A comparison of mutation and amplification-driven resistance mechanisms and their impacts on tumor recurrence

Abstract

Tumor recurrence, driven by the evolution of drug resistance is a major barrier to therapeutic success in cancer. Tumor drug resistance is often caused by genetic alterations such as point mutation, which refers to the modification of a single genomic base pair, or gene amplification, which refers to the duplication of a region of DNA that contains a gene. These mechanisms typically confer varying degrees of resistance, and they tend to occur at vastly different frequencies. Here we investigate the dependence of tumor recurrence dynamics on these mechanisms of resistance, using stochastic multi-type branching process models. We derive tumor extinction probabilities and deterministic estimates for the tumor recurrence time, defined as the time when an initially drug sensitive tumor surpasses its original size after developing resistance. For models of amplification-driven and mutation-driven resistance, we prove law of large numbers results regarding the convergence of the stochastic recurrence times to their mean. Additionally, we prove sufficient and necessary conditions for a tumor to escape extinction under the gene amplification model, discuss behavior under biologically relevant parameters, and compare the recurrence time and tumor composition in the mutation and amplification models both analytically and using simulations. In comparing these mechanisms, we find that the ratio between recurrence times driven by amplification versus mutation depends linearly on the number of amplification events required to acquire the same degree of resistance as a mutation event, and we find that the relative frequency of amplification and mutation events plays a key role in determining the mechanism under which recurrence is more rapid for any specific system. In the amplification-driven resistance model, we also observe that increasing drug concentration leads to a stronger initial reduction in tumor burden, but that the eventual recurrent tumor population is less heterogeneous, more aggressive and harbors higher levels of drug-resistance.

Appendix

1.1 Restriction of mutation and amplification to cell division

To restrict mutations to arise only during cell division, let ${\tilde{\mu }}_n$ be the probability that a cell undergoing division produces a mutated cell. Then the rate at which mutated cells arise in the population is $r_s {\tilde{\mu }}_n X_s(t)$. The bounds in Proposition 4 then become

$$\begin{aligned} a_n&= \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log \left( \frac{ -\lambda _s}{r_s} \right) \right) ,\\ A_n&= \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log \left( \frac{{\lambda _m - \lambda _s}}{r_s} \right) \right) . \end{aligned}$$

However, it is still the case that ${\tilde{u}}_n \rightarrow -\frac{\lambda _s }{\lambda _m }\alpha$ as $n \rightarrow \infty$ so our estimated recurrence time in the large population limit does not change and the convergence in Theorem 2 will still hold.

Similarly, to restrict amplification events to only occur during cell division, let ${\tilde{\nu }}_n$ be the probability that a cell undergoing division produces an amplified cell. Then the rate at which amplified cells with k copies of the gene arise is $r_{k-1} {\tilde{\nu }}_n X_{k-1}(t)$. For the proof of Proposition 1, $s_{n,k}$ will instead be of the form

$$\begin{aligned} s_{n,k} = {\left\{ \begin{array}{ll} \displaystyle {\frac{d_k}{r_k}} + O({\tilde{\nu }}_n) &{} \text { if } k' \le k \le M,\\ \displaystyle {1 - \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} \left( \prod _{i=k}^{k'-1} r_i\right) {\tilde{\nu }}_n^\ell + O ({\tilde{\nu }}_n^{\ell +1})} &{} \text { if } 2< k < k'. \end{array}\right. } \end{aligned}$$

Thus, we will have an extinction probability of

$$\begin{aligned} q = \lim _{n \rightarrow \infty } \left( s_{n,2}\right) ^n = {\left\{ \begin{array}{ll} 0 &{} \text { if } 0< \beta < \frac{1}{k'-2},\\ \displaystyle {\exp \left( -\frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} \left( \prod _{i=k}^{k'-1} r_i\right) \right) } &{} \text { if } \beta = \frac{1}{k'-2}, \\ 1 &{} \text { if } \beta > \frac{1}{k'-2}. \end{array}\right. } \end{aligned}$$

That is, the only difference occurs when $\beta = \frac{1}{k'-2}$.

For Proposition 2, the bounds become

$$\begin{aligned} b_n&= \frac{1}{\lambda _M} \left[ -\lambda _2 \beta (M-2) - \frac{1}{t_n} \log \left[ \frac{ \prod _{i=2}^M -r_i }{D^{M-2}{\tilde{P}}_{M,M}} \left( 1 - \frac{\lambda _M}{\lambda _2} \right) \right] \right] ,\\ B_n&= \frac{1}{\lambda _M} \left[ -\lambda _2 \beta (M-2) - \frac{1}{t_n} \log \left[ \frac{ \prod _{i=2}^M -r_i}{D^{M-2}{\tilde{P}}_{M,M}} \right] \right] . \end{aligned}$$

Hence we still have that ${\tilde{v}}_n \rightarrow -\frac{\lambda _2}{\lambda _M} \beta (M-2)$ as $n \rightarrow \infty$. As in the mutation model, our estimated recurrence time is unchanged and our convergence in Theorem 1 will still hold as well.

1.2 Proof of lemma 1

We know that the first moment of the process generated by a single sensitive cell is $m_1^{s}(t) = e^{\lambda _s t}$. Notice then that

$$\begin{aligned} \phi _s(zt_n) = n m^s_1(zt_n) = ne^{\lambda _s z t_n} = ne^{-z \log n } = n^{1-z}. \end{aligned}$$

By Equation 5 in Chapter III Part 4 of Athreya and Ney (1972), we know that the second moment of the process generated by a single sensitive cell is given by

$$\begin{aligned} m_2^s(t) = u''(1) \frac{e^{2\lambda _s t} - e^{\lambda _s t}}{\lambda _s} + e^{\lambda _s t}, \end{aligned}$$

where $u(x) = d_s + r_s x^2 -x(r_s+d_s)$. Then

$$\begin{aligned} {{\,\textrm{Var}\,}}[X_s(t)]&= n\left( m_2^s(t) - \left( m_1^s(t)\right) ^2\right) = 2 n r_s \frac{e^{2\lambda _s t} - e^{\lambda _s t}}{\lambda _s} + n e^{\lambda _s t} - ne^{2 \lambda _s t}. \end{aligned}$$

Thus,

$$\begin{aligned} \psi _s(zt_n)&= \textrm{Var}[X_s(zt_n)]\\&= 2 n r_s \frac{e^{2\lambda _s zt_n} - e^{\lambda _s zt_n}}{\lambda _s} + n e^{\lambda _s zt_n} - n e^{2 \lambda _s zt_n}\\&=2 r_s \frac{n^{1-2z} - n^{1-z}}{\lambda _s} + n^{1-z} - n^{1-2z}\\&= -\left( n^{1-z} - n^{1-2z}\right) \left( \frac{2r_s}{\lambda _s} - 1\right) \\&= n^{1-z} (1-n^{-z}) \frac{2r_s - r_s +d_s}{-\lambda _s}\\&= n^{1-z} \frac{r_s+d_s}{-\lambda _s} (1-n^{-z}) , \end{aligned}$$

as desired.

By definition, $\mathbb {E}[X_m(t)]$ satisfies

$$\begin{aligned} \frac{d}{dt} \mathbb {E}[X_m(t)] = \lambda _m \mathbb {E}[X_m(t)] + n^{-\alpha } \mathbb {E}[X_s(t)], \end{aligned}$$

with initial condition $\mathbb {E}[X_m(t)] = 0$. Notice then that $\frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m t}}-{e^{\lambda _s t}}\right)$ satisfies the differential equation and initial condition:

$$\begin{aligned} \frac{d}{dt}\left[ \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m t}}-{e^{\lambda _s t}}\right) \right]&= \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s}\left( {\lambda _m e^{\lambda _m t}}-{\lambda _s e^{\lambda _s t}}\right) \\&= \frac{n^{1-\alpha }}{\lambda _m - \lambda _s}\left( {\lambda _m e^{\lambda _m t}}-{\lambda _m e^{\lambda _s t}}\right) +\frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {(\lambda _m-\lambda _s) e^{\lambda _s t}}\right) \\&= \lambda _m \left( \frac{n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m t}}-{e^{\lambda _s t}}\right) \right) + n^{-\alpha } \mathbb {E}[X_s(t)]. \end{aligned}$$

So indeed, $\mathbb {E}[X_m(t)] = \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m t}}-{e^{\lambda _s t}}\right) .$ Thus,

$$\begin{aligned} \phi _m(zt_n) = \mathbb {E}[X_m(zt_n)] = \frac{n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m zt_n}}-{e^{\lambda _s zt_n}}\right) = \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {n^{-\frac{\lambda _m}{\lambda _s} z }}-n^{-z}\right) . \end{aligned}$$

To calculate ${{\,\textrm{Var}\,}}[X_m(t)]$, let B(t) represent the population size at t of a process generated by a single resistant cell, and let N(t) be the number of resistant cells generated by the sensitive cells before time t. Then we have that

$$\begin{aligned}&\mathbb {E}[X_m(t)^2\mid (X_s(s))_{s \le t}] = \sum _{k=0}^\infty \mathbb {P}(N(t)\\&= k\mid (X_s(s))_{s \le t})\cdot \mathbb {E}[X_m(t)^2 \mid N(t) = k, (X_s(s))_{s \le t}]\\&= \sum _{k=0}^\infty \mathbb {P}(N(t) = k\mid (X_s(s))_{s \le t})\cdot \mathbb {E}\left[ \left( \sum _{i = 1}^{N(t)} B(t-\tau _i)\right) ^{\!\!2} \,\bigg | \,N(t) = k, (X_s(s))_{s \le t}\right] , \end{aligned}$$

where $\tau _i$ represents the time that the i-th resistant cell is generated. Notice that

$$\begin{aligned} P(N(t) = k \mid (X_s(s))_{s \le t}) =\frac{\rho ^k e^{-\rho }}{k!}, \end{aligned}$$

where $\rho = \int _0^t X_s(s)\mu \, ds$. Let us consider the conditional expectation on its own. Conditioned on $N(t) = k$, the $\tau _i$’s are distributed as order statistics of $\{T_i\mid 1 \le i \le k\}$, where the $T_i$’s are i.i.d. random variables with PDF

$$\begin{aligned} f(s) = \frac{X_s(s)\mu _n}{\int _0^t X_s(s)\mu _n \,ds} = \frac{1}{\rho } X_s(s)\mu _n, \end{aligned}$$

supported on $s \in (0,t]$. We have that

$$\begin{aligned} \mathbb {E}\left[ \left( \sum _{i = 1}^{N(t)} B(t-\tau _i)\right) ^{\!\!2} \,\bigg | \,N(t) = k, (X_s(s))_{s\le t}\right]&\overset{(d)}{=} \mathbb {E}\left[ \left( \sum _{i = 1}^{k} B(t-T_{(i)})\right) ^{\!\!2}\, \bigg | (X_s(s))_{s \le t} \right] , \end{aligned}$$

and we can reorder terms as necessary and expand the square to write

$$\begin{aligned} \mathbb {E}\left[ \left( \sum _{i = 1}^{k} B(t-T_i)\right) ^{\!\!2} \,\bigg | (X_s(s))_{s \le t}\right]&=\sum _{i = 1}^k \mathbb {E}\left[ B(t-T_i)^2 \,\bigg | (X_s(s))_{s \le t}\right] \\&\quad + 2\sum _{1 \le i < j \le k} \mathbb {E}\left[ B(t-T_i) B(t-T_j) \,\bigg |(X_s(s))_{s \le t}\right] . \end{aligned}$$

In particular, we have that

$$\begin{aligned}{} & {} \mathbb {E}[X_m(t)^2\mid (X_s(s))_{s \le t}] \overset{(d)}{=}\sum _{k=0}^\infty \mathbb {P}(N(t) = k\mid (X_s(s))_{s \le t}) \left( \sum _{i = 1}^k \mathbb {E}\left[ B(t-T_i)^2 \,\bigg | (X_s(s))_{s \le t}\right] \right) \nonumber \\{} & {} \quad +\sum _{k=0}^\infty \mathbb {P}(N(t) = k\mid (X_s(s))_{s \le t}) \left( 2\sum _{1 \le i < j \le k} \mathbb {E}\left[ B(t-T_i) B(t-T_j) \,\bigg |(X_s(s))_{s \le t}\right] \right) . \end{aligned}$$

(2)

Notice that $\mathbb {E}\left[ B(t-T_i)^2 \,\bigg |(X_s(s))_{s \le t}\right] = \int _0^t f(s) \mathbb {E}\left[ B(t-s)^2 \right] ds.$ Following the process above for the sensitive process, we can get that for $s \le t$,

$$\begin{aligned} \mathbb {E}[ B(t-s)^2]&= m_2^{B}(t-s)=\frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)}. \end{aligned}$$

We have then that

$$\begin{aligned} \mathbb {E}\left[ B(t-T_i)^2 \,\bigg |(X_s(s))_{s \le t}\right]&= \int _0^t\frac{1}{\rho } X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds. \end{aligned}$$

Then

$$\begin{aligned}{} & {} \sum _{i = 1}^k \mathbb {E}\left[ B(t-T_i)^2 \,\bigg | (X_s(s))_{s \le t}\right] \nonumber \\{} & {} \quad = k \cdot \frac{1}{\rho } \int _0^t X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds. \end{aligned}$$

(3)

Similarly, by independence

$$\begin{aligned} \mathbb {E}\left[ B(t-T_i) B(t-T_j) \,\bigg | (X_s(s))_{s \le t}\right]&= \mathbb {E}\left[ B(t-T_i) \,\bigg | (X_s(s))_{s \le t}\right] \mathbb {E}\left[ B(t-T_j) \,\bigg | (X_s(s))_{s \le t}\right] \\&= \mathbb {E}\left[ B(t-T_i)\,\bigg | (X_s(s))_{s \le t}\right] ^2\\&= \left( \int _{-\infty }^\infty f(s) \mathbb {E}\left[ B(t-s) \right] ds\right) ^2\\&=\frac{1}{\rho ^2 } \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2. \end{aligned}$$

So

$$\begin{aligned} 2\sum _{1 \le i < j \le k} \mathbb {E}\left[ B(t-T_i) B(t-T_j) \,\bigg |(X_s(s))_{s \le t}\right] = k(k-1) \cdot \frac{1}{\rho ^2 } \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2.\nonumber \\ \end{aligned}$$

(4)

Then inserting (3) and (4) into (2) gets us that

$$\begin{aligned}&\mathbb {E}[X_m(t)^2 \mid (X_s(s))_{s \le t}]\\&\quad \overset{(d)}{=} \sum _{k=0}^\infty \frac{\rho ^k e^{-\rho }}{k!} \left( k \cdot \frac{1}{\rho } \int _0^t X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds\right) \\&\qquad +\sum _{k=0}^\infty \frac{\rho ^k e^{-\rho }}{k!} \left( k(k-1) \cdot \frac{1}{\rho ^2 } \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2\right) \\&\quad = \int _0^t X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds \\&\qquad + \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2. \end{aligned}$$

Repeating the ideas from the above calculation, we can calculate

$$\begin{aligned}&\mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]^2 {=} \left( \sum _{k=0}^\infty \mathbb {P}(N(t) {=} k\mid (X_s(s))_{s \le t}) \cdot \mathbb {E}\left[ \sum _{i {=} 1}^{N(t)} B(t-\tau _i)\,\bigg | N(t) = k, (X_s(s))_{s \le t}\right] \right) ^2\\&\quad \overset{(d)}{=}\left( \sum _{k=0}^\infty \frac{\rho ^k e^{-\rho }}{k!} \cdot \mathbb {E}\left[ \sum _{i = 1}^{k} B(t-T_i)\,\bigg | \, (X_s(s))_{s \le t}\right] \right) ^2\\&\quad = \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2. \end{aligned}$$

Then we have that

$$\begin{aligned} {{\,\textrm{Var}\,}}\left( X_m(t) \bigg |(X_s(s))_{s\le t}\right)&=\mathbb {E}[X_m(t)^2\mid (X_s(s))_{s \le t}] - \mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]^2\\&= \int _0^t X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds. \end{aligned}$$

Taking the expectation and then interchanging the expectation of the integral, we have that

$$\begin{aligned}&\mathbb {E}\left[ {{\,\textrm{Var}\,}}\left( X_m(t) \bigg |(X_s(s))_{s\le t}\right) \right] = \mathbb {E}\left[ \int _0^t X_s(s)\mu _n \left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds\right] \\&\quad = \mu _n \int _0^t ne^{\lambda _s(s)}\left( \frac{2r_m}{\lambda _m} e^{2\lambda _m(t-s)} + \left( \frac{-d_m-r_m}{\lambda _m}\right) e^{\lambda _m(t-s)} \right) ds\\&\quad = n\mu _n \left[ \frac{2r_m e^{2\lambda _m t + (\lambda _s -2\lambda _m)s}}{\lambda _m(\lambda _s -2\lambda _m)} + \frac{(-d_m-r_m)e^{\lambda _m t + (\lambda _s-\lambda _m)s}}{\lambda _m (\lambda _s-\lambda _m)} \right] _{s=0}^t\\&\quad = \frac{n\mu _n}{\lambda _m} \left( \frac{2r_m e^{\lambda _s t}}{\lambda _s -2\lambda _m} + \frac{(-d_m-r_m)e^{\lambda _s t}}{ \lambda _s-\lambda _m} - \frac{2r_m e^{2\lambda _m t}}{\lambda _s -2\lambda _m} - \frac{(-d_m-r_m)e^{\lambda _m t}}{\lambda _s-\lambda _m} \right) . \end{aligned}$$

Let $g(t) = \mathbb {E}\left[ {{\,\textrm{Var}\,}}\left( X_m(t) \bigg |(X_s(s))_{s\le t}\right) \right]$. Notice that since $\lambda _s< 0 < \lambda _m$, we have that

$$\begin{aligned} g(zt_n)&\sim n \mu _n \frac{2r_m e^{2\lambda _m zt_n}}{\lambda _m(2\lambda _m -\lambda _s)} = \frac{2r_m}{\lambda _m(2\lambda _m -\lambda _s)} n^{1-\alpha -2 \lambda _mz/\lambda _s}. \end{aligned}$$

The Law of Total Variance gives that

$$\begin{aligned} {{\,\textrm{Var}\,}}[X_m(t)] = g(t) + h(t), \end{aligned}$$

where $h(t) = {{\,\textrm{Var}\,}}[\mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]]$. Define $h_1(t) = \mathbb {E}[\mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]^2]$ and define $h_2(t) = \mathbb {E}[\mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]]^2$. Then $h(t) = h_1(t) - h_2(t)$. Notice that

$$\begin{aligned} h_2(t)&= \mathbb {E}[\mathbb {E}[X_m(t) \mid (X_s(s))_{s\le t}]]^2= \mathbb {E}[X_m(t)]^2= \left( \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( {e^{\lambda _m t}}-{e^{\lambda _s t}}\right) \right) ^2. \end{aligned}$$

We also have that

$$\begin{aligned} h_1(t)&=\mathbb {E}\left[ \left( \int _0^t X_s(s)\mu _n e^{\lambda _m(t-s)} \,ds\right) ^2\right] \\&=\mu _n^2 \int _0^t\int _0^t \mathbb {E}[X_s(s)X_s(y)] e^{\lambda _m(t-s)} e^{\lambda _m(t-y)}\, dsdy\\&= \mu _n^2 \int _0^t\int _0^y \mathbb {E}[X_s(s)\mathbb {E}[X_s(y)|X_s(s)]] e^{\lambda _m(t-s)} e^{\lambda _m(t-y)}\, dsdy\\&\quad + \mu _n^2 \int _0^t\int _y^t \mathbb {E}[\mathbb {E}[X_s(s)|X_s(y)]X_s(y)] e^{\lambda _m(t-s)} e^{\lambda _m(t-y)}\, dsdy\\&= 2\mu _n^2 \int _0^t\int _y^t \mathbb {E}[X_s(y)^2]e^{\lambda _s(s-y)} e^{\lambda _m(t-s)} e^{\lambda _m(t-y)}\, dsdy. \end{aligned}$$

by symmetry. Note that $\mathbb {E}[X_s(t)^2] = {{\,\textrm{Var}\,}}[X_s(t)] + \mathbb {E}[X_s(t)]^2 =\left( n^2+\frac{r_s+d_s}{\lambda _s}n\right) e^{2\lambda _s t} + \frac{-r_s-d_s}{\lambda _s}ne^{\lambda _s t}.$ Then

$$\begin{aligned}&\int _0^t\int _y^t \mathbb {E}[X_s(y)^2]e^{\lambda _s(s-y)} e^{\lambda _m(t-s)} e^{\lambda _m(t-y)}\, dsdy\\&\quad = \int _0^t\mathbb {E}[X_s(y)^2]e^{\lambda _m(t-y)}\int _y^t e^{\lambda _s(s-y)} e^{\lambda _m(t-s)} \, dsdy\\&\quad =\int _0^t\left( \left( n^2+\frac{r_s+d_s}{\lambda _s}n\right) e^{2\lambda _s y} + \frac{-r_s-d_s}{\lambda _s}ne^{\lambda _s y}\right) \frac{e^{\lambda _m(t-y)}}{\lambda _s-\lambda _m}\left( e^{\lambda _s(t-y)} -e^{\lambda _m(t-y)} \right) \, dy\\&\quad = \int _0^t \left( \frac{n^2}{\lambda _s-\lambda _m}+\frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)}\right) e^{(\lambda _s+\lambda _m)t + (\lambda _s-\lambda _m)y} \\&\qquad + \left( \frac{n^2}{\lambda _m-\lambda _s}+\frac{(r_s+d_s)n}{\lambda _s(\lambda _m-\lambda _s)}\right) e^{2\lambda _mt + (2\lambda _s-2\lambda _m)y} \\&\qquad + \frac{(r_s+d_s)n}{\lambda _s(\lambda _m-\lambda _s)} e^{(\lambda _s+\lambda _m)t - \lambda _m y} + \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)} e^{2\lambda _mt + (\lambda _s-2\lambda _m)y} \, dy\\&\quad =\frac{1}{(\lambda _m-\lambda _s)^2}\left( \left( n^2+\frac{(r_s+d_s)n}{\lambda _s}\right) e^{2 \lambda _s t} + \left( \frac{n^2}{-2}+\frac{(r_s+d_s)n}{-2\lambda _s}\right) e^{2\lambda _s t}\right. \\&\qquad \left. -\left( n^2+\frac{(r_s+d_s)n}{\lambda _s}\right) e^{(\lambda _s+\lambda _m)t}+ \left( \frac{n^2}{2}+\frac{(r_s+d_s)n}{2\lambda _s}\right) e^{2\lambda _mt}\right) + \frac{(r_s+d_s)n}{\lambda _s(-\lambda _m)(\lambda _m-\lambda _s)} e^{\lambda _s t} \\&\qquad + \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{\lambda _s t} + \frac{(r_s+d_s)n}{\lambda _s\lambda _m(\lambda _m-\lambda _s)} e^{(\lambda _s+\lambda _m)t} \\&\qquad -\frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{2\lambda _mt }. \end{aligned}$$

Then

$$\begin{aligned} h_1(t)&= \left( {n^2}+\frac{(r_s+d_s)n}{\lambda _s}\right) \left( \frac{\mu _n (e^{\lambda _m t}- e^{\lambda _s t})}{\lambda _m-\lambda _s}\right) ^2 + 2\mu _n^2\left[ \frac{(r_s+d_s)n}{\lambda _s(-\lambda _m)(\lambda _m-\lambda _s)} e^{\lambda _s t} \right. \\&\quad \left. + \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{\lambda _s t} + \frac{(r_s+d_s)n}{\lambda _s\lambda _m(\lambda _m-\lambda _s)} e^{(\lambda _s+\lambda _m)t} - \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{2\lambda _mt }\right] \\&= h_2(t) +\left( \frac{(r_s+d_s)n}{\lambda _s}\right) \left( \frac{\mu _n (e^{\lambda _m t}- e^{\lambda _s t})}{\lambda _m-\lambda _s}\right) ^2 + 2\mu _n^2\left[ \frac{(r_s+d_s)n}{\lambda _s(-\lambda _m)(\lambda _m-\lambda _s)} e^{\lambda _s t} \right. \\&\quad \left. + \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{\lambda _s t} + \frac{(r_s+d_s)n}{\lambda _s\lambda _m(\lambda _m-\lambda _s)} e^{(\lambda _s+\lambda _m)t} - \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{2\lambda _mt }\right] . \end{aligned}$$

Then

$$\begin{aligned} h(zt_n)&= \left( \frac{(r_s+d_s)n}{\lambda _s}\right) \left( \frac{n^{-\alpha } (e^{\lambda _m zt_n}- e^{\lambda _s zt_n})}{\lambda _m-\lambda _s}\right) ^2 + 2n^{-2\alpha }\left[ \frac{(r_s+d_s)n}{\lambda _s(-\lambda _m)(\lambda _m-\lambda _s)} e^{\lambda _s zt_n} \right. \\&\quad \left. + \frac{(r_s+d_s)n}{\lambda _s\lambda _m(\lambda _m-\lambda _s)} e^{(\lambda _s+\lambda _m)zt_n} - \frac{(r_s+d_s)n}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} e^{2\lambda _m zt_n }\right] \\&= n^{1-2\alpha }\left( \frac{(r_s+d_s)}{\lambda _s}\right) \left( \frac{ (n^{-\lambda _m z/\lambda _2}- n^{-z})}{\lambda _m-\lambda _s}\right) ^2 + 2n^{1-2\alpha }\left[ \frac{(r_s+d_s)}{\lambda _s(-\lambda _m)(\lambda _m-\lambda _s)} n^{-z} \right. \\&\quad \left. + \frac{(r_s+d_s)}{\lambda _s\lambda _m(\lambda _m-\lambda _s)} n^{-(1 +\lambda _m/\lambda _2)z} - \frac{(r_s+d_s)}{\lambda _s(\lambda _s-\lambda _m)(\lambda _s-2\lambda _m)} n^{-2\lambda _m z/\lambda _s}\right] \\&= O(n^{1-2\alpha -2\lambda _mz/\lambda _s}), \end{aligned}$$

because $-\lambda _mz/\lambda _s > 0$. Then since $g(zt_n) \sim \frac{2r_m}{\lambda _m(2\lambda _m-\lambda _s)}n^{1-\alpha -2\lambda _mz/\lambda _s}$, we have that

$$\begin{aligned} \psi _m(zt_n)&= {{\,\textrm{Var}\,}}[X_m(zt_n)]\\&= g(zt_n) + h(zt_n)\\&\sim \frac{2r_m}{\lambda _m(2\lambda _m-\lambda _s)}n^{1-\alpha -2\lambda _mz/\lambda _s}, \end{aligned}$$

as desired.

1.3 Proof of lemma 2

Proof

We will prove this by induction. Let’s start by proving the base case: $k=3$. We need to show that

$$\begin{aligned} \mathbb {E}[X_3(t)] = n^{1-\beta } \left( \frac{e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{e^{\lambda _3 t}}{\lambda _2 - \lambda _3} \right) . \end{aligned}$$

Because $X_3(t)$ is a birth-death process with mutation, its mean is governed by the following ODE:

$$\begin{aligned} \frac{d}{dt} \mathbb {E}[X_3(t)] = \lambda _3 \mathbb {E}[X_3(t)] + n^{-\beta } \mathbb {E}[X_2(t)]. \end{aligned}$$

Sure enough, plugging $\mathbb {E}[X_2(t)]$ and $\mathbb {E}[X_3(t)]$ into the above ODE yields

$$\begin{aligned} \frac{d}{dt} \left[ n^{1-\beta } \left( \frac{e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{e^{\lambda _3 t}}{\lambda _2 - \lambda _3} \right) \right]&= \lambda _3 n^{1-\beta } \left( \frac{e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{e^{\lambda _3 t}}{\lambda _2 - \lambda _3} \right) + n^{-\beta } n e^{\lambda _2 t}\\ n^{1-\beta } \left( \frac{\lambda _2 e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{\lambda _3 e^{\lambda _3 t}}{\lambda _2 - \lambda _3} \right)&= n^{1-\beta } \left( \frac{\lambda _3 e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{\lambda _3 e^{\lambda _3 t}}{\lambda _2 - \lambda _3} + \frac{(\lambda _2 - \lambda _3)e^{\lambda _2 t}}{\lambda _2 - \lambda _3} \right) , \end{aligned}$$

which is clearly true. So we do indeed have $\mathbb {E}[X_3(t)] = n^{1-\beta } \left( \frac{e^{\lambda _2 t}}{\lambda _2 - \lambda _3} - \frac{e^{\lambda _3 t}}{\lambda _2 - \lambda _3} \right)$, as desired. Now assume

$$\begin{aligned} \mathbb {E}[X_\ell (t)] = n^{1-(\ell -2)\beta } (-1)^l S_\ell (t). \end{aligned}$$

We want to show that

$$\begin{aligned} \mathbb {E}[X_{\ell +1}(t)] = n^{1-(\ell -1)\beta } (-1)^{\ell +1} S_{\ell +1}(t). \end{aligned}$$

Because $X_{\ell +1}(t)$ is a birth-death process with mutation, its mean is governed by the following ODE:

$$\begin{aligned} \frac{d}{dt} \mathbb {E}[X_{\ell +1}(t)] = \lambda _{\ell +1} \mathbb {E}[X_{\ell +1}(t)] + n^{-\beta } \mathbb {E}[X_\ell (t)]. \end{aligned}$$

Plugging $\mathbb {E}[X_\ell (t)]$ and $\mathbb {E}[X_{\ell +1}(t)]$ into the above ODE yields

$$\begin{aligned} \frac{d}{dt} \left[ n^{1-(\ell -1)\beta } (-1)^{\ell +1} S_{\ell +1}(t) \right]&= \lambda _{\ell +1} n^{1-(\ell -1)\beta } (-1)^{\ell +1} S_{\ell +1}(t) \\&\hspace{4mm}+ n^{-\beta } n^{1-(\ell -2)\beta } (-1)^l S_\ell (t) \\ n^{1-(\ell -1)\beta } (-1)^{\ell +1} \sum _{i=2}^{\ell +1} \frac{\lambda _i e^{\lambda _i t}}{P_{i, \ell +1}}&= n^{1-(\ell -1)\beta } (-1)^{\ell +1} \left[ \lambda _{\ell +1} S_{\ell +1}(t) - S_\ell (t) \right] . \end{aligned}$$

Dividing both sides by $n^{1-(\ell -1)\beta } (-1)^{\ell +1}$, we get

$$\begin{aligned} \sum _{i=2}^{\ell +1} \frac{\lambda _i e^{\lambda _i t}}{P_{i, \ell +1}}&= \lambda _{\ell +1} S_{\ell +1}(t) - S_\ell (t) \\&= \sum _{i=2}^l \frac{\lambda _{\ell +1} e^{\lambda _i t}}{P_{i,l+1}} + \frac{\lambda _{\ell +1} e^{\lambda _{\ell +1} t}}{P_{\ell +1,l+1}} - \sum _{i=2}^l \frac{(\lambda _{\ell +1} - \lambda _i) e^{\lambda _i t}}{P_{i,l+1}}, \end{aligned}$$

which is clearly true. So we have shown the desired result. $\square$

1.4 Proof of proposition 1

Proof

We first will solve for ${{\hat{s}}}_{n,k}$, the extinction probability of a type k cell under a slightly different model. Consider a model where amplification events involve replacing a type k cell with a type $k+1$ cell. That is, when a new amplified cell arises, a type k cell is removed. Then the following relation will hold

$$\begin{aligned} {{\hat{s}}}_{n,k} = {\left\{ \begin{array}{ll} \displaystyle {{\frac{r_k}{r_k + d_k + \nu _n}({{\hat{s}}}_{n,k})^2 + \frac{d_k}{r_k + d_k + \nu _n} + \frac{\nu _n}{r_k + d_k + \nu _n}{{\hat{s}}}_{n,k+1}}}&{} \text { if } 2 \le k < M,\\ \displaystyle {{\frac{r_M}{r_M + d_M}({{\hat{s}}}_{n,M})^2 + \frac{d_M}{r_M + d_M}}}&{} \text { if } k = M. \end{array}\right. } \end{aligned}$$

(5)

The intuition for why this relation holds is that the terms represent whether the next event is a birth, a death, or an amplification, respectively. If the next event is a birth event, then there are now two independent type k cells each with extinction probability ${{\hat{s}}}_{n,k}$. If the next event is a death, then the particle becomes extinct. If the next event is a gene amplification, then the type k cell becomes a type $k+1$ cell, whose extinction probability is ${{\hat{s}}}_{n,k+1}$. In the case that $k = M$, no further amplification events are possible so only the first two terms exist. By Theorem 2.1 of Hautphenne et al. (2013), for $2 \le k < M$, ${{\hat{s}}}_{n,k}$ is the minimal non-negative solution to (5).

Applying the quadratic formula to 5, the minimal non-negative solution is ${{\hat{s}}}_{n,M} = \frac{d_M}{r_M}$. Similarly, for $2 \le k < M$,

$$\begin{aligned} {{\hat{s}}}_{n,k}&= \frac{r_k+d_k + \nu _n}{2r_k} \pm \frac{1}{2} \sqrt{\frac{(r_k+d_k + \nu _n)^2}{{r_k}^2} - 4 \frac{d_k + \nu _n {{\hat{s}}}_{n, k+1}}{r_k}}\\&= \frac{r_k+d_k + \nu _n}{2r_k} \pm \frac{1}{2r_k} \sqrt{(r_k + d_k)^2 + 2(r_k+d_k)\nu _n + {\nu _n}^2 - 4 r_k d_k - 4 r_k\nu _n {{\hat{s}}}_{n,k+1}}\\&= \frac{r_k+d_k + \nu _n}{2r_k} \pm \frac{|r_k-d_k|}{2r_k} \sqrt{1 + \frac{2(r_k+d_k)}{(r_k-d_k)^2}\nu _n + \frac{1}{{(r_k-d_k)^2}}{\nu _n}^2 - \frac{4 r_k}{{(r_k-d_k)^2}}\nu _n {{\hat{s}}}_{n,k+1}}\\&= \frac{(r_k+d_k)+\nu _n}{2r_k} \pm \frac{|r_k-d_k|}{2r_k} \sqrt{1+ x_k} \end{aligned}$$

where

$$\begin{aligned} x_k = \frac{2(r_k+d_k)}{(r_k-d_k)^2}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 - \frac{4r_k}{(r_k-d_k)^2} {{\hat{s}}}_{n,k+1}\nu _n. \end{aligned}$$

Since $\nu _n = n^{-\beta } \rightarrow 0$ in the large population limit, the Taylor expansion $\sqrt{1+x} = \sum _{i=0}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) x^i,$ gives

$$\begin{aligned} {{\hat{s}}}_{n,k} = \frac{(r_k+d_k)+\nu _n}{2r_k} \pm \frac{|r_k-d_k|}{2r_k} \left( \sum _{i=0}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_k}^i\right) . \end{aligned}$$

Now consider k such that $k' \le k < M$. In particular, $r_k > d_k$ because the cells with $k'$ or more copies are supercritical by definition. Then,

$$\begin{aligned} {{\hat{s}}}_{n,k}&= \frac{(r_k+d_k)+\nu _n}{2r_k} \pm \frac{r_k-d_k}{2r_k} \left( \sum _{i=0}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_k}^i\right) \\&= \frac{(r_k+d_k)+\nu _n}{2r_k} \pm \frac{r_k-d_k}{2r_k} \left( 1 + \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_k}^i\right) \\&= 1 + O(\nu _n) \text { or } \frac{d_k}{r_k} + O(\nu _n), \end{aligned}$$

because $x_k$ is a multiple of $\nu _n$. Since $r_k > d_k$, the minimal non-negative root is thus of the form ${{\hat{s}}}_{n,k} = \frac{d_k}{r_k} + O(\nu _n)$. Intuitively, this aligns with the knowledge that without mutation, the extinction probability would be $\frac{d_k}{r_k}$.

Now consider k such that $2 \le k < k'$. We will show with induction that

$$\begin{aligned} {{\hat{s}}}_{n, k} = 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^\ell + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

where $\ell = k'-k$. Note that in these cases $r_k < d_k$.

For our base case, consider ${{\hat{s}}}_{n, k'-1}$. Then

$$\begin{aligned} {{\hat{s}}}_{n,k'-1}&= \frac{(r_{k'-1}+d_{k'-1})+\nu _n}{2r_{k'-1}} \pm \frac{d_{k'-1}-r_{k'-1}}{2r_{k'-1}} \left( 1+\sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k'-1}}^i\right) \\&=1 + \frac{\nu _n}{2r_{k'-1}} - \frac{d_{k'-1}-r_{k'-1}}{2r_{k'-1}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k'-1}}^i\right) , \\&\text { or } \frac{d_{k'-1}}{r_{k'-1}} + \frac{\nu _n}{2r_{k'-1}} + \frac{d_{k'-1}-r_{k'-1}}{2r_{k'-1}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k'-1}}^i\right) . \end{aligned}$$

Since $r_{k'-1} < d_{k'-1}$, we know that the former must be the correct root so

$$\begin{aligned} {{\hat{s}}}_{n,k'-1} =1 + \frac{\nu _n}{2r_{k'-1}} - \frac{d_{k'-1}-r_{k'-1}}{2r_{k'-1}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k'-1}}^i\right) , \end{aligned}$$

where

$$\begin{aligned} x_{k'-1}&= \frac{2(r_{k'-1}+d_{k'-1})}{(r_{k'-1}-d_{k'-1})^2}\nu _n + \frac{1}{(r_{k'-1}-d_{k'-1})^2} \nu _n^2 - \frac{4r_{k'-1}}{(r_{k'-1}-d_{k'-1})^2} \left( \frac{d_{k'}}{r_{k'}} + O(\nu _n)\right) \nu _n \end{aligned}$$

Intuitively, this makes sense because the lineage of a subcritical cell without mutation is guaranteed to go extinct. Notice that the coefficient of $\nu _n$ in ${{\hat{s}}}_{n, k'-1}$ is

$$\begin{aligned}&\frac{1}{2r_{k'-1}} - \frac{d_{k'-1}-r_{k'-1}}{2r_{k'-1}} \left( {\begin{array}{c}1/2\\ 1\end{array}}\right) \left( \frac{2(r_{k'-1}+d_{k'-1})-4r_{k'-1}\cdot {d_{k'}}/{r_{k'}}}{2(r_{k'-1}-d_{k'-1})^2} \right) \\&\quad = \frac{1}{2r_{k'-1}} \left( 1+ \frac{(r_{k'-1}+d_{k'-1})-2r_{k'-1} \cdot {d_{k'}}/{r_{k'}}}{(r_{k'-1}-d_{k'-1})} \right) \\&\quad =\frac{1}{r_{k'-1}} \left( \frac{2r_{k'-1}-2r_{k'-1} \cdot {d_{k'}}/{r_{k'}}}{(r_{k'-1}-d_{k'-1})} \right) \\&\quad = \frac{1-d_{k'}/r_{k'}}{r_{k'-1}-d_{k'-1}}\\&\quad = -\frac{1-d_{k'}/r_{k'}}{d_{k'-1}-r_{k'-1}}. \end{aligned}$$

Thus, we indeed have that

$$\begin{aligned} {{\hat{s}}}_{n, k'-1} = 1 - \frac{1-d_{k'}/r_{k'}}{d_{k'-1}-r_{k'-1}} \nu _n + O\left( \nu _n^2\right) , \end{aligned}$$

as desired. Now assume that for some $2< k+1 < k'$ that

$$\begin{aligned} {{\hat{s}}}_{n, k+1} = 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell -1} + O\left( \nu _n^{\ell }\right) , \end{aligned}$$

where $\ell = k'-k$. Then we know

$$\begin{aligned} {{\hat{s}}}_{n,k}&= \frac{(r_{k}+d_{k})+\nu _n}{2r_{k}} \pm \frac{d_{k}-r_{k}}{2r_{k}} \left( 1+\sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k}}^i\right) \\&=1 + \frac{\nu _n}{2r_{k}} - \frac{d_{k}-r_{k}}{2r_{k}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k}}^i\right) \text { or } \frac{d_{k}}{r_{k}} + \frac{\nu _n}{2r_{k}} + \frac{d_{k}-r_{k}}{2r_{k}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k}}^i\right) , \end{aligned}$$

and that the former root is the correct one because $r_k < d_k$. So

$$\begin{aligned} {{\hat{s}}}_{n,k} =1 + \frac{\nu _n}{2r_{k}} - \frac{d_{k}-r_{k}}{2r_{k}} \left( \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k}}^i\right) , \end{aligned}$$

where

$$\begin{aligned} x_k&= \frac{2(r_k+d_k)}{(r_k-d_k)^2}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 - \frac{4r_k}{(r_k-d_k)^2} \left( 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell -1} + O\left( \nu _n^{\ell }\right) \right) \nu _n\\&=\frac{2(r_k+d_k)-4r_k}{(r_k-d_k)^2}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 + \frac{4r_k}{(r_k-d_k)^2} \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) \\&= \frac{-2}{(r_k-d_k)}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 + \frac{4r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) . \end{aligned}$$

Notice that if we collect the terms in the sum that arise from the first two terms of $x_k$, we can write

$$\begin{aligned}&\sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) {x_{k}}^i = \sum _{i=1}^\infty \left( {\begin{array}{c}1/2\\ i\end{array}}\right) \left( \frac{-2}{(r_k-d_k)}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 \right) ^i\\&\quad +\left( {\begin{array}{c}1/2\\ 1\end{array}}\right) \frac{4r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) \\&\quad = \sqrt{1+\frac{-2}{(r_k-d_k)}\nu _n + \frac{1}{(r_k-d_k)^2} \nu _n^2 } -1 + \frac{2r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } \\&\qquad + O\left( \nu _n^{\ell +1}\right) \\&\quad = \left| 1-\frac{\nu _n}{r_k-d_k}\right| - 1 + \frac{2r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) \\&\quad = \frac{\nu _n}{d_k-r_k}+ \frac{2r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

because $r_k < d_k$. Then, we have that

$$\begin{aligned} {{\hat{s}}}_{n,k}&=1 + \frac{\nu _n}{2r_{k}} - \frac{d_{k}-r_{k}}{2r_{k}} \left( \frac{\nu _n}{d_k-r_k}+ \frac{2r_k}{(r_k-d_k)^2} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) \right) \\&= 1 + \frac{\nu _n}{2r_k} - \frac{\nu _n}{2r_k} - \frac{1}{d_k-r_k} \cdot \frac{1-d_{k'}/r_{k'}}{\prod _{i=k+1}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) \\&= 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^{\ell } + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

as desired. This completes our inductive step so indeed we have that for $2 \le k < k'$,

$$\begin{aligned} {{\hat{s}}}_{n, k} = 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^\ell + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

where $\ell = k'-k$.

To complete the proof, we will derive an answer for $s_{n,k}$ in terms of ${{\hat{s}}}_{n,k}$. Notice that instead of (5), $s_{n,k}$ satisfies

$$\begin{aligned} s_{n,k} = {\left\{ \begin{array}{ll} \displaystyle {\frac{r_k}{r_k+d_k+\nu _n} (s_{n,k})^2 + \frac{d_k}{r_k+d_k+\nu _n} + \frac{\nu _n}{r_k+d_k+\nu _n}s_{n,k} s_{n,k+1}} &{} \text { if }2 \le k < M,\\ \displaystyle {\frac{r_M}{r_M+d_M}( s_{n,M})^2 + \frac{d_M}{r_M+d_M} }&{} \text { if } k = M. \end{array}\right. }\nonumber \\ \end{aligned}$$

(6)

So $s_{n,M} = \frac{d_k}{r_k}$ as above. Let $k' \le k < M$. Then $r_k > d_k$. Notice that

$$\begin{aligned} 0 = r_k (s_{n,k})^2 - (r_k+d_k) s_{n,k} + d_k + O(\nu _n). \end{aligned}$$

So the minimal non-negative root is of the form $s_{n,k} = \frac{d_k}{r_k} + O(\nu _n)$.

Now let $2 \le k < k'$. Notice that by (5) and (6),

$$\begin{aligned}&(r_k + d_k + \nu _n)({{\hat{s}}}_{n,k} - s_{n,k}) = r_k(({{\hat{s}}}_{n,k})^2 - (s_{n,k})^2) + \nu _n({{\hat{s}}}_{n,k+1} - s_{n,k} s_{n,k+1})\\&\quad =r_k({{\hat{s}}}_{n,k} + s_{n,k})({{\hat{s}}}_{n,k} - s_{n,k}) + \nu _n \left( {{\hat{s}}}_{n,k+1} - {{\hat{s}}}_{n,k}{{\hat{s}}}_{n,k+1} + {{\hat{s}}}_{n,k}{{\hat{s}}}_{n,k+1}\right. \\&\qquad \left. -s_{n,k}{{\hat{s}}}_{n,k+1} + s_{n,k}{{\hat{s}}}_{n,k+1} - s_{n,k} s_{n,k+1}\right) \\&\quad = r_k({{\hat{s}}}_{n,k} + s_{n,k})({{\hat{s}}}_{n,k} - s_{n,k}) \\&\qquad + \nu _n \left( {{\hat{s}}}_{n,k+1}(1 - {{\hat{s}}}_{n,k}) +{{\hat{s}}}_{n,k+1}( {{\hat{s}}}_{n,k} -s_{n,k}) + s_{n,k}({{\hat{s}}}_{n,k+1} - s_{n,k+1})\right) . \end{aligned}$$

Then

$$\begin{aligned}&(r_k - r_k({{\hat{s}}}_{n,k} + s_{n,k}) + d_k + \nu _n(1-s_{n,k+1})) ({{\hat{s}}}_{n,k} - s_{n,k}) \\&\quad = \nu _n \left( {{\hat{s}}}_{n,k+1}(1 - {{\hat{s}}}_{n,k}) + s_{n,k}({{\hat{s}}}_{n,k+1} - s_{n,k+1})\right) . \end{aligned}$$

Then

$$\begin{aligned}&|r_k - r_k({{\hat{s}}}_{n,k} + s_{n,k}) + d_k + \nu _n(1-s_{n,k+1})|\cdot |{{\hat{s}}}_{n,k} - s_{n,k}| \\&= \nu _n \left| {{\hat{s}}}_{n,k+1}(1 - {{\hat{s}}}_{n,k}) + s_{n,k}({{\hat{s}}}_{n,k+1} - s_{n,k+1})\right| \\&\quad \le \nu _n|1-{{\hat{s}}}_{n,k}| + \nu _n|{{\hat{s}}}_{n,k+1} - s_{n,k+1}|, \end{aligned}$$

because $|{{\hat{s}}}_{n,k+1}|, |s_{n,k}| \le 1$. Since ${{\hat{s}}}_{n,k} + s_{n.k} \le 2$ and $1-s_{n,k+1} \ge 0$,

$$\begin{aligned} r_k - r_k({{\hat{s}}}_{n,k} + s_{n,k}) + d_k + \nu _n(1-s_{n,k+1}) \ge d_k - r_k. \end{aligned}$$

Since $k < k'$, we know that $d_k > r_k$. So

$$\begin{aligned} |{{\hat{s}}}_{n,k} - s_{n,k}| \le \frac{\nu _n}{d_k - r_k}\left( |1-{{\hat{s}}}_{n,k}| + |{{\hat{s}}}_{n,k+1} - s_{n,k+1}|\right) . \end{aligned}$$

(7)

Our goal is to show that for $2 \le k < k'$ that $|{{\hat{s}}}_{n,k} - s_{n,k}| = O(\nu _n^{\ell +1})$, where $\ell = k'-k$. Proceed by induction on $\ell$, recall that

$$\begin{aligned} {{\hat{s}}}_{n, k} = {\left\{ \begin{array}{ll} \displaystyle 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^\ell + O\left( \nu _n^{\ell +1}\right) &{} \text { if } 2 \le k < k',\\ \displaystyle \frac{d_k}{r_k} + O(\nu _n)&{} \text { if } \,\, k' \le k \le M. \end{array}\right. } \end{aligned}$$

For our base case, consider $\ell = 1$, that is $k = k'-1$. Then ${{\hat{s}}}_{n,k'-1} = 1+ O(\nu _n)$ so $|1-{{\hat{s}}}_{n,k'-1}| = O(\nu _n)$. We also know from above that ${{\hat{s}}}_{n,k'} = \frac{d_{k'}}{r_{k'}} + O(\nu _n)$ and $s_{n,k'} = \frac{d_{k'}}{r_{k'}} + O(\nu _n)$, so $|{{\hat{s}}}_{n,k'} - s_{n,k'}|= O(\nu _n)$ as well. Then by (7),

$$\begin{aligned} |{{\hat{s}}}_{n,k'-1} - s_{n,k'-1}| \le \frac{\nu _n}{d_k - r_k}\left( |1-{{\hat{s}}}_{n,k'-1}| + |{{\hat{s}}}_{n,k'} - s_{n,k'}|\right) = \frac{\nu _n}{d_k - r_k}O(\nu _n) = O(\nu _n^2). \end{aligned}$$

Now consider $\ell > 1$, i.e. $k < k'-1$. Then by induction, $|{{\hat{s}}}_{n,k+1} - s_{n,k+1}| = O\left( \nu _n^{(\ell -1) + 1}\right) = O(\nu _n^\ell )$. We know ${{\hat{s}}}_{n,k} = 1 + O(\nu _n^\ell )$ so $|1- {{\hat{s}}}_{n,k}| = O(\nu _n^\ell )$ as well. Then by (7),

$$\begin{aligned} |{{\hat{s}}}_{n,k} - s_{n,k}| \le \frac{\nu _n}{d_k - r_k}\left( |1-{{\hat{s}}}_{n,k}| + |{{\hat{s}}}_{n,k+1} - s_{n,k+1}|\right) = \frac{\nu _n}{d_k - r_k}O(\nu _n^\ell ) = O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

as desired. So indeed $|{{\hat{s}}}_{n,k} - s_{n,k}| = O(\nu _n^{\ell +1})$ for $2 \le k < k'$. Then since

$$\begin{aligned} {{\hat{s}}}_{n,k} = 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^\ell + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

we know that

$$\begin{aligned} s_{n,k} = {{\hat{s}}}_{n,k} - ({{\hat{s}}}_{n,k} - s_{n,k}) = 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=k}^{k'-1} d_i-r_i} {\nu _n}^\ell + O\left( \nu _n^{\ell +1}\right) , \end{aligned}$$

as well.

In particular, we have that

$$\begin{aligned} s_{n, 2}&= 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} {\nu _n}^{k'-2} + O\left( \nu _n^{k'-1}\right) \\&= 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} { n^{-\beta (k'-2)}}+ O\left( n^{-\beta (k'-2+1)}\right) . \end{aligned}$$

We know that our extinction probability starting with n particles of of type 2 is $q_n = (s_{n,2})^n$. Notice that since $r_{k'} > d_{k'}$ and $r_i < d_i$ for $2 \le i < k'$, we have $\frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} > 0$. Then

$$\begin{aligned} q&= \lim _{n \rightarrow \infty } {q_n} \\&=\lim _{n \rightarrow \infty } \left( 1- \frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} { n^{-\beta (k'-2)}}+ O\left( n^{-\beta (k'-2+1)}\right) \right) ^n\\&= {\left\{ \begin{array}{ll} 0 &{} \text { if } 0< \beta < \frac{1}{k'-2},\\ \displaystyle \exp \left( -\frac{1-d_{k'}/r_{k'}}{\prod _{i=2}^{k'-1} d_i-r_i} \right) &{} \text { if } \beta = \frac{1}{k'-2}, \\ 1 &{} \text { if } \beta > \frac{1}{k'-2}. \end{array}\right. } \end{aligned}$$

$\square$

1.5 Proof of Proposition 4

Proof

Let ${\bar{f}}_n(z) = n^{z-1}(\phi _s(z t_n) + \phi _m(z t_n) - n)$. Using the definitions of $\phi _s$ and $\phi _m$, we see that

$$\begin{aligned} {\bar{f}}_n(z)&= n^{z-1} \left( n^{1-z} + \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( n^{-\lambda _m z/\lambda _s} - n^{-z} \right) - n \right) \\&= 1 + \frac{1}{n^{\alpha }(\lambda _m - \lambda _s)} n^{z(1 - \lambda _m /\lambda _s)} - \frac{1}{n^{\alpha }(\lambda _m - \lambda _s)} - n^z. \end{aligned}$$

Taking the derivative with respect to z yields

$$\begin{aligned} {\bar{f}}^{\prime }_n(z)&= \frac{1}{n^{\alpha }(\lambda _m - \lambda _s)} \frac{\lambda _s - \lambda _m }{\lambda _s} n^{z(1 - \lambda _m /\lambda _s)} \log n - n^z \log n \\&= n^z \log n \left( \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m z/\lambda _s} - 1 \right) . \end{aligned}$$

Next we set ${\bar{f}}'_n(z) = 0$ and solve for z to find any local maxima or minima of ${\bar{f}}_n(z)$. Since $n^z > 0$ and $\log n > 0$ for sufficiently large n, we have that

$$\begin{aligned} {\bar{f}}'_n(z) = 0&\implies \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m z/\lambda _s} = 1 \\&\implies -\frac{\lambda _m }{\lambda _s} z \log n = \log \left( -{n^{\alpha } \lambda _s} \right) . \end{aligned}$$

Solving the above equation for z yields

$$\begin{aligned} z&= \frac{-\lambda _s}{\lambda _m } \cdot \frac{ \log (-n^{\alpha } \lambda _s)}{ \log n} \\&= \frac{-\lambda _s}{\lambda _m } \left( \alpha + \frac{\log (-\lambda _s)}{\log n} \right) \\&= \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log \left( {-\lambda _s} \right) \right) . \end{aligned}$$

Hence $a_n = \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log (-\lambda _s) \right)$ is the only maximum or minimum of ${\bar{f}}_n(z)$. Moreover, we have that $a_n > 0$ for sufficiently large n.

Next note that ${\bar{f}}'_n(a_n-1) = n^{a_n-1} \log n \left( \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m (a_n-1)/\lambda _s} - 1 \right)$. Since $n^{a_n-1} > 0$ and $\log n > 0$ for sufficiently large n and

$$\begin{aligned} \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m (a_n-1)/\lambda _s} - 1&= \frac{-1}{n^{\alpha } \lambda _s} n^{\frac{\lambda _m }{-\lambda _s} \cdot \frac{1}{\lambda _m } (-\lambda _s \alpha + \frac{1}{t_n} \log ({-\lambda _s}) - \lambda _m )} - 1 \\&= \frac{-1}{n^{\alpha } \lambda _s} n^{\alpha + \frac{1}{\log n} \log ({-\lambda _s}) + \lambda _m /\lambda _s} - 1 \\&= n^{\lambda _m /\lambda _s} - 1 \\&< 0 \hbox { for sufficiently large}\ n, \end{aligned}$$

we have that ${\bar{f}}'_n(a_n-1) < 0$ for sufficiently large n. Hence ${\bar{f}}_n(z)$ is monotonically decreasing for $z < a_n$. Similarly, ${\bar{f}}'_n(a_n+1) = n^{a_n+1} \log n \left( \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m (a_n+1)/\lambda _s} - 1 \right)$. Since $n^{a_n+1} > 0$ and $\log n > 0$ for sufficiently large n and

$$\begin{aligned} \frac{-1}{n^{\alpha } \lambda _s} n^{-\lambda _m (a_n+1)/\lambda _s} - 1&= \frac{-1}{n^{\alpha } \lambda _s} n^{\frac{\lambda _m }{-\lambda _s} \cdot \frac{1}{\lambda _m } (-\lambda _s \alpha + \frac{1}{t_n} \log ({-\lambda _s}) + \lambda _m )} - 1 \\&= \frac{-1}{n^{\alpha } \lambda _s} n^{\alpha + \frac{1}{\log n} \log ({-\lambda _s}) - \lambda _m /\lambda _s} - 1 \\&= n^{-\lambda _m /\lambda _s} - 1 \\&> 0 \hbox { for sufficiently large}\ n, \end{aligned}$$

we have that ${\bar{f}}'_n(a_n+1) > 0$ for sufficiently large n. Hence ${\bar{f}}_n(z)$ is monotonically increasing for $z > a_n$.

Now let $A_n = \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log (\lambda _m - \lambda _s) \right)$. Then

$$\begin{aligned} {\bar{f}}_n(A_n)&= 1 + \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \left( n^{A_n(1 - \lambda _m /\lambda _s)} - 1 \right) - n^{A_n} \\&= 1 + \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \left( n^{\frac{\lambda _m - \lambda _s}{-\lambda _s\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n}\log \left( {\lambda _m - \lambda _s} \right) \right) } - 1 \right) - n^{\frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log \left( {\lambda _m - \lambda _s}\right) \right) } \\&= 1 + \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \left( n^{\alpha (1 - \lambda _s/\lambda _m )} n^{\frac{\lambda _m - \lambda _s}{\lambda _m \log n} \log \left( {\lambda _m - \lambda _s} \right) } - 1 \right) - n^{-\lambda _s\alpha /\lambda _m } n^{\frac{-\lambda _s}{\lambda _m \log n} \log \left( {\lambda _m - \lambda _s}\right) } \\&= 1 + \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \left( n^{\alpha (1 - \lambda _s/\lambda _m )} \left( {\lambda _m - \lambda _s} \right) ^{1 - \lambda _s/\lambda _m } - 1 \right) - n^{-\lambda _s\alpha /\lambda _m } \left( {\lambda _m - \lambda _s}\right) ^{-\lambda _s/\lambda _m } \\&= 1 + n^{-\lambda _s\alpha /\lambda _m } \left( {\lambda _m - \lambda _s}\right) ^{-\lambda _s/\lambda _m } - \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} - n^{-\lambda _s\alpha /\lambda _m } \left( {\lambda _m - \lambda _s} \right) ^{-\lambda _s/\lambda _m } \\&= 1 - \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \\&> 0 \hbox { for sufficiently large}\ n. \end{aligned}$$

Note that since ${\bar{f}}_n(0) = 0$ and ${\bar{f}}_n(z)$ is decreasing for all $z \in (0, a_n)$, we have that ${\bar{f}}_n(a_n) < 0$. But ${\bar{f}}_n(z)$ is increasing for $z > a_n$ and we have that ${\bar{f}}_n(A_n) > 0$. Therefore, there exists a unique ${\tilde{u}}_n \in (a_n, A_n)$ such that ${\bar{f}}_n({\tilde{u}}_n) = 0$ by monotonicity. Using the definition of ${\bar{f}}_n(z)$ and the fact that $n^{z-1} > 0$, this implies that there is a unique ${\tilde{u}}_n > 0$ such that $a_n< {\tilde{u}}_n < A_n$ and $\phi _s({\tilde{u}}_n t_n) + \phi _m({\tilde{u}}_n t_n) = n$. Furthermore, $\lim _{n \rightarrow \infty } a_n = \lim _{n \rightarrow \infty } A_n = \frac{-\lambda _s \alpha }{\lambda _m }$. Hence ${\tilde{u}}_n \rightarrow \frac{-\lambda _s \alpha }{\lambda _m }$ as $n \rightarrow \infty$. $\square$

1.6 Proof of theorem 2

Proof

In order to show the desired result, we must show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} (\tau _n > u_n + \varepsilon ) + \lim _{n \rightarrow \infty } \mathbb {P} (\tau _n < u_n - \varepsilon ) = 0. \end{aligned}$$

Let’s start by proving that $\mathbb {P} (\tau _n < u_n - \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$. Note that

$$\begin{aligned}&\mathbb {P} \left( \tau _n< u_n - \varepsilon \right) = \mathbb {P} \left( \frac{\tau _n}{t_n} < u_n^-(\varepsilon ) \right) \\&\quad \le \mathbb {P} \left( \sup _{z \in [c, u_n^-(\varepsilon )]} \left( X_s(z t_n) + X_m(z t_n) - n \right)> 0 \right) \\&\quad = \mathbb {P} \Biggl ( \sup _{z \in [c, u_n^-(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} (X_s(z t_n) - \phi _s(z t_n) + \phi _s(z t_n) - n \\&\qquad + \phi _m(z t_n) - \phi _m(z t_n) + X_m(z t_n))> 0 \Biggr )\\&\quad \le \mathbb {P} \left( {\bar{B}}_1(n, \varepsilon ) + {\bar{B}}_2(n, \varepsilon ) + {\bar{B}}_3(n, \varepsilon ) > 0 \right) , \end{aligned}$$

where

$$\begin{aligned} {\bar{B}}_1(n, \varepsilon )&= \sup _{z \in [c, u_n^-(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} (X_s(z t_n) - \phi _s(z t_n)),\\ {\bar{B}}_2(n, \varepsilon )&= \sup _{z \in [c, u_n^-(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} (\phi _s(z t_n) + \phi _m(z t_n) - n),\\ {\bar{B}}_3(n, \varepsilon )&= \sup _{z \in [c, u_n^-(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} (X_m(z t_n) - \phi _m(z t_n)). \end{aligned}$$

Note that, for $i \in \{s, m\}$,

$$\begin{aligned}&\sup _{z \in [c, u_n^-(\varepsilon )]} \left| n^{\alpha + \lambda _m z/\lambda _s - 1} \left( X_i(zt_n) - \phi _i(zt_n) \right) \right| \\&\quad \le \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_i(zt_n) - \phi _i(zt_n) \right| , \end{aligned}$$

which converges to zero in probability by Proposition 5. Now we just need to show that ${\bar{B}}_2(n, \varepsilon )$ is negative in the large population limit. Let ${\bar{g}}_n(z) = n^{\alpha + \lambda _m z/\lambda _s - 1} \left( \phi _s(z t_n) + \phi _m(z t_n) - n \right)$. Using the definitions of $\phi _s$ and $\phi _m$, we see that

$$\begin{aligned} {\bar{g}}_n(z)&= n^{\alpha + \lambda _m z/\lambda _s -1} \left( n^{1-z} + \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( n^{-\lambda _m z/\lambda _s} - n^{-z} \right) - n \right) \\&= n^{\alpha + z(\lambda _m /\lambda _s - 1)} + \frac{1}{\lambda _m - \lambda _s} - \frac{1}{\lambda _m - \lambda _s} n^{z(\lambda _m /\lambda _s - 1)} - n^{\alpha + \lambda _m z/\lambda _s}. \\ \end{aligned}$$

Taking the derivative with respect to z yields

$$\begin{aligned} {\bar{g}}'_n(z)&= \frac{\lambda _m - \lambda _s}{\lambda _s} n^{\alpha +z(\lambda _m /\lambda _s-1)} \log n - \frac{1}{\lambda _s} n^{z(\lambda _m /\lambda _s-1)} \log n - \frac{\lambda _m }{\lambda _s} n^{\alpha + \lambda _m z/\lambda _s} \log n \\&= \frac{-1}{\lambda _s} n^{\alpha +z(\lambda _m /\lambda _s-1)} \log n \left[ -\lambda _m + \lambda _s + n^{-\alpha } + \lambda _m n^z \right] . \end{aligned}$$

Next we set ${\bar{g}}'_n(z) = 0$ and solve for z to find any local maxima or minima of ${\bar{g}}_n(z)$. Since $-1/\lambda _s > 0$ and $n^{\alpha -z(1 - \lambda _m /\lambda _s)} > 0$ and $\log n > 0$ for sufficiently large n, we have that

$$\begin{aligned} {\bar{g}}'_n(z) = 0&\implies \lambda _m n^z = \lambda _m - \lambda _s - n^{-\alpha } \\&\implies z \log n = \log \left( \frac{\lambda _m - \lambda _s - n^{-\alpha }}{\lambda _m } \right) \\&\implies z = \frac{1}{\log n} \log \left( \frac{\lambda _m - \lambda _s - n^{-\alpha }}{\lambda _m } \right) . \end{aligned}$$

Hence ${\bar{q}}_n = \frac{1}{\log n} \log \left( \frac{\lambda _m - \lambda _s - n^{-\alpha }}{\lambda _m } \right)$ is the only maximum or minimum of ${\bar{g}}_n(z)$. Next note that ${\bar{g}}'_n({\bar{q}}_n+1) = \frac{-1}{\lambda _s} n^{\alpha -({\bar{q}}_n+1)(1 - \lambda _m /\lambda _s)} \log n \left[ -\lambda _m + \lambda _s + n^{-\alpha } + \lambda _m n^{{\bar{q}}_n+1} \right]$. Since $-1/\lambda _s > 0$ and $n^{\alpha - ({\bar{q}}_n+1)(1 - \lambda _m /\lambda _s)} > 0$ and $\log n > 0$ for sufficiently large n and

$$\begin{aligned} -\lambda _m + \lambda _s + n^{-\alpha } + \lambda _m n^{{\bar{q}}_n+1}&= \lambda _m n \cdot n^{\frac{1}{\log n} \log \left( \frac{\lambda _m - \lambda _s - n^{-\alpha }}{\lambda _m } \right) } - \lambda _m + \lambda _s + n^{-\alpha } \\&= \lambda _m n \left( \frac{\lambda _m - \lambda _s - n^{-\alpha }}{\lambda _m } \right) - \lambda _m + \lambda _s + n^{-\alpha } \\&= (n-1) \left( \lambda _m - \lambda _s - n^{-\alpha } \right) \\&\quad > 0 \hbox { for sufficiently large}\ n, \end{aligned}$$

we have that ${\bar{g}}'_n({\bar{q}}_n+1) > 0$ for sufficiently large n. Hence ${\bar{g}}_n(z)$ is monotonically increasing for $z > {\bar{q}}_n$. Note that the only positive solution to ${\bar{g}}_n(z) = 0$ occurs at $z = {\tilde{u}}_n$ by Proposition 4. Also note that ${\bar{q}}_n< c< u_n^-(\varepsilon ) < {\tilde{u}}_n$ for sufficiently large n. Therefore, we have that ${\bar{B}}_2(n, \varepsilon ) < 0$. Moreover, we may rewrite ${\bar{B}}_2(n, \varepsilon )$ as

$$\begin{aligned} {\bar{B}}_2(n, \varepsilon ) = n^{\alpha + \lambda _m u_n^-(\varepsilon )/\lambda _s - 1} \left( \phi _s \left( u_n^-(\varepsilon ) t_n \right) + \phi _m \left( u_n^-(\varepsilon ) t_n \right) - n \right) . \end{aligned}$$

Then by the definitions of $\phi _s$ and $\phi _m$, we have

$$\begin{aligned} {\bar{B}}_2(n, \varepsilon )&= n^{\alpha + \lambda _m u_n^-(\varepsilon )/\lambda _s - 1} \left( n^{1 - u_n^-(\varepsilon )} + \frac{ n^{1-\alpha }}{\lambda _m - \lambda _s} \left( n^{-\lambda _m u_n^-(\varepsilon )/\lambda _s} - n^{-u_n^-(\varepsilon )} \right) - n \right) \\&= n^{\alpha -u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)} + \frac{1}{\lambda _m - \lambda _s} - \frac{1}{\lambda _m - \lambda _s} n^{-u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)} - n^{\alpha + \lambda _m u_n^-(\varepsilon )/\lambda _s}. \end{aligned}$$

Note that

$$\begin{aligned} n^{\alpha + \lambda _m u_n^-(\varepsilon )/\lambda _s}&= n^{\alpha } n^{\frac{\varepsilon \lambda _m }{\log n}} n^{{\tilde{u}}_n(\lambda _m /\lambda _s)} \\&= e^{\varepsilon \lambda _m } n^{\alpha + {\tilde{u}}_n(\lambda _m /\lambda _s)} \\&\ge e^{\varepsilon \lambda _m } n^{\alpha + A_n(\lambda _m /\lambda _s)} \\&= e^{\varepsilon \lambda _m } n^{\alpha + \frac{\lambda _m }{\lambda _s} \cdot \frac{1}{\lambda _m } \left( -\lambda _s \alpha + \frac{1}{t_n} \log \left( {\lambda _m - \lambda _s}\right) \right) }\\&= e^{\varepsilon \lambda _m } n^{\alpha - \alpha - \frac{1}{\log n} \log \left( {\lambda _m - \lambda _s} \right) } \\&= e^{\varepsilon \lambda _m } \left( \frac{1}{\lambda _m - \lambda _s} \right) . \end{aligned}$$

Therefore, we have that

$$\begin{aligned}&{\bar{B}}_2(n, \varepsilon ) \le n^{\alpha -u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)} + \frac{1}{\lambda _m - \lambda _s} - \frac{1}{\lambda _m - \lambda _s} n^{-u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)} - e^{\varepsilon \lambda _m } \left( \frac{1}{\lambda _m - \lambda _s} \right) \nonumber \\&\quad = \frac{1}{\lambda _m - \lambda _s} \left( 1 - e^{\varepsilon \lambda _m } \right) + n^{\alpha - u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)} \left( 1 - \frac{ n^{-\alpha }}{\lambda _m - \lambda _s} \right) . \end{aligned}$$

(8)

Clearly, $1 - n^{-\alpha }/(\lambda _m - \lambda _s) \rightarrow 1$ as $n \rightarrow \infty$ since $\alpha > 0$. We also have that

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\alpha - u_n^-(\varepsilon )(1 - \lambda _m /\lambda _s)}&= \lim _{n \rightarrow \infty } n^{\alpha } n^{\frac{\varepsilon }{t_n}(1 - \lambda _m /\lambda _s)} n^{-{\tilde{u}}_n(1 - \lambda _m /\lambda _s)} \\&= e^{\varepsilon (\lambda _m - \lambda _s)} \lim _{n \rightarrow \infty } n^{\alpha - {\tilde{u}}_n(1 - \lambda _m /\lambda _s)} \\&= e^{\varepsilon (\lambda _m - \lambda _s)} \lim _{n \rightarrow \infty } e^{[\alpha - {\tilde{u}}_n(1 - \lambda _m /\lambda _s)] \log n}\\&= 0 \end{aligned}$$

since ${\tilde{u}}_n \rightarrow -\alpha \lambda _s / \lambda _m$ as $n \rightarrow \infty$ by Proposition 4. Therefore, the right-hand side of equation (8) converges to $\frac{1}{\lambda _m - \lambda _s}(1 - e^{\varepsilon \lambda _m }) < 0$. So ${\bar{B}}_2(n, \varepsilon )$ is negative in the large population limit. Putting this all together, we have our desired result: $\mathbb {P}({\bar{B}}_1(n, \varepsilon ) + {\bar{B}}_2(n, \varepsilon ) + {\bar{B}}_3(n, \varepsilon ) > 0) \rightarrow 0$ as $n \rightarrow \infty$. Therefore, we have shown $\mathbb {P}(\tau _n < u_n - \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$. The proof that $\mathbb {P}(\tau _n > u_n + \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$ follows using a similar argument. $\square$

1.7 Proof of proposition 5

Proof

Let’s start by proving the $i=s$ case. That is, we want to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_s(zt_n) - \phi _s(zt_n) \right| > \delta \right) = 0. \end{aligned}$$

Because a branching process normalized by its mean is a martingale, we know that

$$\begin{aligned} Z_s(z) = n^{\alpha + z - 1} \left( X_s(zt_n) - \phi _s(zt_n) \right) \end{aligned}$$

is a martingale in z. To prove our desired result for $i=s$, we need to show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\lambda _m z/\lambda _s - z} |Z_s(z)| > \delta \right) = 0. \end{aligned}$$

Note that

$$\begin{aligned} \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\lambda _m z/\lambda _s - z} |Z_s(z)|&\le \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\lambda _m z/\lambda _s - z} \cdot \sup _{z \in [c, u_n^+(\varepsilon )]} |Z_s(z)| \\&= n^{\lambda _m c/\lambda _s - c} \cdot \sup _{z \in [c, u_n^+(\varepsilon )]} |Z_s(z)|. \end{aligned}$$

Therefore, we have that

$$\begin{aligned} \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\lambda _m z/\lambda _s - z} |Z_s(z)|> \delta \right)&\le \mathbb {P} \left( n^{\lambda _m c/\lambda _s - c} \cdot \sup _{z \in [c, u_n^+(\varepsilon )]} |Z_s(z)|> \delta \right) \\&= \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} |Z_s(z)| > \delta \cdot n^{c - \lambda _m c/\lambda _s} \right) \\&\le \frac{1}{\delta ^2} n^{2\lambda _m c/\lambda _s - 2c} \cdot \mathbb {E} \left[ \left( Z_s(u_n^+(\varepsilon )) \right) ^2 \right] \end{aligned}$$

by Doob’s Martingale Inequality. Note that

$$\begin{aligned} \mathbb {E} \left[ \left( Z_s(u_n^+(\varepsilon )) \right) ^2 \right]&= \mathbb {E} \left[ \left( n^{\alpha + u_n^+(\varepsilon ) - 1} X_s(u_n^+(\varepsilon ) t_n) - n^{\alpha + u_n^+(\varepsilon ) - 1} \phi _s(u_n^+(\varepsilon ) t_n) \right) ^2 \right] \\&= \textrm{Var} \left[ n^{\alpha + u_n^+(\varepsilon ) - 1} X_s(u_n^+(\varepsilon ) t_n) \right] . \end{aligned}$$

Therefore, now we have

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\lambda _m z/\lambda _s - z} |Z_s(z)| > \delta \right) \\&\quad \le \frac{1}{\delta ^2} n^{2\lambda _m c/\lambda _s - 2c} \cdot \textrm{Var} \left[ n^{\alpha + u_n^+(\varepsilon ) - 1} X_s(u_n^+(\varepsilon ) t_n) \right] \\&\quad = \frac{1}{\delta ^2} n^{2\lambda _m c/\lambda _s - 2c} n^{2\alpha + 2u_n^+(\varepsilon ) - 2} n^{1 - u_n^+(\varepsilon )} \left( \frac{r_s + d_s}{-\lambda _s} \right) \left( 1 - n^{-u_n^+(\varepsilon )} \right) \\&\quad = \frac{1}{\delta ^2} \left( \frac{r_s + d_s}{-\lambda _s} \right) \left( 1 - n^{-u_n^+(\varepsilon )} \right) n^{2\alpha + 2c(\lambda _m /\lambda _s - 1) + u_n^+(\varepsilon ) - 1}. \end{aligned}$$

Clearly, $\frac{1}{\delta ^2} \left( \frac{r_s + d_s}{-\lambda _s} \right)$ is just a finite constant. Note that

$$\begin{aligned} \lim _{n \rightarrow \infty } \left( 1 - n^{-u_n^+(\varepsilon )} \right)&= 1 - \lim _{n \rightarrow \infty } n^{-\varepsilon /t_n} n^{-{\tilde{u}}_n} \\&= 1 - e^{\varepsilon \lambda _s} \lim _{n \rightarrow \infty } e^{-{\tilde{u}}_n \log n} \\&= 1 \end{aligned}$$

since ${\tilde{u}}_n \rightarrow -\lambda _s \alpha /\lambda _m$ as $n \rightarrow \infty$ by Proposition 4. We also have that

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{2\alpha + 2c(\lambda _m /\lambda _s-1) + u_n^+(\varepsilon ) - 1}&= \lim _{n \rightarrow \infty } n^{2\alpha + \frac{\alpha \lambda _s(\lambda _s - 2\lambda _m )}{\lambda _m (\lambda _m - \lambda _s)} \cdot \frac{\lambda _m - \lambda _s}{\lambda _s} - 1} n^{{\tilde{u}}_n} n^{\varepsilon /t_n} \\&= e^{-\varepsilon \lambda _s} \lim _{n \rightarrow \infty } n^{2\alpha + \alpha (\lambda _s - 2\lambda _m )/\lambda _m - 1 + {\tilde{u}}_n} \\&= e^{-\varepsilon \lambda _s} \lim _{n \rightarrow \infty } e^{[{\tilde{u}}_n + \alpha \lambda _s/\lambda _m - 1] \log n} \\&= 0 \end{aligned}$$

since ${\tilde{u}}_n \rightarrow -\lambda _s \alpha /\lambda _m$ as $n \rightarrow \infty$ by Proposition 4. So we are done with the $i=s$ case.

Next, let’s prove the desired result for $i=m$. As a reminder, we want to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right| > \delta \right) = 0. \end{aligned}$$

Note that

$$\begin{aligned} \phi _m(z t_n)&= \int _0^{zt_n} n^{-\alpha } n e^{\lambda _s s} e^{\lambda _m (zt_n - s)} \, ds\\&= n^{1 -\lambda _m z/\lambda _s - \alpha } \int _0^{zt_n} e^{(\lambda _s - \lambda _m )s} \, ds. \end{aligned}$$

Therefore,

$$\begin{aligned}&n^{\alpha + \lambda _m z/\lambda _s - 1} \left( X_m(zt_n) - \phi _m(zt_n) \right) \\&\quad = n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \int _0^{zt_n} e^{(\lambda _s - \lambda _m )s} \, ds\\&\quad = n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds + \frac{1}{n} \int _0^{zt_n} \left( X_s(s) - ne^{\lambda _ss} \right) e^{-\lambda _m s} \, ds. \end{aligned}$$

Taking the absolute value of both sides and using the triangle inequality, we get

$$\begin{aligned}&n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right| \\&\quad \le \left| n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds \right| + \frac{1}{n} \int _0^{zt_n} \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds. \end{aligned}$$

This implies that

$$\begin{aligned}&\sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right| \\&\quad \le \sup _{z \in [c, u_n^+(\varepsilon )]} \left| n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds \right| \\&\qquad + \sup _{z \in [c, u_n^+(\varepsilon )]} \frac{1}{n} \int _0^{zt_n} \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds. \end{aligned}$$

Hence

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right|> \delta \right) \\&\quad \le \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} \left| n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds \right|> \delta /2 \right) \\&\qquad + \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} \frac{1}{n} \int _0^{zt_n} \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds > \delta /2 \right) . \end{aligned}$$

The process in the second term of the above sum is monotonically increasing in z. So we may simplify the expression above to

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right|> \delta \right) \\&\quad \le \mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} \left| n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds \right|> \delta /2 \right) \\&\qquad + \mathbb {P} \left( \frac{1}{n} \int _0^{u_n^+(\varepsilon ) t_n} \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds > \delta /2 \right) . \end{aligned}$$

By Lemma 1 in Durrett and Moseley (2010), we know that

$$\begin{aligned} e^{-\lambda _m t} X_m(t) - \int _0^t n^{-\alpha } e^{-\lambda _m s} X_s(s) \, ds \end{aligned}$$

is a martingale. Setting $t = zt_n = \frac{z}{-\lambda _s}\log n$, we get that

$$\begin{aligned} n^{\lambda _m z/\lambda _s}X_m(zt_n) - \frac{1}{n^{\alpha }} \int _0^{zt_n} e^{-\lambda _m s} X_s(s) \, ds \end{aligned}$$

is a martingale in z. Since linear combinations of martingales are also martingales,

$$\begin{aligned} n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} e^{-\lambda _m s} X_s(s) \, ds \end{aligned}$$

is also a martingale in z. Therefore,

$$\begin{aligned} \left| n^{\alpha + \lambda _m z/\lambda _s - 1} X_m(zt_n) - \frac{1}{n} \int _0^{zt_n} X_s(s) e^{-\lambda _m s} \, ds \right| \end{aligned}$$

is a non-negative submartingale in z, so we can apply Doob’s Martingale Inequality to get

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [c, u_n^+(\varepsilon )]} n^{\alpha + \lambda _m z/\lambda _s - 1} \left| X_m(zt_n) - \phi _m(zt_n) \right|> \delta \right) \nonumber \\&\quad \le \frac{4}{\delta ^2} \cdot \mathbb {E} \left[ \left( n^{\alpha + \lambda _m u_n^+(\varepsilon )/\lambda _s - 1} X_m (u_n + \varepsilon ) - \frac{1}{n} \int _0^{u_n + \varepsilon } X_s(s) e^{-\lambda _m s} \, ds \right) ^2 \right] \nonumber \\&\qquad + \mathbb {P} \left( \frac{1}{n} \int _0^{u_n + \varepsilon } \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds > \delta /2 \right) . \end{aligned}$$

(9)

Let us start by showing that the second term in (9) converges to zero. Since convergence in mean implies convergence in probability (by Markov’s Inequality), it suffices to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {E} \left[ \frac{1}{n} \int _0^{u_n+\varepsilon } \left| X_s(s) - ne^{\lambda _ss} \right| e^{-\lambda _m s} \, ds \right] = 0, \end{aligned}$$

or, equivalently,

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{1}{n} \int _0^{u_n+\varepsilon } \mathbb {E} \left[ \left| X_s(s) - ne^{\lambda _ss} \right| \right] e^{-\lambda _m s} \, ds = 0. \end{aligned}$$

By the Cauchy-Schwarz Inequality, we know

$$\begin{aligned} \mathbb {E} \left[ \left| X_s(s) - ne^{\lambda _ss} \right| \right]&\le \sqrt{\mathbb {E} \left[ \left| X_s(s) - ne^{\lambda _ss} \right| ^2 \right] }\\&\le \sqrt{ \text {Var} \left[ X_s(s) \right] }\\&= n^{1/2} \left[ \left( \frac{r_s + d_s}{\lambda _s} \right) \left( e^{2\lambda _ss} - e^{\lambda _ss} \right) \right] ^{1/2}. \end{aligned}$$

Multiplying both sides by $\frac{1}{n} e^{-\lambda _m s}$ and integrating yields

$$\begin{aligned}&\frac{1}{n} \int _0^{u_n + \varepsilon } \mathbb {E} \left[ \left| X_s(s) - ne^{\lambda _ss} \right| \right] e^{-\lambda _m s} \, ds \le \frac{1}{n^{1/2}} \int _0^{u_n + \varepsilon } \left[ \left( \frac{r_s + d_s}{\lambda _s} \right) \left( e^{2\lambda _ss} - e^{\lambda _ss} \right) \right] ^{1/2}\\&\qquad \times e^{-\lambda _m s} \, ds\\&\quad = \left( \frac{r_s + d_s}{-\lambda _sn} \right) ^{1/2} \int _0^{u_n + \varepsilon } \left( 1 - e^{\lambda _ss} \right) ^{1/2} e^{\lambda _ss/2} e^{-\lambda _m s} \, ds\\&\quad \le \left( \frac{r_s + d_s}{-\lambda _sn} \right) ^{1/2} \int _0^{u_n + \varepsilon } e^{(\lambda _s/2 - \lambda _m )s} \, ds\\&\quad = \frac{1}{\lambda _s/2 - \lambda _m } \left( \frac{r_s + d_s}{-\lambda _sn} \right) ^{1/2} \left( e^{(\lambda _s/2 - \lambda _m )(u_n + \varepsilon )} - 1 \right) . \end{aligned}$$

Note that, since $u_n = {\tilde{u}}_n t_n = \frac{{\tilde{u}}_n}{-\lambda _s} \log n$,

$$\begin{aligned} e^{(\lambda _s/2 - \lambda _m )(u_n + \varepsilon )}&= e^{(\lambda _s/2 - \lambda _m ) \frac{{\tilde{u}}_n}{-\lambda _s} \log n} e^{(\lambda _s/2 - \lambda _m ) \varepsilon }\\&= n^{(\lambda _m /\lambda _s - 1/2) {\tilde{u}}_n} e^{(\lambda _s/2 - \lambda _m )\varepsilon }, \end{aligned}$$

which converges to zero since ${\tilde{u}}_n \rightarrow \frac{-\lambda _s \alpha }{\lambda _m }$ as $n \rightarrow \infty$ by Proposition 4. Therefore,

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{1}{n} \int _0^{u_n+\varepsilon } \mathbb {E} \left[ \left| X_s(s) - ne^{\lambda _ss} \right| \right] e^{-\lambda _m s} \, ds = 0, \end{aligned}$$

as desired. Next, we will show the first term in (9) converges to zero as well.

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\alpha +\lambda _m u_n^+(\varepsilon )/\lambda _s - 1} X_m(u_n + \varepsilon ) - \frac{1}{n} \int _0^{u_n + \varepsilon } X_s(s) e^{-\lambda _m s} \, ds \right) ^2 \right] \\&\quad = n^{2(\alpha + \lambda _m u_n^+(\varepsilon )/\lambda _s - 1)} \mathbb {E} \left[ X_m(u_n + \varepsilon )^2 \right] \\&\qquad - 2 n^{\alpha + \lambda _m u_n^+(\varepsilon )/\lambda _s - 2} \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_m(u_n + \varepsilon ) \right] e^{-\lambda _m s} \, ds\\&\qquad + \left( \frac{1}{n} \right) ^2 \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m s} e^{-\lambda _m y} \, ds \, dy. \end{aligned}$$

From Lemma 1 in Foo and Leder (2013), we know that

$$\begin{aligned} \mathbb {E} \left[ X_m(u_n + \varepsilon )^2 \right]&= \left( \frac{1}{n^{\alpha }} \right) ^2 \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{\lambda _m (u_n + \varepsilon - s)} e^{\lambda _m (u_n + \varepsilon - y)} \, ds \, dy\\&\hspace{4mm}+ \left( \frac{1}{n^{\alpha }} \right) \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) \right] \mathbb {E} \left[ {\tilde{X}}_m(u_n + \varepsilon - s)^2 \right] \, ds, \\ \mathbb {E} \left[ X_s(s) X_m(u_n + \varepsilon ) \right]&= \left( \frac{1}{n^{\alpha }} \right) \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(y) X_s(s) \right] e^{\lambda _m (u_n + \varepsilon - y)} \, dy, \end{aligned}$$

where ${\tilde{X}}_m$ is a binary branching process starting from size one with birth rate $r_m$ and death rate $d_m$. Substituting these expressions into our equation yields

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\alpha +\lambda _m u_n^+(\varepsilon )/\lambda _s - 1} X_m(u_n + \varepsilon ) - \frac{1}{n} \int _0^{u_n + \varepsilon } X_s(s) e^{-\lambda _m s} \, ds \right) ^2 \right] \\&\quad = n^{2(\lambda _m u_n^+(\varepsilon )/\lambda _s - 1)} \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{\lambda _m (2u_n + 2\varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{\alpha + 2(\lambda _m u_n^+(\varepsilon )/\lambda _s - 1)} \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) \right] \mathbb {E} \left[ {\tilde{X}}_m(u_n + \varepsilon - s)^2 \right] \, ds\\&\qquad - 2 n^{\lambda _m u_n^+(\varepsilon )/\lambda _s - 2} \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{\lambda _m (u_n + \varepsilon - s - y)} \, ds \, dy\\&\qquad + n^{-2} \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m (s+y)} \, ds \, dy. \end{aligned}$$

Then, by the definition of $u_n^+(\varepsilon )$,

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\alpha +\lambda _m u_n^+(\varepsilon )/\lambda _s - 1} X_m(u_n + \varepsilon ) - \frac{1}{n} \int _0^{u_n + \varepsilon } X_s(s) e^{-\lambda _m s} \, ds \right) ^2 \right] \\&\quad = n^{-2 \left( \frac{\lambda _m (u_n + \varepsilon )}{\log n} + 1 \right) } \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{\lambda _m (2u_n + 2\varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{\alpha - 2 \left( \frac{\lambda _m (u_n + \varepsilon )}{\log n} + 1 \right) } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) \right] \mathbb {E} \left[ {\tilde{X}}_m(u_n + \varepsilon - s)^2 \right] \, ds \\&\qquad - 2 n^{-\frac{\lambda _m (u_n + \varepsilon )}{\log n} - 2} \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{\lambda _m (u_n + \varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{-2} \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m (s+y)} \, ds \, dy \\&\quad = \left( \frac{1}{n} \right) ^2 \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m (s+y)} \, ds \, dy \\&\qquad + n^{\alpha - 2} e^{-2\lambda _m (u_n + \varepsilon )} \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) \right] \mathbb {E} \left[ {\tilde{X}}_m(u_n + \varepsilon - s)^2 \right] \, ds\\&\qquad - 2 \left( \frac{1}{n} \right) ^2 \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m (s+y)} \, ds \, dy \\&\qquad + \left( \frac{1}{n} \right) ^2 \int _0^{u_n + \varepsilon } \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) X_s(y) \right] e^{-\lambda _m (s+y)} \, ds \, dy \\&\quad = n^{\alpha - 2} e^{-2 \lambda _m (u_n + \varepsilon )} \int _0^{u_n + \varepsilon } \mathbb {E} \left[ X_s(s) \right] \mathbb {E} \left[ {\tilde{X}}_m(u_n + \varepsilon - s)^2 \right] \, ds. \end{aligned}$$

Note that

$$\begin{aligned} \mathbb {E} \left[ X_s(t) \right]&= ne^{\lambda _st}, \\ \mathbb {E} \left[ {\tilde{X}}_m(t)^2 \right]&= \frac{2r_m}{\lambda _m } e^{2\lambda _m t} - \frac{r_m + d_m}{\lambda _m } e^{\lambda _m t}. \end{aligned}$$

Substituting these into the above expression yields

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\alpha +\lambda _m u_n^+(\varepsilon )/\lambda _s-1} X_m(u_n + \varepsilon ) - \frac{1}{n} \int _0^{u_n + \varepsilon } X_s(s) e^{-\lambda _m s} \, ds \right) ^2 \right] \\&\quad = n^{\alpha - 2} e^{-2\lambda _m (u_n + \varepsilon )} \int _0^{u_n + \varepsilon } ne^{\lambda _ss} \left( \frac{2r_m}{\lambda _m } e^{2\lambda _m (u_n + \varepsilon - s)} - \frac{r_m + d_m}{\lambda _m } e^{\lambda _m (u_n + \varepsilon - s)} \right) \, ds \\&\quad = \frac{2r_m n^{\alpha - 1}}{\lambda _m } \int _0^{u_n + \varepsilon } e^{(\lambda _s - 2\lambda _m )s} \, ds - \frac{ n^{\alpha - 1} (r_m + d_m)}{\lambda _m } e^{-\lambda _m (u_n + \varepsilon )} \int _0^{u_n + \varepsilon } e^{(\lambda _s - \lambda _m )s} \, ds \\&\quad = \frac{2r_m n^{\alpha - 1}}{\lambda _m (\lambda _s - 2\lambda _m )} \left( e^{(\lambda _s - 2\lambda _m )(u_n + \varepsilon )} - 1 \right) \\&\qquad - \frac{ n^{\alpha - 1} (r_m + d_m)}{\lambda _m (\lambda _s - \lambda _m )} \left( e^{(\lambda _s - 2\lambda _m )(u_n + \varepsilon )} - e^{-\lambda _m (u_n + \varepsilon )} \right) . \end{aligned}$$

Since $u_n = {\tilde{u}}_n t_n$ and $t_n = \frac{-1}{\lambda _s} \log n$, the expression above is equivalent to

$$\begin{aligned}&\frac{ n^{\alpha - 1}}{\lambda _m } \Biggl [ e^{(\lambda _s - 2\lambda _m )({\tilde{u}}_n \frac{-1}{\lambda _s} \log n + \varepsilon )} \left( \frac{2r_m}{\lambda _s - 2\lambda _m } - \frac{r_m + d_m}{\lambda _s - \lambda _m } \right) \\&\quad - \frac{2r_m}{\lambda _s - 2\lambda _m } + \frac{r_m + d_m}{\lambda _s - \lambda _m } e^{-\lambda _m ({\tilde{u}}_n \frac{-1}{\lambda _s} \log n + \varepsilon )} \Biggr ]. \end{aligned}$$

Because ${\tilde{u}}_n \rightarrow \frac{-\lambda _s \alpha }{\lambda _m }$ as $n \rightarrow \infty$, $e^{(\lambda _s - 2\lambda _m )({\tilde{u}}_n \frac{-1}{\lambda _s} \log n + \varepsilon )} \rightarrow 0$ and $e^{-\lambda _m ({\tilde{u}}_n \frac{-1}{\lambda _s} \log n + \varepsilon )} \rightarrow 0$. Therefore, the entire expression above converges to zero since $\alpha < 1$, so we are done. $\square$

1.8 Proof of lemma 3

Proof

Since $\mathbb {P} \left( \sum _{k=2}^M X_k(dt_n) - n \le 0 \right) = 1 - \mathbb {P} \left( \sum _{k=2}^M X_k(dt_n) - n > 0 \right)$, it suffices to show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sum _{k=2}^M X_k(dt_n) - n > 0 \right) = 0. \end{aligned}$$

Note that

$$\begin{aligned}&\mathbb {P} \left( \sum _{k=2}^M X_k(dt_n) - n> 0 \right) \\&\quad = \mathbb {P} \left( n^{\beta (M-2) + \lambda _Md/\lambda _2 - 1} \left( \sum _{k=2}^M X_k(dt_n) + \sum _{k=2}^M \phi _k(dt_n) - \sum _{k=2}^M \phi _k(dt_n) - n \right)> 0 \right) \\&\quad = \mathbb {P} \left( \sum _{k=2}^M {\hat{A}}_k(n) + {\hat{A}}_{\phi }(n) > 0 \right) , \end{aligned}$$

where

$$\begin{aligned} {\hat{A}}_k(n)&= n^{\beta (M-2) + \lambda _Md/\lambda _2 - 1} \left( X_k(dt_n) - \phi _k(dt_n) \right) , \\ {\hat{A}}_{\phi }(n)&= n^{\beta (M-2) + \lambda _Md/\lambda _2 - 1} \left( \sum _{k=2}^M \phi _k(dt_n) - n \right) . \end{aligned}$$

Note that

$$\begin{aligned} \left| {\hat{A}}_k(n) \right|&= n^{\beta (M-2) + \lambda _Md/\lambda _2 - 1} \left| X_k(dt_n) - \phi _k(dt_n) \right| \\&\le \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left| X_k(zt_n) - \phi _k(zt_n) \right| , \end{aligned}$$

which converges to zero in probability by Proposition 3, below. Now we just need to show that ${\hat{A}}_{\phi }(n)$ is negative in the large population limit. By the definitions of $\phi _2$ and $\phi _k$, $k \ge 3$,

$$\begin{aligned} {\hat{A}}_{\phi }(n)&= n^{\beta (M-2) + \lambda _Md/\lambda _2 - 1} \left( n^{1-d} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(d) - n \right) \\&= n^{\beta (M-2) + \lambda _Md/\lambda _2} \left( n^{-d} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{-\beta (k-2)} {\tilde{S}}_k(d) - 1 \right) \\&\sim n^{\beta (M-2) + \lambda _Md/\lambda _2} \left( \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \cdot \frac{n^{-\beta (k-2) - \lambda _kd/\lambda _2}}{{\tilde{P}}_{k,k}} - 1 \right) . \end{aligned}$$

For $3 \le k < M$, we obviously have $k-2 < M-2$. Since $d > -\lambda _2 \beta \frac{M-2}{\lambda _M - \lambda _2} = \frac{-\lambda _2 \beta }{D}$ by (1), this means that $Dd/\lambda _2 < -\beta$, and hence $\beta + Dd/\lambda _2 < 0$. Therefore, for all $3 \le k < M$,

$$\begin{aligned} -(k-2)(\beta + Dd/\lambda _2)&< -(M-2)(\beta + Dd/\lambda _2) \\ - d - \beta (k-2) - (k-2)Dd/\lambda _2&< -d - \beta (M-2) - (M-2)Dd/\lambda _2 \\ -\beta (k-2) - [\lambda _2 + (k-2)D]d/\lambda _2&< -\beta (M-2) - [\lambda _2 + (M-2)D]d/\lambda _2 \\ -\beta (k-2) - \lambda _kd/\lambda _2&< -\beta (M-2) - \lambda _Md/\lambda _2. \end{aligned}$$

This implies that

$$\begin{aligned} {\hat{A}}_{\phi }(n)&\sim n^{\beta (M-2) + \lambda _Md/\lambda _2} \left( \frac{ (-1)^M}{D^{M-2} {\tilde{P}}_{M,M}} n^{-\beta (M-2) - \lambda _Md/\lambda _2} - 1 \right) \\&= \frac{ (-1)^M}{D^{M-2} {\tilde{P}}_{M,M}} - n^{\beta (M-2) + \lambda _Md/\lambda _2}. \end{aligned}$$

Since $d < \frac{-\lambda _2 \beta (M-2)}{\lambda _M}$ by (1), we have that $\lambda _M d/\lambda _2 > -\beta (M-2)$, and hence $n^{\beta (M-2) + \lambda _Md/\lambda _2} \rightarrow \infty$ as $n \rightarrow \infty$. Therefore, ${\hat{A}}_{\phi }(n)$ is definitely negative in the large population limit. Putting this all together, we have our desired result: $\mathbb {P} \left( \sum _{k=2}^M {\hat{A}}_k(n) + {\hat{A}}_{\phi }(n) > 0 \right) \rightarrow 0$ as $n \rightarrow \infty$. $\square$

1.9 Proof of proposition 2

Proof

Let ${\hat{f}}_n(z) = \sum _{k=2}^M \phi _k(z t_n) - n$. Taking the first derivative of $\phi _2$ with respect to z, we get

$$\begin{aligned} \frac{d}{dz} \phi _2(z t_n) = -n^{1-z} \log n. \end{aligned}$$

Taking a second derivative yields

$$\begin{aligned} \frac{d^2}{dz^2} \phi _2(z t_n)&= (\log n)^2 n^{1-z} \\&> 0. \end{aligned}$$

Hence $\phi _2(z t_n)$ is concave up everywhere. Now for $k>2$, we have

$$\begin{aligned} \phi _k(z t_n) = \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(z). \end{aligned}$$

Taking the first derivative of this equation with respect to z, we get

$$\begin{aligned} \frac{d}{dz} \phi _k(z t_n) = \frac{ (-1)^k}{D^{k-2}} n^{1 - (k-2)\beta } \log n \left( -\frac{\lambda _i}{\lambda _2} \right) {\tilde{S}}_k(z). \end{aligned}$$

Taking a second derivative yields

$$\begin{aligned} \frac{d^2}{dz^2} \phi _k(zt_n)&= \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } (\log n)^2 \left( \frac{\lambda _i}{\lambda _2} \right) ^2 {\tilde{S}}_k(z) \\&\sim \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } (\log n)^2 \frac{(\lambda _k/\lambda _2)^2 n^{-\lambda _kz/\lambda _2}}{{\tilde{P}}_{k,k}} \\&> 0. \end{aligned}$$

So $\phi _k(z t_n)$ is concave up in the large population limit for all $k \ge 2$. Therefore, $\sum _{k=2}^M \phi _k(z t_n)$ is concave up since a sum of concave up functions is always concave up. Hence ${\hat{f}}_n(z)$ is concave up as well. Since ${\hat{f}}_n(z)$ is clearly differentiable everywhere, this implies that either ${\hat{f}}_n(z)$ is monotonically increasing everywhere, or ${\hat{f}}_n(z)$ is decreasing on the first part of its domain and then increasing on the rest of its domain. Note that ${\hat{f}}_n(0) = 0$. We also have that

$$\begin{aligned} {\hat{f}}_n(b_n)&= n^{1-b_n} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(b_n) - n \\&= n \Biggl [ n^{\frac{\lambda _2}{\lambda _M} \beta (M-2)} n^{\frac{1}{\lambda _M t_n} \log \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \left( \frac{\lambda _2 - \lambda _M}{\lambda _2} \right) \right] } \\&\quad + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \sum _{i=2}^k \frac{n^{-\beta (k-2)} n^{\frac{\lambda _i}{\lambda _M} \beta (M-2)} n^{\frac{\lambda _i}{\lambda _M \lambda _2 t_n} \log \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \left( \frac{\lambda _2 - \lambda _M}{\lambda _2} \right) \right] }}{{\tilde{P}}_{i,k}} - 1 \Biggr ]\\&\quad \sim n \left[ \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \cdot \frac{n^{\beta [ \lambda _k(M-2)/\lambda _M - (k-2)]} \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \left( \frac{\lambda _2 - \lambda _M}{\lambda _2} \right) \right] ^{-\lambda _k/\lambda _M}}{{\tilde{P}}_{k,k}} - 1 \right] \\&= n \Biggl [ \sum _{k=3}^{M-1} \frac{ (-1)^k}{D^{k-2}} \cdot \frac{n^{\beta [ \lambda _k(M-2)/\lambda _M - (k-2)]} \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \left( \frac{\lambda _2 - \lambda _M}{\lambda _2} \right) \right] ^{-\lambda _k/\lambda _M}}{{\tilde{P}}_{k,k}} \\&+ \frac{\lambda _2}{\lambda _2 - \lambda _M} - 1 \Biggr ]. \end{aligned}$$

Note that

$$\begin{aligned} \beta \left[ \frac{\lambda _k(M-2)}{\lambda _M} - (k-2) \right]&= \beta \frac{ \left[ \lambda _2 + (k-2) \frac{\lambda _M - \lambda _2}{M-2} \right] (M-2) - \left[ \lambda _2 + (M-2) \frac{\lambda _M - \lambda _2}{M-2} \right] (k-2)}{\lambda _M} \\&= \beta \frac{ \lambda _2(M-2) + (k-2)(\lambda _M - \lambda _2) - \lambda _2(k-2) - (k-2)(\lambda _M - \lambda _2)}{\lambda _M} \\&= \beta \frac{\lambda _2}{\lambda _M} (M-k), \end{aligned}$$

so we can substitute this to get

$$\begin{aligned} {\hat{f}}_n(b_n)&\sim n \Biggl [ \sum _{k=3}^{M-1} \frac{ (-1)^k \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \left( \frac{\lambda _2 - \lambda _M}{\lambda _2} \right) \right] ^{-\lambda _k/\lambda _M}}{D^{k-2} {\tilde{P}}_{k,k}} n^{\beta \lambda _2(M-k)/\lambda _M} + \frac{\lambda _2}{\lambda _2 - \lambda _M} - 1 \Biggr ] \\&\sim n \left( \frac{\lambda _2}{\lambda _2 - \lambda _M} - 1 \right) \\&< 0. \end{aligned}$$

So ${\hat{f}}_n(b_n) < 0$ in the large population limit. Similarly,

$$\begin{aligned} {\hat{f}}_n(B_n)&= n^{1-B_n} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(B_n) - n \\&= n \Biggl [ n^{\frac{\lambda _2}{\lambda _M} \beta (M-2)} n^{\frac{1}{\lambda _M t_n} \log \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \right] } \\&\quad + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \sum _{i=2}^k \frac{n^{-\beta (k-2)} n^{\frac{\lambda _i}{\lambda _M} \beta (M-2)} n^{\frac{\lambda _i}{\lambda _M \lambda _2 t_n} \log \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \right] }}{{\tilde{P}}_{i,k}} - 1 \Biggr ] \\&\quad \sim n \left[ \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \cdot \frac{n^{\beta [ \lambda _k(M-2)/\lambda _M - (k-2)]} \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \right] ^{-\lambda _k/\lambda _M}}{{\tilde{P}}_{k,k}} - 1 \right] \\&= n \left[ \sum _{k=3}^{M-1} \frac{ (-1)^k \left[ \frac{ (-1)^M}{D^{M-2}{\tilde{P}}_{M,M}} \right] ^{-\lambda _k/\lambda _M}}{D^{k-2} {\tilde{P}}_{k,k}} n^{\beta \lambda _2(M-k)/\lambda _M} \right] \\&> 0. \end{aligned}$$

So ${\hat{f}}_n(B_n) > 0$ in the large population limit. Therefore, we know that ${\hat{f}}_n(z)$ must be monotonically decreasing on the first part of its domain and monotonically increasing on the rest of its domain. This implies that there exists one and only one positive solution ${\tilde{v}}_n$ to the equation ${\hat{f}}_n(z) = 0$. And since ${\hat{f}}_n(b_n) < 0$ and ${\hat{f}}_n(B_n) > 0$ in the large population limit, we must have $b_n< {\tilde{v}}_n < B_n$. Lastly, because $\lim _{n \rightarrow \infty } b_n = \lim _{n \rightarrow \infty } B_n = -\frac{\lambda _2}{\lambda _M}\beta (M-2)$, the solution ${\tilde{v}}_n \rightarrow -\frac{\lambda _2}{\lambda _M}\beta (M-2)$ as $n \rightarrow \infty$ as well. $\square$

1.10 Proof of theorem 1

Proof

In order to show the desired result, we must show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} (\omega _n > v_n + \varepsilon ) + \lim _{n \rightarrow \infty } \mathbb {P} (\omega _n < v_n - \varepsilon ) = 0. \end{aligned}$$

Let’s start by proving that $\mathbb {P} (\omega _n < v_n - \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$. Note that

$$\begin{aligned}&\mathbb {P} (\omega _n< v_n - \varepsilon ) \\&\quad = \mathbb {P} \left( \frac{\omega _n}{t_n} < v_n^-(\varepsilon ) \right) \\&\quad \le \mathbb {P} \left( \sup _{z \in [d, v_n^-(\varepsilon )]} \left( \sum _{k=2}^M X_k(zt_n) - n \right)> 0 \right) \\&\quad = \mathbb {P} \left( \sup _{z \in [d, v_n^-(\varepsilon )]} n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left( \sum _{k=2}^M X_k(zt_n) + \sum _{k=2}^M \phi _k(zt_n) - \sum _{k=2}^M \phi _k(zt_n) - n \right)> 0 \right) \\&\quad \le \mathbb {P} \left( \sum _{k=2}^M {\hat{B}}_k(n, \varepsilon ) + {\hat{B}}_{\phi }(n, \varepsilon ) > 0 \right) , \end{aligned}$$

where

$$\begin{aligned} {\hat{B}}_k(n, \varepsilon )&= \sup _{z \in [d, v_n^-(\varepsilon )]} n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left( X_k(zt_n) - \phi _k(zt_n) \right) , \\ {\hat{B}}_{\phi }(n, \varepsilon )&= \sup _{z \in [d, v_n^-(\varepsilon )]} n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left( \sum _{k=2}^M \phi _k(zt_n) - n \right) . \end{aligned}$$

Note that

$$\begin{aligned}&\sup _{z \in [d, v_n^-(\varepsilon )]} \left| n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left( X_k(zt_n) - \phi _k(zt_n) \right) \right| \\&\le \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left| X_k(zt_n) - \phi _k(zt_n) \right| , \end{aligned}$$

which converges to zero in probability by Proposition 3. Now we just need to show that ${\hat{B}}_{\phi }(n, \varepsilon )$ is negative in the large population limit. Let ${\hat{g}}_n(z) = n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left( \sum _{k=2}^M \phi _k(z t_n) - n \right)$. Using the definitions of $\phi _2$ and $\phi _k$ for $k>2$, we see that

$$\begin{aligned} {\hat{g}}_n(z)&= n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} \left( n^{1-z} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(z) - n \right) \\&= n^{\beta (M-2) + z(\lambda _M/\lambda _2 - 1)} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \sum _{i=2}^k \frac{n^{\beta (M-k) + (\lambda _M - \lambda _i)z/\lambda _2}}{{\tilde{P}}_{i,k}} - n^{\beta (M-2) + \lambda _Mz/\lambda _2} \\&= n^{(M-2)(\beta + Dz /\lambda _2)} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \sum _{i=2}^k \frac{n^{\beta (M-k) + (M-i)Dz/\lambda _2}}{{\tilde{P}}_{i,k}} - n^{\beta (M-2) + \lambda _Mz/\lambda _2}. \end{aligned}$$

Taking the derivative with respect to z yields

$$\begin{aligned} {\hat{g}}_n^{\prime }(z)&= n^{(M-2)(\beta + Dz /\lambda _2)} \log n \cdot \frac{(M-2)D}{\lambda _2} \\&\quad + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} \sum _{i=2}^k \frac{n^{\beta (M-k) + (M-i)Dz/\lambda _2} \log n \cdot \frac{(M-i)D}{\lambda _2}}{{\tilde{P}}_{i,k}} \\&\quad - n^{\beta (M-2)}\log n \cdot \frac{\lambda _M}{\lambda _2} n^{\lambda _Mz/\lambda _2} \end{aligned}$$

Since $\lambda _2 < 0$ we know that the inner sum of the second term will be asymptotically dominated by the $i = k$ term as $n \rightarrow \infty$. Thus,

$$\begin{aligned}&{\hat{g}}_n^{\prime }(z) \sim \frac{\log n}{\lambda _2} \left[ (M-2)D \cdot n^{(M-2)(\beta + Dz /\lambda _2)}\right. \\&\quad \left. + \sum _{k=3}^M \frac{ (-1)^k (M-k)}{D^{k-3} {\tilde{P}}_{k,k}} n^{(M-k)(\beta + Dz /\lambda _2)} - \lambda _M n^{\beta (M-2)+\lambda _Mz/\lambda _2} \right] , \end{aligned}$$

Let $z \in [d, v_n^{-}(\varepsilon )]$. Since $z \ge d$ we have that $z > -\lambda _2 \beta \frac{M-2}{\lambda _M - \lambda _2} = -\lambda _2 \beta /D$ by (1). Therefore, $Dz /\lambda _2 < -\beta$, and hence $\beta + Dz /\lambda _2 < 0$. So $(M-k)(\beta + Dz /\lambda _2) \le 0$ for $2 \le k \le M$. On the other hand, since $z \le v_n^{-}(\varepsilon ) < -\frac{\lambda _2}{\lambda _M} \beta (M-2)$ for sufficiently large n by Proposition 2, we have that $\lambda _M z/\lambda _2 > - \beta (M-2)$, and hence $\beta (M-2) + \lambda _M z /\lambda _2 > 0$. Together, this implies that when $z \in [d, v_n^{-}(\varepsilon )]$,

$$\begin{aligned} {\hat{g}}_n^{\prime }(z)&\sim \frac{\log n}{\lambda _2} \left[ - \lambda _M n^{\beta (M-2)+\lambda _Mz/\lambda _2}\right] \\&= \frac{\lambda _M \log n}{-\lambda _2} n^{\beta (M-2)+\lambda _Mz/\lambda _2}\\&\quad > 0 \hbox { for sufficiently large}\ n. \end{aligned}$$

Hence ${\hat{g}}_n(z)$ is monotonically increasing on the interval $[d, v_n^-(\varepsilon )]$. Therefore, we may rewrite ${\hat{B}}_{\phi }(n, \varepsilon )$ as

$$\begin{aligned} {\hat{B}}_{\phi }(n, \varepsilon ) = n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2 - 1} \left( \sum _{k=2}^M \phi _k(v_n^-(\varepsilon ) t_n) - n \right) . \end{aligned}$$

Then by the definitions of $\phi _2$ and $\phi _k$ for $k>2$, we have

$$\begin{aligned} {\hat{B}}_{\phi }(n, \varepsilon )&= n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2 - 1} \left( n^{1-v_n^-(\varepsilon )} + \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } {\tilde{S}}_k(v_n^-(\varepsilon )) - n \right) \\&= \Biggl ( n^{(M-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \\&\quad + n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2 - 1} \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}} n^{1-(k-2)\beta } \sum _{i=2}^k \frac{n^{-v_n^-(\varepsilon ) - (i-2)D v_n^-(\varepsilon )/\lambda _2}}{{\tilde{P}}_{i,k}}\\&\quad -n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2} \Biggr ) \end{aligned}$$

Similar to the previous calculations, since $\lambda _2 < 0$ we know that the inner sum of the second term will be dominated by the $i = k$ term as $n \rightarrow \infty$. Thus,

$$\begin{aligned} {\hat{B}}_{\phi }(n, \varepsilon )&\sim \Biggl ( n^{(M-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \\&\quad + n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2 - 1} \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}{\tilde{P}}_{k,k}} n^{1-(k-2)\beta -v_n^-(\varepsilon ) - (k-2)D v_n^-(\varepsilon )/\lambda _2} \\&\quad -n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2} \Biggr ) \\&= \Biggl ( n^{(M-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \\&\quad + n^{(M-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}{\tilde{P}}_{k,k}} n^{-(k-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \\&\quad -n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2} \Biggr ) \\&= \Biggl ( n^{(M-2)(\beta + D v_n^-(\varepsilon )/\lambda _2)}\Bigg . \\&\Bigg .+ \sum _{k=3}^M \frac{ (-1)^k}{D^{k-2}{\tilde{P}}_{k,k}} n^{(M-k)(\beta + D v_n^-(\varepsilon )/\lambda _2)} \quad -n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2} \Biggr ) \end{aligned}$$

Since $v_n^-(\varepsilon ) > d$, we have that $v_n^-(\varepsilon ) > -\lambda _2 \beta \frac{M-2}{\lambda _M - \lambda _2} = -\lambda _2 \beta /D$ by (1). Therefore, $Dv_n^-(\varepsilon )/\lambda _2 < -\alpha$, and hence $\beta + Dv_n^-(\varepsilon )/\lambda _2 < 0$. This implies that $(M-k)(\beta + D v_n^-(\varepsilon )/\lambda _2) < 0$ for $2 \le k \le M$. Moreover, since $v_n^-(\varepsilon ) < \frac{-\lambda _2}{\lambda _M}\beta (M-2)$ for sufficiently large n by Proposition 2, we have that $\frac{\lambda _M}{\lambda _2} v_n^-\varepsilon > - \beta (M-2)$ and hence $\beta (M-2) +\frac{\lambda _M}{\lambda _2} v_n^-\varepsilon > 0$. Thus,

$$\begin{aligned} {\hat{B}}_{\phi }(n, \varepsilon )\sim -n^{\beta (M-2) + \lambda _M v_n^-(\varepsilon )/\lambda _2}. \end{aligned}$$

So ${\hat{B}}_{\phi }(n, \varepsilon )$ is definitely negative in the large population limit. Putting this all together, we have that $\mathbb {P} \left( \sum _{k=2}^M {\hat{B}}_k(n, \varepsilon ) + {\hat{B}}_{\phi }(n, \varepsilon ) > 0 \right) \rightarrow 0$ as $n \rightarrow \infty$. Therefore, we have shown $\mathbb {P} (\omega _n < v_n - \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$. The proof that $\mathbb {P}(\omega _n > v_n + \varepsilon ) \rightarrow 0$ as $n \rightarrow \infty$ follows using a similar argument. $\square$

1.11 Proof of Proposition 3

Proof

Let’s start by proving the $k=2$ case. That is, we want to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left| X_2(zt_n) - \phi _2(zt_n) \right| > \delta \right) = 0. \end{aligned}$$

Because a branching process normalized by its mean is a martingale, we know that

$$\begin{aligned} Z_2(z) = n^{\beta (M-2) + z - 1} \left( X_2(zt_n) - \phi _2(zt_n) \right) \end{aligned}$$

is a martingale in z. To prove our desired result for $k=2$, we need to show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(\lambda _M - \lambda _2)z/\lambda _2} |Z_2(z)| > \delta \right) = 0, \end{aligned}$$

or, equivalently,

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(M-2)Dz/\lambda _2} |Z_2(z)| > \delta \right) = 0. \end{aligned}$$

Note that

$$\begin{aligned} \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(M - 2)Dz/\lambda _2} |Z_2(z)|&\le \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(M - 2)Dz/\lambda _2} \cdot \sup _{z \in [d, v_n^+(\varepsilon )]} |Z_2(z)|\\&= n^{(M - 2)Dd/\lambda _2} \cdot \sup _{z \in [d, v_n^+(\varepsilon )]} |Z_2(z)|. \end{aligned}$$

Therefore, we have that

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(M-2)Dz/\lambda _2} |Z_2(z)|> \delta \right) \\&\quad \le \mathbb {P} \left( n^{(M-2)Dd/\lambda _2} \cdot \sup _{z \in [d, v_n^+(\varepsilon )]} |Z_2(z)|> \delta \right) \\&\quad = \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} |Z_2(z)| > \delta \cdot n^{(2-M)Dd/\lambda _2} \right) \\&\quad \le \frac{1}{\delta } n^{(M-2)Dd/\lambda _2} \cdot \mathbb {E} \left[ \left| Z_2(v_n^+(\varepsilon )) \right| \right] \text { by Doob's Martingale Inequality} \\&\quad = \frac{1}{\delta } n^{(M-2)Dd/\lambda _2} \cdot \mathbb {E} \left[ \left| n^{\beta (M-2) + v_n^+(\varepsilon ) - 1} (X_2(v_n^+(\varepsilon ) t_n) - \phi _2(v_n^+(\varepsilon ) t_n)) \right| \right] \\&\quad = \frac{1}{\delta } n^{\beta (M-2) + (M-2)Dd/\lambda _2 + v_n^+(\varepsilon ) - 1} \cdot \mathbb {E} \left[ \left| X_2(v_n^+(\varepsilon ) t_n) - \phi _2(v_n^+(\varepsilon ) t_n) \right| \right] . \end{aligned}$$

Clearly,

$$\begin{aligned} \left| X_2(v_n^+(\varepsilon ) t_n) - \phi _2(v_n^+(\varepsilon ) t_n) \right|&\le \left| X_2(v_n^+(\varepsilon ) t_n) \right| + \left| \phi _2(v_n^+(\varepsilon ) t_n) \right| \\&= X_2(v_n^+(\varepsilon ) t_n) + \phi _2(v_n^+(\varepsilon ) t_n). \end{aligned}$$

This implies that

$$\begin{aligned} \mathbb {E} \left[ \left| X_2(v_n^+(\varepsilon ) t_n) - \phi _2(v_n^+(\varepsilon ) t_n) \right| \right]&\le \mathbb {E} \left[ X_2(v_n^+(\varepsilon ) t_n) + \phi _2(v_n^+(\varepsilon ) t_n) \right] \\&= 2 \phi _2(v_n^+(\varepsilon ) t_n). \end{aligned}$$

And hence

$$\begin{aligned} \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{(M - 2)Dz/\lambda _2} |Z_2(z)| > \delta \right)&\le \frac{2}{\delta } n^{\beta (M-2) + (M - 2)Dd/\lambda _2 + v_n^+(\varepsilon ) - 1} n^{1-v_n^+(\varepsilon )} \\&= \frac{2}{\delta } n^{(M-2)(\beta + Dd/\lambda _2)}. \end{aligned}$$

So it suffices to show that $\beta + Dd/\lambda _2 < 0$. From (1), we have that

$$\begin{aligned} d > -\lambda _2 \beta \frac{M-2}{\lambda _M - \lambda _2} = \frac{-\lambda _2 \beta }{m}, \end{aligned}$$

which implies that $Dd/\lambda _2 < -\beta$, so we are done with the $k=2$ case.

Next, let’s prove the desired result for $k>2$. Note that

$$\begin{aligned} \phi _k(zt_n)&= \int _0^{zt_n} n^{-\beta } \phi _{k-1}(s) e^{\lambda _k(zt_n - s)} \, ds \\&= n^{-\beta -\lambda _k z / \lambda _2} \int _0^{zt_n} \phi _{k-1}(s) e^{-\lambda _ks} \, ds. \end{aligned}$$

Therefore,

$$\begin{aligned}&n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left( X_k(zt_n) - \phi _k(zt_n) \right) \\&\quad = n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (\lambda _M - \lambda _k)z/\lambda _2 - 1} \int _0^{zt_n} \phi _{k-1}(s) e^{-\lambda _k s} \, ds \\&\quad = n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} \phi _{k-1}(s) e^{-\lambda _k s} \, ds \\&\quad = n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _k s} \, ds \\&\qquad + n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} \left( X_{k-1}(s) - \phi _{k-1}(s) \right) e^{-\lambda _k s} \, ds. \end{aligned}$$

Taking the absolute value of both sides and using the triangle inequality, we get

$$\begin{aligned}&n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left| X_k(zt_n) - \phi _k(zt_n) \right| \\&\quad \le \left| n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _k s} \, ds \right| \\&\qquad + n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _k s} \, ds. \end{aligned}$$

This implies that

$$\begin{aligned}&\sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left| X_k(zt_n) - \phi _k(zt_n) \right| \\&\quad \le \sup _{z \in [d, v_n^+(\varepsilon )]} \left| n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \right. \\&\qquad \left. \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _ks} \, ds \right| + \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \\&\qquad \int _0^{zt_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _ks} \, ds. \end{aligned}$$

Hence

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} \left| X_k(zt_n) - \phi _k(zt_n) \right|> \delta \right) \nonumber \\&\quad \le \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} \left| n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _ks} \, ds \right| > \delta /2 \right) \end{aligned}$$

(10)

$$\begin{aligned}&\qquad + \mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _ks} \, ds > \delta /2 \right) . \end{aligned}$$

(11)

As a reminder, our goal is to show that (11) converges to 0 as $n \rightarrow \infty$.

Let’s start with the second term. Note that

$$\begin{aligned}&\sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _ks} \, ds \\&\quad \le \sup _{z \in [d, v_n^+(\varepsilon )]} n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \sup _{z \in [d, v_n^+(\varepsilon )]} \int _0^{zt_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _ks} \, ds \\&\quad = n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n^+(\varepsilon )t_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _ks} \, ds. \end{aligned}$$

So, in order to show the second term in (11) converges to 0, we must show

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {P} \left( n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n^+(\varepsilon ) t_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _k s} \, ds > \delta /2 \right) = 0. \end{aligned}$$

Since convergence in mean implies convergence in probability (by Markov’s Inequality), it suffices to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \mathbb {E} \left( n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n^+(\varepsilon ) t_n} \left| X_{k-1}(s) - \phi _{k-1}(s) \right| e^{-\lambda _k s} \, ds \right) = 0 \end{aligned}$$

or, equivalently,

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ \left| X_{k-1}(s) - \phi _{k-1}(s) \right| \right] e^{-\lambda _k s} \, ds = 0. \end{aligned}$$

Note that

$$\begin{aligned} \left| X_{k-1}(s) - \phi _{k-1}(s) \right|&\le \left| X_{k-1}(s) \right| + \left| \phi _{k-1}(s) \right| \text { by the triangle inequality}\\&= X_{k-1}(s) + \phi _{k-1}(s). \end{aligned}$$

Taking the mean of both sides yields

$$\begin{aligned} \mathbb {E} \left[ \left| X_{k-1}(s) - \phi _{k-1}(s) \right| \right]&\le \mathbb {E} \left[ X_{k-1}(s) + \phi _{k-1}(s) \right] \\&= 2 \phi _{k-1}(s). \end{aligned}$$

Multiplying both sides by $n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} e^{-\lambda _k s}$ and integrating gives us the following inequality:

$$\begin{aligned}&n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ \left| X_{k-1}(s) - \phi _{k-1}(s) \right| \right] e^{-\lambda _k s} \, ds \\&\le n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } 2 \phi _{k-1}(s) e^{-\lambda _k s} \, ds. \end{aligned}$$

So it suffices to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } 2 \phi _{k-1}(s) e^{-\lambda _k s} \, ds = 0 \end{aligned}$$

or, equivalently,

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] e^{-\lambda _k s} \, ds = 0. \end{aligned}$$

Using the definition of $\mathbb {E} \left[ X_{k-1}(s) \right]$, we see that

$$\begin{aligned}&n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] e^{-\lambda _k s} \, ds \\&\quad = n^{\beta (M-3) + (M-k)Dd/\lambda _2 - 1} \int _0^{v_n + \varepsilon } \left[ n^{1-(k-3)\beta } (-1)^{k-1} S_{k-1}(s) \right] e^{-\lambda _k s} \, ds \\&\quad = n^{\beta (M-k) + (M-k)Dd/\lambda _2} (-1)^{k-1} \int _0^{v_n + \varepsilon } \sum _{i=2}^{k-1} \frac{e^{(\lambda _i - \lambda _k)s}}{P_{i,k-1}} \, ds \\&\quad = \left( \frac{1}{m} \right) ^{k-3} n^{(M-k)(\beta + Dd/\lambda _2)} (-1)^{k-1} \sum _{i=2}^{k-1} \int _0^{v_n + \varepsilon } \frac{e^{(i-k)Ds}}{{\tilde{P}}_{i,k-1}} \, ds \\&\quad = \frac{1}{D^{k-2}} n^{(M-k)(\beta + Dd/\lambda _2)} (-1)^k \sum _{i=2}^{k-1} \frac{1}{{\tilde{P}}_{i,k}} \left( e^{(i-k)D(v_n + \varepsilon )} - 1 \right) . \end{aligned}$$

Since $v_n = {\tilde{v}}_n t_n = -\frac{{\tilde{v}}_n}{\lambda _2} \log n$,

$$\begin{aligned} e^{(i-k)D(v_n + \varepsilon )}&= e^{(k-i)D \frac{{\tilde{v}}_n}{\lambda _2} \log n} e^{(i-k)D \varepsilon } \\&= n^{(k-i)D {\tilde{v}}_n / \lambda _2} e^{(i-k)D\varepsilon }. \end{aligned}$$

So we want to show that

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{1}{D^{k-2}} n^{(M-k)(\beta + Dd/\lambda _2)} (-1)^k \sum _{i=2}^{k-1} \frac{1}{{\tilde{P}}_{i,k}} \left( n^{(k-i)D {\tilde{v}}_n / \lambda _2} e^{(i-k)D\varepsilon } - 1 \right) = 0 \end{aligned}$$

or, equivalently,

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{(M-k)(\beta + Dd/\lambda _2)} \sum _{i=2}^{k-1} \frac{1}{{\tilde{P}}_{i,k}} \left( n^{(k-i)D {\tilde{v}}_n / \lambda _2} e^{(i-k)D\varepsilon } - 1 \right) = 0. \end{aligned}$$

Since ${\tilde{v}}_n \rightarrow - \frac{\lambda _2}{\lambda _M} \beta (M-2)$ as $n \rightarrow \infty$, we have that $n^{(k-i)D {\tilde{v}}_n /\lambda _2} \rightarrow 0$ and hence $n^{(k-i)D {\tilde{v}}_n /\lambda _2} e^{(i-k)D\varepsilon } - 1 \rightarrow -1$. Therefore, this problem simplifies to showing that $n^{(M-k)(\beta + Dd/\lambda _2)} \rightarrow 0$ as $n \rightarrow \infty$. But we have already seen that $\beta + Dd/\lambda _2 < 0$ in the $k=2$ case, so we have the desired result.

Now that we have proved the second term in (11) converges to 0 as $n \rightarrow \infty$, all we have left is to show that the first term converges to 0 as well. By Lemma 1 in Durrett and Moseley (2010), we know that

$$\begin{aligned} e^{-\lambda _k t} X_k(t) - \int _0^t n^{-\beta } e^{-\lambda _k s} X_{k-1}(s) \, ds \end{aligned}$$

is a martingale. Setting $t = zt_n = -\frac{z}{\lambda _2} \log n$, we get that

$$\begin{aligned} n^{\lambda _k z/\lambda _2} X_k(zt_n) - n^{-\beta } \int _0^{zt_n} e^{-\lambda _k s} X_{k-1} (s) \, ds \end{aligned}$$

is a martingale in z. Since linear combinations of martingales are also martingales,

$$\begin{aligned}{} & {} n^{\lambda _k z/\lambda _2 + \beta (M-2) + (\lambda _M - \lambda _k) z/\lambda _2 - 1} X_k(zt_n) - n^{-\beta + \beta (M-2) + (\lambda _M - \lambda _k) z/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _k s} \, ds \end{aligned}$$

is also a martingale in z. The expression above simplifies to

$$\begin{aligned} n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)D z/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _k s} \, ds. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| n^{\beta (M-2) + \lambda _M z/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)D z/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _k s} \, ds \right| \end{aligned}$$

is a non-negative submartingale in z, so we can apply Doob’s Martingale Inequality to get

$$\begin{aligned}&\mathbb {P} \left( \sup _{z \in [d, v_n^+(\varepsilon )]} \left| n^{\beta (M-2) + \lambda _Mz/\lambda _2 - 1} X_k(zt_n) - n^{\beta (M-3) + (M-k)Dz/\lambda _2 - 1} \int _0^{zt_n} X_{k-1}(s) e^{-\lambda _ks} \, ds \right| > \delta /2 \right) \\&\quad \le \frac{4}{\delta ^2} \cdot \mathbb {E} \Biggl [ \Biggl ( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) \\&\quad - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1} \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \Biggr )^2 \Biggr ]. \\ \end{aligned}$$

Therefore, it suffices to show that

$$\begin{aligned}{} & {} \lim _{n \rightarrow \infty } \mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1} \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] = 0. \end{aligned}$$

We can expand the quantity above as follows:

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1}\right. \right. \\&\left. \left. \qquad \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad = n^{2\beta (M-2) + 2\lambda _M v_n^+(\varepsilon )/\lambda _2 - 2} \mathbb {E} \left[ X_k(v_n + \varepsilon )^2 \right] \\&\qquad - 2 n^{\beta (2M-5) + (2\lambda _M - \lambda _k) v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_k(v_n + \varepsilon ) \right] e^{-\lambda _k s} \, ds \\&\qquad + n^{2\beta (M-3) + 2(\lambda _M - \lambda _k) v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k s} e^{-\lambda _k y} \, ds \, dy. \end{aligned}$$

By making a slight modification to the proof of Lemma 1 in Foo and Leder (2013), we have that

$$\begin{aligned}&\mathbb {E} \left[ X_k(v_n + \varepsilon )^2 \right] = (n^{-\beta })^2 \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{\lambda _k(v_n + \varepsilon - s)} e^{\lambda _k(v_n + \varepsilon - y)} \, ds \, dy \\&\hspace{4mm}+ n^{-\beta } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right] \, ds \\&\mathbb {E} \left[ X_{k-1}(s) X_k(v_n + \varepsilon ) \right] = n^{-\beta } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(v_n + \varepsilon - y)} \, dy \end{aligned}$$

where ${\tilde{X}}_k$ is a binary branching process starting from size one with birth rate $r_k$ and death rate $d_k$. Substituting these expressions into our equation yields

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1}\right. \right. \\&\left. \left. \qquad \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad = n^{2\beta (M-3) + 2\lambda _M v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{\lambda _k(2v_n + 2\varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{\beta (2M-5) + 2\lambda _M v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right] \, ds\\&\qquad - 2 n^{2\beta (M-3) + (2\lambda _M - \lambda _k) v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{\lambda _k(v_n + \varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{2\beta (M-3) + 2(\lambda _M - \lambda _k) v_n^+(\varepsilon )/\lambda _2 - 2} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(s+y)} \, ds \, dy. \end{aligned}$$

Then, by definition of $v_n^+(\varepsilon )$,

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1}\right. \right. \\&\left. \left. \qquad \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad = n^{2\beta (M-3) - 2} e^{-2\lambda _M(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{\lambda _k(2v_n + 2\varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{\beta (2M-5) - 2} e^{-2\lambda _M(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right] \, ds \\&\qquad - 2 n^{2\beta (M-3) - 2} e^{-(2\lambda _M - \lambda _k)(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{\lambda _k(v_n + \varepsilon - s - y)} \, ds \, dy \\&\qquad + n^{2\beta (M-3) - 2} e^{-2(\lambda _M - \lambda _k)(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(s+y)} \, ds \, dy \\&\quad = n^{2\beta (M-3) - 2} e^{2(\lambda _k - \lambda _M)(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(s+y)} \, ds \, dy \\&\qquad + n^{\beta (2M-5) - 2} e^{-2\lambda _M(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right] \, ds \\&\qquad - 2 n^{2\beta (M-3) - 2} e^{2(\lambda _k - \lambda _M)(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(s+y)} \, ds \, dy \\&\qquad + n^{2\beta (M-3) - 2} e^{2(\lambda _k - \lambda _M)(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) X_{k-1}(y) \right] e^{-\lambda _k(s+y)} \, ds \, dy \\&\qquad = n^{\beta (2M-5) -2} e^{-2\lambda _M(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \mathbb {E} \left[ X_{k-1}(s) \right] \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right] \, ds. \end{aligned}$$

Note that

$$\begin{aligned} \mathbb {E} \left[ X_{k-1}(s) \right]&= n^{1-(k-3)\beta } (-1)^{k-1} S_{k-1}(s), \\ \mathbb {E} \left[ {\tilde{X}}_k(v_n + \varepsilon - s)^2 \right]&= \frac{2r_k}{\lambda _k} e^{2\lambda _k(v_n + \varepsilon - s)} - \frac{r_k + d_k}{\lambda _k} e^{\lambda _k(v_n + \varepsilon - s)}. \end{aligned}$$

Substituting these into the above expression yields

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1} \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad = n^{\beta (2M-5) - 2} e^{-2\lambda _M(v_n + \varepsilon )} \cdot \\&\qquad \int _0^{v_n + \varepsilon } n^{1-(k-3)\beta } (-1)^{k-1} S_{k-1}(s) \left( \frac{2r_k}{\lambda _k} e^{2\lambda _k(v_n + \varepsilon - s)} - \frac{r_k + d_k}{\lambda _k} e^{\lambda _k(v_n + \varepsilon - s)} \right) \, ds \\&\quad = n^{\beta (2M-k-2)-1} e^{-2\lambda _M(v_n + \varepsilon )} (-1)^{k-1} \Biggl [ \frac{2r_k}{\lambda _k} e^{2\lambda _k(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \sum _{i=2}^{k-1} \frac{e^{(\lambda _i - 2\lambda _k)s}}{P_{i,k-1}} \, ds \\&\qquad - \frac{r_k + d_k}{\lambda _k} e^{\lambda _k(v_n + \varepsilon )} \int _0^{v_n + \varepsilon } \sum _{i=2}^{k-1} \frac{e^{(\lambda _i - \lambda _k)s}}{P_{i,k-1}} \, ds \Biggr ] \\&\quad = n^{\beta (2M-k-2)-1} (-1)^{k-1} e^{-2\lambda _M(v_n + \varepsilon )} \frac{1}{\lambda _k} \cdot \\&\quad \sum _{i=2}^{k-1} \left[ \frac{2r_k \left( e^{\lambda _i(v_n + \varepsilon )} - e^{2\lambda _k(v_n + \varepsilon )} \right) }{(\lambda _i - 2\lambda _k) P_{i,k-1}} - \frac{(r_k + d_k) \left( e^{\lambda _i(v_n + \varepsilon )} - e^{\lambda _k(v_n + \varepsilon )} \right) }{(\lambda _i - \lambda _k) P_{i,k-1}} \right] \\&\quad = n^{\beta (2M-k-2)-1} (-1)^{k-1} e^{-2\lambda _M(v_n + \varepsilon )} \frac{1}{\lambda _k} \cdot \\&\quad \sum _{i=2}^{k-1} \frac{2r_k(\lambda _i - \lambda _k) \left( e^{\lambda _i(v_n + \varepsilon )} - e^{2\lambda _k(v_n + \varepsilon )} \right) - (r_k + d_k)(\lambda _i - 2\lambda _k) \left( e^{\lambda _i(v_n + \varepsilon )} - e^{\lambda _k(v_n + \varepsilon )} \right) }{(\lambda _i - \lambda _k)(\lambda _i - 2\lambda _k) P_{i,k-1}}. \end{aligned}$$

Now note that

$$\begin{aligned}&2r_k(\lambda _i - \lambda _k) \left( e^{\lambda _i(v_n + \varepsilon )} - e^{2\lambda _k(v_n + \varepsilon )} \right) - (r_k + d_k)(\lambda _i - 2\lambda _k) \left( e^{\lambda _i(v_n + \varepsilon )} - e^{\lambda _k(v_n + \varepsilon )} \right) \\&\quad = \lambda _k \left( \lambda _i + 2d_k \right) e^{\lambda _i(v_n + \varepsilon )} + \left( r_k + d_k \right) \left( \lambda _i - 2\lambda _k \right) e^{\lambda _k(v_n + \varepsilon )} + 2r_k \left( \lambda _k - \lambda _i \right) e^{2\lambda _k(v_n + \varepsilon )}. \end{aligned}$$

Substituting this back in, we get

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1}\right. \right. \\&\left. \left. \qquad \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad = n^{\beta (2M-k-2)-1} (-1)^{k-1} e^{-2\lambda _M(v_n + \varepsilon )} \frac{1}{\lambda _k} \cdot \\&\quad \sum _{i=2}^{k-1} \frac{\lambda _k \left( \lambda _i + 2d_k \right) e^{\lambda _i(v_n + \varepsilon )} + \left( r_k + d_k \right) \left( \lambda _i - 2\lambda _k \right) e^{\lambda _k(v_n + \varepsilon )} + 2r_k \left( \lambda _k - \lambda _i \right) e^{2\lambda _k(v_n + \varepsilon )}}{(\lambda _i - \lambda _k)(\lambda _i - 2\lambda _k) P_{i,k-1}} \\&\quad \sim \frac{ n^{\beta (2M-k-2) -1} (-1)^{k-1} e^{-2\lambda _M(v_n + \varepsilon )}}{\lambda _k(\lambda _{k-1} - \lambda _k)(\lambda _{k-1} - 2\lambda _k) P_{k-1,k-1}} \cdot \\&\quad \left[ (r_k + d_k)(\lambda _{k-1} - 2\lambda _k) e^{\lambda _k(v_n + \varepsilon )} + 2r_k (\lambda _k - \lambda _{k-1}) e^{2\lambda _k(v_n + \varepsilon )} \right] . \end{aligned}$$

Since $v_n = {\tilde{v}}_n t_n = -\frac{{\tilde{v}}_n}{\lambda _2} \log n$,

$$\begin{aligned}&\mathbb {E} \left[ \left( n^{\beta (M-2) + \lambda _M v_n^+(\varepsilon )/\lambda _2 - 1} X_k(v_n + \varepsilon ) - n^{\beta (M-3) + (M-k)D v_n^+(\varepsilon )/\lambda _2 - 1}\right. \right. \\&\left. \left. \qquad \int _0^{v_n + \varepsilon } X_{k-1} (s) e^{-\lambda _k s} \, ds \right) ^2 \right] \\&\quad \sim \frac{ n^{\beta (2M-k-2)-1} (-1)^{k-1} e^{-2\lambda _M \varepsilon } e^{\frac{2\lambda _M {\tilde{v}}_n}{\lambda _2} \log n}}{\lambda _k (\lambda _{k-1} - \lambda _k) (\lambda _{k-1} - 2\lambda _k) P_{k-1,k-1}} \cdot \\&\quad \left[ (r_k + d_k) (\lambda _{k-1} - 2\lambda _k) e^{\lambda _k \varepsilon } e^{\frac{\lambda _k {\tilde{v}}_n}{-\lambda _2} \log n} + 2r_k (\lambda _k - \lambda _{k-1}) e^{2\lambda _k \varepsilon } e^{\frac{2\lambda _k {\tilde{v}}_n}{-\lambda _2} \log n} \right] \\&\quad = \frac{ (-1)^{k-1} e^{-2\lambda _M \varepsilon }}{\lambda _k(\lambda _{k-1} - \lambda _k)(\lambda _{k-1} - 2\lambda _k) P_{k-1,k-1}} \cdot \\&\hspace{4mm} \Bigl [ (r_k + d_k) (\lambda _{k-1} - 2\lambda _k) e^{\lambda _k \varepsilon } n^{\beta (2M-k-2) - 1 + (2\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} \\&\quad + 2r_k (\lambda _k - \lambda _{k-1}) e^{2\lambda _k \varepsilon } n^{\beta (2M-k-2) - 1 + 2(\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} \Bigr ]. \end{aligned}$$

Note that

$$\begin{aligned} \frac{2\lambda _M - \lambda _k}{\lambda _2}&= \frac{2(\lambda _2 + (M-2)D) - (\lambda _2 + (k-2)D)}{\lambda _2} \\&= 1 + \frac{D(2M-k-2)}{\lambda _2}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + (2\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} = \lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + {\tilde{v}}_n + \frac{D(2M-k-2)}{\lambda _2} {\tilde{v}}_n} \\&\quad = \lim _{n \rightarrow \infty } e^{ \left[ \beta (2M-k-2) - 1 + {\tilde{v}}_n + \frac{D(2M-k-2)}{\lambda _2} {\tilde{v}}_n \right] \log n}. \end{aligned}$$

Then, since

$$\begin{aligned}&\lim _{n \rightarrow \infty } \left[ \beta (2M-k-2) - 1 + {\tilde{v}}_n + \frac{D(2M-k-2)}{\lambda _2} {\tilde{v}}_n \right] \\&\quad = \beta (2M-k-2) - 1 - \frac{\lambda _2}{\lambda _M} \beta (M-2) - \frac{D(2M-k-2)}{\lambda _2} \frac{\lambda _2}{\lambda _M} \beta (M-2) \\&\qquad \text { by Proposition}~2 \\&\quad = \beta (2M-k-2) \left( 1 - \frac{M-2}{\lambda _M} \frac{\lambda _M - \lambda _2}{M-2} \right) - 1 - \frac{\lambda _2}{\lambda _M} \beta (M-2)\\&\quad = \frac{\lambda _2}{\lambda _M} \beta (M-k) - 1 \\&\quad < 0, \end{aligned}$$

we have that

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + (2\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} = 0. \end{aligned}$$

Similarly, note that

$$\begin{aligned} 2 \frac{\lambda _M - \lambda _k}{\lambda _2}&= 2 \frac{(\lambda _2 + (M-2)D) - (\lambda _2 + (k-2)D)}{\lambda _2} \\&= 2D \frac{M-k}{\lambda _2}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + 2(\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} = \lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + 2D \frac{M-k}{\lambda _2} {\tilde{v}}_n} \\&\quad = \lim _{n \rightarrow \infty } e^{ \left[ \beta (2M-k-2) - 1 + 2D \frac{M-k}{\lambda _2} {\tilde{v}}_n \right] \log n}. \end{aligned}$$

We also have that

$$\begin{aligned}&\lim _{n \rightarrow \infty } \left[ \beta (2M-k-2) - 1 + 2D \frac{M-k}{\lambda _2} {\tilde{v}}_n \right] \\&\quad = \beta (2M-k-2) - 1 - 2D \frac{M-k}{\lambda _2} \frac{\lambda _2}{\lambda _M} \beta (M-2) \text { by Proposition}~2 \\&\quad = \beta (M-2) + \beta (M-k) - 1 - 2\beta \left( 1 - \frac{\lambda _2}{\lambda _M} \right) (M-k) \\&\quad = \beta (M-2) - 1 + \beta (M-k) \left( 2 \frac{\lambda _2}{\lambda _M} - 1 \right) . \end{aligned}$$

Since $\beta (M-k) (2\lambda _2/\lambda _M - 1) < 0$ and $\beta < 1/(M-2)$, the above limit is negative, and hence

$$\begin{aligned} \lim _{n \rightarrow \infty } n^{\beta (2M-k-2) - 1 + 2(\lambda _M - \lambda _k) {\tilde{v}}_n/\lambda _2} = 0. \end{aligned}$$

So we are done. $\square$