2.1. Proof of Theorem 1

Proof. For the first part of the statement, we apply Theorem 1.11 in Chen [Reference Chen8] (see also Theorem 2.2 in Tweedie [Reference Tweedie37] for the case $\textbf{m}\le 0$ and Theorem 2.1 in Meyn and Tweedie [Reference Meyn and Tweedie28] for the case $\textbf{m}>0$). The proof of the aforementioned result follows from a localization argument, and its conditions can be written in terms of the localized transition rates without changing the conclusion. More precisely, we consider the particular case where the state space is $\mathbb{N}_0$ and introduce the sequence of sets $E_n=\{0, 1,2, \ldots, n\} $ for $n\in \mathbb{N}$, which increases towards $\mathbb{N}_0$. We also introduce the localized transition rates $(q^{(n)}_{i,j})_{i,j\in \mathbb{N}_0}$ as follows:

\begin{equation*}q^{(n)}_{i,j}=\textbf{1}_{E_n}(i)q_{i,j} \qquad\textrm{with}\qquad-q^{(n)}_{i,i}=\sum_{j\neq i}q^{(n)}_{i,j}\quad \textrm{for } i,j\in \mathbb{N}_0.\end{equation*}

If a nonnegative function $\varphi=(\varphi(i), i\in \mathbb{N}_0)$ satisfies

\begin{equation*}\lim_{n\to\infty}\inf_{i\notin E_n}\varphi(i)=\infty,\qquad\textrm{and}\qquad\sum_{j\in \mathbb{N}_0}q^{(n)}_{i,j}(\varphi(j)-\varphi(i))\le c \varphi(i) \qquad \textrm{for}\,\, i\in E_n,\end{equation*}

where $c\in \mathbb{R}$, then the associated process to $(q_{i,j})_{i,j\in \mathbb{N}_0}$ does not explode, since the arguments from Parts (ii), (iii) and (iv) of the proof of Theorem 1.11 in Chen [Reference Chen8] still hold with the above assumptions. In our particular case, we take

\begin{equation*}\varphi(i)=\left\{\begin{array}{l@{\quad}l}i & \textrm{if } i\in E_n,\\[5pt] n & \textrm{otherwise.}\end{array} \right .\end{equation*}

Figure 1:

This figure exemplifies the long-term behaviour of branching processes with interactions in the subcritical cooperative regime, depending on its parameters. Dotted lines mean that the parameter is different from zero, while continuous lines mean that the parameter is zero. For example, if $a=0$, $d=0$, and $\pi_i=0$ for all $i\in \mathbb{N}$, we take the right edge three times. In this case the process gets absorbed at $\{1\}$. The symbol $\mu$ means that the process has a nondegenerate stationary distribution.

Since $\inf_{i\notin E_n}\varphi(i)=n$ and $\varsigma \le 0$, we deduce

\begin{equation*}\sum_{j\in \mathbb{N}_0}q^{(n)}_{i,j}\Big(\varphi(j)-\varphi(i)\Big)\le i\textbf{m} + i(i-1) \varsigma -\sum_{k=2}^ i (k-1)\binom{i}{k} \lambda_{i,k} \le \varphi(i) \textbf{m} \qquad \textrm{for}\,\, i\in E_n,\end{equation*}

which implies from above that the process Z does not explode (or it is unique or regular, in the terminology of [Reference Chen8]).

For the second statement, we assume that the measure $\Lambda$ is identically zero and that $\varsigma>0$. We first treat the case when $d=\rho=0$. In order to do so, we introduce $A=(A_t, t\in \mathbb{R}_{+,0})$, a compound Poisson process with parameter $a+b+c$ and jump distribution $\eta$ satisfying

\begin{equation*}\eta(i)=\left\{ \begin{array}{ll}\displaystyle\frac{a}{a+b+c} & \textrm{ if } i=-2,\\[12pt] \displaystyle\frac{c}{a+b+c} & \textrm{ if } i=-1,\\[12pt] \displaystyle\frac{b_i}{a+b+c} & \textrm{ if } i\in \mathbb{N}. \end{array}\right .\end{equation*}

We also consider the function $g(i)=i(i-1)$, which is nonnegative for $i\in \mathbb{N}_{0}$, with $g(i)=0$ for $i=0,1$. Hence, from Theorems 6.1.1 and 6.1.3 in Ethier and Kurtz [Reference Ethier and Kurtz11], the random time-changed process

\begin{equation*}Z_{t}=A_{\int_0^t g(Z_s) \textrm{d} s},\qquad t\le \theta_{0,1} \,:\!=\,\inf\{s\,:\, A_s=0 \textrm{ or } 1\},\end{equation*}

defines a Markov process with absorbing states $\{0,1\}$ whose infinitesimal generator is given by

\begin{equation*}Qf(n)=n(n-1)\mathcal{B}f(n),\end{equation*}

where, for $f\in C_0(\mathbb{Z}\cup\{\infty\}, \mathbb{R})$ with $f(i)=0$ for $i\in \{\ldots, -2,-1, 0,1\}$, the infinitesimal generator $\mathcal{B}$ acts as follows:

\begin{equation*}\mathcal{B}f(n)=a[f(n-2)-f(n)] + c[f(n-1)-f(n)]+\sum_{i\in \mathbb{N}} b_i[f(n+i)-f(n)], \quad n\in \mathbb{N}\setminus\{1\}.\end{equation*}

In other words, the process Z has the same law as a branching process with interactions with parameters a, c, and $b_i$ for $i\in \mathbb{N}$.

In the same probability space, we also consider a Poisson process $\widetilde{A}$ with parameter $\delta\in(0, \varsigma)$; then, by using a similar random time change as above, we introduce a branching process with interactions with parameters $c=a=0$, $b_1=\delta>0$, and $b_i=0$ for $i\in\mathbb{N}\setminus\{1\}$. We observe that the process $\widetilde{Z}$, which is the process with these parameters in this probability space, explodes a.s. Indeed, let us define $\widetilde{\tau}^+_i=\inf\{t>0\,:\,\widetilde{Z}_t\geq i\}$ and observe that its lifetime, here denoted by $\widetilde{\tau}_\infty$, satisfies $\widetilde{\tau}_\infty=\sup_{i\in \mathbb{N}}\widetilde{\tau}_i^+ $. Thus, we see that

(2.5) \begin{equation}\widetilde{\textbf{E}}_n[\widetilde{\tau}_\infty]\leq\widetilde{\textbf{E}}_2[\widetilde{\tau}_\infty]=\sum_{i\in\mathbb{N}} \widetilde{\textbf{E}}_i[\widetilde{\tau}^+_{i+1}-\widetilde{\tau}^+_{i}]=\frac{1}{\delta}\sum_{i\in\mathbb{N}\setminus\{1\}} \frac{1}{ i(i-1)}<\infty,\end{equation}

where $\widetilde{\textbf{E}}_n$ denotes the expectation of $\widetilde{Z}$ starting from $n\in\mathbb{N}\setminus\{1\}$. The previous identity implies that $\widetilde{\tau}_{\infty}$ is finite a.s., or equivalently that $\widetilde{Z}$ explodes a.s.

Next, we assume that $A_0=n>m=\widetilde{A}_0$ and define the event $E=\{A_t>\widetilde{A}_t, \text{ for all }t>0\}$. It is straightforward to deduce that the process $S_t=A_t-\widetilde{A}_t$, for $t\ge 0$, is a compound Poisson process starting from $n-m>0$ and drifting to $\infty$, since $\varsigma>\delta$. Hence with positive probability the process S stays strictly positive, implying that $\mathbb{P}(E|S_0=n-m)>0$. In other words, conditionally on E and given that $S_0=n-m>0$, we have that the process $\widetilde{Z}$ is stochastically dominated by Z, and since the former explodes, Z also explodes. This proves our statement for the case $d=\rho=0$.

For the case $d+\rho>0$, our analysis will be based on the following coupling argument. Let us consider two branching processes with interactions, $Z=(Z_t, t\in \mathbb{R}_{+,0})$ and $\overline{Z}=(\overline{Z}_t, t\in \mathbb{R}_{+, 0})$, with respective parameters d, a, c, $b_i$, and $\pi_i$, for $i\in \mathbb{N}$, such that $d+\rho>0$; and $\overline{d}=\sum_{i\in \mathbb{N}}\overline{\pi_i}=0$, $\overline{c}=c+\epsilon d,$ $\overline{a}=a$, and $\overline{b_i}=b_i$ for all $i\in \mathbb{N}$, where $\epsilon$ is chosen positive and such that

(2.6) \begin{equation}-2\overline{a}-\overline{c}+\sum_{i\in \mathbb{N}} i \overline{b_i} =-2a-c-\epsilon d+\sum_{i\in \mathbb{N}} i b_i> 0.\end{equation}

We denote by $\overline{\textbf{P}}_n$ the law of $\overline{Z}$ starting from $\overline{Z}_0=n$. We also observe, from above, that $\overline{Z}$ explodes with positive probability. Moreover, let $n_0=\inf\{n\in \mathbb{N}\,:\, 1< (n-1)\epsilon\}$ and note that for any starting condition $Z_0=\overline{Z}_0=n> n_0$, we necessarily have

\begin{equation*} nd+n(n-1)c<n(n-1)\overline{c}=\epsilon n(n-1)d+n(n-1)c. \end{equation*}

Next, we consider the following coupling between the Markov chains Z and $\overline{Z}$. Let us introduce the process $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ with the following dynamics. Assume that $U_0=\overline{U}_0>n_0$. For all $t<\gamma_{n_0}\,:\!=\,\inf\{t>0\,:\, \overline{U}_t\le n_0\}$, the couple $(U_t,\overline{U}_t)$ has the following transitions rates, for $m\ge n$:

(2.7) \begin{equation}(U_t,\overline{U}_t)\text{ goes from }(n,m)\text{ to}\left\{ \begin{array}{l@{\quad}l}(n+i,m+i), & \textrm{ with rate } m(m-1)b_i,\\[4pt] (n+i,m), & \textrm{ with rate } n\pi_i+[n(n-1)-m(m-1)]b_i,\\[4pt] (n-1,m-1), & \textrm{ with rate } \underline{\Theta}(n,m),\\[4pt] (n,m-1), & \textrm{ with rate } m(m-1)(\epsilon d+c)-\underline{\Theta}(n,m),\\[4pt] (n-1,m), & \textrm{ with rate } n(n-1)c+nd-\underline{\Theta}(n,m),\\[4pt] (n-2,m-2), &\textrm{ with rate } n(n-1)a,\end{array}\right .\end{equation}

where $\underline{\Theta}(n,m)=\min( n(n-1)c+nd,m(m-1)(\epsilon d+c))$. For $t\geq \gamma_{n_0}$, the Markov chains $U_t$ and $\overline{U}_t$ evolve independently with the same transition rates as Z and $\overline{Z}$, respectively. We denote by $\textbf{P}_{(n,m)}$ the law of $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ starting from (n,m). Note from our construction that the Markov chains U and $\overline{U}$ are equal in distribution to Z and $\overline{Z}$, respectively. Furthermore, we necessarily have

\begin{equation*}\textbf{P}_{(n,n)}(U_t\geq \overline{U_t}, \textrm{ for } t<\gamma_{n_0})=1.\end{equation*}

We deduce that for $n>n_0$, $\textbf{P}_{(n, n)}(U_t=\infty)>\textbf{P}_{(n, n)}(\overline{U}_t=\infty,\gamma_{n_0}>t)$, where the right-hand side of the inequality is positive for t sufficiently large. The latter implies that Z is not conservative.

For the proof of Proposition 1, we verify the Foster–Lyapunov conditions for positive recurrence. In order to do so, we consider the jump Markov chain associated to Z, which we denote by $Y=(Y_n, n\in \mathbb{N}_0)$, and whose jump matrix is given by $P=(p_{i,j})_{i,j\in \mathbb{N}_0}$, where

\begin{equation*}p_{i,j}=\left\{ \begin{array}{l@{\quad}l} -q_{i,j}/q_{i,i} & \textrm{if $j\ne i$}\\[7pt] 0 & \textrm{if $j=i$,}\end{array} \right.\end{equation*}

for $i\in \mathbb{N}$ and $p_{0,0}=1$. On the other hand, we recall that if a state is recurrent or transient for the jump chain Y, then the state is also recurrent or transient for the chain Z. (See for instance the monograph of Norris [Reference Norris32] for further details on Markov chains and their associated jump chains.)

Proof of Proposition 1. We first prove Part (ii). Recall that $d=0$ and $\rho>0$ and that under our assumption $c>0$. From the form of the jump matrix P defined above, we observe that the jump chain Y is irreducible in $\mathbb{N}$ and that the state $\{0\}$ is not accessible from any state $i\in\mathbb{N}$.

In order to analyse the class properties of the state space of the jump chain Y, and implicitly for Z, we apply the Foster–Lyapunov criteria (see for instance Proposition 1.3 in Hairer [Reference Hairer17] or Theorem 4.2 in Meyn and Tweedie [Reference Meyn and Tweedie28]). Let us consider the discrete generator associated to Y, which is defined as follows:

\begin{equation*}\mathcal{L}f(i)= Pf(i)-f(i)=-\frac{1}{q_{i,i}}\sum_{j\in \mathbb{N}_0}q_{i,j}f(j),\end{equation*}

for $f\,:\,\mathbb{N}_0\to \mathbb{R}$. We choose the Lyapunov function to be $f(i)=C i$ for $i\in \mathbb{N}_0$ and C a positive constant that we will specify later. According to the Foster–Lyapunov criteria we know that if $\mathcal{L}f(i)\leq -1$ for all but finitely many values of i, then Y is positive recurrent. Hence straightforward computations lead to

(2.8) \begin{equation}\begin{split}\mathcal{L}f(i)&=-\frac{C}{q_{i,i}}\left(i\textbf{m} + i(i-1) \varsigma-\sum_{k=2}^ i (k-1)\binom{i}{k} \lambda_{i,k} \right)\\&\le C\left( \frac{m}{(i-1)(b+c+a)}+\frac{\varsigma}{(b+c+a)}-\frac{\sum_{k=2}^ i (k-1)\binom{i}{k} \lambda_{i,k}}{i(i-1)(b+c+a)}\right)\!.\end{split}\end{equation}

For i sufficiently large, we can clearly choose C in such a way that $\mathcal{L}f(i)\le -1$, implying that the Foster–Lyapunov criteria hold. In other words, Z is positive recurrent in $\mathbb{N}$.

Now we prove Part (i). Recall that $\rho=0$ and $d=0$ and that under our assumption $c>0$. Again from the form of the jump matrix P, we observe that the state $\{1\}$ is absorbing and that $\{0\}$ is not accessible. Moreover, if $b>0$ the class $\{2, 3, \ldots\}$ is irreducible for Z, and if $b=0$, the process Z is decreasing. In any case, we can always stochastically dominate Z with another branching process with interactions, here denoted by $\widetilde{Z}$, with the same parameters except $\pi_1>0$. We refer the reader to Section 2 of López et al. [Reference López, Martínez and Sanz26] for the conditions on the intensities $(q_{i,j})_{i,j\ge 0}$ that guarantee the existence of an order-preserving coupling. Since $\widetilde{Z}$ is recurrent and reaches the state $\{1\}$ in finite time a.s., it follows that Z is absorbed at 1 in finite time a.s.

Finally, we deduce Part (iii). Since in this case $d>0$, from the form of the jump matrix P we observe that the state $\{0\}$ is accessible. If $\rho=0$ and $b=0$, the process Z is decreasing, and if $\rho>0$ or $b>0$, then $\mathbb{N}$ is irreducible for the $\bar Z$ constucted in Part (ii), and again an order-preserving coupling argument guarantees that Z is absorbed at $\{0\}$ in finite time a.s.

2.2. Moment duality property and proof of Theorem 2

The aim of this section is to prove Theorem 2, which characterizes a class of jump-diffusions that fulfils the moment duality property with respect to (sub)critical branching processes with interactions and without annihilation, i.e. $a=0$. This moment duality will be very helpful for our purposes. The proof of Theorem 2 follows Lemmas 1 and 2 below.

In order to prove Theorem 2 we first introduce the class of jump-diffusions via the following SDE:

(2.9) \begin{equation}\textrm{d} X_t=\mu(X_{t})\textrm{d} t+\sigma(X_t)\textrm{d} B_t +\int_{(0,1]}\int_{(0,1]}zg(X_{t-}, u) \widetilde{N}(\textrm{d} t,\textrm{d} z,\textrm{d} u),\end{equation}

where $\widetilde{N}$ is a compensated Poisson random measure on $(0,\infty)\times (0,1]^2$ with intensity $\textrm{d} t z^{-2}\Lambda(\textrm{d} z) \textrm{d} u$, the function g(x,u) is defined by

\begin{equation*}g(x,u)=\textbf{1}_{\{u\le 1\land x \}}-(1\land x)\textbf{1}_{\{x\ge 0\}},\end{equation*}

and the functions $\mu\,:\,\mathbb{R}\rightarrow \mathbb{R}$ and $\sigma\,:\, \mathbb{R}\rightarrow \mathbb{R}$ are continuous and satisfy

\begin{equation*}\mu(x)=d\textbf{1}_{\{x < 0\}}+\left(d(1-x)+\sum_{i\in \mathbb{N}}\pi_{i}(x^{i+1}-x)\right)\textbf{1}_{\{x\in[0,1]\}}\end{equation*}

and

\begin{equation*}\sigma^2(x)=\left(c(x-x^2)+\sum_{i\in \mathbb{N}} b_i(x^{i+2}-x^2)\right)\textbf{1}_{\{x\in[0,1]\}},\end{equation*}

for some positive constants d, c, $b_i$, and $\pi_i$ with $i\in \mathbb{N}$.

Proof. We first observe that if X satisfies (2.9) with $X_0\in [0,1]$, a.s., then $X_t\in[0,1]$ for all $t\in \mathbb{R}_+$, a.s. Suppose that there exists $\epsilon>0$ such that $\tau\,:\!=\, \inf\{t\ge 0\,:\, X_t\le -\epsilon\}$ is finite with positive probability. Then on the event $\{\tau<\infty\}$, we have $X_{\tau}=-\epsilon$ and

\begin{equation*}\tau>\tau_0=\inf\{s<\tau\,:\, X_t\le 0, \forall s\le t\le \tau \}.\end{equation*}

So we take $r\ge 0$ such that $\{\tau_0\le r<\tau\}$ occurs with strictly positive probability, and on that event, we observe

\begin{equation*}X_{t\land \tau}=X_{r\land \tau}+ d(t\land \tau -r\land \tau), \qquad t\ge r,\end{equation*}

implying that $t\mapsto X_{t\land \tau}$ is nondecreasing on $[r,\infty)$. Since $X_r>-\epsilon$ on $\{\tau_0\le r<\tau\}$, we get a contradiction. The previous argument implies that $X_t\ge 0$ a.s. A similar argument proves that $X_t\le 1$ a.s. We leave the details to the reader.

Now, we show that X has a unique strong solution on [0, 1]. In order to do so, we first observe that for $x, y \in [0,1]$,

\begin{equation*} |\mu(x)-\mu(y)|\le \left(d + \sum_{i\in \mathbb{N}}(i+1)\pi_i\right)|x-y|. \end{equation*}

The latter inequality implies that $\mu$ is Lipschitz continuous on [0, 1].

On the other hand, we also observe that $\sigma$ satisfies, for all $x,y\in [0,1]$,

\begin{align*}|\sigma^2(x)-\sigma^2(y)|&\leq \left|c(x-y)+c(y^2-x^2)+\sum_{i\in \mathbb{N}} b_i\Big(x^{2+i}-y^{i+2} -(x^{2}-y^{2})\Big)\right|\\&\leq \left(3c+\sum_{i\in \mathbb{N}} b_i(i+2)+2\sum_{i\in \mathbb{N}} b_i\right)|x-y|,\end{align*}

implying from our assumptions that $\sigma$ is Hölder continuous on [0, 1] with exponent 1/2.

Finally, we observe from Corollary 6.2 in Li and Pu [Reference Li and Pu24] that $x\mapsto x+g(x, u)$ is nondecreasing, and for any $0\le x, y\le 1$,

\begin{equation*} \int_{(0,1]}\frac{\Lambda(\textrm{d} z)}{z^2}\int_0^1 \textrm{d} r\, z^2|g(x,r)-g(y,r)|\le |x-y|\int_{(0,1]}\Lambda(\textrm{d} z). \end{equation*}

Hence the conditions (3.a), (3.b), and (5.a) in Li and Pu [Reference Li and Pu24] are satisfied, implying that there is a unique strong solution to the SDE (2.9).

Let $X=(X_t, t\in \mathbb{R}_{+,0})$ be the unique strong solution of the SDE (2.9). We denote by $\mathcal{A}$ the infinitesimal generator of the process X, which satisfies

\begin{equation*}\mathcal{A}f(x)\,:\!=\,\lim_{t\downarrow 0}\frac{\mathbb{E}_{x}[f(X_t)]-f(x)}{t}\end{equation*}

for any $f\,:\,[0,1]\mapsto\mathbb{R}$ continuous such that the limit exists uniformly on $x\in[0,1]$. If this limit exists, then we say that $f\in\mathcal{D}_{\mathcal{A}}$, the domain of the infinitesimal generator $\mathcal{A}$. Observe from Itô’s formula that for $f\in C^2_b([0,1], \mathbb{R})$ (the set of twice continuously differentiable bounded functions) and $x\in [0,1]$, the infinitesimal generator $\mathcal{A}$ satisfies

\begin{equation*}\begin{split}\mathcal{A}f(x)= &\mu(x)f^\prime(x)+\frac{\sigma^2(x)}{2}f^{\prime\prime}(x)\\ &+\int_{[0,1]}(xf(x(1-z)+z)+(1-x)f(x(1-z))-f(x))\frac{\Lambda(\textrm{d} z)}{z^2},\end{split}\end{equation*}

where $\mu, \sigma\,:\,[0,1]\to\mathbb{R}$ and $\Lambda$ are defined as above. In other words, for $f\in C^2_b([0,1], \mathbb{R})$, the process

\begin{equation*}f(X_t)-\int_0^t\mathcal{A}f(X_s)\textrm{d} s\end{equation*}

is a martingale. On the other hand, we also observe that $C^2_b([0,1], \mathbb{R})$ is dense in $C([0,1], \mathbb{R})$, the set of continuous functions from [0, 1] to $\mathbb{R}$. It is not so difficult to see that the generator $\mathcal{A}$ satisfies the positive maximum principle. Since [0, 1] is a compact space and the space of all polynomials on [0, 1] is dense on $C([0,1], \mathbb{R})$, the Hille–Yosida theorem implies that the closure of $\mathcal{A}$ restricted to $C^2_b([0,1], \mathbb{R})$ generates a strongly continuous, time-homogeneous, positive contraction Feller semigroup on $C([0,1], \mathbb{R})$; see for instance Theorem 4.2.2 and Lemma 4.2.3 in [Reference Ethier and Kurtz11]. Theorem 4.2.7 and Corollary 4.2.8 in [Reference Ethier and Kurtz11] imply that the process X is Feller.

Recall that $\mathbb{P}_x$ and $\textbf{P}_n$ denote the laws of X starting from $x\in[0,1]$ and Z starting from $n\ge 0$, respectively. Moreover, from our assumptions we have that $\textbf{P}_n(Z_t<\infty)=1$ for any $n\in\mathbb{N}_0$ and $t\in\mathbb{R}_{+,0}$.

In practice, an effective methodology to show duality between two Markov processes is via their infinitesimal generators. Indeed, the following result is a special case of Theorem 4.11 and Corollary 4.15 in Ethier and Kurtz [Reference Ethier and Kurtz11] and can also be seen as a small modification of Proposition 1.2 of Jansen and Kurt [Reference Jansen and Kurt19].

Proposition 3. Let $(Y^{(1)}_t, t\in\mathbb{R}_{+,0})$ and $(Y^{(2)}_t, t\in\mathbb{R}_{+,0})$ be two Markov processes taking values on $E_1$ and $ E_2$, respectively. Let $H\,:\,E_1\times E_2\rightarrow \mathbb{R}$ be a bounded and continuous function, and assume that there exist functions $g_i\,:\,E_1\times E_2\mapsto \mathbb{R}$, for $i=1,2$, such that for every $n\in E_1,x\in E_2$ and every $T>0$, the processes $(M_t^{(1)}, 0\le t\le T)$ and $(M_t^{(2)}, 0\le t\le T)$, defined as

(2.10) \begin{eqnarray}M_t^{(1)}\,:\!=\,H\big(Y^{(1)}_t,x\big)&-&\int_0^t g_1\big(Y^{(1)}_s,x\big)\textrm{d} s, \end{eqnarray}

(2.11) \begin{eqnarray}M_t^{(2)}\,:\!=\,H\big(n,Y^{(2)}_t\big)&-&\int_0^t g_2\big(n, Y^{(2)}_s\big)\textrm{d} s, \end{eqnarray}

are martingales with respect to the natural filtrations of $Y^{(1)}_t$ and $Y^{(2)}_t$ respectively. Then, if

\begin{equation*}g_1(n,x)=g_2(n,x)\qquad {for\ all}\qquad n\in E_1, x\in E_2,\end{equation*}

the processes $\big(Y^{(1)}_t, t\in\mathbb{R}_{+,0}\big)$ and $\big(Y^{(2)}, t\in\mathbb{R}_{+,0}\big)$ are dual with respect to H.

This proposition is a direct consequence of Theorem 4.11 in Ethier and Kurtz [Reference Ethier and Kurtz11], with H taken to be bounded and continuous and $\alpha=\beta=0$.

If the function H is such that $H(n,x)=x^n$ for $x\in [0,1]$ and $n\in \mathbb{N}_0$, the above duality relationship is known as moment duality. The moment duality between X and Z reads as follows.

Lemma 2. Assume $|\textbf{m}|<\infty$, $\varsigma\le 0$, and $a=0$. Let X be the unique strong solution of the SDE (2.9); then it is the unique (in distribution) moment dual of Z a (sub)critical branching process with interactions and no annihilation. More precisely, for $x\in [0,1]$ and $n\in\mathbb{N}_{0}$, we have

\begin{equation*}\ensuremath{\mathbb{E}}_x[X_t^n]=\textbf{E}_n\big[x^{Z_t}\big]\qquad {for}\quad t\in\mathbb{R}_{+,0}.\end{equation*}

Proof. We take $H(n,x)=x^n$ for $x\in [0,1]$ and $n\in \mathbb{N}$, and when $n=0$, $H(0,x)\,:\!=\,1$ for $x\in[0,1]$; these are continuous and bounded.

Now, we claim that for fixed $x\in[0,1]$, H(n, x) is in the domain of Q, and for fixed $n\in \mathbb{N}_0$, it is in the domain of $\mathcal{A}$. The latter follows from the fact that $H(n,\cdot)$ is a polynomial on [0, 1], which clearly belongs to $C^2_b([0,1], \mathbb{R})$. Implicitly, we have that

\begin{equation*}H(n,X_t)-\int_0^t\mathcal{A}H(n, X_s)\textrm{d} s\end{equation*}

is a martingale. The first claim follows from recalling that linear combinations of exponential functions as in (1.2) belong to the domain of Q (see Section 1), and the fact that $H(n,x)=e^{n\log x }$ implies that the process

\begin{equation*}H(Z_t,x)-\int_0^t QH(Z_s, x)\textrm{d} s\end{equation*}

is also a martingale.

Next, we observe that for $x\in(0,1)$ and $n\in \mathbb{N}\setminus\{1\}$,

\begin{equation*}\begin{split}\mathcal{A}H(n,x) \ =\ & nd(x^{n-1}-x^n)+n\sum_{i\in \mathbb{N}}\pi_{i}(x^{n+i}-x^n)\\ &+n(n-1)\sum_{i\in \mathbb{N}} b_i(x^{n+i}-x^n)+n(n-1)c(x^{n-1}-x^n)\\ &+\int_{[0,1]}(x(x(1-z)+z)^n+(1-x)(x(1-z))^n-x^n)\frac{\Lambda(\textrm{d} z)}{z^2}.\end{split}\end{equation*}

On the other hand, let B be a Bernoulli random variable taking the value 1 with probability x and 0 with probability $1-x$. Then, for $n\in \mathbb{N}\setminus\{1\}$, we observe that

\begin{equation*}\begin{split}x(x(1-z)+z)^n +(1-x)(x(1-z))^n-x^n & =\mathbb{E}\big[(x(1-z)+Bz)^n\big]-x^n\\[3pt] &=\mathbb{E}\left[\sum_{k=0}^n \dbinom{n}{k}x^{n-k}(1-z)^{n-k}z^k B^k\right]-x^n\\ &=\sum_{k=2}^n \dbinom{n}{k}x^{n-k+1}(1-z)^{n-k}z^k\\ &\quad +x^n(1-z)^n+nx^n(1-z)^{n-1}z-x^n\\ &=\sum_{k=2}^n \dbinom{n}{k}x^{n-k+1}(1-z)^{n-k}z^k\\ &\quad -x^n\big(1-(1-z)^n-n(1-z)^{n-1}z\big)\\ &=\sum_{k=2}^n \dbinom{n}{k}(1-z)^{n-k}z^k\big(x^{n-k+1} -x^n\big).\\\end{split}\end{equation*}

Hence, for $n\in\mathbb{N}\setminus\{1\}$, we deduce that

\begin{equation*}\int_{[0,1]}\big(x(x(1-z)+z)^n+(1-x)(x(1-z))^n-x^n\big)\frac{\Lambda(\textrm{d} z)}{z^2}=\sum_{k=2}^n\binom{n}{k}\lambda_{n,k}[x^{n-k+1}-x^n],\end{equation*}

implying that $\mathcal{A}H(n, x)=QH(n, x)$ for $x\in(0,1)$ and $n\in \mathbb{N}\setminus\{1\}$. For the case $n=1$ and $x\in(0,1)$, we have

\begin{equation*}\mathcal{A}H(1, x)=d(1-x)+\sum_{i\in \mathbb{N}}\pi_{i}(x^{1+i}-x)=QH(1, x).\end{equation*}

Finally, we consider the remaining cases. For $n\in\mathbb{N}$ and $x=0$ or $x=1$, it is clear that $\mathcal{A}H(n, x)=0=QH(n, x)$. Moreover, when $n=0$, we have $QH(0, x)=0=\mathcal{A}H(0, x)$ for $x\in[0,1]$.

Putting all the pieces together, we deduce that $\mathcal{A}H(n, x)=QH(n, x)$ for $x\in[0,1]$ and $n\in \mathbb{N}_0$. Hence we can apply Proposition 3 to deduce the moment duality between X and Z.

We finish this section with the following useful lemma, which provides an upper bound for the expected fixation time of the process X defined by (2.9) with $\mu\equiv0$ and $\Lambda\equiv 0$. We will use this lemma for the proofs of Theorems 3 and 4. In particular, we have that the fixation time for X with $\mu\equiv 0$, $\Lambda\equiv 0$ and starting at $x\in[0,1]$ is finite $\mathbb{P}_x$-a.s.

Lemma 3. Let X be the process defined by the SDE (2.9) with $\varsigma<0$, $\Lambda\equiv 0$, $d=0$, $\pi_i=0$ for all $i\in \mathbb{N}$, and recall that $T_{0,1}=\inf\{t>0\,:\,X_t\in\{0,1\}\}$. Then for any $x\in[0,1]$,

\begin{equation*}\mathbb{E}_x[T_{0,1}]\le -\frac{2}{c-\sum_{i=1}^\infty i b_i }\big((1-x)\ln (1-x) +x\ln x\big).\end{equation*}

Moreover, we have $\mathbb{P}_x(X_{T_{0,1}}=1)=x$.

Proof. Since $\Lambda\equiv 0$, $d=0$, and $\pi_i=0$ for all $i\in \mathbb{N}$, we observe that the process X, under $\mathbb{P}_x$, satisfies the following SDE:

(2.12) \begin{equation}X_t=x+\int_0^t\sigma(X_s)\textrm{d} B_s,\end{equation}

where $\sigma$ is defined as above. The process X is a diffusion taking values on [0, 1] with $\{0,1\}$ as absorbing states. Observe that its associated scale function is given by $S(y)=y+\textbf{c}$, where $\textbf{c}$ is a constant, and its associated speed measure satisfies

\begin{equation*} m(\textrm{d} y)=\frac{2}{\sigma^2(y)}\textrm{d} y. \end{equation*}

Hence, from Corollary VII 3.8 in [Reference Revuz and Yor34], we have

\begin{equation*} \mathbb{E}_x[T_{0,1}]=2\int_0^1 G(x,y)\frac{1}{\sigma^2(y)}\textrm{d} y, \end{equation*}

where

\begin{equation*} G(x,y)=\left\{ \begin{array}{l@{\quad}l} x(1-y) & \textrm{ if } 0\le x\le y\le 1,\\[3pt] y(1-x) & \textrm{ if } 0\le y\le x\le 1,\\[3pt] 0 & \textrm{ otherwise.} \end{array} \right . \end{equation*}

Now, we observe that for $y\in[0,1]$, the following inequality holds:

\begin{equation*}\sigma^2(y)\ge \Big( c-\sum_{i\in \mathbb{N}} i b_i \Big)y (1-y).\end{equation*}

This implies

\begin{equation*}\begin{split}\mathbb{E}_x[T_{0,1}]&\le \frac{2}{\Big( c-\sum_{i\in \mathbb{N}}^\infty i b_i \Big)}\left(\int_0^x \frac{y(1-x)}{y(1-y)}\textrm{d} y +\int_x^1 \frac{x(1-y)}{y(1-y)}\textrm{d} y\right)\\&=- \frac{2}{\Big( c-\sum_{i\in \mathbb{N}} i b_i \Big)}\Big((1-x)\ln (1-x) +x\ln x\Big),\end{split}\end{equation*}

which provides the desired inequality.

Finally, since X is a local martingale taking values on [0, 1], it is a uniformly integrable martingale. Hence from the optional stopping theorem, we deduce that

\begin{equation*}x=\mathbb{E}_1[X_0]=\mathbb{E}_x[X_{ T_{0,1}}]=\mathbb{P}_x(X_{T_{0,1}}=1).\end{equation*}

This completes the proof.

2.3. Proof of Theorem 3 & Corollary 2

For the proof of Theorem 3, we need the following lemma.

Lemma 4. Assume that there is no annihilation, no death, and no branching; i.e. $\varsigma<0$, $a=0$, $d=0$, and $\rho=0$. Then

\begin{equation*}\sup_{n\in \mathbb{N}}\textbf{E}_n\big[\tau_1\big]<\infty,\end{equation*}

where $\tau_{1}=\inf\{t>0\,:\,Z_t=1\}$.

Proof. To prove the statement, it is enough to prove it in the case when there are no catastrophes, since an order-preserving coupling can be established. Indeed, we can stochastically dominate the case with catastrophes with another branching process with interactions having the same parameters but no catastrophes. We refer the reader to Section 2 of López et al. [Reference López, Martínez and Sanz26] for the conditions on the intensities $(q_{i,j})_{i,j\in\mathbb{N}_0}$ that ensure the existence of an order-preserving coupling.

From the moment duality, i.e. Lemma 2, we get

\begin{equation*}\textbf{E}_n\big[x^{Z_t}\big]=\mathbb{E}_x[X^n_{ t}]=\mathbb{E}_x\Big[X^n_{ t}\textbf{1}_{\{T_{0,1}\ge t\}}\Big]+\mathbb{E}_x\Big[X^n_{ T_{0,1}}\textbf{1}_{\{T_{0,1}<t\}}\Big].\end{equation*}

Thus, by letting t go to $\infty$, applying the dominated convergence theorem, and recalling that $T_{0,1}$ is finite $\mathbb{P}_x$-a.s. (see Lemma 2), we obtain

(2.13) \begin{equation}\lim_{t\to \infty}\textbf{E}_n\big[x^{Z_t}\big]=\mathbb{P}_x(X_{T_{0,1}}=1)=x.\end{equation}

On the other hand, again from the moment duality between X and Z, we deduce the following inequality:

\begin{equation*}\textbf{P}_n(Z_t=1)=\frac{\textbf{E}_n\big[x^{Z_t}\big]-\sum_{i=2}^\infty x^{i}\textbf{P}_n(Z_t=i)}{x}\ge \frac{\textbf{E}_n\big[x^{Z_t}\big]}{x}-x.\end{equation*}

Therefore, from the identity (2.13), we get

\begin{equation*}\lim_{t\rightarrow \infty}\textbf{P}_n(Z_t=1)\geq 1-x.\end{equation*}

Implicitly, there are $t_1>0$ and $\epsilon>0$ such that for every $n\in\mathbb{N},$

\begin{equation*}\textbf{P}_n(Z_{t_1}=1)>\epsilon.\end{equation*}

Next we observe, using the Markov property of Z starting from $n\in \mathbb{N}$ at times $(j t_1)_{j\in \mathbb{N}}$, that $\sigma_1$ is stochastically dominated by a geometric random variable with parameter $\epsilon$, here denoted by $\Upsilon_\epsilon$, that bounds the number of steps taken by the discretized Markov chain $(Z_{jt_1}, j\ge 0)$ to reach the state 1. In other words, for any $n,m\in \mathbb{N}$, we have

\begin{equation*}\textbf{P}_n(\tau_1> mt_1)\le \mathbb{P}(\Upsilon_\epsilon > mt_1).\end{equation*}

The latter inequality implies, in particular, that

\begin{equation*}\sup_{n\in \mathbb{N}}\textbf{E}_n\big[\tau_1\big]<\frac{t_1}{\epsilon},\end{equation*}

which completes the proof.

Proof of Theorem 3. We first deal with the case with no catastrophes, i.e. $\Lambda\equiv 0$. From our assumptions, Z has parameters $d=0$, c, $\pi_i$, and $b_i$, for $i\in \mathbb{N}$, such that $\rho=\sum_{i\in \mathbb{N}} \pi_i>0$ and $\varsigma<0$. Our arguments will be based on a coupling argument. More precisely, let us consider two branching processes with interactions, $Z=(Z_t, t\in \mathbb{R}_{+,0})$ and $\overline{Z}=(\overline{Z}_t, t\in \mathbb{R}_{+,0})$, with respective parameters $a=d=0$, c, $b_i$, and $\pi_i$, for $i\in \mathbb{N}$, such that $\rho>0$; and $\overline{d}=\overline{a}=\sum_{i\in \mathbb{N}}\overline{\pi_i}=0$, $\overline{c}=c$, and $\overline{b_i}=b_i+\epsilon \pi_i$ for all $i\in \mathbb{N}$, where $\epsilon$ is chosen positive and such that

(2.14) \begin{equation}-\overline{c}+\sum_{i\in \mathbb{N}} i \overline{b_i} =-c+\sum_{i\in \mathbb{N}} i b_i +\epsilon \sum_{i\in \mathbb{N}} i \pi_i < 0.\end{equation}

We denote by $\overline{\textbf{P}}_n$ the law of $\overline{Z}$ starting from $\overline{Z}_0=n$ and observe, from the first part of the proof and under our assumptions, that $\overline{Z}$ does not explode and gets absorbed in the state $\{1\}$. Moreover, let $n_0=\inf\{n\in \mathbb{N}\,:\, 2< (n-1)\epsilon\}$ and observe that for any starting condition $Z_0=\overline{Z}_0=n> n_0$, we necessarily have

\begin{equation*} n\pi_i+n(n-1)b_i<n(n-1)\overline{b}_i=\epsilon n(n-1)\pi_i+n(n-1)b_i. \end{equation*}

Next, we consider the following coupling between the Markov chains Z and $\overline{Z}$. Let us introduce the process $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ with the following dynamics. For all $t<\gamma_{n_0}\,:\!=\,\inf\{t>0\,:\, U_t\le n_0\}$, the couple $(U_t,\overline{U}_t)$ has the following transition rates, for $m\ge n$:

(2.15) \begin{equation}(U_t,\overline{U}_t)\text{ goes from }(n,m)\text{ to}\left\{ \begin{array}{ll}(n+i,m+i), & \textrm{ with rate } n\pi_i+n(n-1)b_i,\\[3pt] (n,m+i), & \textrm{ with rate } m(m-1)\overline{b}_i-n\pi_i+n(n-1)b_i,\\[3pt] (n-1,m-1), & \textrm{ with rate } n(n-1)c,\\[3pt] (n,m-1), & \textrm{ with rate } m(m-1)c-n(n-1)c.\\\end{array}\right .\end{equation}

For $t\geq \gamma_{n_0}$, the Markov chains $U_t$ and $\overline{U}_t$ evolve independently with the same transition rates as Z and $\overline{Z}$, respectively. We denote by $\textbf{P}_{(n,m)}$ the law of $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ starting from (n, m). Note that from our construction, the Markov chains U and $\overline{U}$ are equal in distribution to Z and $\overline{Z}$, respectively. (We refer the reader to Chapters IV and V in Lindvall [Reference Lindvall25] for further details on this type of coupling and stochastic dominance.)

Furthermore, we necessarily have, for all $n>n_0$,

\begin{equation*}\textbf{P}_{(n,n)}\big(U_t\leq \overline{U_t}, \textrm{ for } t<\gamma_{n_0}\big)=1,\end{equation*}

where $\textbf{P}_{(n,n)}$ denotes the law of $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ starting from (n, n). Hence, from Lemma 4 and the fact that Z and $\overline{Z}$ are skip-down-free (or skip-free to the left), we deduce that

(2.16) \begin{equation}\textbf{E}_n[\tau_{n_0}]=\mathbb{E}[\gamma_{n_0}|U_0=n]\leq \mathbb{E}[\overline{\gamma}_{n_0}|\overline{U}_0=n]=\overline{\textbf{E}}_n[\overline{\tau}_{n_0}]<\overline{\textbf{E}}_n[\overline{\tau}_{1}]<K,\end{equation}

where ${\gamma}_{n_0}\,:\!=\,\inf\{t>0\,:\, {U}_t\le n_0\}$, $\overline{\gamma}_{n_0}\,:\!=\,\inf\{t>0\,:\, \overline{U}_t\le n_0\}$, ${\tau}_{i}\,:\!=\,\inf\{t>0\,:\,{Z}_t= i\}$, $\overline{\tau}_{i}\,:\!=\,\inf\{t>0\,:\,\overline{Z}_t= i\}$, and K is a positive constant that does not depend on n.

Then the process Z is positive recurrent, and we apply Theorem 21.14 of [Reference Levin, Peres and Wilmer23] to conclude that there exists a unique stationary distribution for Z, here denoted by $\nu$. To finish the proof of the first statement, we use Markov’s inequality and (2.16) to deduce that

\begin{equation*}\textbf{P}_n(\tau_{n_0}>a)<\frac{K}{a} \qquad \textrm{for every}\quad n\ge n_0.\end{equation*}

This allows us to compute the total variation distance between $\textbf{P}_n(Z_t=\cdot)$ and $\nu({\cdot})$ by conditioning on the event $\{\tau_{n_0}>t_0\}$, where $t_0\in(0,t)$. In other words,

\begin{align*}||\textbf{P}_n(Z_t=\cdot)-\nu({\cdot})||\ \leq & \ ||\textbf{P}_{n}(Z_t=\cdot , \tau_{n_0}\leq t_0)-\nu({\cdot})||\\[4pt] &+\ \textbf{P}_n(\tau_{n_0}>t_{0})||\textbf{P}_n(Z_t=\cdot|\tau_{n_0}>t_{0})-\nu({\cdot})||\\[4pt] \le &\int_0^{t_0}||\textbf{P}_{n_0}(Z_{t-s}=\cdot)-\nu({\cdot})||\textbf{P}_n(\tau_{n_0}\in \textrm{d} s)+\frac{K}{t_0}.\end{align*}

Therefore, again using Theorem 21.14 of [Reference Levin, Peres and Wilmer23], we observe that

\begin{equation*}||\textbf{P}_{n_0}(Z_{t-s}=\cdot)-\nu({\cdot})||\rightarrow 0 \qquad \textrm{ as }\quad t\rightarrow \infty\end{equation*}

for all $s\in[0,t_0]$, and thus we conclude that

\begin{equation*}\lim_{t\rightarrow \infty} \sup_{n\in \mathbb{N} }||\textbf{P}_n(Z_t=\cdot)-\nu({\cdot})||<\frac{K}{t_0}.\end{equation*}

Since $t_0$ was taken to be arbitrary, we deduce (1.4).

For the case with catastrophes, we use an order-preserving coupling. In other words, we introduce two branching processes Z and $\overline{Z}$ with the same parameters $d=\overline{d}=0,$ $c=\overline{c},$ $\pi_i=\overline{\pi}_i,$ $b_i=\overline{b}_i$, for $i\in \mathbb{N}$, but the former takes catastrophes into account and the latter has no catastrophes. Again, we refer the reader to Section 2 of López et al. [Reference López, Martínez and Sanz26] for the conditions on the intensities $(q_{i,j})_{i,j\in \mathbb{N}_0}$ that ensure the existence of an order-preserving coupling, which are clearly satisfied by Z and $\overline{Z}$. We denote by $\overline{\textbf{P}}_n$ the law of $\overline{Z}$ starting at n. Hence, the order-preserving property and the upper bound in (2.16) imply

\begin{equation*} \textbf{E}_n[\tau_{n_0}]\le \overline{\textbf{E}}_n[\overline{\tau}_{n_0}]\le K \qquad \textrm{for} \quad n\ge n_0,\end{equation*}

where $\overline{\tau}_{i}\,:\!=\,\inf\{t>0\,:\,\overline{Z}_t= i\}$, and K is a positive constant that does not depend on n. Then we proceed similarly as in the case with no catastrophes to prove uniform convergence to stationarity in the total variation norm. We leave the details to the reader.

In order to prove last statement of this theorem, we first assume that $\mathbb{P}_x(X_\infty\in\{0,1\})<1$, where $X_\infty\,:\!=\,\lim_{t\rightarrow \infty}X_t$. This implies that $\lim_{t\rightarrow \infty}\mathbb{E}_x[X_t^{n}]<\lim_{t\rightarrow \infty}\mathbb{E}_x[X_t]$, since for every $x\in (0,1)$, $x^n$ is strictly smaller than x. From the moment duality between Z and X and the fact that $\nu$ is the invariant distribution of Z, we obtain

\begin{equation*}\sum_{i\ge 0} x^i\nu(i)=\lim_{t\rightarrow \infty}\textbf{E}_n\big[x^{Z_t}\big]=\lim_{t\rightarrow \infty}\mathbb{E}_x[X_t^{n}]<\lim_{t\rightarrow \infty}\mathbb{E}_x[X_t]=\lim_{t\rightarrow \infty}\textbf{E}_1[x^{Z_t}]=\sum_{i\ge 0} x^i\mu(i),\end{equation*}

which contradicts our hypothesis. In other words, we conclude that $\mathbb{P}_x(X_\infty\in\{0,1\})=1$ for any starting point $x\in(0,1)$. Finally, we use the left-hand side of the identity (2.13) to deduce that $\nu$ satisfies

\begin{equation*} \sum_{i\ge 0} x^i\nu(i)=\mathbb{P}_x(X_\infty=1) \end{equation*}

for any $x\in (0,1)$. This completes the proof.

Next we prove Corollary 2.

Proof of Corollary 2. From the dominated convergence theorem and the moment duality between Z and X, we observe that for any $x\in (0,1)$ and $n\ge 1$, we have

\begin{equation*}\mathbb{E}_x\left[\lim_{t\rightarrow\infty}X_t^n\right]=\lim_{t\rightarrow\infty}\mathbb{E}_x[X_t^n]=\lim_{t\rightarrow\infty}\textbf{E}_n\big[x^{Z_t}\big]=\textbf{E}_n\left[\lim_{t\rightarrow\infty}x^{Z_t}\right].\end{equation*}

From Theorem 3, we observe that the limit of the generating function of Z does not depend on the starting state n. This implies that all the moments of $X_\infty=\lim_{t\rightarrow\infty}X_t$ must be equal. Since the limit of X must be in [0, 1], from the latter observation we deduce that $\mathbb{P}_x(X_\infty\in\{0,1\})=1$, and thus

\begin{equation*}\mathbb{P}_x\left(X_\infty=1\right)=\mathbb{E}_x\left[\lim_{t\rightarrow\infty}X_t^n\right]=\textbf{E}_n\left[\lim_{t\rightarrow\infty}x^{Z_t}\right].\end{equation*}

The proof is completed once we compute $\textbf{E}_n[\lim_{t\rightarrow\infty}x^{Z_t}]$ in each of the three cases of Theorem 3.

2.4. Proof of Theorem 4

For the proof of Theorem 1, we need Lemma 4 and the following result.

Lemma 5. Assume that $\varsigma<0,$ $a=0$, $d>0$, and $\rho>0$. Then

\begin{equation*}\sup_{n\in \mathbb{N} }\textbf{E}_n\big[\tau_0\big]<\infty,\end{equation*}

where $\tau_{0}=\inf\{t>0\,:\,Z_t=0\}$.

Proof. To prove this lemma, we again use a coupling method. From our assumptions Z has parameters $d>0,$ c, $\pi_i$, $b_i$, for $i\in \mathbb{N}$, and $\lambda_{i,k}, $ for $ k\in\{ 2, \ldots, i\} $. For our purposes we also consider a branching process with interactions $\overline{Z}=(\overline{Z}_t, t\in \mathbb{R}_{+,0})$ with parameters $\overline{d}=0$, $\overline{c}=c$, $\overline{\pi_i}=\pi_i$, $\overline{b_i}=b_i$, for all $i\in \mathbb{N}$, and $\overline{\lambda}_{i,k}=\lambda_{i,k}, $ for $k\in\{ 2, \ldots, i\} $. We denote by $\overline{\textbf{P}}_n$ its law starting from $\overline{Z}_0=n$. Observe that the process $\overline{Z}$ fulfils the conditions of Theorem 3, and thus there exists a unique stationary distribution $\nu$. In particular,

\begin{equation*} \lim_{t\rightarrow \infty}\overline{\textbf{P}}_n(\overline{Z}_t=1)=\nu(\{1\})>0, \end{equation*}

and this is a uniform limit on the variable n. Thus, for every $\epsilon\in(0,\nu(\{1\}))$ there exists $t_0>0$ such that

\begin{equation*} \inf_{n\in \mathbb{N}}\overline{\textbf{P}}_n(\overline{Z}_{t_0}=1)>\epsilon. \end{equation*}

Since the processes differ only in the death parameter, the order-preserving coupling is straightforward. Let us denote this coupling by $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$. In particular, we have that $U_t$ and $\overline{U}_t$ are equal in distribution to $Z_{t}$ and $\overline{Z}_t$, respectively. As we said before, such a coupling preserves order, meaning that

\begin{equation*}U_t\leq \overline{U_t} \qquad \textrm{for } t\in \mathbb{R}_{+,0}, \qquad \textrm{ a.s.}\end{equation*}

On the other hand, we observe that $\mathbb{P}(U_{h}=0|U_0=1)>0$ for $h\in \mathbb{R}_{+}$. Hence we conclude that for $n\in \mathbb{N}$,

\begin{align*}\textbf{P}_n(Z_{t_0+h}=0)=\,&\mathbb{P}(U_{t_0+h}=0|U_0=n)\\ &> \mathbb{P}(U_{t_0+h}=0|U_{t_0}=1)\mathbb{P}(U_{t_0}=1|U_0=n)\\ &> \mathbb{P}(U_{h}=0|U_{0}=1)\mathbb{P}(\overline{U}_{t_0}=1|\overline{U}_0=n)\\ &>\textbf{P}_1(Z_{h}=0)\epsilon.\end{align*}

Again, it is important to note that the right-hand side of the previous equation does not depend on n. From the Markov property we deduce that

\begin{align*}\textbf{P}_n(\tau_0>t)\leq&\textbf{P}_n(\tau_0>\lfloor t/(t_0+h)\rfloor t_0)\\<&(1-\textbf{P}_1(Z_{h}=0)\epsilon)^{\lfloor t/(t_0+h)\rfloor}.\end{align*}

This allow us to conclude that stochastically $\tau_0\leq (t_0+h)G$, where G is a geometric random variable with parameter $\textbf{P}_1(Z_{h}=0)\epsilon$. Note that G is independent of the starting condition n. This leads to the conclusion that, for every $n\in \mathbb{N}$,

\begin{equation*}\textbf{E}_n[\tau_0]<\frac{t_0+h}{\textbf{P}_1(Z_{h}=0)\epsilon}<\infty,\end{equation*}

which completes the proof.

Proof of Theorem 4. In order to prove our result, we first deduce that under our assumptions the sequence $(\textbf{P}_n)_{n\in \mathbb{N}}$ converges weakly in the space of probability measures on $D(\overline{\mathbb{N}}, [0, T])$, the Skorokhod space of càdlàg functions on [0, T] with values in $\overline{\mathbb{N}}$. We follow the tightness argument provided in the proof of Theorem 1 in Donelly [Reference Donnelly10]. We observe that the process does not possess instantaneous states and also that it is stochastically monotone with respect to the starting point. In other words, the condition (A1) of Theorem 1 in [Reference Donnelly10] is satisfied. Moreover, under our assumptions the process does not explode, and the condition (A2) is also satisfied, if we denote by $B_n^N$ the branching process with interactions starting from n and stopped at state N. Then tightness holds and we identify the finite marginal distributions by noticing that for $k\ge 1$, for $t_1, \ldots t_k\in \mathbb{R}_{+,0}$, and for $n_1, \dots, n_ k \in \mathbb{N}$, the probabilities $\textbf{P}_n(Z_{t_1}\le n_1, \cdots, Z_{t_k}\le n_k)$ are nonincreasing with respect to n.

Next, since Z and X are moment duals, for any $t\in \mathbb{R}_+$ we have

(2.17) \begin{equation}\textbf{P}_\infty(Z_t<\infty)=\lim_{x\rightarrow 1}\lim_{n\rightarrow \infty} \textbf{E}_n\big[x^{Z_t}\big]=\lim_{x\rightarrow 1}\lim_{n\rightarrow \infty} \mathbb{E}_x[X_t^n]=\lim_{x\rightarrow 1}\mathbb{P}_x(X_t=1),\end{equation}

where in the first equality we have used that $(\textbf{P}_n)_{n\in\mathbb{N}}$ converges weakly in the space of probability measures. In order to deduce that the previous probability equals 1, we first consider the case $\Lambda\equiv 0$, $d=0$, and $\rho=0$. From Lemma 3, we deduce that

\begin{equation*}\lim_{x\rightarrow 1}\mathbb{P}_x(X_t=1)=\lim_{x\rightarrow 1}\mathbb{P}_x(X_{T_{0,1}}=1, T_{0,1}<t)=1.\end{equation*}

By a comparison result for diffusions (see for instance Theorem 2.2 in Dawson and Li [Reference Dawson and Li9]), we deduce that the latter also holds for the case $\rho>0,$ $\Lambda\equiv 0$, and $d=0$. In other words, when $\Lambda\equiv 0$ and $d=0$, we deduce from (2.17) that

\begin{equation*}\textbf{P}_\infty(Z_t<\infty)=1 \qquad \textrm{for any} \quad t\in\mathbb{R}_+,\end{equation*}

and conclude that the process Z comes down from infinity. Finally, the remaining cases (i.e. $d>0$ and/or catastrophes) follow from an order-preserving coupling.

Next, we study the limit of $\textbf{E}_n[\tau_1]$ as n goes to $\infty$. If the parameters d, c, $\pi_i$, $b_i$, for $i\in \mathbb{N}$, and $\lambda_{i,k}$ for $k\in\{2,\ldots, i\}$ are as in Lemmas 4 and 5, then it is clear that $\lim_{n\to \infty}\textbf{E}_n[\tau_1]<\infty$. If the parameters are as in Theorem 3, then the identity (1.4) implies that there exists a time $t_0$ such that $\sup_{n\in\mathbb{N}}\textbf{P}_n(Z_{t_0}=1)>\nu(1)/2.$ By the Markov property, for every $n\ge 1$, we observe that

\begin{equation*}\textbf{P}_{n}(\tau_1>kt_0 )<\textbf{P}_n(Z_{it_0}\neq 1, \text{ for all }i\in\{1,2,...,k\})<(1-\nu(1)/2)^k.\end{equation*}

This implies that $\textbf{E}_n[\tau_1]<2/\nu(1)<\infty$ for all $n\in\mathbb{N}$; in other words we have $\lim_{n\rightarrow \infty}\textbf{E}_n[\tau_1]<2/\nu(1)$. This completes the proof.

2.5. Proof of Proposition 2

Proof of Proposition 2. We first deduce that for any choice of parameter it holds that

\begin{equation*}\sup_{n\in\mathbb{N}}\textbf{E}_n[\tau_{0,1}]<\infty.\end{equation*}

Let us first assume that $a=0$. If $d>0$, or $d=0$ and $\pi_i=0$ for all $i\in \mathbb{N}$, then the result follows immediately from Lemmas 4 and 5. If $d=0$ and $\pi_i>0$ for some $i\in\mathbb{N}$, then the identity (1.4) guarantees that for every $\epsilon<\nu(1)$ there exists $t_\epsilon>0$ such that $\inf_{n\in\mathbb{N}}\textbf{P}_n(Z_{t_\epsilon}=1)>\epsilon.$ Thus, by the strong Markov property we conclude that

\begin{equation*}\sup_{n\in\mathbb{N}}\textbf{E}_n[\tau_{0,1}]\leq \sup_{n\in\mathbb{N}}\textbf{E}_n[\tau_{1}]\leq\frac{t_\epsilon}{\epsilon}<\infty.\end{equation*}

Next, we assume $a>0$. In this case, we will again use an order-preserving coupling. Let Z be a branching process with interactions with parameters $a>0$, d, c, $\pi_i$, $b_i$, and $\lambda_{i,k}$ for $ k\in\{ 2, \ldots, i\}$, which represents the catastrophes. Recall that $\textbf{P}_n$ denotes its law starting from ${Z}_0=n$. We also consider another branching process with interactions $\overline{Z}=(\overline{Z}_t, t\in \mathbb{R}_{+,0})$, with parameters $\overline{a}=0$, $\overline{d}=d$, $\overline{\pi_i}=\pi_i$, $\overline{c}=c+a$, $\overline{b_i}=b_i$, for all $i\in \mathbb{N}$, and $\overline{\lambda}_{i,k}$ for $ k\in\{ 2, \ldots, i\} $. We denote by $\overline{\textbf{P}}_n$ its law starting from $\overline{Z}_0=n$. Since the only parameters in which the processes Z and $\overline{Z}$ differ are the annihilation and competition parameters, the coupling is straightforward (every time there is an annihilation event in Z there is a competition event in $\overline{Z}$).

Let us denote by $\{(U_t,\overline{U}_t), t\in \mathbb{R}_{+,0}\}$ such a coupling, and observe that

\begin{equation*}U_t\leq \overline{U_t} \qquad \textrm{for } t\in\mathbb{R}_{+,0}, \qquad \textrm{ a.s.}\end{equation*}

Since $\overline{U}$ has no annihilation, we conclude by the previous argument that

\begin{equation*}\sup_{n\in\mathbb{N}}\textbf{E}_n[\tau_{0,1}]<\sup_{n\in\mathbb{N}}\textbf{E}_n[\overline{\tau}_{0,1}]<\infty,\end{equation*}

where $\overline{\tau}_{0,1}$ denotes the first hitting time of the states $\{0,1\}$ of the process $\overline{Z}$. Whenever $d>0$, we have

\begin{equation*}\sup_{n\in\mathbb{N}}\textbf{E}_n[\tau_{0}]<\sup_{n\in\mathbb{N}}\textbf{E}_n[\overline{\tau}_{0}]<\infty.\end{equation*}

This proves Part (ii).

From the first part of the proof, we know that Z visits $\{0,1\}$. Observe that if Z visits the state $\{0\}$ it will get absorbed, but if it visits the state $\{1\}$ it will get absorbed if and only if $d=0$ and $\pi_i=0$ for all ${i\in\mathbb{N}}$. The proofs of Parts (a) and (b) rely on verifying whether the states $\{0\}$ and $\{1\}$ are absorbing states or not.

For Part (a), the assumptions guarantee that for every $n\in\mathbb{N}\setminus\{1\}$ there exists some number $i\in\mathbb{Z}$ such that Z goes from n to $n+2i-1$ at a positive rate. Since $a>0$, we have that Z goes from n to $n-2$ at a positive rate. In other words, the states $\{0\}$ and $\{1\}$ are accessible.

On the other hand, we deduce that $\{1\}$ is an absorbing state if and only if $\sum_{i\in\mathbb{N}}\pi_i=0$. If $\{1\}$ is an absorbing state, then from the first part of the proof, we deduce that $\sup_{{n\in\mathbb{N}}}\textbf{E}_n[\tau_{0,1}]<\infty.$ If $\{1\}$ is not absorbing, then $\{0\}$ is accessible from $\{1\}$, which implies that there exist $\epsilon>0$ and $t_\epsilon\in\mathbb{R}_+$ such that

\begin{equation*}\textbf{P}_1(Z_{t_\epsilon}=0)>\epsilon.\end{equation*}

Using the strong Markov property we conclude that

\begin{equation*}\sup_{n\geq1}\textbf{E}_n[\tau_{0}]<\frac{1}{\epsilon}\Big(\sup_{n\geq1}\textbf{E}_n[\tau_{0,1}]+t_\epsilon\Big)<\infty.\end{equation*}

This completes the proof of Part (a).

For Part (b), we observe that Z only makes jumps of even size, so the support of $\textbf{P}_n(Z_t=\cdot)$ consists of the even or the odd numbers, depending the value of n. If n is even, then $\{1\}$ is not accessible and $\{0\}$ is accessible. In other words, $\tau_{0,1}=\tau_{0}$, and by the first part of our proof, we conclude that $\sup_{{n\in\mathbb{N}}}\textbf{E}_n[\tau_{0}]<\infty$. Similarly, if n is odd then $\{1\}$ is accessible but $\{0\}$ is not. In this case, $\tau_{0,1}=\tau_{1}$, and again from the first part of the proof we conclude that $\sup_{{n\in\mathbb{N}}}\textbf{E}_n[\tau_{1}]<\infty$.

On the other hand, under our assumptions, we deduce that $\{1\}$ is an absorbing state if and only if $\sum_{i\in\mathbb{N}}\pi_i=0$. If $\sum_{{i\in\mathbb{N}}}\pi_i>0$, we have that $\{1\}$ is positive recurrent. The uniform convergence to the stationary distribution in this case follows from arguments similar to those used for Theorem 3, Part (ii). We leave the details to the reader.