---
abstract: 'The stochastic SIS logistic process is a well-known birth-and-death process, often used to model the spread of an epidemic within a population of size $N$. Our focus is on the subcritical regime, where the recovery rate exceeds the infection rate by more than $N^{-1/2}$, and the epidemic dies out rapidly with high probability. Previously, only a first-order estimate for the mean extinction time of the epidemic was known, even in the case where the recovery rate and infection rate are fixed constants: we derive precise asymptotics for the distribution of the extinction time throughout the subcritical regime. In proving our results, we illustrate a technique for approximating certain types of Markov chain by differential equations over long time periods.'
address:
- 'Department of Mathematics, London School of Economics, Houghton Street, London WC2A 2AE, United Kingdom'
- 'School of Mathematics, Queen Mary, University of London'
author:
- Graham Brightwell
- Malwina Luczak
date: 28 December 2013
title: Extinction Times in the Subcritical Stochastic SIS Logistic Epidemic
---

[^1]

Introduction {#Sintro}
============

The [*stochastic SIS logistic process*]{} is defined as follows. Given a “size parameter” $N$, and two further parameters $\lambda$ and $\mu$, let $X^N = (X^N_t)_{t\ge 0}$ be the continuous-time Markov chain with state space $\{ 0, \dots, N\}$, and transitions as follows: $$\begin{aligned}
x &\to& x+1 \quad \mbox{ at rate } \quad \lambda x (1-x/N), \\
x &\to& x-1 \quad \mbox{ at rate } \quad \mu x.\end{aligned}$$ This is the most basic stochastic model of the spread of an SIS (susceptible-infective-susceptible) epidemic within a population of size $N$. In this context, $X^N_t$ represents the number of infective individuals at time $t$. Each infective encounters a random other member of the population at rate $\lambda$; if the other individual is currently susceptible, they become infective. Also, each infective recovers at rate $\mu$; once they are recovered they become susceptible again. The stochastic SIS logistic process is also used as a model for a metapopulation process, where $N$ represents the number of available patches, and $X^N_t$ is the number of patches that are populated at time $t$, $\lambda$ represents the rate at which one existing colony attempts to colonise another patch, and $\mu$ represents the rate at which an entire patch becomes depopulated due to some catastrophe. The model was first introduced by Feller [@Feller] in 1939, and further studied by Bartlett [@Bartlett]. It was rediscovered by Weiss and Dishon [@WD] in 1971, and has since been investigated by many authors. A recent thorough treatment of the model is the book of Nåsell [@Nasell].

Our results in this paper are for instances where $\lambda < \mu$. One such case is where $\lambda$ and $\mu$ are fixed real numbers with $\lambda < \mu$. We shall also be interested in the case where $\lambda=\lambda(N)$ and $\mu=\mu(N)$ are functions of $N$ with $\mu(N)-\lambda(N) \to 0^+$. We shall always assume that $\mu(N)$ and $\lambda(N)$ are bounded away from both 0 and $\infty$.

The Markov chain $(X^N_t)$ will start in a state $X_0^N = x_0=x_0(N)$. One case of natural interest is where $x_0 \simeq z_0N$ for some constant $z_0 \in (0,1]$, but our methods and results will also cover the case where $x_0/N \to 0$, as long as $x_0 (\mu-\lambda) \to \infty$. We set $T_e^{X^N}$ to be the time to extinction (i.e., the hitting time of the absorbing state 0) for the chain $(X_t^N)$, with parameters $\mu = \mu(N)$, $\lambda = \lambda(N)$ and $x_0 = x_0(N)$. Our interest is in the asymptotic distribution of $T_e^{X^N}$, as $N \to \infty$.

There is an exact expression for ${\operatorname{\mathbb E{}}}T_e^{X^N}$ as a double summation, due to Weiss and Dishon [@WD] in the case where $x_0 = N$, and in general to Leigh [@Leigh]: the asymptotics of this expression vary considerably according to the behaviour of the parameters, and it is far from straightforward to extract information from the expression.

The stochastic SIS logistic process is naturally associated with the differential equation $$\label{eq.diff-eq}
\frac{dz}{dt} = \lambda z (1-z) - \mu z = \lambda z(1 - \mu/\lambda - z),$$ where $z(t)$ represents the [*proportion*]{} of infective individuals at time $t$. This equation was first studied by Verhulst [@Verhulst], and it is known as the [*Verhulst equation*]{} or [*logistic equation*]{}. It follows from the general theory of Kurtz [@Kurtz] that $X^N_t/N$ is well concentrated around the solution $z(t)$ of the differential equation with initial condition $z(0) = x_0/N$, uniformly over fixed time intervals. For our purposes, we shall need to show concentration for longer periods, and part of our aim in this paper is to illustrate a technique for proving such a law of large numbers over longer timescales.

When $\lambda$ and $\mu$ are fixed, the behaviour of both the deterministic process $z(t)$ and the Markov chain $(X^N_t)$ vary dramatically depending on whether $\lambda$ is greater than, equal to, or less than $\mu$. In the case where $\lambda > \mu$, there is a stable fixed point of the drift equation (\[eq.diff-eq\]) at $z = 1-\mu/\lambda$. If there are initially a large number of infective individuals at time 0, then with high probability the scaled Markov chain $X^N_t/N$ heads rapidly towards the fixed point, then spends most of its time in the neighbourhood of the fixed point, making excursions into the rest of the state space until eventually one of these excursions reaches the absorbing state 0. Precise results are known about the distribution of the time to extinction, which is exponential in $N$, and about the quasi-stationary distribution, which is centred around the stable fixed point of (\[eq.diff-eq\]). See, for instance [@Barbour76; @KL; @Nasell96; @AD; @Nasell].

If $\lambda \le \mu$, then the differential equation (\[eq.diff-eq\]) has a single stable fixed point at $z=0$, and all its solutions converge to zero as $t\to \infty$. For the corresponding Markov chain, it is also known that the epidemic dies out rapidly with high probability whenever $\lambda \le \mu$. Doering, Sargsyan and Sander [@DSS] give an asymptotic formula for the mean extinction time, in the case where $\lambda < \mu$ are fixed constants and the starting state is $x_0 = z_0N$ with $z_0$ a constant: $$\label{dss}
{\operatorname{\mathbb E{}}}T_e^{X^N} = \frac{1}{\mu-\lambda} (\log N + \log z_0 + O(1)).$$ In the case where $\lambda = \mu$, they obtain: $${\operatorname{\mathbb E{}}}T_e^{X^N} = \left( \frac{\pi}{2} \right)^{3/2} \sqrt N + \log N + \log z_0 + O(1).$$ Doering, Sargsyan and Sander [@DSS] also study the mean time to extinction starting from a state with a single infective individual, and Kessler [@Kessler] extends these results to cover the whole of the “transition region”, where $\mu-\lambda$ is of order $N^{-1/2}$. The main focus of the paper of Kessler [@Kessler] is the mean [*number*]{} of infections in this regime.

However, it appears that deriving the precise asymptotic distribution of the time $T_e^{X^N}$ to extinction has been open until now, even in the case where $\lambda$ and $\mu$ are fixed constants with $\lambda < \mu$. A formula for this asymptotic distribution is presented in Kryscio and Lefèvre [@KL], with a heuristic argument, and is then reproduced in Andersson and Djehiche [@AD]; this formula is similar in nature to (\[eq.inter\]) below, but different in several important respects. However, it was noted by Doering, Sargsyan and Sander [@DSS] that the formula in [@KL] and [@AD] is inconsistent with their result (\[dss\]), and with their numerical results. In his 2011 book, Nåsell [@Nasell] identifies two distinct regimes: one “critical regime”, where $\mu-\lambda$ is of order at most $N^{-1/2}$, and another (subcritical) where $\mu-\lambda$ is constant or tends to zero more slowly than $N^{-1/2}$. For both regimes, Nåsell [@Nasell] poses as an open problem the determination of the mean extinction time ${\operatorname{\mathbb E{}}}T_e^{X^N}$.

In this paper, we obtain the asymptotic distribution of $T_e^{X^N}$ throughout the subcritical regime. Our main result is as follows. We recall that a random variable $W$ has the [*standard Gumbel distribution*]{} if ${\operatorname{\mathbb P{}}}(W \le w) = e^{-e^{-w}}$ for all $w \in {{\mathbb R}}$. The mean of $W$ is equal to Euler’s constant $\gamma \approx 0.5772$.

\[thm.main\] Suppose that $(\mu - \lambda) N^{1/2} \to \infty$, and that $x_0(\mu-\lambda) \to \infty$. We have $$\label{eq.general}
(\mu-\lambda) T_e^{X^N} - \Big( \log x_0 + \log (\mu-\lambda) - \log\big( 1 + \frac{\lambda x_0}{(\mu-\lambda) N}\big) - \log \mu \Big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution. Therefore $${\operatorname{\mathbb E{}}}_{x_0} T_e^{X^N} = \frac{\log x_0 + \log (\mu-\lambda) - \log\big( 1 + \frac{\lambda x_0}{(\mu-\lambda) N}\big) - \log \mu + \gamma + o(1)}{\mu-\lambda},$$ as $N \to \infty$.

We next give versions of (\[eq.general\]) valid when $x_0/N$ lies in certain ranges. If $(\mu(N)-\lambda(N)) N^{1/2} \to \infty$, and $x_0/N \to z_0$, with $z_0 \in (0,1]$, then Theorem \[thm.main\] gives that $$\label{eq.inter}
(\mu-\lambda) T_e^{X^N} - \big( \log N + \log z_0 + 2\log (\mu-\lambda) - \log(\mu -\lambda +\lambda z_0) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution. In general, (\[eq.general\]) is the most that can be said if $x_0/N$ is of the same order as $\mu-\lambda$ (e.g., if both are constants). On either side of this regime, the formula in (\[eq.general\]) can be simplified. If $x_0 = o((\mu-\lambda)N)$ and $x_0(\mu-\lambda) \to \infty$, then we have $$\label{eq.low}
(\mu-\lambda) T_e^{X^N} - \big( \log x_0 + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, whereas if $x_0/(\mu-\lambda)N \to \infty$, then $$\label{eq.high}
(\mu-\lambda) T_e^{X^N} - \big( \log N + 2\log (\mu-\lambda) - \log \mu -\log \lambda \big) \to W,$$ in distribution, as $N \to \infty$, where in both cases $W$ has the standard Gumbel distribution. In (\[eq.high\]), we necessarily have $\mu-\lambda \to 0$, so either of the terms $\log \mu$ and $\log \lambda$ could be replaced by the other.

We observe that the asymptotic formula for the distribution of $T_e^{X^N}$ in (\[eq.low\]) is independent of $N$, while that in (\[eq.high\]) is independent of $x_0$. An explanation for the first of these phenomena becomes apparent in Section \[sec:final\]: we show that, in this regime, the stochastic SIS logistic process behaves essentially identically to a linear birth-and-death chain with birth-rate $\lambda$ and death rate $\mu$, so that the “logistic correction” to the birth rate does not affect the asymptotics.

If $\mu-\lambda \to 0$, then for large enough $x_0$ (e.g., when $\liminf_{N\to \infty} x_0/N > 0$), we are in the regime covered by (\[eq.high\]), and the time to extinction does not depend significantly on the starting state. To explain this, we give an informal description of the typical course of the epidemic in the case where $\mu-\lambda \to 0$, and we start in some large state, say with $x_0 = z_0N$ and $z_0 > 0$. For such a regime, in the initial phase of the epidemic, the number $X^N_t$ of infectives very rapidly drops until it reaches states of the same order as $(\mu-\lambda)N$. The majority of the time of the epidemic – approximately $\log (N(\mu-\lambda)^2))/(\mu-\lambda)$ – is spent in an “intermediate” phase, getting from there to states of the same order as $(\mu-\lambda)^{-1}$; the time taken to cross this “gap” is very well concentrated around the value derived from the approximating differential equation. Most of the variability of the time to extinction comes from the final phase of the epidemic, starting when the state gets down to about order $(\mu-\lambda)^{-1}$; for this final phase, the differential equation is no longer an adequate guide to the extinction time, and instead the Markov chain is well approximated by a linear birth and death chain. Note that this description relies on having $(\mu-\lambda)N \gg (\mu-\lambda)^{-1}$: if $(\mu-\lambda) \ll N^{-1/2}$, then the situation is completely different (in this regime the time to extinction is essentially distributed as in the case $\mu=\lambda$).

We briefly discuss what happens if we drop the assumption that $x_0(\mu-\lambda) \to \infty$ at the end of Section \[sec:final\].

The plan of the paper is as follows. We track the process through three phases, roughly corresponding to the three phases mentioned in the informal description above. For some regimes, not all the phases are necessary, and we tackle them in “reverse order”, starting with the final phase of the epidemic. Our results are stated in terms of a function $\omega(N)$ tending to infinity suitably slowly, which can be taken equal to $(\mu-\lambda)^{1/4}N^{1/8}$.

We treat the final phase in Section \[sec:final\]. Here, we start from a state $x^*$ below $N^{1/2} \omega(N)$, and show that, from this point on, the stochastic SIS logistic process is well approximated by a linear birth and death chain with the same parameters: as the distribution of the extinction time for a linear birth and death chain is known explicitly, this enables us to estimate very precisely the behaviour of the logistic chain. An alternative way to view this phase of the epidemic is to approximate it by a branching process, where each initially infected individual sparks a brief small epidemic within the population, and these various small epidemics do not interact. The time to extinction is then the maximum of the duration of these small epidemics, and this explains the appearance of the standard Gumbel distribution – which typically arises as the maximum of a number of independent samples from a given distribution – in our formulae.

The intermediate phase is covered in Section \[sec:intermediate\]. Here, effectively we prove Theorem \[thm.main\] under the additional assumption that $x_0 \le (\mu-\lambda)N \omega(N)$. In this phase, we use a method, based on a concentration of measure inequality, developed by Luczak [@Luczak] to show that the scaled process $(X^N_t/N)$ stays close to the solution of the differential equation (\[eq.diff-eq\]), for a (deterministic) period of time until there are about $N^{1/2} \omega(N)$ infective individuals (from which point the analysis for the final phase can be invoked). The general method of Kurtz [@Kurtz] for showing that a Markov chain follows a differential equation would not suffice for our purposes: the idea of this approach is to use the Lipschitz constant of the differential equation to bound the rate at which the solution to the differential equation and the Markov chain get further apart: in our setting, this would only show that the Markov chain follows the differential equation for a fixed time, or at best a time given by a small constant times $\log N$. Our method allows us to take advantage of the fact that, if we have two copies of the chain started in different states, we can couple them so that the expected distance between the copies is always decreasing. We believe that our method can often be used to show that Markov chains with this “contractive” property follow differential equations for longer periods. Another illustration of this approach is given in [@BL].

In Section \[sec:initial\], we provide a fairly crude upper bound on the time taken for the initial phase of the chain, starting from any state and reaching a state of order about $N(\mu-\lambda)\omega(N)$. The point is that the time taken for this phase is negligible compared to the overall extinction time. In Section \[sec.proof\], we combine the results of the preceding sections to prove Theorem \[thm.main\].

Our results have some bearing on the “critical regime”, where $|\mu-\lambda| = O(N^{-1/2})$. In particular, the results of Section \[sec:initial\] can be used to show that the expected time taken for the epidemic, started from an arbitrary state, to reach a state of order about $N^{1/2}$ is of order at most $N^{1/2}$. Dolgoarshinnykh and Lalley [@DL] show that, in this regime, the scaled stochastic SIS epidemic process started from a state of order $N^{1/2}$ converges in law to a diffusion process they call the “attenuated Feller diffusion”. One consequence is that the time to extinction from states of order $N^{1/2}$ is of order $N^{1/2}$ (and is not well-concentrated). We discuss the critical regime in the short final section: we make no attempt to provide precise quantitative results.

Throughout the paper, we shall use the notation ${\operatorname{\mathbb P{}}}(\cdot)$ and ${\operatorname{\mathbb E{}}}(\cdot)$ to denote probability in whatever space we are working in. Sometimes we find it convenient to use a subscript $x_0$ to indicate that we are working with a process with starting state $x_0$. We will sometimes treat a real quantity as though it were an integer, so as to avoid any unnecessary clutter with floors and ceilings.

Final phase: approximation by linear birth and death chains {#sec:final}
===========================================================

In this section, we show that the stochastic SIS logistic process is well approximated by a linear birth and death chain, from a state $x^*$ until extinction, provided $x^* \le N^{5/8}(\mu-\lambda)^{1/4}$. We also assume that $x^*(\mu-\lambda) \to \infty$: this assumption is necessary only for expressing the distribution of the extinction time of the approximating linear birth and death chain in a convenient form.

A [*linear birth and death chain*]{} with parameters $(\lambda, \mu)$ is a continuous-time Markov chain $(Y_t)_{t\ge 0}$ with state space ${{\mathbb Z}}_+ = \{ 0, 1,2,\dots\}$ and the following transitions: $$\begin{aligned}
y &\to& y+1 \quad \mbox{ at rate } \quad \lambda y, \\
y &\to& y-1 \quad \mbox{ at rate } \quad \mu y.\end{aligned}$$

Let $T^Y_e$ be the hitting time of the absorbing state $0$ for the Markov chain $(Y_t)$ starting at $x^*$. It is known, see for instance Renshaw [@Renshaw] (2.4.23), that, for $t \ge 0$ and $\mu \not= \lambda$, $$\begin{aligned}
{\operatorname{\mathbb P{}}}(T^Y_e \le t) &=& {\operatorname{\mathbb P{}}}(Y_t = 0) \nonumber \\
&=& \left( \frac{\mu  - \mu e^{-(\mu-\lambda)t}}{\mu - \lambda e^{-(\mu-\lambda)t} } \right)^{x^*} \nonumber \\
&=& \left( 1 - \frac{(\mu-\lambda) e^{-(\mu - \lambda) t}}{\mu - \lambda e^{-(\mu-\lambda)t}} \right)^{x^*}. \label{eq.lbdc}\end{aligned}$$ We shall be interested in comparing the stochastic SIS logistic process with parameters $\lambda = \lambda(N)$, $\mu = \mu(N)$ and $x^*=x^*(N)$ with the corresponding linear birth and death chain. Accordingly, we shall be interested in the case where all the parameters of the Markov chain $(Y_t)$ vary with $N$, and in this case we shall use the notation $(Y^N_t)$ and $T_e^{Y^N}$ to indicate that we are dealing with a sequence of Markov chains with different parameters, indexed by $N$.

We shall use the following routine result several times in the course of the paper.

\[lem.escape\] Let $(Y_t)$ be a birth-and-death chain such that, for any state $x$, the probability that the next step is upwards is at most $\lambda/(\lambda+\mu) < 1/2$. Suppose the chain is started in state $x^*$. For any state $y > x^*$, the probability that the chain $(Y_t)$ reaches $y$ before it reaches 0 is at most $$\frac{(\mu/\lambda)^{x^*} - 1}{(\mu/\lambda)^y - 1} \le \left( \frac{\lambda}{\mu} \right)^{y-x^*} \le \exp \left( - \frac{\mu-\lambda}{\mu} (y-x^*)\right).$$

Suppose that the linear birth and death chain $(Y_t^N)$ is started in a state $x^* = x^*(N)$, where $x^*(\mu-\lambda) \to \infty$ as $N \to \infty$. For each fixed $w \in {{\mathbb R}}$, we set $$t_w = t_w(\mu,\lambda,x^*) = \frac{\log x^* + \log (\mu-\lambda) - \log \mu + w}{\mu-\lambda},$$ and note that $t_w > 0$ for sufficiently large $N$. Restricting to those $N$ for which $t_w$ is indeed positive, we have that $e^{-(\mu-\lambda)t_w} = \mu e^{-w} / (\mu-\lambda)x^*$, and hence $$\begin{aligned}
{\operatorname{\mathbb P{}}}(T^{Y^N}_e \le t_w) &=& \left( 1 - \frac{\mu e^{-w}/x^*}{\mu - \lambda \mu e^{-w}/(\mu-\lambda)x^*} \right)^{x^*} \\
&=& \left( 1 - \frac{e^{-w}}{x^* - \lambda e^{-w}/(\mu-\lambda)} \right)^{x^*} \to e^{-e^{-w}},\end{aligned}$$ as $N \to \infty$, since $x^*(\mu-\lambda) \to \infty$. This can be written as $$\label{eq.lbdc2}
(\mu - \lambda) T^{Y^N}_e - \big( \log x^* + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution (${\operatorname{\mathbb P{}}}(W\le w) = e^{-e^{-w}}$), provided $x^*(\mu-\lambda) \to \infty$.

\[lem.late\] Suppose that $(\mu-\lambda)N^{1/2} \to \infty$, and set $\omega (N) = \big( (\mu-\lambda)N^{1/2} \big)^{1/4}$. Take any function $x^* = x^*(N)$ such that $x^*(\mu-\lambda) \to \infty$ and $x^* \le N^{1/2} \omega(N)$.

Suppose that $(X^N_t)$ is a copy of the stochastic SIS logistic process, started in state $x^*$. Then $$(\mu - \lambda) T_e^{X^N} - \big( \log x^* + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution.

Note that this result is the same as the special case of Theorem \[thm.main\] covered by (\[eq.low\]), with $x_0$ replaced by $x^*$, under the slightly more restrictive hypothesis that $x^* \le N^{1/2} \omega(N) = N^{5/8}(\mu-\lambda)^{1/4}$ instead of $x^* = o((\mu-\lambda)N)$. Later results will supply the conclusion of Theorem \[thm.main\] with no restrictions on the starting state.

The plan of the proof is to sandwich the stochastic SIS logistic process between two linear birth and death chains: the upper bound on $x^*$ is to ensure that the distributions of the extinction times of the two linear birth and death chains are asymptotically equal.

We couple three Markov chains: one is the stochastic SIS logistic process $(X^N_t)$, another is the linear birth and death chain $(Y^N_t)$ with the same parameters $(\lambda,\mu)$ as $(X^N_t)$, and the third is a different linear birth and death chain $(Z^N_t)$, this one with parameters $(\lambda', \mu)$ where $\lambda' = \lambda (1-2x^*/N)$. All three processes start at $x^*$. Let $\tau$ be the first time that either $Y^N_t=0$ or $Y^N_t = 2x^*$. The infection rate of $(X^N_t)$ is $\lambda X^N_t (1- X^N_t/N)$, which for $t < \tau$ is sandwiched between the birth rates of the two linear birth and death chains in the same state. We may thus construct a coupling with $Z^N_t \le X^N_t \le Y^N_t$ for all $t\le \tau$ and all $N$: the rule is that, if any two chains are in the same state, then they make jumps together as far as possible; otherwise two chains in different states make jumps independently according to their given transition rates, and so they a.s. do not jump simultaneously (so they do not cross). Under this coupling, on the event that $Y^N_\tau = 0$ (i.e., the “highest” chain reaches 0 before it reaches the upper boundary $2x^*$), we have $T^{Z^N}_e \le T_e^{X^N} \le T^{Y^N}_e$.

By Lemma \[lem.escape\], the probability that $(Y^N_t)$ escapes to $2x^*$ before it reaches 0 is at most $e^{-(\mu-\lambda)x^*/\mu} = o(1)$, since $(\mu-\lambda)x^* \to \infty$.

For each fixed $w$ and $N$, we choose $v=v(w,N)$ so that $t_w(\mu,\lambda,x^*) = t_v(\mu,\lambda',x^*)$, i.e., $$\frac{\log x^* + \log (\mu-\lambda) - \log \mu + w}{\mu-\lambda} = \frac{\log x^* + \log (\mu-\lambda') - \log \mu + v}{\mu-\lambda'}.$$ This translates to $$v(w,N) - w = \frac{\lambda-\lambda'}{\mu-\lambda}\left( \log x^* + \log(\mu-\lambda) -\log \mu + w\right) -
\log \left(1 + \frac{\lambda-\lambda'}{\mu-\lambda} \right).$$ We observe that $$\frac{\lambda - \lambda'}{\mu-\lambda} = \frac{2x^* \lambda}{N(\mu-\lambda)} \le \frac{2\lambda \omega(N)}{N^{1/2}(\mu-\lambda)}
\le \frac{2\lambda}{\omega(N)^3}.$$ Also we have, for $N$ sufficiently large, $$\log (x^*(\mu-\lambda)) \le (x^*(\mu-\lambda))^{1/2} \le \big( N^{1/2}\omega(N) N^{-1/2}\omega(N)^4 \big)^{1/2} = \omega(N)^{5/2}.$$ Therefore, for each fixed $w$, we have both $$\frac{\lambda-\lambda'}{\mu-\lambda}\left( \log x^* + \log(\mu-\lambda) -\log \mu + w\right) = o(1); \quad
\log \left(1 + \frac{\lambda-\lambda'}{\mu-\lambda} \right) = o(1),$$ and so $|v(w,N)-w| = o(1)$. In other words, $v(w,N) \to w$ as $N\to \infty$, for each fixed $w$. Thus $$\begin{aligned}
\lefteqn{|{\operatorname{\mathbb P{}}}(T^{Z^N}_e \le t_w(\mu,\lambda,x^*)) - e^{-e^{-w}}|} \\
&=& |{\operatorname{\mathbb P{}}}(T^{Z^N}_e \le t_v(\mu,\lambda',x^*)) - e^{-e^{-w}}| \\
&\le& | {\operatorname{\mathbb P{}}}(T^{Z^N}_e \le t_v(\mu,\lambda',x^*)) - e^{-e^{-v(w,N)}}| + | e^{-e^{-v(w,N)}} - e^{-e^{-w}}| \to 0\end{aligned}$$ as $N \to \infty$. Here we used (\[eq.lbdc2\]) applied to $(Z^N_t)$; applying the same formula to $(Y^N_t)$ tells us that $$|{\operatorname{\mathbb P{}}}(T^{Y^N}_e \le t_w(\mu,\lambda,x^*)) - e^{-e^{-w}}| = o(1).$$

As $(X^N_t)$ is sandwiched between $(Y^N_t)$ and $(Z^N_t)$ on the event that $Y^N_\tau = 0$, which occurs with probability $1-o(1)$, we now deduce that $$|{\operatorname{\mathbb P{}}}(T^{X^N}_e \le t_w(\mu,\lambda,x^*)) - e^{-e^{-w}}| = o(1).$$ Equivalently, $$(\mu - \lambda) T_e^{X^N} - \big( \log x^* + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution, as claimed.

In the case where $(\mu-\lambda)x^*$ does not tend to infinity, the same argument tells us that the distribution of the extinction time of the stochastic SIS logistic process is asymptotically the same as that of the linear birth and death chain with the same parameters. The formula (\[eq.lbdc\]) for the distribution of the extinction time of the linear birth and death chain can be approximated for various different parameter regimes. For convenience, we state two asymptotic formulae, doubtless familiar in other contexts.

For instance, it can be seen that, if $(\mu-\lambda)x^* \to 0$ but $x^* \to \infty$, then for any $v \in (0,\infty)$, $${\operatorname{\mathbb P{}}}\big(T_e^{X^N} \le v x^* /\mu\big) \to e^{-1/v}$$ as $N \to \infty$. We can also consider the case where the epidemic starts with a single infective: if $\mu-\lambda \to 0$ and $x_0 = 1$, then, for any $u \in (0, \infty)$, $${\operatorname{\mathbb P{}}}\big(T_e^{X^N} \le u /\mu\big) \to \frac{u}{1+ u}$$ as $N \to \infty$.

Intermediate phase: differential equation approximation {#sec:intermediate}
=======================================================

Fix some function $\omega(N)$ tending to infinity. In this section, we follow the process from a starting state below $(\mu-\lambda) N \omega(N)$, for a deterministic period of time $t^*$, and show that $X^N_{t^*}$ is close to $x^* = x^*(N) = N^{1/2} \omega(N)$ with probability $1-o(1)$ as $N \to \infty$. Our basic plan is to show that the process, suitably scaled, is well approximated by the differential equation (\[eq.diff-eq\]) in this phase.

The technique we use requires us to work with discrete-time Markov chains. Accordingly, we next define a discrete-time version of the logistic process; later we shall convert our results to the continuous-time case. We consider a Markov chain $({\widehat{X}}_k)_{k \in {{\mathbb Z}}_+} = ({\widehat{X}}_k^{N,K})_k$, with state space $\{ 0, \dots, N\}$ and transition probabilities given by $$p_{x,x+1} = \frac{\lambda}{K(\mu+\lambda)N} x (1 - x/N), \qquad p_{x,x-1} = \frac{\mu}{K(\mu+\lambda)N} x,$$ $$p_{x,x} = 1 - p_{x,x+1} - p_{x,x-1},$$ for $x = 0, \dots, N$. In the above, $K \ge 2$ is a constant that may depend on the parameters of the model, but not on $x$. The restriction $K \ge 2$ ensures that $p_{x,x} \ge \frac12$ for all $x$, but in fact $K$ can be set arbitrarily large without affecting the argument, and we shall later let $K$ tend to infinity. (We choose to express the denominators here as $K(\mu+\lambda)N$ so as to make $K$ a “dimensionless” constant.) The idea is that one step of the discrete-time chain approximates the continuous-time chain over a time period of $1/K(\mu+\lambda)N$, in a sense that we will make precise later.

We shall use the following result of Luczak [@Luczak]. For a discrete-time Markov chain with transition matrix $P$, a function $f$ on the state space $S$ of the chain, and a state $x \in S$, we write $(Pf)(x) = \sum_y P(x,y) f(y)$, where the sum is over all states $y$ with $P(x,y) > 0$.

\[thm.concb-general\] Let $P$ be the transition matrix of a discrete-time Markov chain with discrete state space $S$. Let $f: S \to {{\mathbb R}}$ be a function. Suppose the set $S_0$ and numbers $a_{x,i}(y)$ ($x,y  \in S_0$) satisfy the following conditions: for all $i \in {{\mathbb Z}}^+$ and all $x,y \in S_0$ such that $P(x,y) > 0$, $$\label{eq.coupling}
|{\operatorname{\mathbb E{}}}_x [f(X_i)] -  {\operatorname{\mathbb E{}}}_y [f(X_i)]| \le a_{x,i}(y).$$ Let $S_0^0 = \{x \in S_0: y \in S_0 \mbox{ whenever } P(x,y) > 0\}$. Assume that, for some sequence $(\alpha_i)_{i \in {{\mathbb Z}}^+}$ of positive constants, $$\sup_{x \in S_0^0} (P a_{x,i}^2)(x) \le \alpha_i^2.$$ Let $k >0$, and let $\beta = 2\sum_{i=0}^{k-1} \alpha_i^2$. Suppose also that $\alpha$ is such that $$2\sup_{0 \le i \le k -1}  \sup_{x \in S^0_0, y: P(x,y)>0} \alpha_i (x,y)\le \alpha.$$ Finally, let $A_k= \{\omega: X_s(\omega) \in S_0^0: 0 \le s \le k-1\}$. Then, for all $a >0$, $${\operatorname{\mathbb P{}}}_{x_0} \Big ( \left \{ |f(X_k)-{\operatorname{\mathbb E{}}}_{x_0} [f(X_k)] |\ge a \right \} \cap A_k \Big)
\le 2e^{-a^2/(2\beta+2\alpha a/3)}.$$

Our chain $({\widehat{X}}_k)$ is a birth-and-death chain, so that, for $x > 0$, $P(x,y) > 0$ if and only if $|x-y| \le 1$. We shall apply Theorem \[thm.concb-general\] to the function $f(x) = x$. To establish (\[eq.coupling\]), we will analyse a coupling and show that the expected difference between copies of the chain started in adjacent states decreases exponentially in time, so that the quantity $\beta$ is uniformly bounded for all time. The conclusion then gives us a concentration of measure inequality valid for all time (although in practice it is only useful if the expectation of ${\widehat{X}}_k$ is not too small).

Here and in future, for $z\in (0,1]$, we set $$s(z) = \log \left( 1 + \frac{\lambda}{\mu-\lambda} z \right) - \log z.$$ The significance of this function for us is that, for any $z_0 \in (0,1]$, the differential equation (\[eq.diff-eq\]) has an explicit solution $$z(t) = \frac{ z_0 (\mu-\lambda) e^{-(\mu-\lambda)t}}{\mu-\lambda + z_0 \lambda(1- e^{-(\mu-\lambda)t})},$$ for $t\ge 0$, where $z(0) = z_0$. The inverse of the function $z(t)$ is given by, for $0< z \le z_0$, $$t(z) = \frac{s(z) - s(z_0)}{\mu-\lambda}.$$ Our overall aim is to show that the time for the Markov chain $(X^N_t)$ to travel from $x_0 = z_0N$ to $x^* =z^*N$ is well-concentrated around $t(z^*)$: the total time to extinction will then be obtained by adding $t(z^*)$ to the time to extinction from $x^*$, which is covered in Lemma \[lem.late\]. Our first task is to establish the analogous result for the discrete time approximation $({\widehat{X}}_k)$.

\[lem.inter\] Suppose $(\mu-\lambda) N^{1/2} \to \infty$ as $N \to \infty$. Let $\omega(N)$ be any function tending to infinity with $N$, and set $x^* = N^{1/2} \omega(N)$. Let $x_0 = x_0(N)$ be any function such that $x^* \le x_0 \le N (\mu-\lambda) \omega(N)$.

Then, whenever $K \ge N$, $${\operatorname{\mathbb P{}}}_{x_0}\big( | {\widehat{X}}_{\lceil k^*\rceil}^{N,K} - x^* | \ge \omega(N)^{-1/3} x^* \big) = o(1),$$ as $N \to \infty$, where $$k^* = \frac{K(\mu+\lambda)N}{\mu-\lambda} \Big( s\Big(\frac{x^*}{N}\Big) - s\Big(\frac{x_0}{N}\Big) \Big).$$

As mentioned above, we first describe and analyse a coupling between two copies $({\widehat{X}}_k)_{k\in {{\mathbb Z}}_+}$ and $({\widehat{Y}}_k)_{k\in {{\mathbb Z}}_+}$ of the discrete chain. We couple the chains so that, if ${\widehat{X}}_k = {\widehat{Y}}_k$, then ${\widehat{X}}_{k+1} = {\widehat{Y}}_{k+1}$, and otherwise at most one of the chains moves at each step: such a coupling exists since $p_{x,x} \ge 1/2$ for each state $x$.

We claim that, for each $k \ge 0$, $${\operatorname{\mathbb E{}}}\big( | {\widehat{X}}_{k+1} - {\widehat{Y}}_{k+1} | \mid {\mathcal F}_k \big) \le |{\widehat{X}}_k - {\widehat{Y}}_k|
\big(1- \frac{\mu-\lambda}{K(\mu+\lambda)N}\big),$$ where $({\mathcal F}_k)$ is the natural filtration of the coupling. This is certainly true on the event that ${\widehat{X}}_k = {\widehat{Y}}_k$. Otherwise, we assume without loss of generality that ${\widehat{X}}_k > {\widehat{Y}}_k$, and note that any change to one of ${\widehat{X}}$ or ${\widehat{Y}}$ at step $k+1$ either reduces or increases the distance by 1. (The requirement that only one of the chains move unless they have already coalesced is to ensure that the chains do not “cross”.) Then we have $$\begin{aligned}
\lefteqn{{\operatorname{\mathbb E{}}}({\widehat{X}}_{k+1} - {\widehat{Y}}_{k+1} \mid {\mathcal F}_k)} \\
&=& ({\widehat{X}}_k - {\widehat{Y}}_k) + \frac{1}{K(\mu+\lambda)N}
\left( \lambda {\widehat{X}}_k ( 1 - \frac{{\widehat{X}}_k}{N}) - \lambda {\widehat{Y}}_k (1 - \frac{{\widehat{Y}}_k}{N}) - \mu {\widehat{X}}_k + \mu {\widehat{Y}}_k \right) \\
&=& ({\widehat{X}}_k - {\widehat{Y}}_k) \left( 1 + \frac{1}{K(\mu+\lambda)N} \big(\lambda - \mu  - \lambda \frac{{\widehat{X}}_k + {\widehat{Y}}_k}{N} \big) \right) \\
&\le& ({\widehat{X}}_k - {\widehat{Y}}_k) \left( 1 - \frac{\mu-\lambda}{K(\mu+\lambda)N} \right),\end{aligned}$$ as claimed.

By induction, we obtain that, for all $k \ge 0$, $${\operatorname{\mathbb E{}}}( |{\widehat{X}}_k - {\widehat{Y}}_k | \mid {\mathcal F}_0 ) \le |{\widehat{X}}_0 - {\widehat{Y}}_0| \Big(1 - \frac{\mu-\lambda}{K(\mu+\lambda)N} \Big)^k
\le |{\widehat{X}}_0 - {\widehat{Y}}_0| e^{-(\mu-\lambda) k/K(\mu+\lambda)N }.$$

We shall apply Theorem \[thm.concb-general\] with $S_0 = \{ 0, \dots, 2x_0\}$, $f(x) = x$ and $a_{x,i}(y) = e^{-(\mu-\lambda) i/K(\mu+\lambda)N}$ whenever $|x-y| = 1$ (and $a_{x,i}(x) = 0$ for all $x$). We note that $$P a_{x,i}^2 = (p_{x,x+1} + p_{x,x-1}) e^{-2(\mu-\lambda) i /K(\mu+\lambda)N} \le \frac{2x_0}{KN} e^{-2(\mu-\lambda) i/K(\mu+\lambda)N},$$ for all $x \in S_0$. Then we may take $$\alpha_i^2 = \frac{2x_0}{KN} e^{-2(\mu-\lambda) i/K(\mu+\lambda)N},$$ and so, since $(\mu-\lambda)/K(\mu+\lambda)N \le 1$, we have $$\beta = \sum_{i=0}^k \alpha_i^2 \le \frac{2x_0}{KN} \frac{1}{1- e^{-2(\mu-\lambda) /K(\mu+\lambda)N}} \le \frac{2x_0}{KN} \frac{K(\mu+\lambda)N}{\mu-\lambda}
= \frac{2x_0 (\lambda+\mu)}{\mu-\lambda},$$ and $\alpha = 2 \ge 2 \max_{i,x,y} a_{x,i}(y)$. Theorem \[thm.concb-general\] now gives that $${\operatorname{\mathbb P{}}}_{x_0} \Big( \{ | {\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k | \ge a \} \cap A_k \Big) \le 2 \exp\left( - \frac{a^2}{4x_0(\lambda+\mu)/(\mu-\lambda) + \frac43 a} \right),$$ for all $a \ge 0$ and all $k \ge 0$, where $A_k$ is the event that ${\widehat{X}}_\ell < 2x_0$ for all $\ell \in \{ 0, \dots, k\}$.

We see that ${\operatorname{\mathbb P{}}}(\overline{A_k})$ is at most the probability that the chain $({\widehat{X}}_k)$, started at $x_0$, reaches $2x_0$ before it reaches $0$. By Lemma \[lem.escape\], $${\operatorname{\mathbb P{}}}(\overline{A_k}) \le e^{-(\mu-\lambda)x_0/\mu}.$$ Therefore we have $${\operatorname{\mathbb P{}}}_{x_0} \Big( | {\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k | \ge a \Big) \le 2 \exp\left( - \frac{a^2}{4x_0(\lambda+\mu)/(\mu-\lambda) + \frac43 a} \right)
+ e^{-(\mu-\lambda)x_0/\mu},$$ for all $a \ge 0$ and all $k \ge 0$. In particular, for $a \le 2x_0$, we have $$\begin{aligned}
{\operatorname{\mathbb P{}}}_{x_0} \Big( | {\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k | \ge a \Big) &\le& 2 \exp\left( - \frac{a^2(\mu-\lambda)}{8x_0(\lambda+\mu)} \right) + 2 e^{-3a/8} \nonumber \\
&&\mbox{} + \exp\left( -\frac{(\mu-\lambda)a^2}{4 x_0 \mu} \right) \nonumber \\
&\le & 3 \exp\left( - \frac{a^2(\mu-\lambda)}{8x_0(\lambda+\mu)} \right) + 2 e^{-3a/8}. \label{eq.concwXk}\end{aligned}$$

For $a \ge 2x_0$, we note that $0 \le {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k \le x_0$ for all $k \ge 0$, so $${\operatorname{\mathbb P{}}}_{x_0} \Big( | {\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k | \ge a \Big) \le {\operatorname{\mathbb P{}}}_{x_0} ( {\widehat{X}}_k \ge 2x_0 ) \le \exp\left( - \frac{\mu-\lambda}{\mu} (a-x_0) \right),$$ using Lemma \[lem.escape\].

For the calculation that follows, we need a bound on the variance of ${\widehat{X}}_k$, which we can now obtain readily. For sufficiently large $N$, $$\begin{aligned}
\lefteqn{{\operatorname{\mathbb E{}}}_{x_0}({\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k)^2} \\
&\le& \int_{z=0}^\infty {\operatorname{\mathbb P{}}}_{x_0}(|{\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k| \ge \sqrt z\ ) \, dz \\
&\le& \int_{z=0}^\infty 3 e^{-z(\mu-\lambda)/8x_0(\lambda + \mu)} + 2 e^{-3\sqrt z/8} \, dz
+ \int_{z=4x_0^2}^\infty e^{-(\mu-\lambda)(\sqrt z-x_0)/\mu} \, dz \\
&=& \frac{24 x_0 (\lambda + \mu)}{\mu-\lambda} + \frac{256}{9} +
\left( \frac{4\mu x_0}{\mu-\lambda} + \frac{2\mu^2}{(\mu-\lambda)^2} \right) e^{-(\mu-\lambda)x_0/\mu} \\
&\le& \frac{30x_0(\lambda+\mu)}{\mu-\lambda} \le \frac{60x_0 \mu}{\mu-\lambda},\end{aligned}$$ since our hypotheses imply that $(\mu-\lambda)x_0 \ge (\mu-\lambda) x^* \to \infty$ as $N \to \infty$.

We now show that ${\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k/N$ remains close to the appropriately scaled solution $z(t)$ of the differential equation (\[eq.diff-eq\]) with $z(0) = z_0 = x_0/N$ until the chain drops to about $x^*$. Accordingly, we consider $$E_k = \frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} - z\Big(\frac{k}{K(\mu+\lambda)N}\Big).$$ Note that $$\begin{aligned}
\lefteqn{{\operatorname{\mathbb E{}}}_{x_0} ({\widehat{X}}_{k+1} - {\widehat{X}}_k)} \\
&=& {\operatorname{\mathbb E{}}}_{x_0} \big( {\operatorname{\mathbb E{}}}_{x_0} ({\widehat{X}}_{k+1} - {\widehat{X}}_k \mid {\mathcal G}_k) \big) \\
&=& {\operatorname{\mathbb E{}}}_{x_0} \Big( \frac{1}{K(\lambda+\mu)N} \big(\lambda {\widehat{X}}_k \big(1- \frac{{\widehat{X}}_k}{N}\big) - \mu{\widehat{X}}_k\big) \Big) \\
&=& \frac{1}{K(\lambda+\mu)N} \Big( -(\mu - \lambda) {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k - \lambda \frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k^2}{N} \Big) \\
&=& \frac{1}{K(\lambda+\mu)N} \Big( -(\mu - \lambda) {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k - \lambda \frac{({\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k)^2}{N}
- \lambda \frac{{\operatorname{\mathbb E{}}}_{x_0} ({\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k)^2}{N} \Big).\end{aligned}$$ By Taylor’s Theorem, we have that, for some $u \in \big(\frac{k}{K(\mu+\lambda)N}, \frac{k+1}{K(\mu+\lambda)N}\big)$, $$\begin{aligned}
\lefteqn{z\Big( \frac{k+1}{K(\mu+\lambda)N}\Big) - z\Big(\frac{k}{K(\mu+\lambda)N} \Big)} \\
&=& \frac{1}{K(\lambda+\mu)N} z'\Big(\frac{k}{K(\mu+\lambda)N}\Big) + \frac{1}{2K^2(\mu+\lambda)^2N^2} z''(u) \\
&=& \frac{1}{K(\lambda+\mu)N} \Big[ -(\mu-\lambda) z\Big(\frac{k}{K(\mu+\lambda)N}\Big) - \lambda z\Big(\frac{k}{K(\mu+\lambda)N}\Big)^2 \Big] \\
&&\mbox{} + \frac{1}{2K^2(\mu+\lambda)^2N^2} z''(u),\end{aligned}$$ since $z(t)$ satisfies (\[eq.diff-eq\]). Thus we have, for some $u \in \big(\frac{k}{K(\mu+\lambda)N}, \frac{k+1}{K(\mu+\lambda)N}\big)$, $$\begin{aligned}
\lefteqn{E_{k+1} - E_k} \\
&=& \frac{1}{K(\mu+\lambda)N} \Big[ -(\mu-\lambda) \frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} - \lambda \left( \frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} \right) ^2
- \frac{\lambda}{N^2} {\operatorname{\mathbb E{}}}_{x_0} ({\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k)^2 \\
&&\mbox{} + (\mu-\lambda) z\Big(\frac{k}{K(\mu+\lambda)N}\Big) + \lambda z\Big(\frac{k}{K(\mu+\lambda)N}\Big)^2 - \frac{1}{2K(\mu+\lambda)N} z''(u) \Big] \\
&=& \frac{1}{K(\mu+\lambda)N} \Big[ -(\mu-\lambda)E_k - \lambda E_k \Big(\frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} + z\Big(\frac{k}{K(\mu+\lambda)N}\Big) \Big) \\
&&\mbox{} - \frac{\lambda}{N^2} {\operatorname{\mathbb E{}}}_{x_0} ({\widehat{X}}_k - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k)^2 - \frac{1}{2K(\mu+\lambda)N} z''(u) \Big].\end{aligned}$$ One may easily check that $0 \le z''(u) \le \mu(\mu+\lambda)$ for all $u \in [0,1]$, so we may now write $$\begin{aligned}
|E_{k+1}| &\le& |E_k| \Big( 1 - \frac{(\mu-\lambda) + \lambda \frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} + \lambda z\big(\frac{k}{K(\lambda+\mu)N}\big)}{K(\mu+\lambda)N} \Big) \\
&& \mbox{} + \frac{\lambda}{K(\mu+\lambda)N^3} \frac{60 x_0 \mu}{\mu-\lambda}+ \frac{1}{2K^2(\mu+\lambda)^2N^2} \sup_u |z''(u)| \\
&\le& |E_k| \Big( 1 - \frac{\mu-\lambda}{K(\mu+\lambda)N} \Big) + \frac{60\lambda x_0\mu}{(\mu-\lambda)(\mu+\lambda) K N^3} + \frac{\mu}{2K^2(\mu+\lambda)N^2}.\end{aligned}$$

It follows by induction that, for all $k \ge 0$, $$|E_k| \le \frac{60 x_0 \lambda\mu}{N^2 (\mu-\lambda)^2} + \frac{\mu}{2K(\mu-\lambda) N}.$$ Provided $K \ge N$, the second term above is at most $\lambda\mu/N^2(\mu-\lambda)^2$, so we have, for all $k\ge 0$, $$\label{eq.Ek}
\Big|\frac{{\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_k}{N} - z\Big(\frac{k}{K(\mu+\lambda)N}\Big) \Big|= |E_k| \le \frac{70 x_0 \lambda \mu}{N^2 (\mu-\lambda)^2}.$$

Recall that $t(z) = (s(z) - s(z_0))/(\mu-\lambda)$ is the inverse to $z(t)$. We set $t^* = t(x^*/N)$, and $k^* = K (\mu+\lambda)N t^*$, so that $k^*$ is as in the statement of the lemma, and $z\big( \frac{k^*}{K(\mu+\lambda)N} \big) = z(t^*) = x^*/N$. Note that $k^* \ge 0$ since $x_0 \ge x^*$. Also, since $|z'(t)| \le \mu$ for all $t \in [0,1]$, we have $$\left| z \Big(\frac{\lceil k^*\rceil }{K(\mu+\lambda)N} \Big) - z \Big( \frac{k^*}{K(\mu+\lambda)N} \Big) \right| \le \frac{\mu}{K(\mu+\lambda)N} \le \frac{1}{N}.$$ We then obtain from (\[eq.Ek\]) that $$\label{eq.f1}
\left| {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_{\lceil k^*\rceil} - x^* \right| \le \frac{70 x_0\lambda \mu}{N(\mu-\lambda)^2} + 1 \le \frac{70 \lambda \mu \omega(N)}{\mu-\lambda} + 1 = o(x^*),$$ since $(\mu-\lambda)N^{1/2} \to \infty$ and $x^* = N^{1/2} \omega(N)$. Also, from (\[eq.concwXk\]) with $a = \omega(N)^{-1/3} x^*$, we have that $$\begin{aligned}
\lefteqn{{\operatorname{\mathbb P{}}}_{x_0} \left( \left| {\widehat{X}}_{\lceil k^*\rceil} - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_{\lceil k^*\rceil } \right | \ge \omega(N)^{-1/3} x^* \right)} \\
&\le& 3 \exp\left( - \frac{\omega(N)^{-2/3}(x^*)^2(\mu-\lambda)}{8x_0(\lambda+\mu)} \right) + 2 e^{-3\omega(N)^{-1/3} x^*/8} \\
&\le& 3 \exp\left( - \frac{\omega(N)^{1/3}}{8 (\lambda + \mu)} \right) + 2 e^{-3 N^{1/2} \omega(N)^{2/3} /8} = o(1).\end{aligned}$$ Hence $$\label{eq.f2}
{\operatorname{\mathbb P{}}}_{x_0} \left( \left| {\widehat{X}}_{\lceil k^*\rceil} - {\operatorname{\mathbb E{}}}_{x_0} {\widehat{X}}_{\lceil k^*\rceil} \right | \ge \omega(N)^{-1/3} x^* \right) \to 0.$$

From (\[eq.f1\]) and (\[eq.f2\]), we deduce that, provided $K \ge N$, $${\operatorname{\mathbb P{}}}_{x_0} \left( \left| {\widehat{X}}_{\lceil k^*\rceil} - x^* \right| \ge \omega(N)^{-1/3} x^* \right) = o(1),$$ as $N \to \infty$, as desired.

The next result shows how to connect the discrete-time chain $({\widehat{X}}_k)$ and the continuous-time chain $(X^N_t)$.

\[lem.disc-cont\] For each fixed $N$, $K$, and $t_0 \ge 0$, $$d_{TV} \left( \mathcal{L}(X^N_{t_0}), \mathcal{L}({\widehat{X}}^{N,K}_{\lceil K(\mu+\lambda)Nt_0\rceil}) \right) \le \frac{4 (\mu+\lambda)N t_0 + 1}{K},$$

Recall that $(X^N_t)$ is a continuous-time Markov chain with transitions given by $x \to x+1$ at rate $\lambda x (1-x/N)$ and $x \to x-1$ at rate $\mu x$. We fix $K \ge 2$ for the moment, and set $\delta = 1/K(\mu+\lambda)N$. We also assume for the moment that $t_0$ is an integer multiple of $\delta$. Recall now also that $({\widehat{X}}_k) = ({\widehat{X}}^{N,K}_k)$ is a discrete time chain with transition probabilities given by $$p_{x,x+1} = \delta \lambda x (1 - x/N), \qquad p_{x,x-1} = \delta \mu x,$$ $$p_{x,x} = 1 - p_{x,x+1} - p_{x,x-1},$$ for $x = 0, \dots, N$.

We consider the sequence $(Z_k)_k = (X^N_{k\delta})_k$, obtained by observing $(X^N_t)$ at each integer multiple of $\delta$, as a discrete-time Markov chain. Our plan is to show that $(Z_k)$ is well-approximated by $({\widehat{X}}_k)$, for all $k$ up to $t_0/\delta = K(\mu+\lambda)Nt_0$. The idea here is that, for $K$ large, the chain $(X^N_t)$ usually does not jump in an interval of length $\delta$, which means that $(Z_k)$ usually does not move. Moreover, the chain $(X^N_t)$ is very unlikely to take two jumps within the same small interval of length $\delta$ at any time during the period $[0,t_0]$.

To be precise, let $P'$ be the transition matrix of $(Z_k)$. Then we have $$p'_{x,x} \ge e^{-p_{x,x+1} - p_{x,x-1}} \ge 1 - p_{x,x+1} - p_{x,x-1}.$$ To see this, note that $(X^N_t)$, while in state $x$, makes jumps at rate $q_x = \lambda x(1-x/N) + \mu x$. So the probability that the chain makes no jumps in a time interval of length $\delta$, conditional on the chain being in state $x$ at the start of the interval, is equal to the probability that a Poisson random variable of mean $q_x\delta = p_{x,x+1} + p_{x,x-1}$ is equal to zero, which is given by $e^{-p_{x,x+1} - p_{x,x-1}}$.

Now note that the jump rate of $(X_t^N)$ in state $x$ satisfies $q_x \le (\lambda+ \mu) N$ for all $x$. The probability, conditional on the chain $(X^N_t)$ being in state $x$ at the start of an interval of length $\delta$, that the chain makes one jump upwards and no other jumps during that interval, is $$\int_{s=0}^{\delta} e^{-q_xs} e^{-q_{x+1}(\delta - s)} \lambda x (1-x/N) \, ds,$$ where $s$ represents the time of the single jump within the interval. This integral is at least $$\lambda x (1-x/N) \delta e^{-(\lambda+\mu) N \delta} = p_{x,x+1} e^{-1/K}.$$ So we have $$p'_{x,x+1} \ge p_{x,x+1} e^{-1/K} \ge p_{x,x+1} (1-1/K) \ge p_{x,x+1} - \frac{1}{K^2},$$ and similarly $$p'_{x,x-1} \ge p_{x,x-1} - \frac{1}{K^2}$$

As also $p'_{x,x} + p'_{x,x+1} + p'_{x,x-1} \le 1$, these lower bounds also imply corresponding upper bounds: $$p'_{x,x} \le 1 - p'_{x,x+1} - p'_{x,x-1} \le 1 - p_{x,x+1} - p_{x,x-1} - \frac{2}{K^2} = p_{x,x} + \frac{2}{K^2},$$ and similarly $$p'_{x,x+1} \le p_{x,x+1} + \frac{1}{K^2}; \quad p'_{x,x-1} \le p_{x,x-1} + \frac{1}{K^2}.$$

We couple the discrete-time chains $({\widehat{X}}_k)$ and $(Z_k)$ so that they jump together as far as possible. Specifically, if the chains are both in state $x$, they jump up by 1 together with probability $\min(p_{x,x+1},p'_{x,x+1})$, jump down by 1 together with probability $\min(p_{x,x-1},p'_{x,x-1})$, and stay in state $x$ together with probability $\min(p_{x,x},p'_{x,x})$. Then, for each $k$, $${\operatorname{\mathbb P{}}}({\widehat{X}}_{k+1} \not= Z_{k+1} \mid {\widehat{X}}_k = Z_k) \le \frac{4}{K^2}.$$ Hence, by induction, we have ${\operatorname{\mathbb P{}}}({\widehat{X}}_k = Z_k \mbox{ for all } k \le \ell) \ge 1 - \frac{4\ell}{K^2}$, for each $\ell \ge 0$. In particular, for $\ell = K(\mu+\lambda)Nt_0$, we have that $$d_{TV}  \left( \mathcal{L}(X^N_{t_0}), \mathcal{L}({\widehat{X}}^{N,K}_{K(\mu+\lambda)Nt_0}) \right)
\le {\operatorname{\mathbb P{}}}({\widehat{X}}^{N,K}_{K(\mu+\lambda)Nt_0} \not= X^N_{t_0}) \le \frac{4 (\mu+\lambda)N t_0}{K}$$ for all $K$, and all times $t_0$ that are integer multiples of $\delta = 1/K(\mu+\lambda)N$.

For values $t_0$ that are not necessarily integer multiples of $\delta$, we set $t_0^* = \delta \lceil t_0/\delta \rceil$, so that $\lceil K(\mu+\lambda)N t_0 \rceil = K(\mu+\lambda)N t_0^*$. The probability that $X^N_{t_0}$ is not equal to $X^N_{t_0^*}$ is at most $1 - \max_x p'_{x,x} \le 1/K$, so we have, for all values of $N$, $t_0$ and $K$, $$d_{TV} \left( \mathcal{L}(X^N_{t_0}), \mathcal{L}({\widehat{X}}^{N,K}_{\lceil K(\mu+\lambda)Nt_0\rceil}) \right) \le \frac{4 (\mu+\lambda)N t_0 + 1}{K},$$ as claimed.

The following result now follows from Lemmas \[lem.inter\] and \[lem.disc-cont\].

\[lem.slow\] Suppose $(\mu-\lambda) N^{1/2} \to \infty$ as $N\to \infty$. Let $\omega(N)$ be any function tending to infinity, and set $x^*=x^*(N) = N^{1/2} \omega(N)$. Suppose $x_0 = x_0(N) \le \omega(N) (\mu-\lambda) N$. Then $${\operatorname{\mathbb P{}}}_{x_0} \left( \left| X^N_{t^*} - x^* \right| > \omega(N)^{-1/3} x^* \right) = o(1),$$ where $$t^* = t^*(x_0) = \frac{s(x^*/N) - s(x_0/N)}{\mu-\lambda}.$$

If, additionally, $(\mu-\lambda) N^{1/2} / \omega(N) \to \infty$, then we have $$\label{eq.number}
t^*(x_0) = \frac{1}{\mu-\lambda} \left( \log x_0 - \log(x^*) - \log \left(1+ \frac{\lambda}{\mu-\lambda} \frac{x_0}{N} \right) + o(1) \right).$$

For each $N$, choose $K = K(N)$ so that $K(N) \ge N$ and $\frac{4 (\mu+\lambda)N t^* + 1}{K(N)} \le 1/N$. Note that the value $k^*$ in Lemma \[lem.inter\] is equal to $\lceil K(\mu+\lambda)N t^* \rceil$.

Now we have from Lemma \[lem.inter\] and Lemma \[lem.disc-cont\] that $$\begin{aligned}
\lefteqn{{\operatorname{\mathbb P{}}}_{x_0} \left( \left| X^N_{t^*} - x^* \right| > \omega(N)^{-1/3} x^* \right)} \\
&\le& {\operatorname{\mathbb P{}}}_{x_0} \left( \left| {\widehat{X}}_{k^*}^{N,K(N)} - x^* \right| > \omega(N)^{-1/3} x^* \right) + d_{TV} \left( X^N_{t^*}, {\widehat{X}}_{k^*}^{N,K(N)} \right) \\
&\le& o(1) + \frac{1}{N} = o(1),\end{aligned}$$ as $N \to \infty$.

The final assertion in the statement is immediate, since, with the assumptions given, $\log \big(1 + \frac{\lambda}{\mu-\lambda} \frac{x^*}{N} \big) = o(1)$.

Initial Phase: Upper Bounds {#sec:initial}
===========================

In this section, we show that, if our chain $(X^N_t)$ is started in any state and run for a suitably chosen time period $t_0 = o((\mu-\lambda)^{-1})$, then with high probability the chain will have reached a state below $(\mu-\lambda)N \omega(N)$.

An alternative approach to analysing this phase of the process would be to show, using similar methods to those of the previous section, that the process closely follows the differential equation. The argument we give is significantly simpler, since all we need to show is that the time taken for this initial phase does not contribute significantly to the total extinction time.

We first prove a lemma, showing that the expected state of the stochastic process is always bounded above by the appropriately scaled solution of the differential equation (\[eq.diff-eq\]). The lemma below holds without any assumptions on the values of the parameters or on the starting state.

\[lem.inequality\] Let $x_0$ be any starting state for the stochastic logistic chain, and let $z(t)$ be the solution to the differential equation (\[eq.diff-eq\]) with $z(0) = z_0 = x_0/N$. Then, for all $t \ge 0$, $${\operatorname{\mathbb E{}}}_{x_0} X^N_t \le N z(t).$$

We note that, for all $t \ge 0$, $$\begin{aligned}
\frac{d}{dt} {\operatorname{\mathbb E{}}}_{x_0} X^N_t &=& \lambda {\operatorname{\mathbb E{}}}_{x_0} \big( X^N_t (1-X^N_t/N) \big) - \mu {\operatorname{\mathbb E{}}}_{x_0} X^N_t \\
&=& - (\mu-\lambda) {\operatorname{\mathbb E{}}}_{x_0} X^N_t - \frac{\lambda}{N} {\operatorname{\mathbb E{}}}_{x_0} (X^N_t)^2 \\
&=& - (\mu-\lambda) {\operatorname{\mathbb E{}}}_{x_0} X^N_t - \frac{\lambda}{N} ({\operatorname{\mathbb E{}}}_{x_0} X^N_t)^2 - \frac{\lambda}{N} {\operatorname{\mathbb E{}}}_{x_0} (X^N_t - {\operatorname{\mathbb E{}}}_{x_0} X^N_t)^2.\end{aligned}$$ Therefore we have $$\begin{aligned}
\frac{d}{dt} \big( {\operatorname{\mathbb E{}}}_{x_0} X^N_t - N z(t) \big) &=& - (\mu-\lambda) {\operatorname{\mathbb E{}}}_{x_0} X^N_t - \frac{\lambda}{N} ({\operatorname{\mathbb E{}}}_{x_0} X^N_t)^2 \\
&&\mbox{} - \frac{\lambda}{N} {\operatorname{\mathbb E{}}}_{x_0} (X^N_t - {\operatorname{\mathbb E{}}}_{x_0} X^N_t)^2 + N \big( (\mu-\lambda) z(t) + \lambda z(t)^2 \big) \\
&=& \big( {\operatorname{\mathbb E{}}}_{x_0} X^N_t - Nz(t) \big) \left\{ - (\mu-\lambda) - \frac{\lambda}{N} \big({\operatorname{\mathbb E{}}}_{x_0} X^N_t + z(t)\big) \right\} \\
&&\mbox{} - \frac{\lambda}{N} {\operatorname{\mathbb E{}}}_{x_0} (X^N_t - {\operatorname{\mathbb E{}}}_{x_0} X^N_t)^2.\end{aligned}$$ It follows now that, since ${\operatorname{\mathbb E{}}}_{x_0} X^N_t - Nz(t)$ is equal to zero at $t=0$, it is non-positive for all $t\ge 0$.

We now state a version of this result in the specific context we shall use it. (It might be useful to note that the lemma is trivial if $(\mu-\lambda) \omega(N) \ge 1$ fr sufficiently large $N$.)

\[lem.quick\] Let $\omega(N)$ be any function tending to infinity, and set $t_0 = t_0(N) = \frac{1}{\omega(N)^{1/2} \lambda (\mu-\lambda)}$.

For any starting state $x_0=x_0(N)$, ${\operatorname{\mathbb P{}}}_{x_0} \big(X^N_{t_0} \ge N (\mu-\lambda) \omega(N) \big) = o(1)$.

Let $z_0 = x_0 / N \le 1$. We note that $$\begin{aligned}
z(t_0) &=& \frac{ z_0 (\mu-\lambda) e^{- (\mu-\lambda)t_0}}{(\mu-\lambda) + z_0 \lambda (1-e^{-(\mu-\lambda)t_0})}\\
&\le& \frac{(\mu-\lambda) e^{-(\mu-\lambda)t_0}}{\lambda (1- e^{-(\mu-\lambda)t_0})} \\
&\le& \frac{1}{\lambda t_0} = (\mu-\lambda) \omega(N)^{1/2}.\end{aligned}$$ Here we used the elementary inequality $e^{-u} \le (1-e^{-u})/u$, valid for all $u > 0$. Hence, by Lemma \[lem.inequality\], we have ${\operatorname{\mathbb E{}}}X^N_{t_0} \le (\mu-\lambda) N \omega(N)^{1/2}$. Moreover, we have $${\operatorname{\mathbb P{}}}_{x_0} \Big(X^N_{t_0} \ge (\mu-\lambda)N \omega(N) \Big) \le \frac{1}{\omega(N)^{1/2}} = o(1),$$ as desired.

Proof of Theorem \[thm.main\] {#sec.proof}
=============================

In this section, we assemble all the lemmas from the preceding three sections into a proof of Theorem \[thm.main\].

Set $\omega(N) = \big( N^{1/2} (\mu-\lambda) \big)^{1/4}$, and note that $\omega(N)$ tends to infinity as $N \to \infty$.

We now distinguish three ranges for the starting state $x_0 = x_0(N)$, assuming always that $(\mu-\lambda)x_0 \to \infty$: (a) $x_0(N) \le N^{1/2}\omega(N)$, (b) $N^{1/2} \omega(N) < x_0(N) \le (\mu-\lambda)N \omega(N)$, (c) $x_0(N) > (\mu - \lambda)N \omega(N)$. It could be that $x_0(N)$ falls into different ranges for different values of $N$: we partition the set of natural numbers into three sets depending on which of (a), (b), (c) holds. It suffices to prove the result separately for whichever subsequence(s) are infinite, and so we may treat each of the three ranges in turn, working (tacitly) with an infinite sequence of values of $N$ for which the inequalities defining the range hold.

\(a) Suppose $(\mu-\lambda) x_0(N) \to \infty$ and $x_0(N) \le N^{1/2} \omega(N)$. In this case, we have, from Lemma \[lem.late\] with $x^* = x_0$, that $$(\mu - \lambda) T_e^{X^N} - \big( \log x_0 + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, where $W$ has the standard Gumbel distribution. This is equivalent to (\[eq.general\]), since in this range we have $$\begin{aligned}
\log \left( 1 + \frac {\lambda x_0}{(\mu-\lambda)N} \right) \le \log \left( 1 + \frac{\lambda \omega(N)}{(\mu-\lambda) N^{1/2}} \right) \\
= \log \left( 1 + \frac{\lambda}{\big((\mu-\lambda) N^{1/2}\big)^{3/4}} \right)= o(1).\end{aligned}$$

\(b) Set $x^*= x^*(N) = N^{1/2} \omega(N)$, and suppose $x^*(N) < x_0(N) \le (\mu-\lambda)N \omega(N)$. Now we run the chain for a time $t^* = t^*(x_0)$, as in the statement of Lemma \[lem.slow\]. Note that $N^{1/2}(\mu-\lambda)/ \omega(N) = \omega(N)^3 \to \infty$ as $N \to \infty$, so the asymptotic formula (\[eq.number\]) for $t^*$ is valid. Let $A$ be the event that $x^*(1 - \omega(N)^{-1/3}) \le X_{t^*} \le x^*(1 + \omega(N)^{-1/3})$: by Lemma \[lem.slow\] we have that ${\operatorname{\mathbb P{}}}(\overline A) = o(1)$.

Now let $(Y^N_t)$ be a copy of the stochastic SIS logistic process started in state $x^*(1 - \omega(N)^{-1/3})$, and $(Z^N_t)$ be another copy started in state $x^*(1 + \omega(N)^{-1/3})$. We may couple these two copies, as well as the chain $(X^N_t)$ from time $t^*$ onwards, in an obvious way. Then, on the event $A$, under the coupling, we have $Y^N_t \le X^N_{t^*+t} \le Z^N_t$ for all $t \ge 0$. We therefore also have, on the event $A$, under the coupling, that $T_e^{Y^N} + t^* \le T_e^{X^N} \le T_e^{Z^N} + t^*$. By Lemma \[lem.late\], we have $$(\mu - \lambda) T_e^{Y^N} - \big( \log x^* + \log (1- \omega(N)^{-1/3}) + \log (\mu-\lambda) - \log \mu \big) \to W,$$ in distribution, as $N\to \infty$, where $W$ has the standard Gumbel distribution. Therefore, since $\log(1-\omega(N)^{-1/3}) = o(1)$, we have, using the asymptotic formula (\[eq.number\]) for $t^*$, $$(\mu - \lambda) (T_e^{Y^N} + t^*) - \Big( \log x_0 + \log (\mu-\lambda) - \log \big( 1 + \frac{\lambda x_0}{(\mu-\lambda)N}\big) - \log \mu \Big) \to W,$$ in distribution, as $N \to \infty$. The same result holds for $T_e^{Z^N} + t^*$. Hence we also have $$(\mu - \lambda) T_e^{X^N} - \Big( \log x_0 + \log (\mu-\lambda) - \log \big( 1 + \frac{\lambda x_0}{(\mu-\lambda)N}\big) - \log \mu \Big) \to W,$$ in distribution, as $N \to \infty$, as required.

\(c) Suppose $x_0(N) > (\mu - \lambda)N \omega(N)$. Let $(Y^N_t)$ be a copy of the chain started in state $(\mu - \lambda)N \omega(N)$. Then there is a coupling of $(X^N_t)$ and $(Y^N_t)$ so that $T_e^{X^N} \ge T_e^{Y^N}$. Now let $t_0 = 1/\omega(N)^{1/2}(\mu-\lambda)$, as in Lemma \[lem.quick\]: as mentioned earlier, the idea is that $t_0$ is negligible compared to the overall extinction time. Let $B$ be the event that $X^N_{t_0} \le (\mu-\lambda)N \omega(N)$. By Lemma \[lem.quick\], ${\operatorname{\mathbb P{}}}({\overline B}) = o(1)$. There is then a coupling of $(Y^N_t)$ and $(X^N_{t_0+t})$ so that, on the event $B$, $Y^N_t \ge X^N_{t_0+t}$ for all $t \ge 0$, and so $T_e^{X^N} \le T_e^{Y^N} + t_0$.

Now, by the result in part (b), we have $$(\mu - \lambda) T_e^{Y^N} - \big( \log \big((\mu-\lambda)N \omega(N)\big) + \log (\mu-\lambda) - \log \big( 1 + \lambda \omega(N)\big) - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$, which is equivalent to $$(\mu - \lambda) T_e^{Y^N} - \big( \log N + 2 \log (\mu-\lambda) - \log \lambda - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$. Since $(\mu-\lambda) t_0 = o(1)$, it follows that $$(\mu - \lambda) T_e^{X^N} - \big( \log N + 2 \log (\mu-\lambda) - \log \lambda - \log \mu \big) \to W,$$ in distribution, as $N \to \infty$. This is equivalent to (\[eq.general\]), since in this range we have $$\log \big( 1 + \frac{\lambda x_0}{(\mu-\lambda)N} \big) = \log \lambda + \log x_0 - \log (\mu-\lambda) - \log N + o(1).$$ This completes the proof.

The critical regime {#sec.critical}
===================

We finish with a remark about the application of our methods to the critical regime, where $|\mu-\lambda| = O(N^{-1/2})$. In this case there exist constants $\delta, c > 0$ (depending on $\limsup_{N\to \infty} (\lambda-\mu)N^{1/2}$)), such that, for all starting states $x_0$, ${\operatorname{\mathbb P{}}}_{x_0} (T_e^{X^N} \le c N^{1/2}) > \delta$. One way to prove this is to follow our methods: first use Lemma \[lem.inequality\] with $t_0 = N^{1/2}$ and $z(t_0) = O(N^{-1/2})$, to show that, for some constant $c_1$, with positive probability, $X_{t_0} \le c_1 N^{1/2}$. Then compare the chain $(X_t^N)$ with the linear birth-and-death chain with the same parameters, both started in state $c_1 N^{1/2}$, and show that, for some constant $c_2$, with positive probability, the chain $(X_t^{N})$ reaches 0 in a further time $c_2 N^{1/2}$. Once we have this result, it follows that ${\operatorname{\mathbb P{}}}_{x_0} (T_e^{X^N} > c(N) N^{1/2})$ tends to 0 whenever $c(N) \to \infty$. Throughout the critical regime, a lower bound on the extinction time of the form ${\operatorname{\mathbb P{}}}_{N^{1/2}} (T_e^{X^N} \le {\varepsilon}(N) N^{1/2} ) \to 0$ whenever ${\varepsilon}(N)\to 0$ can again be obtained by comparing with a suitable linear birth-and-death chain. Much more precise results concerning the process in the critical regime started from a state of order $N^{1/2}$ are given by Dolgoarshinnykh and Lalley [@DL].

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[^1]: The research of Malwina Luczak is supported by an EPSRC Leadership Fellowship, grant reference EP/J004022/2.
