Critical branching as a pure death process coming down from infinity

Abstract

We consider the critical Galton–Watson process with overlapping generations stemming from a single founder. Assuming that both the variance of the offspring number and the average generation length are finite, we establish the convergence of the finite-dimensional distributions, conditioned on non-extinction at a remote time of observation. The limiting process is identified as a pure death process coming down from infinity.

This result brings a new perspective on Vatutin’s dichotomy, claiming that in the critical regime of age-dependent reproduction, an extant population either contains a large number of short-living individuals or consists of few long-living individuals.

1. Introduction

Consider a self-replicating system evolving in the discrete-time setting according to the following rules:

1. Rule 1: The system is founded by a single individual, the founder, born at time 0.

2. Rule 2: The founder dies at a random age L and gives a random number N of births at random ages $\tau_j$ satisfying $1\le\tau_1\le \ldots\le \tau_N\le L$ .

3. Rule 3: Each new individual lives independently from others according to the same life law as the founder.

An individual that was born at time $t_1$ and dies at time $t_2$ is considered to be alive during the time interval $[t_1,t_2-1]$ . Letting Z(t) stand for the number of individuals alive at time t, we study the random dynamics of the sequence

\begin{equation*}Z(0)=1, Z(1), Z(2),\ldots,\end{equation*}

which is a natural extension of the well-known Galton–Watson process, or GW process for short; see [13]. The process $Z({\cdot})$ is the discrete-time version of what is usually called the Crump–Mode–Jagers process or the general branching process; see [5]. To emphasise the discrete-time setting, we call it a GW process with overlapping generations, or GWO process for short.

Put $b\,:\!=\,\frac{1}{2}\mathrm{var}(N)$ . This paper deals with the GWO processes satisfying

(1) \begin{equation} \mathrm{E}(N)=1,\quad 0<b<\infty.\end{equation}

The condition $\mathrm{E}(N)=1$ says that the reproduction regime is critical, implying $\mathrm{E}(Z(t))\equiv1$ and making extinction inevitable, provided $b>0$ . According to [1, Chapter I.9], given (1), the survival probability

\begin{equation*}Q(t)\,:\!=\,\mathrm{P}(Z(t)>0)\end{equation*}

of a GW process satisfies the asymptotic formula $tQ(t)\to b^{-1}$ as $t\to\infty$ (this was first proven in [6] under a third moment assumption). A direct extension of this classical result for the GWO processes,

\begin{equation*}tQ(ta)\to b^{-1},\quad t\to\infty,\quad a\,:\!=\,\mathrm{E}(\tau_1+\ldots+\tau_N),\end{equation*}

was obtained in [3, 4] under the conditions (1), $a<\infty$ ,

(2) \begin{equation} t^2\mathrm{P}(L>t)\to 0,\quad t\to\infty,\end{equation}

plus an additional condition. (Notice that by our definition, $a\ge1$ , and $a=1$ if and only if $L\equiv1$ , that is, when the GWO process in question is a GW process.) Treating a as the mean generation length (see [5, 8]), we may conclude that the asymptotic behaviour of the critical GWO process with short-living individuals (see the condition (2)) is similar to that of the critical GW process, provided time is counted generation-wise.

New asymptotic patterns for the critical GWO processes are found under the assumption

(3) \begin{equation} t^2\mathrm{P}(L>t)\to d,\quad 0\le d< \infty,\quad t\to\infty,\end{equation}

which, compared to (2), allows the existence of long-living individuals given $d>0$ . The condition (3) was first introduced in the pioneering paper [12] dealing with the Bellman–Harris processes. In the current discrete-time setting, the Bellman–Harris process is a GWO process subject to two restrictions: (a) $\mathrm{P}(\tau_1=\ldots=\tau_N= L)=1$ , so that all births occur at the moment of an individual’s death, and (b) the random variables L and N are independent. For the Bellman–Harris process, the conditions (1) and (3) imply $a=\mathrm{E}(L)$ , $a<\infty$ , and according to [12, Theorem 3], we get

(4) \begin{equation} tQ(t)\to h,\quad t\to\infty,\qquad h\,:\!=\,\frac{a+\sqrt{a^2+4bd}}{2b}.\end{equation}

As was shown in [11, Corollary B] (see also [7, Lemma 3.2] for an adaptation to the discrete-time setting), the relation (4) holds even for the GWO processes satisfying the conditions (1), (3), and $a<\infty$ .

The main result of this paper, Theorem 1 of Section 2, considers a critical GWO process under the above-mentioned set of assumptions (1), (3), $a<\infty$ , and establishes the convergence of the finite-dimensional distributions conditioned on survival at a remote time of observation. A remarkable feature of this result is that its limit process is fully described by a single parameter $c\,:\!=\,4bda^{-2}$ , regardless of complicated mutual dependencies between the random variables $\tau_j, N,L$ .

Our proof of Theorem 1, requiring an intricate asymptotic analysis of multi-dimensional probability generating functions, is split into two sections for the sake of readability. Section 3 presents a new proof of (4) inspired by the proof of [12]. The crucial aspect of this approach, compared to the proof of [7, Lemma 3.2], is that certain essential steps do not rely on the monotonicity of the function Q(t). In Section 4, the technique of Section 3 is further developed to finish the proof of Theorem 1.

We conclude this section by mentioning the illuminating family of GWO processes called the Sevastyanov processes [9]. The Sevastyanov process is a generalised version of the Bellman–Harris process, with possibly dependent L and N. In the critical case, the mean generation length of the Sevastyanov process, $a=\mathrm{E}(L N)$ , can be represented as

\begin{equation*}a=\mathrm{cov}(L,N)+\mathrm{E}(L).\end{equation*}

Thus, if L and N are positively correlated, the average generation length a exceeds the average life length $\mathrm{E}(L)$ .

Turning to a specific example of the Sevastyanov process, take

\begin{equation*}\mathrm{P}(L= t)= p_1 t^{-3}(\!\ln\ln t)^{-1}, \quad \mathrm{P}(N=0|L= t)=1-p_2,\quad \mathrm{P}(N=n_t|L= t)=p_2, \ t\ge2,\end{equation*}

where $n_t\,:\!=\,\lfloor t(\!\ln t)^{-1}\rfloor$ and $(p_1,p_2)$ are such that

\begin{equation*}\sum_{t=2}^\infty \mathrm{P}(L= t)=p_1 \sum_{t=2}^\infty t^{-3}(\!\ln\ln t)^{-1}=1,\quad \mathrm{E}(N)=p_1p_2\sum_{t=2}^\infty n_t t^{-3}(\!\ln\ln t)^{-1}=1.\end{equation*}

In this case, for some positive constant $c_1$ ,

\begin{equation*}\mathrm{E}\big(N^2\big)= p_1p_2\sum_{t=1}^\infty n_t^2 t^{-3}(\!\ln\ln t)^{-1}< c_1\int_2^\infty \frac{d (\!\ln t)}{(\!\ln t)^2\ln\ln t}<\infty,\end{equation*}

implying that the condition (1) is satisfied. Clearly, the condition (3) holds with $d=0$ . At the same time,

\begin{equation*}a=\mathrm{E}(NL)= p_1p_2\sum_{t=1}^\infty n_t t^{-2}(\!\ln\ln t)^{-1}> c_2\int_2^\infty \frac{d (\!\ln t)}{(\!\ln t)(\!\ln\ln t)}=\infty,\end{equation*}

where $c_2$ is a positive constant. This example demonstrates that for the GWO process, unlike for the Bellman–Harris process, the conditions (1) and (3) do not automatically imply the condition $a<\infty$ .

2. The main result

Theorem 1. For a GWO process satisfying (1), (3) and $a<\infty$ , there holds a weak convergence of the finite-dimensional distributions

\begin{align*}(Z(ty),0<y<\infty|Z(t)>0)\stackrel{\rm fdd\,}{\longrightarrow} (\eta(y),0<y<\infty),\quad t\to\infty.\end{align*}

The limiting process is a continuous-time pure death process $(\eta(y),0\le y<\infty)$ , whose evolution law is determined by a single compound parameter $c=4bda^{-2}$ , as specified next.

The finite-dimensional distributions of the limiting process $\eta({\cdot})$ are given below in terms of the k-dimensional probability generating functions $\mathrm{E}\Big(z_1^{\eta(y_1)}\cdots z_k^{\eta(y_k)}\Big)$ , $k\ge1$ , assuming

(5) \begin{align} 0=y_0< y_1< \ldots< y_{j}<1\le y_{j+1}< \ldots< &\ y_k<y_{k+1}=\infty, \nonumber \\ &0\le j\le k,\quad 0\le z_1,\ldots,z_k<1.\end{align}

Here the index j highlights the pivotal value 1 corresponding to the time of observation t of the underlying GWO process.

As will be shown in Section 4.2, if $j=0$ , then

\begin{align*}\mathrm{E}\Big(z_1^{\eta(y_1)}\cdots z_k^{\eta(y_k)}\Big)=1-\frac{1+\sqrt{1+\sum\nolimits_{i=1}^{k}z_1\cdots z_{i-1}(1-z_{i})\Gamma_i}}{\big(1+\sqrt{1+c}\big)y_1},\quad \Gamma_i\,:\!=\,c({y_1}/{y_i} )^2,\end{align*}

and if $j\ge1$ ,

\begin{align*}\mathrm{E} &\Big(z_1^{\eta(y_1)}\cdots z_k^{\eta(y_k)}\Big)\\&=\frac{\sqrt{1+\sum_{i=1}^{j}z_1\cdots z_{i-1}(1-z_{i})\Gamma_i+cz_1\cdots z_{j}y_1^{2} }-\sqrt{1+\sum\nolimits_{i=1}^{k}z_1\cdots z_{i-1}(1-z_{i})\Gamma_i}}{\big(1+\sqrt{1+c}\big)y_1}.\end{align*}

In particular, for $k=1$ , we have

\begin{align*}\mathrm{E}\big(z^{\eta(y)}\big)&= \frac{\sqrt{1+c(1-z)+czy^{2}}-\sqrt{1+c(1-z)}}{\big(1+\sqrt{1+c}\big)y},\quad 0< y<1,\\\mathrm{E}\big(z^{\eta(y)}\big)&= 1-\frac{1+\sqrt{1+c(1-z)}}{\big(1+\sqrt{1+c}\big)y},\quad y\ge1.\end{align*}

It follows that $\mathrm{P}(\eta(y)\ge0)=1$ for $y>0$ , and moreover, putting here first $z=1$ and then $z=0$ yields

\begin{align*}\mathrm{P}(\eta(y)<\infty)&=\frac{\sqrt{1+cy^2}-1}{\big(1+\sqrt{1+c}\big)y}\cdot1_{\{0< y<1\}}+\bigg(1-\frac{2}{\big(1+\sqrt{1+c}\big)y}\bigg)\cdot1_{\{y\ge 1\}},\\\mathrm{P}(\eta(y)=0)&=\frac{y-1}{y}\cdot1_{\{y\ge 1\}},\end{align*}

implying that $\mathrm{P}(\eta(y)=\infty)>0$ for all $y>0$ . In fact, letting $y\to0$ , we may set $\mathrm{P}(\eta(0)= \infty)=1.$

To demonstrate that the process $\eta({\cdot})$ is indeed a pure death process, consider the function

\begin{equation*}\mathrm{E}\Big(z_1^{\eta(y_1)-\eta(y_2)}\cdots z_{k-1}^{\eta(y_{k-1})-\eta(y_{k})}z_k^{\eta(y_k)}\Big)\end{equation*}

determined by

\begin{align*}\mathrm{E}\Big(z_1^{\eta(y_1)-\eta(y_2)}\cdots z_{k-1}^{\eta(y_{k-1})-\eta(y_{k})}z_k^{\eta(y_k)}\Big)&=\mathrm{E}\Big(z_1^{\eta(y_1)}(z_2/z_1)^{\eta(y_2)}\cdots (z_k/z_{k-1})^{\eta(y_k)}\Big).\end{align*}

This function is given by two expressions:

\begin{align*}\frac{\big(1+\sqrt{1+c}\big)y_1-1-\sqrt{1+\sum\nolimits_{i=1}^{k}\! (1-z_{i})\gamma_i}}{\big(1+\sqrt{1+c}\big)y_1}, \quad &\text{for }j=0,\\\frac{\sqrt{1+\sum\nolimits_{i=1}^{j-1}(1-z_{i})\gamma_i+(1-z_{j})\Gamma_j+cz_j y_1^2}-\sqrt{1+\sum\nolimits_{i=1}^{k} \!(1-z_{i})\gamma_i}}{\big(1+\sqrt{1+c}\big)y_1}, \quad &\text{for }j\ge1,\end{align*}

where $\gamma_i\,:\!=\,\Gamma_i-\Gamma_{i+1}$ and $\Gamma_{k+1}=0$ . Setting $k=2$ , $z_1=z$ , and $z_2=1$ , we deduce that the function

(6) \begin{equation} \mathrm{E}\big(z^{\eta(y_1)-\eta(y_2)};\,\eta(y_1)<\infty\big),\quad 0<y_1<y_2,\quad 0\le z\le1,\end{equation}

is given by one of the following three expressions, depending on whether $j=2$ , $j=1$ , or $j=0$ :

\begin{align*}\frac{\sqrt{1+c y_1^2+c(1-z)\big(1-(y_1/y_2)^2\big)}-\sqrt{1+c (1-z)\big(1-(y_1/y_2)^2\big)}}{\big(1+\sqrt{1+c}\big)y_1},\quad &y_2<1, \\[3pt]\frac{\sqrt{1+c y_1^2+c(1-z) \big(1-y_1^2\big)}-\sqrt{1+c(1-z)\big(1-(y_1/y_2)^2\big)}}{\big(1+\sqrt{1+c}\big)y_1},\quad &y_1<1\le y_2, \\[3pt]1- \frac{1+\sqrt{1+c(1-z)\big(1-(y_1/y_2)^2\big)}}{\big(1+\sqrt{1+c}\big)y_1},\quad &1\le y_1.\end{align*}

Since the generating function (6) is finite at $z=0$ , we conclude that

\begin{equation*}\mathrm{P}(\eta(y_1)< \eta(y_2);\, \eta(y_1)< \infty)=0,\quad 0<y_1<y_2.\end{equation*}

This implies

\begin{equation*}\mathrm{P}(\eta(y_2)\le \eta(y_1))=1,\quad 0<y_1<y_2,\end{equation*}

meaning that unless the process $\eta({\cdot})$ is sitting at the infinity state, it evolves by negative integer-valued jumps until it gets absorbed at zero.

Consider now the conditional probability generating function

(7) \begin{equation}\mathrm{E}\big(z^{\eta(y_1)-\eta(y_2)}| \eta(y_1)<\infty\big),\quad 0<y_1<y_2,\quad 0\le z\le1.\end{equation}

In accordance with the three expressions given above for (6), the generating function (7) is specified by the following three expressions:

\begin{align*}\frac{\sqrt{1+c y_1^2+c(1-z)\big(1-(y_1/y_2)^2\big)}-\sqrt{1+c (1-z)\big(1-(y_1/y_2)^2\big)}}{\sqrt{1+c y_1^2}-1},\quad &y_2<1, \\[3pt]\frac{\sqrt{1+c y_1^2+c(1-z) \big(1-y_1^2\big)}-\sqrt{1+c(1-z)\big(1-(y_1/y_2)^2\big)}}{\sqrt{1+c y_1^2}-1},\quad &y_1<1\le y_2, \\[3pt]1- \frac{\sqrt{1+c(1-z)\big(1-(y_1/y_2)^2\big)}-1}{\big(1+\sqrt{1+c}\big)y_1-2},\quad &1\le y_1.\end{align*}

In particular, setting $z=0$ here, we obtain

\begin{equation*}\mathrm{P}(\eta(y_1)-\eta(y_2)=0| \eta(y_1)<\infty)= \left\{\begin{array}{l@{\quad}l@{\quad}r}\frac{\sqrt{1+c\big(1+y_1^2-(y_1/y_2)^2\big)}-\sqrt{1+c\big(1-(y_1/y_2)^2\big)}}{\sqrt{1+c y_1^2}-1} & \!\!\text{for} & \!\! 0<y_1< y_2<1, \\[13pt]\frac{\sqrt{1+c}-\sqrt{1+c\big(1-(y_1/y_2)^2\big)}}{\sqrt{1+c y_1^2}-1} & \!\!\text{for} & \!\! 0<y_1<1\le y_2, \\[13pt]1- \frac{\sqrt{1+c\big(1-(y_1/y_2)^2\big)}-1}{\big(1+\sqrt{1+c}\big)y_1-2} & \!\!\text{for} & \!\! 1\le y_1<y_2.\end{array}\right.\end{equation*}

Notice that given $0<y_1\le1$ ,

\begin{equation*}\mathrm{P}(\eta(y_1)-\eta(y_2)=0| \eta(y_1)<\infty)\to 0,\quad y_2\to\infty,\end{equation*}

which is expected because of $\eta(y_1)\ge\eta(1)\ge1$ and $\eta(y_2)\to0$ as $y_2\to\infty$ .

The random times

\begin{equation*}T=\sup\{u\,:\, \eta(u)=\infty\},\quad T_0=\inf\{u\,:\,\eta(u)=0\}\end{equation*}

are major characteristics of a trajectory of the limit pure death process. Since

\begin{align*}\mathrm{P}(T\le y)=\mathrm{E}\big(z^{\eta(y)}\big)\Big\vert_{z=1},\qquad \mathrm{P}(T_0\le y)=\mathrm{E}\big(z^{\eta(y)}\big)\Big\vert_{z=0},\end{align*}

in accordance with the above-mentioned formulas for $\mathrm{E}\big(z^{\eta(y)}\big)$ , we get the following marginal distributions:

\begin{align*} \mathrm{P}(T\le y)&=\frac{\sqrt{1+cy^2}-1}{\big(1+\sqrt{1+c}\big)y}\cdot1_{\{0\le y<1\}}+\bigg(1-\frac{2}{\big(1+\sqrt{1+c}\big)y}\bigg)\cdot1_{\{y\ge 1\}},\\ \mathrm{P}(T_0\le y)&=\frac{y-1}{y}\cdot1_{\{y\ge 1\}}.\end{align*}

The distribution of $T_0$ is free from the parameter c and has the Pareto probability density function

\begin{equation*}f_0(y)=y^{-2}1_{\{y>1\}}.\end{equation*}

In the special case (2), that is, when (3) holds with $d=0$ , we have $c=0$ and $\mathrm{P}(T=T_0)=1$ . If $d>0$ , then $T\le T_0$ , and the distribution of T has the following probability density function:

\begin{equation*}f(y)=\left\{\begin{array}{l@{\quad}l@{\quad}r}\frac{1}{\big(1+\sqrt{1+c}\big)y^2} \Big(1-\frac{1}{\sqrt{1+cy^2}}\Big)& \text{for} & 0\le y<1, \\[7pt]\frac{2}{\big(1+\sqrt{1+c}\big)y^2} & \text{for} & y\ge1,\end{array}\right.\end{equation*}

which has a positive jump at $y=1$ of size $f(1)-f(1-)=(1+c)^{-1/2}$ ; see Figure 1. Observe that $\frac{f(1-)}{f(1)}\to\frac{1}{2}$ as $c\to\infty$ .

Figure 1. The dashed line is the probability density function of T; the solid line is the probability density function of $T_0$ . The left panel illustrates the case $c=5$ , and the right panel illustrates the case $c=15$ .

Intuitively, the limiting pure death process counts the long-living individuals in the GWO process, that is, those individuals whose life length is of order t. These long-living individuals may have descendants, however none of them would live long enough to be detected by the finite-dimensional distributions at the relevant time scale, see Lemma 2 below. Theorem 1 suggests a new perspective on Vatutin’s dichotomy (see [12]), claiming that the long-term survival of a critical age-dependent branching process is due to either a large number of short-living individuals or a small number of long-living individuals. In terms of the random times $T\le T_0$ , Vatutin’s dichotomy discriminates between two possibilities: if $T>1$ , then $\eta(1)=\infty$ , meaning that the GWO process has survived thanks to a large number of individuals, while if $T\le 1<T_0$ , then $1\le \eta(1)<\infty$ , meaning that the GWO process has survived thanks to a small number of individuals.

3. Proof that $\boldsymbol{{tQ(t)}}\to \boldsymbol{{h}}$

This section deals with the survival probability of the critical GWO process

\begin{equation*}Q(t)=1-P(t),\quad P(t)\,:\!=\,\mathrm{P}(Z(t)=0).\end{equation*}

By its definition, the GWO process can be represented as the sum

(8) \begin{equation} Z(t)=1_{\{L>t\}}+\sum\nolimits_{j=1}^{N} Z_j\left(t-\tau_j\right),\quad t=0,1,\ldots,\end{equation}

involving N independent daughter processes $Z_j({\cdot})$ generated by the founder individual at the birth times $\tau_j$ , $j=1,\ldots,N$ (here it is assumed that $Z_j(t)=0$ for all negative t). The branching property (8) implies the relation

\begin{equation*} 1_{\{Z(t)=0\}}=1_{\{L\le t\}}\prod\nolimits_{j=1}^{N} 1_{\big\{Z_j \left(t-\tau_j\right)=0\big\}},\end{equation*}

which says that the GWO process goes extinct by the time t if, on one hand, the founder is dead at time t and, on the other hand, all daughter processes are extinct by the time t. After taking expectations of both sides, we can write

(9) \begin{equation}P(t)=\mathrm{E}\bigg(\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,L\le t\bigg).\end{equation}

As shown next, this nonlinear equation for $P({\cdot})$ implies the asymptotic formula (4) under the conditions (1), (3), and $a<\infty$ .

3.1. Outline of the proof of (4)

We start by stating four lemmas and two propositions. Let

(10) \begin{equation}\Phi(z) \,:\!=\,\mathrm{E}\big((1-z)^ N-1+Nz\big), \end{equation}

(11) \begin{equation}W(t) \,:\!=\,\big(1-ht^{-1}\big)^{N}+Nht^{-1}-\sum\nolimits_{j=1}^{N}Q\left(t-\tau_j\right)-\prod\nolimits_{j=1}^{N} P\left(t-\tau_j\right), \end{equation}

(12) \begin{equation} D(u,t) \,:\!=\,\mathrm{E}\Big(1-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,u<L\le t\Big)+\mathrm{E}\big(\big(1-ht^{-1}\big)^{N} -1+Nht^{-1};\,L>u\big), \end{equation}

(13) \begin{equation}\mathrm{E}_u(X) \,:\!=\,\mathrm{E}(X;\,L\le u ),\end{equation}

where $0\le z\le 1$ , $u>0$ , $t\ge h$ , and X is an arbitrary random variable.

Lemma 1. Given (10), (11), (12), and (13), assume that $0< u\le t$ and $t\ge h$ . Then

\begin{align*}\Phi\big(ht^{-1}\big)= \mathrm{P}(L> t)+\mathrm{E}_u\bigg(\!\sum\nolimits_{j=1}^{N}Q\left(t-\tau_j\right)\bigg)-Q(t)+\mathrm{E}_u(W(t))+D(u,t).\end{align*}

Lemma 2. If (1) and (3) hold, then $\mathrm{E}(N;\,L>ty)=o\big(t^{-1}\big)$ as $t\to\infty$ for any fixed $y>0$ .

Lemma 3. If (1), (3), and $a<\infty$ hold, then for any fixed $0<y<1$ ,

\begin{align*}\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{1}{t-\tau_j}-\frac{1}{t}\bigg)\bigg)\sim at^{-2},\quad t\to\infty.\end{align*}

Lemma 4. Let $k\ge1$ . If $0\le f_j,g_j\le 1$ for $ j=1,\ldots,k$ , then

\begin{equation*} \prod\nolimits_{j=1}^k\!\big(1-g_j\big)-\prod\nolimits_{j=1}^k\big(1-f_j\big)=\sum\nolimits_{j=1}^k (f_j-g_j)r_j, \end{equation*}

where $0\le r_j\le1$ and

\begin{align*}1-r_j=\sum\nolimits_{i=1}^{j-1}g_i+\sum\nolimits_{i=j+1}^{k}f_i-R_j,\end{align*}

for some $R_j\ge0$ . If moreover $f_j\le q$ and $g_j\le q$ for some $q>0$ , then

\begin{equation*}1- r_j\le(k-1)q,\qquad R_j\le kq,\qquad R_j\le k^2q^2.\end{equation*}

Proposition 1. If (1), (3), and $a<\infty$ hold, then $\limsup_{t\to\infty} tQ(t)<\infty$ .

Proposition 2. If (1), (3), and $a<\infty$ hold, then $\liminf_{t\to\infty} tQ(t)>0$ .

According to these two propositions, there exists a triplet of positive numbers $(q_1,q_2,t_0)$ such that

(14) \begin{equation}q_1\le tQ(t)\le q_2,\quad t\ge t_0,\quad 0<q_1<h<q_2<\infty.\end{equation}

The claim $tQ(t)\to h$ is derived using (14) by accurately removing asymptotically negligible terms from the relation for $Q({\cdot})$ stated in Lemma 1, after setting $u=ty$ with a fixed $0<y<1$ , and then choosing a sufficiently small y. In particular, as an intermediate step, we will show that

(15) \begin{align}Q(t)= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}Q\left(t-\tau_j\right)\bigg)+\mathrm{E}_{ty}(W(t))-aht^{-2}+o\big(t^{-2}\big),\quad t\to\infty. \end{align}

Then, restating our goal as $\phi(t)\to 0$ in terms of the function $\phi(t)$ , defined by

(16) \begin{equation}Q(t)=\frac{h +\phi(t)}{t},\quad t\ge1,\end{equation}

we rewrite (15) as

(17) \begin{align}\frac{h +\phi(t)}{t}&= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{h +\phi\left(t-\tau_j\right)}{t-\tau_j}\bigg)+\mathrm{E}_{ty}(W(t))-aht^{-2}+o\big(t^{-2}\big),\quad t\to\infty. \end{align}

It turns out that the three terms involving h, outside W(t), effectively cancel each other, yielding

(18) \begin{align}\frac{\phi(t)}{t}&= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{\phi\left(t-\tau_j\right)}{t-\tau_j}+W(t)\bigg)+o\big(t^{-2}\big),\quad t\to\infty.\end{align}

Treating W(t) in terms of Lemma 4 yields

(19) \begin{align} \phi(t)&= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\phi\left(t-\tau_j\right) r_j(t)\frac{t}{t-\tau_j}\bigg)+o\big(t^{-1}\big), \end{align}

where $r_j(t)$ is a counterpart of $r_j$ in Lemma 4. To derive from here the desired convergence $\phi(t)\to0$ , we will adapt a clever trick from Chapter 9.1 of [10], which was further developed in [12] for the Bellman–Harris process, with possibly infinite $\mathrm{var}(N)$ . Define a non-negative function m(t) by

(20) \begin{align}m(t)\,:\!=\,|\phi(t)|\, \ln t,\quad t\ge 2. \end{align}

Multiplying (19) by $\ln t$ and using the triangle inequality, we obtain

(21) \begin{align} m(t)\le \mathrm{E}_{ty}\Bigg(\!\sum\nolimits_{j=1}^{N} m\left(t-\tau_j\right)r_j(t) \frac{t\ln t}{\left(t-\tau_j\right)\ln\left(t-\tau_j\right)}\Bigg)+v(t),\end{align}

where $v(t)\ge 0$ and $v(t)=o(t^{-1}\ln t)$ as $t\to\infty$ . It will be shown that this leads to $m(t)=o(\!\ln t)$ , thereby concluding the proof of (4).

3.2. Proof of lemmas and propositions

Proof of Lemma 1.For $0<u\le t$ , the relations (9) and (13) give

(22) \begin{align}P(t)=\mathrm{E}_u\bigg(\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right) \bigg)+\mathrm{E}\bigg(\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,u<L\le t\bigg).\end{align}

On the other hand, for $t\ge h$ ,

\begin{align*}\Phi\big(ht^{-1}\big)&\stackrel{(10)}{=}\mathrm{E}_u\Big(\big(1-ht^{-1}\big)^{N}-1+Nht^{-1}\Big)+\mathrm{E}\Big(\big(1-ht^{-1}\big)^{N}-1 +Nht^{-1};\,L> u\Big).\end{align*}

Adding the latter relation to

\begin{align*}1&=\mathrm{P}(L\le u)+\mathrm{P}(L> t)+\mathrm{P}(u<L\le t)\end{align*}

and subtracting (22) from the sum, we get

\begin{align*}\Phi\big(ht^{-1}\big)+Q(t)=\mathrm{E}_u\bigg(\big(1-ht^{-1}\big)^{N} +Nht^{-1}-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right)\bigg)+\mathrm{P}(L> t)+D(u,t),\end{align*}

with D(u, t) defined by (12). After a rearrangement, we obtain the statement of the lemma.

Proof of Lemma 2.For any fixed $\epsilon>0$ ,

\begin{align*}\mathrm{E}(N;\,L>t)&=\mathrm{E}(N;\,N\le t\epsilon,L>t)+\mathrm{E}\big(N;\,1<N(t\epsilon)^{-1},L>t\big)\\&\le t\epsilon\mathrm{P}(L>t)+(t\epsilon)^{-1}\mathrm{E}\big(N^2;\,L>t\big).\end{align*}

Thus, by (1) and (3),

\begin{align*}\limsup_{t\to\infty} (t\mathrm{E}(N;\,L>t))\le d\epsilon,\end{align*}

and the assertion follows as $\epsilon\to0$ .

Proof of Lemma 3.For $t=1,2,\ldots$ and $y>0$ , put

\begin{align*}B_t(y)&\,:\!=\, t^2\,\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{1}{t-\tau_j}-\frac{1}{t}\bigg)\bigg)-a. \end{align*}

For any $0<u<ty$ , using

\begin{equation*}a=\mathrm{E}_u(\tau_1+\ldots+\tau_N)+A_u,\quad A_u\,:\!=\,\mathrm{E}(\tau_1+\ldots+\tau_N;\,L> u),\end{equation*}

we get

\begin{align*}B_t(y)&= \mathrm{E}_u\bigg(\!\sum\nolimits_{j=1}^{N} \frac{t}{t-\tau_j}\tau_j\bigg)+\mathrm{E}\bigg(\!\sum\nolimits_{j=1}^{N} \frac{t}{t-\tau_j}\tau_j\,;\,u<L\le ty\bigg)\\& \quad -\mathrm{E}_u(\tau_1+\ldots+\tau_N)-A_u\\ &=\mathrm{E}\Bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j}{1-\tau_j/t};\,u<L\le ty\Bigg)+\mathrm{E}_u\Bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j^2}{t-\tau_j}\Bigg)-A_u.\end{align*}

For the first term on the right-hand side, we have $\tau_j\le L\le ty$ , so that

\begin{align*}\mathrm{E}\Bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j}{1-\tau_j/t};\,u<L\le ty\Bigg)\le(1-y)^{-1}A_u.\end{align*}

For the second term, $\tau_j\le L\le u$ and therefore

\begin{align*}\mathrm{E}_u\Bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j^2}{t-\tau_j}\Bigg)\le\frac{u^2}{t-u}\mathrm{E}_u(N)\le\frac{u^2}{t-u}.\end{align*}

This yields

\begin{equation*}-A_u\le B_t(y)\le (1-y)^{-1}A_u+\frac{u^2}{t-u},\quad 0<u<ty<t,\end{equation*}

implying

\begin{equation*}-A_u\le \liminf_{t\to\infty} B_t(y)\le\limsup_{t\to\infty} B_t(y)\le (1-y)^{-1}A_u.\end{equation*}

Since $A_u\to0$ as $u\to\infty$ , we conclude that $B_t(y)\to 0$ as $t\to\infty$ .

Proof of Lemma 4.Let

\begin{equation*}r_j\,:\!=\,\big(1-g_1\big)\ldots \big(1-g_{j-1}\big)\left(1-f_{j+1}\right)\ldots \big(1-f_k\big),\quad 1\le j\le k.\end{equation*}

Then $0\le r_j\le1$ , and the first stated equality is obtained by telescopic summation of

\begin{align*} \big(1-g_1\big)\prod\nolimits_{j=2}^{k}\big(1-f_j\big)-\prod\nolimits_{j=1}^k\big(1-f_j\big)&=(f_1-g_1)r_1,\\ \big(1-g_1\big)\big(1-g_2\big)\prod\nolimits_{j=3}^{k}\big(1-f_j\big)- \big(1-g_1\big)\prod\nolimits_{j=2}^{k}\big(1-f_j\big)&=(f_2-g_2)r_2,\ldots,\\ \prod\nolimits_{j=1}^{k}\big(1-g_j\big)-\prod\nolimits_{j=1}^{k-1}\big(1-g_j\big)\big(1-f_k\big)&=(f_k-g_k)r_k.\end{align*}

The second stated equality is obtained with

\begin{align*}R_j&\,:\!=\,\sum_{i=j+1}^{k}f_i\big(1-\left(1-f_{j+1}\right)\ldots \big(1-f_{i-1}\big)\big)\\& \quad +\sum_{i=1}^{j-1}g_i\big(1-\big(1-g_1\big)\ldots (1-g_{i-1})\left(1-f_{j+1}\right)\ldots \big(1-f_k\big)\big),\end{align*}

by performing telescopic summation of

\begin{align*} 1-\left(1-f_{j+1}\right)&=f_{j+1},\\\left(1-f_{j+1}\right)-\left(1-f_{j+1}\right)(1-f_{j+2})&=f_{j+2}\left(1-f_{j+1}\right),\ldots,\\ \prod\nolimits_{i=j+1}^{k-1}\left(1-f_i\right)- \prod\nolimits_{i=j+1}^{k}\left(1-f_i\right)&=f_k\prod\nolimits_{i=j+1}^{k-1}\left(1-f_i\right),\\ \prod\nolimits_{i=j+1}^{k}\left(1-f_i\right)-\left(1-g_1\right)\prod\nolimits_{i=j+1}^{k}\left(1-f_i\right)&=g_1\prod\nolimits_{i=j+1}^{k}\left(1-f_i\right),\ldots,\\ \prod\nolimits_{i=1}^{j-2}(1-g_i)\prod\nolimits_{i=j+1}^{k}\left(1-f_i\right)- \prod\nolimits_{i=1}^{j-1}(1-g_i)\prod\nolimits_{i=j+1}^{k}\left(1-f_i\right)&=g_{j-1} \prod\nolimits_{i=1}^{j-2}(1-g_i)\prod\nolimits_{i=j+1}^{k}\left(1-f_i\right).\end{align*}

By the above definition of $R_j$ , we have $R_j\ge0$ . Furthermore, given $f_j\le q$ and $g_j\le q$ , we get

\begin{equation*}R_j\le \sum\nolimits_{i=1}^{j-1}g_i+\sum\nolimits_{i=j+1}^{k}f_i\le (k-1)q.\end{equation*}

It remains to observe that

\begin{align*}1-r_j\le 1-(1-q)^{k-1}\le (k-1)q,\end{align*}

and from the definition of $R_j$ ,

\begin{equation*}R_j\le q\sum\nolimits_{i=1}^{k-j-1}(1-(1-q)^{i})+q\sum\nolimits_{i=1}^{j-1}\big(1-(1-q)^{k-j+i-1}\big)\le q^2\sum\nolimits_{i=1}^{k-2}i\le k^2q^2.\end{equation*}

Proof of Proposition 1.By the definition of $\Phi({\cdot})$ , we have

\begin{equation*}\Phi(Q(t))+P(t)=\mathrm{E}_u\big(P(t)^{N} \big)+\mathrm{P}(L> u)-\mathrm{E}\big(1-P(t)^ N;\,L> u\big),\end{equation*}

for any $0<u<t$ . This and (22) yield

(23) \begin{align}\Phi(Q(t))&=\mathrm{E}_u\bigg(P(t)^{N}-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right)\bigg)+\mathrm{P}(L> u) \nonumber\\&\quad -\mathrm{E}\big(1-P(t)^ N;\,L> u\big)-\mathrm{E}\bigg(\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,u<L\le t\bigg). \end{align}

We therefore obtain the upper bound

\begin{align*} \Phi(Q(t))&\le \mathrm{E}_u\Big(P(t)^{N}-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right)\Big)+\mathrm{P}(L> u),\end{align*}

which together with Lemma 4 and the monotonicity of $Q({\cdot})$ implies

(24) \begin{align} \Phi(Q(t))\le \mathrm{E}_u\bigg(\!\sum\nolimits_{j=1}^{N}(Q\left(t-\tau_j\right)-Q(t))\bigg)+\mathrm{P}(L>u).\end{align}

Borrowing an idea from [11], suppose to the contrary that

\begin{equation*}t_n\,:\!=\,\min\{t: tQ(t)\ge n\}\end{equation*}

is finite for any natural n. It follows that

\begin{equation*}Q(t_n)\ge \frac{n}{t_n},\qquad Q(t_n-u)<\frac{n}{t_n-u},\quad 1\le u\le t_n-1.\end{equation*}

Putting $t=t_n$ into (24) and using the monotonicity of $\Phi({\cdot})$ , we find

\begin{eqnarray*} \Phi\big(nt_n^{-1}\big)\le \Phi(Q(t_n))\le \mathrm{E}_u\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{n}{t_n-\tau_j}-\frac{n}{t_n}\bigg)\bigg)+\mathrm{P}(L> u).\end{eqnarray*}

Setting $u=t_n/2$ here and applying Lemma 3 together with (3), we arrive at the relation

\begin{equation*}\Phi\big(nt_n^{-1}\big)=O\big(nt_n^{-2}\big),\quad n\to\infty.\end{equation*}

Observe that under the condition (1), the L’Hospital rule gives

(25) \begin{equation}\Phi(z)\sim bz^2,\quad z\to0.\end{equation}

The resulting contradiction, $n^{2}t_n^{-2}=O\big(nt_n^{-2}\big)$ as $n\to\infty$ , finishes the proof of the proposition.

Proof of Proposition 2.The relation (23) implies

\begin{align*} \Phi(Q(t))\ge \mathrm{E}_u\bigg(P(t)^{N}-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right)\bigg)-\mathrm{E}\big(1-P(t)^ N;\,L> u\big).\end{align*}

By Lemma 4,

\begin{align*}P(t)^{N}-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right)= \sum_{j=1}^{N}\big(Q\left(t-\tau_j\right)-Q(t)\big)r_j^*(t),\end{align*}

where $0\le r_j^*(t)\le 1$ is a counterpart of the term $r_j$ in Lemma 4. By the monotonicity of $P({\cdot})$ , we have, again referring to Lemma 4,

\begin{equation*}1-r_j^*(t)\le (N-1)Q(t-L).\end{equation*}

Thus, for $0<y<1$ ,

(26) \begin{align} \Phi(Q(t))&\ge \mathrm{E}_{ty}\Bigg(\!\sum_{j=1}^{N}(Q\left(t-\tau_j\right)-Q(t))r_j^*(t) \Bigg)-\mathrm{E}\big(1-P(t)^ N;\,L> ty\big). \end{align}

The assertion $\liminf_{t\to\infty} tQ(t)>0$ is proven by contradiction. Assume that $\liminf_{t\to\infty} tQ(t)=0$ , so that

\begin{equation*}t_n\,:\!=\,\min\big\{t: tQ(t)\le n^{-1}\big\}\end{equation*}

is finite for any natural n. Plugging $t=t_n$ into (26) and using

\begin{equation*}Q(t_n)\le \frac{1}{nt_n},\quad Q(t_n-u)-Q(t_n)\ge \frac{1}{n(t_n-u)}-\frac{1}{nt_n},\quad 1\le u\le t_n-1,\end{equation*}

we get

\begin{equation*}\Phi\Big(\frac{1}{nt_n}\Big)\ge n^{-1}\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{1}{t_n-\tau_j}-\frac{1}{t_n}\bigg)r_j^*(t_n)\bigg)-\frac{1}{nt_n}\mathrm{E}(N;\,L> t_ny).\end{equation*}

Given $L\le ty$ , we have

\begin{align*}1-r_j^*(t)\le NQ(t(1-y))\le N\frac{q_2}{t(1-y)}, \end{align*}

where the second inequality is based on the already proven part of (14). Therefore,

\begin{equation*}\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{1}{t_n-\tau_j}-\frac{1}{t_n}\bigg)\big(1-r_j^*(t_n)\big)\bigg)\le \frac{q_2y}{t_n^2(1-y)^2}\mathrm{E}\big(N^2\big),\end{equation*}

and we derive

\begin{align*} nt_n^2\Phi\Big(\frac{1}{nt_n}\Big)&\ge t_n^2\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{1}{t_n-\tau_j}-\frac{1}{t_n}\bigg)\bigg) -\frac{\mathrm{E}\big(N^2\big)q_2y}{(1-y)^2}-t_n\mathrm{E}(N;\,L> t_ny).\end{align*}

Sending $n\to\infty$ and applying (25), Lemma 2, and Lemma 3, we arrive at the inequality

\begin{equation*}0\ge a-yq_2\mathrm{E}\big(N^2\big)(1-y)^{-2},\quad 0<y<1,\end{equation*}

which is false for sufficiently small y.

3.3. Proof of (18) and (19)

Fix an arbitrary $0<y<1$ . Lemma 1 with $u=ty$ gives

(27) \begin{align}\Phi\big(h t^{-1}\big)= \mathrm{P}(L> t)+\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}Q\left(t-\tau_j\right)\bigg)-Q(t)+\mathrm{E}_{ty}(W(t))+D(ty,t). \end{align}

Let us show that

(28) \begin{align}D(ty,t)=o\big(t^{-2}\big),\quad t\to\infty. \end{align}

Using Lemma 2 and (14), we find that for an arbitrarily small $\epsilon>0$ ,

\begin{equation*}\mathrm{E}\Big(1-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,ty<L\le t(1-\epsilon)\Big)=o\big(t^{-2}\big),\quad t\to\infty.\end{equation*}

On the other hand,

\begin{align*}\mathrm{E}\Big(1-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,t(1-\epsilon)<L\le t\Big)\le \mathrm{P}(t(1-\epsilon)<L\le t),\end{align*}

so that in view of (3),

\begin{equation*}\mathrm{E}\Big(1-\prod\nolimits_{j=1}^{N}P\left(t-\tau_j\right);\,ty<L\le t\Big)=o\big(t^{-2}\big),\quad t\to\infty.\end{equation*}

This, (12), and Lemma 2 imply (28).

Observe that

(29) \begin{equation}bh^2=ah+d.\end{equation}

Combining (27), (28), and

\begin{equation*}\mathrm{P}(L> t)-\Phi\big(h t^{-1}\big)\stackrel{(3)(25)}{=}dt^{-2}-bh^2t^{-2}+o\big(t^{-2}\big)\stackrel{(29)}{=}-aht^{-2}+o\big(t^{-2}\big),\quad t\to\infty,\end{equation*}

we derive (15), which in turn gives (17). The latter implies (18) since by Lemmas 2 and 4,

\begin{equation*} \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{h }{t-\tau_j}\bigg)-\frac{h}{t}=\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t-\tau_j}-\frac{h}{t}\bigg)\bigg)-ht^{-1}\mathrm{E}(N;\,L> ty)=aht^{-2}+o\big(t^{-2}\big).\end{equation*}

Turning to the proof of (19), observe that the random variable

\begin{equation*}W(t)=\big(1-h t^{-1}\big)^{N}-\prod\nolimits_{j=1}^{N}\bigg(1-\frac{h +\phi\left(t-\tau_j\right)}{t-\tau_j}\bigg)+\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t}-\frac{h +\phi\left(t-\tau_j\right)}{t-\tau_j}\bigg)\end{equation*}

can be represented in terms of Lemma 4 as

\begin{align*}W(t)&=\prod\nolimits_{j=1}^{N}(1-f_j(t))-\prod\nolimits_{j=1}^{N}(1-g_j(t))+\sum\nolimits_{j=1}^{N}(f_j(t)-g_j(t))\\&=\sum\nolimits_{j=1}^{N}(1-r_j(t))(f_j(t)-g_j(t)),\end{align*}

by assigning

(30) \begin{align}f_j(t)\,:\!=\,h t^{-1},\quad g_j(t)\,:\!=\,\frac{h +\phi\left(t-\tau_j\right)}{t-\tau_j}.\end{align}

Here $0\le r_j(t)\le 1$ , and for sufficiently large t,

(31) \begin{align}1-r_j(t)\stackrel{ (14)}{\le} Nq_2t^{-1}.\end{align}

After plugging into (18) the expression

\begin{equation*}W(t)=\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t}-\frac{h }{t-\tau_j}\Bigg)(1-r_j(t))-\sum\nolimits_{j=1}^{N}\frac{\phi\left(t-\tau_j\right)}{t-\tau_j}(1-r_j(t)),\end{equation*}

we get

\begin{align*}\frac{\phi(t)}{t}&= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{\phi\left(t-\tau_j\right)}{t-\tau_j}r_j(t)\bigg)+\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t-\tau_j}-\frac{h}{t}\bigg)(1-r_j(t) )\bigg)+o\big(t^{-2}\big),\quad\!\! t\to\infty.\end{align*}

The latter expectation is non-negative, and for an arbitrary $\epsilon>0$ , it has the following upper bound:

\begin{align*} \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t-\tau_j}-\frac{h}{t}\bigg)(1-r_j(t) )\bigg)\stackrel{ (31)}{\le} q_2\epsilon\mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\frac{h }{t-\tau_j}-\frac{h}{t} \bigg)\bigg)+\frac{q_2h}{(1-y)t^2}\mathrm{E}\big(N^2;\,N> t\epsilon\big).\end{align*}

Thus, in view of Lemma 3,

\begin{align*}\frac{\phi(t)}{t}&= \mathrm{E}_{ty}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{\phi\left(t-\tau_j\right)}{t-\tau_j}r_j(t)\bigg)+o\big(t^{-2}\big),\quad t\to\infty.\end{align*}

Multiplying this relation by t, we arrive at (19).

3.4. Proof of $\phi(t)\to 0$

Recall (20). If the non-decreasing function

\begin{equation*}M(t)\,:\!=\,\max_{1\le j\le t} m(j)\end{equation*}

is bounded from above, then $\phi(t)=O\big(\frac{1}{\ln t}\big)$ , proving that $\phi(t)\to 0$ as $t\to\infty$ . If $M(t)\to\infty$ as $t\to\infty$ , then there is an integer-valued sequence $0<t_1<t_2<\ldots,$ such that the sequence $M_n\,:\!=\,M(t_n)$ is strictly increasing and converges to infinity. In this case,

(32) \begin{equation}m(t)\le M_{n-1}<M_n,\quad 1\le t< t_n,\quad m(t_n)=M_n,\quad n\ge1.\end{equation}

Since $|\phi(t)|\le \frac{M_{n}}{\ln t_{n}}$ for $t_n\le t<t_{n+1}$ , to finish the proof of $\phi(t)\to 0$ , it remains to verify that

(33) \begin{equation} M_{n}=o(\!\ln t_{n}),\quad n\to\infty.\end{equation}

Fix an arbitrary $y\in(0,1)$ . Putting $t=t_n$ in (21) and using (32), we find

\begin{align*}M_n\le M_n\mathrm{E}_{t_ny}\Bigg(\!\sum\nolimits_{j=1}^{N}r_j(t_n)\frac{t_n\ln t_n}{\big(t_n-\tau_j\big)\ln\big(t_n-\tau_j\big)}\Bigg)+\big(t_n^{-1}\ln t_n\big)o_n.\end{align*}

Here and elsewhere, $o_n$ stands for a non-negative sequence such that $o_n\to0$ as $n\to\infty$ . In different formulas, the sign $o_n$ represents different such sequences. Since

\begin{equation*}0\le \frac{t\ln t}{(t-u)\ln (t-u)}-1\le \frac{u(1+\ln t)}{(t-u)\ln (t-u)},\quad 0\le u< t-1,\end{equation*}

and $r_j(t_n)\in[0,1]$ , it follows that

\begin{align*}M_n-M_n\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}r_j(t_n)\bigg)&\le M_n\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j(1+\ln t_n)}{t_n(1-y)\ln (t_n(1-y))}\bigg)+\big(t_n^{-1}\ln t_n\big)o_n.\end{align*}

Recalling that $a=\mathrm{E}(\!\sum_{j=1}^{N}\tau_j)$ , observe that

\begin{align*}\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}\frac{\tau_j(1+\ln t_n)}{t_n(1-y)\ln (t_n(1-y))}\bigg)\le \frac{a(1+\ln t_n)}{t_n(1-y)\ln (t_n(1-y))}= \big(a(1-y)^{-1}+o_n\big)t_n^{-1}.\end{align*}

Combining the last two relations, we conclude

(34) \begin{align}M_n\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}(1-r_j(t_n))\bigg)&\le a(1-y)^{-1}t_n^{-1}M_n +t_n^{-1}(M_n+\ln t_n)o_n.\end{align}

Now it is time to unpack the term $r_j(t)$ . By Lemma 4 with (30),

\begin{equation*}1-r_j(t)=\sum_{i=1}^{j-1}\frac{h +\phi(t-\tau_i)}{t-\tau_i}+(N-j)\frac{h }{t}-R_j(t),\end{equation*}

where, provided $\tau_j\le ty$ ,

\begin{equation*}0\le R_j(t)\le Nq_2t^{-1}(1-y)^{-1},\quad R_j(t)\le N^2q_2^2t^{-2}(1-y)^{-2},\quad t>t^*,\end{equation*}

for a sufficiently large $t^*$ . This allows us to rewrite (34) in the form

\begin{align*}M_n\mathrm{E}_{t_ny} &\Bigg(\!\sum\nolimits_{j=1}^{N}\Bigg(\!\sum_{i=1}^{j-1}\frac{h +\phi(t_n-\tau_i)}{t_n-\tau_i}+(N-j)\frac{h }{t_n}\Bigg)\Bigg)\\&\le M_n\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N} R_j(t_n)\bigg)+a(1-y)^{-1}t_n^{-1}M_n +t_n^{-1}(M_n+\ln t_n)o_n.\end{align*}

To estimate the last expectation, observe that if $\tau_j\le ty$ , then for any $\epsilon>0$ ,

\begin{equation*}R_j(t)\le Nq_2t^{-1}(1-y)^{-1} 1_{\{N>t\epsilon\}}+ N^2q_2^2t^{-2} (1-y)^{-2}1_{\{N\le t\epsilon\}},\quad t>t^*,\end{equation*}

implying that for sufficiently large n,

\begin{equation*}\mathrm{E}_{t_ny}\bigg(\!\sum\nolimits_{j=1}^{N}R_j(t_n)\bigg)\le q_2t_n^{-1}(1-y)^{-1}\mathrm{E}\big(N^{2} ;\, N> t_n\epsilon\big)+ q_2^2\epsilon t_n^{-1}(1-y)^{-2}\mathrm{E}\big(N^2\big),\end{equation*}

so that

\begin{align*}M_n\mathrm{E}_{t_ny}&\bigg(\!\sum\nolimits_{j=1}^{N}\bigg(\!\sum\nolimits_{i=1}^{j-1}\frac{h +\phi(t_n-\tau_i)}{t_n-\tau_i}+(N-j)\frac{h }{t_n}\bigg)\bigg)\\&\le a(1-y)^{-1}t_n^{-1}M_n +t_n^{-1}(M_n+\ln t_n)o_n.\end{align*}

Since

\begin{equation*}\sum\nolimits_{j=1}^{N}\sum\nolimits_{i=1}^{j-1}\bigg(\frac{h}{t_n-\tau_i}- \frac{h }{t_n}\bigg)\ge0,\end{equation*}

we obtain

\begin{align*}M_n\mathrm{E}_{t_ny}&\Bigg(\!\sum\nolimits_{j=1}^{N}\Bigg(\!\sum_{i=1}^{j-1}\frac{\phi(t_n-\tau_i)}{t_n-\tau_i}+(N-1)\frac{h }{t_n}\Bigg)\Bigg)\\&\le a(1-y)^{-1}t_n^{-1}M_n +t_n^{-1}(M_n+\ln t_n)o_n.\end{align*}

By (16) and (14), we have $\phi(t)\ge q_1-h$ for $t\ge t_0$ . Thus, for $\tau_j\le L\le t_ny$ and sufficiently large n,

\begin{equation*}\frac{\phi(t_n-\tau_i)}{t_n-\tau_i}\stackrel{}{\ge} \frac{q_1-h}{t_n(1-y)}.\end{equation*}

This gives

\begin{equation*}\sum\nolimits_{j=1}^{N}\Bigg(\!\sum_{i=1}^{j-1}\frac{\phi(t_n-\tau_i)}{t_n-\tau_i}+(N-1)\frac{h }{t_n}\Bigg)\ge \bigg(h+\frac{q_1-h }{2(1-y)}\bigg)t_n^{-1}N(N-1),\end{equation*}

which, after multiplying by $t_nM_n$ and taking expectations, yields

\begin{align*}\bigg(h+\frac{q_1-h }{2(1-y)}\bigg)M_n\mathrm{E}_{t_ny}(N(N-1))\le a(1-y)^{-1}M_n +(M_n+\ln t_n)o_n.\end{align*}

Finally, since

\begin{equation*} \mathrm{E}_{t_ny}(N(N-1))\to2b,\quad n\to\infty,\end{equation*}

we derive that for any $0<\epsilon<y<1$ , there is a finite $n_\epsilon$ such that for all $n>n_\epsilon$ ,

\begin{equation*}M_n\big(2bh(1-y)+bq_1-bh-a-\epsilon\big) \le \epsilon\ln t_n.\end{equation*}

By (29), we have $bh\ge a$ , and therefore

\begin{equation*}2bh(1-y)+bq_1-bh-a-\epsilon\ge bq_1-2bhy-y.\end{equation*}

Thus, choosing $y=y_0$ such that $bq_1-2bhy_0-y_0=\frac{bq_1}{2}$ , we see that

\begin{equation*}\limsup_{n\to\infty}\frac{M_n}{\ln t_n} \le \frac{2\epsilon}{bq_1},\end{equation*}

which implies (33) as $\epsilon\to0$ , concluding the proof of $\phi(t)\to 0$ .

4. Proof of Theorem 1

We will use the following notational conventions for the k-dimensional probability generating function

\begin{equation*}\mathrm{E}\Big(z_1^{Z(t_1)}\cdots z_k^{Z(t_k)}\Big)=\sum_{i_1=0}^\infty\ldots\sum_{i_k=0}^\infty\mathrm{P}(Z(t_1)=i_1,\ldots, Z(t_k)=i_k)z_1^{i_1}\cdots z_k^{i_k},\end{equation*}

with $0< t_1\le \ldots\le t_k$ and $z_1,\ldots,z_k\in[0,1]$ . We define

\begin{equation*}P_k\big(\bar t,\bar z\big)\,:\!=\,P_k(t_1,\ldots,t_{n};\,z_1,\ldots,z_{k})\,:\!=\,\mathrm{E}\Big(z_1^{Z(t_1)}\cdots z_k^{Z(t_k)}\Big)\end{equation*}

and write, for $t\ge0$ ,

\begin{equation*}P_k\big(t+\bar t,\bar z\big)\,:\!=\,P_k(t+t_1,\ldots,t+t_{k};\,z_1,\ldots,z_{k}).\end{equation*}

Moreover, for $0< y_1<\ldots<y_k$ , we write

\begin{equation*}P_k(t\bar y,\bar z)\,:\!=\,P_k(ty_1,\ldots,ty_{k};\,z_1,\ldots,z_{k}),\end{equation*}

and assuming $0< y_1<\ldots<y_k<1$ ,

\begin{equation*}P_k^*\big(t,\bar y,\bar z\big)\,:\!=\,\mathrm{E}\Big(z_1^{Z(ty_1)}\cdots z_{k}^{Z(ty_{k})};\,Z(t)=0\Big)=P_{k+1}(ty_1,\ldots,ty_k,t;\,z_1,\ldots,z_k,0).\end{equation*}

These conventions will be similarly applied to the functions

(35) \begin{equation}Q_k\big(\bar t,\bar z\big)\,:\!=\,1-P_k\big(\bar t,\bar z\big),\quad Q_k^*\big(t,\bar y,\bar z\big)\,:\!=\,1-P_k^*\big(t,\bar y,\bar z\big).\end{equation}

Our special interest is in the function

(36) \begin{equation} Q_k(t)\,:\!=\,Q_k\big(t+\bar t,\bar z\big),\quad 0= t_1< \ldots< t_k, \quad z_1,\ldots,z_k\in[0,1),\end{equation}

to be viewed as a counterpart of the function Q(t) treated by Theorem 2. Recalling the compound parameters

\begin{equation*}h=\frac{a+\sqrt{a^2+4bd}}{2b}\end{equation*}

and $c=4bda^{-2}$ , put

(37) \begin{equation}h_k\,:\!=\,h\frac{1+\sqrt{1+cg_k}}{1+\sqrt{1+c}},\quad g_k\,:\!=\, g_k(\bar y,\bar z)\,:\!=\,\sum_{i=1}^{k}z_1\cdots z_{i-1}(1-z_{i})y_{i}^{-2}.\end{equation}

The key step of the proof of Theorem 1 is to show that for any given $1=y_1<y_2<\ldots<y_k$ ,

(38) \begin{equation}tQ_k(t)\to h_k,\quad t_i\,:\!=\,t(y_i-1), \quad i=1,\ldots,k,\quad t\to\infty.\end{equation}

This is done following the steps of our proof of $tQ(t)\to h$ given in Section 3.

Unlike Q(t), the function $Q_k(t)$ is not monotone over t. However, monotonicity of Q(t) was used in the proof of Theorem 2 only for the proof of (14). The corresponding statement

\begin{equation*} 0<q_1\le tQ_k(t)\le q_2<\infty,\quad t\ge t_0,\end{equation*}

follows from the bounds $(1-z_1)Q(t)\le Q_k(t)\le Q(t)$ , which hold by the monotonicity of the underlying generating functions over $z_1,\ldots,z_{n}$ . Indeed,

\begin{equation*}Q_k(t)\le Q_k(t, t+t_2,\ldots,t+t_{k};\,0,\ldots,0)= Q(t),\end{equation*}

and on the other hand,

\begin{equation*}Q_k(t)= Q_k(t,t+t_2,\ldots,t+t_{k};\,z_1,\ldots,z_k)= \mathrm{E}\Big(1-z_1^{Z(t)}z_2^{Z(t+t_2)}\cdots z_k^{Z(t+t_k)}\Big)\ge \mathrm{E}\Big(1-z_1^{Z(t)}\Big),\end{equation*}

where

\begin{equation*} \mathrm{E}\Big(1-z_1^{Z(t)}\Big)\ge \mathrm{E}\Big(1-z_1^{Z(t)};\,Z(t)\ge1\Big)\ge (1-z_1)Q(t).\end{equation*}

4.1. Proof of $\boldsymbol{{tQ}}_{\boldsymbol{{k}}}\boldsymbol{{(t)}}\to \boldsymbol{{h}}_{\boldsymbol{{k}}}$

The branching property (8) of the GWO process gives

\begin{equation*} \prod_{i=1}^{k} z_i^{Z(t_i)}=\prod_{i=1}^{k} z_i^{1_{\{L>t_i\}}}\prod\nolimits_{j=1}^{N} z_i^{Z_j\left(t_i-\tau_j\right)}.\end{equation*}

Given $0< t_1<\ldots<t_k< t_{k+1}=\infty$ , we use

\begin{align*}\prod_{i=1}^{k} z_i^{1_{\{L>t_i\}}}&=1_{\{L\le t_1\}}+\sum_{i=1}^{k}z_1\cdots z_{i}1_{\{t_{i}<L\le t_{i+1}\}}\end{align*}

to deduce the following counterpart of (9):

\begin{align*}P_k\big(\bar t,\bar z\big)&=\mathrm{E}_{t_1}\Bigg(\prod_{j=1}^{N}P_k(\bar t-\tau_j,\bar z)\Bigg)+\sum_{i=1}^{k}z_1\cdots z_{i}\mathrm{E}\Bigg(\prod_{j=1}^{N}P_k(\bar t-\tau_j,\bar z);\,t_{i}<L\le t_{i+1}\Bigg).\end{align*}

This implies

(39) \begin{align}P_k\big(\bar t,\bar z\big)\,=\,&\mathrm{E}_{t_1}\Bigg(\prod_{j=1}^{N}P_k(\bar t-\tau_j,\bar z)\Bigg)+\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{P}(t_{i}<L\le t_{i+1}) \nonumber\\&-\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{E}\Bigg(1-\prod_{j=1}^{N}P_k(\bar t-\tau_j,\bar z);\, t_{i}<L\le t_{i+1}\Bigg).\end{align}

Using this relation we establish the following counterpart of Lemma 1.

Lemma 5. Consider the function (36) and put $P_k(t)\,:\!=\,1-Q_k(t)=P_k\big(t+\bar t,\bar z\big)$ . For $0<u<t$ , the relation

(40) \begin{align}\Phi\big(h_k t^{-1}\big)&= \mathrm{P}(L> t)-\sum_{i=1}^{k}z_1\cdots z_{i}\mathrm{P}\big(t+t_i<L\le t+t_{i+1}\big) \nonumber \\&+\mathrm{E}_u\bigg(\!\sum\nolimits_{j=1}^{N}Q_k\left(t-\tau_j\right)\bigg)-Q_k(t)+\mathrm{E}_u(W_k(t))+D_k(u,t) \end{align}

holds with $t_{k+1}=\infty$ ,

(41) \begin{align} W_k(t)\,:\!=\,\big(1-h_k t^{-1}\big)^{N}+Nh_k t^{-1}-\sum\nolimits_{j=1}^{N}Q_k\left(t-\tau_j\right)-\prod\nolimits_{j=1}^{N}P_k\left(t-\tau_j\right),\end{align}

and

(42) \begin{align} D_k(u,t)\,:\!=\, &\mathrm{E}\Big(1-\prod\nolimits_{j=1}^{N}P_k\left(t-\tau_j\right);\,u<L\le t\Big)+\mathrm{E}\Big(\big(1-h_k t^{-1}\big)^{N} -1+Nh_k t^{-1};\,L>u\Big) \nonumber\\&+\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{E}\Bigg(1-\prod_{j=1}^{N}P_k\left(t-\tau_j\right);\, t+t_{i}<L\le t+t_{i+1}\Bigg).\end{align}

Proof. According to (39),

\begin{align*}P_k(t)&=\mathrm{E}_u\Bigg(\prod_{j=1}^{N}P_k\left(t-\tau_j\right)\Bigg)+\mathrm{E}\bigg(\prod\nolimits_{j=1}^{N}P_k\left(t-\tau_j\right);\,u<L\le t\bigg)\\& \quad +\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{P}\big(t+t_{i}<L\le t+t_{i+1}\big)\\& \quad -\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{E}\Bigg(1-\prod_{j=1}^{N}P_k\left(t-\tau_j\right);\, t+t_{i}<L\le t+t_{i+1}\Bigg).\end{align*}

By the definition of $\Phi({\cdot})$ ,

\begin{align*}\Phi\big(h_k t^{-1}\big)+1\,=\,&\mathrm{E}_u\Big(\big(1-h_k t^{-1}\big)^{N}+Nh_k t^{-1}\Big)+\mathrm{P}(L> t)\\&+\mathrm{E}\Big(\big(1-h_k t^{-1}\big)^{N} -1+Nh_k t^{-1};\,L> u\Big)+\mathrm{P}(u<L\le t),\end{align*}

and after subtracting the two last equations, we get

\begin{align*}\Phi\big(h_k t^{-1}\big)+Q_k(t) \,=\, &\mathrm{E}_u\Big(\big(1-h_k t^{-1}\big)^{N} +Nh_k t^{-1}-\prod\nolimits_{j=1}^{N}P_k\left(t-\tau_j\right)\Big)+\mathrm{P}(L> t)\\&-\sum_{i=1}^{k}z_1\cdots z_{i} \mathrm{P}(t+t_{i}<L\le t+t_{i+1})+D_k(u,t),\end{align*}

with $D_k(u,t)$ satisfying (42). After a rearrangement, the relation (40) follows together with (41).

With Lemma 5 in hand, the convergence (38) is proven by applying almost exactly the same argument as used in the proof of $tQ(t)\to h$ . An important new feature emerges because of the additional term in the asymptotic relation defining the limit $h_k$ . Let $1=y_1<y_2<\ldots<y_k<y_{k+1}=\infty$ . Since

\begin{align*}\sum\nolimits_{i=1}^{k}z_1\cdots z_{i}\mathrm{P}\big(ty_{i}<L\le ty_{i+1}\big)\sim d t^{-2}\sum_{i=1}^{k}z_1\cdots z_{i}\Big(y_{i}^{-2}-y_{i+1}^{-2}\Big),\end{align*}

we see that

\begin{align*}\mathrm{P}(L> t)-\sum\nolimits_{i=1}^{k}z_1\cdots z_{i}\mathrm{P}\big(ty_{i}<L\le ty_{i+1}\big)\sim dg_k t^{-2},\end{align*}

where $g_k$ is defined by (37). Assuming $0\le z_1,\ldots,z_k<1$ , we ensure that $g_k>0$ , and as a result, we arrive at a counterpart of the quadratic equation (29),

\begin{equation*}bh_k^2=ah_k+dg_k,\end{equation*}

which gives

\begin{equation*}h_k=\frac{a+\sqrt{a^2+4bdg_k}}{2b}=h\frac{1+\sqrt{1+cg_k}}{1+\sqrt{1+c}},\end{equation*}

justifying our definition (37). We conclude that for $k\ge1$ ,

(43) \begin{align}\frac{Q_k(t\bar y,\bar z)}{Q(t)}\to &\frac{1+\sqrt{1+c\sum\nolimits_{i=1}^{k}z_1\cdots z_{i-1}(1-z_{i})y_{i}^{-2}}}{1+\sqrt{1+c}}, \nonumber\\& \qquad \qquad 1=y_1<\ldots< y_k,\quad0\le z_1,\ldots,z_k<1.\end{align}

4.2. Conditioned generating functions

To finish the proof of Theorem 1, consider the generating functions conditioned on the survival of the GWO process. Given (5) with $j\ge1$ , we have

\begin{align*}Q(t)\mathrm{E}&\Big(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)}|Z(t)>0\Big)=\mathrm{E}(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)};\,Z(t)>0)\\&=P_k(t\bar y,\bar z)-\mathrm{E}\Big(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)};\,Z(t)=0\Big)\stackrel{(35)}{=}Q_j^*\big(t,\bar y,\bar z\big)-Q_k(t\bar y,\bar z),\end{align*}

and therefore,

\begin{equation*}\mathrm{E}\Big(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)}|Z(t)>0\Big)=\frac{Q_j^*\big(t,\bar y,\bar z\big)}{Q(t)}-\frac{Q_k(t\bar y,\bar z)}{Q(t)}.\end{equation*}

Similarly, if (5) holds with $j=0$ , then

\begin{equation*}\mathrm{E}\Big(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)}|Z(t)>0\Big)=1-\frac{Q_k(t\bar y,\bar z)}{Q(t)}.\end{equation*}

Letting $t^{\prime}=ty_1$ , we get

\begin{equation*}\frac{Q_k(t\bar y,\bar z)}{Q(t)}=\frac{Q_k(t^{\prime},t^{\prime}y_2/y_1,\ldots,t^{\prime}y_k/y_1)}{Q(t^{\prime})}\frac{Q(ty_1)}{Q(t)},\end{equation*}

and applying the relation (43), we have

\begin{equation*}\frac{Q_k(t\bar y,\bar z)}{Q(t)}\to \frac{1+\sqrt{1+\sum\nolimits_{i=1}^{k}z_1\cdots z_{i-1}(1-z_{i})\Gamma_i}}{\big(1+\sqrt{1+c}\big)y_1},\end{equation*}

where $\Gamma_i=c({y_1}/{y_i} )^2$ . On the other hand, since

\begin{equation*}Q_j^*\big(t,\bar y,\bar z\big)=Q_{j+1}(ty_1,\ldots,ty_j,t;\,z_1,\ldots,z_j,0), \quad j\ge1,\end{equation*}

we also get

\begin{equation*}\frac{Q_j^*\big(t,\bar y,\bar z\big)}{Q(t)}\to \frac{1+\sqrt{1+\sum\nolimits_{i=1}^{j}z_1\cdots z_{i-1}(1-z_{i})\Gamma_i+cz_1\cdots z_{j}y_1^2}}{\big(1+\sqrt{1+c}\big)y_1}.\end{equation*}

We conclude that as stated in Section 2,

\begin{align*}\mathrm{E}\Big(z_1^{Z(ty_1)}\cdots z_k^{Z(ty_k)}|Z(t)>0\Big)\to \mathrm{E}\Big(z_1^{\eta(y_1)}\cdots z_k^{\eta(y_k)}\Big).\end{align*}