Invasion probabilities, hitting times, and some fluctuation theory for the stochastic logistic process

Abstract

We consider excursions for a class of stochastic processes describing a population of discrete individuals experiencing density-limited growth, such that the population has a finite carrying capacity and behaves qualitatively like the classical logistic model Verhulst (Corresp Math Phys 10:113–121, 1838) when the carrying capacity is large. Being discrete and stochastic, however, our population nonetheless goes extinct in finite time. We present results concerning the maximum of the population prior to extinction in the large population limit, from which we obtain establishment probabilities and upper bounds for the process, as well as estimates for the waiting time to establishment and extinction. As a consequence, we show that conditional upon establishment, the stochastic logistic process will with high probability greatly exceed carrying capacity an arbitrary number of times prior to extinction.

Appendices

Appendices

Some useful lemmas

For the sake of completeness, we include some simple and well-known lemmas that we have used in the main text. The proofs are elementary and are thus omitted.

Lemma A.1

• (i) Let f be an non-decreasing function. Then,

  $$\begin{aligned} 0 \le \frac{1}{n} \sum _{j=a+1}^{b} f\left( \frac{j}{n}\right) - \int _{\frac{a}{n}}^{\frac{b}{n}} f(x)\, dx \le \frac{1}{n} \left( f\left( \frac{b}{n}\right) - f\left( \frac{a}{n}\right) \right) \end{aligned}$$

  and

  $$\begin{aligned} 0 \le \int _{\frac{a}{n}}^{\frac{b}{n}} f(x)\, dx - \frac{1}{n} \sum _{j=a}^{b-1} f\left( \frac{j}{n}\right) \le \frac{1}{n} \left( f\left( \frac{b}{n}\right) - f\left( \frac{a}{n}\right) \right) . \end{aligned}$$

• (ii) If f is differentiable and $|f'(x)|$ is bounded by M on $\left[ \frac{a}{n},\frac{b}{n}\right]$, then

  $$\begin{aligned} \left| \int _{\frac{a}{n}}^{\frac{b}{n}} f(x)\, dx - \frac{1}{n} \sum _{j=a}^{b-1} f\left( \frac{j}{n}\right) \right| < \frac{M(b-a)}{2 n^{2}}. \end{aligned}$$

• (iii) Further, if f is twice differentiable and $|f''(x)|$ is bounded by M on $\left[ \frac{a}{n},\frac{b}{n}\right]$, then

  $$\begin{aligned} \left| \int _{\frac{a}{n}}^{\frac{b}{n}} f(x)\, dx - \frac{1}{n} \sum _{j=a}^{b-1} f\left( \frac{j}{n}\right) -\frac{1}{2n}\left( f\left( \frac{b}{n}\right) - f\left( \frac{a}{n}\right) \right) \right| < \frac{M(b-a)}{4 n^{3}}. \end{aligned}$$

Lemma A.2

(Dominated convergence theorem for series) Suppose that $a_{m,n}$ and $b_{m}$ are sequences such that

• (i) $a_{m,n} \rightarrow a_{m}$ as $n \rightarrow \infty$,

• (ii) $|a_{m,n}| \le b_{m}$ for all n, and

• (iii) $\sum _{m=0}^{\infty } b_{m} < \infty$.

Then,

$$\begin{aligned} \lim _{n \rightarrow \infty } \sum _{m = 0}^{\infty } a_{m,n} = \sum _{m = 0}^{\infty } a_{m}, \end{aligned}$$

i.e., the sums on the left and right are convergent and one can interchange sum and limit.

Lemma A.3

Suppose we have sequences $a_{n,k}, b_{n,k} \ge 0$ for all n, k, such that

$$\begin{aligned} \lim _{k \rightarrow \infty } \sup _{n} \frac{a_{n,k}}{b_{n,k}} = 1, \end{aligned}$$

and that

$$\begin{aligned} \sum _{k = 1}^{\infty } b_{n,k} = \infty . \end{aligned}$$

Then,

$$\begin{aligned} \lim _{N \rightarrow \infty } \frac{\sum _{k = 1}^{N} a_{n,k}}{\sum _{k = 1}^{N} b_{n,k}} = 1. \end{aligned}$$

Proofs

1.1 Proofs for Sect. 2.2

Proof of Proposition 1

We will prove the result for $\gamma = \alpha$ and $\gamma \in (\alpha ,\beta )$. The other cases follow similarly.

Fix $\varepsilon > 0$ such that $\psi '(\alpha ) + \varepsilon < 0$. Using Taylor’s theorem, we may write

$$\begin{aligned} \psi (x) = \psi (y) + \psi '(y)(x-y) + R(x,y)(x-y), \end{aligned}$$

where $R(x,y) \rightarrow 0$ as $|x-y| \rightarrow 0$. Fix $\delta > 0$ such that

$$\begin{aligned} |\psi '(x) - \psi '(\alpha )| < \frac{\varepsilon }{2} \end{aligned}$$

for all x such that $|x - \alpha | < \delta$ and

$$\begin{aligned} |R(x,\alpha )|< \frac{\varepsilon }{2} \quad \text {and} \quad |g(x) - g(\alpha )| < \frac{\varepsilon }{2} \end{aligned}$$

for all x such that $x - \alpha < 3\delta$, and choose $\eta > 0$ such that $\psi (x) < \psi (\alpha ) - \eta$ for all x such that $x - \alpha \ge \delta$. Fix M and m such that $|g(x)| < M$ and $|h(x)| < m$ for all $x \in [\alpha -\delta ,\beta +\delta ]$ and all n. Since $\frac{a_{n}}{n} \rightarrow \alpha$ and $\frac{b_{n}}{n} \rightarrow \beta$, without loss of generality, we may assume that $|\frac{a_{n}}{n} - \alpha | < \delta$, $|\frac{b_{n}}{n} - \beta | < \delta$ and $|\epsilon _{n}| < \varepsilon$ for all n.

Then,

$$\begin{aligned}&\sum _{k = a_{n}}^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \psi \left( \frac{k}{n}\right) }\\&\quad = e^{n \psi \left( \frac{a_{n}}{n}\right) } \sum _{k = a_{n}}^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{a_{n}}{n}\right) \right) }\\&\quad = e^{n \psi \left( \frac{a_{n}}{n}\right) } \left( \sum _{k = a_{n}}^{a_{n} + \lceil 2n \delta \rceil -1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{a_{n}}{n}\right) \right) }\right. \\&\qquad + \left. \sum _{k = a_{n} + \lceil 2n \delta \rceil }^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{a_{n}}{n}\right) \right) } \right) , \end{aligned}$$

and, provided $k \ge a_{n} + \lceil 2n \delta \rceil$, then $\frac{k}{n} \ge \alpha + \delta$, and

$$\begin{aligned} \left| \sum _{k = a_{n} + \lceil 2n \delta \rceil }^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{a_{n}}{n}\right) \right) } \right| \le (b_{n} - a_{n}) M(1+\varepsilon ) e^{-n \eta } \rightarrow 0 \end{aligned}$$

as $n \rightarrow \infty$, whereas if $k < a_{n} + \lceil 2n \delta \rceil$, then $\frac{k}{n} < \alpha + 3\delta$, and

$$\begin{aligned}&(1-\varepsilon )\left( g\left( \frac{a_{n}}{n}\right) - \varepsilon \right) \sum _{k = a_{n}}^{a_{n} + \lceil 2n \delta \rceil -1} e^{n (\psi '\left( \frac{a_{n}}{n}\right) -\frac{\varepsilon }{2})\left( \frac{k}{n}-\frac{a_{n}}{n}\right) }\\&\quad \le \sum _{k = a_{n}}^{a_{n} + \lceil 2n \delta \rceil -1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{a_{n}}{n}\right) \right) }\\&\quad \le (1+\varepsilon )\left( g\left( \frac{a_{n}}{n}\right) + \varepsilon \right) \sum _{k = a_{n}}^{a_{n} + \lceil 2n \delta \rceil -1} e^{n(\psi '\left( \frac{a_{n}}{n}\right) +\frac{\varepsilon }{2})\left( \frac{k}{n}-\frac{a_{n}}{n}\right) }, \end{aligned}$$

and

$$\begin{aligned}&\sum _{k = a_{n}}^{a_{n} + \lceil 2n \delta \rceil -1} e^{n (\psi '\left( \frac{a_{n}}{n}\right) -\varepsilon )\left( \frac{k}{n}-\frac{a_{n}}{n}\right) } = \sum _{k = 0}^{\lceil 2n \delta \rceil -1} e^{(\psi '\left( \frac{a_{n}}{n}\right) -\varepsilon )k} = \frac{e^{(\psi '\left( \frac{a_{n}}{n}\right) -\varepsilon )\lceil 2n \delta \rceil }-1}{e^{(\psi '\left( \frac{a_{n}}{n}\right) -\varepsilon )}-1}. \end{aligned}$$

We now observe that $|\psi '\left( \frac{a_{n}}{n}\right) - \psi '(\alpha )| < \frac{\varepsilon }{2}$, so $\psi '\left( \frac{a_{n}}{n}\right) +\frac{\varepsilon }{2}< \psi '(\alpha ) + \varepsilon < 0$ and

$$\begin{aligned} e^{(\psi '\left( \frac{a_{n}}{n}\right) + \frac{\varepsilon }{2})\lceil 2n \delta \rceil -1} \rightarrow 0 \end{aligned}$$

as $n \rightarrow \infty$. Proceeding similarly we obtain a lower bound. Since $\varepsilon > 0$ can be chosen arbitrarily small, the result follows.

To prove the case when $\psi '(\gamma ) = 0$, we proceed as previously and write

$$\begin{aligned} \psi (x) = \psi (y) + \psi '(y)\left( x-y\right) + \left( \frac{1}{2}\psi ''(y)+ R(x,y)\right) (x-y)^{2} \end{aligned}$$

where $R(x,y) \rightarrow 0$ as $|x-y| \rightarrow 0$. Fix $\varepsilon > 0$ sufficiently small that $\psi ''(\gamma ) + \varepsilon <0$, and choose $\delta > 0$ sufficiently small that $|R(x,y)| < \varepsilon$ and $|g(x)-g(y)| < \varepsilon$ for all $|x-y| < 2\delta$. As before, suppose that $|g(x)| < M$ and $|h(x)| < m$ for $x \in [\alpha -\delta ,\beta +\delta ]$, that $\psi (\gamma ) > \psi (x) + \eta$ for $|\gamma - x| > \delta$ and that $\epsilon _{n} < \frac{\varepsilon }{2m}$ for all n. Then, setting $c_{n} = \lfloor \gamma n \rfloor$ if $\gamma \in (\alpha ,\beta )$, $c_{n} = a_{n}$ if $\gamma = \alpha$, and $c_{n} = b_{n}$ if $\gamma = \beta$,

$$\begin{aligned}&\sum _{k = a_{n}}^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \psi \left( \frac{k}{n}\right) }\\&\quad = e^{n \psi \left( \frac{c_{n}}{n}\right) } \sum _{k = a_{n}}^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{c_{n}}{n}\right) \right) }\\&\quad = e^{n \psi \left( \frac{c_{n}}{n}\right) } \left( \sum _{k = a_{n}}^{c_{n} - \lceil n \delta \rceil -1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{c_{n}}{n}\right) \right) }\right. \\&\qquad +\left. \sum _{k = c_{n} - \lceil n \delta \rceil } ^{c_{n} + \lceil n \delta \rceil } (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{c_{n}}{n}\right) \right) }\right. \\&\qquad \left. + \sum _{k = c_{n} + \lceil n \delta \rceil +1}^{b_{n}-1} (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{c_{n}}{n}\right) \right) } \right) , \end{aligned}$$

where, as before, the first and last sums are bounded above by $(b_{n} - a_{n}) (1+\varepsilon )M e^{-n \eta }$ and

$$\begin{aligned}&(1-\varepsilon )\left( g\left( \frac{c_{n}}{n}\right) - \varepsilon \right) \sum _{k = c_{n}-\lceil n \delta \rceil } ^{c_{n} + \lceil n \delta \rceil } e^{n\left( \psi '\left( \frac{c_{n}}{n}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) + \left( \frac{1}{2}\psi ''\left( \frac{c_{n}}{n}\right) -\frac{\varepsilon }{2}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) ^{2}\right) }\\&\quad \le \sum _{k = c_{n}-\lceil n \delta \rceil } ^{c_{n} + \lceil n \delta \rceil } (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \left( \psi \left( \frac{k}{n}\right) -\psi \left( \frac{c_{n}}{n}\right) \right) }\\&\quad \le (1+\varepsilon )\left( g\left( \frac{c_{n}}{n}\right) + \varepsilon \right) \sum _{k = c_{n}-\lceil n \delta \rceil }^{c_{n} + \lceil n \delta \rceil } e^{n\left( \psi '\left( \frac{c_{n}}{n}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) + \left( \frac{1}{2}\psi ''\left( \frac{c_{n}}{n}\right) +\frac{\varepsilon }{2}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) ^{2}\right) }, \end{aligned}$$

As previously, we will show that for n sufficiently large, the upper sum has an upper bound arbitrarily close to $g(\gamma ) e^{n \psi (\gamma )}\sqrt{\frac{2n\pi }{|\psi ''(\gamma )|}}$. The lower sum is treated identically. Proceeding, we have

$$\begin{aligned}&\sum _{k = c_{n}-\lceil 2n \delta \rceil }^{c_{n} + \lceil 2n \delta \rceil } e^{n\left( \psi '\left( \frac{c_{n}}{n}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) + \left( \frac{1}{2}\psi ''\left( \frac{c_{n}}{n}\right) +\frac{\varepsilon }{2}\right) \left( \frac{k}{n}-\frac{c_{n}}{n}\right) ^{2}\right) }\\&\quad = \sum _{k = -\lceil 2n \delta \rceil }^{\lceil 2n \delta \rceil } e^{\psi '\left( \frac{c_{n}}{n}\right) k + \frac{\psi ''\left( \frac{c_{n}}{n}\right) +\frac{\varepsilon }{2}}{2n} k^{2}} = e^{z_{n} u_{n}^{2}} \sum _{k = -\lceil 2n \delta \rceil }^{\lceil 2n \delta \rceil } e^{-z_{n} \left( k-u_{n}\right) ^{2}}, \end{aligned}$$

where $z_{n} = \left| \frac{\psi ''\left( \frac{c_{n}}{n}\right) +\frac{\varepsilon }{2}}{2n}\right|$ and $u_{n} = \frac{\psi '\left( \frac{c_{n}}{n}\right) }{2 z_{n}}$.

We analyze the latter sum via Poisson’s summation formula (Katznelson 1976), which tells us that for an integrable function f with Fourier transform $\hat{f}$,

$$\begin{aligned} \sum _{k=-\infty }^{\infty } f(k) = \sum _{k=-\infty }^{\infty } \hat{f}(k). \end{aligned}$$

Applying this with $f(x) = e^{-z_{n} (x-u_{n})^{2}}$ gives

$$\begin{aligned} \sum _{k=-\infty }^{\infty } e^{-z_{n} (k-u_{n})^{2}} = \sum _{k=-\infty }^{\infty } \sqrt{\frac{\pi }{z_{n}}} e^{-i u_{n} k + \frac{k^{2}}{4z_{n}}}. \end{aligned}$$

Now, for $k \ne 0$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \left| \sqrt{\frac{\pi }{z_{n}}} e^{-i u_{n} k + \frac{k^{2}}{4z_{n}}}\right| = 0, \end{aligned}$$

whereas, since $\psi$ is twice continuously differentiable (and thus $\psi '$ is Lipschitz), there exists a constant L such that

$$\begin{aligned} |z_{n} u_{n}^{2}| = \frac{4 n |\psi '\left( \frac{c_{n}}{n}\right) - \psi '(\gamma )|}{\left| \psi ''\left( \frac{c_{n}}{n}\right) +\frac{\varepsilon }{2}\right| } \le 4L n \left| \frac{c_{n}}{n} - \gamma \right| ^{2}, \end{aligned}$$

which thus vanishes as $n \rightarrow \infty$, provided $\left| \frac{c_{n}}{n} - \gamma \right| \ll \frac{1}{\sqrt{n}}$.

Since $\varepsilon$ and $\delta$ may be chosen arbitrarily small, the result follows provided

$$\begin{aligned} \sum _{k = - \lfloor \delta n \rfloor }^{\lfloor \delta n \rfloor } e^{-z_{n} (k-u_{n})^{2}} \sim \sum _{k=-\infty }^{\infty } e^{-z_{n} (k-u_{n})^{2}}. \end{aligned}$$

To see this, we first observe that

$$\begin{aligned} 0\le & {} \sum _{k=\lfloor \delta n \rfloor +1}^{\infty } e^{-z_{n} (k-u_{n})^{2}} = \sum _{k=1}^{\infty } e^{-z_{n} \left( \lfloor \delta n \rfloor ^{2} + (k-u_{n})^{2} + 2\lfloor \delta n \rfloor (k-u_{n})\right) }\\< & {} e^{-z_{n} \lfloor \delta n \rfloor ^{2}+2 z_{n} u_{n}\lfloor \delta n \rfloor } \sum _{k=1}^{\infty } e^{-z_{n} (k-u_{n})^{2}} \end{aligned}$$

and, similarly,

$$\begin{aligned} 0 \le \sum _{k=-\infty }^{-\lfloor \delta n \rfloor -1} e^{-z_{n} (k-u_{n})^{2}} < e^{-z_{n} \lfloor \delta n \rfloor ^{2}-2 z_{n} u_{n}\lfloor \delta n \rfloor } \sum _{k=-\infty }^{-1} e^{-z_{n} (k-u_{n})^{2}} \end{aligned}$$

Thus,

$$\begin{aligned} 0\le & {} \sum _{k=-\infty }^{-\lfloor \delta n \rfloor -1} e^{-z_{n} (k-u_{n})^{2}} + \sum _{k=\lfloor \delta n \rfloor +1}^{\infty } e^{-z_{n} (k-u_{n})^{2}} \\< & {} e^{-z_{n} \lfloor \delta n \rfloor ^{2} + 2 z_{n} |u_{n}|\lfloor \delta n \rfloor } \left( \sum _{k=-\infty }^{\infty } e^{-z_{n} (k-u_{n})^{2}} -1 \right) \end{aligned}$$

so that

$$\begin{aligned} 0 \le 1 - \frac{\sum _{k = - \lfloor \delta n \rfloor }^{\lfloor \delta n \rfloor } e^{-z_{n} (k-u_{n})^{2}}}{\sum _{k=-\infty }^{\infty } e^{-z_{n} (k-u_{n})^{2}}} < e^{-z_{n} \lfloor \delta n \rfloor ^{2} + 2 z_{n} |u_{n}|\lfloor \delta n \rfloor } \left( 1 - \frac{1}{\sum _{k=-\infty }^{\infty } e^{-z_{n} (k-u_{n})^{2}}}\right) , \end{aligned}$$

and, since $z_{n} n^{2} \ll n$, whereas $z_{n} |u_{n}| n \ll n \left| \frac{c_{n}}{n} - \gamma \right| = o(\sqrt{n})$, the right hand side tends to 0 as $n \rightarrow \infty$. $\square$

Remark 4

We note that provided $\alpha < \beta$ (resp. $\alpha< \gamma < \beta$) and the function g is bounded on some fixed interval containing $[\alpha ,\beta ]$, then the error in (i) and (ii) is independent of $b_{n}$ and $\beta$ or $a_{n}$ and $\alpha$ respectively. Similarly, the bound in (iii) is independent of either endpoint.

Proof of Corollary 1

Let $\psi (x) = V(x)$, $g(x) = \sqrt{\frac{\mu (x)}{\lambda (x)}\frac{\lambda (0)}{\mu (0)}}$ and

$$\begin{aligned} \epsilon _{n}(k) = e^{n V^{(n)}(k) - n V\left( \frac{k}{n}\right) - \frac{1}{2}\left( f\left( \frac{k}{n}\right) -f(0)\right) }-1. \end{aligned}$$

Then g(x) is continuous,

$$\begin{aligned} e^{n V^{(n)}(k)} = (1+\epsilon _{n}(k)) g\left( \frac{k}{n}\right) e^{n \psi \left( \frac{k}{n}\right) }, \end{aligned}$$

and, from Lemma A.1, for any positive integers $a < b$,

$$\begin{aligned} |\epsilon _{n}(k)| < \frac{\sup _{x \in \left[ \frac{a}{n},\frac{b}{n}\right] } |f''(x)| (b-a)}{n^{3}}, \end{aligned}$$

(B.1)

which we note is uniform in k. The first two assertions then follow from the corresponding parts of the Proposition.

The third statement follows immediately upon observing that

$$\begin{aligned} e^{n V'\left( \frac{a_{n}}{n}\right) } \sim e^{V'(0)a_{n}} = \left( \frac{\mu (0)}{\lambda (0)}\right) ^{a_{n}}. \end{aligned}$$

$\square$

1.2 Proofs for Sect. 2.3

Proof of Proposition 4

Since the process can only change by increments of $\pm 1$, for any m and j, we have

$$\begin{aligned} \mathbb {P}_{m}\left\{ T^{(n)}_{j}< T^{(n)}_{m+} \right\}&= {\left\{ \begin{array}{ll} \frac{\mu _{m}}{\lambda _{m}+\mu _{m}}\mathbb {P}_{m-1}\left\{ T^{(n)}_{j}< T^{(n)}_{m}\right\} &{} \text {if }j< m,\text { and}\\ \frac{\lambda _{m}}{\lambda _{m}+\mu _{m}}\mathbb {P}_{m+1}\left\{ T^{(n)}_{j}< T^{(n)}_{m}\right\} &{} \text {if }j> m.\\ \end{array}\right. }\\&= {\left\{ \begin{array}{ll} \frac{\mu _{m}}{\lambda _{m}+\mu _{m}} \frac{e^{n V^{(n)}(m-1)}}{\sum _{k=j}^{m-1} e^{n V^{(n)}(k)}} &{} \text {if }j < m,\text { and}\\ \frac{\lambda _{m}}{\lambda _{m}+\mu _{m}} \frac{e^{n V^{(n)}(m)}}{\sum _{k=m}^{m-1} e^{n V^{(n)}(k)}} &{} \text {if }j > m.\\ \end{array}\right. } \end{aligned}$$

Taking $m = \lfloor \nu n \rfloor$ and $j = \lfloor \xi n \rfloor$, we are thus left with the task of estimating the sums

$$\begin{aligned} \sum _{k=\lfloor \xi n \rfloor }^{\lfloor \nu n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))} \quad \text {and} \quad \sum _{k=\lfloor \nu n \rfloor }^{\lfloor \xi n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))}, \end{aligned}$$

using Corollary 1, where $V(x) - V(\nu )$ finds its maximum at either $\xi$ or $\nu$ , and this maximum occurs at either the right or left side of the interval of interest, $[\xi ,\nu ]$ or $[\nu ,\xi ]$, depending on where $\xi$ lies.

Now, given that V(x) is convex with a minimum at $\kappa$, and $V(0) = 0$, there exists a unique $\nu ' \ne \nu$ such that $V(\nu ') = V(\nu )$. If $\xi< \nu< \kappa < \nu '$ or $\xi< \nu '< \kappa < \nu$, the interval is $[\xi ,\nu ]$, the maximum occurs at $x = \xi$ and

$$\begin{aligned} \sum _{k=\lfloor \xi n \rfloor }^{\lfloor \nu n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))} \sim \frac{\sqrt{\frac{\mu (\xi )}{\lambda (\xi )}\frac{\lambda (\nu )}{\mu (\nu )}} e^{n \left( V\left( \frac{\lfloor \xi \rfloor }{n}\right) -V\left( \frac{\lfloor \nu \rfloor }{n}\right) \right) }}{1-\frac{\mu (\xi )}{\lambda (\xi )}}, \end{aligned}$$

whereas if $\nu< \kappa< \nu ' < \xi$ or $\nu '< \kappa< \nu <\xi$, the interval is $[\nu ,\xi ]$, the maximum occurs at $x = \xi$ and

$$\begin{aligned} \sum _{k=\lfloor \nu n \rfloor }^{\lfloor \xi n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))} \sim \frac{\sqrt{\frac{\mu (\xi )}{\lambda (\xi )}\frac{\lambda (\nu )}{\mu (\nu )}} e^{n \left( V\left( \frac{\lfloor \xi \rfloor }{n}\right) -V\left( \frac{\lfloor \nu \rfloor }{n}\right) \right) }}{1-\frac{\lambda (\xi )}{\mu (\xi )}}. \end{aligned}$$

If if $\nu< \xi < \nu '$ or $\nu '< \xi < \nu$, the maximum is at $x = \nu$ whereas the interval is $[\nu ,\xi ]$ or $[\xi ,\nu ]$ respectively, and one has

$$\begin{aligned} \sum _{k=\lfloor \xi n \rfloor }^{\lfloor \nu n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))} \sim \frac{1}{1-\frac{\mu (\nu )}{\lambda (\nu )}}, \end{aligned}$$

and

$$\begin{aligned} \sum _{k=\lfloor \nu n \rfloor }^{\lfloor \xi n \rfloor -1} e^{n (V^{(n)}(k)-V^{(n)}(\lfloor \nu n \rfloor -1))} \sim \frac{1}{1-\frac{\lambda (\nu )}{\mu (\nu )}}, \end{aligned}$$

respectively. $\square$

Proof of Proposition 5

Since the process $X^{(n)}(t)$ is Markov, for any integer k, each excursion from k is an independent renewal, and thus the number of returns prior to hitting zero has a geometric distribution with success parameter $\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\}$:

$$\begin{aligned} \mathbb {P}_{m}\left\{ N^{(n)}_{k}(T^{(n)}_{0}) = l \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0} \right\} = \mathbb {P}_{k}\left\{ T^{(n)}_{0}< T^{(n)}_{k+}\right\} \left( 1-\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\} \right) ^{l-1} \end{aligned}$$

with mean

$$\begin{aligned} \mathbb {E}_{m}\left[ N^{(n)}_{k}(T^{(n)}_{0}) \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0} \right] = \frac{1}{\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\} }. \end{aligned}$$

The result follows taking $k = \lfloor \nu n \rfloor$, and using the asymptotic for $\mathbb {P}_{\lfloor \nu n \rfloor }\left\{ T^{(n)}_{0} {<} T^{(n)}_{k+}\right\}$ from the previous proposition with $\xi = 0$, recalling that $V(0)=0$. $\square$

1.3 Proofs for Sect. 2.4

Proof of Proposition 6

The proof presented here is based upon the treatment given for the Moran model in Durrett (2009). We first observe that the hitting time of 0 or $\lfloor \kappa n\rfloor$ is the sum of the time spent in all in-between states prior to $T^{(n)}_{\lfloor \kappa n\rfloor }$, so that

$$\begin{aligned} \mathbb {E}_{m}\left[ T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right]= & {} \sum _{k = 1}^{\lfloor \kappa n\rfloor - 1} \mathbb {E}_{m}\left[ S^{(n)}_{k}(T^{(n)}_{\lfloor \kappa n\rfloor } ) \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right] \\= & {} \frac{1}{\lambda _{k}+\mu _{k}} \mathbb {E}_{m}\left[ N^{(n)}_{k}(T^{(n)}_{\lfloor \kappa n\rfloor } ) \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right] , \end{aligned}$$

as $\frac{1}{\lambda _{k}+\mu _{k}}$ is the expected time spent in state k, which is exponentially distributed with parameter $\lambda _{k}+\mu _{k}$.

Now, $N^{(n)}_{m}(T^{(n)}_{\lfloor \kappa n\rfloor } )$ has a modified geometric distribution:

$$\begin{aligned}&\mathbb {P}_{m}\left\{ N^{(n)}_{k}(T^{(n)}_{\lfloor \kappa n\rfloor } ) = l \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right\} \\&\quad = {\left\{ \begin{array}{ll} \mathbb {P}_{m}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0} \right\} &{} \text {if l = 0, and}\\ \begin{aligned} &{}\mathbb {P}_{m}\left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0} \right\} \\ &{}\times \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k+} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right\} \\ &{}\times \left( 1-\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k+} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\} \right) ^{l-1} \end{aligned} &{} \text {if }l\ge 1, \end{array}\right. } \end{aligned}$$

which has mean

$$\begin{aligned} \frac{\mathbb {P}_{m}\left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right\} }{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k+} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\} }. \end{aligned}$$

Now, if we specialize to the case when $m = 1$, then the process must pass through k en route to $\lfloor \kappa n\rfloor$, so

$$\begin{aligned} \mathbb {P}_{1}\left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0} \right\} = 1 \end{aligned}$$

Moreover, conditional on $T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}$, starting from k, $T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{k+}$ if and only if a birth occurs and the process hits $\lfloor \kappa n\rfloor$ prior to k:

$$\begin{aligned} \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k+} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right\} = \frac{\lambda _{k}}{\lambda _{k}+\mu _{k}} \mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\} \end{aligned}$$

so that

$$\begin{aligned} \mathbb {E}_{1}\left[ T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right] = \sum _{k = 1}^{\lfloor \kappa n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\} } \end{aligned}$$

whereas

$$\begin{aligned} \mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right\} = \frac{\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k}\right\} }{\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\} }, \end{aligned}$$

since $\left\{ T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{k}\right\} \subseteq \left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\}$, as to reach 0 from $k+1$, the process must pass via k.

Now, Proposition 2 and its proof tell us that $\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{0}\right\}$ tends to 1 as $k \rightarrow \infty$, and, moreover, that this convergence is uniform in n. We may thus apply Lemma A.3, to conclude that

$$\begin{aligned} \mathbb {E}_{1}\left[ T^{(n)}_{\lfloor \kappa n\rfloor } \Big \vert T^{(n)}_{\lfloor \kappa n\rfloor }< T^{(n)}_{0}\right] \sim \sum _{k = 1}^{\lfloor \kappa n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{k}\right\} } \end{aligned}$$

We now observe that

$$\begin{aligned} \mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{k}\right\} = h^{(n)}_{\lfloor \kappa n\rfloor ,k}(k+1) = \frac{e^{nV^{(n)}(k)}}{\sum _{j = k}^{\lfloor \kappa n\rfloor -1} e^{nV^{(n)}(j)}}, \end{aligned}$$

which, by Corollary 1 is asymptotically equivalent to $1-\frac{\mu \left( \frac{k}{n}\right) }{\lambda \left( \frac{k}{n}\right) }$, so recalling that $\lambda _{k} = \lambda \left( \frac{k}{n}\right) k$,

$$\begin{aligned} \sum _{k = 1}^{\lfloor \kappa n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \kappa n\rfloor } < T^{(n)}_{k}\right\} } \sim \sum _{k = 1}^{\lfloor \kappa n\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k}. \end{aligned}$$

The latter is the Riemann sum for the integral of $\frac{1}{(\lambda (x)-\mu (x))x}$ over $[0,\kappa ]$, but this integral diverges at both endpoints. To deal with this, first observe that, using Taylor’s theorem, we may write

$$\begin{aligned} \lambda (x)-\mu (x) = \lambda (0)-\mu (0) + (\lambda '(0)-\mu '(0) + r(x))x, \end{aligned}$$

where $r(x) \rightarrow 0$ as $x \rightarrow 0$ and

$$\begin{aligned} \lambda (x)-\mu (x) = (\lambda '(\kappa )-\mu '(\kappa ))(x-\kappa ) +(\lambda ''(\kappa )-\mu ''(\kappa )+R(x))(x-\kappa )^{2}, \end{aligned}$$

for a continuous function R(x) such that $R(x) \rightarrow 0$ as $x \rightarrow \kappa$. Then, for arbitrary $\varepsilon > 0$, we can choose n sufficiently large that

$$\begin{aligned} {\textstyle \lambda (0)-\mu (0)< \lambda \left( \frac{k}{n}\right) -\mu \left( \frac{k}{n}\right) < \lambda (0)-\mu (0) + \varepsilon } \end{aligned}$$

for all $k \le \frac{n}{\ln {n}}$ and

$$\begin{aligned} {\textstyle (\lambda '(\kappa )-\mu '(\kappa ))\left( \frac{k}{n}-\kappa \right) - \varepsilon< \lambda \left( \frac{k}{n}\right) -\mu \left( \frac{k}{n}\right) < (\lambda '(\kappa )-\mu '(\kappa ))\left( \frac{k}{n}-\kappa \right) + \varepsilon } \end{aligned}$$

for all $\lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor \le k < \lfloor \kappa n\rfloor$, and split the sum in three:

$$\begin{aligned}&\sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k} + \sum _{k = \left\lfloor \frac{n}{\ln {n}} \right\rfloor +1}^{\lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k} \\&\quad + \sum _{k = \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor }^{\lfloor \kappa n\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k}. \end{aligned}$$

For the first sum, we have that

$$\begin{aligned} \frac{1}{\lambda (0)-\mu (0) + \varepsilon } \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k} \le \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k} \le \frac{1}{\lambda (0)-\mu (0)} \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k}, \end{aligned}$$

whereas

$$\begin{aligned} \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k} = \ln {\left\lfloor \frac{n}{\ln {n}} \right\rfloor } + \gamma + \epsilon _{\left\lfloor \frac{n}{\ln {n}} \right\rfloor }, \end{aligned}$$

where $\gamma$ is the Euler-Mascheroni constant and $\epsilon _{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \sim \frac{1}{2\left\lfloor \frac{n}{\ln {n}} \right\rfloor }$.

Similarly,

$$\begin{aligned}&\frac{1}{(\lambda '(\kappa )-\mu '(\kappa )) + \varepsilon )\kappa } \sum _{k = \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor }^{\lfloor \kappa n\rfloor - 1} \frac{1}{k-\kappa n} \le \sum _{k =\lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor }^{\lfloor \kappa n\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k}\\&\quad \le \frac{1}{(\lambda '(\kappa )-\mu '(\kappa )) - \varepsilon )(\kappa - \delta )} \sum _{k = \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor }^{\lfloor \kappa n\rfloor - 1} \frac{1}{k-\kappa n} \end{aligned}$$

and,

$$\begin{aligned} \sum _{k = \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor }^{\lfloor \kappa n\rfloor - 1} \frac{1}{k-\kappa n} = - \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k+\kappa n-\lfloor \kappa n\rfloor }. \end{aligned}$$

Since $0 \le \kappa n - \lfloor \kappa n\rfloor < 1$,

$$\begin{aligned} \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k+1} < \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k+\kappa n-\lfloor \kappa n\rfloor } \le \sum _{k = 1}^{\left\lfloor \frac{n}{\ln {n}} \right\rfloor } \frac{1}{k} \end{aligned}$$

Finally, to deal with the middle sum, we first note that

$$\begin{aligned} \frac{d}{dx} \frac{1}{(\lambda (x)-\mu (x))x} = -\frac{(\lambda '(x)-\mu '(x))x + \lambda (x)-\mu (x)}{(\lambda (x)-\mu (x))^{2}x^{2}} \end{aligned}$$

is bounded on any closed interval in $(0,\kappa )$ and tends to $+\infty$ at 0, where it is decreasing, and at $\kappa$, where it is increasing; in particular, on $\left[ \frac{1}{\ln {n}},\kappa - \frac{1}{\ln {n}}\right]$ the derivative is bounded above by its values at the endpoints, which are bounded above by

$$\begin{aligned} \frac{\sup _{x \in [0,\kappa ]} -(\lambda '(x)-\mu '(x))x}{\min \{\lambda (0)-\mu (0),(\lambda '(\kappa )-\mu '(\kappa ))\kappa \}} (\ln {n})^{2}. \end{aligned}$$

Thus, applying Lemma A.1, we have that

$$\begin{aligned}&\left| \sum _{k = \left\lfloor \frac{n}{\ln {n}} \right\rfloor +1}^{\lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) - \mu \left( \frac{k}{n}\right) \right) k} - \int _{\frac{1}{n}\left( \left\lfloor \frac{n}{\ln {n}} \right\rfloor +1\right) } ^{\frac{1}{n}\left( \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor \right) } \frac{dx}{(\lambda (x)-\mu (x))x}\right| \\&\quad \le \frac{\sup _{x \in [0,\kappa ]} -(\lambda '(x)-\mu '(x))x}{\min \{\lambda (0)-\mu (0),(\lambda '(\kappa )-\mu '(\kappa ))\kappa \}} \frac{(\ln {n})^{2}}{2n} \end{aligned}$$

Moreover,

$$\begin{aligned} 0 \le \int _{\frac{1}{n}\left( \left\lfloor \frac{n}{\ln {n}} \right\rfloor +1\right) } ^{\frac{1}{n}\left( \lfloor \kappa n\rfloor - \left\lfloor \frac{n}{\ln {n}} \right\rfloor \right) } \frac{dx}{(\lambda (x)-\mu (x))x} \le \int _{\frac{1}{\ln {n}}}^{ \kappa - \frac{1}{\ln {n}}} \frac{dx}{(\lambda (x)-\mu (x))x}, \end{aligned}$$

and, since r(x) and R(x) are continuous, and thus bounded on $[0,\kappa ]$,

$$\begin{aligned} \int _{\frac{1}{\ln {n}}}^{\frac{\kappa }{2}} \frac{dx}{(\lambda (x)-\mu (x))x} - \int _{\frac{1}{\ln {n}}}^{\frac{\kappa }{2}} \frac{dx}{(\lambda (0)-\mu (0))x} = \int _{\frac{1}{\ln {n}}}^{\frac{\kappa }{2}} \frac{\lambda '(0)-\mu '(0) + r(x)}{ (\lambda (0)-\mu (0))(\lambda (x)-\mu (x))}\,dx \end{aligned}$$

and

$$\begin{aligned}&\int _{\frac{\kappa }{2}}^{\kappa -\frac{1}{\ln {n}}} \frac{dx}{(\lambda (x)-\mu (x))x} - \int _{\frac{\kappa }{2}}^{\kappa -\frac{1}{\ln {n}}} \frac{dx}{(\lambda '(\kappa )-\mu '(\kappa ))\kappa (x-\kappa )} \\&\quad = \int _{\frac{1}{\ln {n}}}^{\frac{\kappa }{2}} \frac{\lambda ''(\kappa )-\mu ''(\kappa )+R(x)}{ (\lambda '(\kappa )-\mu '(\kappa ))\kappa h(x)}\,dx \end{aligned}$$

are bounded, where

$$\begin{aligned} h(x) = {\left\{ \begin{array}{ll} \frac{(\lambda (x)-\mu (x))x}{(x-\kappa )} &{} \text {for }x \ne \kappa ,\text { and}\\ (\lambda '(\kappa )-\mu '(\kappa ))\kappa &{} \text {for }x = \kappa . \end{array}\right. } \end{aligned}$$

Finally, we observe that

$$\begin{aligned} \int _{\frac{1}{\ln {n}}}^{\frac{\kappa }{2}} \frac{dx}{(\lambda (0)-\mu (0))x} + \frac{1}{\lambda (0)-\mu (0)}\left( \ln {\frac{\kappa }{2}}-\ln {\left( \frac{1}{\ln {n}}\right) }\right) \end{aligned}$$

and

$$\begin{aligned}&\int _{\frac{\kappa }{2}}^{\kappa -\frac{1}{\ln {n}}} \frac{dx}{(\lambda '(\kappa )-\mu '(\kappa ))\kappa (x-\kappa )}\\&\quad = \frac{1}{(\lambda '(\kappa )-\mu '(\kappa ))\kappa } \left( \ln {\left( \frac{1}{\ln {n}}\right) }-\ln {\frac{\kappa }{2}}\right) , \end{aligned}$$

so that the middle sum is $\ll \ln {\ln {n}}$.

Since the choice of $\varepsilon$ is arbitrary, the result follows. $\square$

Proof of Proposition 7

We begin with a pair of lemmas:

Lemma B.1

The logistic process conditioned on the event $T^{(n)}_{0} < T^{(n)}_{M}$ is a Markov birth and death process with transition rates

$$\begin{aligned} \tilde{\lambda }^{(n)}_{k} = \lambda ^{(n)}_{k}\frac{h^{(n)}_{0,M}(k+1)}{h^{(n)}_{0,M}(k)} \quad \text {and} \quad \tilde{\mu }^{(n)}_{k} = \mu ^{(n)}_{k}\frac{h^{(n)}_{0,M}(k-1)}{h^{(n)}_{0,M}(k)}, \end{aligned}$$

In particular, taking $M = \lfloor \nu n \rfloor$ for $\kappa< \nu < \eta$, we have that

$$\begin{aligned} \lim _{n \rightarrow \infty } \tilde{\lambda }^{(n)}_{k} = \mu (0) k \quad \text {and} \quad \lim _{n \rightarrow \infty } \tilde{\mu }^{(n)}_{k} = \lambda (0) k \end{aligned}$$

Proof

This is a special case of Doob’s h-transform (Doob 1957).

Lemma B.2

Let

$$\begin{aligned} \tau ^{(n)}_{M}(m) = \mathbb {E}_{m}\left[ T^{(n)}_{0} \Big \vert T^{(n)}_{0} < T^{(n)}_{M}\right] . \end{aligned}$$

Then,

$$\begin{aligned} \tau ^{(n)}_{M}(m) = \sum _{k=1}^{m} \sum _{j=k}^{M-1} \frac{1}{\tilde{\lambda }^{(n)}_{j}} \prod _{l=k}^{j} \frac{\tilde{\lambda }^{(n)}_{l} }{\tilde{\mu }^{(n)}_{l}} \end{aligned}$$

Proof

For $m < M$ the function $\tau ^{(n)}_{M}$ satisfies the recurrence relation

$$\begin{aligned} \tau ^{(n)}_{M}(m) = \frac{1}{\tilde{\lambda }^{(n)}_{m} + \tilde{\mu }^{(n)}_{m}} + \frac{\tilde{\lambda }^{(n)}_{m}}{\tilde{\lambda }^{(n)}_{m} + \tilde{\mu }^{(n)}_{m}} \tau ^{(n)}_{M}(m+1) + \frac{\tilde{\mu }^{(n)}_{m}}{\tilde{\lambda }^{(n)}_{m} + \tilde{\mu }^{(n)}_{m}} \tau ^{(n)}_{M}(m-1), \end{aligned}$$

with boundary $\tau ^{(n)}_{M}(0) = 0$, whilst

$$\begin{aligned} \tau ^{(n)}_{M}(M-1) = \frac{1}{\tilde{\mu }^{(n)}_{M-1}} + \tau ^{(n)}_{M}(M-2). \end{aligned}$$

Solving the recurrence equation gives the result. As previously, we refer to Karlin and Taylor (1975) for a detailed treatment. $\square$

The proof consists in showing that the sum

$$\begin{aligned} \sum _{j=i}^{\lfloor \kappa n\rfloor -1} \frac{1}{\tilde{\lambda }^{(n)}_{j}} \prod _{k=i}^{j} \frac{\tilde{\lambda }^{(n)}_{k} }{\tilde{\mu }^{(n)}_{k}} \end{aligned}$$

is uniformly bounded in n, so that we can apply Lemma A.2 to interchange sum and limit to obtain

$$\begin{aligned} \lim _{n \rightarrow \infty } \tau ^{(n)}_{\lfloor \kappa n\rfloor }(m)= & {} \sum _{k=1}^{m} \sum _{j=k}^{\infty } \frac{1}{\mu (0) j} \left( \frac{\mu (0)}{\lambda (0)}\right) ^{j} = \frac{1}{\mu (0)} \sum _{k=1}^{m} \sum _{j=k}^{\infty } \int _{0}^{\frac{\mu (0)}{\lambda (0)}} x^{j-1}\, dx\\= & {} \frac{1}{\mu (0)} \sum _{k=1}^{m} \int _{0}^{\frac{\mu (0)}{\lambda (0)}} \frac{x^{k-1}}{1-x}\, dx = \frac{1}{\mu (0)} \int _{0}^{\frac{\mu (0)}{\lambda (0)}} \frac{1-x^{m}}{(1-x)^{2}}\, dx. \end{aligned}$$

First,

$$\begin{aligned} \prod _{l=k}^{j} \frac{\tilde{\lambda }^{(n)}_{l} }{\tilde{\mu }^{(n)}_{l}}= & {} \prod _{l=k}^{j} \frac{\lambda ^{(n)}_{l} }{\mu ^{(n)}_{l}} \prod _{l=k}^{j} \frac{h^{(n)}_{0,\lfloor \kappa n\rfloor }(l+1)}{h^{(n)}_{0,\lfloor \kappa n\rfloor }(k-1)}\\= & {} \prod _{l=1}^{k-1} \frac{\mu ^{(n)}_{l}}{\lambda ^{(n)}_{l}} \prod _{l=1}^{j} \frac{\lambda ^{(n)}_{l} }{\mu ^{(n)}_{l}} \frac{h^{(n)}_{0,\lfloor \kappa n\rfloor }(j+1)h^{(n)}_{0,\lfloor \kappa n\rfloor }(j)}{h^{(n)}_{0,\lfloor \kappa n\rfloor }(k-1)h^{(n)}_{0,\lfloor \kappa n\rfloor }(k-1)}, \end{aligned}$$

so, since $i \le m$, we can ignore terms in i and consider only the sum

$$\begin{aligned}&\sum _{j=1}^{\lfloor \kappa n\rfloor -1} \frac{h^{(n)}_{0,\lfloor \kappa n\rfloor }(j+1)h^{(n)}_{0,\lfloor \kappa n\rfloor }(j)}{\tilde{\lambda }^{(n)}_{j}} \prod _{k=1}^{j} \frac{\lambda ^{(n)}_{k} }{\mu ^{(n)}_{k}} = \sum _{j=1}^{\lfloor \kappa n\rfloor -1} \frac{(h^{(n)}_{0,\lfloor \kappa n\rfloor }(j))^{2}}{\lambda ^{(n)}_{j}} e^{-n V^{(n)}(j)}\\&\quad \le \frac{2}{\left( \sum _{k=0}^{\lfloor \kappa n\rfloor -1} e^{n V^{(n)}(k)}\right) ^{2}} \sum _{j=1}^{\lfloor \kappa n\rfloor -1} \frac{1}{\lambda ^{(n)}_{j}} \sum _{k=j}^{\lfloor \kappa n\rfloor -1} e^{n(2V^{(n)}(k)-V^{(n)}(j))}\\&\quad \le \frac{2}{\lambda (0)\left( \sum _{k=0}^{\lfloor \kappa n\rfloor -1} e^{n V^{(n)}(k)}\right) ^{2}} \sum _{j=1}^{\lfloor \kappa n\rfloor -1} \sum _{k=j}^{\lfloor \kappa n\rfloor -1} e^{n V^{(n)}(k)}. \end{aligned}$$

Now, by Lemma A.1, $n V^{(n)}(k) \le n \int _{0}^{\frac{k}{n}} f(x)\, dx + f\left( \frac{k}{n}\right) - f(0)$, whereas, by the intermediate value theorem for integrals, we have

$$\begin{aligned} n \int _{0}^{\frac{k}{n}} f(x)\, dx = f(z_{k,n}) i \end{aligned}$$

for some $z_{k,n} \in [0,\frac{k}{n}]$. Now, fix $0< \varepsilon < \kappa$. Provided $k \le \lfloor \nu n \rfloor$ for $0< \nu < \eta$, either $\frac{k}{n} < \varepsilon$, in which case $f(z_{k,n})< f(\varepsilon ) < 0$, or

$$\begin{aligned} f(z_{k,n}) \varepsilon< f(z_{k,n}) \frac{k}{n} = \int _{0}^{\frac{k}{n}} f(x)\, dx< \min \left\{ \int _{0}^{\varepsilon } f(x)\, dx, \int _{0}^{\nu } f(x)\, dx\right\} < 0, \end{aligned}$$

and thus, $\rho {:}{=}\sup _{n} f(z_{k,n}) < 0$.

Now $0 \le f\left( \frac{k}{n}\right) - f(0) \le f(\nu ) - f(0)$, so if

$$\begin{aligned} a_{n,k} = {\left\{ \begin{array}{ll} \prod _{j=1}^{k} \frac{\mu _{j}}{\lambda _{j}} &{} \text {if }k \le \lfloor \nu n \rfloor ,\text { and}\\ 0 &{} \text {otherwise} \end{array}\right. }, \end{aligned}$$

then $a_{n,k} \le e^{f(\nu ) - f(0)} e^{\rho i}$, and $e^{\rho } < 1$, so

$$\begin{aligned} 1 \le \sum _{k=j}^{\lfloor \nu n\rfloor -1} e^{n V^{(n)}(k)} \le \sum _{k = j}^{\infty } e^{f(\nu ) - f(0)} e^{\rho k} = \frac{e^{f(\nu ) - f(0)}}{1-e^{\rho }} e^{\rho j}, \end{aligned}$$

and the sum above is bounded, independently of n. $\square$

Proof of Proposition 8

We first observe that the time to extinction is simply the time spent in all states $k > 0$:

$$\begin{aligned} \mathbb {E}_{m}\left[ T^{(n)}_{0}\right]= & {} \sum _{k = 1}^{\infty } \mathbb {E}_{m}\left[ S^{(n)}_{k}(T^{(n)}_{0})\right] \\= & {} \sum _{k = 1}^{\infty } \frac{1}{\lambda _{k}+\mu _{k}} \mathbb {E}_{m}\left[ N^{(n)}_{k}(T^{(n)}_{0})\right] , \end{aligned}$$

as $\frac{1}{\lambda _{k}+\mu _{k}}$ is the expected time spent in state k per visit, and, by definition, $N^{(n)}_{k}(T^{(n)}_{0})$ is the total number of visits to k prior to extinction. As before, $N^{(n)}_{k}(T^{(n)}_{0})$ has a modified geometric distribution with mean

$$\begin{aligned} \frac{\mathbb {P}_{m}\left\{ T^{(n)}_{k}< T^{(n)}_{0}\right\} }{\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\} }. \end{aligned}$$

For the denominator, the process can only fail to return to k if the next event is a death and the process hits 0 prior to hitting k:

$$\begin{aligned} \mathbb {P}_{k}\left\{ T^{(n)}_{0}< T^{(n)}_{k+}\right\} = \frac{\mu _{k}}{\lambda _{k}+\mu _{k}} \mathbb {P}_{k-1}\left\{ T^{(n)}_{0} < T^{(n)}_{k}\right\} = \frac{\mu _{k}}{\lambda _{k}+\mu _{k}} \frac{e^{n V^{(n)}(k-1)}}{\sum _{j=0}^{k-1} e^{n V^{(n)}(j)}}. \end{aligned}$$

We first note that in light of (B.1), there is a bound $\epsilon _{n}$ that tends to 0 as $n \rightarrow \infty$ such that

$$\begin{aligned} \left| \frac{e^{n V^{(n)}(k-1)}}{\sqrt{\frac{\mu \left( \frac{k-1}{n}\right) }{\mu (0)} \frac{\lambda (0)}{\lambda \left( \frac{k-1}{n}\right) }} e^{n V\left( \frac{k-1}{n}\right) } }-1 \right| < \epsilon _{n} \end{aligned}$$

uniformly in k.

Now, fix some small $\delta >0$ such that $V(\delta ) > V(\kappa )$. We consider the sum in k in two parts, $k \le \lfloor \delta n \rfloor$, and $k > \lfloor \delta n \rfloor$. We first consider the latter.

As we observed in Remark 4, since $V(0) > V(\frac{k-1}{n})$ and $k > \lfloor \delta n \rfloor$,

$$\begin{aligned} \left| \frac{\sum _{j=0}^{k-1} e^{n V^{(n)}(j)}}{\frac{1}{1-\frac{\mu (0)}{\lambda (0)}}}-1 \right| < \eta _{n}, \end{aligned}$$

where $\eta _{n} \rightarrow 0$ as $n \rightarrow \infty$, independently of k. Thus,

$$\begin{aligned} \mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\} \sim \left( 1-\frac{\mu (0)}{\lambda (0)}\right) \sqrt{\frac{\mu \left( \frac{k-1}{n}\right) }{\mu (0)} \frac{\lambda (0)}{\lambda \left( \frac{k-1}{n}\right) }} e^{n V\left( \frac{k-1}{n}\right) } \end{aligned}$$

uniformly in i.

For the numerator, from Proposition 2, we know that

$$\begin{aligned} \mathbb {P}_{m}\left\{ T^{(n)}_{k} < T^{(n)}_{0}\right\} \sim 1-\left( \frac{\mu (0)}{\lambda (0)}\right) ^{m} \end{aligned}$$

if $m < \eta n$, whereas

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P}_{m}\left\{ T^{(n)}_{k} < T^{(n)}_{0}\right\} =0 \end{aligned}$$

otherwise, again uniformly in k.

Thus, since $\mu _{k} = n \mu \left( \frac{k}{n}\right) \frac{k}{n}$, $-V(x)$ is maximized at $x = \kappa$, and $\mu (\kappa ) = \lambda (\kappa )$, from Proposition 1 we have

$$\begin{aligned}&\sum _{k = \lfloor \delta n \rfloor +1}^{\infty } \frac{1}{\lambda _{k}+\mu _{k}} \mathbb {E}_{m}\left[ N^{(n)}_{k}(T^{(n)}_{0})\right] \\&\quad \sim \frac{1-\left( \frac{\mu (0)}{\lambda (0)}\right) ^{m}}{1-\frac{\mu (0)}{\lambda (0)}} \sum _{k = \lfloor \delta n \rfloor + 1}^{\lfloor \eta n \rfloor } \frac{e^{-n V\left( \frac{k-1}{n}\right) }}{n \mu \left( \frac{k}{n}\right) \frac{k}{n} \sqrt{\frac{\mu \left( \frac{k-1}{n}\right) }{\lambda \left( \frac{k-1}{n}\right) }\frac{\lambda (0)}{\mu (0)}}}\\&\quad \sim \sqrt{\frac{2\pi }{n\left( \frac{\mu '(\kappa )}{\mu (\kappa )}-\frac{\lambda '(\kappa )}{\lambda (\kappa )}\right) } \frac{\mu (0)}{\lambda (0)}} \frac{1-\left( \frac{\mu (0)}{\lambda (0)}\right) ^{m}}{1-\left( \frac{\mu (0)}{\lambda (0)}\right) } \frac{e^{-n V(\kappa )}}{\mu (\kappa ) \kappa \left( 1-\frac{\lambda (0)}{\mu (0)}\right) }. \end{aligned}$$

Finally, we observe that for $k \le \lfloor \delta n \rfloor$,

$$\begin{aligned} \mathbb {P}_{m}\left\{ T^{(n)}_{k} < T^{(n)}_{0}\right\} \le 1, \end{aligned}$$

whereas, since $V^{(n)}(j) < 0$,

$$\begin{aligned} \sum _{j=0}^{k-1} e^{n V^{(n)}(j)} \le k-1 \le \delta n. \end{aligned}$$

Moreover, since $\lambda$ and $\mu$ are, respectively, decreasing and increasing,

$$\begin{aligned} e^{n V^{(n)}(k-1)} \ge \sqrt{\frac{\mu \left( \frac{k-1}{n}\right) }{\mu (0)} \frac{\lambda (0)}{\lambda \left( \frac{k-1}{n}\right) }} e^{n V\left( \frac{k-1}{n}\right) } (1-\epsilon _{n}) \ge e^{n V(\delta )} (1-\epsilon _{n}), \end{aligned}$$

so that

$$\begin{aligned} \sum _{k = 1}^{\lfloor \delta n \rfloor } \frac{1}{\lambda _{k}+\mu _{k}} \mathbb {E}_{m}\left[ N^{(n)}_{k}(T^{(n)}_{0})\right] \le \frac{(\delta n)^{2}}{\mu (0)(1-\epsilon _{n})} e^{-n V(\delta )}, \end{aligned}$$

which is asymptotically smaller than the sum over $k > \lfloor \delta n \rfloor$. $\square$

Proof of Corollary 4

By the strong Markov property, each excursion starting from state k is independent. Thus, conditional on $n_{k}$ visits to k, the time spent in k after each return is a sum of $n_{k}$ independent exponentially distributed random variables with rate $\lambda _{k} + \mu _{k}$ i.e., a gamma-distributed with shape and rate parameters $n_{k}$ and $\lambda _{k} + \mu _{k}$: the probability that the total time is in $[t,t+dt)$ is

$$\begin{aligned}&\int _{0}^{t}\int _{0}^{t-t_{1}} \cdots \int _{0}^{t-t_{1}-t_{2}-\cdots -t_{n_{k}-2}} \prod _{j=1}^{n_{k}-1} (\lambda _{k} + \mu _{k}) e^{-(\lambda _{k} + \mu _{k})t_{j}}\\&\qquad \times (\lambda _{k} + \mu _{k}) e^{-(\lambda _{k} + \mu _{k})(t-t_{1}-t_{2}-\cdots -t_{n_{k}-1})} \, dt_{1}dt_{2}\cdots dt_{n_{k}-1}\\&\quad = \frac{(\lambda _{k} + \mu _{k})^{n_{k}}}{(n_{k}-1)!} t^{n_{k}-1} e^{-(\lambda _{k} + \mu _{k})t}. \end{aligned}$$

Now, we observed above that the number of visits to k prior to extinction, $N^{(n)}_{k}(T^{(n)}_{0})$, has a modified geometric distribution, with probability $\mathbb {P}_{m}\left\{ T^{(n)}_{k} < T^{(n)}_{0}\right\}$ of reaching k prior to extinction, and return probability $\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\}$. The former gives the probability that $L^{(n)}_{k}(T^{(n)}_{0}) > 0$, whereas summing over the distribution of $N^{(n)}_{k}(T^{(n)}_{0})$ the probability that $L^{(n)}_{k}(T^{(n)}_{0}) \in [t,t+dt)$ is

$$\begin{aligned}&\mathbb {P}_{m}\left\{ T^{(n)}_{k}< T^{(n)}_{0}\right\} \left( 1-\mathbb {P}_{k}\left\{ T^{(n)}_{0}< T^{(n)}_{k+}\right\} \right) \\&\qquad \times \sum _{n_{k} = 1}^{\infty } \frac{(\lambda _{k} + \mu _{k})^{n_{k}}}{(n_{k}-1)!} t^{n_{k}-1} e^{-(\lambda _{k} + \mu _{k})t}\mathbb {P}_{k}\left\{ T^{(n)}_{0}< T^{(n)}_{k+}\right\} ^{n_{k}-1}\\&\quad = \mathbb {P}_{m}\left\{ T^{(n)}_{k}< T^{(n)}_{0}\right\} (\lambda _{k} + \mu _{k}) \left( 1-\mathbb {P}_{k}\left\{ T^{(n)}_{0}< T^{(n)}_{k+}\right\} \right) e^{-(\lambda _{k} + \mu _{k}) \left( 1-\mathbb {P}_{k}\left\{ T^{(n)}_{0} < T^{(n)}_{k+}\right\} \right) t}. \end{aligned}$$

The result then follows using the asymptotic approximations of the previous proof. $\square$

Proof of Proposition 9

Proceeding as previously, we have that

$$\begin{aligned} \mathbb {E}_{1}\left[ T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right] \sim \sum _{k = 1}^{\lfloor \nu n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} } \end{aligned}$$

and

$$\begin{aligned} \mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} = \frac{1}{\sum _{j = k}^{\lfloor \nu n\rfloor -1} e^{n(V^{(n)}(j)-V^{(n)}(k))}}. \end{aligned}$$

Recall, $\nu ' \ne \nu$ is the unique value such that $V(\nu ') = V(\nu )$. Then, for $k < \lfloor \nu ' n\rfloor$, $V\left( \frac{j}{n}\right) - V\left( \frac{k}{n}\right)$ is maximized at $j = k$, whereas for $\lfloor \nu ' n\rfloor< k < \lfloor \nu n\rfloor$, it is maximized at $j = \lfloor \nu n\rfloor - 1$.

We thus have

$$\begin{aligned} \mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} \sim 1-\frac{\mu \left( \frac{k}{n}\right) }{\lambda \left( \frac{k}{n}\right) } \end{aligned}$$

for $k < \lfloor \nu ' n\rfloor$, whereas for $\lfloor \nu ' n\rfloor< k < \lfloor \nu n\rfloor$,

$$\begin{aligned} \frac{1}{\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} } \sim \frac{\sqrt{\frac{\mu (\nu )\lambda \left( \frac{k}{n}\right) }{\lambda (\nu )\mu \left( \frac{k}{n}\right) }} e^{n\left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) - V\left( \frac{k}{n}\right) \right) }}{1-\frac{\lambda (\nu )}{\mu (\nu )}} \end{aligned}$$

We now split the sum over k at $\lfloor \nu ' n \rfloor$. Then,

$$\begin{aligned} \sum _{k = 1}^{\lfloor \nu ' n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} } \sim \sum _{k = 1}^{\lfloor \nu ' n\rfloor - 1} \frac{1}{\left( \lambda \left( \frac{k}{n}\right) -\mu \left( \frac{k}{n}\right) \right) i}, \end{aligned}$$

whereas

$$\begin{aligned} \lambda (0)-\mu (0) \le \lambda \left( \frac{k}{n}\right) -\mu \left( \frac{k}{n}\right) \le \lambda (\nu )-\mu (\nu ), \end{aligned}$$

so, as previously,

$$\begin{aligned} \frac{1}{\lambda (\nu )-\mu (\nu )}\le & {} \liminf _{n \rightarrow \infty } \frac{1}{\ln (\nu ' n)} \sum _{k = 1}^{\lfloor \nu ' n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor }< T^{(n)}_{k}\right\} }\\\le & {} \limsup _{n \rightarrow \infty } \frac{1}{\ln (\nu ' n)} \sum _{k = 1}^{\lfloor \nu ' n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} } \le \frac{1}{\lambda (0)-\mu (0)}. \end{aligned}$$

On the other hand, we observe that for $x \in [\nu ',\nu ]$, $V(\nu ) - V(x)$ is maximized at $x = \kappa$, so that, applying Proposition 1, we have

$$\begin{aligned}&\sum _{k = \lfloor \nu ' n\rfloor }^{\lfloor \nu n\rfloor - 1} \frac{1}{\lambda _{k}\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n\rfloor } < T^{(n)}_{k}\right\} } \sim \sum _{k = \lfloor \nu ' n\rfloor }^{\lfloor \nu n\rfloor - 1} \frac{e^{n\left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) - V\left( \frac{k}{n}\right) \right) }}{n \lambda \left( \frac{k}{n}\right) \left( \frac{k}{n}\right) \left( 1-\frac{\lambda (\nu )}{\mu (\nu )}\right) }\\&\quad \sim \sqrt{\frac{2\pi }{n\left( \frac{\mu '(\kappa )}{\mu (\kappa )}-\frac{\lambda '(\kappa )}{\lambda (\kappa )}\right) }} \frac{\sqrt{\frac{\mu (\nu )}{\lambda (\nu )}\frac{\lambda (\kappa )}{\mu (\kappa )}} e^{n\left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V(\kappa )\right) }}{\lambda (\kappa ) \kappa \left( 1-\frac{\lambda (\nu )}{\mu (\nu )}\right) }. \end{aligned}$$

The result follows on observing that $\lambda (\kappa ) = \mu (\kappa )$. $\square$

Proof of Proposition 10

As previously, we have that

$$\begin{aligned} \mathbb {E}_{\lfloor \nu n \rfloor }\left[ T^{(n)}_{\lfloor \kappa n \rfloor }\right] = \sum _{k = \lfloor \kappa n \rfloor + 1}^{\infty } \frac{1}{n (\lambda _{k}+\mu _{k})} \frac{\mathbb {P}_{\lfloor \nu n \rfloor } \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n \rfloor }\right\} }{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k+}\right\} }. \end{aligned}$$

Now,

$$\begin{aligned} \mathbb {P}_{\lfloor \nu n \rfloor } \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n \rfloor }\right\} = {\left\{ \begin{array}{ll} 1 &{} \text {if }\lfloor \kappa n \rfloor< k \le \lfloor \nu n \rfloor ,\text { and}\\ \frac{\sum _{j = \lfloor \kappa n \rfloor }^{\lfloor \nu n \rfloor -1} e^{n V^{(n)}(j)}}{\sum _{j = \lfloor \kappa n \rfloor }^{k-1} e^{n V^{(n)}(j)}} &{} \text {if }\lfloor \nu n \rfloor < k, \end{array}\right. } \end{aligned}$$

whereas

$$\begin{aligned} \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor }< T^{(n)}_{k+}\right\} = \frac{\mu _{k}}{\lambda _{k}+\mu _{k}} \mathbb {P}_{k-1}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k}\right\} \end{aligned}$$

and

$$\begin{aligned} \mathbb {P}_{k-1}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k}\right\} = \frac{e^{n V^{(n)}(k-1)}}{\sum _{j = \lfloor \kappa n \rfloor }^{k-1} e^{n V^{(n)}(j)}}. \end{aligned}$$

Thus, for $k > \lfloor \nu n \rfloor$,

$$\begin{aligned}&\frac{1}{n (\lambda _{k}+\mu _{k})} \frac{\mathbb {P}_{\lfloor \nu n \rfloor } \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n \rfloor }\right\} }{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k+}\right\} } = \frac{\sum _{j = \lfloor \kappa n \rfloor }^{\lfloor \nu n \rfloor -1} e^{n V^{(n)}(j)}}{n \mu _{k} e^{n V^{(n)}(k-1)}} \\&\quad \sim \frac{\sqrt{\frac{\mu (\nu )\lambda \left( \frac{k-1}{n}\right) }{\lambda (\nu )\mu \left( \frac{k-1}{n}\right) }}e^{n \left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V\left( \frac{k-1}{n}\right) \right) }}{\mu \left( \frac{k-1}{n}\right) k \left( 1-\frac{\lambda (\nu )}{\mu (\nu )}\right) } \end{aligned}$$

since V is minimized at $\kappa$. Moreover, as $\mu (x)$ and $\lambda (x)$ are, respectively, increasing and decreasing, the latter is bounded above by

$$\begin{aligned} \frac{e^{n \left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V\left( \frac{k-1}{n}\right) \right) }}{ (\mu (\nu )-\lambda (\nu ))\nu }. \end{aligned}$$

Further,

$$\begin{aligned}&V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V\left( \frac{k-1}{n}\right) = - V'(z)\left( \frac{k-1}{n}-\frac{\lfloor \nu n\rfloor }{n}\right) \\&\quad < -V'\left( \frac{\lfloor \nu n\rfloor }{n}\right) \left( \frac{k-1}{n}-\frac{\lfloor \nu n\rfloor }{n}L\right) \end{aligned}$$

for some $z \in \left[ \frac{\lfloor \nu n\rfloor }{n},\frac{k-1}{n}\right]$; the inequality follows since $V''(x) > 0$ for all x. Thus,

$$\begin{aligned} \sum _{k = \lfloor \nu n \rfloor }^{\infty } e^{n \left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V\left( \frac{k-1}{n}\right) \right) } < e^{V'\left( \frac{\lfloor \nu n\rfloor }{n}\right) } \sum _{k = 0}^{\infty } e^{-V'\left( \frac{\lfloor \nu n\rfloor }{n}\right) k} = \frac{e^{V'\left( \frac{\lfloor \nu n\rfloor }{n}\right) }}{1-e^{-V'\left( \frac{\lfloor \nu n\rfloor }{n}\right) }}, \end{aligned}$$

since $\nu > \kappa$ and $V'(\kappa ) = 0$. In particular, we see that the sum

$$\begin{aligned} \sum _{k = \lfloor \nu n \rfloor + 1}^{\infty } \frac{1}{n (\lambda _{k}+\mu _{k})} \frac{\mathbb {P}_{\lfloor \nu n \rfloor } \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \kappa n \rfloor }\right\} }{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k+}\right\} } \end{aligned}$$

is bounded above.

Finally, consider

$$\begin{aligned} \sum _{k = \lfloor \kappa n \rfloor + 1}^{\lfloor \nu n \rfloor } \frac{1}{n \mu _{k} \mathbb {P}_{k-1}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k}\right\} }. \end{aligned}$$

Arguing as above,

$$\begin{aligned} \mathbb {P}_{k-1}\left\{ T^{(n)}_{\lfloor \kappa n \rfloor } < T^{(n)}_{k}\right\} \sim \frac{1}{1 - \frac{\lambda \left( \frac{k-1}{n}\right) }{\mu \left( \frac{k-1}{n}\right) }} \sim \frac{1}{1 - \frac{\lambda \left( \frac{k}{n}\right) }{\mu \left( \frac{k}{n}\right) }}, \end{aligned}$$

and, proceeding as in Proposition 6, one can show that

$$\begin{aligned} \sum _{k = \lfloor \kappa n \rfloor + 1}^{\lfloor \nu n \rfloor } \frac{1}{\left( \lambda \left( \frac{k}{n}\right) -\mu \left( \frac{k}{n}\right) \right) k} \sim - \frac{1}{(\lambda '(\kappa )-\mu '(\kappa ))\kappa }\ln {n}. \end{aligned}$$

$\square$

Proof of Proposition 11

To begin, we decompose the expectation according to whether, starting from $\lfloor \nu n \rfloor$, the next event is a birth or a death:

$$\begin{aligned}&\mathbb {E}_{\lfloor \nu n \rfloor } \left[ T^{(n)}_{\lfloor \nu n \rfloor +} \Big \vert T^{(n)}_{\lfloor \nu n \rfloor +}< T^{(n)}_{0}\right] \\&\quad = \frac{\lambda _{\lfloor \nu n \rfloor }}{\lambda _{\lfloor \nu n \rfloor } + \mu _{\lfloor \nu n \rfloor }} \mathbb {E}_{\lfloor \nu n \rfloor + 1} \left[ T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right] \\&\qquad + \frac{\mu _{\lfloor \nu n \rfloor }}{\lambda _{\lfloor \nu n \rfloor } + \mu _{\lfloor \nu n \rfloor }} \mathbb {E}_{\lfloor \nu n \rfloor - 1} \left[ T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right] \\&\quad = \frac{\lambda _{\lfloor \nu n \rfloor }}{\lambda _{\lfloor \nu n \rfloor } + \mu _{\lfloor \nu n \rfloor }} \sum _{k = \lfloor \nu n \rfloor + 1}^{\infty } \frac{\mathbb {P}_{\lfloor \nu n \rfloor + 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }{n \mu _{k} \mathbb {P}_{k - 1} \left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }\\&\qquad + \frac{\mu _{\lfloor \nu n \rfloor }}{\lambda _{\lfloor \nu n \rfloor } + \mu _{\lfloor \nu n \rfloor }} \sum _{k=1}^{\lfloor \nu n \rfloor - 1} \frac{\mathbb {P}_{\lfloor \nu n \rfloor - 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }{n \lambda _{k} \mathbb {P}_{k + 1} \left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} }. \end{aligned}$$

For the first sum, we observe that for any $k \ge \lfloor \nu n \rfloor$,

$$\begin{aligned} \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} = 1, \end{aligned}$$

and we may thus replace the conditional probabilities with the unconditional ones. Then, using (5),

$$\begin{aligned} \frac{\mathbb {P}_{\lfloor \nu n \rfloor + 1}\left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor }\right\} }{\mathbb {P}_{k - 1}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{k} \right\} } = e^{n \left( V^{(n)}(\lfloor \nu n \rfloor ) - V^{(n)}(k)\right) }, \end{aligned}$$

so that, using Lemma 1, the first sum is asymptotic to

$$\begin{aligned} \frac{\mu (\nu )}{\mu (\nu ) + \lambda (\nu )} \frac{1}{(\mu (\nu ) - \lambda (\nu ))\nu }. \end{aligned}$$

For the second sum, we observe that

$$\begin{aligned} \left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{k}\right\} \cap \left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} = \left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{k}\right\} , \end{aligned}$$

whereas

$$\begin{aligned} \mathbb {P}_{\lfloor \nu n \rfloor - 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor }, T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} = \mathbb {P}_{\lfloor \nu n \rfloor - 1}\left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor }\right\} \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} , \end{aligned}$$

so that, applying Bayes’ theorem,

$$\begin{aligned} \frac{\mathbb {P}_{\lfloor \nu n \rfloor - 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor } \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }{\mathbb {P}_{k + 1} \left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{k} \Big \vert T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} } = \frac{\mathbb {P}_{\lfloor \nu n \rfloor - 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor }\right\} \mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }{\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{k}\right\} }. \end{aligned}$$

Again, from (5), we see that

$$\begin{aligned} \frac{\mathbb {P}_{\lfloor \nu n \rfloor - 1} \left\{ T^{(n)}_{k}< T^{(n)}_{\lfloor \nu n \rfloor }\right\} }{\mathbb {P}_{k+1}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{k}\right\} } = e^{n V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))}, \end{aligned}$$

so this sum reduces to

$$\begin{aligned} \sum _{k=1}^{\lfloor \nu n \rfloor - 1} \frac{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} }{n \lambda _{k}} e^{n V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))}. \end{aligned}$$

To evaluate the sum, it is useful to consider it in two pieces. To do so, we first re-introduce $\nu ' < \kappa$ such that $V(\nu ') = V(\nu )$, and then consider

$$\begin{aligned}&\sum _{k=1}^{\lfloor \nu ' n \rfloor - 1} \frac{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor }< T^{(n)}_{0}\right\} }{n \lambda _{k}} e^{n V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))}\\&\quad + \sum _{k=\lfloor \nu ' n \rfloor }^{\lfloor \nu n \rfloor - 1} \frac{\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} }{n \lambda _{k}} e^{n V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))}. \end{aligned}$$

For the former, $V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))$ is maximized at $k = \lfloor \nu ' n \rfloor -1$, whereas $\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\}$ us bounded above by 1. Using Lemma 1, the first sum is asymptotically bounded above by

$$\begin{aligned} \frac{1}{\left( \lambda \left( \frac{\lfloor \nu ' n \rfloor -1}{n}\right) - \mu \left( \frac{\lfloor \nu ' n \rfloor -1}{n}\right) \right) (\lfloor \nu ' n \rfloor -1)} \sqrt{\frac{\mu \left( \frac{\lfloor \nu n \rfloor -1}{n}\right) \lambda \left( \frac{\lfloor \nu ' n \rfloor -1}{n}\right) }{\lambda \left( \frac{\lfloor \nu n \rfloor -1}{n}\right) \mu \left( \frac{\lfloor \nu ' n \rfloor -1}{n}\right) }}. \end{aligned}$$

For the second piece, we note that for $\lfloor \nu ' n \rfloor \le k < \lfloor \nu ' n \rfloor$, $\mathbb {P}_{k}\left\{ T^{(n)}_{\lfloor \nu n \rfloor } < T^{(n)}_{0}\right\} \sim 1$, whereas $V^{(n)}(\lfloor \nu n \rfloor -1) - V^{(n)}(k))$ is maximized at $\lfloor \kappa n \rfloor$, so appealing to Lemma 1, it is asymptotically equivalent to

$$\begin{aligned} \sqrt{\frac{2\pi }{n\left( \frac{\mu '(\kappa )}{\mu (\kappa )}-\frac{\lambda '(\kappa )}{\lambda (\kappa )}\right) } \frac{\mu (\nu )}{\lambda (\nu )}} \frac{e^{n\left( V\left( \frac{\lfloor \nu n\rfloor }{n}\right) -V(\kappa )\right) }}{\lambda (\kappa ) \kappa } \end{aligned}$$

The result follows. $\square$