2. An age-structured branching process

In this section we introduce the age-structured branching process and give some basic characterizations of its transition probabilities. Most of the results presented here are essentially known, so we only sketch the proofs; see, e.g., [Reference Bose and Kaj6, Reference Kaj and Sagitov19, Reference Li21, Reference Li23].

Let $\alpha\in C^1\!\big(\mathbb{R}_+\big)^{+}$ be a function bounded away from zero. For each $x\in \mathbb{R}_+$ let $\{p(x,i) \,:\, i\in \mathbb{N}\}$ be a discrete probability distribution with generating function $g(x,z) = \sum_{i=0}^\infty p(x,i)z^i$ , $z\in [0,1]$ . We assume that $p(\cdot,i)\in C^1\big(\mathbb{R}_+\big)^{+}$ for every $i\in \mathbb{N}$ , and that $\|\partial_zg(\cdot,1{-})\| = \sup_{x\ge 0}\sum_{i=1}^\infty p(x,i)i< \infty$ , where $\partial_z$ denotes the first derivative with respect to z. A branching particle system is characterized by the following properties:

1. (i) The ages of the particles increase at unit speed, i.e. they move according to realizations of the deterministic process $\xi= (\xi_t)_{t\ge 0}$ in $\mathbb{R}_+$ defined by $\xi_t= \xi_0+t$ .

2. (ii) For a particle which is alive at time $r\ge 0$ with age $x\ge 0$ , the conditional probability of survival in the time interval [r, t) is $\exp\!\big\{{-}\int_0^{t-r} \alpha(x+s) \, \text{d} s\big\}$ .

3. (iii) When a particle dies at age $x\ge 0$ , it gives birth to a random number of offspring with age zero according to the probability law $\{p(x, i) \,:\, i = 0, 1, \ldots\}$ determined by the generating function $g(x,\cdot)$ .

We assume that the lifetimes and the offspring productions of different particles are independent. Let $X_t(B)$ denote the number of particles alive at time $t\ge 0$ with ages belonging to the Borel set $B\subset \mathbb{R}_+$ . If we assume $X_0\big(\mathbb{R}_+\big)< \infty$ , then $\{X_t \,:\, t\ge 0\}$ is a Markov process with state space $\mathfrak{N}\big(\mathbb{R}_+\big)$ . We refer to [Reference Li21, Section 4.3] for the formulation of general branching particle systems.

Let $\sigma\in \mathfrak{N}\big(\mathbb{R}_+\big)$ and let $\{X_t^\sigma \,:\, t\ge 0\}$ be the above system with initial value $X^\sigma_0 = \sigma$ . Let $\delta_x$ denote the unit measure concentrated at $x\in \mathbb{R}_+$ . Suppose that the process is defined on a probability space $(\Omega, \mathscr{G}, \mathbb{P})$ . Properties (i)–(iii) imply that, for $f\in B\big(\mathbb{R}_+\big)^+$ ,

\begin{equation*} \mathbb{E}\left[\exp\!\big\{-\big\langle X_t^\sigma, f\big\rangle \big\}\right] = \exp\{-\langle \sigma, u_t\,f\rangle \}, \qquad u_t\,f(x) = -\log \mathbb{E}\big[\exp\!\big\{-\big\langle X_t^{\delta_x}, f\big\rangle \big\}\big].\end{equation*}

From properties (i)–(iii) we derive, as in [Reference Kaj and Sagitov19, Section 3] and [Reference Li21, Section 4.3], the following renewal equation:

(2.1) \begin{align} \text{e}^{-u_t\,f(x)} = & \exp\!\bigg\{{-}f(x+t)-\int_0^t \alpha(x+s) \, \text{d} s\bigg\}\nonumber\\[3pt]& + \int_0^t \exp\!\bigg\{-\int_0^s\alpha(x+r) \, \text{d} r\bigg\}\alpha(x+s) g\big(x+s, \text{e}^{-u_{t-s}f(0)}\big) \, \text{d} s.\end{align}

From [Reference Li21, Proposition 2.9], the above equation implies

(2.2) \begin{eqnarray} \text{e}^{-u_t\,f(x)} = \text{e}^{-f(x+t)} + \int_0^t \alpha(x+t-s)\big[g\big(x+t-s, \text{e}^{-u_sf(0)}\big) - \text{e}^{-u_sf(x+t-s)}\big] \, \text{d} s.\end{eqnarray}

The uniqueness of the solution to (2.1) and (2.2) follows by Gronwall’s inequality. The two equations are therefore equivalent.

We call any Markov process $(X_t \,:\, t\ge 0)$ with state space $\mathfrak{N}\big(\mathbb{R}_+\big)$ an $(\alpha,g)$ -age-structured branching process if it has a transition semigroup $(Q_t)_{t\ge 0}$ defined by

(2.3) \begin{eqnarray} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)}\text{e}^{-\langle \nu, f\rangle }Q_t(\sigma,d\nu) = \exp\!\big\{-\langle \sigma, u_t\,f\rangle \big\}, \qquad f\in B\big(\mathbb{R}_+\big)^+,\end{eqnarray}

where $u_t\,f(x)$ is the unique solution to (2.2).

Proposition 2.1. For any $x\ge 0$ , $t\ge 0$ , and $f\in C^1_0\big(\mathbb{R}_+\big)^+$ , we have

(2.4) \begin{eqnarray} \partial_tu_t\,f(x) = \partial_xu_t\,f(x) + \alpha(x)\big[1 - \text{e}^{u_t\,f(x)}g\big(x,\text{e}^{-u_t\,f(0)}\big)\big].\end{eqnarray}

Proof.For any $t\ge 0$ let $T_t$ be the operator on the Banach space $C_0\big(\mathbb{R}_+\big)$ defined by $T_tf(x)= f(x+t)$ . By (2.2) it is easy to see that $U_t\,f(x)= 1-\text{e}^{-u_t\,f(x)}$ solves the evolution integral equation

(2.5) \begin{eqnarray}U_t\,f(x) = T_tU_0f(x) - \int_0^t T_{t-s}\phi(\cdot,U_sf)(x) \, \text{d} s,\end{eqnarray}

where $\phi(x,f) = \alpha(x)\big[g(x,1-f(0))-1+f(x)\big]$ . By a general result on semi-linear evolution equations, we know that $t\mapsto U_t\,f(x)$ is continuously differentiable and solves the differential evolution equation $\partial_tU_t\,f(x)= \partial_xU_t\,f(x) - \phi(x,U_t\,f)$ , $U_0f(x)= 1-\text{e}^{-f(x)}$ ; see, e.g., [Reference Pazy27, Theorem 6.1.5, p. 187]. Then $t\mapsto \partial_tu_t\,f(x)$ is also continuously differentiable. By differentiating both sides of (2.2) we have

\begin{align*} \text{e}^{-u_t\,f(x)}\partial_tu_t\,f(x) & = \partial_x \text{e}^{-f(x+t)} \\ & \quad + \int_0^t \partial_x\big[\alpha(x+t-s)\big(g\big(x+t-s,\text{e}^{-u_sf(0)}\big) - \text{e}^{-u_sf(x+t-s)}\big)\big] \, \text{d} s \\ & \quad + \alpha(x)\big[\text{e}^{-u_t\,f(x)}-g\big(x, \text{e}^{-u_t\,f(0)}\big)\big] \\ & = \text{e}^{-u_t\,f(x)}\partial_xu_t\,f(x) + \alpha(x)\big[\text{e}^{-u_t\,f(x)} - g\big(x, \text{e}^{-u_t\,f(0)}\big)\big] , \end{align*}

which proves (2.4).

Proposition 2.2. For any $t\ge 0$ and $\sigma\in \mathfrak{N}\big(\mathbb{R}_+\big)$ we have

(2.6) \begin{eqnarray} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)}\langle \nu,f\rangle Q_t(\sigma, \text{d}\nu) = \langle \sigma, \pi_tf\rangle , \qquad f\in B\big(\mathbb{R}_+\big),\end{eqnarray}

where $(\pi_t)_{t\ge 0}$ is the semigroup of bounded kernels on $\mathbb{R}_+$ defined by

(2.7) \begin{eqnarray} { \pi_tf(x) = f(x+t) + \int_0^t \alpha(x+s)\big[\partial_z g(x+s,1{-})\pi_{t-s}f(0) - \pi_{t-s}f(x + s)\big] \, \text{d} s.}\end{eqnarray}

Proposition 2.4. Let $c_1= \sup_{y\ge 0}\alpha(y)$ . Then, for any $t\ge 0$ and $x\ge 0$ ,

(2.8) \begin{eqnarray} \pi_tf(x)\ge u_t\,f(x)\ge \big(1-\text{e}^{-f(x+t)}\big)\text{e}^{-c_1t}, \qquad f\in B\big(\mathbb{R}_+\big)^+.\end{eqnarray}

Proof.By taking $\sigma= \delta_x$ in (2.3) and (2.6) and using Jensen’s inequality we get the first inequality in (2.8). Let $U_t\,f(x)$ be as in the proof of Proposition 2.1. By (2.5) and a comparison theorem we have $U_t\,f(x)\ge v_tf(x)$ , where $(t,x)\mapsto v_tf(x)$ solves

\begin{eqnarray*}v_tf(x) = 1-\text{e}^{-f(x+t)} - \int_0^t \alpha(x+s)v_{t-s}f(x+s) \, \text{d} s.\end{eqnarray*}

The unique locally bounded solution to the above equation is given by

\begin{eqnarray*}v_tf(x) = \big(1-\text{e}^{-f(x+t)}\big)\exp\!\bigg\{-\int_0^t \alpha(x+s) \, \text{d} s\bigg\}.\end{eqnarray*}

Then we have the estimate (2.8).

3. Construction by stochastic equations

In this section we give a construction of the age-structured branching process by solving a stochastic equation driven by a Poisson random measure. Recall that $\mathfrak{D}\big(\mathbb{R}_+\big)$ denotes the set of bounded positive right-continuous increasing functions f on $\mathbb{R}$ satisfying $f(x)= 0$ for $x< 0$ . For $\mu\in \mathfrak{D}\big(\mathbb{R}_+\big)$ and $\alpha\in B\big(\mathbb{R}_+\big)^+$ define $A_\alpha(\mu,y)=\inf\{z\ge 0 \,:\, \langle \mu,\alpha\textbf{1}_{[0,z]}\rangle > \langle \mu,\alpha\rangle y\}$ , $0\le y\le 1$ , with $\inf\emptyset= \infty$ by convention. Then $\langle \mu,\alpha\rangle = 0$ implies $A_\alpha(\mu,y)= \infty$ for all $0\le y\le 1$ . By an elementary result in probability theory, we have the following lemma.

Suppose that $(\Omega, \mathscr{F}, \mathscr{F}_t, \mathbb{P})$ is a filtered probability space satisfying the usual hypotheses. Let $M(\text{d} t, \text{d} u, \text{d} y, \text{d} z, \text{d} v)$ be an $(\mathscr{F}_t)$ -Poisson random measure on $(0,\infty)^2\times (0, 1]\times \mathbb{N}\times (0, 1]$ with intensity $\text{d} t \, \text{d} u \, \text{d} y \, \pi(\text{d} z) \, \text{d} v$ , where $\pi(\text{d} z)$ denotes the counting measure on $\mathbb{N}$ . Given an $\mathscr{F}_0$ -measurable random function $X_0\in \mathfrak{D}\big(\mathbb{R}_+\big)$ , we consider the stochastic integral equation, for $t\ge 0$ and $x\ge 0$ ,

(3.1) \begin{align} X_t(x) =\, & X_0(x-t) + \int_0^t\int_0^{\langle X_{s-},\alpha\rangle }\!\! \int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)}\!\! z\textbf{1}_{\{t-s\le x\}} M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v)\nonumber\\[3pt]& - \int_0^t\int_0^{\langle X_{s-},\alpha\rangle }\!\! \int_0^1\int_{\mathbb{N}} \int_0^{p(A_\alpha(X_{s-},y),z)}\!\! \textbf{1}_{\{A_\alpha(X_{s-},y)+t-s\le x\}} M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align}

Heuristically, the left-hand side $X_t(x)$ is the number of individuals at time t with ages less than x. On the right-hand side, the first term $X_0(x-t)$ counts the number of individuals having ages less than $x-t$ at time 0, and thus having ages less than x at time t. A death in the population occurs at time $s\in [0,t]$ at rate $\langle X_{s-}, \alpha\rangle \text{d} s$ . In that case, the age of the dying individual is distributed according to the probability measure $\langle X_{s-}, \alpha\rangle ^{-1}\alpha(x)X_{s-}(\text{d} x)$ and is realized as $A_\alpha(X_{s-},y)$ , where $y\in (0, 1]$ is taken according to the uniform distribution by the Poisson random measure. The number of offspring of the individual takes the value $z\in \mathbb{N}$ with probability $p(A_\alpha(X_{s-},y),z)$ and contributes to the number $X_t(x)$ if and only if $t-s\le x$ , which is recorded by the second term. The death of the individual affects $X_t(x)$ if and only if $A_\alpha(X_{s-},y) + t-s\le x$ , which is recorded by the third term.

Let $\zeta_a(x) = \textbf{1}_{\{a\le x\}}$ for $a, x\in \mathbb{R}$ . Given a function f on $\mathbb{R}$ define $f\circ\theta_t(x)= f(x + t)$ for $x, t\in \mathbb{R}$ . Then we may rewrite (3.1) equivalently, for $t\ge 0$ and $x\ge 0$ , as

(3.2) \begin{align} X_t(x) = \, & X_0\circ\theta_{-t}(x) + \int_0^t\int_0^{\langle X_{s-},\alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)} \big[z \zeta_0\circ\theta_{s-t}(x)\nonumber\\[4pt]& - \zeta_{A_\alpha(X_{s-},y)}\circ\theta_{s-t}(x)\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align}

A pathwise solution to (3.2) is constructed by unscrambling the equation as follows. Let $\tau_0 = 0$ . Given $\tau_{k-1}\ge 0$ and $X_{\tau_{k-1}}\in \mathfrak{D}\big(\mathbb{R}_+\big)$ , we first define $\tau_k = \tau_{k-1} + \inf\big\{t>0 \,:\, M\big((\tau_{k-1},\tau_{k-1}+t]\times (0,\langle X_{\tau_{k-1}}, \alpha\rangle ]\times H_k\big)> 0\big\}$ , where $H_k= \big\{(y,z,v) \,:\, y\in (0, 1], z\in\mathbb{N},0< v\le p\big(A_\alpha\big(X_{\tau_{k-1}},y\big),z\big)\big\}$ , and

(3.3) \begin{eqnarray} X_t(x)= X_{\tau_{k-1}}\circ\theta_{\tau_{k-1}-t}(x), \qquad { \tau_{k-1}\le t< \tau_k,\ x\ge 0.}\end{eqnarray}

Then we define

(3.4) \begin{eqnarray} X_{\tau_k}(x)= X_{\tau_k-}(x) + z_k\zeta_0(x) - \zeta_{A_\alpha(X_{\tau_k-},y_k)}(x), \qquad x\ge 0,\end{eqnarray}

where $X_{\tau_k-}(x)= X_{\tau_{k-1}}\circ\theta_{\tau_{k-1}-\tau_k}(x)$ , and $(u_k,y_k,z_k,v_k)\in (0,\infty)\times (0, 1]\times \mathbb{N}\times (0, 1]$ is the point such that $(\tau_k, u_k,y_k,z_k,v_k)\in \text{supp}(M)$ . Since $A_\alpha\big(X_{\tau_k-},y_k\big)\in \text{supp}\big(X_{\tau_k-}\big)$ , we have $X_{\tau_k}\in \mathfrak{D}\big(\mathbb{R}_+\big)$ . Equation (3.4) means that at time $\tau_k$ an individual at age $A_\alpha\big(X_{\tau_k-},y_k\big)$ dies and gives birth to $z_k$ offspring with starting age $0\in \mathbb{R}_+$ .

It is clear that (3.3) and (3.4) uniquely determine the behavior of the trajectory $t\mapsto X_t$ on the time intervals $[\tau_{k-1},\tau_k]$ , $k= 1,2,\ldots$ Let $\tau= \lim_{k\to \infty}\tau_k$ and let $X_t= \infty$ for $t\ge \tau$ . Then $\{X_t \,:\, t \ge 0\}$ is the pathwise-unique solution to (3.2) up to the lifetime $\tau$ . More precisely, the equations hold almost surely with t replaced by $t\land \tau_k$ for every $k= 1,2,\ldots$ Let $n(t)= \sup\{k\ge 0 \,:\, \tau_k\le t\}$ for $t\ge 0$ , and $\beta= \|\alpha \partial_zg(\cdot,1{-})\|< \infty$ .

Lemma 3.2. Suppose that $\mathbb{E}[X_0(\infty)]< \infty$ . Then, for any $k\ge 1$ ,

(3.5) \begin{eqnarray}\mathbb{E}\big[{\textstyle \sup_{0\le s\le t\land\tau_k}} X_s(\infty)\big]\le\mathbb{E}[X_0(\infty)]\text{e}^{\beta t}, \qquad t\ge 0.\end{eqnarray}

Proof.Recall that $X_t(\infty)= \lim_{x\to \infty} X_t(x)= X_t\big(\mathbb{R}_+\big)$ . Let $\eta_i= \inf\{s\ge 0 \,:\, X_s(\infty)\ge i\}$ for $i\ge 1$ . It is clear that $\lim_{i\to \infty}\eta_i= \tau$ . Let $\zeta_{i,k}= \eta_i\land \tau_k$ . In view of (3.2), we have

\begin{equation*} X_t(\infty)= X_0(\infty) + \int_0^t\int_0^{\langle X_{s-},\alpha\rangle } \int_0^1\int_{\mathbb{N}}\int_0^{p\big(A_\alpha(X_{s-},y),z\big)} (z-1) M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{equation*}

It follows that

\begin{align*} \mathbb{E}\big[{\textstyle \sup_{0\le s\le t\land\zeta_{i,k}}} X_s(\infty)\big] & \le \mathbb{E}[X_0(\infty)] + \sum_{z\in \mathbb{N}} \mathbb{E}\bigg[\int_0^{t\land\zeta_{i,k}} \!\! \langle X_{s-},\alpha\rangle \text{d} s \!\! \int_0^1 \!\! p(A_\alpha(X_{s-},y),z) z \,\text{d} y\bigg] \\ & = \mathbb{E}[X_0(\infty)] + \sum_{z\in \mathbb{N}} \mathbb{E}\bigg[\int_0^{t\land\zeta_{i,k}} \text{d} s \int_{\mathbb{R}_+}p(y,z)z \alpha(y) \, \text{d} X_{s-}(y)\bigg] \\ & = \mathbb{E}[X_0(\infty)] + \mathbb{E}\bigg[\int_0^{t\land\zeta_{i,k}} \text{d} s \int_{\mathbb{R}_+} \alpha(y)\partial_zg(y,1{-}) \, \text{d} X_{s-}(y)\bigg] \\ & \le \mathbb{E}[X_0(\infty)] + \beta\mathbb{E}\bigg[\int_0^{t\land\zeta_{i,k}} X_{s-}(\infty) \, \text{d} s\bigg]. \end{align*}

Clearly, we have $X_{s-}(\infty)\le i$ for $0< s\le t\land\zeta_{i,k}$ . Then $\mathbb{E}\big[{\sup}_{0\le s\le t\land\zeta_{i,k}}X_s(\infty)\big]$ is locally bounded in $t\ge 0$ . Since the trajectory $s\mapsto X_s(\infty)$ has at most countably many jumps, it follows that

\begin{align*} \mathbb{E}\big[{\textstyle \sup_{0\le s\le t\land\zeta_{i,k}}} X_s(\infty)\big] & \le \mathbb{E}[X_0(\infty)] + \beta\mathbb{E}\bigg[\int_0^{t\land\zeta_{i,k}} X_s(\infty) \, \text{d} s\bigg] \cr & \le \mathbb{E}[X_0(\infty)] + \beta\int_0^t \mathbb{E}[X_{s\land\zeta_{i,k}}(\infty)] \, \text{d} s \cr & \le \mathbb{E}[X_0(\infty)] + \beta\int_0^t { \mathbb{E}\big[{\textstyle \sup_{0\le r\le s\land\zeta_{i,k}}} X_r(\infty)\big]} \, \text{d} s. \end{align*}

By Gronwall’s inequality we have $\mathbb{E}\big[{\sup}_{0\le s\le t\land\zeta_{i,k}}X_s(\infty)\big] \le \mathbb{E}[X_0(\infty)]\text{e}^{\beta t}$ . Then, letting $i\to \infty$ and using Fatou’s lemma, we obtain (3.5).

Proposition 3.1. Suppose that $\mathbb{E}[X_0(\infty)]< \infty$ . Then $\mathbb{P}\{\tau= \infty\}= 1$ and

(3.6) \begin{eqnarray}\mathbb{E}[n(t)]\le \|\alpha\|\mathbb{E}[X_0(\infty)] \int_0^t \text{e}^{\beta s} \, \text{d} s, \qquad t\ge 0.\end{eqnarray}

Proof.By (3.2) and monotone convergence we have

\begin{align*} \mathbb{E}[n(t)] & = \lim_{k\to \infty}\mathbb{E}\bigg[\int_0^{t\land\tau_k}\int_0^{\langle X_{s-},\alpha\rangle } \int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)} M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v)\bigg] \\ & = \lim_{k\to \infty}\mathbb{E}\bigg[\sum_{z\in \mathbb{N}} \int_0^{t\land\tau_k} \langle X_{s-},\alpha\rangle \, \text{d} s \int_0^1 p(A_\alpha(X_{s-},y),z) \, \text{d} y\bigg] \\ & = \lim_{k\to \infty}\mathbb{E}\bigg[\sum_{z\in \mathbb{N}} \int_0^{t\land\tau_k} \text{d} s \int_{\mathbb{R}_+} p(y,z)\alpha(y) \, \text{d} X_{s-}(y)\bigg] \\ & \le \lim_{k\to \infty}\|\alpha\|\mathbb{E}\bigg[\int_0^{t\land\tau_k} X_s(\infty) \, \text{d} s\bigg] = \lim_{k\to \infty}\|\alpha\|\int_0^t \mathbb{E}[X_{s\land\tau_k}(\infty)] \, \text{d} s. \end{align*}

Then (3.6) follows by (3.5). In particular, we have $\mathbb{P}\{\tau> t\}= \mathbb{P}\{n(t)< \infty\}= 1$ for every $t\ge 0$ , which implies $\mathbb{P}\{\tau= \infty\}= 1$ .

By Proposition 3.1 the solution of (3.1) or (3.2) has infinite lifetime and determines a measure-valued strong Markov process $\{X_t \,:\, t\ge 0\}$ . The following propositions give some useful characterization of the process.

Proposition 3.3. For any $t\ge 0$ and $f \in B\big(\mathbb{R}_+\big)$ ,

(3.7) \begin{align} \langle X_t,f\rangle =\, & \langle X_0,f\circ \theta_t\rangle + \int_0^t\int_0^{\langle X_{s-},\alpha\rangle }\int_0^1 \int_{\mathbb{N}} \int_0^{p(A_\alpha(X_{s-},y),z)} \big[z f\circ\theta_{t-s}(0)\nonumber\\& - f\circ\theta_{t-s}(A_\alpha(X_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align}

Proof.Let $C^1_0\big(\mathbb{R}_+\big)$ denote the subspace of $C^1\big(\mathbb{R}_+\big)$ consisting of functions vanishing at infinity. For any fixed integer $n\ge 1$ , let $x_i = in/2^n$ with $i = 0, 1, \ldots, 2^n$ . By (3.2), almost surely for any $f \in C^1_0\big(\mathbb{R}_+\big)$ ,

\begin{align*} \sum_{i = 1}^{2^n}f^{\prime}(x_i)X_t(x_i) & - \sum_{i = 1}^{2^n}f^{\prime}(x_i)X_0\circ\theta_{-t}(x_i) \\& = \int_0^t\int_0^{\langle X_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)}\sum_{i = 1}^{2^n}f^{\prime}(x_i)\big[z \zeta_0\circ\theta_{s-t}(x_i) \\ &\quad - \zeta_{A_\alpha(X_{s-},y)}\circ\theta_{s- t}(x_i)\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align*}

Then we multiply the above equation by $2^{-n}$ and let $n\to \infty$ to get, almost surely,

(3.8) \begin{align} \int_0^\infty f^{\prime}(x)X_t(x) \, \text{d} x & - \int_0^\infty f^{\prime}(x)X_0\circ\theta_{-t}(x) \, \text{d} x \nonumber \\ & = \int_0^t\int_0^{\langle X_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)} \bigg\{\!\int_0^\infty f^{\prime}(x) \big[z\zeta_0\circ\theta_{s-t}(x) \nonumber \\ & \quad - \zeta_{A_\alpha(X_{s-},y)}\circ\theta_{s-t}(x)\big] \, \text{d} x\bigg\} M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \nonumber \\ & = -\int_0^t\int_0^{\langle X_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y), z)} \big[z f\circ\theta_{t-s}(0) \nonumber \\ & \quad - f\circ\theta_{t-s}(A_\alpha(X_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align}

By integration by parts we have $\langle X_t, f\rangle = - \int_0^\infty f^{\prime}(x)X_t(x) \, \text{d} x$ . From this and (3.8) we see that (3.7) holds for any $f \in C_0^1\big(\mathbb{R}_+\big)$ . Then the relation also holds for any $f\in B\big(\mathbb{R}_+\big)$ by a monotone class argument.

Proposition 3.4. For any $t\ge 0$ and $f\in C^1\big(\mathbb{R}_+\big)$ ,

(3.9) \begin{align} \langle X_t, f\rangle =\, & \langle X_0, f\rangle + \int_0^t \langle X_{s-}, f^{\prime}\rangle \, \text{d} s + \int_0^t \int_0^{\langle X_{s-},\alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)}\big[z f(0)\nonumber\\[4pt]& - f(A_\alpha(X_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \qquad\end{align}

Proof.For $n\ge 1$ we consider a partition $\Delta_n = \{0 = t_0 < t_1 < \cdots < t_n = t\}$ of [0, t]. Notice that $ \partial_t(f\circ \theta_t)(x) = f^{\prime}(x+t).$ By (3.7) we have

\begin{align*} \langle X_t, f\rangle & = \langle X_0, f\rangle + \sum_{i=1}^n \left[\big\langle X_{t_i}, f\big\rangle - \big\langle X_{t_{i-1}}, f\circ\theta_{t_{i} - t_{i-1}}\big\rangle \right] \\ & \quad + \sum_{i=1}^n \left[\big\langle X_{t_{i-1}}, f\circ\theta_{t_{i} - t_{i-1}}\big\rangle - \big\langle X_{t_{i-1}},f\big\rangle \right] \\ & = \langle X_0, f\rangle + \sum_{i=1}^n \int_{t_{i-1}}^{t_i} \int_0^{\langle X_{s-},\alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y), z)} \big[z\, f\circ\theta_{t_i-s}(0) \\ & \quad - f\circ\theta_{t_i - s}(A_\alpha(X_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \\ & \quad + \sum_{i=1}^n \int_{t_{i-1}}^{t_i}\big\langle X_{t_{i-1}}, f^{\prime}\circ\theta_{s - t_{i-1}}\big\rangle \, \text{d} s. \end{align*}

By letting $|\Delta_n|\,:\!=\, \max_{1\le i\le n} (t_i-t_{i-1})\to 0$ and using the right continuity of $s\to X_s$ and the continuity of $s\to f\circ\theta_s$ , we obtain (3.9).

Proposition 3.5. For $f, G \in C^1\big(\mathbb{R}_+\big)$ let $G_f(\mu)= G(\langle \mu,f\rangle )$ , and let

\begin{align*} \mathscr{L}_0G_f(\mu) & = \big\langle \mu,f^{\prime}\big\rangle G^{\prime}(\langle \mu,f\rangle ) \\ & \quad - \sum_{z \in \mathbb{N}}\int_{\mathbb{R}_+} \alpha(y)p(y,z)\big[G(\langle \mu, f\rangle ) - G(\langle \mu,f\rangle +zf(0)-f(y))\big] \mu(\text{d} y). \end{align*}

Then we have

(3.10) \begin{eqnarray} G_f(X_t) = G_f(X_0) + \int_0^t { \mathscr{L}_0G_f(X_s)} \, \text{d} s + {mart.} \end{eqnarray}

Proof.Let $\tilde{M}$ denote the compensated measure of M. Since the process $s\mapsto X_s$ has at most countably many jumps, by Proposition 3.4 and Itô’s formula we have

\begin{align*} G(\langle X_t,f\rangle ) & = G(\langle X_0,f\rangle ) + \int_0^t G^{\prime}(\langle X_s,f\rangle ) \langle X_s, f^{\prime}\rangle \text{d} s \\ & \quad - \int_0^t \int_0^{\langle X_{s-},\alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y),z)} \big[G(\langle X_{s-},f\rangle ) \\ & \quad - G(\langle X_{s-},f\rangle + zf(0) - f(A_\alpha(X_{s-},y)))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \\ & = G(\langle X_0,f\rangle ) + \int_0^t G^{\prime}(\langle X_s, f\rangle ) \langle X_s, f^{\prime}\rangle \, \text{d} s - M_t^G(f) \\ & \quad - \int_0^t \text{d} s \int_{\mathbb{R}_+} \alpha(y)\sum_{z\in \mathbb{N}}p(y,z) \big[G(\langle X_{s-},f\rangle ) \\ & \quad - G(\langle X_{s-},f\rangle + zf(0) - f(y))\big] X_{s-}(\text{d} y) \\ & = G(\langle X_0,f\rangle ) + \int_0^t \mathscr{L}_0G_f(X_s) \, \text{d} s - M_t^G(f), \end{align*}

where

\begin{align*} M_t^G(f) = \int_0^t \int_0^{\langle X_{s-},\alpha\rangle }\int_0^1&\int_{\mathbb{N}}\int_0^{p(A_\alpha(X_{s-},y), z)}\big[G(\langle X_{s-}, f\rangle ) \\ &\ \ - G(\langle X_{s-}, f\rangle + z f(0) - f(A_\alpha(X_{s-},y)))\big] \tilde{M}(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v). \end{align*}

By Proposition 3.2 we can check that $\big\{M_t^G(f) \,:\, t\ge 0\big\}$ is a martingale.

Proof.Recall that $\{X_t \,:\, t \ge 0\}$ is a càdlàg process. Let $e_f(\mu)= \text{e}^{-\langle \mu,f\rangle }$ and let $\mathscr{L}_0$ be defined as in Proposition 3.5. It is elementary to see that

\begin{align*} \mathscr{L}_0e_f(\mu) & = -\big\langle \mu,f^{\prime}\big\rangle \text{e}^{-\langle \mu,f\rangle } - \text{e}^{-\langle \mu,f\rangle }\int_{\mathbb{R}_+} \alpha(y)\bigg[1 - \text{e}^{f(y)}\sum_{z\in \mathbb{N}} p(y,z)\text{e}^{-zf(0)}\bigg] \mu(\text{d} y) \\ & = - \text{e}^{-\langle \mu,f\rangle }\big\langle \mu,f^{\prime}\big\rangle - \text{e}^{-\langle \mu,f\rangle }\int_{\mathbb{R}_+} \alpha(y)\big(1 - \text{e}^{f(y)} g\big(y,\text{e}^{-f(0)}\big)\big) \mu(\text{d} y) . \end{align*}

Let $\mathscr{F}_t= \sigma\{X_s \,:\, 0\le s\le t\}$ . By (2.4), (3.10), and the mean-value theorem, we have

\begin{align*} \text{e}^{-\big\langle X_t,u_{T-t}f\big\rangle } & = \text{e}^{-\big\langle X_0,u_Tf\big\rangle } + \sum_{i=0}^\infty \bigg[\text{e}^{-\big\langle X_{t\wedge(i+1)/k},u_{T-t\wedge i/k}f\big\rangle } - \text{e}^{-\big\langle X_{t\wedge i/k},u_{T-t\wedge i/k}f\big\rangle }\bigg] \\ & \quad + \sum_{i=0}^\infty \bigg[\text{e}^{-\big\langle X_{t\wedge (i+1)/k},u_{T-t\wedge (i+1)/k}f\big\rangle } - \text{e}^{-\big\langle X_{t\wedge (i+1)/k},u_{T-t\wedge i/k}f\big\rangle }\bigg] \end{align*}

\begin{align*} & = \text{e}^{-\big\langle X_0,u_Tf\big\rangle } - \sum_{i=0}^\infty\int_{t\wedge i/k}^{t\wedge (i+1)/k} \text{e}^{-\big\langle X_s,u_{T-t\wedge i/k}f\big\rangle } \bigg[\big\langle X_s,\partial_xu_{T-t\wedge i/k}f\big\rangle \\ & \quad + \int_{\mathbb{R}_+} \alpha(y)\big[1 - \text{e}^{u_{T-t\wedge i/k}f(y)}g\big(y,\text{e}^{-u_{T-t\wedge i/k}f(0)}\big)\big] X_s(\text{d} y)\bigg] \, \text{d} s \\ & \quad + M_k(t) + \sum_{i=0}^\infty \text{e}^{-\xi_k(t)}\big\langle X_{t\wedge (i+1)/k}, u_{T-t\wedge i/k}f-u_{T-t\wedge (i+1)/k}f\big\rangle \\ & = \text{e}^{-\big\langle X_0,u_Tf\big\rangle } - \sum_{i=0}^\infty\int_{t\wedge i/k}^{t\wedge (i+1)/k} \text{e}^{-\big\langle X_s,u_{T-t\wedge i/k}f\big\rangle } \bigg[\big\langle X_s,\partial_xu_{T-t\wedge i/k}f\big\rangle \\ & \quad + \int_{\mathbb{R}_+} \alpha(y)\big[1 - \text{e}^{u_{T-t\wedge i/k}f(y)}g\big(y,\text{e}^{-u_{T-t\wedge i/k}f(0)}\big)\big] X_s(\text{d} y)\bigg] \, \text{d} s \\ & \quad + M_k(t) + \sum_{i=0}^\infty\int_{t\wedge i/k}^{t\wedge (i+1)/k} \text{e}^{-\xi_k(t)}\bigg[\big\langle X_{t\wedge (i+1)/k}, \partial_xu_{T-s}f\big\rangle \\ & \quad + \int_{\mathbb{R}_+} \alpha(y)\big[1 - \text{e}^{u_{T-s}f(y)}g\big(y,\text{e}^{-u_{T-s}f(0)}\big)\big] X_{t\wedge (i+1)/k}(d y)\bigg] \, \text{d} s, \end{align*}

where $t\mapsto M_k(t)$ is an $(\mathscr{F}_t)$ -martingale and

\begin{eqnarray*} \big\langle X_{t\wedge (i+1)/k}, u_{T-t\wedge i/k}f\big\rangle \le \xi_k(t)\le \big\langle X_{t\wedge (i+1)/k},u_{T-t\wedge (i+1)/k}f\big\rangle \end{eqnarray*}

or

\begin{eqnarray*} \big\langle X_{t\wedge (i+1)/k}, u_{T-t\wedge (i+1)/k}f\big\rangle \le \xi_k(t)\le \big\langle X_{t\wedge (i+1)/k},u_{T-t\wedge i/k}f\big\rangle . \end{eqnarray*}

By letting $k\to \infty$ we see that $t\mapsto \text{e}^{-\langle X_t,u_{T-t}f\rangle }$ is an $(\mathscr{F}_t)$ -martingale. In particular, we have $\mathbb{E}\big[\text{e}^{-\langle X_T,f\rangle } \mid \mathscr{F}_t\big]= \text{e}^{-\langle X_t,u_{T-t}f\rangle }$ , $T\ge t\ge 0$ . Then $\{X_t \,:\, t \ge 0\}$ is a Markov process with transition semigroup $(Q_t)_{t\ge 0}$ given by (2.3).

A calculation of the generator for an age-structured birth–death process was given in [Reference Bose4, (3.1)]; see also [Reference Bose and Kaj5]. A stochastic equation for a similar model was proposed in [Reference Tran28, (2.5)], which assumed that the death rate of an individual may depend on the whole population and also calculated the generator of the model.

4. The branching process with immigration

In this section we introduce an age-structured branching process with immigration and discuss its ergodicity. Let $\psi$ be a functional on $B\big(\mathbb{R}_+\big)^+$ given by $\psi(f) = \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} (1 - \text{e}^{-\langle \nu, f\rangle })L (\text{d}\nu)$ , $f\in B\big(\mathbb{R}_+\big)^+$ , where $L(\text{d}\nu)$ is a finite measure on $\mathfrak{N}\big(\mathbb{R}_+\big)^\circ \,:\!=\, \mathfrak{N}\big(\mathbb{R}_+\big)\setminus 0$ , and 0 denotes the null measure.

A Markov process $Y= (Y_t \,:\, t\ge 0)$ with state space $\mathfrak{N}\big(\mathbb{R}_+\big)$ is called an $(\alpha,g, \psi)$ -age-structured branching process with immigration if it has the transition semigroup $(P_t)_{t\ge 0}$ defined by

(4.1) \begin{eqnarray} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)}\text{e}^{-\langle \nu, f\rangle}P_t(\sigma, \text{d}\nu) = \exp\!\bigg\{{-}\langle \sigma, u_t\,f\rangle -\int_0^t\psi(u_sf) \, \text{d} s\bigg\},\end{eqnarray}

where $u_t\,f(x)$ is the unique solution to (2.2). Such a process is characterized by the properties (i)–(iii) given in Section 2, along with the following:

1. (iv) The immigrants come according to a Poisson random measure on $(0,\infty)\times \mathfrak{N}\big(\mathbb{R}_+\big)^\circ$ with intensity $d sL(d\nu)$ .

Suppose that $(\Omega, \mathscr{F}, \mathscr{F}_t, \mathbb{P})$ is a filtered probability space satisfying the usual hypotheses. Let $M(\text{d} t,\text{d} u,\text{d} y,\text{d} z,\text{d} v)$ be a Poisson random measure as in Section 3, and let $N(\text{d} t,\text{d} \nu)$ be an $(\mathscr{F}_t)$ -Poisson random measure on $(0, \infty)\times \mathfrak{N}\big(\mathbb{R}_+\big)^\circ$ with intensity $\text{d} t L(\text{d}\nu)$ . We assume that the two random measures are independent of each other. Consider the stochastic integral equation

(4.2) \begin{align} Y_t(x) & = Y_0\circ\theta_{-t}(x) + \int_0^t\int_0^{\langle Y_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}} \int_0^{p(A_\alpha(Y_{s-},y), z)} \big[z\zeta_0 \nonumber \\ & \quad - \zeta_{A_\alpha(Y_{s-},y)}\big]\circ\theta_{s-t}(x) M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \nonumber \\ & \quad + \int_0^t\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \nu\circ\theta_{s-t}(x)N(\text{d} s,\text{d} \nu).\end{align}

Here, the first two terms on the right-hand side are as explained for (3.1) and the last term represents the immigration. A pathwise-unique solution to (4.2) is constructed as follows. Let $\sigma_0 = 0$ . Given $\sigma_{k-1}$ , we define $\sigma_k = \sigma_{k-1} + \inf\big\{t>0 \,:\, N\big(\big(\sigma_{k-1},\sigma_{k-1}+t\big]\times N\big(\mathbb{R}_+^\circ\big)\big)> 0\big\}$ and $Y_t(x) = X_{k,t-\sigma_{k-1}}(x)$ , $\sigma_{k-1} \le t < \sigma_k$ , where $\{X_{k,t}(x) \,:\, t\ge 0\}$ is the pathwise-unique solution to the following equation:

\begin{align*} X_t(x) =& Y_{\sigma_{k-1}}\circ\theta_{-t}(x) + \int_0^t\int_0^{\langle X_{s-},\alpha\rangle }\int_0^1\int_{\mathbb{N}} \int_0^{p(A_\alpha(X_{s-},y), z)} \big[z \zeta_0\circ\theta_{s-t}(x) \\& - \zeta_{A_\alpha(X_{s-},y)}\circ\theta_{s-t}(x)\big] M(\sigma_{k-1}+\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v).\end{align*}

Then we define $Y_{\sigma_k}(x) = Y_{\sigma_k-}(x) + \int_{\{\sigma_k\}} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \nu(x) N(\text{d} s, \text{d}\nu)$ . It is easy to see that $\lim_{k\to \infty}\sigma_k = \infty$ and $\{Y_t \,:\, t \ge 0\}$ is the pathwise-unique solution to (4.2).

The solution to (4.2) determines a measure-valued strong Markov process $\{Y_t \,:\, t \ge 0\}$ with state space $\mathfrak{N}\big(\mathbb{R}_+\big)$ . We omit the proofs of some of the following results since the arguments are similar to those for the corresponding results in Section 3.

Proposition 4.1. For any $f \in B\big(\mathbb{R}_+\big)^+$ ,

\begin{align*} \langle Y_t, f\rangle & = \langle Y_0, f\circ \theta_t\rangle + \int_0^t\int_0^{\langle Y_{s-}, \alpha\rangle }\int_{[0, 1]}\int_{\mathbb{N}}\int_0^{p(A_\alpha(Y_{s-},y), z)}\big[z f\circ\theta_{t - s}(0) \\ & \quad - f\circ\theta_{t - s}(A_\alpha(Y_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \\ & \quad + \int_0^t\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \langle \nu, f\circ\theta_{t - s}\rangle N(\text{d} s, \text{d} \nu). \end{align*}

Proposition 4.2. For any $t \ge 0$ and $f \in C^1\big(\mathbb{R}_+\big)^+$ ,

(4.3) \begin{align} \langle Y_t, f\rangle & = \langle Y_0, f\rangle + \int_0^t \langle Y_{s-}, f^{\prime}\rangle \, \text{d} s + \int_0^t\int_0^{\langle Y_{s-}, \alpha\rangle }\int_{[0, 1]}\int_{\mathbb{N}}\int_0^{p(A_\alpha(Y_{s-},y),z)}\big[z f(0) \nonumber \\ & \quad - f(A_\alpha(Y_{s-},y))\big] M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \nonumber \\ & \quad + \int_0^t\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \langle \nu, f\rangle N(\text{d} s, \text{d} \nu). \end{align}

Proposition 4.3. For any $f, G\in C^1\big(\mathbb{R}_+\big)$ , let $G_f(\mu)= G(\langle \mu,f\rangle )$ and

\begin{align*} \mathscr{L}\,G_f(\mu) & = \big\langle \mu,f^{\prime}\big\rangle G^{\prime}(\langle \mu,f\rangle ) - \int_{\mathbb{R}_+} \alpha(y) \sum_{z \in \mathbb{N}}p(y,z)\big[G(\langle \mu, f\rangle ) \\ & \quad - G\big(\langle \mu, f\rangle + zf(0) - f(y)\big)\big] \mu(\text{d} y) \\ & \quad + \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}\big[G(\langle \mu, f\rangle + \langle \nu, f\rangle ) - G(\langle \mu, f\rangle )\big]L(\text{d}\nu). \end{align*}

Then $G(\langle Y_t, f\rangle ) = G(\langle Y_0, f\rangle ) + \int_0^t \mathscr{L}\,G_f(Y_s) \, \text{d} s + {mart.}$

Proof.Let $\tilde{M}$ and $\tilde{N}$ denote the compensated measure of M and N, respectively. By (4.3) and Itô’s formula we have

\begin{align*} G(\langle Y_t,f\rangle ) & = G(\langle Y_0,f\rangle ) + \int_0^t G^{\prime}(\langle Y_s, f\rangle )\langle Y_s, f^{\prime}\rangle \, \text{d} s \\ & \quad + \int_0^t\int_0^{\langle Y_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(Y_{s-},y), z)}\big[G\big(\langle Y_{s-},f\rangle + zf(0) \\ & \quad - f(A_\alpha(Y_{s-},y))\big) - G(\langle Y_{s-},f\rangle ) \big]M(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \\ & \quad + \int_0^t\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}\big[G\big(\langle Y_{s-},f\rangle + \langle \nu, f\rangle \big) - G\big(\langle Y_{s-},f\rangle \big)\big]N(\text{d} s, \text{d} \nu) \\ & = G(\langle Y_0,f\rangle ) + \int_0^t G^{\prime}(\langle Y_s, f\rangle )\langle Y_s, f^{\prime}\rangle \, \text{d} s + N_t^G(f) \\ & \quad + \int_0^t \text{d} s\int_{\mathbb{R}_+} \alpha(y)\sum_{z\in \mathbb{N}}p(y,z) \big[G\big(\langle Y_{s-},f\rangle + zf(0) - f(y)\big) \\ & \quad - G(\langle Y_{s-},f\rangle ) \big]Y_{s-}(\text{d} y) \\ & \quad + \int_0^t \text{d} s\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}\big[G\big(\langle Y_{s-},f\rangle + \langle \nu, f\rangle \big) - G(\langle Y_{s-},f\rangle )\big]L(\text{d}\nu), \end{align*}

where

\begin{align*} N_t^G(f) & = \int_0^t\int_0^{\langle Y_{s-}, \alpha\rangle }\int_0^1\int_{\mathbb{N}}\int_0^{p(A_\alpha(Y_{s-},y),z)} \big[G(\langle Y_{s-},f\rangle + zf(0) - f(A_\alpha(Y_{s-},y))) \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \ - G(\langle Y_{s-},f\rangle ) \big]\tilde{M}(\text{d} s,\text{d} u,\text{d} y,\text{d} z,\text{d} v) \\ & \quad + \int_0^t\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}\big[G\big(\langle Y_{s-},f\rangle + \langle \nu, f\rangle \big) - G\big(\langle Y_{s-},f\rangle \big)\big]\tilde{N}(\text{d} s, \text{d} \nu). \end{align*}

A first moment estimate can check that $\big\{N_t^G(f) \,:\, t\ge 0\big\}$ is a martingale.

A necessary and sufficient condition for the ergodicity of the $(\alpha, g, \psi)$ -age-structured branching process with immigration is given in the next theorem.

Theorem 4.2. Suppose that $c_*= \inf_{y\ge 0}\alpha(y)[1-\partial_zg(y,1{-})]> 0$ . Then $P_t(\sigma, \cdot)$ converges as $t\to \infty$ to a probability measure $\eta$ on $\mathfrak{N}\big(\mathbb{R}_+\big)$ for every $\sigma \in \mathfrak{N}\big(\mathbb{R}_+\big)$ if and only if

(4.4) \begin{eqnarray} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)} \textbf{1}_{\{\langle \nu,1\rangle \ge 1\}}\log\langle \nu,1\rangle L(\text{d}\nu) < \infty. \end{eqnarray}

In this case, the Laplace transform of $\eta$ is given by

(4.5) \begin{eqnarray} \int_{\mathfrak{N}\big(\mathbb{R}_+\big)} \text{e}^{-\langle \nu, f\rangle } \eta(\text{d}\nu) = \exp\!\bigg\{{-} \int_0^\infty \psi(u_sf)\, \text{d} s\bigg\}. \end{eqnarray}

Proof.First, for $f\in B\big(\mathbb{R}_+\big)^+$ let $f_*= \inf_{x\ge 0} f(x)$ . From Propositions 2.3 and 2.4 we have

(4.6) \begin{eqnarray} \big(1-\text{e}^{-f_*}\big)\text{e}^{-c_1t}\le u_t\,f(x) \le \|u_t\,f\| \le \|\pi_tf\| \le \text{e}^{-c_*t}\|f\|, \qquad x\ge 0. \end{eqnarray}

Moreover, for $a,c>0$ it is elementary to see that

(4.7) \begin{align} \int_0^\infty \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-ae^{-cs}\langle \nu,1\rangle }\big)L(\text{d}\nu) & = \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}L(\text{d}\nu) \int_0^\infty \big(1-\text{e}^{-a\text{e}^{-cs}\langle \nu,1\rangle }\big) \, \text{d} s \nonumber \\[3pt] & = c^{-1}\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}L(\text{d}\nu) \int_0^{a\langle \nu,1\rangle }\big(1-\text{e}^{-z}\big)z^{-1} \, \text{d} z. \end{align}

Second, suppose that (4.4) holds. For any $f\in B\big(\mathbb{R}_+\big)^+$ we see from (4.6) that $\sigma(u_t\,f)\to 0$ as $t\to \infty$ . Take any $a>\|f\|$ . By (4.6) and (4.7) we have

\begin{align*} \int_0^\infty \psi(u_sf) \, \text{d} s & = \int_0^\infty \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-\langle \nu,u_sf\rangle }\big)L(\text{d}\nu) \\[3pt] & \le \int_0^\infty \text{d} s\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-a\text{e}^{-c_*s}\langle \nu,1\rangle }\big)L(\text{d}\nu) \\[3pt] & \le c_*^{-1}\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \bigg[1 + \int_1^{1\vee (a\langle \nu,1\rangle )} z^{-1} \, \text{d} z\bigg]L(\text{d}\nu) \\[3pt] & \le c_*^{-1}\int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ}\big[1 + 0\vee\log(a\langle \nu,1\rangle )\big]L(\text{d}\nu)< \infty. \end{align*}

Then, as $t\to \infty$ ,

\begin{align*} \int_0^t \psi(u_sf) \, \text{d} s & = \int_0^t \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-\langle \nu,u_sf\rangle }\big)L(\text{d}\nu) \\[3pt] &\qquad\qquad\quad\!\!\! \to \int_0^\infty \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-\langle \nu,u_sf\rangle }\big)L(\text{d}\nu) = \int_0^\infty \psi(u_sf) \, \text{d} s< \infty. \end{align*}

From (4.1) we infer that $P_t(\sigma,\cdot)$ converges as $t \rightarrow \infty$ to a probability measure $\eta$ defined by (4.5); see, e.g., [Reference Li21, Theorem 1.20].

Third, suppose that $P_t(\sigma, \cdot)$ converges as $t\to \infty$ to a probability measure $\eta$ on $\mathfrak{N}\big(\mathbb{R}_+\big)$ for every $\sigma \in \mathfrak{N}\big(\mathbb{R}_+\big)$ . By (4.1) we see that $\eta$ has a Laplace functional given by (4.5); see, e.g., [Reference Li21, Theorems 1.17 and 1.18]. In particular, we have $\int_0^\infty \psi(u_sf) \, \text{d} s< \infty$ for every $f\in B\big(\mathbb{R}_+\big)^+$ . Let $a=(1-\text{e}^{-1})$ . By (4.6) and (4.7) we have

\begin{align*} \int_0^\infty \psi(u_s1) \, \text{d} s & = \int_0^\infty \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-\langle \nu,u_s1\rangle }\big)L(\text{d}\nu) \\ & \ge \int_0^\infty \text{d} s \int_{\mathfrak{N}\big(\mathbb{R}_+\big)^\circ} \big(1-\text{e}^{-a\text{e}^{-c_1s}\langle \nu,1\rangle }\big)L(\text{d}\nu) \\ & \ge ac_1^{-1}\int_{\{a\langle \nu,1\rangle \ge 1\}}L(\text{d}\nu) \int_1^{a\langle \nu,1\rangle } z^{-1} \, \text{d} z \\ & \ge ac_1^{-1}\int_{\{a\langle \nu,1\rangle \ge 1\}}\big[\log a + \log\langle \nu,1\rangle \big]L(\text{d}\nu). \end{align*}

This gives us (4.4).