2. Phase diagram and main results

Let us formally state the setting and the assumptions that hold throughout the paper. Given $n \in \mathbb{N}$ , $r_n\in[0,\infty)$ , and $X_n(0) \in \{0,\dots,n\}$ , $X_n$ denotes the Markov chain $(X_n(t))_{t \ge 0}$ with initial value $X_n(0)$ and transition rates

(2) \begin{align} X_n \to \begin{cases}X_n+1 & \textrm{at rate} \quad r_nX_n(1-X_n/n) , \\ X_n-1 & \textrm{at rate} \quad X_n. \end{cases}\end{align}

The extinction time of $X_n$ is $\tau_n=\inf\{t\colon X_n(t)=0\}$ . We fix a sequence of values $(X_n(0))_{n \ge 1}$ and $(r_n)_{n\ge 1}$ and seek deterministic sequences $(t_n)$ and $(s_n)$ and a random variable T such that $(\tau_n-t_n)/s_n$ converges in distribution to T as $n\to\infty$ ; we refer to this as a distributional limit for $(\tau_n)$ , subject to the following assumptions.

The behaviour of $\tau_n$ depends on $r_n$ and $X_n(0)$ , or equivalently $r_n$ and $x_n(0)=X_n(0)/n$ . So, we can organize our results into a phase diagram for $(r,x) \in \mathbb{R}_+ \times [0,1]$ . As observed in [Reference Nasell18], the parameter values are effectively partitioned into three phases:

1. 1. Subcritical: $c_n \to -\infty$ as $n\to\infty$

2. 2. Critical: $c_n \to c_\infty \in \mathbb{R}$ as $n\to\infty$

3. 3. Supercritical: $c_n \to \infty$ as $n\to\infty$ .

It is important to note that since $c_n=\sqrt{n}\,(r_n-1)$ , the critical phase has width $O\big(\frac{1}{\sqrt{n}\,}\big)=o(1)$ around the point $r=1$ . Along with parameter values, there is a dependence on the initial value $X_n(0)$ . The following more refined partition into six regions delineates the different regimes for the limit behaviour of $\tau_n$ .

1. 1. Discrete: $X_\infty(0) \in \mathbb{N}$ and $r_\infty \le 1$

2. 2. Linear diffusive: $X_\infty(0)=\infty$ and either

   1. (a) subcritical phase and $a_\infty<\infty$ , or

   2. (b) critical/supercritical phase and $a_\infty= 0$

3. 3. Subthreshold cutoff: subcritical phase and $a_\infty = \infty$

4. 4. Non-linear diffusive: critical phase and $a_\infty>0$

5. 5. Threshold: supercritical phase and $0<a_\infty<\infty$

6. 6. Metastable: supercritical phase and $a_\infty=\infty$ .

The following methods are used in each region to establish the distributional limit for $(\tau_n)$ .

1. 1. Discrete: As observed in [Reference Andersson and Djehiche1], in this region, for each n, $X_n$ can be coupled to the linear birth and death process $Z_n$ with $Z_n(0)=X_n(0)$ and transition rates

   (3) \begin{align}Z_n \to \begin{cases}Z_n+1 & \text{at rate} \quad r_n\, Z_n, \\ Z_n-1 & \text{at rate} \quad Z_n \end{cases}\end{align}
   in such a way that $\mathbb{P}(X_n(t)=Z_n(t)$ for all $t \in [0,\tau_n] \, ) = 1-o(1)$ .

2. 2. Linear diffusive: As observed in [Reference Dolgoarshinnykh and Lalley7], in this region $X_n(X_n(0)\, t)/X_n(0)$ converges in distribution to the diffusion Y with $Y(0)=1$ and

   (4) \begin{align} {\mathrm{d}} Y = -a_\infty \, Y {\mathrm{d}} t + \sqrt{2Y}{\mathrm{d}} B, \end{align}
   where B is standard Brownian motion, and it is not hard to strengthen this result to show that $\tau_n/X_n(0)$ converges in distribution to $\inf\{t \colon Y(t)=0\}$ . This explains nicely the form of the limit. The method we will use instead, since it applies across several regions, is that of [Reference Brightwell, House and Luczak5], which is essentially the same coupling method as [Reference Andersson and Djehiche1] but extends the approach by not requiring that $X_n(t)=Z_n(t)$ for all $t\le \tau_n$ with high probability, but instead using both upper and lower bounds $(Z_n)$ and $(Z^{\prime}_n)$ , both linear birth and death processes, such that the extinction time of both is comparable to $(\tau_n)$ . To define the lower bound $(Z^{\prime}_n)$ , a ceiling $M_n$ is defined such that the coupling is valid so long as $X_n \le M_n$ and such that $X_n$ hits 0 before $M_n$ with high probability.

3. 3. Subthreshold cutoff: As observed in [Reference Brightwell, House and Luczak5], $x_n$ is well approximated by the logistic ODE $x'=r_\infty\, x\,(1-x)$ until $|\delta_n|\, X_n =O(1)$ , at which point $X_n$ enters the linear diffusive regime and the above coupling method can be used. In practice, [Reference Brightwell, House and Luczak5] shows the coupling method can be applied once $|\delta_n|\, X_n \le \omega_n$ for some $(\omega_n)$ tending to $\infty$ slowly enough as $n\to\infty$ as a function of $(c_n)$ .

4. 4. Non-linear diffusive: As observed in [Reference Dolgoarshinnykh and Lalley7], the process $X_n(\sqrt{n}\, t)/\sqrt{n}$ converges in distribution to the diffusion Y with $Y(0)=\lim_{n\to\infty} X_n(0)/\sqrt{n}$ and ${\mathrm{d}} Y = (c_\infty Y - Y^2)\,{\mathrm{d}} t + \sqrt{2Y}{\mathrm{d}} B$ . In this case we make use of this diffusion limit combined with a result that says $\limsup_n \mathbb{P}\big(\frac{\tau_n}{\sqrt{n}\,} >t \mid X_n(0) \le \alpha \sqrt{n}\big) \to 0$ for each $t>0$ as $\alpha>0$ to establish convergence in distribution of $\frac{\tau_n}{\sqrt{n}}$ to $\inf\{t\colon Y(t)=0\}$ .

5. 5. Threshold: Let $x_n^\star=1-1/r_n$ , $X_n^\star = \lfloor n x_n^\star \rfloor$ , and $\tau_n^\star=\inf\{t\colon X_n(t) \in \{0,X_n^\star\}\}$ . Then, as observed in [Reference Andersson and Djehiche1] when $r_n=r_\infty>1$ and $X_n(0)=X_\infty(0)$ for all n, and extended in this article to the whole supercritical phase,

   \[\lim_{n\to\infty}\mathbb{P}(X_n(\tau_n^\star) =0 )=\begin{cases} r_\infty^{-X_\infty(0)} & \text{if} \ X_\infty(0)<\infty,\\{\mathrm{e}}^{-a_\infty} & \text{if} \ X_\infty(0)=\infty. \end{cases}\]
   In particular, the limiting probability is in (0,1). Conditioned on $\{X_n(\tau_n^\star)=0\}$ , the process behaves effectively as though it is in the linear diffusive regime with parameter $1-1/r_\infty$ , and the same approximation applies. On the event $\{X_n(\tau_n^\star)=X_n^\star \}$ , the process enters the metastable regime. This dichotomy appears to have been first observed in [Reference Andersson and Djehiche1].

6. 6. Metastable: In this case, the process $X_n$ reaches $X_n^\star$ relatively quickly, with typical fluctuations described as follows: the process $(X_n(t/\delta_n)-X_n^\star)/\sqrt{n}$ converges in distribution to the Ornstein–Uhlenbeck process Y described by ${\mathrm{d}} Y = -Y{\mathrm{d}} t + \sqrt{2/r_\star}{\mathrm{d}} B$ . General results of this type can be found in [Reference Kurtz17]. As observed in [Reference Andersson and Djehiche1], when $r_n=r_\infty>1$ for each n, on each excursion from $X_n^\star$ the probability of hitting 0 can be computed quite precisely using well-known formulae for birth and death processes. In this article we extend these calculations to the whole supercritical phase, using essentially the same method but exercising greater care in the barely supercritical ( $c_\infty=\infty$ and $r_\infty=1$ ) case.

In stating the results for the subcritical phase it is helpful to define $\gamma_n=1-r_n=-\delta_n$ . We now give precise statements of the results, taking care to cite previous work in the relevant cases.

Theorem 2 is proved by first studying the extinction time of the linear birth and death process $Z_n$ from (3), then transferring the result to $X_n$ using a coupling with the property that $\mathbb{P}(X_n(t)=Z_n(t)$ for all $t \in [0,\tau_n])=1-o(1)$ .

The next result appears not to have been proved yet in any of the listed references, though it follows straightforwardly from the coupling method already discussed from [Reference Brightwell, House and Luczak5].

Theorem 2. (Linear diffusive.) Suppose $r_\infty \le 1$ , $X_\infty(0)=\infty$ , and either $c_\infty=-\infty$ and $a_\infty<\infty$ , or $c_\infty>-\infty$ and $a_\infty=0$ . Then $\tau_n/X_n(0) \stackrel{({\mathrm{d}})}{\to} H_{a_\infty}$ with

\[\mathbb{P}(H_{a_\infty}\le w) = \begin{cases} {\mathrm{e}}^{-1/w} & \text{if} \quad a_\infty=0, \\ {\mathrm{e}}^{-a_\infty/({\mathrm{e}}^{a_\infty w}-1)} & \text{if} \quad a_\infty>0. \end{cases}\]

Theorem 2 is proved in a similar way to Theorem 1. In this case, we cannot enforce $X_n(t)=Z_n(t)$ for all $t\in [0,\tau_n]$ with high probability, so instead we bound $X_n$ above and below by linear birth and death processes with different parameters. This ‘sandwiching’ idea appears in [Reference Brightwell, House and Luczak5] to prove the final stage of Theorem 3 (see below). A particular case of Theorem 2, where $r=1+Cn^{-\alpha}$ for $\alpha <1/2$ , is discussed in [Reference Dolgoarshinnykh and Lalley7]; they prove convergence to the diffusion limit in (4), but not the convergence of $\tau_n$ . Although we do not do it here, we could also prove Theorem 2 via the diffusion limit (4) by the same methods we use to prove Theorem 4.

The next result is the main result of [Reference Brightwell, House and Luczak5].

Theorem 3. (Subthreshold cutoff, Theorem 1.1 in [Reference Brightwell, House and Luczak5].) Suppose that $c_\infty=-\infty$ and $a_\infty=\infty$ . Let G denote the standard Gumbel: $\mathbb{P}(G\le w)=\exp(-{\mathrm{e}}^{-w})$ . Then $\gamma_n \, \tau_n - g_n(X_n(0)) \stackrel{({\mathrm{d}})}{\rightarrow} G$ as $n\to\infty$ , where $g_n(X) = \log (\gamma_n^2 n) - \log (r_n + \gamma_n n/X)$ . In particular,

\[g_n(X_n(0)) = \begin{cases} \log(\gamma_n X_n(0)) + o(1) & \text{if} \quad X_n(0)=o(\gamma_n n), \\ \\[-7pt] \log(\gamma_n^2 n) - \log(r_\infty+1/b_\star) & \text{if} \quad X_n(0)/\gamma_n n \to b_\star \in (0,\infty].\end{cases}\]

In [Reference Brightwell, House and Luczak5] the parameter r is denoted $\lambda$ , and the rate of $X\to X-1$ is equal to $\mu X$ instead of X; since $\mu$ can be set to 1 by a uniform time change, no generality is lost. In [Reference Brightwell, House and Luczak5], Theorem 3 is proved in three stages, which we describe in the present notation.

Initial stage: A differential inequality for $x_n$ (incidentally, the same one originally proved in [Reference Feller10]) is used to show that from any initial value, $x_n$ drops to $\gamma_n |c_n|^\epsilon$ for some $\epsilon>0$ within $o(1/\gamma_n)$ amount of time (see [Reference Brightwell, House and Luczak5, Lemma 4.2] and note that since we take $\mu=1$ , in our notation $\mu-\lambda$ is just $\gamma_n$ , and since $\mu-\lambda \to 0$ , $\lambda \to 1$ as $n\to\infty$ ).

Intermediate stage: From any initial value $x_n(0) \le \gamma_n |c_n|^\epsilon$ until the first time that $x_n \le \gamma_n |c_n|^{-\epsilon}$ , $x_n$ remains close to the solution of the corresponding logistic equation $x'=rx(1-x)-x$ .

Final stage: From any initial value $x_n(0)\le \gamma_n |c_n|^{-\epsilon}$ , a coupling argument is used to bound $X_n$ both above and below by linear birth and death processes with different parameters, as discussed earlier.

Our proof of Theorem 3 in the initial and final stages is basically identical to [Reference Brightwell, House and Luczak5]; in the intermediate stage we take a somewhat different approach as discussed in Section 6.

Theorem 4. (Non-linear diffusive.) Suppose $c_\infty \in \mathbb{R}$ and $a_\infty>0$ , and let $Y_n$ denote the rescaled process $Y_n(t) = X_n(\sqrt{n}\, t)/\sqrt{n}$ . Let $y=\lim_{n\to\infty}Y_n(0)\in (0,\infty]$ . Then, $Y_n$ converges in distribution to the diffusion Y that solves the SDE

\[{\mathrm{d}} Y = Y(c_\infty-Y)\, {\mathrm{d}} t + \sqrt{2Y}{\mathrm{d}} B, \qquad Y(0) = y,\]

and $\tau_n/\sqrt{n} \stackrel{({\mathrm{d}})}{\rightarrow} T$ , where $T=\inf\{t:Y(t)=0\}$ .

The convergence of $X_n(\sqrt{n} \, t)/\sqrt{n}$ to the diffusion limit Y is proved in [Reference Dolgoarshinnykh and Lalley7], although they do not prove convergence of $\tau_n$ . To obtain the latter we employ the continuous mapping theorem to obtain convergence of the hitting time of $\epsilon$ for small $\epsilon>0$ , then show that once $Y_n \le \epsilon$ , $Y_n$ is likely to hit 0 in a short time.

The following result is proved in [Reference Andersson and Djehiche1] in the particular case where $r_n=r_\infty>1$ for all n, although the rapid extinction result is expressed without conditioning on hitting 0 before $X_n^\star$ , which is just a difference in the presentation. Our proof takes the same basic approach as [Reference Andersson and Djehiche1], with the caveats that the calculations become more delicate when $r_\infty =1$ , requiring a finer analysis, and that we have found a couple of gaps in their proof of the exponential limit (see the discussion below the statement of Theorem 5), so we took the steps described below to bridge those gaps.

Having access to the prefactors (the stuff in front of the exponential function) in the expected time to extinction allows us to see how it blends into the non-linear diffusive limit as $c_n$ approaches O(1). For $r_n$ near 1, a Taylor expansion gives $\log r_n+1/r_n-1 \sim \delta_n^2/2$ , so the exponential term is $\exp((1+o(1))n\delta_n^2/2) = \exp((1+o(1))c_n^2/2)$ since $c_n=\sqrt{n}\delta_n$ , and the prefactor is $\sqrt{2\pi}r_n/(\sqrt{n}\delta_n^2) = \sqrt{2\pi n}r_n/c_n^2$ . Thus, when $r_\infty=1$ ,

\[\mathbb{E}[\tau_n \mid A_n^\star] \sim \frac{\sqrt{2\pi n}}{c_n^2} {\mathrm{e}}^{(1+o(1))c_n^2/2},\]

which is of order $\sqrt{n}$ when $c_n=O(1)$ .

Much of the proof of Theorem 5 revolves around precise estimation of various sums that all seem to involve the function $\nu(\,j,k) \,:\!=\, \prod_{i=j+1}^k q_-(i)/q_+(i)$ , where $q_-(i) = i$ , $q_+(i) = rX(1-X/n)$ is the rate at which $X_t$ decreases, respectively increases, by 1 when $X_t=i$ . We discuss the basic approach for each part:

1. Probability of rapid extinction: An explicit formula for $h_+(\,j)\,:\!=\, \mathbb{P}(A_n^\star \mid X_0=j)$ is available in terms of the transition rates of the process, and can be estimated. Here is where we first encounter $\nu(\,j,k)$ .

2. Scaling on $\tau_n$ on rapid extinction: Using the Doob h-transform applied to the function $h_-\,:\!=\,1-h_+$ , we study $X_n$ conditioned on $B_n^\star$ . The transition rates can be written using $h_-$ . By estimating transition rates we show the conditioned process corresponds to the setting of Theorem 1 or Theorem 2.

3. Scaling of $\tau_n$ at metastability: Conditioning on $A_n^\star$ , we break up $\tau$ into three epochs: the time to hit $X_n^\star$ (approach), the time spent on excursions that return to $X_n^\star$ (sojourn), and the last excursion from $X_n^\star$ to 0 (fall). We show that the approach and fall time are small compared to the sojourn time, in expectation; so far this is the same approach as in [Reference Andersson and Djehiche1]. We then use a coupling argument that shows the process is ‘forgetful’ in order to derive the exponential limit for the sojourn time; this method is a departure from [Reference Andersson and Djehiche1].

4. We have also found the following gaps in the proofs in [Reference Andersson and Djehiche1]:

5. Exponential limit: The exponential limit for $\tau_n/\mathbb{E}[\tau_n]$ is obtained in [Reference Andersson and Djehiche1] via the same approach taken here: showing that the approach and fall time are small and that the sojourn time converges to exponential. Using their notation, for each n the sojourn time can be written as the sum $\sum_{m=1}^{K_n} \sigma_n(m)$ , where $(\sigma_n(m))_{m=1}^{\infty}$ are independent and identically distributed (i.i.d.) and $K_n$ is geometric independent of $(\sigma_n(m))$ , with parameter $1/\mathbb{E}[K_n]$ that $\to 0$ as $n\to\infty$ . In claiming that the normalized sojourn time has an exponential limit, [Reference Andersson and Djehiche1] appeals to [Reference Keilson15, Theorem 8.1A]. However, the theory discussed there, including that result, pertains only to a single i.i.d. sequence $(T_m)$ of random variables. On the other hand, in this case the fact that the process itself depends on n implies that a doubly indexed sequence must be considered. I did some calculations (not included) that suggest that uniform integrability of $\sigma_1(n)$ with respect to n is sufficient in order to adapt the result from [Reference Keilson15]. However, in [Reference Andersson and Djehiche1] this is not done.

6. Expected excursion length: When expressing the time to extinction as the sum of the approach, sojourn, and fall times, it must be assumed that during the sojourn the process necessarily returns to $X_n^\star$ before hitting zero, which requires conditioning on $A_n^\star$ . However, the formula used to compute the expected duration of each excursion from $X_n^\star$ , namely $\mathbb{E}[\sigma_n(1)]$ , in the proof of [Reference Andersson and Djehiche1, Lemma 2], does not take account at all of conditioning on returning to $X_\star$ . The answer comes out correct, since the conditioning does not have much effect, but it cannot be ignored outright.

The rest of the paper is organized as follows. In Section 3 we discuss some basic coupling results that are used a few times throughout the paper. Section 4 treats the linear birth and death process, essentially proving Theorems 1 and 2 for the linearization of $X_n$ . In Section 6 we prove Theorems 1 and 2, and the ‘final stage’ of Theorem 3. In Section 7 we prove the rest of Theorem 3. In Section 6 we prove Theorem 4. Finally, in Section 8 we prove Theorem 5. In most of our proofs we suppress the dependence on n, to avoid writing subscripts everywhere; I will point this out as we go along.

3. Coupling of birth and death processes

Recall that a birth and death (b–d) process is a right-continuous, continuous-time Markov chain with state space $\mathbb{N} = \{0,1,2,\dots\}$ that jumps by $\pm 1$ at each transition. There is a natural way to construct such a process, or even multiple such processes with different transition rates, from all initial conditions on a single probability space. We will focus on b–d processes with state space $\{0,\dots,N\}$ for some N, since this is all that is needed for this paper. We begin with the case of a single process.

3.1. Natural coupling for a single process

A b–d process is defined by its transition rates. For $x \in \{0,\dots,N\}$ let b(x) and d(x) be the transition rates from x to $x+1$ and $x-1$ respectively, assume that $d(0)=0$ and $b(N)=0$ , and define independent Poisson point processes B(x) and D(x) on $\mathbb{R}_+$ with respective rates b(x) and d(x). As a function of the collection $(B(x),D(x)\colon x \in \{0,\dots,N\})$ we shall define, for each $x \in \{0,\dots,N\}$ , a process $(\Phi(x,t)\colon t \in [0,\infty))$ which is a copy of the b–d process with the given transition rates, and with initial value x at time 0. Let $\Phi(x,0)=x$ , and define $(t_i(x))_{i \le I(x)}$ and $(\Phi(x,t_i(x)))_{i \le I(x)}$ recursively by

\begin{align*}&t_0(x)=0, \quad t_{i+1}(x)= \inf \{t>t_i(x) \colon t \in B(\Phi(x,t_i(x))) \cup D(\Phi(x,t_i(x))), \\ \\[-7pt] &\Phi(x,t_{i+1}(x)) = \begin{cases} \Phi(x,t_i(x))+1 & \ \text{if} \ \ t_{i+1}(x) \in B(\Phi(x,t_i(x))) , \\[3pt] \Phi(x,t_i(x))-1 & \ \text{if} \ \ t_{i+1}(x) \in D(\Phi(x,t_i(x))) , \end{cases} \\ & I(x) = \inf\{i \colon b(\Phi(x,t_i(x)))+d(\Phi(x,t_i(x)))=0\}.\end{align*}

Then, for $i<I(x)$ let $\Phi(x,t) = \Phi(x,t_i(x))$ for $t \in [t_i(x),t_{i+1}(x))$ , and if $I(x)<\infty$ let $\Phi(x,t)=\Phi(x,t_{I(x)}(x))$ for $t \in [t_{I(x)}(x),\infty)$ . This defines the process on $[0,\zeta(x))$ where

\[\zeta(x)\,:\!=\,\begin{cases}\lim_{i\to\infty} t_i(x)) & \text{if} \ \ I(x)=\infty , \\ \\[-7pt] \infty & \text{if} \ \ I(x)<\infty.\end{cases}\]

To verify that $\zeta(x)=\infty$ when $I(x)=\infty$ , note that $\{t_i(x)\colon i \ge 1\} \subset U\,:\!=\, \bigcup_{y \le N}B(y) \cup D(y)$ . Since U is a Poisson point process with finite total intensity $\sum_{y \le N}b(y)+d(y)$ , with probability 1 $U_N \cap [0,T]$ is finite for each fixed $T>0$ , which implies that $\lim_{i\to\infty}t_i$ must be infinite.

Having constructed the coupling, we verify the following desirable properties.

Lemma 1. Let $\Phi(x,t)$ be as defined above. If $\Phi(x,s)=\Phi(y,s)$ then $\Phi(x,t)=\Phi(y,t)$ for all $t\ge s$ , and if $x \le y$ then $\Phi(x,t) \le \Phi(y,t)$ for all $t\ge 0$ .

Proof. Suppose $\Phi(x,s)=\Phi(y,s)$ , which we denote by z. Then, for some i,j, $s\ge \max(t_i(x),t_j(y))$ and $\Phi(x,t_i(x))=\Phi(y,t_j(y))=z$ . If $b(z) + d(z)=0$ then $I(x)=i$ and $I(y)=j$ , and $\Phi(x,t) = \Phi(y,t) = z$ for all $t\ge s$ . Otherwise, $I(x)>i$ , $I(y)>j$ , and $\min(t_{i+1}(x),$ $ t_{j+1}(y))>s$ . We then have $t_{i+1}(x)=t_{j+1}(y)$ since both are equal to $\inf\{t>s \colon t \in B(z) \cup D(z)\}$ , and the construction ensures that $\Phi(x,t)=\Phi(y,t)$ for all $t\ge s$ .

For the second statement, first note that since $(B(x)\cup D(x)\colon x \in \{0,\dots,N\})$ are independent, from standard properties of Poisson point processes it follows that if $x \ne y$ then $B(x)\cup D(x)$ and $B(y) \cup D(y)$ are disjoint. It follows from the construction that if $\Phi(x,t^-) \ne \Phi(y,t^-)$ then either $\Phi(x,t)=\Phi(x,t^-)$ or $\Phi(y,t) = \Phi(y,t^-)$ , i.e. both cannot jump simultaneously. Since $\Phi(x,0)-\Phi(y,0)$ is integer valued, and since

• • for any x, $t\mapsto \Phi(x,t)$ is piecewise constant,

• • for any x and fixed $T>0$ , the set $\{t \in [0,T]\colon \Phi(x,t) \ne \Phi(x,t^-)\}$ is almost surely (a.s.) finite, and

• • for any x,t, $|\Phi(x,t)-\Phi(x,t^-)| \in \{-1,0,1\}$ ,

if $x \le y$ and $\Phi(x,t)>\Phi(y,t)$ then, since $\Phi(x,0)=x$ and $\Phi(y,0)=y$ , for some $s<t$ , $\Phi(x,s)=\Phi(y,s)$ . Using the first statement, we then have $\Phi(x,t)=\Phi(y,t)$ , contradicting $\Phi(x,t)>\Phi(y,t)$ . □

4. Linear birth and death process

The linear birth and death process $\mathbb{Z}_+$ with parameter r is defined by the transitions

\[Z \to \begin{cases}Z+1 & \text{at rate} \quad r Z , \\ Z-1 & \text{at rate} \quad Z. \end{cases}\]

Z can be thought of as the number of cells in a process in which each cell independently dies at rate 1 and splits into two cells at rate r. For this process we are interested in how the extinction time $\tau=\inf\{t:Z(t)=0\}$ scales with r and $Z_0$ in cases where extinction is asymptotically certain, i.e. $\mathbb{P}(\tau<\infty)\to 1$ . The results are by now routine, but as I could not find a reference that states them all together, and since they are easy to prove, I have included the proof.

Proof of Theorem 6. For compactness of notation we will suppress the dependence on n in all variables and write $Z_t$ instead of Z(t). In [Reference Athreya and Ney2, Chapter III] a more general model is considered in which each particle independently dies at some rate $\alpha$ (they call it a but that notation is already in use for us) and is replaced with k particles with probability $p_k$ ; the present model corresponds to $\alpha =1+r$ and $p_0=1/(1+r)$ , $p_2=r/(1+r)$ . Using the Kolmogorov backward equation of the process [Reference Athreya and Ney2, equation (4) of III.2], for $\rho(t)=\mathbb{P}(Z_t=0 \mid Z_0=1)$ the differential equation $\rho' = \alpha (-\rho + \sum_{k=0}^{\infty}p_k \rho^k)$ is obtained [Reference Athreya and Ney2, equation (2) of III.4]. In our case this gives $\rho' = -(1+r)\rho + 1 + r\rho^2$ , which can be conveniently factored to give

(5) \begin{align}\rho' = (1-r\rho)(1-\rho), \qquad \rho_0=0.\end{align}

Separating variables,

\[\frac{{\mathrm{d}}\rho}{(1-r\rho)(1-\rho)} = {\mathrm{d}} t.\]

Letting $\gamma=1-r$ we note that

\[\frac{1}{1-\rho} - \frac{r}{1-r\rho} = \frac{\gamma}{(1-r\rho)(1-\rho)}.\]

Solving the differential equation and using $\rho(0)=0$ ,

\[\gamma \, t = \log\left(\frac{1-r\rho(t)}{1-\rho(t)}\right).\]

If $\gamma \ne 0$ , solving for $\rho(t)$ then gives

(6) \begin{align}\rho(t)^{-1} = \frac{{\mathrm{e}}^{\gamma t}-r}{{\mathrm{e}}^{\gamma t}-1} = 1 + \gamma / ({\mathrm{e}}^{\gamma t}-1).\end{align}

Part 1 of Theorem 6 then follows easily; the case $r_\infty=1$ follows by taking the limit of $\rho(t)$ as $\gamma \to 0$ , or it could be computed directly by solving the ODE in the special case $r=1$ . We now tackle the second part of Theorem 6, using (6) to determine $\mathbb{P}(\tau \le t \mid Z_0) = \rho(t)^{Z_0}$ when $Z_0 \to \infty$ under various limits of $\gamma Z_0$ . As $Z_0\to\infty$ , $\rho(t)^{Z_0}\to {\mathrm{e}}^{-c}$ if $\gamma/({\mathrm{e}}^{\gamma t}-1)=c/Z_0$ . Solving this gives $\gamma \, t = \log (1 + \gamma Z_0/c)$ . Recalling that $a=\gamma Z_0$ , there are then two cases.

Case 1: $a \to a_\infty \in [0,\infty)$ . In this case, if $\rho(t)^{Z_0} \to {\mathrm{e}}^{-c}$ then $t = (Z_0/a)(\log(1 + a/c))$ . Setting $w=\frac{1}{a}\log(1+a/c)$ gives $c = a/({\mathrm{e}}^{aw}-1)$ . If $a \to 0$ then $a/({\mathrm{e}}^{aw}-1) \to 1/w$ and $\mathbb{P}(\tau \le w Z_0) \to {\mathrm{e}}^{-1/w}$ , while if $a \to a_\infty>0$ then $\mathbb{P}(\tau \le w Z_0) \to {\mathrm{e}}^{-a_\infty/({\mathrm{e}}^{a_\infty w}-1)}$ .

Case 2: $a\to\infty$ . Taking $\gamma \, t = \log a - \log c$ gives $\rho(t)^{Z_0} \to {\mathrm{e}}^{-c}$ as $Z_0\to\infty$ . Letting $w=-\log c$ so that $c={\mathrm{e}}^{-w}$ , $\mathbb{P}(\tau \le \gamma^{-1}(\log \gamma Z_0 + w) ) \to {\mathrm{e}}^{-{\mathrm{e}}^{-w}}$ as $Z_0 \to \infty$ . In other words, $\gamma \tau - \log \gamma Z_0$ has a standard Gumbel distribution.

The following short lemma concludes the proof of case 2, and thus the proof of Theorem 6. □

Lemma 3. Let $Z_n$ be a sequence of copies of the linear birth and death process with initial value $Z_n(0)\to\infty$ and parameter $r_n=1-a_n/Z_n(0)$ with $a_n\to\infty$ . Let $(b_n)$ be a sequence with $b_n\to\infty$ and let $\tau_n=\inf\{t:Z_n(t)=0\}$ . Then

\[\lim_{n\to\infty}\mathbb{P}\left(\tau_n \le \frac{Z_n(0)}{b_n}(\log b_n + w)\right) \to {\mathrm{e}}^{-{\mathrm{e}}^{-w}}\]

for all $w \in \mathbb{R}$ if and only if $a_n-b_n = o(b_n/\log b_n)$ .

Proof. As before, suppress the dependence on n. Fix $w \in \mathbb{R}$ and let v be such that $\frac{1}{b}( \log b + w) = \frac{1}{a}(\log a + v)$ . Since $\log a + v = \displaystyle\frac{a}{b}(\log b + u)$ , subtracting $\log a+w$ from both sides gives

(7) \begin{align}v-w=\log(b/a) + \frac{a-b}{b}\log b + \frac{a-b}{b}w.\end{align}

If $a-b=o(b/\log b)$ the second term is o(1). Since $b\to\infty$ , $a = b + o(b)$ so $\log(b/a)=o(1)$ , $\displaystyle\frac{a-b}{b}w=o(1)$ , and so $v-w=o(1)$ . On the other hand, suppose that $v-w=o(1)$ for every $w\in \mathbb{R}$ . Setting $w=0$ gives

(8) \begin{align}\log(b/a) + \frac{a-b}{b}\log b = o(1).\end{align}

Then, setting $w=1$ and using (7) and (8) gives $\displaystyle\frac{a-b}{b}=o(1)$ . This in turn implies that $\log(b/a)=o(1)$ . Using (8) once more gives $\displaystyle\frac{a-b}{b}\log b=o(1)$ , as required. □

5. Approximation by linear birth and death processes

In this section we use the method of [Reference Brightwell, House and Luczak5], namely approximation by linear birth and death processes, to prove Theorem 1, Theorem 2, and the ‘final stage’ part of Theorem 3. As in the previous section, we suppress dependence on n and write $X_t$ , $Z_t$ , etc.

For $M > X_0>0$ (both depending on n) to be determined, let $r'=r(1-M/n)$ and let Z and Z’ be linear birth and death processes with respective parameters r and r’ and common initial value $X_0$ . Let $\tau_M = \inf\{t:X_t \in \{0,M\}\}$ . Then, applying the natural coupling of Section 3 to Z’,X,Z, labelled 1,2,3 respectively, and using Lemma 2 gives $Z^{\prime}_t \le X_t \le Z_t$ for $t\le \tau_M$ . Let $\tau_Z = \inf\{t:Z_t=0\}$ and $\tau_{Z'}= \inf\{t:Z_t'=0\}$ . The goal is to take M large enough that $\mathbb{P}(X_{\tau_M}=M) = o(1)$ , and small enough that $\mathbb{P}(X_t=Z_t$ for $t\le \tau_M)=1-o(1)$ if $X_0$ is fixed and $\tau_Z$ and $\tau_{Z'}$ have a common rescaled limit if $X_0\to\infty$ .

We begin with a general observation. Let $p_-(X)$ and $p_+(X)$ denote the probability that $X\to X-1$ or $X\to X+1$ in the embedded discrete-time Markov chain, or jump chain, of $X_t$ . Then $p_-(X)=1/(1+r(1-X/n))$ and $p_+(X)=r(1-X/n)/(1+r(1-X/n))$ ; in particular, $p_-(X)/p_+(X) = 1/(r(1-X/n))\ge 1/r$ for all X, so it follows easily that $r^{-X_t}$ is a supermartingale. Recall that $\tau = \inf\{t \colon X_t=0\}$ . Using optional stopping, which is applicable since X is bounded and $\mathbb{P}(\tau<\infty)=1$ , it is easy to show that

(9) \begin{align}\mathbb{P}(\tau \ne \tau_M \mid X_0) = \mathbb{P}(X_{\tau_M} = M \mid X_0) \le \frac{r^{-X_0} - 1}{r^{-M}-1}.\end{align}

If $r < 1$ then using the estimate $a/b \le (a+c)/(b+c)$ that holds for $0<a\le b$ and $c>0$ , we find that if $X_0,M>0$ then

(10) \begin{align}\frac{r^{-X_0}-1}{r^{-M}-1} \le \frac{r^{-X_0}}{r^{-M}} = (1-\gamma)^{M-X_0} \to 0 \qquad \text{if } \gamma \, (M-X_0)\to\infty.\end{align}

Recall that $\gamma = 1-r$ . If $\gamma \to 0$ and if, a fortiori, $\gamma M \to 0$ then

(11) \begin{align}\frac{r^{-X_0}-1}{r^{-M}-1} = \frac{(1-\gamma)^{-X_0}-1}{(1-\gamma)^{-M}-1} \sim \frac{X_0}{M} \to 0 \qquad \text{if } X_0=o(M).\end{align}

We proceed by cases. The first case is Theorem 1.

Proof of Theorem 1. In this setting, $\gamma\to\gamma_\infty\ge 0$ and we can assume $X_0$ is constant. For any $t \ge 0$ ,

\begin{equation*} |\mathbb{P}(\tau \le t \mid X_0) - \mathbb{P}(\tau_Z \le t \mid X_0)| \le \mathbb{P}(\tau \ne \tau_M \mid X_0) + \mathbb{P}(\tau = \tau_M , \tau_M \wedge t \ne \tau_Z \wedge t \mid X_0).\end{equation*}

Theorem 1 then follows from Theorem 6 if M can be chosen so that, for any $t\ge 0$ , the above $\to 0$ as $n\to\infty$ .

We let $p_1 = \mathbb{P}(\tau \ne \tau_M\mid X_0)$ and $p_2 = \mathbb{P}(\tau = \tau_M , \tau_M \wedge t \ne \tau_Z \wedge t \mid X_0)$ . We first find M such that $p_1 \to 0$ . If $\gamma_\infty > 0$ , using (10) we find that $p_1 \to 0$ if $M\to\infty$ . If $\gamma_\infty = 0$ , then using (11) we find that $p_1 \to 0$ if $M \to \infty$ and $\gamma M \to 0$ .

We next find M such that $p_2\to 0$ . If $\tau=\tau_M$ then since X and Z are both absorbed at 0, $p_2 \le \mathbb{P}(X_s \ne Z_s$ for some $s \le t \wedge \tau_M \mid X_0)$ . If $X,Z \in [0,M]$ their transition rates differ by at most $(r-r')M = rM^2/n$ , so using the exponential distribution and noting that ${\mathrm{e}}^{-x} \ge 1-x$ we find that $\mathbb{P}(X_s \ne Z_s$ for some $s \le t \wedge \tau_M \mid X_0) \le 1 - {\mathrm{e}}^{-rM^2t/n} \le rM^2t/n$ . Since r is bounded in n, the right-hand side $\to 0$ for fixed t provided $M=o(\sqrt{n})$ .

Thus, both $p_1,p_2 \to 0$ if we take $M\to\infty$ sufficiently slowly. □

Next we prove Theorem 2.

Proof of Theorem 2. We first reinterpret somewhat the conditions of the theorem.

First option: $c_\infty=-\infty$ and $a_\infty<\infty$ . Since $c_\infty=-\infty$ , $\frac{1}{\sqrt{n}\,}=o(|\delta_n|)$ so $a_\infty = \lim_{n\to\infty} |\delta_n| X_n(0)$ . Since $a_\infty<\infty$ , $X_n(0)=O(1/|\delta_n|) = o(\sqrt{n})$ . In shorthand: $\gamma X_0 \to a_\infty\in [0,\infty)$ and $X_0=o(\sqrt{n})$ .

Second option: $c_\infty>-\infty$ and $a_\infty=0$ . If $c_\infty \in \mathbb{R}$ then $a_\infty=0$ is equivalent to $X_n(0)/\sqrt{n} \to 0$ , and implies $|\delta_n| X_n(0) \to 0$ . If $c_\infty=\infty$ then, in the same way as above, $a_\infty =\lim_{n\to\infty} |\delta_n| X_n(0)$ and $X_n(0)=o(\sqrt{n})$ . So, in this case as well, $\gamma X_0 \to a_\infty \in [0,\infty)$ and $X_0=o(\sqrt{n})$ .

Thus the theorem is proved if it can be proved when $\gamma X_0 \to a_\infty \in [0,\infty)$ and $X_0=o(\sqrt{n})$ . First note that, with $p_1$ as in the proof of Theorem 1,

(12) \begin{align}|\mathbb{P}(\tau^{\prime}_Z \le \tau \le \tau_Z \mid X_0)| \ge 1 - p_1.\end{align}

Since $X_0\to\infty$ and $\gamma X_0 \to a_\infty<\infty$ , $\gamma \to 0$ . For $W \in \{X,Z,Z'\}$ , and writing $\tau$ as $\tau_X$ , let $F_W(t) = \mathbb{P}(\tau_W/X_0 \le t)$ . From (12),

(13) \begin{align}F_Z(t) - p_1 \le F_X(t) \le F_{Z'}(t)+p_1.\end{align}

By Theorem 6, if $\gamma X_0, \, \gamma' X_0 \to a_\infty$ then, for all $t\ge 0$ , $F_Z(t),F_{Z'}(t) \to \mathbb{P}(H_{a_\infty} \le t)$ . Since $\gamma \, X_0 \to a_\infty$ by assumption, it is enough to find M such that (i) $p_1\to 0$ and (ii) $(\gamma' - \gamma)\, X_0 \to 0$ .

For (i), using (10) and (11) we need either $\gamma\, M \to \infty$ or both $X_0/M \to 0$ and $\gamma\, M \to 0$ . For (ii) we compute $(\gamma'-\gamma)X_0 = (r-r')X_0 = rX_0M/n$ , so it is enough that $X_0M/n \to 0$ .

Subcase 1: $a_\infty=0$ . Define $\beta = |\gamma| \vee \frac{1}{\sqrt{n}\,}$ and let $M = \beta^{-1}(\beta X_0)^{1/2}$ . Since $\gamma X_0 \to 0$ and $X_0=o(\sqrt{n})$ , $\beta X_0 = \max(|\gamma| X_0,\,X_0/\sqrt{n}) \to 0$ , so $X_0/M = (\beta X_0)^{1/2} \to 0$ and $\gamma \, M \le \beta \, M = (\beta X_0)^{1/2} \to 0$ , satisfying (i). Since $\beta \ge \frac{1}{\sqrt{n}\,}$ , $\beta^{-1} \le \sqrt{n}$ so $M=o(\sqrt{n})$ and $X_0M/n \to 0$ , satisfying (ii).

Subcase 2: $a_\infty>0$ . Since $X_0=o(\sqrt{n})$ and $\gamma X_0$ has a positive limit, $\sqrt{n} \gamma \to \infty$ . Let $M = \gamma^{-1}(\sqrt{n}\gamma)^{1/2}$ . Then $\gamma M = (\sqrt{n}\gamma)^{1/2}\to\infty$ , satisfying (i). Since $X_0=o(\sqrt{n})$ and $M = (\sqrt{n}/\gamma)^{1/2} = \sqrt{n}/(\sqrt{n}\gamma)^{1/2} = o(\sqrt{n})$ , $X_0M/n \to 0$ , satisfying (ii). □

Now we prove Theorem 3 in the ‘final stage’, i.e. when $X_n(0)\le n\gamma_n |c_n|^{-\epsilon}$ for some $\epsilon>0$ . The proof is broadly the same as in [Reference Brightwell, House and Luczak5], although here the result is applicable to a somewhat larger set of initial values.

Proof of Theorem 3 when $X_n(0) \le n\gamma_n |c_n|^{-\epsilon}$ for some $\epsilon>0$ .

Using our abbreviated notation, we shall prove the result under the slightly less restrictive condition $X_0\log(\gamma X_0) = o(\gamma n)$ . To see that this is less restrictive, suppose $X_0 \le \gamma n|c|^{-\epsilon}$ , so that $X_0|c|^\epsilon \le \gamma n$ . By assumption in Theorem 3, $|c|\to\infty$ and $a=\gamma X_0 \to \infty$ . Using $|c|\to\infty$ , $\log|c|=o(|c|^\epsilon)$ so $X_0\log|c| = o(X_0|c|^\epsilon)= o(\gamma n)$ . Using $a\to\infty$ and $X_0=o(\gamma n)$ , $\log(\gamma X_0) = o(\log(\gamma^2 n)$ . Since $|c|=\gamma\sqrt{n}$ , $\log(\gamma X_0)=o(\log(|c|)$ . Therefore, $X_0 \log (\gamma X_0) = o(X_0\log|c|) = o(\gamma n)$ as desired.

If $X_0\log(\gamma X_0)=o(\gamma n)$ then, since $\gamma X_0 \to \infty$ , $X_0=o(\gamma n)$ . As noted in the statement of Theorem 3, in this case $g = \log(\gamma X_0)+o(1)$ . So, we want to show that $\gamma \tau - \log(\gamma X_0)$ converges to standard Gumbel.

Let $a =\gamma X_0$ and for $W \in \{X,Z,Z'\}$ let $F_W(t) = \mathbb{P}(\gamma \, \tau_W - \log a \le t)$ . Then, (13) also holds for this choice of $F_W(t)$ . According to Theorem 6, for all $t\ge 0$ $F_Z(t) \to \mathbb{P}(G\le t)$ , and if $(\gamma'-\gamma)X_0 = o(a/\log a)$ then, for all $t\ge 0$ , $F_{Z'}(t) \to \mathbb{P}(G \le t)$ . Following again the logic of the proof of Theorem 2, it is enough to find M such that (i) $p_1\to 0$ and (ii) $(\gamma' - \gamma)\, X_0 = o(a/\log a)$ .

For (i) we need $\gamma \, (M-X_0)\to\infty$ , for which $M = 2X_0$ suffices (and which, assuming only $\gamma X_0\to\infty$ , cannot be improved beyond a factor of 2). For (ii) we need $X_0M/n = o(\gamma X_0/\log(\gamma X_0))$ . Using $M=2X_0$ , the condition becomes $X_0 = o(\gamma n/\log(\gamma X_0))$ , which is the condition given. □

6. Subthreshold cutoff

In this section we prove Theorem 3 in the case where $X_n(0) \ge n\gamma_n |c_n|^{-\epsilon}$ for any $\epsilon>0$ . This corresponds to the intermediate and initial stages as described in [Reference Brightwell, House and Luczak5]; the final stage, where $X_n(0)\le n\gamma_n |c_n|^{-\epsilon}$ , was proved in the previous section. The proof for the initial stage is just about identical to their proof. For the intermediate stage, the setup of the problem follows their approach, then I use a different method to handle the error term between the process and its deterministic approximation (denoted $e_N$ in [Reference Brightwell, House and Luczak5] and W here). In [Reference Brightwell, House and Luczak5], an auxiliary result (Lemma 3.2 there) is used to show that the maximum of the error term is (deterministically) bounded by twice the maximum of the compensator of X, then the latter is estimated using the corresponding exponential martingale. Here, we compute the drift and diffusivity of the error term, then after making a time change we use a so-called drift barrier estimate, Lemma 19, which is proved in [Reference Basak, Durrett and Foxall4], to show that the error term remains small on the desired time interval. Neither approach seems to be strictly simpler or more efficient than the other.

Proof of Theorem 3 when $X_n(0) \ge n\gamma_n |c_n|^{-\epsilon}$ . As before, suppress n from the notation and write $X_t$ , $Y_t$ , etc. when it is convenient. For this proof only, let $c=\sqrt{n}\gamma$ which amounts to a change of sign; this saves us from always writing $|c|$ . Fix a small $\epsilon>0$ , to be determined, and let $Y_t=X(t/\gamma)/(\gamma n)$ . Then $\gamma X_0=c^2Y_0$ . Let us rewrite $g_n(X_n(0))$ from the statement of Theorem 3 in terms of Y. Abusing notation slightly, we have $g(Y_0) = 2\log c - \log(r + 1/Y_0)$ .

Let $t_\star = \inf\{t \colon Y_t \le c^{-\epsilon}\} = \inf\{t\colon X(t/\gamma) \le n\gamma c^{-\epsilon}\}$ . The final stage corresponds to $t_\star=0$ . Since X jumps by $\pm 1$ , Y jumps by $\pm 1/\gamma n = \pm 1/c\sqrt{n} = o(1/c)$ , so $g(Y_0)-g(Y(t_\star)) = -\log(r+1/Y_0) + \log(c^{-\epsilon} \pm o(1/c)) = -\log(r+1/Y_0) - \epsilon \log c + o(1)$ . Thus, to prove the result it remains to show that

(14) \begin{align}t_\star - (-\log(r+1/Y_0)-\epsilon \log c) \to 0.\end{align}

For the intermediate stage, suppose $c^{-\epsilon} \le Y_0 \le c^{\epsilon}$ . We will use the notation for drift and diffusivity discussed in the Appendix. Using the transition rates from (2) in (54) and noting that $\gamma=-(r-1)$ , for $X_t$ we compute $\mu(X) = rX(1-X/n)-X = -\gamma X - rX^2/n$ and $\sigma^2(X) = (1+r(1-X/n))X \le (1+r)X$ . From (54) we infer that if $Y(t)=\alpha X(\beta t)$ then $\mu(Y) = \alpha\beta \mu(X)$ and $\sigma^2(Y) = \alpha^2\beta \sigma^2(X)$ . So,

\[\mu(Y) = \frac{1}{\gamma^2 n}\mu(X) = -\frac{1}{\gamma n}X - r\frac{1}{(\gamma n)^2}X^2 = - Y - rY^2\]

and, using $\gamma^2n=c^2$ ,

\[\sigma^2(Y) = \frac{1}{\gamma^3n^2}\sigma^2(X) \le \frac{1}{\gamma c^2}(1+r)X = \frac{1}{c^2}(1+r)Y.\]

As in [Reference Brightwell, House and Luczak5], we directly estimate the distance between Y and its deterministic approximation. Let y(t) denote the solution to the initial value problem $y' = -y - ry^2$ , $y(0)=Y_0$ , and let $t^{\pm} = \inf\{t : y(t) = (1 \pm c^{-\epsilon})c^{-\epsilon} \}$ . Solving by separation of variables,

\begin{align*}t^{\pm} &= \log \frac{Y_0}{1+rY_0} - \log\frac{(1 \pm c^{-\epsilon})c^{-\epsilon}}{1+r(1 \pm c^{-\epsilon})c^{-\epsilon}} \\[3pt] &= -\log(r+1/Y_0) - \epsilon \log c + o(1).\end{align*}

Thus, $t^\pm$ both have the desired limit for $t_\star$ as in (14). The result will be proved if we show that $t^- \le t_\star \le t^+$ with probability $1-o(1)$ . Define the error process $W_t=Y_t-y(t)$ . If $\sup_{t \le t^-}|W_t| \le c^{-2\epsilon}$ then $t^- \le t_\star \le t^+$ , so we will show that the former has probability $1-o(1)$ . We first compute the drift and diffusivity of W. Factoring the difference of squares, $\mu(W) = \mu(Y) - y' = -Y-rY^2+y-ry^2 = -W(1+r(Y+y))$ . Since (y(t)) is continuous and has finite variation it has zero quadratic variation, so

\[\sigma^2(W) = \sigma^2(Y) \le \frac{1}{\gamma^2}(1+r)Y.\]

Since $r,y,Y>0$ ,

(15) \begin{equation}\mathrm{sgn}(\mu(W)) = -\mathrm{sgn}(W) , \qquad \frac{|\mu(W)|}{\sigma^2(W)} \ge \frac{c^2r}{1+r}|W|.\end{equation}

Next, we change time from t to s such that $\mu_s(W) = -W_s$ . To do so, let $s(t) = \int_0^t (1 + r(Y_r+y(r))\,{\mathrm{d}} r$ . Since $\mu(W)$ and $\sigma^2(W)$ are both scaled by ${\mathrm{d}} t/{\mathrm{d}} s \in (0,1]$ , (15) remains valid after the time change, so $\mu_s(W) = -W_s$ and $\sigma^2_s(W) = O(1/c^2)$ . Also, if $|W_s| \le c^{-2\epsilon}$ then $Y_s \le y(s) + c^{-2\epsilon} \le y(0) + c^{-2\epsilon} \le c^\epsilon+c^{-2\epsilon}$ , so $s'(t) \le 1 + r(2c^{\epsilon} + c^{-2\epsilon}) = O(c^{\epsilon})$ . Since $t^- = O(\log c)$ , if $\sup_{s \le s(t^-)}|W_t| \le c^{-2\epsilon}$ then $s(t^-) = O(c^{\epsilon}\log c)=o(c)$ so it is enough to show that $\sup_{s \le c}|W_s| \le c^{-2\epsilon}$ .

We first give an upper bound on W. Next we use Lemma 19, which is proved in [Reference Basak, Durrett and Foxall4]. In the notation of Lemma 19, let $x = c^{-2\epsilon}/2$ , $X = W - x$ , $\Delta_\infty(X) = 1/\gamma n$ , $\mu_\star = x$ , $\sigma^2_\star=C/c^2$ for some constant C, $C_\mu = 2x$ , and, since $\Delta_\infty(X)\mu_\star/\sigma_\star^2 = c^2 x/C\gamma n = \gamma^2n x/C\gamma n \le \gamma/C=O(1)$ , take $C_\Delta$ to be some large enough constant. Then, $\Gamma = \exp(x^2/16\sigma^2_\star) = \exp(c^{2-2\epsilon}/16 C) \ge c$ for large enough c, and since $c\to\infty$ by assumption, with probability $1-o(1)$ , $X_s \le x$ or equivalently $W_s \le 2x = c^{-2\epsilon}$ for all $s \le c$ . A matching bound for $-W$ is proved in the same way.

For the initial stage, suppose $Y_0 \ge c^{\epsilon}$ and let $t^\star = \inf\{t:Y_t \le c^{\epsilon}\}$ . If $Y_0 \ge c^\epsilon$ then $Y_0 \to \infty$ so $g(Y_0)=2\log c - \log r+o(1)$ . Since the result is proved for $Y_0\le c^{\epsilon}$ it is enough to show that $t^\star = o(1)$ . From the drift of Y and Jensen’s inequality applied to $\mathbb{E}[Y_t^2]$ we find that $u(t) = \mathbb{E}[Y_t]$ satisfies the differential inequality $u' \le -u - ru^2 \le -ru^2$ . Integrating the inequality, $u(t) \le ( u(0)^{-1} + rt)^{-1} \le 1/rt$ . By Markov’s inequality, $Y_t \le c^{\epsilon}$ or equivalently $t^\star \le t$ with high probability if $u(t) = o(c^{\epsilon})$ , which is the case if $t=c^{-\epsilon/2}$ . Since $c^{-\epsilon/2}=o(1)$ , the result is proved. □

7. Non-linear diffusive

Proof of Theorem 4. The first part of the proof is to show convergence to the limiting diffusion, which is done in [Reference Dolgoarshinnykh and Lalley7] by convergence of generators; here, we do it using a general result, Lemma 20, that requires convergence of drift and diffusion coefficients and vanishing jump size. The second part of the proof is to show the extinction time is short when the initial value is small, relative to the space and time scale of the limiting diffusion.

For this section let $Y_n(t)=X_n(\sqrt{n}t)/\sqrt{n}$ and let $c_n=\sqrt{n}(r_n-1)$ as in the statement of the theorem. There are two cases to cover: $Y_n(0) \to y \in (0,\infty)$ and $Y_n(0) \to \infty$ .

Case 1: $Y_n(0) \to y\in (0,\infty)$ . First we use Lemma 9 in the Appendix, which is a result from [Reference Ethier and Kurtz9], to show that if $Y_n(0) \to y$ then, for all but countably many R, $Y_n(\cdot\wedge \tau_n^R) \stackrel{({\mathrm{d}})}{\rightarrow} Y(\cdot\wedge \tau^R)$ , where $\tau_n^R$ and $\tau^R$ are the exit times of $Y_n$ and Y from $\big(\frac{1}{R},R\big)$ as described in Lemma 20. Recall that $Y_n(t)$ has transitions

\[Y_n \to \begin{cases}Y_n+\frac{1}{\sqrt{n}\,} & \text{at rate} \quad nY + \sqrt{n}Y(c_n-Y) - c_nY, \\[4pt] Y_n-\frac{1}{\sqrt{n}\,} & \text{at rate} \quad nY, \end{cases}\]

so $Y_n$ has jump size $\frac{1}{\sqrt{n}\,}=o(1)$ and $\mu(Y_n) = Y(c_n-Y) - c_nY/\sqrt{n}$ , $\sigma^2(Y_n) = 2Y + Y(c_n-Y)/\sqrt{n} - c_nY/n$ . For $|y| \le R$ , $\mu(y) \to b(y)$ and $\sigma^2(y) \to a(y)$ uniformly, where $b(y) = y(c_\infty-y)$ and $a(y) = 2y$ . Note that b and $\sqrt{a}$ are Lipschitz on compact subsets of $(0,\infty)$ . By Lemma 20, the desired convergence holds.

For $y>0$ define the mapping $T_y(f) = \inf\{t:f(t) \le y\}$ from cÀdlÀg functions $f:\mathbb{R}_+\to \mathbb{R}_+$ with the topology of uniform convergence on compacts. If $f_i \to f$ and f is continuous then, since $\inf\{f(t):t \in [0,T_y(f)-\epsilon]\}>y$ for any $\epsilon>0$ , it follows that $\liminf_i T_y(f_i) \ge T_y(f)$ . On the other hand, if $y>0$ then, for any $\epsilon>0$ , $Y(T(y)+\epsilon)$ intersects [0,y) since its diffusion coefficient at y is non-zero. In other words, $\mathbb{Q}(f\colon \text{if} \ f_i \to f \ \text{then} \ \limsup_i T_y(f_i) \le T_y(f))=1$ , where $\mathbb{Q}$ is the law of Y. Combining the two, the discontinuity points of $T_y$ have $\mathbb{Q}$ -measure 0.

Let $T_n(y) = T_y(Y_n)$ and $T(y) = T_y(Y)$ , and let $T_n=T_n(0)$ and $T=T(0)$ . By the continuous mapping theorem and convergence of $Y_n$ to Y, $T_n(y) \stackrel{({\mathrm{d}})}{\rightarrow} T(y)$ for $y >0$ . Let $\mathbb{P}_n$ be the law of $Y_n$ . Since $T_n \ge \lim_{y\to 0^+}T_n(y)$ and $T = \lim_{y\to 0^+}T(y)$ ,

\[\limsup_n \mathbb{P}_n(T_n \le t) \le \limsup_{y\to 0^+}\lim_n \mathbb{P}_n(T_n(y) \le t) = \lim_{y\to 0^+} \mathbb{Q}(T(y) \le t) = \mathbb{Q}(T \le t).\]

To obtain the opposite inequality it is enough to show that for any $\epsilon,t>0$ there are $\alpha,n_0$ such that $\mathbb{P}_n(T_n >t \mid Y_n(0)\le \alpha)\le \epsilon$ for $n\ge n_0$ , since then

\begin{align*}\liminf_n \mathbb{P}_n(T_n \le t) & \ge \liminf_n \mathbb{P}_n(T_n(\alpha) \le t)\inf_{x \le \alpha}\mathbb{P}_n(T_n \le t \mid Y_n(0) =x) \\[3pt]& \ge (1-\epsilon)\lim_{n\to\infty}\mathbb{P}_n(T_n(\alpha) \le t) \\& = (1-\epsilon)\mathbb{Q}(T(\alpha) \le t) \ge (1-\epsilon)\mathbb{Q}(T \le t).\end{align*}

Returning to the original time scale, $X_n$ is dominated by the linear birth and death process Z with parameter r and $Z_0=X_n(0)$ , so, using $\rho(t)$ as in the proof of Theorem 6, $\mathbb{P}(X_n(\sqrt{n} t)>0 \mid X_n(0) \le \sqrt{n}\alpha ) \le 1-(\rho(\sqrt{n}t))^{\sqrt{n}\alpha}$ . If $c_n\le 0$ we can take $r=1$ which has $\rho(t) = 1-1/(1+t)$ , so $1-(\rho(\sqrt{n}t))^{\sqrt{n}\alpha} \le \alpha/t$ is at most $\epsilon$ if $\alpha \le \epsilon/t$ .

If $c_n>0$ , rewrite $\rho(t)$ from (6) with $\delta=r-1$ instead of $\gamma = 1-r$ to obtain

\[\rho(t)^{-1} = \frac{r {\mathrm{e}}^{\delta t}-1}{{\mathrm{e}}^{\delta t}-1} = 1 + \frac{\delta}{1-{\mathrm{e}}^{-\delta t}}.\]

Using $\delta_n=c_n/\sqrt{n}$ ,

\[(\rho(\sqrt{n}t))^{\sqrt{n}\alpha} = \left(1 + \frac{c_n/\sqrt{n}}{1-{\mathrm{e}}^{-c_nt}}\right)^{-\sqrt{n}\alpha} \to \exp(-c_\infty\alpha/(1-{\mathrm{e}}^{-c_\infty t})) \qquad \text{as} \ n\to\infty,\]

and since for fixed t the limit $\to 1$ uniformly as $\alpha \to 0$ , we are done.

Case 2: $Y_n(0) \to \infty$ . It remains to show the results of Step 2 are true for $y=\infty$ . First we need to make sense of T when $Y(0)=\infty$ . Let $T(y,w) = \inf\{t:Y(t) \le w \mid Y(0)=y\}$ . Since Y is continuous, and using the strong Markov property,

(16) \begin{equation}T(y,0) \stackrel{({\mathrm{d}})}{=} T(y,w) + T(w,0),\end{equation}

where the last two are independent. In particular, T(y,0) dominates T(w,0) for $y>w$ . On the other hand, letting $U=Y-c_\infty$ , $\mu(U)= -U(U+c_\infty) \le -U^2$ so, integrating and using Jensen’s inequality, $\mathbb{E}[U(t)] \le (1/\mathbb{E}[U_0] + t)^{-1} \le 1/t$ . Using Markov’s inequality, if $y>w>c_\infty$ then $\mathbb{P}(T(y,w) > t) \le \mathbb{P}(U(t) > w-c_\infty \mid Y(0) = y) \le ((w-c_\infty)t)^{-1}$ . It follows that $T(y,w) \stackrel{(\mathrm{p})}{\rightarrow} 0$ uniformly over $y \in [w,\infty)$ as $w\to\infty$ . Combining with (16), there exists $T(\infty,0)$ such that $T(y,0) \stackrel{({\mathrm{d}})}{\rightarrow} T(\infty,0)$ as $y\to\infty$ . A similar argument shows that for $T_n(y,0) = \inf\{t:Y_n(t)=0 \mid Y_n(0) = y\}$ there is a limit $T_n(\infty,0)$ . By Step 2, $T_n(y,0) \stackrel{({\mathrm{d}})}{\rightarrow} T(y,0)$ for each $y>0$ , so it follows that $T_n(\infty,0) \stackrel{({\mathrm{d}})}{\rightarrow} T(\infty,0)$ . □

8. Threshold and metastable

In this section we prove Theorem 5. As we’ve done so far, we’ll avoid writing subscript n on everything. So, for example $X_n^\star$ is simply denoted $X_\star$ , etc. Since we need some additional decorations later, we move the $\star$ from $\tau_n^\star$ into the subscript, so $\tau_\star = \inf\{t \colon X_t \in \{0,X_\star\}\}$ . We begin with some basic theory and estimation of an important function. Let $q_+(\,j)=rj(1-j/n)$ and $q_-(\,j)=j$ denote the transition rates of X, and for later use let $q(\,j)=q_+(\,j)+q_-(\,j)$ . For integer j, define $h_+(\,j) = \mathbb{P}(X_{\tau_\star}=X_\star \mid X_0=j)$ and $h_-(\,j) = \mathbb{P}(X_{\tau_\star}=0 \mid X_0=j) = 1-h_+(\,j)$ .

By definition, $h_+(X(t\wedge \tau_\star))$ is a martingale, as is $h_-(X(t\wedge \tau_\star))$ . Using the generator of the process, it follows that $q_+(\,j)(h_+(\,j+1)-h_+(\,j))+q_-(\,j)(h_+(\,j-1)-h_+(\,j))=0$ and similarly for $h_-$ . Let $\nu(0)=1$ , and for $j \ge 1$ let $\nu(\,j)= \prod_{i=1}^j q_-(i)/q_+(i)$ . Using the linear equations for $h_+,h_-$ and the boundary conditions $h_+(0)=0$ , $h_+(X_\star)=1$ , $h_-(0)=1$ , $h_-(X_\star)=0$ , we can solve to find

(17) \begin{align}h_+(\,j) = \frac{\sum_{k=0}^{j-1}\nu(k)}{\sum_{k=0}^{X_\star-1}\nu(k)} , \qquad h_-(\,j) = \frac{\sum_{k=j}^{X_\star-1} \nu(k)}{\sum_{k=0}^{X_\star-1}\nu(k)}.\end{align}

The solution of the above linear equations for $h_+,h_-$ to obtain (17) is not hard; it can be found, for example, in [Reference Durrett8, Example 5.3.9]. We begin by estimating $\nu(k)$ . Since it is no more difficult to estimate, and since we will need it later, we estimate the more general $\smash{\nu(\,j,k) = \prod_{i=j+1}^k q_-(i)/q_+(i)}$ , defined for $0\le j<k <n$ ; we recover from it $\nu(k)= \nu(0,k)$ . It will be helpful to have both a general upper bound and a precise estimate.

Lemma 4. Let $V(x) = x(\log r - 1) - (1-x)\log(1-x)$ . Then, for integer na and nb with $0 \le a < b \le 1$ , $\nu(na,nb) \le \exp(-n(V(b+1/n)-V(a+1/n)))$ and

\[\nu(na,nb) = \sqrt{\frac{1-a}{1-b}}\exp(-n(V(b)-V(a))E_n(a,b),\]

with $|\log E_n(a,b)| \le (12n(1-b)^2(b-a))^{-1}$ .

Proof. Since $q_+(i)/q_-(i) = r(1-i/n)$ ,

(18) \begin{align}-\log \nu(\,j,k) = \sum_{i=j+1}^k \left(\log r + \log(1-i/n) \right) = (k-j)\log r - \sum_{i=j+1}^k f(i/n),\end{align}

where $f(x)=-\log(1-x)$ is positive and increasing for $x \in (0,1)$ . Since f(x) has the antiderivative $x + (1-x)\log (1-x)$ , the upper bound follows. Using a trapezoidal approximation with $k-j$ subintervals of size $1/n$ and writing the approximation as an upper Riemann sum minus a telescoping triangular correction,

(19) \begin{align}\int_{j/n}^{k/n} f(x) \, {\mathrm{d}} x = \frac{1}{n}\sum_{i=j+1}^{k} f(i/n) - \frac{1}{2n}(f(k/n)-f(\,j/n)) + R_n(\,j,k),\end{align}

where the error term (see [Reference Talvila and Wiersma21] for a simple proof) has the bound

\[|R_n(\,j,k)| \le \frac{\max_{x \in [j/n,k/n]}|f''(x)|}{12(k/n-j/n)n^2} = \frac{1}{(1-k/n)^2}\frac{1}{12(k/n-j/n)n^2}.\]

Using the antiderivative of f together with (18) and (19),

\[-\frac{1}{n}\log \nu(na,nb) = V(b)-V(a)+\frac{1}{2n}\log\frac{1-b}{1-a}+R_n(na,nb),\]

and the precise estimate follows. □

Note that V(x) has $V(0) = 0$ , $V'(x) = \log(r(1-x))$ , and $V''(x) = -1/(1-x)$ . In particular, it is concave on [0,1) and has $V'(x_\star)=\log 1 = 0$ , so is increasing and positive on $(0,x_\star)$ and decreasing on $(x_\star,1)$ , with maximum $V_\star = V(x_\star) = \log r + 1/r - 1>0$ , and has $V''(x_\star) = -1/(1-(1-1/r)) = -r$ . If $\delta_\infty = \lim_n \delta>0$ then $V_\star$ has a positive limit, while if $\delta \to 0$ then $V_\star = \log(1+\delta)+1/(1+\delta)-1 = \delta - \delta^2/2 - \delta + \delta^2 + O(\delta^3) \sim \delta^2/2$ .

8.1. Extinction probability

In this section we prove the estimate of extinction probability in Theorem 2. The result in the case $\delta_\infty>0$ belongs to [Reference Andersson and Djehiche1, (A2) of Theorem 1], while in the case $\delta \to 0$ it is new. The approach is to estimate $h_-(X_0)$ for $h_-$ as in (17), and the proof works by estimating the values of $\nu(k)$ separately for small and large k.

Lemma 5. Suppose $\delta X_0 \to a_\infty \in (0,\infty)$ . Then,

\[\mathbb{P}(X_{\tau_\star}=0 ) \to \begin{cases} {\mathrm{e}}^{-a_\infty} & \text{if} \quad \delta \to 0 , \\(1+\delta_\infty)^{-X_0} & \text{if} \quad \delta \to \delta_\infty>0.\end{cases}\]

If $\delta X_0 \to \infty$ then $\mathbb{P}(X_{\tau_\star}=X_\star ) \to 1$ .

Proof. The quantity of interest is

\[\mathbb{P}(X_{\tau_\star}=0 ) = h_-(X_0) = \frac{\sum_{k=X_0}^{X_\star-1}\nu(k)}{\sum_{k=0}^{X_\star-1}\nu(k)}.\]

Since $q_-(k)/q_+(k) = (r(1-k/n))^{-1}$ , for any $M \ge 1$ and $k \in \{1,\dots,M\}$ , $r^{-k} \le \nu(k) \le (r(1-M/n))^{-k}$ and so

(20) \begin{align}r^{-j}\frac{1-r^{-(M-j)}}{1-1/r} \le \sum_{k=j}^{M-1}\nu(k) \le r^{-j}\frac{1}{1-(r(1-M/n))^{-1}}.\end{align}

Case 1: $\delta\to 0$ . Suppose M is taken large enough that $\delta(M-j) \to\infty$ , and small enough that $M = o(\delta n)$ . Then $r^{-(M-j)} = (1+\delta)^{-(M-j)} \le {\mathrm{e}}^{-\delta(M-j)} \to 0$ and $r(1-M/n) = (1+\delta)(1-M/n) = 1 + \delta + o(\delta)$ , so that $1-1/(r(1-M/n)) = 1-1/(1+\delta+o(\delta)) = 1-(1-\delta+o(\delta)) \sim \delta$ . Similarly, $1-1/r \sim \delta$ . Since $\delta\to 0$ , $r^{-j} = (1+\delta)^{-j} \sim {\mathrm{e}}^{-\delta j}$ , so

(21) \begin{align}\sum_{k=j}^{M-1}\nu(k) \sim {\mathrm{e}}^{-\delta j}/\delta.\end{align}

We have $V'(0) = \log r = \delta-o(\delta)$ and $x_\star = \delta/r=o(1)$ , so $V(x) \sim x$ uniformly over $x \in [0,x_\star]$ as $n\to\infty$ . Using the upper bound from Lemma 4 with $a=0$ and nb in the range $\{M,\dots,X_\star-1\}$ , for large n, $\nu(nb) = \nu(0,nb) \le \exp(-\delta nb/2)$ . Summing over nb,

\[\sum_{k=M}^{X_\star-1}\nu(k) \le \frac{{\mathrm{e}}^{-\delta M/2}}{1-{\mathrm{e}}^{-\delta/2}} = o(1/\delta),\]

since $\delta M\to\infty$ by assumption and $1-{\mathrm{e}}^{-\delta/2} \sim \delta/2$ . Noting that $1-1/r \sim \delta$ and combining with (21),

(22) \begin{align}\sum_{k=j}^{X_\star-1}\nu(k) \sim {\mathrm{e}}^{-\delta j}/\delta.\end{align}

Using the values $j=0$ and $j=X_0$ , we conclude that if $\delta M - \delta X_0 \to\infty$ (which also implies $\delta M \to\infty$ ) and $M=o(\delta n)$ , then

\[h_-(X_0) \sim \frac{{\mathrm{e}}^{-\delta X_0}/\delta}{{\mathrm{e}}^{-0}/\delta} = {\mathrm{e}}^{-\delta X_0} \to {\mathrm{e}}^{-a_\infty}.\]

If $a_\infty<\infty$ , since $\delta X_0$ has a finite limit and $\sqrt{n}\delta\to\infty$ it is easy to check that $M=\sqrt{n}(\sqrt{n}\delta)^{1/2}$ satisfies the conditions. If $a_\infty=\infty$ , since ${\mathrm{e}}^{-a_\infty}=0$ and $h_-(X_0)$ decreases with $X_0$ it is enough to consider the case where $\delta X_0\to\infty$ arbitrarily slowly; thus, to satisfy the condition $\delta M - \delta X_0 \to\infty$ it is sufficient that $\delta M \to \infty$ , for which the above choice of M suffices.

Case 2: $\delta \to \delta_\infty >0$ . Note that $r\to r_\infty = (1+\delta_\infty)>1$ . Also, the condition $\delta X_0\to a_\infty$ is equivalent to $X_0$ eventually being constant if $a_\infty<\infty$ , and to $X_0\to\infty$ if $a_\infty=\infty$ . Suppose that $(M-j)\to\infty$ and $M=o(n)$ . Then, $r^{-(M-j)} \to 0$ and $1-M/n \to 1$ . From (20) we find

(23) \begin{align}\sum_{k=j}^{M-1}\nu(k) \sim r^{-j}/(1-1/r_\infty).\end{align}

Since $V(0)=0$ , V is concave and both $x_\star$ and $V(x_\star)$ have a positive limit; for some constant $C_1>0$ , eventually $V(x) \ge C_1x$ for $x \in [0,x_\star]$ . Using Lemma 2 as before with $a=0$ and nb in the range $\{M,\dots,X_\star-1\}$ , since $\delta$ is bounded by assumption, $1-b \ge 1-x_\star$ , which has a positive lower bound, and $nb \ge nM \to\infty$ , so uniformly $|\log E_n(a,b)| \to 0$ and for large n and some constant $C_2>0$ , $\nu(nb) = \nu(0,nb) \le C_2{\mathrm{e}}^{-C_1 nb}$ . Again, summing over nb,

\[\sum_{k=M}^{X_\star-1}\nu(k) \le C_2\frac{{\mathrm{e}}^{-C_1 M}}{1-{\mathrm{e}}^{-C_1}} \to 0.\]

Using the values $j=0$ and $j=X_0$ for constant $X_0<\infty$ and letting $M=\sqrt{n}$ , combining the above with (23) we find

\[h_-(X_0) = \frac{r^{-X_0}/(1-1/r_\infty) + o(1)}{1-1/r_\infty + o(1)} \to r_\infty^{-X_0}.\]

If $X_0\to\infty$ we may again assume it does so arbitrarily slowly, in which case $M=\sqrt{n}$ again suffices. Using $j=0$ and $j=X_0$ as above we find the numerator $\to 0$ while the denominator $\to 1-1/r_\infty>0$ , so $h_-(X_0) \to 0$ as desired. □

8.2. Rapid extinction

Next we prove the results on rapid extinction from Theorem 5. Define the probability measures $P^\star$ and $P^0$ for events E by

(24) \begin{align}P^\star(E) = \mathbb{P}(E \mid X_{\tau_\star}=X_\star) , \qquad P^0(E) = \mathbb{P}(E \mid X_{\tau_\star} = 0).\end{align}

Using the well-known Doob h-transform (which can be found by computing the generator of the conditioned process), we find that with respect to $P^\star$ and $P^0$ respectively, for $t<\tau_\star$ , X is a continuous-time Markov chain with transition rates

(25) \begin{align} \begin{split} q_+^\star(\,j) & = q_+(\,j)\frac{h_+(\,j+1)}{h_+(\,j)} , \qquad q_-^\star(\,j) = q_-(\,j) \frac{h_+(\,j-1)}{h_+(\,j)} ; \\[3pt] q_+^0(\,j) & = q_+(\,j)\frac{h_-(\,j+1)}{h_-(\,j)} , \qquad q_-^0(\,j) = q_-(\,j) \frac{h_-(\,j-1)}{h_-(\,j)}. \end{split}\end{align}

The following lemma is an equivalent formulation of the rapid extinction results of Theorem 5.

Lemma 6. Suppose $\delta X_0 \to a_\infty \in (0,\infty)$ and let $P^0$ be as in (24).

1. 1. If $\delta \to 0$ , then, for $w\ge 0$ , $P^0( \tau \le w X_0) \to \exp(-a_\infty/({\mathrm{e}}^{-a_\infty w}-1))$ .

2. 2. If $\delta \to \delta_\infty >0$ , then, letting $\gamma_\infty=\delta_\infty/r_\infty$ , $P^0(\tau \le t/r_\infty) \to (1+ \gamma_\infty/({\mathrm{e}}^{-\gamma_\infty t}-1))^{-X_0}$ .

Proof. We first estimate the transition rates (25) with respect to $P^0$ , then approximate by a linear birth and death process as in the proof of Theorems 1 and 2 to obtain the scaling of $\tau$ .

(26) \begin{align}\frac{h_-(\,j+1)}{h_-(\,j)} = \frac{\sum_{k=j+1}^{X_\star-1}\nu(k)}{\sum_{k=j}^{X_\star-1}\nu(k)} = 1 - \frac{\nu(\,j)}{\sum_{k=j}^{X_\star-1}\nu(k)} = 1 - \frac{1}{\sum_{k=j}^{X_\star-1}\nu(\,j,k)} ;\end{align}

similarly,

(27) \begin{align}\frac{h_-(\,j-1)}{h_-(\,j)} = 1 + \frac{1}{\sum_{k=j}^{X_\star-1}\nu(\,j-1,k)}.\end{align}

Again, we divide by cases.

Case 1: $\delta\to 0$ . Let $c=\sqrt{n}\delta$ so that $c\to\infty$ and $c=o(\sqrt{n})$ by assumption, let $M = \lfloor 2\sqrt{n/c} \rfloor$ , so that $M/\sqrt{n} =o(1)$ , and let $M' = \lfloor \sqrt{n} \rfloor$ . For $j \le M$ , write

(28) \begin{align}\sum_{k=j}^{X_\star-1}\nu(\,j,k) = \sum_{k=j}^{M'-1} \nu(\,j,k) + \sum_{k= M'}^{X_\star-1}\nu(\,j,k).\end{align}

If $j \le k < M'$ then $r^{-(k-j)} \le \nu(\,j,k)\le (r(1-M'/n))^{-(k-j)}$ , so, uniformly over $j \le M$ ,

(29) \begin{align}\frac{1-r^{-(M'-M)}}{1-1/r} \le \sum_{k=j}^{M'-1} \nu(\,j,k) \le \frac{1}{1-1/(r(1-M'/n))}.\end{align}

Thus, the above sum $\sim 1/(1-1/r) \sim 1/\delta$ uniformly over $j \le M$ provided $r^{-(M'-M)} \to 0$ and $M'/n = o(\delta)$ . The second point is clear, since $M'/n \le \frac{1}{\sqrt{n}\,} =o(\delta)$ . To check the first point, since $r^{-(M'-M)}=(1+\delta)^{-(M'-M)} \le {\mathrm{e}}^{-\delta(M'-M)}$ it is enough that $\delta(M'-M)\to\infty$ . Since $c=\sqrt{n}\delta \to \infty$ and $\delta=O(1)$ , $\delta(M'-M) = c - 2\sqrt{c} + O(\delta) \to \infty$ . To estimate the second sum on the right-hand side of (28), we note that since $V'(0)=\delta-o(\delta)$ and $x_\star=o(1)$ , for large n, $V(x)-V(y)\ge \delta(x-y)/2$ if $0 \le y<x\le x_\star$ . With $na=j \le M$ and $M' \le nb \le X_\star-1$ , $1-a,1-b \ge 1-x_\star\to 1$ and $nb-na \ge M'-M \to \infty$ , so $\log|E_n(a,b)| \to 0$ uniformly over a and b; thus, for large n and $j\le M$ , $M' \le k \le X_\star-1$ , $\nu(\,j,k) \le 2\exp(-\delta(k-j)/2)$ . Summing over k, the second term in (28) is $\le 2{\mathrm{e}}^{-\delta(M'-M)/2}/(1-{\mathrm{e}}^{-\delta/2}) \sim 4{\mathrm{e}}^{-\delta(M'-M)/2}/\delta=o(1/\delta)$ . Combining the two estimates, it follows that the sum on the left-hand side of (28) $\sim 1/\delta$ uniformly over $j\le M$ .

Since $\nu(\,j-1,k)=\nu(\,j-1,j)\nu(\,j,k)$ and $1/r \le \nu(\,j-1,j) \le 1/(r(1-M/n))$ which $\to 1$ uniformly over $j\le M$ , using (26), (27), and (25) we find that, uniformly over $j \le M$ ,

(30) \begin{align}q_+^0(\,j) \sim r \, j \, (1-j/n) \, (1 - \delta) , \qquad q_-^0(\,j) \sim j \, (1+\delta) .\end{align}

Let $\tilde X$ denote the process with $\tilde X_0=X_0$ and transition rates $\tilde q_-(\,j) = j$ and $\tilde q_+(\,j) = q_+^0(\,j)( j / q_-(\,j))$ , which are the same as for X with respect to the measure $P^0$ except multiplied by the factor $j/q_-^0(\,j)$ at each non-zero j. Then X with respect to $P^0$ is obtained from $\tilde X$ as $X_t = \tilde X_{s(t)}$ for $t\le \tau$ , where s(t) is the inverse of the function

\[t(s) = \int_0^s \frac{\tilde X_u}{q_-^0(\tilde X_u)} \, {\mathrm{d}} u.\]

From the estimate of (30) we have $\sup_{t \le \tau_M}s(t)/t \sim 1+\delta \to 1$ , so to obtain the desired result for X with respect to $P^0$ it is enough to show it for $\tilde X$ . Since $M=o(\sqrt{n})$ , $M/\delta n = o(1/\sqrt{n}\delta) = o(1)$ . Using $r=1+\delta$ , uniformly over $j\le M$ we have $\tilde q_+(\,j)/j \sim (1+\delta)(1-o(\delta))(1-\delta)/(1+\delta) = 1-\delta+o(\delta)$ . Let $\gamma,\gamma'$ be lower and upper bounds on $1-\tilde q_+(\,j)/j$ , respectively, and construct linear birth and death processes Z,Z’ with parameters $1-\gamma$ and $1-\gamma'$ , and initial value $X_0$ , so that $Z^{\prime}_t \le X_t \le Z_t$ for $t\le \tau_M$ . Following the proof of Theorem 2, combining (9) and (10) we have $\mathbb{P}(\tau \ne \tau_M \mid X_0) \le (1-\gamma)^{M-X_0}$ , which $\to 0$ if $X_0 \le \sqrt{n/c}$ since then $\gamma \sim \delta$ and $\delta (M-X_0) \ge \delta \sqrt{n/c} = \sqrt{c} \to \infty$ . As in the proof of Theorem 2, it only remains to check that $(\gamma-\delta)X_0,(\gamma'-\delta)X_0 \to 0$ . This follows easily from the fact that $\gamma,\gamma' \sim \delta$ .

Case 2: $\delta \to \delta_\infty >0$ . Let $r_\infty=1+\delta_\infty$ . Since $\delta X_0$ converges, we may assume $X_0$ is constant. We follow the same approach as before, only with different M. So, let $M'=\lfloor \sqrt{n} \rfloor$ and let $M\to \infty$ slowly. Then $M'/n \to 0$ , $M'-M\to\infty$ , and, since $r \to r_\infty>1$ , $r^{-(M'-M)} \to 0$ . Thus, from (29) the first sum on the right-hand side of (28) $\sim 1/(1-1/r) \to r_\infty/\delta_\infty$ . To estimate the second sum, note that if $j\le M$ and $M' \le k \le X_\star-1$ then $\nu(\,j,k) = \nu(\,j,M')\nu(M',X_\star-1) \le \nu(M,M') \le (r(1-M'/n))^{-(M'-M)}$ . We may assume that $M=o(\sqrt{n})$ ; then, for large n, the right-hand side above is $\le {\mathrm{e}}^{-\delta_\infty\sqrt{n}/2}$ . Summing over at most n such terms, the second sum on the right-hand side of (28) is $\le n{\mathrm{e}}^{-\delta_\infty \sqrt{n}/2} = o(1)$ so, combining the two, the left-hand side of (28) $\to r_\infty/\delta_\infty$ uniformly for $j\le M$ .

Recall that $\nu(\,j-1,k)=\nu(\,j-1,j)\nu(\,j,k)$ . Uniformly for $j\le M$ , $\nu(\,j-1,j)\to 1/r_\infty$ and $(1-j/n) \to 1$ . In addition, $1-\delta_\infty/r_\infty = 1/r_\infty$ . Thus, uniformly over $j\le M$ ,

\[q_+^0(\,j) \sim r_\infty j(1-\delta_\infty/r_\infty) = j , \qquad q_-^0(\,j) \sim j(1+\delta_\infty) = r_\infty j.\]

Let $\tilde X$ be as before, which has $\tilde q_-(\,j) = j$ and $\tilde q_+(\,j) \sim j/r_\infty$ . In this case, $X_t = \tilde X_{s(t)}$ with $\sup_{t \le \tau_m}s(t)/t \sim r_\infty$ . Thus, the result for X is obtained from the one for $\tilde X$ by changing time by the factor $r_\infty$ . For $\tilde X$ , compare to the linear birth and death process Z with parameter $1/r_\infty$ . Following the proof of Theorem 2 it suffices to show that, for fixed $t>0$ , $\mathbb{P}(\tau \ne \tau_M),\mathbb{P}(\tilde X_s \ne Z_s$ for some $s \le t \wedge \tau_M)=o(1)$ . Since $\tilde q_+(\,j) \le \tilde q_-(\,j)$ , $\tilde X$ is a supermartingale, so the first statement is true provided $M\to\infty$ . For $t< \tau_m$ the difference in rates between X and Z, when they take the same value, is at most $\sup_{j \le M}\tilde |q_+(\,j) - j/r_\infty| = o(M)$ , so is o(1) if $M\to\infty$ slowly enough. Thus, for constant $X_0$ and fixed $t>0$ , $\mathbb{P}(X_s \ne Z_s$ for some $s \le t \wedge \tau_M \mid X_0) \le 1-{\mathrm{e}}^{-o(1)t} = o(1)$ , which shows the second statement and completes the proof. □

8.3. Metastability

In this section we prove the metastability results of Theorem 5. In particular, in this section we condition on $A_\star = \{X_{\tau_\star} = X_\star\}$ , the event denoted $A_n^\star$ in Theorem 5. On $A_\star$ , $\tau_\star$ is the time of the first visit to $X_\star$ , and the time to extinction can be broken into three epochs. Define the time of the last visit to $X_\star$ as $\tau_\star^\mathrm{o} = \sup\{t\colon X_t=X_\star\}$ , setting $\tau_\star^\mathrm{o}=-\infty$ if X never reaches $X_\star$ . On $A_\star$ , the time to extinction $\tau=\inf\{t \colon X_t=0\}$ is the sum of the approach time $\tau_\star$ , the sojourn time $\tau_\star^\mathrm{o} - \tau_\star$ , and the fall time $\tau-\tau_\star^\mathrm{o}$ . We proceed as follows:

1. 1. Estimate the expected sojourn time $E_\star^\mathrm{o} \,:\!=\, \mathbb{E}[ \tau_\star^\mathrm{o}-\tau_\star \mid X_{\tau_\star}=X_\star ]$ .

2. 2. Show that the expected approach time and fall time are $o(E_\star^\mathrm{o})$ .

3. 3. With a coupling argument, show that the rescaled sojourn time $(\tau_\star^\mathrm{o}-\tau_\star)/E_\star^\mathrm{o}$ , conditioned on $X_{\tau_\star}=X_\star$ , converges in distribution to exponential with mean 1.

Let us make the formal statements that will be the goal of this section.

Lemma 7. (Expected sojourn time.) $\displaystyle E_\star^\mathrm{o} \sim \sqrt{\frac{2\pi}{n}}\frac{r}{\delta^2}\exp(n(\log r + 1/r-1))$ .

Lemma 8. (Approach time.) $\max_{j \in \{1,\dots,n\}}\mathbb{E}[\tau_\star \mid X_{\tau_\star}=X_\star, \, X_0=j] = o(E_\star^\mathrm{o})$ .

Lemma 9. (Fall time.) $\mathbb{E}[\tau-\tau_\star^\mathrm{o}] = o(E_\star^\mathrm{o})$ .

Lemma 10. (Exponential limit.) For each $t\ge 0$ , $\mathbb{P}((\tau_\star^\mathrm{o}-\tau_\star)/E_\star^\mathrm{o} > t) \to {\mathrm{e}}^{-t}$ .

Note that the expected sojourn time and fall time do not depend on $X_0$ ; we have emphasized the uniformity of the estimate with respect to $X_0$ only in Lemma 8. Before proving these results, we use them to prove the rest of Theorem 5.

Proof of metastability results of Theorem 5 The extinction time is the sum of the approach, sojourn, and fall times: $\tau = \tau_\star + (\tau_\star^\mathrm{o}-\tau_\star) + (\tau-\tau\star^\mathrm{o})$ . Combining Lemmas 7, 8, and 9, $\mathbb{E}[\tau \mid X_{\tau_\star}=X_\star] \sim E_\star^\mathrm{o}$ , which is the desired estimate of the expected time to extinction.

Using Lemmas 8 and 9 and Markov’s inequality, conditioned on $X_{\tau_\star}=X_\star$ , with probability $1-o(1)$ , $\tau/E_\star^\mathrm{o} = (\tau_\star^\mathrm{o}-\tau_\star)/E_\star^\mathrm{o} + o(1)$ . The exponential limit for $\tau$ then follows from Lemma 10. □

We begin by deriving some formulas for the expected time to hit $j+1$ or $j-1$ , starting from j.

8.3.1. Crossing times

Let $T_+(\,j) = \inf\{t \colon X_t = j + 1\}$ and $T_-(\,j) = \inf\{t \colon X_t=j-1\}$ , and let $S_+(\,j) = \mathbb{E}[T_+(\,j) \mid X_0 =j]$ and $S_-(\,j) = \mathbb{E}[T_-(\,j) \mid X_0=j]$ . Define also the conditioned versions, $S_\pm^\star(\,j) = \mathbb{E}[T_\pm(\,j) \mid X_0=j \ \text{and} \ X_{\tau_\star}=X_\star]$ and $S_\pm^0(\,j) = \mathbb{E}[T_\pm(\,j) \mid X_0=j \ \text{and} \ X_{\tau_\star}=0]$ . We will need to estimate the following quantities: (i) $S_+^\star(\,j)$ for $j<X_\star$ , (ii) $S_-(\,j)$ for $1 \le j \le n$ , and (iii) $S_-^0(\,j)$ for $j<X_\star$ .

Quantity (i): For $1<j<X_\star$ a first step analysis gives

\[S_+^\star(\,j) = \frac{1}{q_+^\star(\,j)+ q_-^\star(\,j)} + \frac{q_-^\star(\,j)}{q_+^\star(\,j) + q_-^\star(\,j)}(S_+^\star(\,j-1) + S_+^\star(\,j)),\]

and solving gives $q_+^\star(\,j)S_+^\star(\,j) = 1 + q_-^\star(\,j)S_+^\star(\,j-1)$ . Following [Reference Keilson15, Section 5.2] we let $\pi(i,i) = 1$ and $\pi(i,j) = \prod_{k=i}^{j-1} q_+^\star(k)/q_-^\star(k+1)$ , and multiply through by $\pi(1,j)$ to obtain $q_+^\star(\,j)\pi(1,j)S_+^\star(\,j) = \pi(1,j) + q_+^\star(\,j-1)\pi(1,j-1)S_+^\star(\,j-1)$ . Then, since $q_+^\star(1)S_1^+ = 1$ , we solve to obtain

\[S_+^\star(\,j) = \frac{1}{q_+^\star(\,j)\pi(1,j)}\sum_{i=1}^j \pi(1,i) = \frac{1}{q_+^\star(\,j)}\sum_{i=1}^j \frac{1}{\pi(i,j)}.\]

Using (25), for $i < j$

\begin{align*}\frac{1}{\pi(i,j)} &= \prod_{k=i}^{j-1} \frac{q_-(k+1)}{q_+(k)}\frac{h_+(k)/h_+(k+1)}{h_+(k+1)/h_+(k)} \\[3pt] &= \left(\frac{h_+(i)}{h_+(\,j)}\right)^2 \frac{q_-(\,j)}{q_+(i)} \ \nu(i,j-1).\end{align*}

This gives

(31) \begin{align}S_+^\star(\,j) = \frac{q_-(\,j)}{q_+^\star(\,j)}\sum_{i=1}^j \nu(i,j-1) \frac{h_+(i)^2}{h_+(\,j)^2} \frac{1}{q_+(i)}.\end{align}

Quantity (ii): For $1\le j< n$ , a first step analysis gives the recursion $q_-(\,j)S_-(\,j) = 1 + q_+(\,j)S_-(\,j+1)$ . Multiplying through by $1/\pi(\,j,n)$ gives

\[\frac{q_-(\,j)}{\pi(\,j,n)}S_-(\,j) = \frac{1}{\pi(\,j,n)} + \frac{q_-(\,j+1)}{\pi(\,j+1,n)}S_-(\,j+1);\]

since $S_-(n) = 1/q_-(n)$ , the solution is

\[S_-(\,j) = \frac{\pi(\,j,n)}{q_-(\,j)}\sum_{i=j}^n \frac{1}{\pi(i,n)} = \frac{1}{q_-(\,j)}\sum_{i=j}^n \pi(\,j,i).\]

Writing in terms of $\nu$ we obtain

(32) \begin{align}S_-(\,j) = \frac{q_+(\,j)}{q_-(\,j)}\sum_{i=j}^n \nu(i-1,j)\frac{1}{q_-(i)} .\end{align}

Quantity (iii): For $1 \le j < X_\star$ a first step analysis gives $q_-^0(\,j)S_-^0(\,j) = 1 + q_+^0(\,j)S_-^0(\,j+1)$ . In this case it is better to express the solution using the function $\nu^0$ , defined like $\nu$ except with $q_\pm^0$ in place of $q_\pm$ . Since $S_-^0(X_\star-1) = 1/q_-^0(X_\star-1)$ , following the same approach as above we find that

(33) \begin{align}S_-^0(\,j) = \frac{q_+^0(\,j)}{q_-^0(\,j)}\sum_{i=j}^{X_\star-1}\nu^0(i-1,j)\frac{1}{q_-^0(i)}.\end{align}

8.3.2. Sojourn time

Here we prove Lemma 7. We use the same basic approach as in [Reference Andersson and Djehiche1]; the calculations become more delicate when $\delta\to 0$ . Define recursively $\tau_\star(0)=\tau_\star$ and, for $0<k<K= \min\{ k >0 \colon X_{\tau_\star(k)}=0\}$ , $\tau_\star(k+1)=\inf\{t > \tau_\star(k) \colon X_t \in \{0,X_\star\}$ and $X_s \ne X_\star$ for some $\tau_\star(k)<s<t\}$ . For $k=1,\dots,K$ let $\rho_k = \tau_\star(k)-\tau_\star(k-1)$ . Then, conditioned on $X_{\tau_\star} = X_\star$ ,

(34) \begin{align}\tau_\star^0 - \tau_\star = \sum_{k=1}^{K-1} \rho_k.\end{align}

By the strong Markov property, K is geometric with success probability $p_\star =\mathbb{P}(X_{\tau_\star}= 0 \mid X_0=X_\star)$ and, defining $\tau_\star^+ = \inf\{t \colon X_t=X_\star$ and $X_s \ne X_\star$ for some $0 < s < t \}$ , $\mathbb{E}[\rho_k \mid K>k] = \mathbb{E}[\tau_\star^+ \mid X_0=X_{\tau_\star^+}=X_\star]$ . Let $L_\star$ denote this expectation. Applying Wald’s equation to (34), $E_\star^\mathrm{o} = L_\star(1/p_\star - 1)$ . Recall $V(x) = x(\log r - 1) - (1-x)\log(1-x)$ defined in Lemma 4, and $V_\star \,:\!=\, V(x_\star)=\log r + 1/r-1$ . Lemma 7 follows immediately from the following estimates on $p_\star$ and $L_\star$ .

Lemma 11. $p_\star \to 0$ and $\displaystyle p_\star \sim \frac{\delta}{2\sqrt{r}}\exp(-nV_\star)$ .

Lemma 12. $\displaystyle L_\star \sim \frac{1}{\sqrt{n}\,\delta}\,\sqrt{\frac{\pi r}{2}}$ .

First we prove Lemma 11.

Proof of Lemma 11. First we show that if $p_\star \sim (\delta/2\sqrt{r})\exp(-nV_\star)$ then $p_\star \to 0$ ; then we establish the estimate. Since $r\ge 1$ , $V_\star \ge 0$ . If $\delta \to 0$ it follows that $p_\star \to 0$ . If $\delta \to \delta_\infty>0$ then $V_\star \to V_\infty = \log r_\infty + 1/r_\infty -1>0$ , so $\exp(-nV_\star) \to 0$ and again $p_\star \to 0$ .

If the first jump of X is to $X_\star+1$ then $X_{\tau_\star}=X_\star$ . So, conditioning on the first jump,

\[p_\star = \frac{q_-(X_\star)}{q(X_\star)}h_-(X_\star -1).\]

By the definition of $X_\star$ , $q_+(X_\star)\sim q_-(X_\star)$ , so if $X_0=X_\star$ then its first jump is to $X_\star-1$ with probability $1/2+o(1)$ . Thus, in the notation of (17) it is enough to show that $h_-(X_\star-1) \sim \frac{\delta}{\sqrt{r}}\exp(-nV_\star)$ . We have

\[h_-(X_\star-1) = \frac{\nu(X_\star-1)}{\sum_{k=0}^{X_\star-1} \nu(k)},\]

and we begin by estimating the numerator using Lemma 4 with $a=0$ and $b = (X_\star-1)/n$ . Notice that $|b-x_\star| \le 2/n$ and that $x_\star = \delta/r = \delta/(1+\delta)$ , so the assumption $\limsup_n \delta<\infty$ implies $\limsup_n b < 1$ . Since $V'(x_\star)=0$ , $b<x_\star$ , and $x \mapsto |V''(x)|$ is increasing, it follows that $|V(b)-V(x_\star)| \le \frac{1}{2}V''(x_\star)(2/n)^2 = O(1/n^2)$ . Since $n b \ge n(x_\star-2/n) = n \delta/r-2 \to \infty$ and $\limsup_n 1-b >0$ , it follows that $n(1-b)^2(b-a) \to\infty$ . Since $1/r=1-x_\star$ , $1-b = 1-x_\star+O(1/n) = (1 - x_\star)(1+o(1)) \sim 1/r$ . Putting it together, $\nu(X_\star-1) \sim \sqrt{r}\exp(-nV(x_\star))$ . The denominator is estimated in the proof of Lemma 5; in both cases,

\[\sum_{k=0}^{X_\star-1} \nu(k) \sim \frac{1}{1-1/r} = r/\delta\]

and the result follows. □

In order to estimate $L_\star$ we will need additional information about the function $\nu$ . First, extend the domain of $\nu$ by defining $\nu(k,j)=1/\nu(\,j,k)$ for $0\le j< k<n$ . An equivalent, unifying definition is given by the formula

\[\nu(\,j,k) = \frac{\prod_{i=j+1}^nq_-(i)/q_+(i)}{\prod_{i=k+1}^nq_-(i)/q_+(i)}.\]

Say that $f(n,\lambda) \sim g(n,\lambda)$ uniformly over $\lambda \in A$ if $\lim_{n\to\infty} \sup_{\lambda \in A} \left|\log \left( \frac{f(n,\lambda)}{g(n,\lambda)} \right)\right| = 0$ .

Lemma 13. Uniformly over $|\sigma| \le n^{1/8}$ , $\nu(X_\star-\sigma \sqrt{n},X_\star) \sim \exp(-\sigma^2r/2)$ . Moreover, for $1 \le \sigma \le n^{1/8}$ ,

\[\sum_{0 \le j \le n-1 \colon |j-X_\star| \ge \sigma \sqrt{n}}\nu(\,j,X_\star) \le (2+o(1))\exp(-\sigma^2r/2)\frac{\sqrt{n}}{\sigma r}.\]

Proof. Since $r(1-x_\star)=1$ and $x\mapsto \log(r(1-x))$ is differentiable at $x_\star$ , it follows that, uniformly over j such that $|j-X_\star| \le n^{1/4}$ , $\log(r(1-j/n)) = O(n^{-3/4})$ . Summing at most $n^{1/4}$ terms, $\log \nu(\,j,X_\star) = O\big(\frac{1}{\sqrt{n}\,}\big)$ . This proves the first statement restricted to $|\sigma| \le n^{-1/4}$ .

If $n^{1/4} < |j-X_\star| \le n^{5/8}$ , then since $\limsup_n x_\star <1$ , $1-j/n \sim 1-X_\star/n$ . Since $|j-X_\star| > n^{1/4}\to \infty$ , by Lemma 4, uniformly over such j, $E_n(\,j/n,X_\star/n) \to 1$ if $j<X_\star$ and $E_n(X_\star/n,j/n) \to 1$ if $j \ge X_\star$ , so

(35) \begin{equation}\nu(\,j,X_\star) \sim \exp(-n(V(X_\star/n)-V(\,j/n));\end{equation}

note that this expression is valid not only for $j<X_\star$ but also for $j\ge X_\star$ under the extended definition of $\nu$ . Since $\limsup_n 1-x_\star>0$ , $V'''(x)=1/(1-x)^2$ is bounded on $[0,x_\star +o(1))]$ , and recall that $V'(x_\star)=0$ and $V''(x_\star) = -r$ . Thus, if $|\sigma| \le n^{1/8}$ , using a second-order Taylor approximation we find that

(36) \begin{equation}V(x_\star + \sigma /\sqrt{n}) - V(x_\star) = -\sigma^2 r/2n + O(n^{-9/8}).\end{equation}

In particular, $V(X_\star/n)-V(x_\star) = O(n^{-9/8})$ . Combining this with (35) and (36), the first statement is proved for the remaining values of $|\sigma|$ , namely, $(n^{-1/4},n^{1/8}]$ .

Next, for $j<k<X_\star$ , since $\log(r(1-x_\star))=0$ and $x\mapsto \log(r(1-x))$ is decreasing,

\[-\log \nu(\,j,k) = \sum_{i=j+1}^k \log(r(1-i/n)) \ge (k-j)\log(r(1-k/n))\]

and thus $\nu(\,j,k) \le (r(1-k/n))^{-(k-j)}$ . Fix $\sigma$ with $1 \le \sigma \le n^{1/8}$ and observe that, for any $j,k,\ell$ , $\nu(\,j,\ell)=\nu(\,j,k)\nu(k,\ell)$ . Using this property with $k=X_\star-\sqrt{n}\sigma$ and $\ell=X_\star$ and then bounding the sum by a geometric series,

\[\sum_{j=0}^{X_\star-\sqrt{n}\sigma}\nu(\,j,X_\star) \le \frac{(1+o(1))\exp(-\sigma^2r/2)}{1-(r(1-(X_\star-\sqrt{n}\sigma)/n))^{-1}}.\]

Since $r(1-X_\star/n) = r(1-x_\star)+O(1/n) = 1 + O(1/n)$ , the denominator is

\[1-(1 + O(1/n) + \sigma r / \sqrt{n})^{-1} = 1-(1-(1+o(1))\sigma r/\sqrt{n}) = (1+o(1))n^{-1/2}\sigma r,\]

so

\[\sum_{j=0}^{X_\star-\sqrt{n}\sigma-1}\nu(\,j,X_\star) \le (1+o(1))\exp(-\sigma^2r/2)\frac{\sqrt{n}}{\sigma r}.\]

From the other end, if $X_\star<k<j$ then since $\log(r(1-i/n)) \le \log(r(1-x_\star))=0$ if $i>X_\star$ , we have

\[\log \nu(\,j,k) = -\log \nu(k,j) = \sum_{i=k+1}^j \log(r(1-i/n)) \le (\,j-k)\log(r(1-k/n))\]

and thus $\nu(\,j,k) \le (r(1-k/n))^{j-k}$ . By an analogous argument we find that

\[\sum_{j=X_\star+\sqrt{n}\sigma}^{n-1}\nu(\,j,X_\star) \le \frac{(1+o(1))\exp(-\sigma^2r/2)}{1-r(1-(X_\star+\sqrt{n}\sigma)/n)} \le(1+o(1))\exp(-\sigma^2r/2)\frac{\sqrt{n}}{\sigma r}.\]

Combining the two estimates completes the proof. □

We are now ready to prove Lemma 12.

Proof of Lemma 12. Let $S_+^\star(\,j)$ , $S_-(\,j)$ be as in Section 8.3.1. Conditioning on the first step,

\[L_\star = \frac{1}{q_+(X_\star) + q_-(X_\star)}\left( 1 + q_-(X_\star) S_+^\star(X_\star-1) + q_+(X_\star)S_-(X_\star+1) \right).\]

We have $q_-(X_\star)=X_\star$ and, since $r(1-X_\star/n) = 1 + O(1/n)$ , $q_+(X_\star) \sim X_\star$ , so $L_\star \sim \frac{1}{2}\left(1/X_\star + S_+^\star(X_\star-1) + S_-(X_\star+1) \right)$ . Since $h_+(X_\star)=1$ and $h_+(X_\star-1) \to 1$ , $q_+^\star(X_\star -1) \sim q_+(X_\star-1)$ . By the definition of $X_\star$ , $q_+(X_\star-1) \sim q_-(X_\star-1)$ . In addition, by Lemma 13, $\nu(i,X_\star)=\nu(i,X_\star-2)\nu(X_\star-2,X_\star) \sim \nu(i,X_\star-2)$ uniformly over i. Thus, using (31) with $j=X_\star-1$ ,

(37) \begin{align}S_+^\star(X_\star-1) \sim \sum_{i=1}^{X_\star -1} \nu(i,X_\star)\frac{h_+(i)^2}{q_+(i)}.\end{align}

We will estimate the bulk of the sum in (37), finding that it tends to a, then show that the rest of the sum, as well as $S_-(X_\star+1)$ , are negligible in comparison. Recall the notation $c=\sqrt{n}\delta$ , noting that $c\to\infty$ by assumption. We then have $X_\star \sim n\delta = \sqrt{n}c$ . Since $r(1-x_\star)=1$ , $q_+(i) = ri(1-i/n) = i(1+r(x_\star-i/n))$ , so $q_+(i) \sim X_\star$ uniformly over $|i-X_\star| \le C$ provided $C=o(X_\star)$ , or equivalently $C=o(\sqrt{n}c)$ . By Lemma 5, $h_+(i) \to 1$ uniformly over $i\ge M$ provided $\delta M \to \infty$ . If $M = X_\star - C$ with $C=o(\sqrt{n}c)$ then $\delta M \sim \delta X_\star \sim c^2/r \to \infty$ . Thus, if we define $\Sigma = \sqrt{c} \wedge n^{1/8}$ , then since $\sqrt{nc} = o(\sqrt{n}c)$ ,

\[\sum_{i= \lfloor X_\star - \Sigma\sqrt{n} \rfloor +1}^{X_\star-1} \nu(i,X_\star)\frac{h_+(i)^2}{q_+(i)} \ \sim \ \frac{1}{X_\star} \sum_{i= \lfloor X_\star - \Sigma\sqrt{n} \rfloor +1 }^{X_\star-1} \nu(i,X_\star).\]

Using Lemma 13 and the fact that $\Sigma \to\infty$ and $\Sigma \le n^{1/8}$ ,

\[\sum_{i= \lfloor X_\star - \Sigma\sqrt{n} \rfloor +1}^{X_\star-1} \nu(i,X_\star) \sim \sqrt{n}\int_{-\infty}^0 {\mathrm{e}}^{-\sigma^2r/2} \, {\mathrm{d}}\sigma = \sqrt{\frac{n\pi}{2r}}.\]

Assuming the rest of the sum is negligible in comparison, since $X_\star \sim n\delta/r$ we then have

(38) \begin{align}S_+^\star(X_\star-1) \sim \frac{\sqrt{n}}{X_\star}\frac{\pi}{2r} = \frac{1}{\sqrt{n}\delta}\sqrt{\frac{\pi r}{2}}.\end{align}

So, we now show the rest of the sum in the brackets in (37) is $o(\sqrt{n})$ ; we may of course ignore the 1. Using the trivial estimate $q_+(i) \ge 1$ and $h_+(i) \le 1$ for $i\ge 1$ , as well as $X_\star \le n\delta$ and Lemma 13 with $\sigma=n^{1/8}$ , if $\sqrt{c} \ge n^{1/8}$ then the rest of the sum is

\[\sum_{i=1}^{\lfloor X_\star - n^{5/8} \rfloor}\nu(i,X_\star-2) \frac{h_+(i)^2}{q_+(i)}X_\star \le (2+o(1)){\mathrm{e}}^{-n^{1/4}r/2}\frac{n^{3/8}}{r} \, \frac{n\delta}{r} =O(1)= o(\sqrt{n}).\]

If $\sqrt{c} < n^{1/8}$ we treat the remainder in two parts, beginning with $i \le 1/\delta = \sqrt{n}/c$ . X is dominated by the linear birth and death process Z with $Z_0=X_0$ and parameter r, and using (5), since 0 is absorbing for Z, $\mathbb{P}(Z_t=0$ for some $t>0 \mid Z_0) = \lim_{t\to\infty}\rho(t)^{Z_0} = \min(1,1/r)^{Z_0} = r^{-Z_0}$ since $r>1$ by assumption in this section. If $Z_t = i$ for $i>0$ , then with probability $p_i>0$ , Z hits 0 before it returns to i. Thus, Z visits any $i>0$ at most a geometric $(p_i)$ number of times, which is almost surely finite and implies that a.s., $\lim_{t\to\infty}Z_t \in \{0,\infty\}$ . Therefore, $\mathbb{P}(\lim_{t\to\infty}Z_t =\infty \mid Z_0) = 1 - r^{-Z_0} = 1-(1+\delta)^{-Z_0} \le \delta Z_0$ . The last inequality follows from $(1+\delta)^{-Z_0} \ge e^{-\delta Z_0} \ge 1-\delta Z_0$ , where we used the estimate $1+u \le {\mathrm{e}}^u$ for $u \in \mathbb{R}$ which follows from the convexity of $u\mapsto {\mathrm{e}}^u$ . Since Z dominates X,

(39) \begin{align}h_+(i) \le \mathbb{P}(\lim_{t\to\infty}Z_t =\infty \mid Z_0=i) \le \delta i.\end{align}

Since $h_+(i) \le 1$ , $h_+(i)^2 \le h_+(i)$ , and since $X_\star \le n\delta$ and $q_+(i) \ge i$ for $i<X_\star$ , $h_+(i)^2 X_\star /q_+(i) \le \delta^2n = c^2$ . Using Lemma 13 with $\sigma = (X_\star - \lfloor \sqrt{n}/c \rfloor)/\sqrt{n} \sim c-1/c = c-o(1)$ ,

\[\sum_{i=1}^{\lfloor \sqrt{n}/c \rfloor} \nu(i,X_\star)\frac{h_+(i)^2}{q_+(i)}X_\star \le (2+o(1)){\mathrm{e}}^{-\sigma^2 r/2}\frac{\sqrt{n}}{\sigma r} c^2 \sim 2{\mathrm{e}}^{-(c-o(1))^2r/2} \frac{c}{r} \sqrt{n} = o(\sqrt{n}),\]

since $c\to\infty$ and thus $(2c/r){\mathrm{e}}^{-(c-o(1))^2r/2} \to 0$ as $n\to\infty$ . Using again the trivial estimate $h_+(i)\le 1$ and $q_+(i) \ge i$ , and also $\nu(i,X_\star) \le \nu(X_\star-\sqrt{nc},X_\star) \sim {\mathrm{e}}^{-c r/2}$ for $i \le X_\star-\sqrt{nc}$ ,

\begin{align*}\sum_{i=\lfloor \sqrt{n}/c \rfloor +1}^{\lfloor X_\star-\sqrt{nc} \rfloor}\nu(i,X_\star)\frac{h_+(i)^2}{q_+(i)}X_\star& \le (1+o(1)){\mathrm{e}}^{-cr/2}\sqrt{n}c\sum_{i=\lfloor \sqrt{n}/c \rfloor+1}^{\lfloor X_\star-\sqrt{nc} \rfloor}\frac{1}{i}, \\[3pt] & \sim \sqrt{n}c{\mathrm{e}}^{-cr/2}(\log(\sqrt{n}c)-\log(\sqrt{n}/c)) \\[3pt] & \sim \sqrt{n}(2c\log(c){\mathrm{e}}^{-cr/2}) = o(\sqrt{n}).\end{align*}

This completes the estimation of the sum from (37).

For $S_-(X_\star+1)$ , using (32) with $j=X_\star+1$ and simplifying as before,

\[S_-(X_\star+1) \sim \sum_{i=X_\star+1}^n \nu(i-1,X_\star)\frac{1}{q_-(i)}.\]

Since this case is similar to the one before, we just give an outline. Breaking up the sum in the same way, the bulk of the sum is estimated in the same way as before and gives the same result. To bound the remainder, it suffices to note that $q_-(i)=i \ge X_\star$ for $i \ge X_\star$ , then use Lemma 13 directly, noting that $2{\mathrm{e}}^{-\Sigma^2 r/2}/(\Sigma r) \to 0$ . Since $L_\star$ scales like the average of the two values, the result follows from (38). □

8.3.3. Approach time

We now prove Lemma 8. For this and for Lemma 8, we first derive a more concrete lower bound on the expected sojourn time $E_\star^\mathrm{o}$ , using the formula $E_\star^\mathrm{o} \sim \sqrt{2\pi/n}(r/\delta^2)\exp(n V_\star)$ of Lemma 7, where $V_\star=\log r + 1/r-1$ .

If $\delta \to 0$ then $V_\star = \log(1+\delta)+1/(1+\delta)-1 = \delta-\delta^2/2 + 1-\delta + \delta^2-1+O(\delta^3) = \delta^2/2 + O(\delta^3) \sim \delta^2/2$ and, since $c=\sqrt{n}\delta$ and $r\to 1$ , $\sqrt{2\pi/n}(r/\delta^2) = \sqrt{2\pi}(r/c \delta)$ and $nV_\star \sim c^2/2$ . If $\delta\to\delta_\infty>0$ then $V_\star \to V_\infty = \log r_\infty + 1/r_\infty -1>0$ . Thus in either case,

(40) \begin{align}E_\star^\mathrm{o} \ge \begin{cases} \displaystyle(1-o(1))\frac{\sqrt{2\pi}}{\delta}\,\frac{1}{c} \exp((1-o(1))c^2/2) & \text{if} \ \delta \to 0, \\ \\[-7pt] \sqrt{2\pi/n}(r_\infty/\delta_\infty^2){\mathrm{e}}^{(1-o(1))V_\infty n} & \text{if} \ \delta \to \delta_\infty>0.\end{cases}\end{align}

Using (40), the following result implies Lemma 8, since $c^2\log c = o((1/c)\exp((1-o(1))c^2/2))$ in the case $\delta \to 0$ and $n\log n = o(\exp((1-o(1))V_\infty n))$ in the case $\delta \to \delta_\infty>0$ .

Lemma 14.

\[\max_{j \in \{1,\dots,n\}}\mathbb{E}[\tau_\star \mid X_{\tau_\star}=X_\star, \ X_0=j] =\begin{cases} O((1/\delta)c^2\log(c)) & \text{if} \ \delta \to 0 \\O(n\log(n)) & \text{if} \ \delta \to \delta_\infty>0.\end{cases}\]

Proof. Using the natural coupling, by Lemma 1 it is enough to consider the initial values $X_0=1$ and $X_0=n$ ; we begin with $X_0=n$ . We break up the travel time to $X_\star$ into three checkpoints: $2nx_\star$ , $nx_\star+\sqrt{n}$ , and $X_\star$ .

First checkpoint: If $x>x_\star$ then $\mu(x-x_\star) = \mu(x) = x(r(1-x)-1) = rx(x_\star-x) \le -r(x-x_\star)^2 \le -(x-x_\star)^2$ . The differential equation $y'=-y^2$ has solution flow $\phi(t,y) = 1/(1/y+t)$ , so letting $\tau_1=\inf\{t\colon x_t \le 2x_\star\}$ and defining the continued process $\tilde x$ by $\tilde x_t = \phi(t - t\wedge \tau_1,x_{t \wedge \tau_1})$ we have $\mu_t(\tilde x) \le -\tilde x_t^2$ for all $t\ge 0$ . Taking expectations and using Jensen’s inequality,

\[\frac{{\mathrm{d}}}{{\mathrm{d}} t}\mathbb{E}[\tilde x_t] \le - \mathbb{E}[\tilde x_t^2] \le -(\mathbb{E}[ \tilde x_t])^2,\]

which gives $\mathbb{E}[\tilde x_t] \le \phi(t,x_0) = 1/(1/x_0+t)$ . Since $x_0\le 1$ , Markov’s inequality then gives

\[\mathbb{P}(\tau_1>t) \le \mathbb{P}(\tilde x_t > 2x_\star) \le \frac{1}{2x_\star}\frac{1}{1+t} = \frac{r}{2\delta(1+t)}.\]

Letting $t=1/\delta$ , the above is at most $1/2$ . Using the Markov property and iterating, $\mathbb{P}(\tau_1 > k/\delta) \le 2^{-k}$ , so $\mathbb{E}[\tau_1] \le (1/\delta) \sum_{k \ge 0}\mathbb{P}(\tau_1 > k/\delta) \le 2/\delta$ .

Second checkpoint: To get from $2x_\star$ to $x_\star+\frac{1}{\sqrt{n}\,}$ , let $\tau_2=\inf\big\{t \colon x_t \le x_\star+\frac{1}{\sqrt{n}\,}\big\}$ and note that if $x>x_\star$ then $\mu(x-x_\star) = -rx(x-x_\star) \le -rx_\star(x-x_\star) = -\delta(x-x_\star)$ . Thus, $\xi_t = {\mathrm{e}}^{\delta (t\wedge \tau_2)}(x_{t \wedge \tau_2}-x_\star)$ is a supermartingale and if $x_0 \le 2x_\star$ then $\xi_0 \le x_\star$ and $\mathbb{P}(\tau_2 > t \mid x_0 \le 2x_\star) \le \mathbb{P}(\xi_t \ge {\mathrm{e}}^{\delta t}/\sqrt{n}) \le {\mathrm{e}}^{-\delta}\sqrt{n}x_\star \le {\mathrm{e}}^{-\delta t}\sqrt{n}\delta = {\mathrm{e}}^{-\delta t}c$ . Thus, $\mathbb{E}[\tau_2 \mid x_0 \le 2x_\star] = \int_0^{\infty}\mathbb{P}(\tau_2>t\mid x_0\le 2x_\star) \, {\mathrm{d}} t \le c/\delta$ .

Third checkpoint: Finally we estimate $\tau_3=\inf\{t \colon X_t=X_\star\}$ , assuming $X_0 \le nx_\star+\sqrt{n}$ . With $S_-(\,j)$ as in (32),

(41) \begin{align}\mathbb{E}[\tau_3 \mid X_0 \le nx_\star + \sqrt{n} ] \le \sum_{j=X_\star+1}^{X_\star+\sqrt{n}+1}S_-(\,j).\end{align}

Since $h_+(i)=h_+(\,j)=1$ , $q_+(\,j) \le q_-(\,j)=q_-^\star(\,j)$ , and $q_-(i)=i \ge X_\star$ for $i,j>X_\star$ ,

\[S_-(\,j) \le \frac{1}{X_\star}\sum_{i=j}^n \nu(i-1,j).\]

If $X_\star<j \le X_\star+\sqrt{n}+1$ then, using Lemma 13 with $\sigma \le 1 + 1/\sqrt{n}$ ,

\[\nu(i-1,j) = \frac{\nu(i-1,X_\star)}{\nu(\,j,X_\star)} \le (1+o(1)){\mathrm{e}}^{r/2}\nu(i-1,X_\star).\]

Using Lemma 13 again and approximating the sum by a Gaussian integral, we obtain

\[S_-(\,j) \le \frac{1}{X_\star}(1+o(1))e^{r/2}\sqrt{\frac{n\pi}{2r}}=(1+o(1))\sqrt{\frac{\pi}{2r}}{\mathrm{e}}^{r/2}\frac{\sqrt{n}}{n\delta} = O(1/\sqrt{n}\delta),\]

since r is bounded by assumption. Summing over $\sqrt{n}+1$ terms of the same size and using (41), we find $\mathbb{E}[\tau_3 \mid X_0 \le nx_\star + \sqrt{n} ] = O(\sqrt{n}/\sqrt{n}\delta) = O(1/\delta)$ . In all three cases the expected travel time is $O(c/\delta)$ , which satisfies the stated estimates; for the case $\delta \to \delta_\infty>0$ , note that $c/\delta = \sqrt{n}$ .

Next, we consider the case $X_0=1$ . From (31), and since $q_-(\,j) \le q_+(\,j) \le q_+^\star(\,j)$ and $q_+(i) \ge i$ for $i,j<X_\star$ ,

\[S_+^\star(\,j) \le \sum_{i=1}^j \nu(i,j-1) \frac{h_+(i)^2}{h_+(\,j)^2}\,\frac{1}{i}.\]

Let $s_{ij}$ denote the above summands. Then,

\[\mathbb{E}[\tau_\star \mid X_0=1, \, X_{\tau_\star}=X_\star] = \sum_{j=1}^{X_\star-1}S_+^\star(\,j) \le \sum_{j=1}^{X_\star-1}\sum_{i=1}^j s_{ij}.\]

If $i\le j <X_\star$ then $\nu(i,j-1) \le r$ and $h_+(i)/h_+(\,j) \le 1$ , which we use below. In order to obtain good enough estimates, we need to be a bit more precise. We treat the cases $\delta \to 0$ and $\delta \to \delta_\infty>0$ separately.

Case 1: $\delta \to 0$ . Let $c=\sqrt{n}\delta$ , so $c\to\infty$ and $c=o(\sqrt{n})$ . We treat the sum in three parts: (i) $1 \le i \le j \le 1/\delta$ , (ii) $1 \le i \le 1/\delta < j \le X_\star-1$ , and (iii) $1/\delta < i \le j < X_\star$ .

Part (i): From (39), $h_+(i) \le \delta i$ , and since the denominator $\sim 1/(1-1/r)$ ,

\[h_+(\,j) = \frac{\sum_{k=0}^{X_0-1}\nu(k)}{\sum_{k=0}^{X_\star-1}\nu(k)} \ge \frac{(1-r^{-j})/(1-1/r)}{(1+o(1)/(1-1/r)} = (1-o(1))(1-r^{-j}).\]

Since $\delta\to 0$ , $(1+\delta)^{-j} = ((1+\delta)^{1/\delta})^{-\delta j} \to {\mathrm{e}}^{-\delta j}$ uniformly over $\delta j \le 1$ . Since ${\mathrm{e}}^{-x} \le 1-(1-1/{\mathrm{e}})x$ for $x \in [0,1]$ , if $\delta j \le 1$ then

(42) \begin{align}h_+(\,j) \ge (1-o(1))(1-{\mathrm{e}}^{-\delta j}) \ge (1-o(1))(1-1/{\mathrm{e}})\delta j ,\end{align}

which is at least $\delta j/2$ for large n, since $1-1/{\mathrm{e}} > 1/2$ . Thus, $s_{ij} \le r(i^2/(\,j/2)^2)(1/i) = 4ri/j^2$ , so

\[\sum_{j=1}^{\lfloor 1/\delta \rfloor}\sum_{i=1}^j s_{ij} \le \sum_{j=1}^{\lfloor 1/\delta \rfloor}\frac{4r}{j^2}\sum_{i=1}^j i \le \sum_{j=1}^{\lfloor 1/\delta \rfloor} 2r \le 2r/\delta.\]

Part (ii): Since $q_-(k)/q_+(k)\ge 1/r$ , $\nu(0,i) \ge r^{-i}=(1+\delta)^{-i} \ge {\mathrm{e}}^{-\delta i}$ for each i. Since $q_-(k)\le q_+(k)$ for $k<X_\star$ and $i\le \lfloor 1/\delta \rfloor$ , $\nu(i,j-1) \le \nu(\lfloor 1/\delta \rfloor,j-1)$ , so

\[\nu(i,j-1) = \frac{\nu(0,j-1)}{\nu(0,i)} \le {\mathrm{e}}^{\delta i}\nu(0,j-1).\]

Thus, if $i \le 1/\delta < j < X_\star$ then

\[\nu(i,j-1) = \frac{\nu(0,j-1)}{\nu(0,i)} \le {\mathrm{e}}^1\nu(0,j-1).\]

Since $j>\lfloor 1/\delta \rfloor$ and $j\mapsto h_+(\,j)$ is non-decreasing, using (42), $h_+(\,j) \ge (1-o(1))(1-1/{\mathrm{e}})\delta \lfloor 1/\delta \rfloor$ is at least $1/2$ for large n, since $\delta \to 0$ implies $\delta\lfloor 1/\delta \rfloor \to 1$ . Again using $h_+(i) \le \delta i$ and combining,

\[\sum_{j=\lfloor 1/\delta \rfloor+1}^{X_\star-1}\sum_{i=1}^{\lfloor 1/\delta \rfloor} s_{ij} \le \sum_{j=\lfloor 1/\delta \rfloor+1}^{X_\star-1} {\mathrm{e}}^1 \nu(0,j-1) \sum_{i=1}^{\lfloor 1/\delta \rfloor} \frac{(\delta i)^2}{1/4}\frac{1}{i}.\]

We easily estimate

\[\sum_{i=1}^{\lfloor 1/\delta \rfloor} \frac{(\delta i)^2}{1/4}\frac{1}{i} \le 4\delta^2\sum_{i=1}^{\lfloor 1/\delta \rfloor}i \le 4\delta^2\frac{(1/\delta)^2}{2} = 2.\]

Using (22),

\[\sum_{j=\lfloor 1/\delta \rfloor +1}^{X_\star -1}{\mathrm{e}}^1\nu(0,j-1) \le (1+o(1))/\delta.\]

Combining the two, the sum is at most $(2+o(1))/\delta$ .

Part (iii): This part is the easiest; we simply use $\nu(i,j-1) \le r$ , $h_+(i)/h_+(\,j) \le 1$ , and $1/q_+(i) \le 1/i$ and, treating the sum as a right-endpoint Riemann sum,

\[\sum_{j=\lfloor 1/\delta \rfloor +1}^{X_\star-1}\sum_{i=\lfloor 1/\delta \rfloor +1}^j \frac{1}{i} \le \sum_{j=\lfloor 1/\delta \rfloor +1}^{X_\star-1}r(\log(\,j)-\log(1/\delta)).\]

We can combine the logs as $\log(\delta j)$ , which is increasing in j. Treating the sum as a left-endpoint Riemann sum of the function $\log(x)$ with interval widths $\delta$ , and noting that $\delta X_\star \le n\delta^2 = c^2$ , the sum is at most $\frac{1}{\delta}(\delta X_\star(\log(\delta X_\star) - 1) - \delta(1/\delta)(\log(\delta(1/\delta))-1) \le \frac{1}{\delta}(c^2(\log(c^2)-1) + 1)$ . Combining all three parts, we find $\mathbb{E}[\tau_\star \mid X_0=1, \ X_{\tau_\star}=X_\star] \le \frac{1}{\delta}(c^2\log(c^2) - c^2 + O(1)) \le \frac{1}{\delta} 2c^2\log(c)$ for large n, since $c\to\infty$ .

Case 2: $\delta \to \delta_\infty >0$ . Since $1/\delta=O(1)$ in this case, the whole sum can be treated as in part (iii) above. Since $s_{ij} \le r/i$ ,

\[\sum_{j=1}^{X_\star-1}\sum_{i=1}^j s_{ij} \le \sum_{j=1}^{X_\star-1}(1+\log j).\]

Treating the sum as a left-endpoint Riemann sum, it is at most $X_\star \log(X_\star) - 1\log(1) \le n\delta \log(n \delta) = O(n\log(n))$ . □

8.3.4. Fall time

Here we show that the time to hit zero after the last visit to $X_\star$ is small compared to the sojourn time. The following is an equivalent formulation of Lemma 9.

Lemma 15. $\mathbb{E}[\tau \mid X_0=X_\star, X_{\tau_\star}=0] = o(E_\star^\mathrm{o})$ .

Proof. Let $L_\star^0$ denote the above expectation. With $S_-^0(\,j)$ as in (33),

\[L_\star^0 = \sum_{j=1}^{X_\star}S_-^0(\,j).\]

Since we condition on $X_{\tau_\star}=0$ , the initial jump off $X_\star$ is to $X_\star-1$ with rate $q_+(X_\star)+q_-(X_\star)$ , which we denote $q_-^0(X_\star)$ , after which we use the rates $q_\pm^0$ given by (25). Thus, $S_{X_\star} = 1/q_-^0(X_\star)$ . For $j\le X_\star-1$ , $q_+^0(\,j) \le q_+(\,j)$ and $q_-^0(\,j) \ge q_-(\,j)$ , so $q_+^0(\,j)/q_-^0(\,j) \le q_+(\,j)/q_-(\,j) \le r$ . Moreover, $q_-^0(i) \ge q_-(i)=i$ , so

\[S_-^0(\,j) \le r\sum_{i=j}^{X_\star-1}\nu^0(i-1,j)\, \frac{1}{i}.\]

Let $s_{ij}$ denote the summands. Summing over j and exchanging the order of summation,

\[L_\star^0 = \frac{1}{q_-^0(X_\star)} + r\sum_{i=1}^{X_\star-1}\sum_{j=1}^i s_{ij}.\]

The first term is at most $1/X_\star$ , which is clearly $o(E_\star^\mathrm{o})$ . To estimate the sum we need more information about $\nu^0$ , so we first estimate the ratios

(43) \begin{align}\frac{q_+^0(\,j)}{q_-^0(\,j)} = \frac{q_+(\,j)}{q_-(\,j)} \, \frac{h_-(\,j+1)/h_-(\,j)}{h_-(\,j-1)/h_-(\,j)}\end{align}

of the conditioned rates given by (25). To do so we use the formulas (26) and (27). Since $\nu(\,j-1,k) \le \nu(\,j,k)$ for $j < X_\star$ , from (27),

(44) \begin{align}\frac{h_-(\,j-1)}{h_-(\,j)} \ge 1 + \frac{1}{\sum_{k=j}^{X_\star-1}\nu(\,j,k)} ,\end{align}

which simplifies some calculations. Define $\sigma_j=(X_\star-j)/\sqrt{n}$ and similarly for $\sigma_k$ , and let $\Sigma=n^{1/8}\wedge \sqrt{c}$ .

Estimation for $\sigma_j \le \Sigma$ : By Lemma 13, uniformly over $0 \le \sigma_k \le \sigma_j \le n^{1/8}$ , $\nu(\,j,k) = \nu(\,j,X_\star)/\nu(k,X_\star) \sim {\mathrm{e}}^{(-\sigma_j^2 + \sigma_k^2)r/2}$ , and so

\[\sum_{k=j}^{X_\star-1}\nu(\,j,k) \sim \sqrt{n}\int_0^{\sigma_j} {\mathrm{e}}^{-(\sigma_j-\sigma_k)(\sigma_j+\sigma_k)r/2} \, {\mathrm{d}}\sigma_k.\]

Changing variables to $u=\sigma_j-\sigma_k$ , $\sigma_j+\sigma_k = 2\sigma_j-u$ and the integral becomes

\[\int_0^{\sigma_j} {\mathrm{e}}^{-u(2\sigma_j-u)r/2} \, {\mathrm{d}} u \le \int_0^{\sigma_j} {\mathrm{e}}^{-u\sigma_j r/2} \, {\mathrm{d}} u \le \frac{2}{\sigma_j r}.\]

Letting $b_j = (1-\epsilon_n)\sigma_j r/\sqrt{n}$ with $\epsilon_n\to 0$ sufficiently slowly, and using (26) and (44), it follows that

\begin{equation*}\frac{h_-(\,j-1)}{h_-(\,j)} \ge 1 + b_j/2 , \qquad \frac{h_-(\,j+1)}{h_-(\,j)} \le 1 - b_j/2.\end{equation*}

Since $\sigma_j \le n^{1/8}$ , $b_j=o(1)$ so $h_-(\,j+1)/h_-(\,j-1) \le 1-b_j+o(b_j)$ . On the other hand,

(45) \begin{align}\frac{q_+(\,j)}{q_-(\,j)} &= r(1-j/n) = (1 + r(x_\star-j/n)) \nonumber \\&= 1 + r\sigma_j/\sqrt{n} + O(1/n) = 1 + (1+o(1))b_j+O(1/n).\end{align}

Using (43) and the above estimates,

(46) \begin{equation}\frac{q_+^0(\,j)}{q_-^0(\,j)} \le (1+(1+o(1))b_j+ O(1/n))(1-b_j+O(b_j^2)) = 1-b_j^2 + o(b_j^2) + O(1/n).\end{equation}

Estimation for $\sigma_j \ge \Sigma$ : Recall the upper bound from Lemma 4,

\[\nu(\,j,k) \le \exp(-n(V((k+1)/n)-V((\,j+1)/n)).\]

Using the fact that V is non-decreasing and V’ is non-increasing on $[0,x_\star]$ , $n(V((k+1)/n)-V((\,j+1)/n)) \ge ((k - j ) \wedge \sqrt{n})V'((\,j+1+\sqrt{n})/n)$ . With this bound,

(47) \begin{equation}\sum_{k=j}^{X_\star-1}\nu(\,j,k) \le \frac{1}{1- {\mathrm{e}}^{-V'((\,j+1+\sqrt{n})/n)}} + (X_\star-j){\mathrm{e}}^{-\sqrt{n}V'(\,j+1+\sqrt{n}/n)},\end{equation}

the first at most $\sqrt{n}$ terms forming a partial geometric series, and the last at most $X_\star-j$ terms each contributing at most a constant. Since $V'(x)=\log(r(1-x))$ and $r(1-x) = 1 + r(x_\star-x)$ ,

\begin{align*}{\mathrm{e}}^{V'((\,j+1+\sqrt{n})/n)} &= 1 + r(x_\star-(\,j+1+\sqrt{n})/n) \\&= 1 + r\big((X_\star-j)/n - \frac{1}{\sqrt{n}\,} + O(1/n)) \\&\ge 1 + r(\sigma_j-2)/\sqrt{n}\end{align*}

for large n. This easily gives the bound $\sqrt{n}/(r(\sigma_j-2))$ on the first term on the right-hand side of (47). To bound the second term, note that $\sigma_j \le \sqrt{n}$ and $r\ge 1$ , and that $1+x\ge {\mathrm{e}}^{x/2}$ for $x\in[0,1]$ , so $1+r(\sigma_j-2)/\sqrt{n} \ge {\mathrm{e}}^{(\sigma_j-2)/2\sqrt{n}}$ . Using this on the second term on the right-hand side of (47) and combining the two estimates, for large n,

\[\sum_{k=j}^{X_\star-1}\nu(\,j,k) \le \frac{\sqrt{n}}{r(\sigma_j-2)} + \sqrt{n}\sigma_j{\mathrm{e}}^{-(\sigma_j-2)/2}.\]

Since $x{\mathrm{e}}^{-x/2} \to 0$ faster than $\frac{1}{x}$ as $x\to\infty$ , using $b_j=(1-\epsilon_n)r\sigma_j/\sqrt{n}$ with $\epsilon_n\to 0$ slowly enough, since $\Sigma\to\infty$ it follows that, uniformly over $\sigma_j \ge \Sigma$ , $\sum_{k=j}^{X_\star-1}\nu(\,j,k) \le 1/b_j$ , and so

\[\frac{h_-(\,j+1)}{h_-(\,j-1)} \le \frac{1-b_j}{1+b_j}.\]

Since $\sigma_j \ge \Sigma\to\infty$ , $b_j=\omega\big(\frac{1}{n}\big)$ , and using (45), $q_+(\,j)/q_-(\,j) = 1 + (1+o(1))b_j$ . Using (43), uniformly over $j \le X_\star-\Sigma\sqrt{n}$ ,

(48) \begin{align}\frac{q_+^0(\,j)}{q_-^0(\,j)} \le (1+(1+o(1))b_j)\frac{1-b_j}{1+b_j} = 1-b_j + o(b_j).\end{align}

Case 1: $\delta \to 0$ . Since $\nu^0(i,i-1) \to 1$ uniformly over i, we can work with $\nu^0(i,j)$ instead of $\nu^0(i-1,j)$ . Again, we break the sum into parts; the decomposition is similar to the one in the second half of the proof of Lemma 14, except that the third part has been further subdivided into three parts, for a total of five: (i) $1 \le j \le i \le 1/\delta$ , (ii) $1 \le j \le 1/\delta < i \le X_\star-\Sigma\sqrt{n}$ , (iii) $1/\delta < j \le i \le X_\star-\Sigma\sqrt{n}$ , (iv) $1/\delta < j \le X_\star-\Sigma\sqrt{n} < i \le X_\star-1$ , and (v) $X_\star-\Sigma\sqrt{n} < j \le i \le X_\star-1$ . Note that $\sigma_i \ge \Sigma$ in parts (i)–(iii), and $\sigma_j \ge \Sigma$ in parts (i)–(iv).

Part (i): Note that if $j = o(X_\star)$ , which is the case if $j\le 1/\delta$ , then $\sigma_j \sim \sqrt{n}x_\star$ and $b_j \sim \delta$ , so $\nu^0(i,j) \le (1-\delta+o(\delta))^{i-j} \le {\mathrm{e}}^{-(1+o(1))\delta(i-j)}$ , and treating it as a partial geometric sum,

\[\sum_{j=1}^i\nu^0(i,j) \le \frac{1-{\mathrm{e}}^{-(1+o(1))\delta i}}{1-{\mathrm{e}}^{-(1+o(1))\delta}} \le \frac{(1+o(1))\delta i}{\delta}=(1+o(1))i.\]

Thus,

\[\sum_{i=1}^{\lfloor 1/\delta \rfloor}\sum_{j=1}^i s_{ij} \le \sum_{i=1}^{\lfloor 1/\delta \rfloor}\frac{1}{i}(1+o(1))i \le (1+o(1))/\delta.\]

Part (ii): If $i>1/\delta$ then, from the above,

\[\sum_{j=1}^{\lfloor 1/\delta \rfloor}\nu^0(i,j) \le \frac{1}{1-{\mathrm{e}}^{-(1+o(1))\delta}} \le (1+o(1))/\delta.\]

Noting that $X_\star \le n\delta = \sqrt{n}c$ and $1/\delta = \sqrt{n}/c$ ,

\[\sum_{i=\lfloor 1/\delta \rfloor+1}^{X_\star-\Sigma\sqrt{n}}\sum_{j=1}^{\lfloor 1/\delta \rfloor} s_{ij} \le \frac{1+o(1)}{\delta}(\log(\sqrt{n}c)-\log(\sqrt{n}/c)) = \frac{2+o(1)}{\delta}\log(c).\]

Part (iii): Since $i\ge j$ , $b_j = (1-\epsilon_n)r(X_\star-j)/n \le (1-\epsilon_n)r(X_\star-i)/n = b_i$ , $\nu^0(i,j) \le {\mathrm{e}}^{-(1-o(1))b_i(i-j)}$ , and $\sum_{j \le i}\nu^0(i,j) \le 1/(1-{\mathrm{e}}^{-(1-o(1))b_i}) \sim 1/b_i = n/(X_\star -i)$ uniformly over i, since $b_i \le \delta$ and $\delta \to 0$ . This gives

\[\sum_{i =\lfloor 1/\delta \rfloor+1}^{\lfloor X_\star - \Sigma\sqrt{n} \rfloor}\sum_{j=\lfloor 1/\delta \rfloor+1}^i s_{ij}\le \sum_{i =\lfloor 1/\delta \rfloor+1}^{\lfloor X_\star - \Sigma\sqrt{n} \rfloor}\frac{n}{i(X_\star-i)}\le \frac{n}{X_\star}\sum_{i =\lfloor 1/\delta \rfloor+1}^{\lfloor X_\star - \Sigma\sqrt{n} \rfloor}\bigg(\frac{1}{i} + \frac{1}{X_\star-i}\bigg).\]

Treating the sums as Riemann sums and noting that $n/X_\star \le n/(n\delta-1) \sim \frac{1}{\delta}$ , $\delta X_\star\le n\delta^2=c^2$ and $X_\star/(\Sigma\sqrt{n})\le \sqrt{n}\delta/\Sigma = o(\sqrt{c})$ , this is at most

\[\frac{1+o(1)}{\delta}\left(\log(X_\star)-\log(1/\delta) + \log(X_\star) - \log(\Sigma\sqrt{n})\right) \le \frac{1+o(1)}{\delta}(c^2 + o(\sqrt{c})).\]

Part (iv): Writing as a product and using (46) on the first term, then proceeding as in part (iii) on the sum, for $i\ge X_\star-\Sigma\sqrt{n} \ge j$ ,

\begin{align*}\sum_{j \le X_\star-\Sigma\sqrt{n}}\nu^0(i,j)&= \nu^0(i,X_\star-\Sigma\sqrt{n})\sum_{j \le X_\star-\Sigma\sqrt{n}}\nu^0(X_\star-\Sigma\sqrt{n},j) \\&\le {\mathrm{e}}^{n^{5/8}O(1/n)}/(1 - {\mathrm{e}}^{-(1-o(1))\Sigma/\sqrt{n})} = (1+o(1))\sqrt{n}/\Sigma.\end{align*}

Since $\Sigma=o(c)=o(X_\star/\sqrt{n})$ , $1/i \sim 1/X_\star$ and

\begin{align*}\sum_{i=X_\star-\Sigma\sqrt{n}+1}^{X_\star-1}\sum_{j=\lfloor 1/\delta \rfloor+1}^{X_\star-\Sigma\sqrt{n}}s_{ij}&\le (1+o(1))\frac{\sqrt{n}}{\Sigma}\sum_{i=X_\star-\Sigma\sqrt{n}+1}^{X_\star-1}\frac{1}{i} \\&\le (1+o(1))\frac{\sqrt{n}}{\Sigma}\frac{\Sigma\sqrt{n}}{X_\star} \sim \frac{n}{X_\star} \le \frac{1}{\delta}.\end{align*}

Part (v): Using (46) as in part (iv), $\nu^0(i,j) \le {\mathrm{e}}^{\Sigma\sqrt{n}O(1/n)}=1+o(1)$ . Again, $1/i \sim 1/X_\star$ . Since there are at most $\Sigma^2n$ terms in the sum and $\Sigma \le \sqrt{c}$ , it is bounded by

\[(1+o(1))\frac{\Sigma^2n}{X_\star} = (1+o(1))\frac{c}{\delta}.\]

In all five parts, the sum is $O(c^2/\delta)$ ; referring to (40), this is $o(E_\star^\mathrm{o})$ .

Case 2: $\delta \to \delta_\infty>0$ . Since, by (48), $q_+^-(\,j)/q_-^0(\,j) \le 1$ for $j\le X_\star-\Sigma\sqrt{n}$ and, by (46), $q_+^0(\,j)/q_-^0(\,j) \le 1 + O(1/n)\le {\mathrm{e}}^{O(1/n)}$ for $X_\star-\Sigma\sqrt{n} \le j < X_\star$ , and since $\Sigma \le n^{1/8}$ , $\nu^0(i,j) \le {\mathrm{e}}^{n^{5/8}O(1/n)} = 1+o(1)$ for $j \le i < X_\star$ . Thus,

\[\sum_{i=1}^{X_\star-1}\sum_{j=1}^i s_{ij} \le (1+o(1))\sum_{i=1}^{X_\star-1}\frac{1}{i} \, i \sim X_\star,\]

and $X_\star \le n\delta = o(E_\star^\mathrm{o})$ , again by (40). □

8.3.5. Exponential limit

Here we prove Lemma 10. By the strong Markov property, this is equivalent to showing that $\tau_\star^\mathrm{o}/E_\star^\mathrm{o}$ converges in distribution to exponential with mean 1, assuming $X_0=X_\star$ . Let $\Phi$ denote the natural coupling, so that for each $j\in \{0,\dots,N\}$ , $((\Phi(\,j,t))_{t\ge 0}$ is a copy of the logistic process with initial value j, and by Lemma 1, $\Phi(i,t)\le \Phi(\,j,t)$ for all t if $i\le j$ ; let $\tau_\star(\,j)=\inf\{t>0 \colon \Phi(\,j,t)\in \{0,X_\star\}\}$ and $\tau_\star^\mathrm{o}(\,j) = \sup\{t>0 \colon \Phi(\,j,t)=X_\star\}$ . We give a sufficient condition for $(\tau_\star^\mathrm{o}-\tau_\star)/E_\star^\mathrm{o}$ to have an exponential limit.

Lemma 16. Let $\overline E_\star^\mathrm{o} = \mathbb{E}[\tau_\star^\mathrm{o}(n)]$ and assume that (i) $\mathbb{P}(\tau_\star^\mathrm{o}(\,j) = \tau_\star^\mathrm{o}(n) \mid \Phi(\,j,\tau_\star(\,j))=X_\star) = 1-o(1)$ uniformly over $j \in \{1,\dots,n\}$ , and (ii) uniformly over $j \in \{1,\dots,n\}$ and $t>0$ , $\mathbb{P}(\tau_\star^\mathrm{o}(n)> t \overline E_\star^\mathrm{o} \mid \Phi(\,j,\tau_\star(\,j))=X_\star) \ge \mathbb{P}(\tau_\star^\mathrm{o}(n)> t \overline E_\star^\mathrm{o})-o(1)$ . Then, for each $t>0$ , $\mathbb{P}(\tau_\star^\mathrm{o}(X_\star) > tE_\star^\mathrm{o}) \to {\mathrm{e}}^{-t}$ .

Proof. By definition, $\Phi(X_\star,\tau_\star^\mathrm{o}(X_\star))=X_\star$ , so using assumption (i) it is enough to show that $\mathbb{P}(\tau_\star^\mathrm{o}(n) > tE_\star^\mathrm{o}) \to {\mathrm{e}}^{-t}$ for $t>0$ . By the definition of $\tau_\star^\mathrm{o}(n)$ , $\overline E_\star^\mathrm{o} = \mathbb{E}[\tau_\star^\mathrm{o} \mid X_0=n]$ , so using the Markov property and then Lemma 8 we find that $\overline E_\star^\mathrm{o} = E_\star^\mathrm{o} + \mathbb{E}[\tau_\star \mid X_0=n] \sim E_\star^\mathrm{o}$ . Thus, it is enough to show that $p_n(t) \,:\!=\, \mathbb{P}(\tau_\star^\mathrm{o}(n) > t\overline E_\star^\mathrm{o}) \to {\mathrm{e}}^{-t}$ for $t>0$ .

Using the natural coupling of Section 3, if $j\le n$ then, for any $t>0$ , $\Phi(\,j,t) \le \Phi(n,t)$ , which implies $\tau_\star^\mathrm{o}(\,j) \le \tau_\star^\mathrm{o}(n)$ . Conditioning on the value of $\Phi(n,t)$ and using the Markov property, it follows that $p_n(t+s) = p_n(t)\sum_j \mathbb{P}(\tau_\star^\mathrm{o}(\,j) > s\overline E_\star^\mathrm{o})\mathbb{P}(\Phi(n,t \overline E_\star^\mathrm{o})=j) \le p_n(t)p_n(s)$ , i.e. $t\mapsto p_n(t)$ is submultiplicative for each n. Given $t,s>0$ , conditioning on $\Phi(n,t \overline E_\star^\mathrm{o})$ , using the Markov property, then using assumption (i) followed by assumption (ii) and the law of total probability,

\begin{align*}\mathbb{P}(\tau_\star^\mathrm{o}(n)>(t+s)\overline E_\star^\mathrm{o} & \mid \tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o}) \\[3pt]&= \sum_j \mathbb{P}(\tau_\star^\mathrm{o}(\,j)>s \overline E_\star^\mathrm{o}) \mid \Phi(\,j,\tau_\star(\,j))=X_\star)\mathbb{P}(\Phi(n,t \overline E_\star^\mathrm{o})=j) \\[3pt]&\ge \sum_j \mathbb{P}(\tau_\star^\mathrm{o}(n)>s \overline E_\star^\mathrm{o}) \mid \Phi(\,j,\tau_\star(\,j))=X_\star)\mathbb{P}(\Phi(n,t \overline E_\star^\mathrm{o})=j)-o(1) \\[3pt]&\ge \sum_j \mathbb{P}(\tau_\star^\mathrm{o}(n)>s \overline E_\star^\mathrm{o}))\mathbb{P}(\Phi(n,t \overline E_\star^\mathrm{o})=j)-o(1) \\[3pt]&= p_n(s)-o(1)\end{align*}

uniformly over t and s. Since the above left-hand side is just $p_n(t+s)/p_n(t)$ , we obtain $p_n(t+s) \ge p_n(t)p_n(s)-o(1)$ . Combining with $p_n(t+s)\le p_n(t)p_n(s)$ , it follows easily that for rational t, $p_n(t) = p_n(1)^t + o(1)$ , and since $t\mapsto p_n(t)$ is non-increasing, the same holds for real $t>0$ by rational approximation. Thus, it remains only to show that $p_n(1)\to \frac{1}{{\mathrm{e}}}$ as $n\to\infty$ .

Let $T(n) = \tau_\star^\mathrm{o}(n)/\overline E_\star^\mathrm{o}$ , so that $\mathbb{E}[T(n)]=1$ for each n. By Markov’s inequality, $p_n(2) \le \frac{1}{2}$ , and $p_n(2j) \le p_n(2)^j = 2^{-j}$ for integer $j\ge 1$ , so $\mathbb{E}[\, T(n) \mathbf{1}(T(n) > 2k) \, ] \le \sum_{j \ge k}2\mathbb{P}(T_\star^\mathrm{o} > 2j) \le 2^{-(k-2)}$ , which $\to 0$ as $k\to\infty$ uniformly in n. Since $\mathbb{P}(T(n)>t) = \mathbb{P}(T(n)>1)^t + o(1)$ for each t, an easy approximation argument using the monotonicity of $t\mapsto \mathbb{P}(T(n)>t)$ then shows that $\mathbb{E}[ \, T(n) \, ] = \int_0^{\infty} \mathbb{P}(T(n)>1)^t \, {\mathrm{d}} t + o(1)$ . Since $\mathbb{E}[ \, T(n) \, ] = 1$ for all n, it follows that $\mathbb{P}(T(n)>1) \to \frac{1}{e}$ as $n\to\infty$ , as desired. □

It remains to show that assumptions (i) and (ii) of Lemma 16 are satisfied. Let $\tau_\mathrm{c}(\,j) = \inf\{t \colon \Phi(\,j,t)=\Phi(n,t)\}$ denote the coupling time of the two trajectories, which is a.s. finite since both eventually hit 0. We begin by extracting a further sufficient condition, which we then prove.

Lemma 17. Let $\tau_\mathrm{c}^\star(\,j) = \inf\{t \ge \tau_\mathrm{c}(\,j) \colon \Phi(\,j,t) = X_\star\}$ . Suppose that

(49) \begin{align}\min_{j \in \{1,\dots,n\}}\mathbb{P}(\tau_\mathrm{c}^\star(\,j) <\infty \mid \Phi(\,j,\tau_\star(\,j))=X_\star) \to 1.\end{align}

Then assumptions (i) and (ii) of Lemma 16 are satisfied.

Proof. Assumption (i): Since $\Phi(\,j,t)=\Phi(n,t)$ for all $t\ge \tau_\mathrm{c}(\,j)$ (see Lemma 3.1), the event $\tau_\star^\mathrm{o}(\,j) = \tau_\star^\mathrm{o}(n)$ is equivalent to the event that $\Phi(\,j,t)=\Phi(n,t)=X_\star$ for some $t>0$ , which in turn is equivalent to the event $\tau_\mathrm{c}^\star(\,j)<\infty$ , and assumption (i) follows directly from (49).

Assumption (ii): First note that, since $\mathbb{P}(\Phi(X_\star,\tau_\star(X_\star)) = X_\star)=1$ , it follows from the above that

(50) \begin{align}\mathbb{P}(\tau_\star^\mathrm{o}(X_\star)=\tau_\star^\mathrm{o}(n)) = 1-o(1).\end{align}

Next, using the strong Markov property, $\tau_\star^\mathrm{o}(n)$ conditioned on $\tau_\mathrm{c}^\star(\,j)<\infty$ is equal in distribution to $\tau_\mathrm{c}^\star(\,j)$ conditioned on $\tau_\mathrm{c}^\star(\,j)<\infty$ , plus an independent copy of $\tau_\star^\mathrm{o}(X_\star)$ . In particular, $\tau_\star^\mathrm{o}(n)$ , conditioned on $\tau_\mathrm{c}^\star(\,j)<\infty$ , dominates $\tau_\star^\mathrm{o}(X_\star)$ (with no conditioning). Since $\tau_\mathrm{c}^\star(\,j)<\infty$ implies $\Phi(\,j,\tau_\star(\,j))=X_\star$ , using (49), then the above observation, then (50), it follows that, uniformly over j,

\begin{align*}\mathbb{P}(\tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o} & \mid \Phi(\,j,\tau_\star(\,j)) = X_\star) \\[3pt]& = \mathbb{P}(\tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o} \mid \tau_\mathrm{c}^\star(\,j)<\infty )\mathbb{P}(\tau_\mathrm{c}^\star(\,j)<\infty \mid \Phi(\,j,\tau_\star(\,j))=X_\star) \\[3pt]&= \mathbb{P}(\tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o} \mid \tau_\mathrm{c}^\star(\,j)<\infty )(1-o(1)) \\[3pt]& \ge \mathbb{P}(\tau_\star^\mathrm{o}(X_\star) > t \overline E_\star^\mathrm{o})(1-o(1)) \\[3pt]&= (\mathbb{P}(\tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o}) - o(1))(1-o(1)) \\[3pt]&= \mathbb{P}(\tau_\star^\mathrm{o}(n) > t \overline E_\star^\mathrm{o}) - o(1).\\[-40pt] \end{align*}

□

Finally, we prove the hypothesis of Lemma 17. To do so we show that, within a short time after the paths started from j and from n reach $X_\star$ , they meet (if they have not met already), and then with probability $1-o(1)$ their common trajectory hits $X_\star$ at least once more before going to 0.

Lemma 18. As $n\to\infty$ , $\min_{j \in \{1,\dots,n\}}\mathbb{P}(\tau_\mathrm{c}^\star(\,j) <\infty \mid \Phi(\,j,\tau_\star(\,j))=X_\star) \to 1$ .

Proof. Let $X^j,X^n$ denote the processes $(\Phi(\,j,t))_{t\ge 0},(\Phi(n,t))_{t\ge 0}$ . As we will see, when $X^j,X^n>(1+\epsilon)X_\star/2$ , the drift tends to push them together. Let $\tau_b(\,j) = \inf\{t>\tau_\star(\,j) \vee \tau_\star(n)\colon \max(|X^j_t-nx_\star|,|X^n_t -nx_\star|) \ge nx_\star/4\}$ . If $\tau_\mathrm{c}(\,j) < \tau_\star(\,j) \vee \tau_\star(n)$ then $X_t^j=X_t^n=X_\star$ with $t=\tau_\star(\,j) \vee \tau_\star(n)$ , which implies $\tau_\mathrm{c}^\star(\,j)<\infty$ . On the other hand, if $\tau_\star(\,j) \vee \tau_\star(n) < \tau_\mathrm{c}(\,j) < \tau_b(\,j)$ then $X_{\tau_\mathrm{c}(\,j)}^j=X_{\tau_\mathrm{c}(\,j)}^n \ge 3X_\star/4$ . Using the strong Markov property and the fact that $X^j$ and $X^n$ remain together once they meet, on the latter event it follows from Lemma 8.1 that $\tau_\mathrm{c}^\star(\,j)<\infty$ with probability $1-o(1)$ uniformly over j. Thus, it is enough to show that

(51) \begin{align}\max_j \mathbb{P}(\tau_b(\,j) \wedge \tau_\mathrm{c}(\,j) < \tau_b(\,j) \mid \Phi(\,j,\tau_\star(\,j))=X_\star) \to 1.\end{align}

We begin with a lower bound on $\tau_b(\,j)$ , that ensures both $X^j$ and $X^n$ remain fairly close to $X_\star$ for a while after they hit it. Then, we estimate the drift and diffusivity of $X^j-X^n$ assuming that both are at least $3X_\star/4$ , and with the help of the lower bound, deduce (51).

Lower bound on $\tau_b(\,j)$ : Let $W=X-nx_\star$ . Then $\mu(W) = \mu(X) = X(r(1-X/n)-1) = rX(x_\star-X/n) = -rXW/n$ and $\sigma^2(W) = \sigma^2(X) = X(r(1-X/n)+1) \le (1+r)X$ . Since W jumps by $\pm 1$ , if $|W| \ge 1$ then $\mu(|W|) = \mathrm{sgn}(W)\mu(W)$ and $\sigma^2(|W|) = \sigma^2(W)$ . Suppose $nx_\star/8 \le |W| \le nx_\star/4$ , noting that $1\le nx_\star/8$ for large n. Since $3nx_\star/4 \le X \le 5nx_\star/4$ , $\sigma^2(|W|) = O(nx_\star) = O(n\delta)$ , $\mu(|W|) \le -(3rx_\star/4)(nx_\star/8) = -3n\delta^2/32r$ , and $|\mu(|W|)| \le (5rx_\star/4)nx_\star/4 = O(n\delta^2)$ . We shall use Lemma 19 (with apologies for overloading notation). In the notation of Lemma 19, let $X=|W|-nx_\star/8$ , $x=nx_\star/8 = n\delta/8r$ , $\mu_\star = 3n\delta^2/32r = 3c^2/32r$ , $\sigma^2_\star = Cn\delta$ , and $C_{\mu_\star}=Cn\delta^2$ for some $C>0$ and $C_\Delta=\frac{1}{2}$ . Since $\Delta_\infty(X)=1$ , $\Delta_\infty(X)\mu_\star/\sigma^2_\star = 3\delta/32Cr$ is at most $\frac{1}{2}$ if $C>0$ is chosen large enough. Then, $\Gamma = \exp(\Omega(c^2))$ and $x/16C_{\mu_\star}=\Omega(1/\delta)$ , so $\mathbb{P}(\sup_{t \le (1/\delta)\exp(\Omega(c^2))}|W_t| > nx_\star/4 \ \big| \ |W_0| \le nx_\star/8) = o(1)$ . Applying this bound to $X^j$ , $X^n$ from time $\tau_\star(\,j)$ , respectively $\tau_\star(n)$ , we find that

(52) \begin{align}\mathbb{P}(\tau_b(\,j) \le (1/\delta)\exp(\Omega(c^2)) \mid \Phi(\,j,\tau_\star(\,j))=X_\star)=o(1)\end{align}

uniformly over $j\in \{1,\dots,n\}$ .

Upper bound on $\tau_b(\,j) \wedge \tau_\mathrm{c}(\,j)$ : Let $F(x) = x(r(1-x)-1) = rx(x_\star-x)$ and $G(x)=x(r(1-x)+1) \ge x$ , so that $\mu(X)=nF(X/n)$ and $\sigma^2(X) = nG(X/n) \ge X$ . We have $F'(x) = r(x_\star-2x)$ , so if $3x_\star/4 \le x \le 5x_\star/4$ then $F'(x) \in [F'(5x_\star/4),F'(3x_\star/4)]=[-3rx_\star/2,-rx_\star/2] = [-3\delta/2,-\delta/2]$ . If $X^j,X^n \ge 3nx_\star/4$ and $X^j \ne X^n$ then, letting $U=X^j-X^n$ , by the mean value theorem, $\mu(U)/(U) = n(F(X^j/n)-F(X^n/n))/(X^j-X^n) \in [-3\delta/2,-\delta/2]$ and, since $X^j$ and $X^n$ evolve independently until they collide, $\sigma^2(U) = n(G(X^j/n) + G(X^n/n)) \ge 3nx_\star/2 \ge n\delta/r$ . Since U jumps by $\pm 1$ and takes values in $\mathbb{Z}$ , if $U\ne 0$ then $\mu(|U|)=\mathrm{sgn}(U)\mu(U)$ and $\sigma^2(|U|) = \sigma^2(U)$ , so letting $V=|U|$ , the above implies that, conditional on $\Phi(\,j,\tau_\star(\,j))=X_\star$ , $\mu_t(V)\in [-(3\delta/2)V_t,-(\delta /2)V_t]$ and $\sigma^2_t(V) \ge n\delta/r$ for all $\tau_\star(\,j) \vee \tau_\star(n)\le t<\tau_\mathrm{c}(\,j)\wedge \tau_b(\,j)$ . If this interval is empty, then $\tau_\mathrm{c}(\,j)\wedge \tau_b(\,j)<\tau_\star(\,j)\vee \tau_\star(n)\le \tau_b(\,j)$ so there is nothing to show. Otherwise, since on this time interval $\mathrm{sgn}(U_t)$ is fixed, then, given $\mathrm{sgn}(U_{\tau_\star(\,j)})$ , on the same time interval V is a Markov chain with state space a subset of $\mathbb{Z}$ . Let $u = \lfloor \sqrt{n} \rfloor$ , $t_0 = \tau_\star(\,j)$ , and $t_1 = \tau_b(\,j) \wedge \inf\{t>t_0 \colon V_t = u\}$ , and define recursively $t_i = \tau_b(\,j) \wedge \inf\{t>t_{i-1}\colon V_t \in \{0,u,2u\} \setminus V_{t_{i-1}}\}$ . Let $\rho_i = t_i-t_{i-1}$ for $i=1,\dots,N = \min\{i\colon V_{t_i}=0$ or $t_i=\tau_b(\,j)\} = \min\{i \colon t_i = \tau_\mathrm{c}(\,j) \wedge \tau_b(\,j)\}$ . By the a priori bound, we may assume that $|V_{t_0}| \le nx_\star/2 = n\delta/2r$ . Then, $\xi_t= {\mathrm{e}}^{\delta (t_0 + t \wedge \rho_1)/2}V_{t_0 + t \wedge \rho_1}$ is a supermartingale with $\xi_0 \le n\delta/2r$ , so

\[\mathbb{P}(\rho_1>t) \le \mathbb{P}(\xi_t > {\mathrm{e}}^{\delta t/2}u) \le {\mathrm{e}}^{-2\delta t}\frac{n\delta/2r}{\sqrt{n}-1} \sim {\mathrm{e}}^{-2\delta t}c/2r.\]

This probability is $O(1/c)=o(1)$ if $t=(1/\delta)\log c$ . Using a similar estimate with $\xi_0=2u$ , $\mathbb{P}(\rho_i>t \mid V_{\rho_{i-1}}=2u) \le {\mathrm{e}}^{-2\delta t}\frac{2u}{u} \le 2{\mathrm{e}}^{-2\delta t}$ and, integrating over t, $\mathbb{E}[\rho_i \mid V_{\rho_{i-1}}=2u] \le \frac{1}{\delta}$ . To estimate $\rho_2$ , we note that, for $\alpha>0$ , $\mu_t(\alpha V^2) = 2 \alpha V_t\mu(V_t) + \alpha^2\sigma^2_t(V) \ge -3\alpha \delta V_t^2 + \alpha^2n\delta/r$ so, for $t_1\le t < t_2$ , since $V_t \le 2u \le 2\sqrt{n}$ , choosing $\alpha=13r$ we have $\mu_t(\alpha V^2) \ge \alpha n\delta$ . Thus, $V_{t_1 + t \wedge \rho_2}^2-n\delta(t \wedge \rho_2)$ is a submartingale. Since $V_{t_1}=u$ and $V_{t_2} \le 2u$ , using optional stopping,

\[\mathbb{E}[\rho_2] \le \frac{1}{n\delta}(\mathbb{E}[V_{t_2}^2-V_{t_1}^2]) \le \frac{1}{n\delta}(4u^2-u^2) \le \frac{3n}{n\delta} = \frac{3}{\delta}.\]

By the Markov property, the same estimate holds for $\mathbb{E}[\rho_i \mid V_{t_{i-1}}=u]$ . Using simply that $\mu_t(V) \le 0$ and optional stopping, $\mathbb{P}(V_{t_i}=2u \mid V_{t_{i-1}}=u) \le 1/2$ . Summarizing, on the time interval $[\tau_\star(\,j)\vee \tau_\star(n),\tau_\mathrm{c}(\,j) \wedge \tau_b(\,j)]$ , V hits u, then goes to 2u and back to u at most geometric $(1/2)$ number of times before either $V_{t_i}=0$ or $t_i=\tau_b(\,j)$ . As shown above, $\rho_1$ , the time to first hit u, is at most $\frac{1}{\delta}\log c$ with probability $1-o(1)$ , and the expected time to go from u to either 0 or to 2u and back to u is at most $(4/\delta)$ . Using Wald’s lemma and $\mathbb{E}[\text{geometric}(1/2)=2]$ , $\sum_{i=2}^{N-1} \rho_i \le 8/\delta$ . Using Markov’s inequality, the sum is at most $\frac{1}{\delta}\log c$ with probability $1-o(1)$ , so combining with the estimate on $\rho_1$ , we find that, uniformly over $j \in \{1,\dots,n\}$ , $\mathbb{P}(\tau_b(\,j) \wedge \tau_\mathrm{c}(\,j) - \tau_\star(\,j) \vee \tau_\star(n) > (2/\delta)\log c \mid \Phi(\,j,\tau_\star(\,j))=X_\star)=o(1)$ . From Lemma 14, if $\delta \to 0$ then $\mathbb{E}[\tau_\star(\,j)],\mathbb{E}[\tau_\star(n)]=O(c^2\log(c)/\delta)$ uniformly over $j\in \{1,\dots,n\}$ , so using Markov’s inequality, with probability $1-o(1)$ uniformly in j, $\tau_\star(\,j)\vee \tau_\star(n) = o({\mathrm{e}}^{\epsilon c^2}/\delta)$ for any fixed $\epsilon>0$ . Summing the two and combining with (52), we obtain (51). □