The multiple birth properties of multi-type Markov branching processes

Abstract

The main purpose of this paper is to consider the multiple birth properties for multi-type Markov branching processes. We first construct a new multi-dimensional Markov process based on the multi-type Markov branching process, which can reveal the multiple birth characteristics. Then the joint probability distribution of multiple birth of multi-type Markov branching process until any time t is obtained by using the new process. Furthermore, the probability distribution of multiple birth until the extinction of the process is also given.

1 Introduction

Markov branching processes play an important role in the research and application of stochastic processes. Standard references are Anderson [1], Harris [2], Athreya & Ney [3], Asmussen & Hering [4], Athreya & Jagers [5] and others.

The basic property governing the evolution of a Markov branching process is the branching property, i.e., different individuals act independently when giving offsprings. The classical Markov branching processes are well studied, some related references are Harris [2], Athreya & Ney [3], Asmussen & Hering [4], and Athreya & Jagers [5]. Based on the branching structure, there are many references concentrating on generalization of ordinary Markov branching processes. For example, Vatutin [6], Li, Chen & Pakes [7] considered the branching processes with state-independent immigration. Chen, Li & Ramesh [8] and Chen, Pollet, Zhang & Li [9] considered weighted Markov branching processes, Li & Chen [10] considered generalized Markov interacting branching processes, Li & Wang [11, 12] and Meng & Li [13] considered n-type branching processes with or without immigration. Recently, Li & Li [14, 15] considered down/up crossing properties of weighted Markov collision processes and one-dimensional Markov branching processes.

In this paper, we mainly discuss the multiple birth properties of multi-type Markov branching processes. Different from the one-type case, the number of individuals of other types may change when an individual splits.

For convenience of our discussion, we make the following notations throughout of this paper. Let $\mathbf{Z}_{+}$ be the set of non-negative integers.

(C-1) $\mathbf{Z}_{+} ^{d} :=\{\boldsymbol{i}=({i_{1}},\ldots ,{i_{d}}):{i_{1}}, \ldots ,{i_{d}} \in {\mathbf{Z}_{+}}\}$, and for any $\boldsymbol{i}=(i_{1},\ldots ,i_{d}) \in \mathbf{Z}_{+}^{d}$, denote $\mid \boldsymbol{i}\mid =\sum \limits _{k=1}^{d} i_{k}$.

(C-2) ${[0,1]^{d}} = \{\boldsymbol{x}=({x_{1}},\ldots ,{x_{d}}):0 \leq {x_{1}},\ldots ,{x_{d}} \leq 1\}$.

(C-3) ${\chi _{{\mathbf{Z}_{+}^{d}}}}(\cdot )$ is the indicator of $\mathbf{Z}_{+}^{d}$

(C-4) $\boldsymbol{0}=(0,\ldots ,0)$, $\boldsymbol{1}= (1,\ldots ,1)$, ${\boldsymbol{e}_{k}} = (0,\ldots ,{1_{k}},\ldots ,0)$ are vectors in ${[0,1]^{d}}$.

(C-5) For any $\boldsymbol{x},\boldsymbol{y}\in [0,1]^{d}$, $\boldsymbol{x}\leq \boldsymbol{y}$ means $x_{k} \leq y_{k}$ for all $k= 1,\ldots ,d$. $\boldsymbol{x}<\boldsymbol{y}$ means $x_{k} \leq y_{k}$ for all $k= 1,\ldots ,d$, and $x_{k}< y_{k}$ for at least one k.

(C-6) For any $\boldsymbol{x}\in [0,1]^{d}$, denote $\| \boldsymbol{x}\|_{1} = \sum \limits _{k=1}^{d} |x_{k}|$.

A d-type Markov branching process can be intuitively described as follows:

(1) Consider a system involving d types of individuals. The life length of a type-k individual is exponentially distributed with mean ${\boldsymbol{\theta}_{k}}\ (k= 1,\ldots ,d)$.

(2) Individuals in the system split independently. When a type-k individual dies after a random time, it is replaced by ${j_{1}}$ individuals of type-1, ⋯ , and $j_{d}$ individuals of type-d, with probability $p^{(a)}_{\boldsymbol{j}}$, here $\boldsymbol{j}=(j_{1},\ldots ,j_{d})$. Without loss of generality, we can assume $p^{(k)}_{\boldsymbol{e}_{k}}=0\ (k=1,\ldots ,d)$, since such split does not change the state of the system.

(3) When this system is empty, it stops, i.e., 0 is an absorbing state.

We now define the infinitesimal generator of d-type Markov branching processes, i.e., the Q-matrix.

Definition 1.1

A Q-matrix $Q = (q_{\boldsymbol{ij}}:\boldsymbol{i},\boldsymbol{j} \in \mathbf{Z}_{+}^{d})$ is called a d-type Markov branching Q-matrix (henceforth referred to as a dTMB Q-matrix), if

$$\begin{aligned} q_{\boldsymbol{ij}}= \textstyle\begin{cases} \sum \limits _{k=1}^{d} i_{k}b_{\boldsymbol{j}-\boldsymbol{i}- \boldsymbol{e}_{k}}^{(k)},\ & if \mid \boldsymbol{i}\mid > 0, \\ 0,\ &\ otherwise, \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $b^{(k)}_{\boldsymbol{j}}=0$ for $\boldsymbol{j}\notin \mathbf{Z}_{+}^{d}$ and

$$\begin{aligned} b_{\boldsymbol{j}}^{(k)}=\theta _{k}p_{\boldsymbol{j}}^{(k)} \geq 0 \ (\ \boldsymbol{j}\neq \boldsymbol{e}_{k}), \quad b^{(k)}_{ \boldsymbol{e}_{k}}=-\sum \limits _{\boldsymbol{j}\neq \boldsymbol{e}_{k}} b_{\boldsymbol{j}}^{(k)}\ (k=1,\ldots ,d). \end{aligned}$$

(1.2)

Definition 1.2

A d-type Markov branching process (henceforth referred to as dTMBP) is a continuous-time Markov chain with state space $\mathbf{Z}_{+} ^{d}$ whose transition probability function $P(t)=(p_{\boldsymbol{ij}}(t):\boldsymbol{i},\boldsymbol{j}\in \mathbf{Z}_{+}^{d})$ satisfies the Kolmogorov forward equation

$$\begin{aligned} P'(t)=P(t)Q, \end{aligned}$$

where Q is given in (1.1)–(1.2),

2 Preliminaries

In this section, we make some preliminaries related to the problem considered in this paper. For $k=1,\ldots , d$, let $R_{k}\subset \mathbf{Z}_{+}^{d}$ be finite subsets. Since if $b^{(k)}_{\boldsymbol{j}_{0}}=0$ for some $\boldsymbol{j}_{0}\in R_{k}$, then there is no individual may giving $\boldsymbol{j}_{0}$-birth, therefore, we assume $b^{(k)}_{\boldsymbol{j}}>0$ for all $\boldsymbol{j}\in R_{k}$. Also, let $r_{k}$ denote the number of elements in $R_{k}$ and $r=r_{1}+\cdots +r_{d}$. This paper is devoted to considering the probability distribution property of the number of type-k individuals giving $R_{k}$-birth until time t.

For convenience of our discussion, we only discuss the case of 2-type Markov branching process. The general case of the d-type $(d \geq 3)$ can be studied analogously.

Define

$$\begin{aligned} B_{k}(\boldsymbol{x})=\sum \limits _{\boldsymbol{j}\in {\mathbf{Z}}_{+}^{2}}b^{(k)}_{ \boldsymbol{j}}\boldsymbol{x} ^{\boldsymbol{j}},\quad \boldsymbol{x}\in [0,1]^{2},\ \ k=1,2, \end{aligned}$$

(2.1)

and

$$\begin{aligned} B_{ij}(\boldsymbol{x})= \frac{\partial B_{i}(\boldsymbol{x})}{\partial x_{j}},\quad \boldsymbol{x}\in [0,1]^{2},\ \ i,j=1,2. \end{aligned}$$

In order to avoid some trivial cases, we assume the following conditions hold.

(A-1) $(B_{1}(\boldsymbol{x}),B_{2}(\boldsymbol{x}))$ is nonsingular, i.e., there is no $2\times 2$-matrix M such that $(B_{1}(\boldsymbol{x}), B_{2}(\boldsymbol{x}))=\boldsymbol{x}M$;

(A-2) $B_{ij}(1,1)<\infty$, $i,j=1,2$;

(A-3) The matrix $(B_{ij}(1,1):i,j=1,2)$ is positively regular, i.e., there exists an integer m such that $(B_{ij}(1,1):i,j=1,2)^{m}>0$ in sense of all the elements are positive.

(A-1) guarantees that the model under consideration is not trivial. (A-2) guarantees the regularity of the process. (A-3) guarantees different type of individuals can exchange.

For any $\boldsymbol{x}\in [0,1]^{2}$, the maximal eigenvalue of $(B_{ij}(\boldsymbol{x}):i,j=1,2)$ is denoted by $\rho (\boldsymbol{x})$. The following lemma is due to Li & Wang [12], we only state it without proof.

Lemma 2.1

The system of equations

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\boldsymbol{x})=0, \\ B_{2}(\boldsymbol{x})=0, \end{cases}\displaystyle \end{aligned}$$

(2.2)

has at most two solutions in $[0,1]^{2}$. Let $\boldsymbol{q}=(q_{1},q_{2})$ denote the smallest nonnegative solution to (2.2). Then,

(i) $q_{i}$ is the extinction probability when the Feller minimal process starts at state $\boldsymbol{e}_{i}\ (i=1,2)$. Moreover, if $\rho (\boldsymbol{1})\leq 0$, then $\boldsymbol{q}=\boldsymbol{1}$; while if $\rho (\boldsymbol{1})>0$, then $\boldsymbol{q}<\boldsymbol{1}$, i.e., $q_{1}, q_{2}<1$.

(ii) $\rho (\boldsymbol{q})\leq 0$.

The following result is well known and reveals the basic property of 2-type Markov branching processes.

Lemma 2.2

Let $P(t)=(p_{\boldsymbol{i}\boldsymbol{j}}(t):\boldsymbol{i},\boldsymbol{j}\in \mathbf{Z}_{+}^{2})$ be the transition function with Q-matrix Q given in (1.1)–(1.2). Then,

$$\begin{aligned} \frac{\partial F_{\boldsymbol{i}}(t,\boldsymbol{x})}{\partial t} =B_{1}( \boldsymbol{x}) \frac{\partial F_{\boldsymbol{i}}(t,\boldsymbol{x})}{\partial x_{1}}+B_{2}( \boldsymbol{x}) \frac{\partial F_{\boldsymbol{i}}(t,\boldsymbol{x})}{\partial x_{2}}, \end{aligned}$$

where $F_{\boldsymbol{i}}(t,\boldsymbol{x})=\sum \limits _{\boldsymbol{j}\in \mathbf{Z}_{+}^{2}}p_{ \boldsymbol{i}\boldsymbol{j}}(t)\boldsymbol{x} ^{\boldsymbol{j}}$ with $\boldsymbol{x}^{\boldsymbol{j}}=x_{1}^{j_{1}}x_{2}^{j_{2}}$.

Li & Meng [16] derived the regularity criteria for 2-type Markov branching processes. Assumption (A-1) guarantees the regularity of the process.

Let $\boldsymbol{Y}(t)=(Y_{\boldsymbol{k}}(t):\boldsymbol{k}\in R_{1})$ be the number of type-1 individuals giving $R_{1}$-birth until time t and $\boldsymbol{Z}(t)=(Z_{\boldsymbol{k}}(t):\boldsymbol{k}\in R_{2})$ be the number of type-2 individuals giving $R_{2}$-birth until time t. We will discuss the probability distribution property of $(\boldsymbol{Y}(t),\boldsymbol{Z}(t))$. For this end, we define

$$\begin{aligned}& B_{1}(\boldsymbol{x},\boldsymbol{y}) = \sum \limits _{ \boldsymbol{j}\in R_{1}} b_{\boldsymbol{j}}^{(1)}\boldsymbol{x}^{ \boldsymbol{j}} y_{\boldsymbol{j}} ,\quad \bar{B}_{1}( \boldsymbol{x})= \sum \limits _{\boldsymbol{j}\in R_{1}^{c}} b_{ \boldsymbol{j}}^{(1)}\boldsymbol{x}^{\boldsymbol{j}}, \end{aligned}$$

(2.3)

$$\begin{aligned}& B_{2}(\boldsymbol{x},\boldsymbol{z}) = \sum \limits _{ \boldsymbol{j}\in R_{2}} b_{\boldsymbol{j}}^{(2)}\boldsymbol{x}^{ \boldsymbol{j}} z_{\boldsymbol{j}} ,\quad \bar{B}_{2}( \boldsymbol{x}) =\sum \limits _{\boldsymbol{j}\in R_{2}^{c}} b_{ \boldsymbol{j}}^{(2)}\boldsymbol{x}^{\boldsymbol{j}}, \end{aligned}$$

(2.4)

where $\boldsymbol{x}=(x_{1},x_{2})\in \mathbf{Z}_{+}^{2}$; $\boldsymbol{y}=(y_{\boldsymbol{j}}:\boldsymbol{j}\in R_{1})$, $\boldsymbol{z}=(z_{\boldsymbol{j}}:\boldsymbol{j}\in R_{2})$. It is obvious that $\bar{B}_{1}(\boldsymbol{x})$, $\bar{B}_{2}(\boldsymbol{x})$ are well defined at least on ${[0,1]^{2}}$. $B_{1}(\boldsymbol{x},\boldsymbol{y})$, $B_{2}(\boldsymbol{x},\boldsymbol{z})$ are well defined at least on $[0,1]^{2+r_{1}}$ and $[0,1]^{2+r_{2}}$, respectively.

Since the 2-type branching process itself cannot directly reveal the detailed multi-birth, we define a new Q-matrix $\tilde{Q} =(q_{{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}: (\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}})$ as follows:

$$\begin{aligned} q_{{(\boldsymbol{i},\boldsymbol{k},\tilde{\boldsymbol{k}}), ( \boldsymbol{j},\boldsymbol{l},\tilde{\boldsymbol{l}})}}= \textstyle\begin{cases} \sum \limits _{a=1}^{2} i_{a}b^{(a)}_{\boldsymbol{j}- \boldsymbol{i}+ \boldsymbol{e}_{a}},& if\ |~\boldsymbol{i}~|>0, \boldsymbol{l}=\boldsymbol{k}+I_{{R_{1}}}(\boldsymbol{j} \!-\! \boldsymbol{i}\!+\!\boldsymbol{e}_{1})\varepsilon _{ \boldsymbol{j} \!-\boldsymbol{i}\!+\boldsymbol{e}_{1}},\\ &\hphantom{if\ }\tilde{\boldsymbol{l}}=\tilde{\boldsymbol{k}}+I_{{R_{2}}} ( \boldsymbol{j} \!-\!\boldsymbol{i}\!+\!\boldsymbol{e}_{2}) \tilde{\varepsilon} _{\boldsymbol{j} \!-\boldsymbol{i}\!+ \boldsymbol{e}_{2}}, \\ 0,& otherwise, \end{cases}\displaystyle \end{aligned}$$

(2.5)

where $\varepsilon _{\boldsymbol{k}}\ (\boldsymbol{k}\in R_{1})$ denotes the vector in $\mathbf{Z}_{+}^{r_{1}}$ with the k’th element being 1 and the others being 0. $\tilde{\varepsilon}_{\tilde{\boldsymbol{k}}}\ ( \tilde{\boldsymbol{k}}\in R_{2})$ denotes the vector in $\mathbf{Z}_{+}^{r_{2}}$ with the $\tilde{\boldsymbol{k}}$’th element being 1 and the others being 0. $I_{{R_{1}}}$ and $I_{{R_{2}}}$ are the indicators of $R_{1}$ and $R_{2}$ respectively. It follows from the definition of Q̃, we can see that $\boldsymbol{l}=\boldsymbol{k}+\varepsilon _{\boldsymbol{j} \!- \boldsymbol{i}\!+\boldsymbol{e}_{1}}$ if and only if $\boldsymbol{j} \!-\boldsymbol{i}\!+\boldsymbol{e}_{1}\in R_{1}$, $\tilde{\boldsymbol{l}}=\tilde{\boldsymbol{k}}+ \tilde{\varepsilon}_{\boldsymbol{j} \!-\boldsymbol{i}\!+ \boldsymbol{e}_{2}}$ if and only if $\boldsymbol{j} \!-\boldsymbol{i}\!+\boldsymbol{e}_{2}\in R_{2}$. Hence, Q̃ counts the multi-birth.

It is obvious that the Q-matrix Q̃ defined in (2.5) determines a $(2+r_{1}+r_{2})$-dimensional continuous-time Markov chain $(\boldsymbol{X}(t),\boldsymbol{Y}(t),\boldsymbol{Z}(t))$, where $\boldsymbol{X}(t)$ is the 2-type Markov branching process, $\boldsymbol{Y}(t) =(Y_{\boldsymbol{k}}(t):\boldsymbol{k} \in R_{1})$ (or $\boldsymbol{Z}(t) =(Z_{\boldsymbol{k}}(t):\boldsymbol{k} \in R_{2})$) counts the number of type-1 (or type-2) individuals giving $R_{1}$-birth (or $R_{2}$-birth) until time t. We assume that $Y_{\boldsymbol{k}}(0)=0$ and $Z_{\boldsymbol{k}}(0)=0$ for all $\boldsymbol{k}\in R_{1}$ and $\boldsymbol{k}\in R_{2}$. In particular,

(1) if $R_{1}=\{\boldsymbol{0}\}$ (or $R_{2}=\{\boldsymbol{0}\}$), then $Y_{\boldsymbol{0}}(t)$ (or $Z_{\boldsymbol{0}}(t)$) counts the pure death number of type-1 (or type-2) individuals until time t.

(2) If $R_{1}=\{(n_{1},n_{2})\}$, then $Y_{(n_{1},n_{2})}(t)$ counts the $(n_{1},n_{2})$-birth number of type-1 individuals until time t.

(3) If $R_{2}=\{(n_{1},n_{2})\}$, then $Z_{(n_{1},n_{2})}(t)$ counts the $(n_{1},n_{2})$-birth number of type-2 individuals until time t.

Let $\tilde{P}(t):=(\tilde{p}_{{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t): (\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}})$ be the transition probability of $(\boldsymbol{X}(t), \boldsymbol{Y}(t),\boldsymbol{Z}(t))$. Define

$$ F_{{\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}}}(t, \boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}) =\sum \limits _{( \boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}}\tilde{p}_{{(\boldsymbol{i}, \boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j}, \boldsymbol{l},\tilde{\boldsymbol{l}})}}(t) \boldsymbol{x}^{ \boldsymbol{j}}\boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}},\quad (\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})\in [0,1]^{2+r_{1}+r_{2}}, $$

where $\boldsymbol{x}^{\boldsymbol{j}}=x_{1}^{j_{1}}x_{2}^{j_{2}}$, $\boldsymbol{y}^{\boldsymbol{l}}=\prod \limits _{\boldsymbol{m} \in R_{1}}y_{\boldsymbol{m}}^{l_{\boldsymbol{m}}}$ and $\boldsymbol{z}^{\tilde{\boldsymbol{l}}}=\prod \limits _{ \boldsymbol{m}\in R_{2}}z_{\boldsymbol{m}}^{\tilde{l}_{ \boldsymbol{m}}}$.

Lemma 2.3

Let $\tilde{P}(t)=(\tilde{p}_{{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l},\tilde{\boldsymbol{l}})}}(t): (\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l},\tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}})$ be the transition probability of $(\boldsymbol{X}(t),\boldsymbol{Y}(t),\boldsymbol{Z}(t))$. Then,

(1) for any $(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}) \in [0,1]^{2+r_{1}+r_{2}}$,

$$\begin{aligned} & \frac{\partial F_{\boldsymbol{i},\boldsymbol{0},\tilde{\boldsymbol{0}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})}{\partial t} \\ =&[B_{1}(\boldsymbol{x},\boldsymbol{y})+\bar{B} _{1}(\boldsymbol{x})] \frac{\partial F_{\boldsymbol{i},\boldsymbol{0},\tilde{\boldsymbol{0}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})}{{\partial {x_{1}}}}+[B_{2}(\boldsymbol{x},\boldsymbol{z})+\bar{B} _{2}( \boldsymbol{x})] \frac{\partial F_{\boldsymbol{i},\boldsymbol{0},\tilde{\boldsymbol{0}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})}{{\partial x_{2}}}, \end{aligned}$$

(2.6)

where $B_{1}(\boldsymbol{x},\boldsymbol{y})$, $B_{2}(\boldsymbol{x},\boldsymbol{z})$, $\bar{B} _{1}(\boldsymbol{x})$ and $\bar{B} _{2}(\boldsymbol{x})$ are defined in (2.1), (2.3)–(2.4).

(2) For any $(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}) \in [0,1]^{2+r_{1}+r_{2}}$ and $(\boldsymbol{i},\boldsymbol{k},\tilde{\boldsymbol{k}}) \in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}$,

$$\begin{aligned} F_{{\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}) =\boldsymbol{y}^{\boldsymbol{k}}\boldsymbol{z}^{\tilde{\boldsymbol{k}}} [ \boldsymbol{F}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})]^{\boldsymbol{i}}, \end{aligned}$$

(2.7)

where $\boldsymbol{F}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})=(F_{1}(t,\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}),F_{2}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}))$ with $F_{k}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})=F_{\boldsymbol{e}_{k},\boldsymbol{0}, \boldsymbol{0}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})\ (k=1,2)$.

Proof

(1) By the Kolmogorov forward equation, for any $(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}),( \boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}$,

$$ \tilde{p}'_{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}(t) =\sum \limits _{(\boldsymbol{a}, \boldsymbol{m},\tilde{\boldsymbol{m}}) \in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}} \tilde{p}_{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{a},\boldsymbol{m}, \tilde{\boldsymbol{m}})}(t) q_{(\boldsymbol{a},\boldsymbol{m}, \tilde{\boldsymbol{m}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}. $$

Multiplying $\boldsymbol{x}^{\boldsymbol{j}}\boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}}$ on both sides of the above equation and summing over $(\boldsymbol{j},\boldsymbol{l},\tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}$ yield (2.6).

(2) Let $\boldsymbol{X}_{a,k}(t)$ denote the offsprings at time t of the k’th individual of type-a at initial, $\boldsymbol{Y}_{a,k}(t)$ denote the number of $R_{1}$-birth individuals of $\boldsymbol{X}_{a,k}(t)\ (a=1,2)$ and $\boldsymbol{Z}_{a,k}(t)$ denote the number of $R_{2}$-birth individuals of $\boldsymbol{X}_{a,k}(t)\ (a=1,2)$. Then, $\{(\boldsymbol{X}_{a,k}(t),\boldsymbol{Y}_{a,k}(t), \boldsymbol{Z}_{a,k}(t)): k=1,\ldots , i_{a}; a=1,2\}$ are independent. Moreover, for $a=1,2$, $(\boldsymbol{X}_{a,k}(t),\boldsymbol{Y}_{a,k}(t), \boldsymbol{Z}_{a,k}(t))$ has the common distribution of $(\boldsymbol{X}(t),\boldsymbol{Y}(t), \boldsymbol{Z}(t))$ starting at $(\boldsymbol{e}_{a},\boldsymbol{0},\boldsymbol{0})$. Thus,

$$ \textstyle\begin{array}{l} \cr E[\boldsymbol{x}^{\boldsymbol{X}(t)}\boldsymbol{y} ^{ \boldsymbol{Y}(t)}\boldsymbol{z}^{\boldsymbol{Z}(t)}\mid ( \boldsymbol{X}(0),\boldsymbol{Y}(0),\boldsymbol{Z}(0)) =( \boldsymbol{i},\boldsymbol{k},\tilde{\boldsymbol{k}})] \\ \cr =E[\boldsymbol{x}^{\sum \limits _{a=1}^{2}\sum \limits _{k=1}^{i_{a}} \boldsymbol{X}_{a,k}(t)}\boldsymbol{y} ^{\boldsymbol{k}+\sum \limits _{a=1}^{2}\sum \limits _{k=1}^{i_{a}} \boldsymbol{Y}_{a,k}(t)} \boldsymbol{z}^{\tilde{\boldsymbol{k}} +\sum \limits _{a=1}^{2} \sum \limits _{k=1}^{i_{a}} \boldsymbol{Z}_{a,k}(t)}] \\ \cr =\boldsymbol{y}^{\boldsymbol{k}}\boldsymbol{z} ^{ \tilde{\boldsymbol{k}}}E[\prod \limits _{k=1}^{i_{1}} \boldsymbol{x} ^{\boldsymbol{X}_{1,k}(t)}\prod \limits _{k=1}^{i_{1}} \boldsymbol{y} ^{\boldsymbol{Y}_{1,k}(t)}\prod \limits _{k=1}^{i_{1}} \boldsymbol{z}^{ \boldsymbol{Z}_{1,k}(t)}\cdot \prod \limits _{k=1}^{i_{2}} \boldsymbol{x} ^{\boldsymbol{X}_{2,k}(t)}\prod \limits _{k=1}^{i_{2}} \boldsymbol{y} ^{\boldsymbol{Y}_{2,k}(t)}\prod \limits _{k=1}^{i_{2}} \boldsymbol{z}^{ \boldsymbol{Z}_{2,k}(t)}] \\ \cr =\boldsymbol{y}^{\boldsymbol{k}}\boldsymbol{z} ^{ \tilde{\boldsymbol{k}}}(E[\boldsymbol{x} ^{\boldsymbol{X}_{1,1}(t)} \boldsymbol{y} ^{\boldsymbol{Y}_{1,1}(t)}\boldsymbol{z}^{ \boldsymbol{Z}_{1,1}(t)}])^{i_{1}}\cdot (E[\boldsymbol{x} ^{ \boldsymbol{X}_{2,1}(t)}\boldsymbol{y} ^{\boldsymbol{Y}_{2,1}(t)} \boldsymbol{z}^{ \boldsymbol{Z}_{2,1}(t)}])^{i_{2}} \\ \cr =\boldsymbol{y}^{\boldsymbol{k}}\boldsymbol{z} ^{ \tilde{\boldsymbol{k}}}[\boldsymbol{F}(t,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z})]^{\boldsymbol{i}}. \end{array} $$

The proof is complete. □

The functions $B_{1}(\boldsymbol{x},\boldsymbol{y}) +\bar{B}_{1}( \boldsymbol{x})$ and $B_{2}(\boldsymbol{x},\boldsymbol{z}) +\bar{B}_{2}( \boldsymbol{x})$ will play a significant role in the later discussion. The following theorem reveals their properties.

Theorem 2.1

(1) For any $\boldsymbol{y}\in [0,1)^{r_{1}}$, $\boldsymbol{z}\in [0,1)^{r_{2}}$,

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\boldsymbol{x},\boldsymbol{y})+\bar{B}_{1}(\boldsymbol{x})=0, \\ B_{2}(\boldsymbol{x},\boldsymbol{z})+\bar{B}_{2}(\boldsymbol{x})=0 \end{cases}\displaystyle \end{aligned}$$

(2.8)

possesses exact one root in ${[0,1]^{2}}$, denoted by $\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z}):=(q_{1}(\boldsymbol{y},\boldsymbol{z}), q_{2}( \boldsymbol{y},\boldsymbol{z}))$. Moreover, $\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z})\leq \boldsymbol{q}$, where $\boldsymbol{q}=(q_{1},q_{2})$ is the minimal nonnegative solution of (2.2) given in Lemma 2.1.

(2) $q_{k}(\boldsymbol{y},\boldsymbol{z})\in {C^{\infty }}([0,1)^{r_{1}+r_{2}}) \ (k=1,2)$, and $q_{k}(\boldsymbol{y},\boldsymbol{z})$ can be expanded as a multivariate nonnegative Taylor series

$$\begin{aligned} q_{k}(\boldsymbol{y},\boldsymbol{z})=\sum \limits _{(\boldsymbol{k,l})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}} \beta ^{(a)}_{\boldsymbol{k,l}}\boldsymbol{y}^{\boldsymbol{k}} \boldsymbol{z}^{\boldsymbol{l}}, \quad (\boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}},\ \ k=1,2. \end{aligned}$$

Proof

Note that $B_{1}(\boldsymbol{1},\boldsymbol{y})+\bar{B}_{1}(\boldsymbol{1})<0$ and $B_{2}(\boldsymbol{1},\boldsymbol{z})+\bar{B}_{2}(\boldsymbol{1})<0$, by a similar argument as Lemma 2.8 in Li & Wang [12], we can prove that (2.8) possesses exact one root in ${[0,1]^{2}}$. Note that

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\boldsymbol{x},\boldsymbol{y}) +\bar{B}_{1}(\boldsymbol{x}) \leq B_{1}(\boldsymbol{x}), \\ B_{2}(\boldsymbol{x},\boldsymbol{z}) +\bar{B}_{2}(\boldsymbol{x}) \leq B_{2}(\boldsymbol{x}), \end{cases}\displaystyle \end{aligned}$$

we further know that $\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z})\leq \boldsymbol{q}$.

Next to prove (2). Integrating (2.6) yields that for $k=1,2$,

$$\begin{aligned} &\sum \limits _{(\boldsymbol{j},\boldsymbol{k}, \tilde{\boldsymbol{k}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}} \tilde{p}_{{(\boldsymbol{e}_{k},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t) \boldsymbol{x}^{\boldsymbol{j}} \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}}-\boldsymbol{x} ^{\boldsymbol{e}_{k}} \\ =&[B_{1}(\boldsymbol{x},\boldsymbol{y})+\bar{B} _{1}( \boldsymbol{x})]\int _{0}^{t} \frac{\partial F_{\boldsymbol{e}_{k},\boldsymbol{0},\boldsymbol{0}} (u,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z})}{{\partial {x_{1}}}}du \\ &+[B_{2}( \boldsymbol{x},\boldsymbol{z})+\bar{B} _{2}(\boldsymbol{x})]\int _{0}^{t} \frac{\partial F_{\boldsymbol{e}_{k},\boldsymbol{0},\boldsymbol{0}}(u,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z})}{{\partial {x_{2}}}}du. \end{aligned}$$

Since all the states $(\boldsymbol{i},\boldsymbol{l},\tilde{\boldsymbol{l}})$ with $|~\boldsymbol{i}~|>0$ are transient and all the states $(\boldsymbol{0},\boldsymbol{l},\tilde{\boldsymbol{l}})$ are absorbing, letting $\boldsymbol{x}=\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z})$ in the above equality and then letting $t\rightarrow \infty$ yield that

$$\begin{aligned} q_{k}(\boldsymbol{y}, \boldsymbol{z})=\sum \limits _{( \boldsymbol{k},\tilde{\boldsymbol{k}}) \in \mathbf{Z}_{+}^{r_{1}+r_{2}}} \tilde{p}_{{(\boldsymbol{e}_{k},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{0},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(+\infty ) \boldsymbol{y}^{ \boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}}, \quad k=1,2. \end{aligned}$$

The proof is complete. □

3 Multiple birth property

Having prepared some preliminaries in the previous section, we now consider the multiple birth property of 2-type Markov branching processes.

We first give the following theorem, which will play a key role in discussing the multiple birth property of 2-type Markov branching processes.

Theorem 3.1

Suppose that $\boldsymbol{x}\in [0,1]^{2}$, $\boldsymbol{y}\in [0,1)^{r_{1}}, [0,1)^{r_{2}}$.

(1) The differential equation

$$ \textstyle\begin{cases} \frac{\partial u_{1}}{\partial t}=B_{1}(\boldsymbol{u},\boldsymbol{y}) +\bar{B}_{1}( \boldsymbol{u}), \\ \frac{\partial u_{2}}{\partial t}=B_{2}(\boldsymbol{u},\boldsymbol{z}) +\bar{B}_{2}( \boldsymbol{u}), \\ \boldsymbol{u}(0)=\boldsymbol{x} \end{cases} $$

(3.1)

has unique solution $\boldsymbol{u}(t)=\boldsymbol{G}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})$, where

$$ \boldsymbol{u}(t)=(u_{1}(t),u_{2}(t)), \quad \boldsymbol{G}(t,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z}) =(g_{1}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}), g_{2}(t, \boldsymbol{x},\boldsymbol{y},\boldsymbol{z})). $$

(2) $\lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z})=\boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z})$, where $\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z})$ is given in Theorem 2.1.

Proof

We first prove (1). For fixed $(\boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}}$, denote

$$ \textstyle\begin{cases} H_{1}(\boldsymbol{u})=B_{1}(\boldsymbol{u},\boldsymbol{y}) + \bar{B}_{1}(\boldsymbol{u})-b_{\boldsymbol{e}_{1}}^{(1)}u_{1}, \\ H_{2}(\boldsymbol{u})=B_{2}(\boldsymbol{u},\boldsymbol{z})+ \bar{B}_{2}(\boldsymbol{u})-b_{\boldsymbol{e}_{2}}^{(2)}u_{2}. \end{cases} $$

By the assumption (A-2), we know that $H_{k}(\boldsymbol{u})$ satisfies Lipchitz condition, i.e., there exists a constant L such that for any $\boldsymbol{u}=(u_{1},u_{2})$, $\tilde{\boldsymbol{u}}=(\tilde{u}_{1}, \tilde{u}_{2})\in [0,1]^{2}$,

$$\begin{aligned} |H_{k}(\boldsymbol{u})-H_{k}(\tilde{\boldsymbol{u}})|\leq L\| \boldsymbol{u}-\tilde{\boldsymbol{u}}\|_{1},\quad k=1,2, \end{aligned}$$

For $\boldsymbol{x}\in [0,1]^{2}$, define $u_{k}^{(0)}(t)=x_{k}e^{b_{\boldsymbol{e}_{k}}^{(k)}t}\ (k=1,2)$ and

$$ u_{k}^{(n)}(t)=e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}+\int _{0}^{t} e^{-b_{ \boldsymbol{e}_{k}}^{(k)}s}H_{k}(\boldsymbol{u}^{(n-1)}(s))ds], \quad n\geq 1,\ \ k=1,2. $$

We can prove that

$$ 0\leq u^{(n)}_{k}(t)\leq 1,\quad t\geq 0, n\geq 1,\ k=1,2 $$

(3.2)

and

$$\begin{aligned} \|\boldsymbol{u}^{(n+1)}(t)-\boldsymbol{u}^{(n)}(t)\|_{1}\leq \frac{M(2L)^{n}}{(n+1)!}t^{n+1},\quad t\geq 0,\ n\geq 1. \end{aligned}$$

(3.3)

where $M:=|b^{(1)}_{\boldsymbol{e}_{1}}|+|b^{(2)}_{\boldsymbol{e}_{2}}|$. Indeed, it is obvious that $0\leq u^{(0)}_{k}(t)=x_{k}e^{b_{\boldsymbol{e}_{k}}^{(k)}t}\leq 1\ (k=1,2)$. Assume that

$$ 0\leq u^{(n)}_{k}(t)\leq 1,\ \quad t\geq 0,\ k=1,2. $$

Then it is obvious that $u^{(n+1)}_{k}(t)\geq 0$, since $H_{k}(\boldsymbol{u})\geq 0$ for all $\boldsymbol{u}\in [0,1)^{2}$. On the other hand, for $k=1,2$,

$$\begin{aligned} u_{k}^{(n+1)}(t) =& e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}+\int _{0}^{t} e^{-b_{\boldsymbol{e}_{k}}^{(k)}s}H_{k}(\boldsymbol{u}^{(n)}(s))ds] \\ \leq &e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}+\int _{0}^{t} e^{-b_{ \boldsymbol{e}_{k}}^{(k)}s}H_{k}(\boldsymbol{1})ds] \\ =&e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}-b_{\boldsymbol{e}_{k}}^{(k)} \int _{0}^{t} e^{-b_{\boldsymbol{e}_{k}}^{(k)}s}ds] \\ =&e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}+ e^{-b_{\boldsymbol{e}_{k}}^{(k)}t}-1] \\ \leq &1. \end{aligned}$$

(3.2) is proved. As for (3.3), by the definition of $\boldsymbol{u}^{(n)}(t)$,

$$\begin{aligned} |u^{(n+1)}_{k}(t)-u^{(n)}_{k}(t)| \leq & e^{b_{\boldsymbol{e}_{k}}^{(k)}t} \int _{0}^{t} e^{-b_{\boldsymbol{e}_{k}}^{(k)}s}|H_{k}( \boldsymbol{u}^{(n)}(s)) -H_{k}(\boldsymbol{u}^{(n-1)}(s))|\ ds \\ \leq & L\int _{0}^{t}\|\boldsymbol{u}^{(n)}(s)- \tilde{\boldsymbol{u}}^{(n-1)}(s)\|_{1}\ ds,\quad n\geq 1,\ k=1,2. \end{aligned}$$

Hence,

$$\begin{aligned} \|\boldsymbol{u}^{(n+1)}(t)-\boldsymbol{u}^{(n)}(t)\|_{1} \leq & 2L \int _{0}^{t}\|\boldsymbol{u}^{(n)}(s)-\tilde{\boldsymbol{u}}^{(n-1)}(s) \|_{1}\ ds,\quad n\geq 1. \end{aligned}$$

(3.4)

Note that

$$\begin{aligned} |u^{(1)}_{k}(t)-u^{(0)}_{k}(t)|=e^{b_{\boldsymbol{e}_{k}}^{(k)}t} \int _{0}^{t} e^{-b_{\boldsymbol{e}_{k}}^{(k)}s}H_{k}( \boldsymbol{u}^{(0)}(s))ds \leq |b^{(k)}_{\boldsymbol{e}_{k}}|t, \quad k=1,2, \end{aligned}$$

we know that

$$\begin{aligned} \|\boldsymbol{u}_{1}(t)-\boldsymbol{u}_{0}(t)\|_{1}\leq Mt, \end{aligned}$$

(3.5)

It follows from (3.4), (3.5) and mathematical induction that (3.3) holds.

Since

$$ u^{(n)}_{k}(t)=u^{(0)}_{k}(t)+\sum \limits _{j=1}^{n}(u^{(j)}_{k}(t)-u^{(j-1)}_{k}(t)), \quad k=1,2, $$

by (3.3), we know that $u^{(n)}_{k}(t)\ (k=1,2)$ converges uniformly in any finite interval $[0,T]$. Therefore, $u_{k}(t):=\lim \limits _{n\rightarrow \infty}u^{(n)}_{k}(t)$ exists and it can be easily checked that $\boldsymbol{u}(t)=(u_{1}(t),u_{2}(t))$ is a solution of (3.1). On the other hand, since $B_{1}(\boldsymbol{u},\boldsymbol{y})$, $\bar{B}_{1}( \boldsymbol{u})$, $B_{2}(\boldsymbol{u},\boldsymbol{z})$ and $\bar{B}_{2}(\boldsymbol{u})$ satisfy Lipchitz condition, by the differential equations theory, we know that (3.1) has unique solution. The unique solution of (3.1) is denoted by $\boldsymbol{G}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})$.

We now prove (2). For fixed $(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})\in [0,1]^{2} \times [0,1)^{r_{1}+r_{2}}$, denote

$$\begin{aligned} &f_{1}(\boldsymbol{u}):=B_{1}(\boldsymbol{u},\boldsymbol{y})+ \bar{B}_{1} (\boldsymbol{u}), \\ &f_{2}(\boldsymbol{u}):=B_{2}(\boldsymbol{u},\boldsymbol{z})+ \bar{B}_{2} (\boldsymbol{u}), \\ & \boldsymbol{G}(t)=(g_{1}(t),g_{2}(t)):=\boldsymbol{G}(t, \boldsymbol{x}, \boldsymbol{y},\boldsymbol{z}) \end{aligned}$$

for a moment.

(a) Suppose that $f_{1}(\boldsymbol{x})\geq 0$, $f_{2}(\boldsymbol{x})\geq 0$. We prove that

$$\begin{aligned} \omega :=\inf \limits _{t\geq 0}\{\min (f_{1}(\boldsymbol{G}(t)),f_{2}( \boldsymbol{G}(t)))\}\geq 0. \end{aligned}$$

Indeed, suppose that $\omega <0$. Then by the continuity of $f_{1}$, $f_{2}$ and $\boldsymbol{G}(t)$, there exist $\tilde{t}<+\infty$ and $\delta >0$ such that

$$\begin{aligned} \min (f_{1}(\boldsymbol{G}(\tilde{t})),f_{2}(\boldsymbol{G} ( \tilde{t})))=0,\quad \min (f_{1}(\boldsymbol{G}(\tilde{t})+s),f_{2}( \boldsymbol{G} (\tilde{t}+s)))< 0,\ \ \forall s\in (0,\delta ). \end{aligned}$$

(3.6)

We can assume $f_{1}(\boldsymbol{G}(\tilde{t}))=0$ without loss of generality. If $f_{2}(\boldsymbol{G}(\tilde{t}))>0$, then there exists $\tilde{\delta}\in (0,\delta )$ such that

$$\begin{aligned} f_{1}(\boldsymbol{G}(\tilde{t}+s))< 0, \quad f_{2}(\boldsymbol{G}( \tilde{t}+s))>0, \quad s\in (0,\tilde{\delta}), \end{aligned}$$

which, by (3.1), implies that

$$\begin{aligned} g_{1}(\boldsymbol{G}(\tilde{t}+s))< g_{1}(\boldsymbol{G}(\tilde{t})), \quad g_{2}(\boldsymbol{G}(\tilde{t}+s))>g_{2}(\boldsymbol{G}( \tilde{t})),\quad s\in (0,\tilde{\delta}). \end{aligned}$$

Therefore,

$$\begin{aligned} f_{1}(g_{1}(\boldsymbol{G}(\tilde{t}+s)),g_{2}(\boldsymbol{G} ( \tilde{t})))\leq f_{1}(\boldsymbol{G}(\tilde{t}+s))< 0, \quad s\in (0, \tilde{\delta}). \end{aligned}$$

(3.7)

However, it is well known that $u=g_{1}(\boldsymbol{G} (\tilde{t}))$ is the unique root of $f_{1}(u,g_{2}(\boldsymbol{G} (\tilde{t})))=0$ in $[0,1]$ with $f_{1}(u,g_{2}(\boldsymbol{G} (\tilde{t})))>0$ for $u\in [0,g_{1}(\boldsymbol{G} (\tilde{t})))$, which contradicts with (3.7). Therefore,

$$\begin{aligned} f_{1}(\boldsymbol{G}(\tilde{t}))=0,\quad f_{2}(\boldsymbol{G}( \tilde{t}))=0. \end{aligned}$$

By Theorem 2.1, $\boldsymbol{G}(\tilde{t})=\boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z})$. Hence, by (1), we know that $\boldsymbol{G}(t)=\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z})$ for $t\geq \tilde{t}$. Thus,

$$\begin{aligned} f_{1}(\boldsymbol{G}(\tilde{t}+s))=f_{2}(\boldsymbol{G}(\tilde{t}+s))=0, \quad s\geq 0, \end{aligned}$$

which contradicts with (3.6). Therefore, we have $\omega \geq 0$. Hence, $\boldsymbol{G}(t)$ is increasing in $t\geq 0$. By (3.1),

$$\begin{aligned} g_{k}(t)=e^{b_{\boldsymbol{e}_{k}}^{(k)}t}[x_{k}+\int _{0}^{t} e^{-b_{ \boldsymbol{e}_{k}}^{(k)}s}H_{k}(\boldsymbol{G}(s))ds],\quad k=1,2. \end{aligned}$$

(3.8)

Letting $t\rightarrow \infty$ in the above equality yields

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t), \boldsymbol{y}) +\bar{B}_{1}(\lim \limits _{t\rightarrow \infty} \boldsymbol{G}(t))=0, \\ B_{2}(\lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t), \boldsymbol{z}) +\bar{B}_{2}(\lim \limits _{t\rightarrow \infty} \boldsymbol{G}(t))=0. \end{cases}\displaystyle \end{aligned}$$

Therefore,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t)= \boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

(b) Suppose that $f_{1}(\boldsymbol{x})\leq 0$, $f_{2}(\boldsymbol{x})\leq 0$. We can prove that

$$\begin{aligned} \omega :=\sup \limits _{t\geq 0}\{\min (f_{1}(\boldsymbol{G}(t)),f_{2}( \boldsymbol{G}(t)))\}\leq 0. \end{aligned}$$

By a similar argument as in (a), it can be proved that $\boldsymbol{G}(t)$ is decreasing in $t\geq 0$ and

$$\begin{aligned} \lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t)= \boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

(c) Suppose that $f_{1}(\boldsymbol{x})\geq 0$, $f_{2}(\boldsymbol{x})< 0$. Let

$$\begin{aligned} \sigma =\inf \{t\geq 0: f_{1}(\boldsymbol{G}(t))\leq 0\ {\mathrm{{or}}}\ f_{2}( \boldsymbol{G}(t))\geq 0\}. \end{aligned}$$

If $\sigma <+\infty$, then $g_{1}(\boldsymbol{G}(t))$ is increasing and $g_{2}(\boldsymbol{G}(t))$ is decreasing in $[0,\sigma )$. It can be easily checked that $\boldsymbol{G}(\sigma +t)$ is the solution of (3.1) with initial condition $\boldsymbol{G}(\sigma )$. Furthermore, we have that $f_{1}(\boldsymbol{G}(\sigma ))\geq 0$, $f_{2}(\boldsymbol{G}( \sigma ))=0$ or that $f_{1}(\boldsymbol{G}(\sigma ))=0$, $f_{2}(\boldsymbol{G}(\sigma ))<0$. In the case that $f_{1}(\boldsymbol{G}(\sigma ))\geq 0$, $f_{2}(\boldsymbol{G}( \sigma ))=0$, by (a), we know that $g_{1}(\boldsymbol{G}(t))$ and $g_{2}(\boldsymbol{G}(t))$ are both increasing in $t\in [\sigma ,+\infty )$ and

$$\begin{aligned} \lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t)= \boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

while in the case that $f_{1}(\boldsymbol{G}(\sigma ))=0$, $f_{2}(\boldsymbol{G}(\sigma ))<0$, by (b), we know that $g_{1}(\boldsymbol{G}(t))$ and $g_{2}(\boldsymbol{G}(t))$ are both decreasing in $t\in [\sigma ,+\infty )$ and

$$\begin{aligned} \lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t)= \boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

If $\sigma =+\infty$, then $g_{1}(\boldsymbol{G}(t))$ is increasing and $g_{2}(\boldsymbol{G}(t))$ is decreasing in $t\geq 0$. By (3.8), we still have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty}\boldsymbol{G}(t)= \boldsymbol{q}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

(d) Suppose that $f_{1}(\boldsymbol{x})<0$, $f_{2}(\boldsymbol{x})\geq 0$. Let

$$\begin{aligned} \sigma =\inf \{t\geq 0: f_{1}(\boldsymbol{G}(t))\geq 0\ {\mathrm{{or}}}\ f_{2}( \boldsymbol{G}(t))\leq 0\}. \end{aligned}$$

A similar argument as in (c) yields the conclusion. The proof is complete. □

The following theorem gives the joint probability generating function of $(\boldsymbol{Y}(t),\boldsymbol{Z}(t))$.

Theorem 3.2

Suppose that $\{\boldsymbol{X}(t):t \geq 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0) = \boldsymbol{e}_{k}$, $(k=1\ {\mathrm{{or}}}\ 2)$. $\boldsymbol{G}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}) = (g_{1}(t,\boldsymbol{x}, \boldsymbol{y},\boldsymbol{z}),g_{2}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}))$ is the unique solution of (3.1). Then, the joint probability generating function of $(\boldsymbol{Y}(t),\boldsymbol{Z}(t))$ is given by

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(t)}\boldsymbol{z}^{\boldsymbol{Z}(t)}\mid \boldsymbol{X}(0)= \boldsymbol{e}_{k}]=g_{k}(t,\boldsymbol{1},\boldsymbol{y},\boldsymbol{z}), \quad ( \boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}},\ \ k=1,2. $$

(3.9)

In particular, the joint probability generating function of $\boldsymbol{Y}(t)$ and $\boldsymbol{Z}(t)$ are given by

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]=g_{k}(t, \boldsymbol{1},\boldsymbol{y},\boldsymbol{1}), \quad \boldsymbol{y}\in [0,1)^{r_{1}},\ \ k=1,2. $$

(3.10)

and

$$ E[\boldsymbol{z}^{\boldsymbol{Z}(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]=g_{k}(t, \boldsymbol{1},\boldsymbol{1},\boldsymbol{z}), \quad \boldsymbol{z}\in [0,1)^{r_{2}},\ \ k=1,2, $$

(3.11)

respectively.

Proof

Let $\tilde{P}(t)=(\tilde{p}_{{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t): (\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}})$ be the transition probability of $(\boldsymbol{X}(t),\boldsymbol{Y}(t),\boldsymbol{Z}(t))$. We need to prove that for any fixed $(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}) \in [0,1]^{2+r_{1}+r_{2}}$,

$$ g_{k}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})= F_{k}(t, \boldsymbol{x},\boldsymbol{y},\boldsymbol{z}),\quad k=1,2, $$

(3.12)

where $F_{k}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})\ (k=1,2)$ are given in Lemma 2.3. It is sufficient to prove that for any $(\boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}}$,

$$ u_{k}(t,\boldsymbol{x}):=F_{k}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}),\quad k=1,2. $$

is a solution of (3.1). Indeed, suppose $k=1$ without loss of generality, by Kolmogorov backward equation, for any $t \ge 0$, we have,

$$ \tilde{p}'_{{(\boldsymbol{e}_{1},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t) =\sum \limits _{(\boldsymbol{i}, \boldsymbol{k},\tilde{\boldsymbol{k}}) \in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}}q_{{( \boldsymbol{e}_{1},\boldsymbol{0}, \tilde{\boldsymbol{0}}), ( \boldsymbol{i},\boldsymbol{k},\tilde{\boldsymbol{k}})}} \tilde{p}_{{(\boldsymbol{i},\boldsymbol{k}, \tilde{\boldsymbol{k}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t). $$

Multiply $\boldsymbol{x}^{\boldsymbol{j}}\boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}}$ on both sides of the above equality and take summation over $(\boldsymbol{j},\boldsymbol{l},\tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}$, we get

$$\begin{aligned} &\sum \limits _{(\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}}) \in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}} \tilde{p}'_{{(\boldsymbol{e}_{1},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t) \boldsymbol{x}^{\boldsymbol{j}} \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}} = \sum \limits _{\boldsymbol{i}\in R_{1}}b^{(1)}_{ \boldsymbol{i}}F_{{\boldsymbol{i},\varepsilon _{ \boldsymbol{i}}, \tilde{\boldsymbol{0}}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})+\sum \limits _{\boldsymbol{i}\in R_{1}^{c}}b^{(1)}_{ \boldsymbol{i}}F_{{\boldsymbol{i}, \boldsymbol{0}, \tilde{\boldsymbol{0}}}}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}) \end{aligned}$$

By (2.7),

$$\begin{aligned} \frac{\partial F_{1}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})}{\partial t} =B_{1}(\boldsymbol{F}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}), \boldsymbol{y})+\bar{B}_{1}(\boldsymbol{F}(t, \boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})). \end{aligned}$$

By a similar argument, we have

$$\begin{aligned} \frac{\partial F_{2}(t,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})}{\partial t} =B_{2}(\boldsymbol{F}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z}), \boldsymbol{y})+\bar{B}_{2}(\boldsymbol{F}(t, \boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})). \end{aligned}$$

Note that $F_{k}(0,\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})=x_{k}\ (k=1,2)$, we know that $u_{k}(t,\boldsymbol{x})=F_{k}(t,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})\ (k=1,2)$ is a solution of (3.1).

Therefore, (3.12) and hence (3.9) hold. Finally, (3.10) and (3.11) follow directly from (3.9). The proof is complete. □

The following proposition presents the probability generating function of $(\boldsymbol{Y}(t),\boldsymbol{Z}(t))$ when the process t starts at $\boldsymbol{X}(0)=\boldsymbol{i}$.

Proposition 3.1

Suppose that $\{\boldsymbol{X}(t):t \geq 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{i}$. Then,

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(t)}\boldsymbol{z}^{\boldsymbol{Z}(t)}\mid \boldsymbol{X}(0)= \boldsymbol{i}]=[\boldsymbol{G}(t,\boldsymbol{1},\boldsymbol{y},\boldsymbol{z})] ^{\boldsymbol{i}}, \quad (\boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}}. $$

(3.13)

In particular,

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(t)}\mid \boldsymbol{X}(0)=\boldsymbol{i}]=[\boldsymbol{G}(t, \boldsymbol{1},\boldsymbol{y},\boldsymbol{1})] ^{\boldsymbol{i}}, \quad \boldsymbol{y}\in [0,1)^{r_{1}}. $$

(3.14)

and

$$ E[\boldsymbol{z}^{\boldsymbol{Z}(t)}\mid \boldsymbol{X}(0)=\boldsymbol{i}]=[\boldsymbol{G}(t, \boldsymbol{1},\boldsymbol{1},\boldsymbol{z})] ^{\boldsymbol{i}}, \quad \boldsymbol{z}\in [0,1)^{r_{2}}. $$

(3.15)

Proof

Since $E[\boldsymbol{y}^{\boldsymbol{Y}(t)}\boldsymbol{z}^{ \boldsymbol{Z}(t)}\mid \boldsymbol{X}(0)=\boldsymbol{i}]=F_{ \boldsymbol{i}, \boldsymbol{0}, \tilde{\boldsymbol{0}}}(t, \boldsymbol{1},\boldsymbol{y}, \boldsymbol{z})$, by (2.7) and Theorem 3.2, we immediately obtain (3.13). (3.14) and (3.15) follows directly from (3.13). The proof is complete. □

As direct consequences of Theorem 3.2, the following corollaries give the probability generating functions of the pure death number of type-k individuals and twins-birth number of type-k individuals.

Corollary 3.1

Suppose that $\{\boldsymbol{X}(t):t \ge 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{e}_{k}\ (k=1,2)$, $Y(t)$ and $Z(t)$ are the pure death numbers of type-1 and type-2 individuals, respectively. Then,

$$ E[y^{Y(t)}z^{Z(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,y,z), \quad y,z\in [0,1),\ k=1,2. $$

(3.16)

In particular,

$$ E[y^{Y(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,y,1),\quad y\in [0,1), \ k=1,2 $$

(3.17)

and

$$ E[z^{Z(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,1,z),\quad z\in [0,1), \ k=1,2, $$

(3.18)

where $(g_{1}(t,y,z),g_{2}(t,y,z))$ is the unique solution of the equation

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u_{1}}{\partial t}=B_{1}(u_{1},u_{2})- b_{\boldsymbol{0}}^{(1)}(1-y), \\ \frac{\partial u_{2}}{\partial t}=B_{2}(u_{1},u_{2})- b_{\boldsymbol{0}}^{(2)}(1-z), \\ u_{1}(0)=u_{2}(0)=1. \end{cases}\displaystyle \end{aligned}$$

Proof

Take $R_{1}=R_{2}=\{\boldsymbol{0}\}\subset \mathbf{Z}_{+}^{2}$. Then, we have

$$\begin{aligned} &B_{1}(\boldsymbol{u},y)+\bar{B}_{1}(\boldsymbol{u})=B_{1}( \boldsymbol{u}) -b^{(1)}_{\boldsymbol{0}}(1-y), \\ &B_{2}(\boldsymbol{u},z)+\bar{B}_{2}(\boldsymbol{u})=B_{2}( \boldsymbol{u}) -b^{(1)}_{\boldsymbol{0}}(1-z). \end{aligned}$$

By Theorem 3.2, we immediately obtain (3.16). (3.17) and (3.18) follows directly from (3.16). The proof is complete. □

Corollary 3.2

Suppose that $\{\boldsymbol{X}(t):t \ge 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{e}_{k}\ (k=1,2)$, $Y(t)$ is the $2\boldsymbol{e}_{1}$-birth numbers of type-1 individuals and $Z(t)$ is the $2\boldsymbol{e}_{2}$-birth numbers of type-2 individuals. Then,

$$ E[y^{Y(t)}z^{Z(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,y,z), \quad y,z\in [0,1),\ k=1,2. $$

In particular,

$$ E[y^{Y(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,y,1),\quad y\in [0,1), \ k=1,2 $$

and

$$ E[z^{Z(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= g_{k}(t,1,z),\quad z\in [0,1), \ k=1,2, $$

where $(g_{1}(t,y,z),g_{2}(t,y,z))$ is the unique solution of the equation

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u_{1}}{\partial t}=B_{1}(u_{1},u_{2})- b_{2\boldsymbol{e}_{1}}^{(1)}(1-y)u_{1}^{2}, \\ \frac{\partial u_{2}}{\partial t}=B_{2}(u_{1},u_{2})- b_{2\boldsymbol{e}_{2}}^{(2)}(1-z)u_{2}^{2}, \\ u_{1}(0)=u_{2}(0)=1. \end{cases}\displaystyle \end{aligned}$$

Proof

Take $R_{1}=\{2\boldsymbol{e}_{1}\}\subset \mathbf{Z}_{+}^{2}$ and $R_{2}=\{2\boldsymbol{e}_{2}\}\subset \mathbf{Z}_{+}^{2}$. Then we have

$$\begin{aligned} &B_{1}(\boldsymbol{u},y)+\bar{B}_{1}(\boldsymbol{u})=B_{1}( \boldsymbol{u}) -b^{(1)}_{2\boldsymbol{e}_{1}}(1-y)u_{1}^{2}, \\ &B_{2}(\boldsymbol{u},z)+\bar{B}_{2}(\boldsymbol{u})=B_{2}( \boldsymbol{u}) -b^{(2)}_{2\boldsymbol{e}_{2}}(1-z)u_{2}^{2}. \end{aligned}$$

By Theorem 3.2, we immediately obtain all the conclusions. The proof is complete. □

Since 0 is the absorbing state of $\{\boldsymbol{X}(t):t \geq 0\}$, now we consider the multiple birth property until the extinction of the system. Let

$$ \tau =\inf \{t\geq 0: \boldsymbol{X}(t)=\boldsymbol{0}\} $$

be the extinction time of $\{\boldsymbol{X}(t):t \geq 0\}$.

The following theorem gives the joint probability generating function of multi-birth number of individuals until the extinction of the system.

Theorem 3.3

Suppose that $\{\boldsymbol{X}(t):t \geq 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{e}_{k}\ (k=1,2)$.

(i) If $\rho (\boldsymbol{1})\leq 0$, then the probability generating function of $(\boldsymbol{Y}(\tau ),\boldsymbol{Z}(\tau ))$ is given by

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(\tau )}\boldsymbol{z}^{\boldsymbol{Z}(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]=q_{k}(\boldsymbol{y},\boldsymbol{z}),\quad ( \boldsymbol{y},\boldsymbol{z})\in [0,1)^{r_{1}+r_{2}},\ k=1,2, $$

where $(q_{1}(\boldsymbol{y},\boldsymbol{z}),q_{2}(\boldsymbol{y},\boldsymbol{z}))$ is the unique solution of

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\boldsymbol{u},\boldsymbol{y})+\bar{B}_{1}(\boldsymbol{u})=0, \\ B_{2}(\boldsymbol{u},\boldsymbol{z})+\bar{B}_{2}(\boldsymbol{u})=0. \end{cases}\displaystyle \end{aligned}$$

(ii) If $\rho (\boldsymbol{1})>0$, then the probability generating function of $(\boldsymbol{Y}(\tau ),\boldsymbol{Z}(\tau ))$ conditioned on $\tau <\infty$ is given by

$$ E[\boldsymbol{y}^{\boldsymbol{Y}(\tau )}\boldsymbol{z}^{\boldsymbol{Z}(\tau )}\mid \tau < \infty ,\boldsymbol{X}(0)=\boldsymbol{e}_{k}]= \frac{q_{k}(\boldsymbol{y},\boldsymbol{z})}{q_{k}},\quad (\boldsymbol{y}, \boldsymbol{z})\in [0,1)_{+}^{r_{1}+r_{2}},\ k=1,2, $$

where $(q_{1},q_{2})$ is the minimal non-negative solution of

$$\begin{aligned} \textstyle\begin{cases} B_{1}(\boldsymbol{u})=0, \\ B_{2}(\boldsymbol{u})=0. \end{cases}\displaystyle \end{aligned}$$

Proof

We first prove (i). It follows from Lemma 2.3(i) that for $k=1,2$ and any $(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}) \in [0,1]^{2} \times [0,1)^{r_{1}+r_{2}}$,

$$\begin{aligned} &\sum \limits _{(\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{2+r_{1}+r_{2}}} \tilde{p}_{{(\boldsymbol{e}_{k},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{j},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t) \boldsymbol{x}^{\boldsymbol{j}} \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}}-x_{k} \\ =&[B_{1}(\boldsymbol{x},\boldsymbol{y})+\bar{B} _{1}( \boldsymbol{x})]\int _{0}^{t} \frac{\partial F_{\boldsymbol{e}_{k},\boldsymbol{0},\tilde{\boldsymbol{0}}} (s,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})}{{\partial {x_{1}}}}ds+[B_{2}(\boldsymbol{x}, \boldsymbol{z})+\bar{B} _{2}(\boldsymbol{x})]\int _{0}^{t} \frac{\partial F_{\boldsymbol{e}_{k},\boldsymbol{0},\tilde{\boldsymbol{0}}}(s,\boldsymbol{x},\boldsymbol{y}, \boldsymbol{z})}{{\partial x_{2}}}ds. \end{aligned}$$

Letting $\boldsymbol{x}=\boldsymbol{q}(\boldsymbol{y},\boldsymbol{z}) =(q_{1}( \boldsymbol{y},\boldsymbol{z}),q_{2}(\boldsymbol{y}, \boldsymbol{z}))$ in the above equality and then letting $t\rightarrow \infty$ yield that

$$\begin{aligned} \sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\tilde{p}_{{(\boldsymbol{e}_{k}, \boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{0}, \boldsymbol{l},\tilde{\boldsymbol{l}})}}(\infty ) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}}-q_{k}(\boldsymbol{y}, \boldsymbol{z})=0. \end{aligned}$$

If $\rho (\boldsymbol{1})\leq 0$, then $q_{k}=P(\tau <\infty \mid \boldsymbol{X}(0)=\boldsymbol{e}_{k})=1$. Therefore, noting that $(\boldsymbol{0},\boldsymbol{l},\tilde{\boldsymbol{l}})$ is absorbing state, we have

$$\begin{aligned} &E[\boldsymbol{y}^{\boldsymbol{Y}(\tau )}\boldsymbol{z} ^{ \boldsymbol{Z} (\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}] \\ =&\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}P((\boldsymbol{Y}(\tau ), \boldsymbol{Z}(\tau )) =(\boldsymbol{l},\tilde{\boldsymbol{l}}) \mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}) \boldsymbol{y}^{ \boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =&\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\lim \limits _{t\rightarrow \infty}P(( \boldsymbol{Y}(\tau ),\boldsymbol{Z}(\tau )) =(\boldsymbol{l}, \tilde{\boldsymbol{l}}), \tau < t\mid \boldsymbol{X}(0)= \boldsymbol{e}_{k}) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =&\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\lim \limits _{t\rightarrow \infty}P(( \boldsymbol{Y}(t),\boldsymbol{Z}(t)) =(\boldsymbol{l}, \tilde{\boldsymbol{l}}), \tau < t\mid \boldsymbol{X}(0)= \boldsymbol{e}_{k}) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =&\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\lim \limits _{t\rightarrow \infty} \tilde{p}_{{(\boldsymbol{e}_{k},\boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{0},\boldsymbol{l}, \tilde{\boldsymbol{l}})}}(t) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =&\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\tilde{p}_{{(\boldsymbol{e}_{k}, \boldsymbol{0}, \tilde{\boldsymbol{0}}), (\boldsymbol{0}, \boldsymbol{l},\tilde{\boldsymbol{l}})}}(\infty ) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{ \tilde{\boldsymbol{l}}} \\ =& q_{k}(\boldsymbol{y}, \boldsymbol{z}). \end{aligned}$$

(i) is proved.

Next we prove (ii). If $\rho (\boldsymbol{1})> 0$, then $q_{k}=P(\tau <\infty \mid \boldsymbol{X}(0)=\boldsymbol{e}_{k})<1$. Therefore, similarly as the above argument, we have

$$\begin{aligned} &E[\boldsymbol{y}^{\boldsymbol{Y}(\tau )}\boldsymbol{z} ^{ \boldsymbol{Z} (\tau )}\mid \tau < \infty , \boldsymbol{X}(0)= \boldsymbol{e}_{k}] \\ =&q_{k}^{-1}\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}P(( \boldsymbol{Y}(\tau ),\boldsymbol{Z}(\tau )) =(\boldsymbol{l}, \tilde{\boldsymbol{l}}),\tau < \infty \mid \boldsymbol{X}(0)= \boldsymbol{e}_{k}) \boldsymbol{y}^{\boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =&q_{k}^{-1}\sum \limits _{(\boldsymbol{l}, \tilde{\boldsymbol{l}})\in \mathbf{Z}_{+}^{r_{1}+r_{2}}}\lim \limits _{t\rightarrow \infty}P((\boldsymbol{Y}(\tau ), \boldsymbol{Z}(\tau )) =(\boldsymbol{l},\tilde{\boldsymbol{l}}), \tau < t\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}) \boldsymbol{y}^{ \boldsymbol{l}} \boldsymbol{z}^{\tilde{\boldsymbol{l}}} \\ =& \frac{q_{k}(\boldsymbol{y},\boldsymbol{z})}{q_{k}}. \end{aligned}$$

The proof is complete. □

By Theorem 3.3, we immediately obtain the following corollaries, which gives the probability generating functions of the pure death number of type-k individuals until the extinction of the system and twins-birth number of type-k individuals until the extinction of the system.

Corollary 3.3

Suppose that $\{\boldsymbol{X}(t):t \ge 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{e}_{k}\ (k=1,2)$, $Y(t)$ and $Z(t)$ are the pure death numbers of type-1 and type-2 individuals, respectively. If $\rho (\boldsymbol{1})\leq 0$, then

$$ E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= q_{k}(y,z), \quad y,z\in [0,1),\ k=1,2. $$

If $\rho (\boldsymbol{1})>0$, then

$$ E[y^{Y(\tau )}z^{Z(\tau )}\mid \tau < \infty ,\boldsymbol{X}(0)=\boldsymbol{e}_{k}]= \frac{q_{k}(y,z)}{q_{k}},\quad y,z\in [0,1),\ k=1,2, $$

where $(q_{1}(y,z),q_{2}(y,z))$ is the unique solution of the equation

$$\begin{aligned} \textstyle\begin{cases} B_{1}(u_{1},u_{2})- b_{\boldsymbol{0}}^{(1)}(1-y)=0, \\ B_{2}(u_{1},u_{2})- b_{\boldsymbol{0}}^{(2)}(1-z)=0. \end{cases}\displaystyle \end{aligned}$$

Proof

Note $R_{1}=R_{2}=\{\boldsymbol{0}\}$, we immediately get the conclusions. □

Corollary 3.4

Suppose that $\{\boldsymbol{X}(t):t \ge 0\}$ is a 2-type Markov branching process with $\boldsymbol{X}(0)=\boldsymbol{e}_{k}\ (k=1,2)$, $Y(t)$ is the $2\boldsymbol{e}_{1}$-birth numbers of type-1 individuals and $Z(t)$ is the $2\boldsymbol{e}_{2}$-birth numbers of type-2 individuals. If $\rho (\boldsymbol{1})\leq 0$, then

$$ E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= q_{k}(y,z), \quad y,z\in [0,1),\ k=1,2. $$

If $\rho (\boldsymbol{1})>0$, then

$$ E[y^{Y(\tau )}z^{Z(\tau )}\mid \tau < \infty ,\boldsymbol{X}(0)=\boldsymbol{e}_{k}]= \frac{q_{k}(y,z)}{q_{k}},\quad y,z\in [0,1),\ k=1,2, $$

where $(q_{1}(y,z),q_{2}(y,z))$ is the unique solution of the equation

$$\begin{aligned} \textstyle\begin{cases} B_{1}(u_{1},u_{2})- b_{2\boldsymbol{e}_{1}}^{(1)}(1-y)u_{1}^{2}=0, \\ B_{2}(u_{1},u_{2})- b_{2\boldsymbol{e}_{2}}^{(2)}(1-z)u_{2}^{2}=0. \end{cases}\displaystyle \end{aligned}$$

Proof

Note $R_{1}=\{2\boldsymbol{e}_{1}\}$ and $R_{2}=\{2\boldsymbol{e}_{2}\}$, we immediately get the conclusions. □

Finally, we give an example to illustrate the main results obtained.

Example 3.1

Suppose that $\{\boldsymbol{X}(t):t\geq 0\}$ is a 2-type birth-death branching process with

$$ B_{1}(\boldsymbol{x})=p-x_{1}+qx_{2}^{2},\quad B_{2}( \boldsymbol{x})=\alpha -x_{2}+\beta x_{1}, $$

where $p,\ \alpha \in (0,1)$, $q=1-p$, $\beta =1-\alpha$. $Y(t)$ is the pure death number of type-1 individuals until time t and $Z(t)$ is the pure death number of type-2 individuals until time t. By Corollary 3.1, we know that

$$\begin{aligned} E[y^{Y(t)}z^{Z(t)}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{k}]= \textstyle\begin{cases} u(t,y,z),\ & k=1, \\ v(t,y,z),\ & k=2, \end{cases}\displaystyle \quad y,z\in [0,1), \end{aligned}$$

where $(u(t,y,z),v(t,y,z))$ is the unique solution of

$$\begin{aligned} \textstyle\begin{cases} \frac{\partial u}{\partial t}=qv^{2}-u+py, \\ \frac{\partial v}{\partial t}=\beta u-v+\alpha z, \\ u(0)=v(0)=1. \end{cases}\displaystyle \end{aligned}$$

It is easy to see that the maximum eigenvalue of $(B_{ij}(\boldsymbol{1}):i,j=1,2)$ is $\rho (\boldsymbol{1})=\sqrt{2q\beta}-1$. For $y,z\in [0,1)$, solving the equation

$$\begin{aligned} \textstyle\begin{cases} qv^{2}-u+py=0, \\ \beta u-v+\alpha z=0, \end{cases}\displaystyle \end{aligned}$$

yields that

$$\begin{aligned} &u=u(y,z)=\frac{1}{2q\beta ^{2}}[1- \sqrt{1-4q\beta (p\beta y+\alpha z)}]-\frac{\alpha z}{\beta}, \\ &v=v(y,z)=\frac{1}{2q\beta}[1-\sqrt{1-4q\beta (p\beta y+\alpha z)}]. \end{aligned}$$

By Corollary 3.3, if $2q\beta \leq 1$, then

$$\begin{aligned}& E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{1}] = \frac{1-\sqrt{1-4q\beta (p\beta y+\alpha z)}-2q\beta \alpha z}{2q\beta ^{2}}, \quad y,z\in [0,1),\\& E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{2}] =\frac{1-\sqrt{1-4q\beta (p\beta y+\alpha z)}}{2q\beta},\quad y,z\in [0,1), \end{aligned}$$

If $2q\beta >1$, then

$$\begin{aligned}& E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{1}] = \frac{1-\sqrt{1-4q\beta (p\beta y+\alpha z)}-2q\beta \alpha z}{2(1-2q\beta +q\beta ^{2})}, \quad y,z\in [0,1),\\& E[y^{Y(\tau )}z^{Z(\tau )}\mid \boldsymbol{X}(0)=\boldsymbol{e}_{2}] =\frac{1-\sqrt{1-4q\beta (p\beta y+\alpha z)}}{2(1-q\beta )},\quad y,z \in [0,1). \end{aligned}$$