambda_{i},1\leq i\leq A^{N}(t)\\}$
is a Markov process with values in $\cup_{k=1}^{N}(\mathbb{R}\times D)^{k}$.
$\bar{\mathfrak{i}}^{N}(t)$ evolves as above, and each new $\lambda_{i}$ is a
random element of $D$ with the same law, independent of everything else.

We expect to write and study the equation for the measure–valued Markov
process $\bar{\mathfrak{i}}^{N}(t)$ in a future work.

######  Remark 3.3.

Recall the special case in Remark 2.3 with
$\lambda_{i}(t)=\tilde{\lambda}(t){\mathbf{1}}_{t<\eta_{i}}$, where
$\tilde{\lambda}(t)$ is a deterministic function. Then
$\bar{\lambda}(t)=\tilde{\lambda}(t)F^{c}(t)$, and
$\mathbb{E}\big{[}\lambda^{0}(t)|\tilde{\tau}^{N}_{0}=y\big{]}=\tilde{\lambda}(t+y)\frac{F^{c}(t+y)}{F^{c}(y)}$.
In that case, the boundary condition in (3.2) becomes

$\bar{\mathfrak{i}}(t,0)=\bar{S}(t)\int_{0}^{t+\bar{x}}\tilde{\lambda}(x)\bar{\mathfrak{i}}(t,x)dx$

This is usually how the boundary condition is imposed in the literature of PDE
epidemic models (see, e.g., [15, equation (2.5)], [22, equation (1.1)] and
[12, equation (2)]). This expression has clearly a very intuitive
interpretation. $\bar{\mathfrak{i}}(t,0)$ is the instantaneous rate for an
individual to get infected at time $t$ (resulting in a newly infectious
individual with a zero age of infection), while the right hand side is the
instantaneous infection rate by the existing infectious population at time
$t$, which depends on all the infectious individuals with all ages of
infection. This of course includes time $t=0$, which formulates a constraint
on the initial condition $\\{\bar{\mathfrak{I}}(0,x)\\}_{0\leq x\leq\bar{x}}$.

###### Proof.

By the fact that $F$ has a density, we see that the two partial derivatives of
$\bar{\mathfrak{I}}$ exist $(t,x)$ a.e. From (2.15), they satisfy

$\displaystyle\bar{\mathfrak{I}}_{t}(t,x)$ $\displaystyle=-{\mathbf{1}}_{x\geq
t}\frac{F^{c}(x)}{F^{c}(x-t)}\bar{\mathfrak{I}}_{x}(0,x-t)-\int_{0}^{(x-t)^{+}}\frac{f(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}_{x}(0,y)dy$
$\displaystyle\qquad+\bar{\mathfrak{I}}_{x}(t,0)-{\mathbf{1}}_{t>x}F^{c}(x)\bar{\mathfrak{I}}_{x}(t-x,0)-\int_{(t-x)^{+}}^{t}f(t-s)\bar{\mathfrak{I}}_{x}(s,0)ds,$
(3.7)

and

$\displaystyle\bar{\mathfrak{I}}_{x}(t,x)$ $\displaystyle={\mathbf{1}}_{x\geq
t}\frac{F^{c}(x)}{F^{c}(x-t)}\bar{\mathfrak{I}}_{x}(0,x-t)+{\mathbf{1}}_{t>x}F^{c}(x)\bar{\mathfrak{I}}_{x}(t-x,0).$
(3.8)

Thus, summing up (3.1) and (3.8), we obtain for $t>0$ and $x>0$,

$\displaystyle\bar{\mathfrak{I}}_{t}(t,x)+\bar{\mathfrak{I}}_{x}(t,x)$
$\displaystyle=-\int_{0}^{(x-t)^{+}}\frac{f(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}_{x}(0,y)dy+\bar{\mathfrak{I}}_{x}(t,0)-\int_{(t-x)^{+}}^{t}f(t-s)\bar{\mathfrak{I}}_{x}(s,0)ds\,.$
(3.9)

Denote
$\bar{\mathfrak{I}}_{x,t}(t,x)=\frac{\partial^{2}\bar{\mathfrak{I}}(t,x)}{\partial
x\partial t}=\frac{\partial}{\partial x}\bar{\mathfrak{I}}_{t}(t,x)$ and
$\bar{\mathfrak{I}}_{x,x}(t,x)=\frac{\partial^{2}\bar{\mathfrak{I}}(t,x)}{\partial
x\partial x}$. By taking the derivative on both sides of (3.9) with respect to
$x$ (possibly in the distributional sense for each term on the left), we
obtain for $t>0$ and $x>0$,

$\displaystyle\bar{\mathfrak{I}}_{x,t}(t,x)+\bar{\mathfrak{I}}_{x,x}(t,x)$
$\displaystyle=-{\mathbf{1}}_{x\geq
t}\frac{f(x)}{F^{c}(x-t)}\bar{\mathfrak{I}}_{x}(0,x-t)-{\mathbf{1}}_{t>x}f(x)\bar{\mathfrak{I}}_{x}(t-x,0).$
(3.10)

Since $\frac{\partial^{2}\bar{\mathfrak{I}}(t,x)}{\partial x\partial
t}=\frac{\partial^{2}\bar{\mathfrak{I}}(t,x)}{\partial t\partial x}$, we
obtain the expression

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}$
$\displaystyle=-{\mathbf{1}}_{x\geq
t}\frac{f(x)}{F^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)-{\mathbf{1}}_{t>x}f(x)\bar{\mathfrak{i}}(t-x,0)\,.$
(3.11)

As concerns the boundary condition, we note that, given (3.3), (3.4) and
(3.5), (3.2) and (3.6) are equivalent. Hence we will establish (3.6), (3.3),
(3.4) and (3.5).

For the boundary condition $\bar{\mathfrak{i}}(t,0)$, by (2.14) and (2.17), we
have

$\displaystyle\bar{\mathfrak{i}}(t,0)=\bar{S}(t)\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right),$

where by (2.13),

$\bar{S}(t)=\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\,.$

Thus we obtain the expression in (3.6). We next prove that equation (3.6) has
a unique non–negative solution. Observe that $u(t)=\bar{\mathfrak{i}}(t,0)$ is
also a solution to

$u(t)=\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)u(s)ds\right)\left(\bar{S}(0)-\int_{0}^{t}u(s)ds\right)^{+},$
(3.12)

and any non–negative solution of (3.6) solves (3.12).

First, note that since for any $t\geq 0$,
$0\leq\bar{\lambda}(t)\leq\lambda^{\ast}$,

$0\leq\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy\leq\lambda^{\ast}\bar{I}(0),$
(3.13)

from which we conclude that $\int_{0}^{t}u(s)ds\leq\bar{S}(0)$. Indeed, if
that were not the case, there would exist a time $T_{\bar{S}(0)}<t$ such that
$\int_{0}^{T_{\bar{S}(0)}}u(s)ds=\bar{S}(0)$, hence
$\int_{0}^{t}u(s)ds\geq\bar{S}(0)$ and from (3.12), we would have $u(t)=0$ for
any $t\geq T_{\bar{S}(0)}$, so that indeed $\int_{0}^{t}u(s)ds\leq\bar{S}(0)$.

Under Assumption 2.2, using (3.13), if $u_{1}(t)$ and $u_{2}(t)$ are two
nonnegative integrable solutions, then

$\displaystyle|u_{1}(t)-u_{2}(t)|$
$\displaystyle\leq\bar{S}(0)\int_{0}^{t}\bar{\lambda}(t-s)|u_{1}(s)-u_{2}(s)|ds+\lambda^{*}(\bar{I}(0)+\bar{S}(0))\int_{0}^{t}|u_{1}(s)-u_{2}(s)|ds$
$\displaystyle\leq 2\lambda^{*}\int_{0}^{t}|u_{1}(s)-u_{2}(s)|ds\,,$

which, combined with Gronwall’s Lemma, implies that $u_{1}\equiv u_{2}$. Now
existence is provided by the fact that the function $\bar{\mathfrak{i}}(t,0)$
is a non–negative solution of (3.12).

Note also that clearly, using a combination of an argument similar to that
used for uniqueness, and of the classical estimate on Picard iterations for
ODEs, one could establish that the sequence defined by $u^{(0)}(t)\equiv 0$
and for $n\geq 0$,

$\displaystyle
u^{(n+1)}(t)=\left(\bar{S}(0)-\int_{0}^{t}u^{(n)}(s)ds\right)\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)u^{(n)}(s)ds\right),$

given $\bar{\mathfrak{i}}(0,\cdot)$, is a Cauchy sequence in
$C({\mathbb{R}}_{+})$, hence existence.

We next derive the explicit solution expressions in (3.4) and (3.5). It
follows from (3.11)Â that for $x\geq t$, $0\leq s\leq t$,

$\frac{\partial\bar{\mathfrak{i}}}{\partial
s}(s,x-t+s)=-\frac{f(x-t+s)}{F^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t),$

while for $t>x$, $0\leq s\leq x$,

$\frac{\partial\bar{\mathfrak{i}}}{\partial
s}(t-x+s,s)=-f(s)\bar{\mathfrak{i}}(t-x,0)\,.$

Integrating the first identity from $s=0$ to $s=t$, we deduce that for $x\geq
t$,

$\displaystyle\bar{\mathfrak{i}}(t,x)$
$\displaystyle=\bar{\mathfrak{i}}(0,x-t)-\frac{\bar{\mathfrak{i}}(0,x-t)}{F^{c}(x-t)}\int_{0}^{t}f(x-t+s)ds$
$\displaystyle=\bar{\mathfrak{i}}(0,x-t)\left(1-\frac{F(x)-F(x-t)}{F^{c}(x-t)}\right)$
$\displaystyle=\frac{F^{c}(x)}{F^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)\,,$

so that for $x\geq t$,

$\frac{f(x)}{F^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)=\frac{f(x)}{F^{c}(x)}\bar{\mathfrak{i}}((t,x)\,.$
(3.14)

Now for $t>x$, we integrate the second identity from $s=0$ to $s=x$, and get

$\displaystyle\bar{\mathfrak{i}}(t,x)$
$\displaystyle=\bar{\mathfrak{i}}(t-x,0)-\bar{\mathfrak{i}}(t-x,0)\int_{0}^{x}f(s)ds$
$\displaystyle=F^{c}(x)\bar{\mathfrak{i}}(t-x,0)\,.$

Hence for $t>x$,

$f(x)\bar{\mathfrak{i}}(t-x,0)=\frac{f(x)}{F^{c}(x)}\bar{\mathfrak{i}}(t,x)\,.$
(3.15)

Clearly, (3.4) is equivalent to (3.14), (3.5) is equivalent to (3.15), and
(3.1) follows from (3.11), (3.14) and (3.15). ∎

######  Corollary 3.1.

The formula (2.14) for $\overline{{\mathcal{I}}}(t)$ can be rewritten

$\overline{{\mathcal{I}}}(t)=\int_{0}^{t+\bar{x}}\frac{\bar{\lambda}(y)}{\frac{F^{c}(y)}{F^{c}(y-t)}}\bar{\mathfrak{I}}(t,dy),$
(3.16)

where $F^{c}(z)=1$, for $z\leq 0$.

###### Proof.

We first deduce from (2.15) that for $t>x$, $x\mapsto\bar{\mathfrak{I}}(t,x)$
is differentiable, and $\bar{\mathfrak{I}}(t,dx)=F^{c}(x)\bar{\Upsilon}(t-x)$,
and for fixed $t$, on $[0,\bar{x}]$, the function
$y\to\bar{\mathfrak{I}}(t,t+y)$ is of finite total variation and satisfies
$\bar{\mathfrak{I}}(t,t+dy)=\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy)$.
Inserting the resulting formulas for $\bar{\mathfrak{I}}(0,dy)$ and
$\bar{\Upsilon}$ in the first and second integrals of the right hand side of
(2.14), we obtain

$\displaystyle\overline{{\mathcal{I}}}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(t+y)\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}\frac{\bar{\lambda}(x)}{F^{c}(x)}\bar{\mathfrak{I}}(t,dx)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(t+y)\frac{F^{c}(y)}{F^{c}(t+y)}\bar{\mathfrak{I}}(t,t+dy)+\int_{0}^{t}\frac{\bar{\lambda}(x)}{F^{c}(x)}\bar{\mathfrak{I}}(t,dx)\,,$

from which the result follows. ∎

######  Remark 3.4.

In the special case
$\lambda_{i}(t)=\tilde{\lambda}(t){\mathbf{1}}_{t<\eta_{i}}$ as discussed in
Remark 3.3, (3.16) reduces to the very simple formula

$\overline{{\mathcal{I}}}(t)=\int_{0}^{t+\bar{x}}\tilde{\lambda}(y)\bar{\mathfrak{I}}(t,dy)\,.$
(3.17)

A similar formula holds if we replace the deterministic function
$\tilde{\lambda}(t)$ by a copy $\lambda_{i}(t)$ of a random function, which is
independent of $\eta_{i}$, as discussed in Remark 2.3, and whose expectation
is $\tilde{\lambda}(t)$. Then, we have

$\displaystyle\bar{\Upsilon}(t)=\bar{S}(t)\overline{\mathcal{I}}(t)$
$\displaystyle=\bar{S}(t)\int_{0}^{t+\bar{x}}\tilde{\lambda}(x)\bar{\mathfrak{I}}(t,dx)$
$\displaystyle=\bar{S}(t)\int_{0}^{t+\bar{x}}\tilde{\lambda}(x)\bar{\mathfrak{i}}(t,x)dx$
(3.18)

Since $\bar{\mathfrak{i}}(t,0)=\bar{\Upsilon}(t)$, the results above can be
stated using this expression of $\bar{\Upsilon}$.

In the special case of exponentially distributed infectious periods, i.e.
$\mu(x)\equiv\mu$, we obtain the following well known results, see, e.g., [28,
16, 22].

######  Corollary 3.2.

If the c.d.f. $F(t)=1-e^{-\mu t}$, we have for $t>0$ and $x>0$,

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
x}=-\mu\bar{\mathfrak{i}}(t,x)$ (3.19)

with the initial condition $\bar{\mathfrak{i}}(0,x)$ given for
$x\in[0,\bar{x}]$ and the boundary condition for $\bar{\mathfrak{i}}(t,0)$ as
given in (3.6).

###### Proof.

In this case, the above proof simplifies. Indeed, we have for $t\geq 0$ and
$x\geq 0$,

$\displaystyle\bar{\mathfrak{I}}(t,x)$ $\displaystyle={\mathbf{1}}_{x\geq
t}e^{-\mu
t}\bar{\mathfrak{I}}(0,x-t)+\int_{(t-x)^{+}}^{t}e^{-\mu(t-s)}\bar{\Upsilon}(s)ds.$

By taking derivative with respect to $x$ when $t>0$ and $x>0$, we obtain that
equation (3.8) becomes

$\displaystyle\bar{\mathfrak{i}}(t,x)$ $\displaystyle={\mathbf{1}}_{x\geq
t}\bar{\mathfrak{i}}(0,x-t)e^{-\mu t}+{\mathbf{1}}_{x<t}e^{-\mu
x}\bar{\mathfrak{I}}_{x}(t-x,0).$

Taking derivatives of this equation with respect to $t$ and $x$, we obtain for
$t>0$ and $x>0$,

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}$
$\displaystyle=-{\mathbf{1}}_{x\geq t}\bar{\mathfrak{i}}(0,x-t)\mu e^{-\mu
t}-\mu e^{-\mu x}\bar{\mathfrak{I}}_{x}(t-x,0)$
$\displaystyle=-\mu\bar{\mathfrak{i}}(t,x).$

The boundary conditions follow in the same way as in the general model. ∎

######  Remark 3.5.

If the remaining infectious periods of the initially infectious individuals
$\\{\eta^{0}_{j},j=1,\dots,I^{N}(0)\\}$ are i.i.d. with c.d.f. $F_{0}$ instead
of depending on the infection age in (2.1), then we obtain the limits

$\displaystyle\bar{\mathfrak{I}}(t,x)$
$\displaystyle=\bar{\mathfrak{I}}(0,(x-t)^{+})F_{0}^{c}(t)+\int_{(t-x)^{+}}^{t}F^{c}(t-s)\bar{\Upsilon}(s)ds,$
$\displaystyle\bar{R}(t)$
$\displaystyle=\bar{I}(0)F_{0}(t)+\int_{0}^{t}F(t-s)\bar{\Upsilon}(s)ds,$

(noting that they are not continuous unless $F_{0}$ is continuous), and
assuming the density functions exist, we obtain the PDE:

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}=-\left({\bf
1}_{x>t}\mu_{0}(t)+{\bf 1}_{t>x}\mu(x)\right)\bar{\mathfrak{i}}(t,x),$

where $\mu_{0}(t)=f_{0}(t)/F^{c}_{0}(t)$. As in the proof of Proposition 3.1,
we obtain that the PDE (3.1) has a unique solution which is given as follows.
For $x\geq t$,

$\bar{\mathfrak{i}}(t,x)=F_{0}^{c}(t)\bar{\mathfrak{i}}(0,x-t)\,,$

while for $t>x$,

$\bar{\mathfrak{i}}(t,x)=F^{c}(x)\bar{\mathfrak{i}}(t-x,0)\,,$

and the boundary function is the unique solution of the integral equation

$\bar{\mathfrak{i}}(t,0)=\left(\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\right)\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right)\,.$

###  3.2. The general case

We now generalize the result of Proposition 3.1 to the case where the
distribution $F$ is not absolutely continuous. We denote below by $\nu$ the
law of $\eta$, i.e. the measure whose distribution function is $F$. For
reasons which will be explained in Remark 3.6 below, we shall in this
subsection use the left continuous versions of $F$ and $F^{c}$. In order to
simplify notations, we define

$G(t)=F(t^{-}),\ \ \ G^{c}(t)=1-G(t)=F^{c}(t^{-})\,.$

######  Proposition 3.2.

Suppose that $\bar{\mathfrak{I}}(0,x)$ is differentiable with respect to $x$,
with the density function $\bar{\mathfrak{i}}(0,x)$. Then for $t>0$, the
increasing function $\bar{\mathfrak{I}}(t,\cdot)$ is absolutely continuous,
and the following identity holds:

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}$
$\displaystyle=-\frac{\bar{\mathfrak{i}}(t,x)}{G^{c}(x)}\nu(dx)\,,$ (3.20)

(i.e., the distribution which appears on the left hand side of (3.20) equals
the measure which has the density $-\frac{\bar{\mathfrak{i}}(t,x)}{G^{c}(x)}$
with respect to the measure $\nu$) with the initial condition
$\bar{\mathfrak{i}}(0,x)=\bar{\mathfrak{I}}_{x}(0,x)$ for $x\in[0,\bar{x}]$,
and the boundary condition

$\bar{\mathfrak{i}}(t,0)=\bar{S}(t)\int_{0}^{t+\bar{x}}\frac{\bar{\lambda}(x)}{\frac{G^{c}(x)}{G^{c}(x-t)}}\bar{\mathfrak{i}}(t,x)dx\,,$
(3.21)

with the convention that $G^{c}=1$ on $\mathbb{R}_{-}$, and that the integrand
in (3.21) is zero whenever $G^{c}(x)=0$.

In addition,

$\bar{S}^{\prime}(t)=-\bar{\mathfrak{i}}(t,0),\quad\text{and }\
\bar{S}(0)=1-\bar{I}(0)\,.$ (3.22)

Moreover, the PDE (3.20) has a unique solution which is given as follows. For
$x\geq t$,

$\bar{\mathfrak{i}}(t,x)=\frac{G^{c}(x)}{G^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)\,,$
(3.23)

while for $t>x$,

$\bar{\mathfrak{i}}(t,x)=G^{c}(x)\bar{\mathfrak{i}}(t-x,0)\,,$ (3.24)

and the boundary function is the unique solution of the integral equation

$\bar{\mathfrak{i}}(t,0)=\left(\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\right)\left(\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right)\,.$
(3.25)

######  Remark 3.6.

The product $\frac{\bar{\mathfrak{i}}(t,x)}{G^{c}(x)}\nu(dx)$ can also be
rewritten as

$\bar{\mathfrak{i}}(t,x)\times\frac{\nu(dx)}{G^{c}(x)}\,,$

where the second factor can be thought of as the “hazard measure”, i.e., the
generalization of the hazard function, of the r.v. $\eta$. The reason why we
want to have $G^{c}(x)$ in the denominator, and not $F^{c}(x)$ is the
following. If the support of $\nu$ is $[0,x_{max}]$, and
$\nu(\\{x_{max}\\})>0$, then $F^{c}(x_{max})=0$, while $G^{c}(x_{max})>0$ and
we need a positive denominator at the point $x_{max}$, since
$\nu(\\{x_{max}\\})>0$.

For consistency, in the present subsection we always choose the left
continuous version $G$ (resp. $G^{c}$) of $F$ (resp. $F^{c}$). Of course, in
the case where $F$ is absolutely continuous this makes no difference.

######  Remark 3.7.

Remark 3.2 can be extended to the present case of a general distribution
function $F$, replacing the infinite dimensional system of ODE-PDE
(2.13)–(3.1)-(3.2) by (2.13)–(3.20)-(3.21).

###### Proof.

We first rewrite equation (2.15) as

$\displaystyle\bar{\mathfrak{I}}(t,x)=\int_{0}^{(x-t)^{+}}\frac{\bar{\mathfrak{I}}_{x}(0,y)}{G^{c}(y)}G^{c}(t+y)dy+\int_{(t-x)^{+}}^{t}\bar{\mathfrak{I}}_{x}(s,0)G^{c}(t-s)ds\,.$

Differentiating $\bar{\mathfrak{I}}(t,x)$ in $x$ can be done exactly as in the
proof of Proposition 3.1. Concerning the differentiation in $t$, the
differentiation with respect to $t$ appearing in the integrands $G^{c}(t+y)$
and $G^{c}(t-s)$ is now a bit more delicate: those functions have not been
assumed to be differentiable. Their derivatives in the distributional sense is
a measure, whose bracket with a measurable bounded function makes sense, so
that

$\displaystyle\frac{d}{dt}\int_{a}^{b}\frac{\bar{\mathfrak{I}}_{x}(0,y)}{G^{c}(y)}G^{c}(t+y)dy$
$\displaystyle=\int_{a}^{b}\frac{\bar{\mathfrak{I}}_{x}(0,y)}{G^{c}(y)}\nu(t+dy)$
$\displaystyle=\int_{a+t}^{b+t}\frac{\bar{\mathfrak{I}}_{x}(0,z-t)}{G^{c}(z-t)}\nu(dz),$
$\displaystyle\frac{d}{dt}\int_{a}^{b}G^{c}(t-s)\bar{\mathfrak{I}}_{x}(s,0)ds$
$\displaystyle=\int_{a}^{b}\bar{\mathfrak{I}}_{x}(s,0)\nu(t-ds)$
$\displaystyle=\int_{t-b}^{t-a}\bar{\mathfrak{I}}_{x}(t-r,0)\nu(dr)\,.$

As a consequence, the above modification in the proof of Proposition 3.1
yields

$\displaystyle\bar{\mathfrak{I}}_{t}(t,x)+\bar{\mathfrak{I}}_{x}(t,x)=\bar{\mathfrak{I}}_{x}(t,0)-\int_{t}^{x\vee
t}\frac{\bar{\mathfrak{I}}_{x}(0,z-t)}{G^{c}(z-t)}\nu(dz)-\int_{0}^{x\wedge
t}\bar{\mathfrak{I}}_{x}(t-r,0)\nu(dr)\,.$

Differentiating with respect to $x$ finally yields

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}=-{\bf 1}_{x\geq
t}\frac{\bar{\mathfrak{i}}(0,x-t)}{G^{c}(x-t)}\nu(dx)-{\bf
1}_{x<t}\bar{\mathfrak{i}}(t-x,0)\nu(dx)\,.$

We thus deduce that for $x\geq t$, $0\leq s\leq t$,

$\frac{\partial\bar{\mathfrak{i}}}{\partial
s}(s,x-t+s)=-\frac{\bar{\mathfrak{i}}(0,x-t)}{G^{c}(x-t)}\nu(x-t+ds),$

while for $t>x$, $0\leq s\leq x$,

$\frac{\partial\bar{\mathfrak{i}}}{\partial
s}(t-x+s,s)=-\bar{\mathfrak{i}}(t-x,0)\nu(ds)\,.$

Let us integrate the first identity on the interval $[0,t)$. We get

$\displaystyle\bar{\mathfrak{i}}(t,x)$
$\displaystyle=\bar{\mathfrak{i}}(0,x-t)\left(1-\frac{\int_{[0,t)}\nu(x-t+ds)}{G^{c}(x-t)}\right)$
$\displaystyle=\bar{\mathfrak{i}}(0,x-t)\left(1-\frac{G(x)-G(x-t)}{G^{c}(x-t)}\right)$
$\displaystyle=\frac{G^{c}(x)}{G^{c}(x-t)}\bar{\mathfrak{i}}(0,x-t)\,.$

We conclude that for $x\geq t$,

$\frac{\bar{\mathfrak{i}}(0,x-t)}{G^{c}(x-t)}\nu(dx)=\frac{\bar{\mathfrak{i}}(t,x)}{G^{c}(x)}\nu(dx)\,.$

We finally consider the case $t>x$, and integrate the second identity on the
interval $[0,x)$, yielding:

$\displaystyle\bar{\mathfrak{i}}(t,x)$
$\displaystyle=\bar{\mathfrak{i}}(t-x,0)\left(1-\int_{[0,x)}\nu(ds)\right)$
$\displaystyle=G^{c}(x)\bar{\mathfrak{i}}(t-x,0),$

so that, for $t>x$,

$\bar{\mathfrak{i}}(t-x,0)\nu(dx)=\frac{\bar{\mathfrak{i}}(t,x)}{G^{c}(x)}\nu(dx),$

and we have established (3.23), (3.24), as well as (3.20). The rest of the
proof is the same as that of Proposition 3.1. ∎

The case of a deterministic duration $\eta$ is a particular case of the last
Proposition.

######  Corollary 3.3.

Suppose that the infectious periods are deterministic and equal to $t_{i}$,
i.e., $F(t)={\mathbf{1}}_{t\geq t_{i}}$, $G(t)={\mathbf{1}}_{t>t_{i}}$. Then
we have

$\displaystyle\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial
t}+\frac{\partial\bar{\mathfrak{i}}(t,x)}{\partial x}$
$\displaystyle=-\delta_{t_{i}}(x)\bar{\mathfrak{i}}(t,x)\,,$ (3.26)

with $\delta_{t_{i}}(x)$ being the Dirac measure at $t_{i}$, with the initial
condition $\bar{\mathfrak{i}}(0,x)=\partial_{x}\bar{\mathfrak{I}}(0,x)$ for
$x\in[0,t_{i}]$, and the boundary condition

$\bar{\mathfrak{i}}(t,0)=\bar{S}(t)\int_{0}^{(t+t_{i})\wedge
t_{i}}\bar{\lambda}(x)\bar{\mathfrak{i}}(t,x)dx\,,$ (3.27)

Note also that the boundary function $\bar{\mathfrak{i}}(t,0)$ solves the
following Volterra equation: if $0<t<t_{i}$,

$\bar{\mathfrak{i}}(t,0)=\left(\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\right)\left(\int_{t}^{t_{i}}\bar{\lambda}(y)\bar{\mathfrak{i}}(0,y-t)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right),$
(3.28)

and if $t\geq t_{i}$,

$\bar{\mathfrak{i}}(t,0)=\left(\bar{S}(0)-\int_{0}^{t}\bar{\mathfrak{i}}(s,0)ds\right)\int_{0}^{t_{i}}\bar{\lambda}(y)\bar{\mathfrak{i}}(t-y,0)dy\,,$
(3.29)

The PDE (3.26) has a unique solution $\bar{\mathfrak{i}}(t,x)$, which is given
as follows. $\bar{\mathfrak{i}}(t,x)=0$ if $x\geq t_{i}$. For $t\leq x<t_{i}$,

$\bar{\mathfrak{i}}(t,x)=\bar{\mathfrak{i}}(0,x-t)\,,$ (3.30)

while for $x<t\wedge t_{i}$,

$\bar{\mathfrak{i}}(t,x)=\bar{\mathfrak{i}}(t-x,0).$ (3.31)

######  Remark 3.8.

The total fraction of the population infected during the epidemic is given by

$\displaystyle\Phi=\int_{0}^{\infty}\bar{\mathfrak{i}}(t,0)dt$

where $\bar{\mathfrak{i}}(t,0)$ is the solution to (3.6). We also refer the
reader to equation (12) in Kaplan [17], based on his constructed “Scratch”
model.

##  4\. On the SIS model with infection-age dependent infectivity

In the SIS model, the infectious individuals become susceptible once they
recover. Since $S^{N}(t)+I^{N}(t)=N$ for each $t\geq 0$ with a population size
$N$, the epidemic dynamics is determined by the process $I^{N}(t)$ alone, and
we have the same representations of the processes $\mathfrak{I}^{N}_{0}(t,x)$
and $\mathfrak{I}^{N}_{1}(t,x)$ in (2.5) and (2.1), respectively, while in the
representations of $A^{N}$ in (2.4) and $\Upsilon^{N}$ in (2.3), the process
$S^{N}(t)$ is replaced by $S^{N}(t)=N-I^{N}(t)$. The aggregate infectivity
process ${\mathcal{I}}^{N}(t)$ is still given by (2.2). The two processes
$(\mathfrak{I}^{N},{\mathcal{I}}^{N})$ determine the dynamics of the SIS
epidemic model. Under Assumptions 2.1 and 2.2,

$\displaystyle(\overline{{\mathcal{I}}}^{N},\bar{\mathfrak{I}}^{N})\to(\overline{{\mathcal{I}}},\bar{\mathfrak{I}})\
\text{in probability, locally uniformly in $t$ and $x$},\quad\mbox{as}\quad
N\to\infty,$ (4.1)

where

$\displaystyle\overline{{\mathcal{I}}}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}\bar{\lambda}(t-s)\big{(}1-\bar{\mathfrak{I}}(s,\infty)\big{)}\bar{{\mathcal{I}}}(s)ds\,,$
(4.2) $\displaystyle\bar{\mathfrak{I}}(t,x)$
$\displaystyle=\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy)+\int_{(t-x)^{+}}^{t}F^{c}(t-s)\big{(}1-\bar{\mathfrak{I}}(s,\infty)\big{)}\bar{{\mathcal{I}}}(s)ds\,,$
(4.3)

for $t,x\geq 0$. If $\mathfrak{I}(0,x)$ is differentiable and $F$ is
absolutely continuous, then the density function
$\bar{\mathfrak{i}}(t,x)=\frac{\partial\bar{\mathfrak{I}}(t,x)}{\partial x}$
exists and satisfies again (3.1). The same calculations as in the case of the
SIR model lead to (3.4), (3.5) and (3.2). However, the formula for
$\bar{S}(t)$ is different in the case of the SIS model, that is, (3.3) does
not hold. Instead, we have

$\displaystyle\bar{S}(t)$
$\displaystyle=1-\bar{I}(t)=1-\int_{0}^{\bar{x}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{i}}(0,y)dy-\int_{0}^{t}F^{c}(t-s)\bar{\mathfrak{i}}(s,0)ds\,.$
(4.4)

Thus, the Volterra equation on the boundary reads

$\begin{split}\bar{\mathfrak{i}}(t,0)&=\left(\int_{0}^{\bar{x}}\bar{\lambda}(t+y)\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)\bar{\mathfrak{i}}(s,0)ds\right)\\\
&\quad\times\left(1-\int_{0}^{\bar{x}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{i}}(0,y)dy-\int_{0}^{t}F^{c}(t-s)\bar{\mathfrak{i}}(s,0)ds\right)\,,\end{split}$
(4.5)

whose form is similar to the one for the SIR model.

It is also clear that if the c.d.f. $F(t)=1-e^{-\beta t}$, we have the same
PDE for $\bar{\mathfrak{i}}(t,x)$ as given in (3.19) with $\mu(x)=\beta$ and
the boundary condition:

$\begin{split}\bar{\mathfrak{i}}(t,0)&=\left(\int_{0}^{\bar{x}}\bar{\lambda}(t+y)e^{-\beta
y}\,\bar{\mathfrak{i}}(0,y)dy+\int_{0}^{t}\bar{\lambda}(t-s)e^{-\beta(t-s)}\,\bar{\mathfrak{i}}(s,0)ds\right)\\\
&\quad\times\left(1-\int_{0}^{\bar{x}}e^{-\beta
t}\,\bar{\mathfrak{i}}(0,y)dy-\int_{0}^{t}e^{-\beta(t-s)}\,\bar{\mathfrak{i}}(s,0)ds\right)\,.\end{split}$

If the c.d.f. $F$ of the infectious period is not absolutely continuous, but
$\mathfrak{I}(0,x)$ is differentiable, then we have essentially the same
result as in Proposition 3.2, except that (3.22) is replaced by (4.4), and
(3.25) by (4.5).

Recall that the standard SIS model has a nontrivial equilibrium point
$\bar{I}^{*}=1-\beta/\lambda$ if $\beta<\lambda$, where $\lambda$ is the
infection rate (the bar over $\lambda$ is dropped for convenience), and
$1/\beta$ is the mean of the infectious periods. See Section 4.3 in [25] for
the account of the SIS model with general infectious periods. Here we consider
the model in the generality of infection-age dependent infectivity. Note that
we provide the explicit expressions for the equilibria below assuming they
exist. We do not prove the existence of the limit of $\bar{\mathfrak{I}}(t,x)$
as $t\to\infty$, which we leave as future work.

######  Proposition 4.1.

Suppose that $\lim\bar{\mathfrak{I}}(t,x)\to\bar{\mathfrak{I}}^{\ast}(x)$
exists as $t\to\infty$ and $\bar{I}^{\ast}=\bar{\mathfrak{I}}^{\ast}(\infty)$.
If $R_{0}=\int_{0}^{\infty}\bar{\lambda}(y)dy\leq 1$, $\bar{I}^{\ast}=0$ (the
disease free equilibrium). In the complementary case,
$R_{0}=\int_{0}^{\infty}\bar{\lambda}(y)dy>1$, if
$\bar{\mathfrak{I}}(0,\bar{x})>0$,

$\bar{I}^{\ast}=1-\left(\int_{0}^{\infty}\bar{\lambda}(y)dy\right)^{-1}=1-\frac{1}{R_{0}}\,.$
(4.6)

The density function $\bar{\mathfrak{i}}(t,x)$ has an equilibrium
$\bar{\mathfrak{i}}^{*}(x)$ in the age of infection $x$, given by

$\displaystyle\bar{\mathfrak{i}}^{*}(x)=\frac{d\bar{\mathfrak{I}}^{*}(x)}{dx}=\bar{I}^{*}\beta
F^{c}(x),$ (4.7)

where $\beta^{-1}=\int_{0}^{\infty}F^{c}(t)dt\in(0,\infty)$ is the expectation
of the duration of the infectious period. If $F$ has a density $f$, then the
equilibrium density $\bar{\mathfrak{i}}^{*}(x)$ satisfies

$\displaystyle\frac{d\,\bar{\mathfrak{i}}^{*}(x)}{dx}=-\bar{I}^{*}\beta
f(x),\quad\bar{\mathfrak{i}}^{*}(0)=\bar{I}^{*}\beta.$

###### Proof.

The fact that $\bar{I}^{\ast}=0$ if $R_{0}\leq 1$ and $>0$ if $R_{0}>1$
follows from branching process arguments, and the fact that the start of the
epidemic can be approximated by a branching process, see e.g. section 1.3 in
[5]. Assume that the equilibrium
$\bar{\mathfrak{I}}^{*}(x):=\bar{\mathfrak{I}}(\infty,x)$ exists. We deduce
from (4.3), combined with (3.16), that $\bar{\mathfrak{I}}^{*}(x)$ must
satisfy

$\displaystyle\bar{\mathfrak{I}}^{\ast}(x)$
$\displaystyle=(1-\bar{\mathfrak{I}}^{\ast}(\infty))\int_{0}^{x}F^{c}(u)du\int_{0}^{\infty}\frac{\bar{\lambda}(y)}{F^{c}(y)}\bar{\mathfrak{I}}^{\ast}(dy)$
$\displaystyle=(1-\bar{I}^{\ast})\beta^{-1}F_{e}(x)\int_{0}^{\infty}\frac{\bar{\lambda}(y)}{F^{c}(y)}\bar{\mathfrak{I}}^{\ast}(dy),$

where $F_{e}(x)=\beta\int_{0}^{x}F^{c}(s)ds$, the equilibrium (stationary
excess) distribution. Letting $x\to\infty$ in this formula, we deduce

$\bar{I}^{\ast}=(1-\bar{I}^{\ast})\beta^{-1}\int_{0}^{\infty}\frac{\bar{\lambda}(y)}{F^{c}(y)}\bar{\mathfrak{I}}^{\ast}(dy)\,.$
(4.8)

Combining the last two equations, we obtain

$\bar{\mathfrak{I}}^{\ast}(x)=\bar{I}^{\ast}F_{e}(x)\,.$ (4.9)

Plugging this formula in the previous identity, we deduce that

$\bar{I}^{\ast}=(1-\bar{I}^{\ast})\bar{I}^{\ast}\int_{0}^{\infty}\bar{\lambda}(y)dy\,.$

Then the formula (4.6) can be directly deduced from this equation. The formula
(4.7) follows by taking the derivative with respect to $x$ in (4.9). ∎

######  Remark 4.1.

If the distribution $F$ is exponential, that is, $F(x)=1-e^{-\beta x}$, then
we obtain

$\bar{\mathfrak{I}}^{*}(x)=\bar{I}^{*}(1-e^{-\beta
x}),\quad\bar{\mathfrak{i}}^{*}(x)=\bar{I}^{*}\beta e^{-\beta
x},\quad\mbox{and}\quad\frac{d\,\bar{\mathfrak{i}}^{*}(x)}{dx}=-\bar{I}^{*}\beta^{2}e^{-\beta
x}=-\beta\bar{\mathfrak{i}}^{*}(x),$

where $\bar{I}^{*}$ is given in (4.6).

######  Remark 4.2.

Suppose that $\lambda_{i}(t)=\lambda(t){\mathbf{1}}_{t<\eta_{i}}$, where
$\lambda(t)$ is a deterministic function, as in Remark 3.4. Then
$\bar{\lambda}(t)=\lambda(t)F^{c}(t)$. If $\lambda(t)\equiv\lambda$ is a
constant and $F$ has mean $\beta^{-1}$, then $\bar{I}^{\ast}$ in (4.6)

$\bar{I}^{\ast}=1-\left(\lambda\int_{0}^{\infty}F^{c}(y)dy\right)^{-1}=1-\beta/\lambda=1-\frac{1}{R_{0}}\,,$
(4.10)

which reduces to the well known result for the standard SIS model with
constant rates, assuming $\beta<\lambda$.

##  5\. Proof of the FLLN

In this section, we prove Theorem 2.1. We will need the following theorem. A
similar pre-tightness criterion can be found in Theorem 3.5.1 in Chapter 6 of
[21], which extends that in the Corollary on page 83 of [4] to the space
$C([0,1]^{k},{\mathbb{R}})$. Those proofs can be easily extended to the space
$D_{D}$. For the convenience of the reader, we give a proof of the following
result in Section 6 below.

######  Theorem 5.1.

Let $\\{X^{N}:N\geq 1\\}$ be a sequence of random elements in $D_{D}$. If the
following two conditions are satisfied: for any $T,S>0$,

  * (i)

for any $\epsilon>0$,
$\sup_{t\in[0,T]}\sup_{s\in[0,S]}\mathbb{P}\big{(}|X^{N}(t,s)|>\epsilon\big{)}\to
0$ as $N\to\infty$, and

  * (ii)

for any $\epsilon>0$, as $\delta\to 0$,

$\displaystyle\limsup_{N\to\infty}\sup_{t\in[0,T]}\frac{1}{\delta}\mathbb{P}\bigg{(}\sup_{u\in[0,\delta]}\sup_{s\in[0,S]}|X^{N}(t+u,s)-X^{N}(t,s)|>\epsilon\bigg{)}\to
0,$
$\displaystyle\limsup_{N\to\infty}\sup_{s\in[0,S]}\frac{1}{\delta}\mathbb{P}\bigg{(}\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}|X^{N}(t,s+v)-X^{N}(t,s)|>\epsilon\bigg{)}\to
0,$

then $X^{N}(t,s)\to 0$ in probability, locally uniformly in $t$ and $s$, as
$N\to\infty$.

We shall also use repeatedly the following Lemma.

######  Lemma 5.1.

Let $f\in D(\mathbb{R}_{+})$ and $\\{g_{N}\\}_{N\geq 1}$ be a sequence of
elements of $D_{\uparrow}(\mathbb{R}_{+})$ which is such that $g_{N}\to g$
locally uniformly, where $g\in C_{\uparrow}(\mathbb{R}_{+})$. Then for any
$T>0$,

$\int_{[0,T]}f(t)g_{N}(dt)\to\int_{[0,T]}f(t)g(dt)\,.$

###### Proof.

The assumption implies that the sequence of measures $g_{N}(dt)$ converges
weakly, as $N\to\infty$, towards the measure $g(dt)$. Since moreover $f$ is
bounded, and the set of discontinuities of $f$ is of $g(dt)$ measure $0$, this
is essentially a minor improvement of the Portmanteau theorem, see [4]. ∎

###  5.1. Convergence of $\mathfrak{I}^{N}_{0}(t,x)$

We first treat the process $\mathfrak{I}^{N}_{0}(t,x)$ in (2.5).

######  Lemma 5.2.

Under Assumption 2.1,

$\bar{\mathfrak{I}}^{N}_{0}(t,x)\to\bar{\mathfrak{I}}_{0}(t,x)\quad\mbox{in}\quad
D_{D}\quad\mbox{as}\quad N\to\infty,$ (5.1)

in probability, where the limit $\bar{\mathfrak{I}}_{0}(t,x)$ is given by

$\bar{\mathfrak{I}}_{0}(t,x):=\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy),\quad
t,x\geq 0.$ (5.2)

###### Proof.

Recall that

$\displaystyle\bar{\mathfrak{I}}^{N}_{0}(t,x)=N^{-1}\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\eta_{j}^{0}>t}{\mathbf{1}}_{\tilde{\tau}_{j,0}^{N}\leq(x-t)^{+}}=N^{-1}\sum_{j=1}^{\mathfrak{I}^{N}(0,(x-t)^{+})}{\mathbf{1}}_{\eta_{j}^{0}>t}\,.$

Note that the pair of variables $(\tilde{\tau}_{j,0}^{N},\eta^{0}_{j})$
satisfies (2.1), and $\mathfrak{I}^{N}(0,(x-t)^{+})=\max\\{j\geq
1:\tilde{\tau}_{j,0}^{N}\leq(x-t)^{+}\\}$. Let

$\displaystyle\widetilde{\mathfrak{I}}^{N}_{0}(t,x)=N^{-1}\sum_{j=1}^{\mathfrak{I}^{N}(0,(x-t)^{+})}\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}=\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)\,.$
(5.3)

We will first show that
$\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\to\bar{\mathfrak{I}}_{0}(t,x)$ (this
will be step 1 of the proof), and then that
$\bar{\mathfrak{I}}^{N}_{0}(t,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\to 0$
(this will be step 2 of the proof), both in probability, locally uniformly in
$t$ and $x$, as $N\to\infty$.

Step 1 We show that, as $N\to\infty$,

$\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\to\bar{\mathfrak{I}}_{0}(t,x)\ \text{
in probability, locally uniformly in $t$ and $x$}.$ (5.4)

From Lemma 5.1, Assumption 2.1 and the continuous mapping theorem, we deduce
that for any $t,x\geq 0$,
$\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\to\bar{\mathfrak{I}}_{0}(t,x)$ in
probability, as $N\to\infty$. It thus remains to show that the sequence
$\\{X^{N}:=\widetilde{\mathfrak{I}}^{N}_{0}-\bar{\mathfrak{I}}_{0},\ N\geq
1\\}$ satisfies condition (ii) in Theorem 5.1. In fact, it is easily seen that
it is sufficient to verify condition (ii) with
$X^{N}=\widetilde{\mathfrak{I}}^{N}_{0}$. Indeed, both

$\sup_{0\leq u\leq\delta}\sup_{0\leq
x\leq\bar{x}}|\bar{\mathfrak{I}}_{0}(t+u,x)-\bar{\mathfrak{I}}_{0}(t,x)|\
\text{ and }\sup_{0\leq v\leq\delta}\sup_{0\leq t\leq
T}|\bar{\mathfrak{I}}_{0}(t,x+v)-\bar{\mathfrak{I}}_{0}(t,x)|$

tend to $0$, as $\delta\to 0$, which is an easy consequence of the
computations which follow. Let us now consider
$X^{N}=\widetilde{\mathfrak{I}}^{N}_{0}$. We have

$\displaystyle\widetilde{\mathfrak{I}}^{N}_{0}(t+u,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)$
$\displaystyle=\int_{0}^{(x-t-u)^{+}}\frac{F^{c}(t+u+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)-\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)$
$\displaystyle=\int_{0}^{(x-t-u)^{+}}\frac{F^{c}(t+u+y)-F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)-\int_{(x-t-u)^{+}}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)$

which gives

$\displaystyle\big{|}\widetilde{\mathfrak{I}}^{N}_{0}(t+u,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\big{|}$
$\displaystyle\leq\int_{0}^{(x-t-u)^{+}}\frac{F^{c}(t+y)-F^{c}(t+u+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)+\int_{(x-t-u)^{+}}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)\,.$

Consequently,

$\displaystyle\sup_{0\leq u\leq\delta,0\leq
x\leq\bar{x}}|\widetilde{\mathfrak{I}}^{N}_{0}(t+u,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)|$
$\displaystyle\leq\int_{0}^{(\bar{x}-t)^{+}}\frac{F^{c}(t+y)-F^{c}(t+\delta+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)$
$\displaystyle\quad+\sup_{0\leq
x\leq\bar{x}}\int_{(x-t-\delta)^{+}}^{(x-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)\,.$

The limit in probability of the first term on the right of the last inequality
equals

$\int_{0}^{(\bar{x}-t)^{+}}\frac{F^{c}(t+y)-F^{c}(t+\delta+y)}{F^{c}(y)}\bar{\mathfrak{I}}(0,dy),$

which tends to $0$ as $\delta\to 0$, since $F^{c}$ is continuous on the right
and the integrand is between $0$ and $1$. The second term on the right of the
above inequality is nonnegative and upper bounded by

$\sup_{0\leq
x\leq\bar{x}}\left(\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t-\delta)^{+})\right),$

which converges in probability towards

$\sup_{0\leq
x\leq\bar{x}}\left(\bar{\mathfrak{I}}(0,(x-t)^{+})-\bar{\mathfrak{I}}(0,(x-t-\delta)^{+})\right),$

and this last expression tends to $0$ as $\delta\to 0$. Combining the above
arguments, we deduce that for $\epsilon>0$, if $\delta>0$ is small enough,

$\limsup_{N}\mathbb{P}\left(\sup_{0\leq u\leq\delta,0\leq
x\leq\bar{x}}\big{|}\widetilde{\mathfrak{I}}^{N}_{0}(t+u,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\big{|}>\epsilon\right)=0\,.$

We next consider

$\displaystyle\widetilde{\mathfrak{I}}^{N}_{0}(t,x+v)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)$
$\displaystyle=\int_{(x-t)^{+}}^{(x+v-t)^{+}}\frac{F^{c}(t+y)}{F^{c}(y)}\bar{\mathfrak{I}}^{N}(0,dy)$
$\displaystyle\leq\bar{\mathfrak{I}}^{N}(0,(x+v-t)^{+})-\mathfrak{I}^{N}(0,(x-t)^{+})\,.$

Hence,

$\displaystyle\sup_{0\leq v\leq\delta,0\leq t\leq
T}\big{|}\widetilde{\mathfrak{I}}^{N}_{0}(t,x+v)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\big{|}=\sup_{0\leq
t\leq
T}\left(\bar{\mathfrak{I}}^{N}(0,(x+\delta-t)^{+})-\mathfrak{I}^{N}(0,(x-t)^{+})\right)\,.$

The term on the right of the last inequality converges in probability as
$N\to\infty$, towards

$\sup_{0\leq t\leq
T}\left(\bar{\mathfrak{I}}(0,(x+\delta-t)^{+})-\mathfrak{I}(0,(x-t)^{+})\right),$

which tends to $0$ as $\delta$ tends to $0$. Again we easily deduce from these
computations that for any $\epsilon>0$, if $\delta>0$ is small enough,

$\limsup_{N}\mathbb{P}\left(\sup_{0\leq v\leq\delta,0\leq t\leq
T}\big{|}\widetilde{\mathfrak{I}}^{N}_{0}(t,x+v)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)\big{|}>\epsilon\right)=0\,.$

We have established (5.4).

Step 2 We finally show that
$V^{N}(t,x):=\bar{\mathfrak{I}}^{N}_{0}(t,x)-\widetilde{\mathfrak{I}}^{N}_{0}(t,x)$
satisfies the two conditions of Theorem 5.1. We have

$\displaystyle
V^{N}(t,x)=N^{-1}\sum_{j=1}^{\mathfrak{I}^{N}(0,(x-t)^{+})}\left({\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\right).$

We first check condition (i) from Theorem 5.1. We have

$\displaystyle\mathbb{E}\big{[}V^{N}(t,x)^{2}\big{]}$
$\displaystyle=\mathbb{E}\Bigg{[}N^{-2}\sum_{j=1}^{\mathfrak{I}^{N}(0,(x-t)^{+})}\left({\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\right)^{2}\Bigg{]}$
$\displaystyle\quad+\mathbb{E}\Bigg{[}N^{-2}\sum_{j,j^{\prime}=1,\,j\neq
j^{\prime}}^{\mathfrak{I}^{N}(0,(x-t)^{+})}\left({\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\right)\left({\mathbf{1}}_{\eta_{j^{\prime}}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j^{\prime},0}^{N})}{F^{c}(\tilde{\tau}_{j^{\prime},0}^{N})}\right)\Bigg{]}$
$\displaystyle=N^{-1}\mathbb{E}\Bigg{[}\int_{0}^{(x-t)^{+}}\frac{F^{c}(t+s)}{F^{c}(s)}\bigg{(}1-\frac{F^{c}(t+s)}{F^{c}(s)}\bigg{)}\bar{\mathfrak{I}}^{N}(0,ds)\Bigg{]}$

where the second term in the first equality is equal to zero by the
independence of $\eta^{0}_{j}$ and $\eta^{0}_{j^{\prime}}$ given the times
$\tilde{\tau}_{j,0}^{N}$ and $\tilde{\tau}_{j^{\prime},0}^{N}$ and by using a
conditioning argument. This implies that as $N\to\infty$,

$\sup_{t\geq 0}\sup_{x\geq 0}\mathbb{E}\big{[}V^{N}(t,x)^{2}\big{]}\to 0,$

and thus condition (i) in Theorem 5.1 holds.

We next show condition (ii) from Theorem 5.1, that is, for any $\epsilon>0$,
as $\delta\to 0$,

$\displaystyle\limsup_{N}\sup_{t\in[0,T]}\frac{1}{\delta}\mathbb{P}\left(\sup_{u\in[0,\delta]}\sup_{x\in[0,T^{\prime}]}\big{|}V^{N}(t+u,x)-V^{N}(t,x)\big{|}>\epsilon\right)\to
0,$ (5.5)

and

$\displaystyle\limsup_{N}\sup_{x\in[0,T^{\prime}]}\frac{1}{\delta}\mathbb{P}\left(\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}V^{N}(t,x+v)-V^{N}(t,x)\big{|}>\epsilon\right)\to
0.$ (5.6)

We first prove (5.5). We have

$\displaystyle\big{|}V^{N}(t+u,x)-V^{N}(t,x)\big{|}$
$\displaystyle=\Bigg{|}N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(x-t-u)^{+})}\bigg{(}{\mathbf{1}}_{\eta_{j}^{0}>t+u}-\frac{F^{c}(t+u+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\bigg{)}-N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(x-t)^{+})}\bigg{(}{\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\bigg{)}\Bigg{|}$
$\displaystyle\leq\Bigg{|}N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(x-t-u)^{+})}\bigg{(}{\mathbf{1}}_{t<\eta_{j}^{0}\leq
t+u}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})-F^{c}(t+u+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\bigg{)}\Bigg{|}$
$\displaystyle\quad+\Bigg{|}N^{-1}\sum_{j={\mathfrak{I}}^{N}(0,(x-t-u)^{+})+1}^{{\mathfrak{I}}^{N}(0,(x-t)^{+})}\bigg{(}{\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\bigg{)}\Bigg{|}$
$\displaystyle\leq
N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(x-t-u)^{+})}{\mathbf{1}}_{t<\eta_{j}^{0}\leq
t+u}+\int_{0}^{(x-t-u)^{+}}\frac{F^{c}(t+s)-F^{c}(t+u+s)}{F^{c}(s)}\bar{\mathfrak{I}}^{N}(0,ds)$
$\displaystyle\qquad+\big{|}\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t-u)^{+})\big{|}\,.$
(5.7)

For the first term,

$\displaystyle\mathbb{P}\Bigg{(}\sup_{u\in[0,\delta]}\sup_{x\in[0,T^{\prime}]}N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(x-t-u)^{+})}{\mathbf{1}}_{t<\eta_{j}^{0}\leq
t+u}>\epsilon/3\Bigg{)}$
$\displaystyle\leq\mathbb{P}\Bigg{(}N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(T^{\prime}-t)^{+})}{\mathbf{1}}_{t<\eta_{j}^{0}\leq
t+\delta}>\epsilon/3\Bigg{)}$
$\displaystyle\leq\mathbb{P}\Bigg{(}N^{-1}\sum_{j=1}^{{\mathfrak{I}}^{N}(0,(T^{\prime}-t)^{+})}\bigg{[}{\mathbf{1}}_{t<\eta_{j}^{0}\leq
t+\delta}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})-F^{c}(t+\delta+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\bigg{]}>\epsilon/6\Bigg{)}$
$\displaystyle\quad+\mathbb{P}\Bigg{(}\int_{0}^{(T^{\prime}-t)^{+}}\frac{F^{c}(t+s)-F^{c}(t+\delta+s)}{F^{c}(s)}\bar{\mathfrak{I}}^{N}(0,ds)>\epsilon/6\Bigg{)}$
(5.8)

By the conditional independence of the $\eta^{0}_{j}$’s, the first term on the
right of (5.1) is bounded by

$\displaystyle\frac{36}{\epsilon^{2}}N^{-1}\mathbb{E}\left[\int_{0}^{(T^{\prime}-t)^{+}}\frac{F^{c}(t+s)-F^{c}(t+\delta+s)}{F^{c}(s)}\left(1-\frac{F^{c}(t+s)-F^{c}(t+\delta+s)}{F^{c}(s)}\right)\bar{\mathfrak{I}}^{N}(0,ds)\right]$
$\displaystyle\leq\frac{36}{\epsilon^{2}}N^{-1}\mathbb{E}\left[\int_{0}^{(T^{\prime}-t)^{+}}\frac{F^{c}(t+s)-F^{c}(t+\delta+s)}{F^{c}(s)}\bar{\mathfrak{I}}^{N}(0,ds)\right],$

which converges to zero as $N\to\infty$. Since by Assumption 2.1
$\bar{\mathfrak{I}}(0,\cdot)$ is continuous, thanks to Lemma 5.1,
$\limsup_{N}$ of the second term is upper bounded by

${\bf
1}\bigg{\\{}\int_{0}^{(T^{\prime}-t)^{+}}\frac{F^{c}(t+s)-F^{c}(t+\delta+s)}{F^{c}(s)}\bar{\mathfrak{I}}(0,ds)\geq\epsilon/6\bigg{\\}},$

which is zero for $\delta>0$ small enough (clearly uniformly over
$t\in[0,T]$).

The second term on the right of (5.1) is treated exactly as the last term we
have just analyzed. Finally for the third term, we note that

$\displaystyle\mathbb{P}\left(\sup_{u\in[0,\delta]}\sup_{x\in[0,T^{\prime}]}\big{|}\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t-u)^{+})\big{|}>\epsilon/3\right)$
$\displaystyle=\mathbb{P}\left(\sup_{x\in[0,T^{\prime}]}\big{|}\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t-\delta)^{+})\big{|}>\epsilon/3\right).$

Thanks to Assumption 2.1, the $\limsup_{N}$ of this probability is upper
bounded by

${\bf
1}\left\\{\sup_{x\in[0,T^{\prime}]}\big{|}\bar{\mathfrak{I}}(0,(x-t)^{+})-\bar{\mathfrak{I}}(0,(x-t-\delta)^{+})\big{|}\geq\epsilon/3\right\\}$

which is zero for $\delta>0$ small enough, since $\bar{\mathfrak{I}}(0,\cdot)$
is continuous. The uniformity over $t\in[0,T]$ is obvious. Thus we have shown
(5.5).

We next prove (5.6). Observe that

$V^{N}(t,x+v)-V^{N}(t,x)=N^{-1}\sum_{j={\mathfrak{I}}^{N}(0,(x-t)^{+})+1}^{{\mathfrak{I}}^{N}(0,(x+v-t)^{+})}\left({\mathbf{1}}_{\eta_{j}^{0}>t}-\frac{F^{c}(t+\tilde{\tau}_{j,0}^{N})}{F^{c}(\tilde{\tau}_{j,0}^{N})}\right)$

from which we obtain

$\displaystyle\mathbb{P}\left(\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}V^{N}(t,x+v)-V^{N}(t,x)\big{|}>\epsilon\right)$
$\displaystyle\leq\mathbb{P}\left(\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}\bar{\mathfrak{I}}^{N}(0,(x+v-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})\big{|}>\epsilon\right)$
$\displaystyle\leq\mathbb{P}\left(\sup_{t\in[0,T]}\big{|}\bar{\mathfrak{I}}^{N}(0,(x+\delta-t)^{+})-\bar{\mathfrak{I}}^{N}(0,(x-t)^{+})\big{|}>\epsilon\right)\,.$

Then following the same argument as for the second term on the right of (5.1),
we can conclude (5.6). ∎

###  5.2. Convergence of $\bar{\mathfrak{I}}^{N}_{1}$

We first write the process $A^{N}$ as

$\displaystyle A^{N}(t)=M_{A}^{N}(t)+\Lambda^{N}(t),$ (5.9)

where

$\displaystyle\Lambda^{N}(t):=\int_{0}^{t}\Upsilon^{N}(s)ds,$ (5.10)

and

$\displaystyle
M_{A}^{N}(t)=\int_{0}^{t}\int_{0}^{\infty}{\mathbf{1}}_{u\leq\Upsilon^{N}(s^{-})}\overline{Q}(ds,du),$
(5.11)

where $\overline{Q}(ds,du)=Q(ds,du)-dsdu$ is the compensated PRM.

######  Lemma 5.3.

Under Assumption 2.2, the process $\\{M^{N}_{A}(t):t\geq 0\\}$ is a square-
integrable martingale with respect to the filtration
${\mathcal{F}}^{N}_{A}=\\{{\mathcal{F}}^{N}_{A}(t):t\geq 0\\}$ where

$\displaystyle{\mathcal{F}}^{N}_{A}(t)$
$\displaystyle:=\sigma\big{\\{}I^{N}(0),\tilde{\tau}^{N}_{j}:j=1,\dots,I^{N}(0)\big{\\}}\vee\sigma\big{\\{}\lambda^{0}_{j}(\cdot)_{j\geq
1},\lambda_{i}(\cdot)_{i\geq 1}\big{\\}}$
$\displaystyle\qquad\vee\sigma\bigg{\\{}\int_{0}^{t^{\prime}}\int_{0}^{\infty}{\mathbf{1}}_{u\leq\Upsilon^{N}(s^{-})}Q(ds,du):0\leq
t^{\prime}\leq t\bigg{\\}}.$

The quadratic variation of $M^{N}_{A}(t)$ is given by

$\langle M^{N}_{A}\rangle(t)=\Lambda^{N}(t),\quad t\geq 0.$ (5.12)

###### Proof.

It is clear that $M^{N}_{A}(t)\in{\mathcal{F}}^{N}_{A}(t)$, and
$\mathbb{E}[|M_{A}^{N}(t)|]\leq 2\mathbb{E}[\Lambda^{N}(t)]\leq
2\lambda^{*}Nt<\infty$ for each $t\geq 0$, under Assumption 2.2. It suffices
to verify the martingale property: for $t_{2}>t_{1}\geq 0$,

$\mathbb{E}\big{[}M^{N}_{A}(t_{2})-M^{N}_{A}(t_{1})\big{|}{\mathcal{F}}^{N}_{A}(t_{1})\big{]}=0$

which can be checked using the above definition of the filtration. In
addition,
$\mathbb{E}[(M_{A}^{N}(t))^{2}]=\mathbb{E}[\Lambda^{N}(t)]\leq\lambda^{*}Nt<\infty$
for each $t\geq 0$. The rest is standard. ∎

Recall that
$(\bar{A}^{N},\bar{S}^{N},\bar{\Upsilon}^{N}):=N^{-1}(A^{N},S^{N},\Upsilon^{N})$.

######  Lemma 5.4.

Under Assumptions 2.1 and 2.2, the sequence of processes
$\\{(\bar{A}^{N},\bar{S}^{N}):N\in{\mathbb{N}}\\}$ is tight in $D^{2}$. The
limit of each convergence subsequence of $\\{\bar{A}^{N}\\}$, denoted by
$\bar{A}$, satisfies

$\bar{A}=\lim_{N\to\infty}\bar{A}^{N}=\lim_{N\to\infty}\int_{0}^{\cdot}\bar{\Upsilon}^{N}(u)du,$
(5.13)

and

$0\leq\int_{s}^{t}\bar{\Upsilon}^{N}(u)du\leq\lambda^{*}(t-s),\quad\text{w.p.\,1}\quad\mbox{for}\quad
0\leq s\leq t.$ (5.14)

###### Proof.

It is clear that under Assumption 2.2, if
$\bar{\Lambda}^{N}(t):=\int_{0}^{t}\bar{\Upsilon}^{N}(u)du$,
$\bar{\Lambda}^{N}(0)=0$ and

$\displaystyle
0\leq\bar{\Lambda}^{N}(t)-\bar{\Lambda}^{N}(s)\leq\lambda^{*}(t-s),\quad\text{w.p.\,1}\quad\mbox{for}\quad
0\leq s\leq t.$ (5.15)

Since

$\langle\bar{M}^{N}_{A}\rangle_{t}\leq N^{-1}\lambda^{\ast}t,$

it follows from Doob’s inequality that $\bar{M}^{N}_{A}(t)$ tends to $0$ in
probability, locally uniformly in $t$. The tightness of
$\\{\bar{A}^{N}:N\in{\mathbb{N}}\\}$ in $D$ follows. Since
$\bar{S}^{N}=\bar{S}^{N}(0)-\bar{A}^{N}$ and
$\bar{S}^{N}(0)\Rightarrow\bar{S}(0)$ from Assumption 2.1, we obtain the
tightness of $\\{\bar{S}^{N}:N\in{\mathbb{N}}\\}$ in $D$, and thus the claim
of the lemma. ∎

In the following of this section, we consider a convergent subsequence of
$\bar{A}^{N}$.

Recall that

$\displaystyle\bar{\mathfrak{I}}^{N}_{1}(t,x)=N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t)}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t},\quad
t,x\geq 0.$

######  Lemma 5.5.

Under Assumptions 2.1 and 2.2, along a subsequence of $\bar{A}^{N}$ which
converges weakly to $\bar{A}$,

$\bar{\mathfrak{I}}^{N}_{1}(t,x)\Rightarrow\bar{\mathfrak{I}}_{1}(t,x)\quad\mbox{in}\quad
D_{D}\quad\mbox{as}\quad N\to\infty,$ (5.16)

where the limit $\bar{\mathfrak{I}}_{1}(t,x)$ is given by

$\bar{\mathfrak{I}}_{1}(t,x):=\int_{(t-x)^{+}}^{t}F^{c}(t-s)d\bar{A}(s),\quad
t,x\geq 0.$ (5.17)

###### Proof.

Let

$\breve{\mathfrak{I}}^{N}_{1}(t,x):=N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t)}F^{c}(t-\tau^{N}_{i}),\quad
t,x\geq 0.$

We can write (from now on, $\int_{a}^{b}$ stands for $\int_{(a,b]}$)

$\breve{\mathfrak{I}}^{N}_{1}(t,x)=\int_{(t-x)^{+}}^{t}F^{c}(t-s)d\bar{A}^{N}(s).$
(5.18)

Then from Lemma 5.1, we deduce that for any $t,x\geq 0$,

$\breve{\mathfrak{I}}^{N}_{1}(t,x)\Rightarrow\bar{\mathfrak{I}}_{1}(t,x)\quad\mbox{as}\quad
N\to\infty\,.$ (5.19)

We will next show that for any $\epsilon>0$, there exists $\delta>0$ such that
the following holds for any $(t,x)$:

$\displaystyle\limsup_{N}\mathbb{P}\left(\sup_{t\leq t^{\prime}\leq
t+\delta,x\leq x^{\prime}\leq
x+\delta}\left|\breve{\mathfrak{I}}^{N}_{1}(t,x)-\breve{\mathfrak{I}}^{N}_{1}(t^{\prime},x^{\prime})\right|>\epsilon\right)=0,$
(5.20)

It is not hard to deduce from (5.19) and (5.20), by a two–dimensional
extension of the argument of the Corollary on page 83 of [4], that as
$N\to\infty$,
$\breve{\mathfrak{I}}^{N}_{1}(t,x)\Rightarrow\bar{\mathfrak{I}}_{1}(t,x)$
locally uniformly in $t$ and $x$. Whenever $t\leq t^{\prime}\leq t+\delta$ and
$x\leq x^{\prime}\leq x+\delta$, we have

$\displaystyle\big{|}\breve{\mathfrak{I}}_{1}^{N}(t,x)-\breve{\mathfrak{I}}_{1}^{N}(t^{\prime},x^{\prime})\big{|}$
$\displaystyle\leq\int_{(t-x)^{+}}^{t}\big{[}F^{c}(t-s)-F^{c}(t^{\prime}-s)\big{]}d\bar{A}^{N}(s)+2\sup_{0\leq
t_{2}-t_{1}\leq
2\delta}\big{[}\bar{A}^{N}(t_{2})-\bar{A}^{N}(t_{1})\big{]}\,.$

Since $\bar{A}^{N}(t)\Rightarrow\int_{0}^{t}\bar{\Upsilon}(s)ds$ locally
uniformly in $t$, and $\bar{\Upsilon}(s)\leq\lambda^{\ast}$, the limit in law
of the right hand side of the last inequality is bounded by

$\lambda^{\ast}\int_{(t-x)^{+}}^{t}\sup_{t\leq t^{\prime}\leq
t+\delta}\big{[}F^{c}(t-s)-F^{c}(t^{\prime}-s)\big{]}ds+4\lambda^{\ast}\delta,$

which is less than $\epsilon$ for $\delta>0$ small enough. Hence, (5.20)
follows.

Let now

$Y^{N}(t,x):=\bar{\mathfrak{I}}^{N}_{1}(t,x)-\breve{\mathfrak{I}}^{N}_{1}(t,x)=N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t)}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)},\quad
t,x\geq 0.$

To prove (5.16), it remains to show that, as $N\to\infty$,

$Y^{N}\to 0\quad\mbox{in}\quad D_{D}\quad\text{in probability}.$ (5.21)

We apply Theorem 5.1. By Markov’s inequality and the decomposition of
$A^{N}(t)$ in (5.9) with $\mathbb{E}[M_{A}^{N}(t)]=0$, we obtain

$\displaystyle\mathbb{P}\left(Y^{N}(t,x)>\epsilon\right)$
$\displaystyle\leq\frac{1}{\epsilon^{2}}\mathbb{E}\left[Y^{N}(t,x)^{2}\right]$
$\displaystyle=\frac{1}{\epsilon^{2}N}\mathbb{E}\left[\int_{(t-x)^{+}}^{t}F(t-s)F^{c}(t-s)d\bar{A}^{N}(s)\right]$
$\displaystyle=\frac{1}{\epsilon^{2}N}\mathbb{E}\left[\int_{(t-x)^{+}}^{t}F(t-s)F^{c}(t-s)\bar{\Upsilon}^{N}(s)ds\right]$
$\displaystyle\leq\frac{1}{\epsilon^{2}N}\lambda^{*}\int_{(t-x)^{+}}^{t}F(t-s)F^{c}(t-s)ds$
$\displaystyle\sup_{t\in[0,T],x\in[0,T^{\prime}]}\mathbb{P}\left(Y^{N}(t,x)>\epsilon\right)$
$\displaystyle\to 0\quad\mbox{as}\quad N\to\infty.$

The result then follows from the next two lemmas. ∎

######  Lemma 5.6.

Under the assumptions of Lemma 5.5, for $\epsilon>0$, as $\delta\to 0$,

$\displaystyle\limsup_{N}\sup_{t\in[0,T]}\frac{1}{\delta}\mathbb{P}\bigg{(}\sup_{u\in[0,\delta]}\sup_{x\in[0,T^{\prime}]}\big{|}Y^{N}(t+u,x)-Y^{N}(t,x)\big{|}>\epsilon\bigg{)}\to
0,$ (5.22)

###### Proof.

We have

$\displaystyle\big{|}Y^{N}(t+u,x)-Y^{N}(t,x)\big{|}$
$\displaystyle=\Bigg{|}N^{-1}\sum_{i=A^{N}((t+u-x)^{+})}^{A^{N}(t+u)}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t+u}-F^{c}(t+u-\tau^{N}_{i})\big{)}-N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t)}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)}\Bigg{|}$
$\displaystyle=\Bigg{|}N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u)}\Big{[}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t+u}-F^{c}(t+u-\tau^{N}_{i})\big{)}-\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)}\Big{]}$
$\displaystyle\quad-N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u-x)^{+}-1}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t+u}-F^{c}(t+u-\tau^{N}_{i})\big{)}+N^{-1}\sum_{i=A^{N}(t)+1}^{A^{N}(t+u)}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)}\Bigg{|}$
$\displaystyle\leq
N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u)}\Big{[}{\mathbf{1}}_{t<\tau^{N}_{i}+\eta_{i}\leq
t+u}+\big{(}F^{c}(t-\tau^{N}_{i})-F^{c}(t+u-\tau^{N}_{i})\big{)}\Big{]}$
$\displaystyle\quad+\Bigg{|}N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u-x)^{+}-1}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t+u}-F^{c}(t+u-\tau^{N}_{i})\big{)}\Bigg{|}+\Bigg{|}N^{-1}\sum_{i=A^{N}(t)+1}^{A^{N}(t+u)}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)}\Bigg{|}$
$\displaystyle\leq
N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u)}{\mathbf{1}}_{t<\tau^{N}_{i}+\eta_{i}\leq
t+u}+N^{-1}\sum_{i=A^{N}((t-x)^{+})}^{A^{N}(t+u)}\big{(}F^{c}(t-\tau^{N}_{i})-F^{c}(t+u-\tau^{N}_{i})\big{)}$
$\displaystyle\quad+\Big{(}\bar{A}^{N}((t+u-x)^{+})-\bar{A}^{N}((t-x)^{+})\Big{)}+\Big{(}\bar{A}^{N}(t+u)-\bar{A}^{N}(t)\Big{)}.$

Then we obtain

$\displaystyle\mathbb{P}\left(\sup_{u\in[0,\delta]}\sup_{x\in[0,T^{\prime}]}\big{|}Y^{N}(t+u,x)-Y^{N}(t,x)\big{|}>\epsilon\right)$
$\displaystyle\leq\mathbb{P}\left(N^{-1}\sum_{i=A^{N}((t-T^{\prime})^{+})}^{A^{N}(t+\delta)}{\mathbf{1}}_{t<\tau^{N}_{i}+\eta_{i}\leq
t+\delta}>\epsilon/3\right)$
$\displaystyle\quad+\mathbb{P}\left(N^{-1}\sum_{i=A^{N}((t-T^{\prime})^{+})}^{A^{N}(t+\delta)}\big{(}F^{c}(t-\tau^{N}_{i})-F^{c}(t+\delta-\tau^{N}_{i})\big{)}>\epsilon/3\right)$
$\displaystyle\quad+2\mathbb{P}\left(\sup_{0\leq s\leq
T}\big{|}\bar{A}^{N}(s+\delta)-\bar{A}^{N}(s)\big{|}>\epsilon/6\right)\,.$
(5.23)

Let $\breve{Q}(ds,dr,dz)$ denote a PRM on ${\mathbb{R}}_{+}^{3}$ with mean
measure $\nu(ds,dr,dz)=dsdrF(dz)$ and $\widetilde{Q}$ denote the associated
compensated PRM. By the Markov inequality, we obtain the first term is bounded
by $9\epsilon^{-2}$ times

$\displaystyle\mathbb{E}\left[\left(N^{-1}\sum_{i=A^{N}((t-T^{\prime})^{+})}^{A^{N}(t+\delta)}{\mathbf{1}}_{t<\tau^{N}_{i}+\eta_{i}\leq
t+\delta}\right)^{2}\right]$
$\displaystyle=\mathbb{E}\left[\left(N^{-1}\int_{(t-T^{\prime})^{+}}^{t+\delta}\int_{0}^{\infty}\int_{t-s}^{t+\delta-s}{\mathbf{1}}_{r\leq{\Upsilon}^{N}(s^{-})}\breve{Q}(ds,dr,dz)\right)^{2}\right]$
$\displaystyle\leq
2\mathbb{E}\left[\left(N^{-1}\int_{(t-T^{\prime})^{+}}^{t+\delta}\int_{0}^{\infty}\int_{t-s}^{t+\delta-s}{\mathbf{1}}_{r\leq
N\bar{\Upsilon}^{N}(s^{-})}\widetilde{Q}(ds,dr,dz)\right)^{2}\right]$
$\displaystyle\quad+2\mathbb{E}\left[\left(\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}\bar{\Upsilon}^{N}(s)ds\right)^{2}\right]$
$\displaystyle=2N^{-1}\mathbb{E}\left[\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}\bar{\Upsilon}^{N}(s)ds\right]$
$\displaystyle\quad+2\mathbb{E}\left[\left(\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}\bar{\Upsilon}^{N}(s)ds\right)^{2}\right]$
$\displaystyle\leq
2\lambda^{*}N^{-1}\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}ds$
$\displaystyle\quad+2\left(\lambda^{*}\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}ds\right)^{2},$

where the last inequality follows from (5.14). The first term converges to
zero as $N\to\infty$, and we note that

$\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}ds=\int_{t-(t-T^{\prime})^{+}}^{t-(t-T^{\prime})^{+}+\delta}F(r)dr-\int_{0}^{\delta}F(r)dr\leq\delta\,,$

since $F(r)\leq 1$. Hence

$\displaystyle\frac{1}{\delta}\left(\lambda^{*}\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F(t+\delta-s)-F(t-s)\big{)}ds\right)^{2}\leq\lambda^{\ast}\delta\to
0,\quad\mbox{as}\quad\delta\to 0.$ (5.24)

For the second term in (5.2), we have

$\displaystyle\mathbb{E}\left[\left(\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F^{c}(t-s)-F^{c}(t+\delta-s)\big{)}d\bar{A}^{N}(s)\right)^{2}\right]$
$\displaystyle\leq
2\mathbb{E}\left[\left(\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F^{c}(t-s)-F^{c}(t+\delta-s)\big{)}d\bar{M}_{A}^{N}(s)\right)^{2}\right]$
$\displaystyle\quad+2\mathbb{E}\left[\left(\int_{(t-T^{\prime})^{+}}^{t+\delta}\big{(}F^{c}(t-s)-F^{c}(t+\delta-s)\big{)}\bar{\Upsilon}^{N}(s)ds\right)^{2}\right],$

where the first term converges to zero as $N\to\infty$ by the convergence
$\bar{M}^{N}_{A}(t)\to 0$ in probability, locally uniformly in $t$, while the
second term is bounded as in (5.24).

For the last term in (5.2), we use the martingale decomposition of
$\bar{A}^{N}$ and the bound for $\bar{\Upsilon}^{N}$ in (5.14), and obtain

$\displaystyle\sup_{0\leq t\leq
T}\big{|}\bar{A}^{N}(t+\delta)-\bar{A}^{N}(t)\big{|}\leq 2\sup_{0\leq t\leq
T+\delta}\big{|}\bar{M}_{A}^{N}(t)\big{|}+\lambda^{*}\delta,$ (5.25)

which, since $\bar{M}_{A}^{N}(t)\to 0$ locally uniformly in $t$, implies that,
provided $\delta<\epsilon/\lambda^{\ast}$,

$\displaystyle\limsup_{N\to\infty}\mathbb{P}\left(\sup_{0\leq t\leq
T}\big{|}\bar{A}^{N}(t+\delta)-\bar{A}^{N}(t)\big{|}\geq\epsilon\right)=0\,.$

Thus we have shown that (5.22) holds. ∎

######  Lemma 5.7.

Under the assumptions of Lemma 5.5, for $\epsilon>0$, as $\delta\to 0$,

$\displaystyle\limsup_{N}\sup_{x\in[0,T^{\prime}]}\frac{1}{\delta}\mathbb{P}\bigg{(}\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}Y^{N}(t,x+v)-Y^{N}(t,x)\big{|}>\epsilon\bigg{)}\to
0.$ (5.26)

###### Proof.

Observe that

$\displaystyle
Y^{N}(t,x+v)-Y^{N}(t,x)=N^{-1}\sum_{i=A^{N}((t-x-v)^{+})}^{A^{N}((t-x)^{+})}\big{(}{\mathbf{1}}_{\tau^{N}_{i}+\eta_{i}>t}-F^{c}(t-\tau^{N}_{i})\big{)},$

from which we obtain

$\displaystyle\mathbb{P}\bigg{(}\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}Y^{N}(t,x+v)-Y^{N}(t,x)\big{|}>\epsilon\bigg{)}$
$\displaystyle\leq\mathbb{P}\bigg{(}\sup_{v\in[0,\delta]}\sup_{t\in[0,T]}\big{|}\bar{A}^{N}((t-x)^{+})-\bar{A}^{N}((t-x-v)^{+})\big{|}>\epsilon\bigg{)}$
$\displaystyle\leq\mathbb{P}\bigg{(}\sup_{t\in[0,T]}\big{|}\bar{A}^{N}((t-x)^{+})-\bar{A}^{N}((t-x-\delta)^{+})\big{|}>\epsilon\bigg{)}\,.$

Then the claim follows from the same argument as the one used to treat the
last term in (5.2) in the end of the proof of the previous lemma. ∎

###  5.3. Convergence of the aggregate infectivity process

Recall $\mathcal{I}^{N}$ in (2.2), and let
$\overline{\mathcal{I}}^{N}:=N^{-1}\mathcal{I}^{N}$. Define

$\displaystyle\widetilde{\mathcal{I}}^{N}(t)$
$\displaystyle:=N^{-1}\sum_{j=1}^{I^{N}(0)}\bar{\lambda}(\tilde{\tau}^{N}_{j,0}+t)+N^{-1}\sum_{i=1}^{A^{N}(t)}\bar{\lambda}(t-\tau^{N}_{i}),\quad
t\geq 0.$ (5.27)

######  Lemma 5.8.

Under Assumptions 2.1 and 2.2, along a convergent subsequence of $\bar{A}^{N}$
which converges weakly to $\bar{A}$, we have in probability,

$\overline{\mathcal{I}}^{N}-\widetilde{\mathcal{I}}^{N}\to
0\quad\mbox{in}\quad D\quad\mbox{as}\quad N\to\infty.$

###### Proof.

We write

$\overline{\mathcal{I}}^{N}(t)-\widetilde{\mathcal{I}}^{N}(t)=\overline{\Xi}^{N}_{0}(t)+\overline{\Xi}^{N}_{1}(t),$

where

$\displaystyle\overline{\Xi}^{N}_{0}(t)$
$\displaystyle=N^{-1}\sum_{j=1}^{I^{N}(0)}\big{(}\lambda_{j}^{0}(\tilde{\tau}^{N}_{j,0}+t)-\bar{\lambda}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\,\,,$
$\displaystyle\overline{\Xi}^{N}_{1}(t)$
$\displaystyle=N^{-1}\sum_{i=1}^{A^{N}(t)}\big{(}\lambda_{i}(t-\tau^{N}_{i})-\bar{\lambda}(t-\tau^{N}_{i})\big{)}\,\,.$

We first consider $\overline{\Xi}^{N}_{0}(t)$. For each fixed $t$, by
conditioning on $\sigma\\{I^{N}(0,y):0\leq
y\leq\bar{x}\\}=\sigma\\{\tilde{\tau}^{N}_{j,0},j=1,\dots,I^{N}(0)\\}$, we
obtain

$\displaystyle\mathbb{E}\big{[}\big{(}\overline{\Xi}^{N}_{0}(t)\big{)}^{2}\big{]}$
$\displaystyle=N^{-1}\mathbb{E}\bigg{[}\int_{0}^{\bar{x}}v(y+t)d\bar{\mathfrak{I}}^{N}(0,y)\bigg{]}\to
0\quad\mbox{as}\quad N\to\infty.$

We then have for $t,u>0$,

$\displaystyle\big{|}\overline{\Xi}^{N}_{0}(t+u)-\overline{\Xi}^{N}_{0}(t)\big{|}$
$\displaystyle\leq
N^{-1}\sum_{j=1}^{I^{N}(0)}\Big{|}\lambda_{j}^{0}(\tilde{\tau}^{N}_{j,0}+t+u)-\lambda_{j}^{0}(\tilde{\tau}^{N}_{j,0}+t)\Big{|}$
$\displaystyle\quad+N^{-1}\sum_{j=1}^{I^{N}(0)}\Big{|}\bar{\lambda}(\tilde{\tau}^{N}_{j,0}+t+u)-\bar{\lambda}(\tilde{\tau}^{N}_{j,0}+t)\Big{|}$
$\displaystyle=:\Delta_{0}^{N,1}(t,u)+\Delta_{0}^{N,2}(t,u)\,.$

Then by Assumption 2.2, writing
$\lambda^{0}_{j}(t)=\sum_{\ell=1}^{k}\lambda^{0,\ell}_{j}(t){\mathbf{1}}_{[\zeta_{j}^{\ell-1},\zeta_{j}^{\ell})}(t)$,
we have

$\displaystyle\Delta_{0}^{N,1}(t,u)$
$\displaystyle=N^{-1}\sum_{j=1}^{I^{N}(0)}\bigg{|}\sum_{\ell=1}^{k}\lambda_{j}^{0,\ell}(\tilde{\tau}^{N}_{j,0}+t+u){\mathbf{1}}_{[\zeta_{j}^{\ell-1},\zeta_{j}^{\ell})}(\tilde{\tau}^{N}_{j,0}+t+u)$
$\displaystyle\qquad\qquad\qquad-\sum_{\ell=1}^{k}\lambda_{j}^{0,\ell}(\tilde{\tau}^{N}_{j,0}+t){\mathbf{1}}_{[\zeta_{j}^{\ell-1},\zeta_{j}^{\ell})}(\tilde{\tau}^{N}_{j,0}+t)\bigg{|}$
$\displaystyle\leq
N^{-1}\sum_{j=1}^{I^{N}(0)}\sum_{\ell=1}^{k}|\lambda_{j}^{0,\ell}(\tilde{\tau}^{N}_{j,0}+t+u)-\lambda_{j}^{0,\ell}(\tilde{\tau}^{N}_{j,0}+t)|{\mathbf{1}}_{\zeta_{j}^{\ell-1}\leq\tilde{\tau}^{N}_{j,0}+t\leq\tilde{\tau}^{N}_{j,0}+t+u\leq\zeta_{j}^{\ell}}$
$\displaystyle\quad+\lambda^{*}N^{-1}\sum_{j=1}^{I^{N}(0)}\sum_{\ell=1}^{k}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+u}$
$\displaystyle\leq\varphi(u)\bar{I}^{N}(0)+\lambda^{*}N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+u}\,\,.$
(5.28)

Both terms on the right hand side are increasing in $u$, and thus, we have

$\displaystyle\sup_{0\leq
u\leq\delta}\Delta_{0}^{N,1}(t,u)\leq\varphi(\delta)\bar{I}^{N}(0)+\lambda^{*}N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}\,\,.$

Here for the second term, we have

$\displaystyle
N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}$
$\displaystyle=N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}\bigg{[}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}-\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\big{)}\bigg{]}$
$\displaystyle\quad+N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\big{)},$

hence

$\displaystyle\mathbb{P}$
$\displaystyle\left(N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}>\epsilon\right)$
$\displaystyle\leq\sum_{\ell=1}^{k}\mathbb{P}\left(N^{-1}\sum_{j=1}^{I^{N}(0)}\bigg{[}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}-\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\big{)}\bigg{]}>\epsilon/2k\right)$
$\displaystyle\quad+\sum_{\ell=1}^{k}\mathbb{P}\left(N^{-1}\sum_{j=1}^{I^{N}(0)}\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\big{)}>\epsilon/2k\right)\,.$
(5.29)

The first term on the right of (5.3) tends to $0$ as $N\to\infty$, since by
conditioning on $\sigma\\{\mathfrak{I}^{N}(0,y):0\leq
y\leq\bar{x}\\}=\sigma\\{\tilde{\tau}^{N}_{j,0},j=1,\dots,I^{N}(0)\\}$, and
since the $\zeta_{j}^{\ell}$’s are mutually independent and globally
independent of the $\tilde{\tau}^{N}_{j,0}$’s, we obtain

$\displaystyle\mathbb{E}\bigg{[}\bigg{(}N^{-1}\sum_{j=1}^{I^{N}(0)}\big{(}{\mathbf{1}}_{\tilde{\tau}^{N}_{j,0}+t\leq\zeta_{j}^{\ell}\leq\tilde{\tau}^{N}_{j,0}+t+\delta}-\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\big{)}\bigg{)}^{2}\bigg{]}$
$\displaystyle=\mathbb{E}\bigg{[}N^{-1}\int_{0}^{\bar{x}}\big{(}F_{\ell}(y+t+\delta)-F_{\ell}(y+t)\big{)}\big{)}\big{[}1-\big{(}F_{\ell}(y+t+\delta)-F_{\ell}(y+t)\big{)}\big{)}\big{]}\bar{\mathfrak{I}}^{N}(0,dy)\bigg{]}\,.$

The second term on the right of (5.3) equals

$\sum_{\ell=1}^{k}\mathbb{P}\left(\int_{0}^{\bar{x}}\big{(}F_{\ell}(y+t+\delta)-F_{\ell}(y+t)\big{)}\big{)}\bar{\mathfrak{I}}^{N}(0,dy)>\epsilon/2k\right),$

whose limsup as $N\to\infty$ is bounded from above by

$\sum_{\ell=1}^{k}{\bf
1}\left\\{\int_{0}^{\bar{x}}\big{(}F_{\ell}(y+t+\delta)-F_{\ell}(y+t)\big{)}\big{)}\bar{\mathfrak{I}}(0,dy)\geq\epsilon/2k\right\\}\,.$

Since for each $1\leq\ell\leq k$,

$
\delta\mapsto\int_{0}^{\bar{x}}\big{(}F_{\ell}(y+t+\delta)-F_{\ell}(y+t)\big{)}\big{)}\bar{\mathfrak{I}}(0,dy)$

is continuous and equals $0$ at $\delta=0$, for any $\epsilon>0$, there exists
$\delta>0$ small enough such that the above quantity vanishes. Thus, we have
shown that

$\displaystyle\limsup_{N\to\infty}\sup_{t\in[0,T]}\frac{1}{\delta}\mathbb{P}\left(\sup_{0\leq
u\leq\delta}\Delta_{0}^{N,1}(t,u)>\epsilon/3\right)\to
0,\quad\mbox{as}\quad\delta\to 0.$ (5.30)

Next, consider $\Delta_{0}^{N,2}(t,u)$, which is $\Delta_{0}^{N,1}(t,u)$, with
the $j$–th term in the absolute value being replaced by its conditional
expectation given $\tilde{\tau}^{N}_{j,0}$. The computations which led above
to (5.3) give

$\displaystyle\sup_{0\leq
u\leq\delta}\Delta_{0}^{N,2}(t,u)\leq\varphi(\delta)\bar{I}^{N}(0)+\lambda^{*}N^{-1}\sum_{\ell=1}^{k}\sum_{j=1}^{I^{N}(0)}\big{(}F_{\ell}(\tilde{\tau}^{N}_{j,0}+t+\delta)-F_{\ell}(\tilde{\tau}^{N}_{j,0}+t)\big{)}\,\,.$

So the same arguments as those used above yield that (5.30) holds with
$\Delta_{0}^{N,1}(t,u)$ replaced by $\Delta_{0}^{N,2}(t,u)$.

Thus we have shown that in probability, $\overline{\Xi}^{N}_{0}\to 0$ in $D$
as $N\to\infty$. The convergence $\overline{\Xi}^{N}_{1}\to 0$ in $D$ in
probability follows from the proof of Lemma 4.6 in [10]. In fact, the above
proof of $\overline{\Xi}^{N}_{0}\to 0$ can be adapted to that proof by
observing the similar roles of $A^{N}$ and $\mathfrak{I}^{N}(0,\cdot)$. This
completes the proof. ∎

######  Lemma 5.9.

Under Assumptions 2.1 and 2.2, along a convergent subsequence of $\bar{A}^{N}$
which converges weakly to $\bar{A}$,

$\overline{{\mathcal{I}}}^{N}\Rightarrow\widetilde{{\mathcal{I}}}\quad\mbox{in}\quad
D\quad\mbox{as}\quad N\to\infty,$ (5.31)

where $\widetilde{{\mathcal{I}}}(t)$ is given by

$\displaystyle\widetilde{{\mathcal{I}}}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}(0,dy)+\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}(s)\,,\quad
t\geq 0.$ (5.32)

###### Proof.

By the above lemma, it suffices to show that

$\widetilde{{\mathcal{I}}}^{N}\Rightarrow\widetilde{{\mathcal{I}}}\quad\mbox{in}\quad
D\quad\mbox{as}\quad N\to\infty.$ (5.33)

The expression of $\widetilde{{\mathcal{I}}}^{N}$ in (5.27) can be rewritten
as

$\displaystyle\widetilde{\mathcal{I}}^{N}(t)$
$\displaystyle=\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}^{N}(0,dy)+\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}^{N}(s).$
(5.34)

It follows from Lemma 5.1 that for any $t>0$, as $N\to\infty$,
$\widetilde{{\mathcal{I}}}^{N}(t)\Rightarrow\widetilde{{\mathcal{I}}}(t)$. It
remains to show that the sequence $\widetilde{{\mathcal{I}}}^{N}$ is tight in
$D$. For that purpose, exploiting the Corollary on page 83 of [4], it suffices
to show that for any $\epsilon>0$,

$\displaystyle\lim_{\delta\to
0}\limsup_{N}\frac{1}{\delta}\mathbb{P}\left(\sup_{0\leq
u\leq\delta}\left|\int_{0}^{\bar{x}}\bar{\lambda}(y+t+u)\bar{\mathfrak{I}}^{N}(0,dy)-\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}^{N}(0,dy)\right|>\epsilon\right)=0,$
(5.35) $\displaystyle\lim_{\delta\to
0}\limsup_{N}\frac{1}{\delta}\mathbb{P}\left(\sup_{0\leq
u\leq\delta}\left|\int_{0}^{t+u}\bar{\lambda}(t+u-s)d\bar{A}^{N}(s)-\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}^{N}(s)\right|>\epsilon\right)=0\,.$
(5.36)

(5.35) follows from the fact that, with $G_{\delta}(s):=\sup_{0\leq
u\leq\delta}|\bar{\lambda}(s+u)-\bar{\lambda}(s)|$,

$\displaystyle\limsup_{N}\mathbb{P}\left(\sup_{0\leq
u\leq\delta}\left|\int_{0}^{\bar{x}}(\bar{\lambda}(y+t+u)-\bar{\lambda}(y+t))\bar{\mathfrak{I}}^{N}(0,dy)\right|>\epsilon\right)\leq{\bf
1}\left\\{\int_{0}^{\bar{x}}G_{\delta}(y+t)\bar{\mathfrak{I}}(0,dy)>\epsilon\right\\}\,.$

Now $\bar{\mathfrak{I}}(0,dy)$ a.e., $G_{\delta}(y+t)\to 0$, and since $0\leq
G_{\delta}(y+t)\leq\lambda^{\ast}$, it follows from Lebesgue’s dominated
convergence that $\int_{0}^{\bar{x}}G_{\delta}(y+t)\bar{\mathfrak{I}}(0,dy)\to
0$, as $\delta\to 0$,hence for $\delta>0$ small enough, this quantity is less
than $\epsilon$, and the indicator vanishes.

It remains to establish (5.36). We have

$\displaystyle\int_{0}^{t+u}\bar{\lambda}(t+u-s)d\bar{A}^{N}(s)-\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}^{N}(s)$
$\displaystyle=\int_{t}^{t+u}\bar{\lambda}(t+u-s)d\bar{A}^{N}(s)$
$\displaystyle\quad+\int_{0}^{t}[\bar{\lambda}(t+u-s)-\bar{\lambda}(t-s)]d\bar{A}^{N}(s),$

hence

$\displaystyle\sup_{0\leq
u\leq\delta}\left|\int_{0}^{t+u}\bar{\lambda}(t+u-s)d\bar{A}^{N}(s)-\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}^{N}(s)\right|$
$\displaystyle\leq(\lambda^{\ast})^{2}\delta+\lambda^{\ast}\int_{0}^{t}G_{\delta}(t-s)ds+\lambda^{\ast}\big{|}\bar{M}_{A}^{N}(t+\delta)-\bar{M}_{A}^{N}(t)\big{|}+\left|\int_{0}^{t}G_{\delta}(t-s)d\bar{M}^{N}_{A}(s)\right|$

The result follows since the sum of the two first terms on the right are less
than $\epsilon/2$ for $\delta>0$ small enough, while the two last terms tend
to $0$, as $N\to\infty$. ∎

###  5.4. Completing the proof of Theorem 2.1

By Lemmas 5.2 and 5.5, we have that, along a subsequence,

$\displaystyle\bar{\mathfrak{I}}^{N}(t,x)=\bar{\mathfrak{I}}^{N}_{0}(t,x)+\bar{\mathfrak{I}}^{N}_{1}(t,x)\Rightarrow\widetilde{\mathfrak{I}}(t,x)=\bar{\mathfrak{I}}_{0}(t,x)+\bar{\mathfrak{I}}_{1}(t,x)\in
D_{D}\quad\mbox{as}\quad N\to\infty,$

where $\bar{\mathfrak{I}}_{0}(t,x)$ and $\bar{\mathfrak{I}}_{1}(t,x)$ are
given in (5.2) and (5.17), respectively. Also recall that
$\bar{S}^{N}=\bar{S}^{N}(0)-\bar{A}^{N}$ by (2.9). We need to show the joint
convergence

$(\bar{S}^{N},\bar{\mathfrak{I}}^{N},\overline{{\mathcal{I}}}^{N})\Rightarrow(\bar{S},\widetilde{\mathfrak{I}},\widetilde{{\mathcal{I}}})\quad\mbox{in}\quad
D\times D_{D}\times D\quad\mbox{as}\quad N\to\infty.$

or equivalently,

$(\bar{A}^{N},\bar{\mathfrak{I}}^{N},\overline{{\mathcal{I}}}^{N})\Rightarrow(\bar{A},\widetilde{\mathfrak{I}},\widetilde{{\mathcal{I}}})\quad\mbox{in}\quad
D\times D_{D}\times D\quad\mbox{as}\quad N\to\infty.$ (5.37)

Indeed, first thanks to Lemma 5.8, we can replace
$\overline{{\mathcal{I}}}^{N}$ by $\widetilde{{\mathcal{I}}}^{N}$. Next we
have the decompositions

$\displaystyle\bar{\mathfrak{I}}^{N}$
$\displaystyle=\bar{\mathfrak{I}}_{0}^{N}+\bar{\mathfrak{I}}_{1}^{N},$
$\displaystyle\widetilde{{\mathcal{I}}}^{N}$
$\displaystyle=\widetilde{{\mathcal{I}}}_{0}^{N}+\widetilde{{\mathcal{I}}}_{1}^{N},$

where $\widetilde{{\mathcal{I}}}_{0}^{N}$ and
$\widetilde{{\mathcal{I}}}_{1}^{N}$ are respectively the first and the second
term on the right of the identity (5.34). By the independence of the
quantities associated with initially and newly infected individuals, it
suffices to prove the joint convergence of the processes
$(\bar{\mathfrak{I}}_{0}^{N},\widetilde{{\mathcal{I}}}_{0}^{N})$ and that of
the processes
$(\bar{A}^{N},\bar{\mathfrak{I}}_{1}^{N},\widetilde{{\mathcal{I}}}_{1}^{N})$
separately. We have proved in Lemma 5.2 that
$\bar{\mathfrak{I}}_{0}^{N}\to\bar{\mathfrak{I}}_{0}$ in $D_{D}$ in
probability, and it follows from the arguments in the proof of Lemma 5.9 that
$\widetilde{{\mathcal{I}}}_{0}^{N}\to\widetilde{{\mathcal{I}}}_{0}$ in $D$ in
probability, where $\widetilde{{\mathcal{I}}}_{0}$ is the first term on the
right of the identity (5.32). Hence, the joint convergence
$(\bar{\mathfrak{I}}_{0}^{N},\widetilde{{\mathcal{I}}}_{0}^{N})\to(\bar{\mathfrak{I}}_{0},\widetilde{{\mathcal{I}}}_{0})$
in $D^{2}$ in probability is immediate.

Exploiting again (5.21), we see that the joint convergence
$(\bar{A}^{N},\bar{\mathfrak{I}}_{1}^{N},\widetilde{{\mathcal{I}}}_{1}^{N})\Rightarrow(\bar{A},\bar{\mathfrak{I}}_{1},\widetilde{{\mathcal{I}}}_{1})$
will be a consequence of

$(\bar{A}^{N},\breve{\mathfrak{I}}^{N}_{1},\widetilde{{\mathcal{I}}}_{1}^{N})\Rightarrow(\bar{A},\bar{\mathfrak{I}}_{1},\widetilde{{\mathcal{I}}}_{1})\quad\mbox{in}\quad
D\times D_{D}\times D\quad\mbox{as}\quad N\to\infty.$ (5.38)

where $\widetilde{{\mathcal{I}}}_{1}$ denotes the second term on the right of
the identity (5.32). Since
$\breve{\mathfrak{I}}^{N}_{1}(t,x)=\int_{(t-x)^{+}}^{t}F^{c}(t-s)d\bar{A}^{N}(s)$
and
$\widetilde{{\mathcal{I}}}_{1}^{N}(t)=\int_{0}^{t}\bar{\lambda}(t-s)d\bar{A}^{N}(s)$,
the joint finite dimensional convergence is a consequence of the continuous
mapping theorem and Lemma 5.1. Hence the result follows from tightness. We
have proved the joint convergence property in (5.37).

Recall the expression of
$\overline{\Upsilon}^{N}(t)=\bar{S}^{N}(t)\overline{\mathcal{I}}^{N}(t)$.
Applying the continuous mapping theorem again, we obtain that

$\overline{\Upsilon}^{N}(t)\Rightarrow\bar{\Upsilon}(t)=\bar{S}(t)\widetilde{{\mathcal{I}}}(t)\quad\mbox{in}\quad
D\quad\mbox{as}\quad N\to\infty.$

Thus by (5.13), we conclude that

$\bar{A}^{N}\Rightarrow\bar{A}=\int_{0}^{\cdot}\bar{\Upsilon}(s)ds=\int_{0}^{\cdot}\bar{S}(s)\widetilde{{\mathcal{I}}}(s)ds\quad\mbox{in}\quad
D\quad\mbox{as}\quad N\to\infty.$

Therefore, the limit $(\bar{S},\widetilde{{\mathcal{I}}})$ satisfies the set
of integral equations in (2.13), (2.14) and the limit
$\widetilde{{\mathcal{I}}}$ coincides with $\overline{{\mathcal{I}}}$ defined
by (2.14). Then, the limit $\widetilde{\mathfrak{I}}$ coincides with
$\bar{\mathfrak{I}}$ in (2.15). The limits $\bar{I}$ in (2.18) and $\bar{R}$
in (2.16) then follow immediately. The set of integral equations has a unique
deterministic solution. Indeed, it is easy to see that the system of equations
(2.13) and (2.14) (together with the first part of (2.17)) has a unique
solution $(\bar{S},\overline{{\mathcal{I}}})$, given the initial values
$\bar{\mathfrak{I}}(0,\cdot)$. The other processes
$\bar{\mathfrak{I}},\bar{I},\bar{R}$ are then uniquely determined. Hence the
whole sequence converges in probability.

From (2.15), we deduce that for all $t>0$,

$\displaystyle\bar{\mathfrak{I}}_{x}(t,0)$ $\displaystyle=\lim_{x\to
0}\frac{\bar{\mathfrak{I}}(t,x)-\bar{\mathfrak{I}}(t,0)}{x}=\lim_{x\to
0}\frac{\bar{\mathfrak{I}}(t,x)}{x}=\bar{\Upsilon}(t).$

This prove the second equality in (2.17).

It remains to prove the continuity. The continuity in $t$ of $\bar{S}(t)$ is
clear. Let us prove that $t\mapsto\overline{\mathcal{I}}(t)$ is continuous.
Since $\lambda_{i}$ is càdlàg and bounded, it is easily checked that
$t\mapsto\bar{\lambda}(t)=\mathbb{E}[\lambda(t)]$ is also càdlàg. In fact it
is continuous if all the $F_{\ell}$’s for $1\leq\ell\leq k$ are continuous.
The points of discontinuity of $\bar{\lambda}(t)$ are the points where one of
the laws of the $\zeta^{\ell}$ has some mass. The set of those points is at
most countable. Consequently, if $t_{n}\to t$, the set of $y$’s where
$\bar{\lambda}(t_{n}+y)$ may not converge to $\bar{\lambda}(t+y)$ is at most
countable, and this is a set of zero $\bar{\mathfrak{I}}(0,dy)$ measure. Since
moreover $0\leq\bar{\lambda}(t_{n}+y)\leq\lambda^{\ast}$,
$t\to\int_{0}^{\bar{x}}\bar{\lambda}(y+t)\bar{\mathfrak{I}}(0,dy)$ is
continuous. Let us now consider the second term in (2.14). We first note that
since $\bar{\lambda}(t-s)\leq\lambda^{\ast}$ and $\bar{S}(t)\leq 1$, it
follows from (2.14), (2.17) and Gronwall’s Lemma that
$\overline{\mathcal{I}}(t)\leq\lambda^{\ast}e^{\lambda^{\ast}t}$. Let
$t_{n}\to t$. We have

$\displaystyle\left|\int_{0}^{t}\bar{\lambda}(t-s)\bar{\Upsilon}(s)ds-\int_{0}^{t_{n}}\bar{\lambda}(t_{n}-s)\bar{\Upsilon}(s)ds\right|$
$\displaystyle\leq\int_{0}^{t}|\bar{\lambda}(t-s)-\bar{\lambda}(t_{n}-s)|\Upsilon(s)ds+(\lambda^{\ast})^{2}e^{\lambda^{\ast}(t\vee
t_{n})}|t-t_{n}|.$

Clearly the above right hand side tends to $0$, as $n\to\infty$. A similar
argument shows that $\bar{R}$ and $\bar{I}$ are continuous, and that
$(t,x)\mapsto\bar{\mathfrak{I}}(t,x)$ is continuous. Finally, since the
convergence holds in $D\times D\times D_{D}\times D$ and the limits are
continuous, the convergence is locally uniform in $t$ and $x$. This completes
the proof of Theorem 2.1.

##  6\. Appendix: Proof of Theorem 5.1

Given $\delta>0$, we define the two sets

$\displaystyle\Gamma_{T,\delta}$
$\displaystyle:=\Big{\\{}0,\delta,2\delta,\ldots,\lfloor\frac{T}{\delta}\rfloor\delta\Big{\\}},$
$\displaystyle\Gamma_{S,\delta}$
$\displaystyle:=\Big{\\{}0,\delta,2\delta,\ldots,\lfloor\frac{S}{\delta}\rfloor\delta\Big{\\}}\,.$

For any $t\in[0,T]$, we define $\gamma_{T,\delta}(t)$ to be the element of
$\Gamma_{T,\delta}$ such that $\gamma_{T,\delta}(t)\leq
t<\gamma_{T,\delta}(t)+\delta$, and for any $s\in[0,S]$, we define
$\gamma_{S,\delta}(s)$ to be the element of $\Gamma_{S,\delta}$ such that
$\gamma_{S,\delta}(s)\leq s<\gamma_{S,\delta}(s)+\delta$.

Let $(t,s)$ and $(t^{\prime},s^{\prime})$ be two points in $[0,T]\times[0,S]$
such that $|t-t^{\prime}|\vee|s-s^{\prime}|\leq\delta$. We have

$\displaystyle X^{N}(t,s)-X^{N}(t^{\prime},s^{\prime})$
$\displaystyle=X^{N}(t,s)-X^{N}(t,\gamma_{S,\delta}(s))+X^{N}(t,\gamma_{S,\delta}(s))-X^{N}(\gamma_{T,\delta}(t),\gamma_{S,\delta}(s))$
$\displaystyle\quad+X^{N}(\gamma_{T,\delta}(t),\gamma_{S,\delta}(s))-X^{N}(\gamma_{T,\delta}(t^{\prime}),\gamma_{S,\delta}(s))$
$\displaystyle\quad+X^{N}(\gamma_{T,\delta}(t^{\prime}),\gamma_{S,\delta}(s))-X^{N}(\gamma_{T,\delta}(t^{\prime}),\gamma_{S,\delta}(s^{\prime}))$
$\displaystyle\quad+X^{N}(\gamma_{T,\delta}(t^{\prime}),\gamma_{S,\delta}(s^{\prime}))-X^{N}(t^{\prime},\gamma_{S,\delta}(s^{\prime}))+X^{N}(t^{\prime},\gamma_{S,\delta}(s^{\prime}))-X^{N}(t^{\prime},s^{\prime}).$

Hence

$\displaystyle\mathbb{P}$ $\displaystyle\left(\sup_{0\leq t,t^{\prime}\leq
T;0\leq s,s^{\prime}\leq
S;|t-t^{\prime}|\vee|s-s^{\prime}|\leq\delta}|X^{N}(t,s)-X^{N}(t^{\prime},s^{\prime})|>\epsilon\right)$
$\displaystyle\leq 3\sum_{s\in\Gamma_{S,\delta}}\mathbb{P}\left(\sup_{0\leq
t\leq T,u\in[0,\delta]}|X^{N}(t,s+u)-X^{N}(t,s)|>\epsilon/6\right)$
$\displaystyle\quad+3\sum_{t\in\Gamma_{T,\delta}}\mathbb{P}\left(\sup_{0\leq
s\leq S,u\in[0,\delta]}|X^{N}(t+u,s)-X^{N}(t,s)|>\epsilon/6\right)$
$\displaystyle\leq 3\left(\frac{1}{\delta}+1\right)\sup_{0\leq s\leq
S}\mathbb{P}\left(\sup_{0\leq t\leq
T,u\in[0,\delta]}|X^{N}(t,s+u)-X^{N}(t,s)|>\epsilon/6\right)$
$\displaystyle\quad+3\left(\frac{1}{\delta}+1\right)\sup_{0\leq t\leq
T}\mathbb{P}\left(\sup_{0\leq s\leq
T,u\in[0,\delta]}|X^{N}(t+u,s)-X^{N}(t,s)|>\epsilon/6\right).$

It then follows from (ii) that, as $\delta\to 0$,

$\limsup_{N}\mathbb{P}\left(\sup_{0\leq t,t^{\prime}\leq T;0\leq
s,s^{\prime}\leq
S;|t-t^{\prime}|\vee|s-s^{\prime}|\leq\delta}|X^{N}(t,s)-X^{N}(t^{\prime},s^{\prime})|>\epsilon\right)\to
0.$

This, combined with (i), implies the result. $\Box$

## Acknowledgement

We thank the reviewers on the helpful comments that have improved the
exposition of our paper. Guodong Pang is partly supported by the NSF grant
DMS-2216765.

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