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\begin{document}

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\title[从单生到单死]{从单生到单死}

\author[张余辉]{张余辉}

\date[2017年7月]{\wuhao (随机过程暑期学校, 2017年7月2日-9日) }

\institute[北师大]{\sihao 北京师范大学}

\frame{\titlepage}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%            目录            %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section*{目录}
  %  \frame{\frametitle{目录}\tableofcontents}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         背景及问题        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\large
%----------------------------------------------------------------------------------------2

%----------------------------------------------------------------------------------------2

\frame{
 \frametitle{Outline}\small
第一部分: 单生过程I
\vskip0.4cm

第二部分: 单生过程II

\vskip0.4cm
第三部分: 单死过程
}

\frame{
 \frametitle{Outline}\small
第一部分: 单生过程I
\begin{enumerate}
\item Background
\item Minimal nonnegative solution
\item Classical problems
\item Criterion on uniqueness
\item Criterion on recurrence
\item Criteria on ergodicity and strong ergodicity
\item Stationary distribution
\item Criterion on $\ell$-ergodicity
\end{enumerate}
}


%----------------------------------------------------------------------------------------4
\frame{ \frametitle{1. Background}\small
\begin{itemize}
\item $Q$-process: on $\mathbb Z_+:=\{0, 1, 2, \dots\}$, suppose that sub-Markov transition probability matrix $P(t)=(p_{ij}(t), i, j\in\mathbb Z_+)$ satisfies

(1) Normal condition: $p_{ij}(t)\ge 0,\ \sum_jp_{ij}(t)\le 1,\  i, j\in\mathbb Z_+, t\ge 0$.

(2) Chapman-Kolmogorov equation: $p_{ij}(t+s)=\sum_kp_{ik}(t)p_{kj}(s)$.

(3) Jump condition: $\lim_{t\to 0}p_{ij}(t)=\delta_{ij}, \  i, j\in\mathbb Z_+.$

(4) $Q$-condition: $\lim_{t\to 0}(p_{ij}(t)-\delta_{ij})/t=q_{ij}, \  i, j\in\mathbb Z_+,$ i.e. the $Q$-matrix $Q=(q_{ij})$ is the derivative matrix at time $0$ of $P(t)$,
 $$0\leqslant q_{ij}<\infty,\ i\ne j;\ \sum_{j\ne i}q_{ij}\leqslant -q_{ii}=:q_i\leqslant \infty,\ i\in \mathbb Z_+.$$
\item totally stable: $q_i<\infty, \ i\in\mathbb Z_+$.
\item conservative: $q_i=\sum_{j\ne i}q_{ij},\ i\in\mathbb Z_+$, i.e. $\sum_{j\in\mathbb Z_+}q_{ij}=0$.
\end{itemize}}
%----------------------------------------------------------------------------------------5
\frame{\frametitle{1. Background }\small
Single birth process: $Q=(q_{ij}:i,j\ge 0)$: $q_{i,i+1}>0,\ q_{i,i+j}=0,\ i\ge 0,j\ge 2$.
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\vskip0.3cm
$$Q=\left(%
\begin{array}{ccccc}
  - & + & 0 & 0 & \cdots \\
  * & - & + & 0 & \cdots \\
  * & * & - & + & \cdots\\
  \vdots & \vdots & \vdots & \vdots & \ddots \\
  \end{array}%
\right)$$
}
%----------------------------------------------------------------------------------------6
\frame{\frametitle{1. Background}\small

\begin{enumerate}
\item The exit boundary of the process consists at most one single extremal point
\vskip0.1cm
\item Fundamental comparison tool in studying more complex processes, such as
infinite-dimensional reaction-diffusion processes
\vskip0.1cm
\item Usually the single birth process is non-symmetric (irreversible)
\item Applications in other fields
\end{enumerate}
}
%----------------------------------------------------------------------------------------7
\frame{\frametitle{1. Background}\small
\begin{enumerate}
\item Upwardly skip-free processes (W.J. Anderson, 1991);
\item generalized birth and death processes (J.K. Zhang, 1984);
\item Population processes: birth and death processes with catastrophes; birth, death and catastrophe processes
$$\aligned i&\to i+1\\
&\to i-1\\
&\to i-2\\
&\to\cdots\\
&\to 0\endaligned
\quad
\aligned \hbox{at rate}\quad & {\blue b_i}\\
& {\blue a_i}+{ c_i}{\red f_{i,i-1}}\\
& { c_i}{\red f_{i,i-2}}\\
&\cdots\\
& { c_i}{\red f_{i0}}\endaligned$$
where $\sum_{j=0}^{i-1}{\red f_{ij}}=1$.
\end{enumerate}
}
%----------------------------------------------------------------------------------------8
\frame{\frametitle{1. Background}\small

Brockwell, Gani, Resnick and Pakes et al (1982-1986) for some special
catastrophes $f_{ij}$ ($b_i=b+\lambda i$, $a_i=0$, $c_i=ci$, $i\ge 0$).

\vskip0.2cm
geometric: $f_{ij}=p(1-p)^{i-1-j}\,(1\le j<i); f_{i0}=(1-p)^{i-1}$;
\vskip0.2cm
uniform: $f_{ij}=1/i\,(0\le j<i)$;
\vskip0.2cm
binomial: $f_{ij}=\binom{i-1}{j}p^j(1-p)^{i-1-j}\,(0\le j<i)$.

\vskip0.3cm
Aim: extinction times and probability of extinction.
\vskip0.3cm
Keys: generating function of $Q$-resolvent.
}
%----------------------------------------------------------------------------------------9
\frame{\frametitle{1. Background}\small

\vskip1cm
B. Cairns \& P. Pollett (2004).
$$\aligned i&\to i+1\\
&\to i-1\\
&\to\cdots\\
&\to 1\\
&\to 0\endaligned
\quad
\aligned \hbox{at rate}\quad & {\blue g_i} b\\
& {\blue g_i}{\red d_{1}}\\
&\cdots\\
& {\blue g_i}{\red d_{i-1}}\\
& {\blue g_i}\hbox{\red $\sum_{k\ge i}d_k$}\endaligned$$
where $b+\sum_{k\ge 1}d_k=1$.
}
%----------------------------------------------------------------------------------------10
\frame{\frametitle{1. Background}\small
single birth process:

 S.J. Yan \& M.F. Chen(1986). Multidimensional $Q$-processes, Chinese Ann. Math.,
\textbf{7(B)}(1), 90-110.

\vskip0.2cm
M.F. Chen(2004). From Markov Chains to Non-Equilibrium Particle Systems, Second Edition,
World Scientific, Singapore.

\vskip0.2cm
陈木法, 毛永华(2007). 随机过程导论, 高等教育出版社, 北京.

\vskip0.4cm
Population processes:

W.J. Anderson(1991). Continuous-Time Markov Chains, Springer-Verlag, New York.
}

%-----------------------------------------------------------6

\frame{\small
\begin{itemize}
\item Uniqueness. J.K. Zhang(1984), S.J. Yan \& M.F. Chen(1986).
\item Recurrence. Yan \& Chen(1986).
\item Extinction/return probability. P.J. Brockwell(1986), W.J. Anderson(1991), Z.T. Hou
\& Q.F. Guo(1978, 1988). Chen \& Z. (2014)
\item Ergodicity. Yan \& Chen(1986).
\item Strong ergodicity. Z.(2001).
\item Polynomial moments and stationary dist. Z.(2003, 2004, 2013)
\item Exponential moments. Z.(2003)
\item Laplace transform. Brockwell(1986), Anderson(1991)
\item Poisson equation. Chen \& Z.(2014)
\item Mean occupation time. P.S. Li \& Z.(2017)
\end{itemize}
}

%----------------------------------------------------------------------------------------11
\frame{\frametitle{2. Minimal nonnegative solution}\small
For $a_{ij}\ge 0, b_i\ge 0$, consider the equation
$$x_i=\sum_{k}a_{ik}x_k+b_i,\ i\in\mathbb Z_+.$$

\begin{block}{Theorem[the first successive approximation scheme]} The minimal solution $(x_i^*)$ always exists uniquely. Furthermore, it can be obtained by the following procedure
$$x^{(0)}_i=0,\ x^{(n+1)}_i=\sum_{k}a_{ik}x^{(n)}_k+b_i,\ i\in\mathbb Z_+,n\ge 0,$$
then $x_i^{(n)}\uparrow x_i^*$ as $n\to\infty$.
\end{block}
}
%----------------------------------------------------------------------------------------12
\frame{%\frametitle{2. Minimal nonnegative solution}
\small

\begin{block}{Comparison Theorem} Assume that $\tilde a_{ij}\ge a_{ij}, \tilde b_i\ge b_i$. Then for any solution $(\tilde x_i)$ to
$$\tilde x_i=\sum_{k}\tilde a_{ik}\tilde x_k+\tilde b_i,\ i\in\mathbb Z_+,$$
we have $\tilde x_i\ge x_i^*$.
\end{block}

\begin{block}{Theorem[the second successive approximation scheme]} Assume that $b^{(n)}_i\ge 0$. Define
$$\tilde x^{(1)}_i=b^{(1)}_i,\ \tilde x^{(n+1)}_i=\sum_ka_{ik}\tilde x^{(n)}_k+b^{(n+1)}_i,\ i\in\mathbb Z_+,n\ge 1.$$
If $b_i^{(n)}\uparrow b_i$ (or, $\sum_{n=1}^\infty b_i^{(n)}=b_i$), then $\tilde x_i^{(n)}\uparrow x_i^*$ (or $x_i^*=\sum_{n=1}^\infty \tilde x_i^{(n)}$). In particular,
if we set  $\tilde x_i^{(1)}=b_i,\ \tilde x^{(n+1)}_i=\sum_ka_{ik}\tilde x^{(n)}_k$,
then $x_i^*=\sum_{n=1}^\infty \tilde x_i^{(n)}$.
\end{block}
}
%----------------------------------------------------------------------------------------13
\frame{%\frametitle{2. Minimal nonnegative solution}
\small

\begin{block}{Localization Theorem} Let $G\subset\mathbb Z_+$ and $(\tilde x^*_i: i\in G (\text{resp.} i\in\mathbb Z_+))$ be the minimal solution to
$$x_i=\sum_{k\in G}a_{ik}x_k+\sum_{k\in G^c}a_{ik}x_k^*+b_i,\ i\in G (\text{resp.} i\in\mathbb Z_+).$$
Then we have $\tilde x_i^*= x_i^*, i\in G (\text{resp.} i\in\mathbb Z_+)$.
\end{block}

\begin{block}{Theorem} Assume that $\sum_ja_{ij}+b_i\le 1$. Then the maximal solution to
$$ x_i=\sum_{k} a_{ik} x_k+ b_i,\ 0\le x_i\le 1, i\in\mathbb Z_+,$$
can be obtained by the following procedure
$$x^{(0)}_i=1,\ x^{(n+1)}_i=\sum_{k}a_{ik}x^{(n)}_k+b_i,\ i\in\mathbb Z_+,n\ge 0,$$
then $x_i^{(n)}\downarrow x_i^*$ as $n\to\infty$.
\end{block}
}

%----------------------------------------------------------------------------------------13
\frame{%\frametitle{2. Minimal nonnegative solution}
\small

{\blue Z.T. Hou \& Q.F. Guo. Homogeneous Denumerable Markov Processes. Beijing: Science
Press, 1978(in Chinese); English translation, Beijing: Science Press and Springer, 1988}
}
%----------------------------------------------------------------------------------------14
\frame{\frametitle{3. Classical problems}\small

$Q$-process: $P'(t)|_{t=0}=Q$.

Backward Kolmogorov equation: $P'(t)=QP(t)$.

Forward Kolmogorov equation: $P'(t)=P(t)Q$.

$$p_{ij}(t)=\sum_{k\ne i}\int_0^tq_{ik}e^{-q_i(t-s)}p_{kj}(s)ds+\delta_{ij}e^{-q_it},$$
$$p_{ij}(t)=\sum_{k\ne j}\int_0^tp_{ik}(s)q_{kj}e^{-q_j(t-s)}ds+\delta_{ij}e^{-q_jt},$$

\begin{itemize}
\item $Q$-process exist? unique?
\end{itemize}

\begin{block}{Theorem} Let $Q=(q_{ij})$ be totally stable and conservative. Every $Q$-process satisfies the backward Kolmogorov equation.
\end{block}
}
%----------------------------------------------------------------------------------------15
\frame{\frametitle{3. Classical problems}\small


\begin{block}{Existence Theorem} Given a totally stable and conservative $Q$-matrix, there always exist a $Q$-process. In details, the minimal solution is a $Q$-process. Moreover, BE and FE have the same minimal solution.
\end{block}

Laplace transform: $p_{ij}(\lambda)=\int_0^\infty e^{-\lambda t}p_{ij}(t)dt$.
$$p_{ij}(\lambda)=\sum_{k\ne i}\frac{q_{ik}}{\lambda+q_i}p_{kj}(\lambda)+\frac{\delta_{ij}}{\lambda+q_i},$$
$$p_{ij}(\lambda)=\sum_{k\ne j}p_{ik}(\lambda)\frac{q_{kj}}{\lambda+q_j}+\frac{\delta_{ij}}{\lambda+q_i}.$$

Denote the minimal process by $P^{\text{min}}(\lambda)$. Consider $P(\lambda)-P^{\text{min}}(\lambda)$:
}
%----------------------------------------------------------------------------------------16

\frame{%\frametitle{3. Classical problems}
\small
$$u_i=\sum_{k\ne i}\frac{q_{ik}}{\lambda+q_i}u_{k}, \ 0\le u_i\le 1, i\in\mathbb Z_+.$$

\begin{block}{Uniqueness Theorem[Feller(1957), Reuter(1957)]} For a given totally stable and conservative $Q$-matrix $Q=(q_{ij})$, the $Q$-process is unique iff the equation has only the trivial solution $u_i\equiv0$ for some (equivalently, for all)
$\lambda>0$.
\end{block}
}
%----------------------------------------------------------------------------------------17
\frame{%\frametitle{2. Poisson equation}
\small
\begin{itemize}
\item $P(t)$ is recurrent if for each $h>0$, $P(h)$ is recurrent. Equivalently, $\int_0^\infty p_{ii}(t)dt=\infty$.
\end{itemize}

embedding chain $(\Pi_{ij})$: $\displaystyle\Pi_{ij}=I_{\{q_i\ne0\}}(1-\delta_{ij})\frac{q_{ij}}{q_i}+I_{\{q_i=0\}}\delta_{ij}.$

\begin{block}{Recurrence Theorem[Feller(1957)]} For a given totally stable and conservative $Q$-matrix $Q=(q_ij)$,
$$\int_0^\infty p^{\text{min}}_{ij}(t)dt=\sum_{n=0}^\infty \frac{\Pi_{ij}^{(n)}}{q_j}.$$
In particular, if $Q$ is irreducible and regular, then $P(t)$ is recurrent iff so is its embedding chain.
\end{block}

{\blue regular: totally stable, conservative and $Q$-process is unique.}
}
%----------------------------------------------------------------------------------------18
\frame{%\frametitle{2. Poisson equation}
\small
\begin{block}{Theorem} For a regular and irreducible $Q=(q_{ij})$,  $P(t)$ is recurrent iff for some
(equivalently, all) $j_0$,
$$
x_i=\sum_{j\ne j_0,i}\frac{q_{ij}}{q_i}x_j,\qquad 0\le x_i\le 1,\,i\ge 0
$$
has only a trivial solution.\end{block}

For embedding chain $(\Pi_{ij})$,
$(f_{i,j_0}=\mathbb P_i(\sigma_{j_0}<\infty):i\in\mathbb Z_+)$ is the minimal solution to the equation
$$x_i=\sum_{k\ne j_0}\Pi_{ik}x_k+\Pi_{i,j_0},\ i\in\mathbb Z_+.$$

$(\Pi_{ij})$ is recurrent iff $f_{i,j_0}\equiv 1$.
}

%----------------------------------------------------------------------------------------19
\frame{
\small

Let $Q=(q_{ij})$ be a regular irreducible $Q$-matrix. Then the limit
$$\lim_{t\to\infty}p_{ij}(t)=:\pi_j$$
exists and it is independent of $i$. Moreover, we have either
$\sum_j\pi_j=1$ or $\sum_j\pi_j=0$.

\begin{itemize}
\item $P(t)$ is positive recurrent or ergodic if so is $P(h)$ for every $h>0$. Equivalently, $\lim_{t\to\infty}p_{ii}(t)=\pi_i>0$.
\item $P(t)$ is strongly ergodic or uniformly ergodic if
$$\lim_{t\to\infty}\sup_i|p_{ij}(t)-\pi_j|=0.$$
\end{itemize}
}
%----------------------------------------------------------------------------------------20
\frame{\small
Let $Q=(q_{ij})$ be a regular irreducible $Q$-matrix and $H$ be a non-empty finite subset of $\mathbb Z_+$. Define
$\sigma_H=\inf\{t\ge\text{\,the first jump}: X(t)\in H\}$.

\begin{block}{Theorem[Issacson \& Arnold(1978)]}

(1) The $Q$-process is ergodic iff $\mathbb E_i\sigma_H<\infty$ for all $i\in H$.

(2) The $Q$-process is strongly ergodic iff $\sup_i\mathbb E_i\sigma_H<\infty$.
\end{block}

\begin{block}{Theorem[Tweedie(1981)]}
 The $Q$-process is ergodic (resp. strongly erergodic) iff the equation
 $$\begin{cases}\sum_jq_{ij}y_j\le -1,\ i\notin H\\\sum_{i\in H}\sum_{j\ne i}q_{ij}y_j<\infty
 \end{cases}$$
 has a finite (resp. bounded) nonnegative solution.
\end{block}

\begin{block}{Theorem($y_i=1/c(i\ne k), y_k=0$)} Given a regular and irreducible $Q$-matrix $Q=(q_{ij})$. If there exists some $k$ so that $c:=\inf_{i\ne k}q_{ik}>0$, then
the $Q$-process is str. erg.
\end{block}

}
%----------------------------------------------------------------------------------------21
\frame{\small
Assume that $Q$ is irreducible. $\pi=(\pi_i)$ is stationary distribution.
\begin{itemize}
\item Ordinary ergodicity: $\displaystyle\lim_{t\to\infty}|p_{ij}(t)-\pi_j|=0$.

\item Exponential ergodicity: $\displaystyle\lim_{t\to\infty}{\red e^{\alpha t}}|p_{ij}(t)-\pi_j|=0$.

\item Strong ergodicity: $\displaystyle\lim_{t\to\infty}\sup_i|p_{ij}(t)-\pi_j|=0$

\hskip3cm
$\Longleftrightarrow\displaystyle\lim_{t\to\infty}{\red e^{\beta t}}\sup_i|p_{ij}(t)-\pi_j|=0$.

\item Strong ergodicity $\Rightarrow$ Exponential ergodicity $\Rightarrow$
Ordinary ergodicity.
\end{itemize}
}

%----------------------------------------------------------------------------------------22
\frame{\frametitle{4. Criterion on uniqueness of SBP}\small
For $0\leq k<n$, define {\blue $q_n^{(k)}=\sum_{j=0}^k q_{nj}$} and
{\blue $$F_i^{(i)}=1, \ F_n^{(i)}=\frac 1{q_{n,n+1}} \sum_{k=i}^{n-1}q_n^{(k)}F_k^{(i)},\ 0\le i<n.$$}
Define
$$m_n
=\sum_{k=0}^n\frac{F_n^{(k)}}{q_{k,k+1}},\qquad n\ge
0.$$

\begin{block}{Uniqueness[J.K. Zhang(1984); S.J. Yan \& M.F. Chen(1986)]}
Given a totally stable and conservative single birth $Q$-matrix $Q=(q_{ij})$, the process is unique $($non-explosive$)$ iff
$\sum_{n=0}^\infty m_n=\infty$.
\end{block}
\begin{block}{}
$$(\lambda+q_i)u_i=\sum_{k\ne i}q_{ik}u_{k}, \ 0\le u_i\le 1, i\in\mathbb Z_+.$$
\end{block}
}
%----------------------------------------------------------------------------------------23
\frame{\small
For birth-death $(a_i,b_i)$: $a_i=q_{i,i-1}>0, i\ge 1; b_i=q_{i,i+1}>0, i\ge 0$.
$$m_n=\frac 1{\mu_nb_n}\mu[0,n],\qquad n\ge 0,$$
where
$$\mu_0=1, \qquad \mu_i=\frac{b_0b_1\cdots b_{i-1}}{a_1a_2\cdots a_i},\quad i\ge 1,$$
and $\mu[i,k]=\sum_{j=i}^k\mu_j$.

\begin{block}{Uniqueness Criterion for BDP}
Given a totally stable and conservative birth-death $Q$~matrix $(a_i, b_i)$. The the birth-death is unique iff
$R:=\sum_{n=0}^\infty \frac 1{\mu_nb_n}\mu[0,n]=\infty$.\end{block}

}
%----------------------------------------------------------------------------------------24
\frame{\frametitle{5. Criterion on recurrence of SBP}\small


\begin{block}{Recurrence[Yan \& Chen(1986)]} Assume the single birth $Q$-matrix $Q=(q_{ij})$ is regular and irreducible.
Then the process is recurrent iff $\sum_{n=0}^\infty F_n^{(0)}=\infty$.
\end{block}
\begin{block}{}
$$
x_i=\sum_{j\ne 0,i}\frac{q_{ij}}{q_i}x_j,\qquad 0\le x_i\le 1,\,i\ge 0
$$
\end{block}

For birth-death $Q$~matrix $(a_i,b_i)$:

$$F_n^{(0)}=\frac{b_0}{\mu_nb_n},\ n\ge 0.$$

\begin{block}{Recurrence Criterion for BDP}
Given a regular birth-death $Q$-matrix $(a_i, b_i)$. The the birth-death is recurrent iff
$\sum_{n=0}^\infty \frac{1}{\mu_nb_n}=\infty$.
\end{block}
}
%----------------------------------------------------------------------------------------23
\frame{\frametitle{6. Criteria on ergodicity and strong ergodicity}\small
Define
$$d_n=\sum_{k=1}^n\frac{F_n^{(k)}}{q_{k,k+1}},\qquad n\ge 0.$$

\begin{block}{Ergodicity[Yan \& Chen(1986); Z.(2001)]}
Assume the single birth $Q$-matrix $Q=(q_{ij})$ is regular and irreducible. Then
the process is ergodic iff
$d:=\sup_{k\ge 0}\frac{\sum_{n=0}^k d_n}{\sum_{n=0}^k F_n^{(0)}}<\infty$;
the process is strongly ergodic iff
$S:=\sup_{k\ge 0}\sum_{n=0}^k(F^{(0)}_nd-d_n)<\infty$.
\end{block}
\begin{block}{$H=\{0\}$}
$$\sum_{j\ne i} q_{ij}(y_j-y_i)+1\le 0,\ i\ge 1.$$
\end{block}
}

%----------------------------------------------------------------------------------------28
\frame{\small
For birth-death $Q$~matrix $(a_i,b_i)$:
$$d_n=\frac 1{\mu_nb_n}\mu[1,n],\qquad n\ge 0.$$
\begin{block}{Ergodicity Criteria for BDP}
Given a regular birth-death $Q$-matrix $(a_i, b_i)$. The the birth-death chain is ergodic iff
$\mu[0,\infty)<\infty$; it is strongly ergodic iff $S:=\sum_{n=0}^\infty\frac 1{\mu_n b_n}\sum_{k=n+1}^\infty\mu_k
<\infty$.
\end{block}

H.J. Zhang, X. Lin \& Z.T. Hou, 2000.
}
%----------------------------------------------------------------------------------------29
\frame{\frametitle{7. Stationary distribution}\small
\begin{block}{Stationary distribution(Y.H. Zhang, 2004)} Assume that the process is ergodic. Define
$$c_k:=\varlimsup_{i\to\infty}\frac{\sum_{j=k}^i m_j}{\sum_{j=k}^iF_j^{(k)}}.\quad \text{Then}\quad
\pi_k=\frac 1{q_{k,k+1}c_k},\quad k\ge 0.$$
\end{block}
\begin{block}{Example 1} $b_i=q_{i,i+1}>0\,(i\ge 0)$, $a_i=q_{i,i-1}>0\,(i\ge 1)$.
Define $\mu_0=1$, $\mu_i=b_0\cdots b_{i-1}/a_1\cdots a_i\,(i\ge 1)$ and $\mu=\sum_{i=0}^\infty\mu_i$.
Then $c_i=\mu/(\mu_ib_i)$, $\pi_i=\mu_i/\mu\,(i\ge 0)$.
\end{block}
\begin{block}{Example 2} $q_{i,i+1}=1\,(i\ge 0)$ and $q_{i0}=1\,(i\ge 1)$.
Then $c_k=2^{k+1}$, $\pi_k=2^{-k-1}\,(k\ge 0)$.\end{block}
}
%----------------------------------------------------------------------------------------30
\frame{\small

\begin{block}{Example 3} $q_{i,i+1}=1\,(i\ge 0)$, $q_{10}=1$ and $q_{i,i-2}=1\,(i\ge 2)$. Then $\{F_j^{(k)}\}_{j=k}^\infty$ is a Fibonacci sequence:
$F_{k+n}^{(k)}=\frac 1{\sqrt 5}[A^{n+1}-(-B)^{n+1}],\ n\ge 0, k\ge 0$,
where $A=(\sqrt 5 +1)/2, B=(\sqrt 5 -1)/2$.
Then $c_k=A^{k+2}$, $\pi_k=B^{k+2}\,(k\ge 0)$.\end{block}

\begin{block}{Example 4(uniform catastrophes)} $q_{i,i+1}=\lambda_i:=a+\lambda i\,(i\ge 0)$,
$q_{ij}=\beta\,(0\le j<i)$, where $a>0$, $\lambda>0$, $\beta>0$. Then
$$\pi_0=\frac{\beta}{a+\beta},\ \pi_k= \frac{(k+1)\beta}{a+\beta}\prod_{j=0}^{k-1}
\frac{\lambda_j}{\lambda_{j+1}+(j+2)\beta},\ k\ge1.$$
In particular, when $a=\lambda$, then $\pi_k=\beta a^k/(a+\beta)^{k+1}\,(k\ge0)$.
Furthermore, when $a=\lambda=\beta=1$, then $\pi_k=2^{-k-1}\,(k\ge0)$.
\end{block}
}
%----------------------------------------------------------------------------------------31
\frame{\small

Define $\sigma_j=\inf\{t\ge$ the first jumping time: $X(t)=j\}$ and
$$\tau_j=\inf\{t\ge 0: X(t)=j\}.$$
\begin{block}{Theorem[Z.(2004)]} Fix $i_0\ge 0$. For recurrent single process, we have
$$
\mathbb E_i\tau_{i_0}=\sum_{j=i}^{i_0-1}m_j,\ i<i_0;$$
$$
\mathbb E_i\tau_{i_0}=\sum_{j=i_0}^{i-1}(F_j^{(i_0)}c_{i_0}-m_j),\  i\ge i_0+1.
$$
\end{block}
}
%----------------------------------------------------------------------------------------32
\frame{\small
\begin{block}{} For recurrent $Q$-process, $(\mathbb E_i\tau_{i_0}:i\in\mathbb Z_+)$ is the minimal solution to the equation
$$x_{i_0}=0, \ x_i=\sum_{j\ne i,i_0}\frac{q_{ij}}{q_i}x_j+\frac 1{q_i},\ i\ne i_0.$$
\end{block}


$$\mathbb E_i\sigma_i=\frac{q_{i,i+1}c_i}{q_i},\ i\ge 0.$$

$$\pi_i q_i\mathbb E_i\sigma_i=1\Rightarrow\pi_i=\frac 1{q_{i,i+1}c_i}.$$
\begin{block}{}
$$
\mathbb E_1\tau_0=d,\ \mathbb E_i\tau_0=\sum_{j=0}^{i-1}(F_j^{(0)}d-d_j),
\ i\ge 1.$$
$$
\mathbb E_0\sigma_0=q_{01}^{-1}+\mathbb E_1\tau_0=q_{01}^{-1}+d.
$$
\end{block}
}

%----------------------------------------------------------------------------------------33
\frame{\small
Probabilistic meaning:

\begin{itemize}

\item Uniqueness. $R=\mathbb E_0\sigma_\infty=\infty$.

\item Recurrence. $\mathbb P_0(\sigma_0<\infty)=1-(\sum_{n=0}^\infty F_n^{(0)})^{-1}=1$.

\item Ergodicity. $d=\mathbb E_1\tau_0<\infty\Leftrightarrow\mathbb E_0\sigma_0<\infty$.

\item Strong ergodicity. $S=\sup_{i\ge 1}\mathbb E_i\tau_0<\infty$.\end{itemize}
}
%----------------------------------------------------------------------------------------34

\frame{\frametitle{8. Criterion on $\ell$-ergodicity}\small
\begin{block}{}
For a positive integer $\ell$, the recurrent chain $P(t)$ is said to be
$\ell$-ergodic if {\red
$\mathbb E_j\sigma_j^\ell<\infty$} for some (and hence for all) $j\in E$.
\end{block}
\begin{block}{}
{\blue $1$}-ergodic $=$ positive recurrent (ergodic), {\blue $0$}-ergodic $=$ null recurrent.
\end{block}

Discrete time:
\begin{itemize}
\item J.G. Kemeny, J.L. Snell and A.W. Knapp, 1976
\item Y.H. Mao, 2003
\item Z.T. Hou and Y.Y. Liu, 2003. Queue Theory
\end{itemize}

Continuous time:
\begin{itemize}
\item P. Coolen-Schrijner and E.A. van Doorn, 2002
\item Y.H. Mao, 2004
\end{itemize}
}
%----------------------------------------------------------------------------------------35
\frame{\small

\begin{itemize}
\item $\ell$-ergodicity provides an {\red algebraic} convergence rate:
$p_{ij}(t)-\pi_j=o({\blue t^{-(\ell-1)}})$ as $t\to\infty$.
\item Given a regular birth-death $(a_i,b_i)$. Assume that $P(t)$ is recurrent.
Then $P(t)$ is $\ell$-ergodic iff $\sum_{i=1}^\infty\mu_i\mathbb E_{i}\sigma_0^{\ell-1}<\infty$,

$$
\mathbb E_{i}\sigma_0^{n}=n\sum_{j=0}^{i-1}\frac1{\mu_jb_j}\sum_{k=j+1}^\infty\mu_k\mathbb E_{k}\sigma_0^{n-1},\ i\ge 1,\,n\ge 1.
$$
\end{itemize}

\begin{block}{}
$Q$-process is $\ell$-ergodic iff $\sum_{i=1}^\infty\pi_i \mathbb E_{i}\sigma_0^{\ell-1}<\infty$.
\end{block}

\begin{block}{}
$Q$-process is $\ell$-regodic if and only if for some (and then for all) $j\in E$,
the following system has a nonnegative finite solution
$$\begin{cases}
\sum_k q_{ik}y_k\le-\ell \mathbb E_{i}\sigma_j^{\ell-1},&i\ne j,\\
\sum_{k\ne j}q_{jk}y_k<\infty.
\end{cases}$$

\end{block}
}
%----------------------------------------------------------------------------------------36
\frame{\small
Define
$$d^{(\ell)}_n=\sum_{k=1}^n\frac{F_n^{(k)}\mathbb E_{k}\sigma_0^{\ell-1}}{q_{k,k+1}},
\ n\ge 0; d^{(\ell)}:=\varlimsup_{i\to\infty }\frac{\sum_{j=0}^id^{(\ell)}_j}{\sum_{j=0}^iF_j^{(0)}}.$$

\begin{block}{Theorem} Given a regular irreducible single birth $Q$-matrix $Q=(q_{ij})$. Assume that the $Q$-process is $(\ell-1)$-ergodic for some $\ell\ge 1$. Then
the single birth process is $\ell$-ergodic if and only if $d^{(\ell)}<\infty$.
\end{block}

\begin{block}{} Given a regular irreducible $Q$-matrix $Q=(q_{ij})$. Assume that the $Q$-process is $(n-1)$-ergodic
Then $(\mathbb E_{i}\sigma_0^{n})$ is the minimal solution to the equation:
$$
x_0=0, \ x_i=\sum_{j\ne i}\frac{q_{ij}}{q_i}x_j+\frac n{q_i} \mathbb E_{i}\sigma_0^{n-1},\ i\ge 1.
$$
\end{block}
}
%----------------------------------------------------------------------------------------38
\frame{\small
第二部分: 单生过程II

\begin{enumerate}
\item Poisson equation
\item Uniqueness
\item Recurrence, extinction/return probability
\item Ergodicity, strong ergodicity, and the first moment of return time
\item Polynomial moments of hitting time and life time
\item Exponential ergodicity and Laplace transform of return time
\item Examples
\item Related topics and applications
\end{enumerate}
}
%----------------------------------------------------------------------------------------39
\frame{\frametitle{1. Poisson equation}\small
Given a function $c$, define
$${\blue \Omega g=Qg+cg,} \quad \text{where}\quad (Qg)_i=\sum_{j}q_{ij}(g_j-g_i).$$
Clearly, if $c\le 0$, then $\Omega$ is an operator corresponding to a single birth
process with killing rates $(-c_i)$. Define
{\blue $$
\widetilde F_i^{(i)}=1,\qquad  \widetilde F_n^{(i)}=\frac 1{q_{n,n+1}}
\sum_{k=i}^{n-1}\tilde q_n^{(k)}\widetilde F_k^{(i)},\qquad n>i\ge 0,$$
$$\tilde q_n^{(k)}=q_n^{(k)}-c_n:=\sum_{j=0}^kq_{nj}-c_n,\qquad 0\le k<n.$$}
}

%----------------------------------------------------------------------------------------40
\frame{\small

\begin{block}{Theorem 1} Given a single-birth $Q$-matrix $Q=(q_{ij})$ and functions $c$ and $f$,
the solution $g$ to the Poisson equation
$$
\Omega g=f
$$
has the following representation:
$$
g_n=g_0+\sum_{0\le k\le n-1}\sum_{0\le j\leq k}\frac{{\widetilde F}_k^{(j)}(f_j-c_jg_0)}{q_{j,j+1}},\quad n\ge 0.
$$
In particular, the harmonic function of $\Omega$ can be represented as
$$g_n=g_0\Bigg(1-\sum_{0\le k\le n-1}\sum_{0\le j\le k}\frac{\widetilde F_k^{(j)}c_j}{q_{j,j+1}}\Bigg),\quad n\ge 0.$$\end{block}
}
%---------------------------------------------------------41
\frame{\small
\noindent{\it Proof}\quad For each $i\ge 0$, we have
\begin{align}
(\Omega g)_i&=q_{i,i+1}(g_{i+1}-g_i)-\sum_{0\le j\le i-1} q_{ij}\sum_{k=j}^{i-1}(g_{k+1}-g_k)+c_ig_i\nonumber\\
&=q_{i,i+1}(g_{i+1}-g_i)-\sum_{0\le j\le i-1}\sum_{j=0}^{k}q_{ij}(g_{k+1}-g_k)+c_ig_i\nonumber\\
&=q_{i,i+1}(g_{i+1}-g_i)-\sum_{0\le j\le i-1}\Bigg(\sum_{j=0}^{k}q_{ij}-c_i\Bigg)(g_{k+1}-g_k)+c_ig_0\nonumber\\
&=q_{i,i+1}(g_{i+1}-g_i)-\sum_{0\le j\le i-1}\tilde q_i^{(k)}(g_{k+1}-g_k)+c_ig_0.
\nonumber\end{align}
$$g_{i+1}-g_i=\frac 1{q_{i,i+1}}\Bigg(\sum_{0\le k \le i-1}\tilde q_{i}^{(k)}(g_{k+1}-g_k)+f_i-c_ig_0\Bigg).$$
}

%--------------------------------------------------------------------42
\frame{\small
$$g_{i+1}-g_i=\sum_{j=0}^i\frac{\widetilde F_i^{(j)}(f_j-c_jg_0)}{q_{j,j+1}},\qquad i\ge 0.$$

\begin{block}{Proposition1} For given $f$, the sequence $(h_n)$ defined successively by
$$h_n=\frac{1}{q_{n, n+1}}\bigg(f_n+\sum_{i\le k \le n-1}{\tilde q}_n^{(k)} h_k\bigg),\qquad n\ge i$$
has an unified expression as follows
$$h_n=\sum_{k=i}^n \frac{{\widetilde F}_n^{(k)}f_k}{q_{k, k+1}} ,\qquad n\ge i. $$
\end{block}
}
%----------------------------------------------------------------------------43
\frame{\frametitle{1. Poisson equation}\small
$${\widetilde F}_i^{(i)}=1,\quad  {\widetilde F}_n^{(i)}=\sum_{k=i+1}^n
\frac{{\widetilde F}_n^{(k)}{\tilde q}_k^{(i)}}{q_{k, k+1}},\quad n\ge i+1. $$

Roughly speaking, the unified treatment consists of the following three steps.
\begin{itemize}
\item[(a)] Find out the Poisson equation corresponding to the problem .
\item[(b)] Apply Theorem 1 to get the solution to Poisson equation.
\item[(c)] Work out a criterion for the problem using the solution obtained in (b).
\end{itemize}

{\blue M.F. Chen \& Z.(2014): Unified representation of formulas for single birth processes, Frontiers of Mathematics in China,
9(4): 761-796.}
}
%---------------------------------------------44
\frame{\small\vskip0.5cm
\begin{center}{\renewcommand{\arraystretch}{1.2}
\begin{tabular}{|c|c|c|}
  \hline
  Problem                  & $c_i\in {\mathbb R}$                   & $f_i\in {\mathbb R}$
  \\ \hline
  Harmonic function               & $c_i\in {\mathbb R}$   &
  $f_i\equiv 0$ \\ \hline
  Uniqueness                & $c_i\equiv -\lambda<0$   & $f_i\equiv 0$ \\ \hline
  Recurrence                & $c_i\equiv 0$            & $f_i=q_{i0}(1-\delta_{i0})$\\ \hline
  Ext./return probability    &$c_i\equiv 0$             & $f_i=q_{i0}(1-\delta_{i0})(g_0-1)$ \\ \hline
  Ergodicity                & $c_i\equiv 0$            & $f_i=q_{i0}(1-\delta_{i0})g_0-1$\\ \hline
  Strong ergodicity         & $c_i\equiv 0$            & $f_i=q_{i0}(1-\delta_{i0})g_0-1$\\ \hline
  Polynomial moment         & $c_i\equiv 0$            & $f_i^{(\ell)}$\\ \hline
  Exp. moment/ergodicity        & $c_i\equiv\lambda>0$     & $f_i=q_{i0}(1-\delta_{i0})(g_0-1)$ \\ \hline
  Laplace transform  & $c_i\equiv -\lambda<0$ & $f_i=q_{i0}(1-\delta_{i0})(g_0-1)$ \\ \hline
\end{tabular}}\end{center}
where $f_i^{(\ell)}=q_{ii_0}(1-\delta_{ii_0})g_{i_0}-\ell\mathbb E_i\sigma_{i_0}^{\ell-1}$.
}
%----------------------------------------------------------------------------------45
\frame{\frametitle{2. Uniqueness}\small
Define
$${\widetilde m}_n
=\sum_{k=0}^n\frac{{\widetilde F}_n^{(k)}}{q_{k,k+1}},\qquad n\ge 0.$$


\begin{block}{Proposition}
Given a totally stable and single birth conservative $Q$-matrix $Q=(q_{ij})$, the process is unique $($non-explosive$)$ iff
$\sum_{n=0}^\infty m_n=\infty$.\end{block}
}

%----------------------------------------------------------46
\frame{\small
Proof.  unique iff the solution $(u_i)$ to the equation
$$(\lambda +q_i)u_i=\sum_{j\ne i}q_{ij}u_j, \qquad i\ge 0;\qquad u_0=1
$$
is unbounded for some (equivalently for all) $\lambda>0$. Rewrite it as
$$\Omega u=Qu- \lambda u=0;\qquad u_0=1.$$
Applying Theorem 1 to $c_i\equiv -\lambda$ and $f_i\equiv 0$, we obtain:
$$u_n=1+\lambda\sum_{0\le k \le n-1}\sum_{j=0}^k\frac{\widetilde F_k^{(j)}}{q_{j,j+1}}=1+\lambda\sum_{0\le k \le n-1}\widetilde m_k,\qquad n\ge 0.$$
Clearly, $u_n$ is increasing in $n$ and then is unbounded iff $\sum_n {\widetilde  m}_n=\infty$. Thus,
it remains to show that $\sum_n {\widetilde  m}_n=\infty$
iff $\sum_n {m}_n=\infty$.
}
%----------------------------------------------------------47
\frame{\frametitle{3. Recurrence, extinction/return probability}\small


\begin{block}{Proposition} Assume the single birth $Q$-matrix $Q=(q_{ij})$ is non-explosive and irreducible. Then the process is recurrent iff $\sum_{n=0}^\infty F_n^{(0)}=\infty$.\end{block}

Proof. recurrent iff the equation
$$
x_i=\sum_{k\ne 0}\Pi_{ik}x_k,\qquad 0\le x_i\le 1,\qquad i\ge 0
$$
has only zero solution, where $\Pi_{ik}=(1-\delta_{ik})q_{ik}/q_i$. It is easily seen that the equation has a non-trivial solution iff the equation
$$x_i=\sum_{k\ne 0}\Pi_{ik}x_k,\qquad i\ge 0; \qquad x_0=1$$
has a nonnegative bounded solution.
}

%----------------------------------------------------------48
\frame{\small

Rewrite the previous equation as
$$(Qx)_0=0,\qquad (Qx)_i=q_{i0},\qquad i\ge 1;\qquad x_0=1.$$
Applying Theorem 1 to $c_i\equiv 0$ and $f_i=q_{i0}(1-\delta_{i0})$, we obtain
the unique solution as follows
$$x_0=1,\quad x_n=1+\sum_{k=1}^{n-1}F_k^{(0)}=\sum_{k=0}^{n-1}F_k^{(0)},\quad n\ge 1.$$
Clearly, $(x_n)$ is bounded iff $\sum_{k=0}^\infty F_k^{(0)}<\infty$. In other words,
the equation has only a trivial solution iff $\sum_{k=0}^\infty F_k^{(0)}=\infty$.
}
%----------------------------------------------------------49
\frame{\small

P. J. Brockwell(1986), W. J. Anderson(1991). Hou \& Guo(1978, 1988)

\begin{block}{Proposition}Let the single birth $Q$-matrix $Q=(q_{ij})$ be non-explosive and irreducible. Then the return/extinction probability is as follows:
$$\mathbb P_0(\sigma_0<\infty)=\frac{\sum_{k=1}^{\infty}F_k^{(0)}}{\sum_{k=0}^\infty F_k^{(0)}},\quad \mathbb P_n(\sigma_0<\infty)=\frac{\sum_{k=n}^{\infty}F_k^{(0)}}{\sum_{k=0}^\infty F_k^{(0)}},\quad n\ge 1.$$
Furthermore, $\mathbb P_n(\sigma_0<\infty)=1$ for all $n\ge 0$ iff $\mathbb P_0(\sigma_0<\infty)=1$, equivalently iff $\sum_{n=0}^\infty F_n^{(0)}=\infty$.
\end{block}
}
%----------------------------------------------------------50
\frame{\small
Proof. $(\mathbb P_i(\sigma_0<\infty): i\in E)$ is the minimal nonnegative solution to the equation
$$x_i=\sum_{k\ne 0, i}\frac{q_{ik}}{q_i}x_k + \frac{q_{i0}}{q_i}(1-\delta_{i0}),\qquad i\in E.$$
The last equation is equivalent to
$$(Qx)_i=q_{i0}(1-\delta_{i0})(x_0-1),\qquad i\ge 0.$$
Applying Theorem 1 to $c_i\equiv 0$ and $f_i= q_{i0}(1-\delta_{i0})(x_0-1)$,
we obtain the solution to the last equation:
$$ x_n=x_0\sum_{0\le k\le n-1} F_k^{(0)}-\sum_{1\le k\le n-1}F_k^{(0)},\qquad n\ge 0.$$
$$x_0^*=1-\frac 1{\sum_{k=0}^\infty F_k^{(0)}},\qquad x_n^*=1-\frac{\sum_{k=0}^{n-1}F_k^{(0)}}{\sum_{k=0}^\infty F_k^{(0)}},\qquad n\ge 1.$$
}
%----------------------------------------------------------------21
\frame{\small
Consider the occupation time of the state $j$:
 $$T_j:=\int_0^\infty \mathbf 1_{\{X(t)=j\}}\textrm{d}\,t.$$
  By the return probability and the embedding chain, we get

 \begin{block}{Proposition[P.S. Li \& Z.(2017)]} Let the single birth $Q$-matrix $Q=(q_{ij})$ be regular and irreducible. Then the mean occupation time
$$\mathbf E_iT_j= \frac{1}{q_{j,j+1}}\sum^\infty_{n=i\vee j} F_n^{(j)}.$$
\end{block}
}
%----------------------------------------------------------51
\frame{\frametitle{\small 4. Ergodicity, strong ergodicity, and the first moment of return time}\small
Define
$$\tilde d_n=\sum_{1\le j\le n}\frac{\widetilde F_n^{(j)}}{q_{j,j+1}},\quad n\ge 0.$$

\begin{block}{Proposition[Yan \& Chen(1986); Z.(2001, 2003)]} Assume that the single birth $Q$-matrix $Q=(q_{ij})$ is irreducible and corresponding process is recurrent. Then
$$\mathbb E_0\sigma_0=\frac 1{q_{01}}+d,\qquad \mathbb E_n\sigma_0=\sum_{k=0}^{n-1}\big(F_k^{(0)}d-d_k\big),\quad n\ge 1,$$\end{block}
}

%----------------------------------------------------------52
\frame{\small
\begin{block}{Proposition(contin.)}
where
$$d=\varlimsup_{k\to\infty} \frac{\sum_{n=0}^k d_n}{\sum_{n=0}^k F_n^{(0)}}
=\lim_{n\to\infty} \frac{d_n}{F_n^{(0)}}\;\;\text{\rm if the limit exists}.$$
Furthermore, the process is ergodic $($i.e. positive recurrent$)$ iff $d<\infty$;
and it is strongly ergodic iff $\sup_{k\in E} \sum_{n=0}^k\big(F^{(0)}_nd-d_n\big)<\infty$.
Actually, for the last conclusion, the recurrence assumption can be replaced by the
uniqueness one.\end{block}
}

%----------------------------------------------------------53
\frame{\small

Proof. $({\mathbb E}_i \sigma_0: i\in E)$ is
the minimal nonnegative solution $(x_i^*)$ to the equation
$$
x_i=\frac{1}{q_i}\sum_{k\notin  \{0, i\}}
  q_{ik}x_k+\frac{1}{q_i}, \qquad i\in E. $$
Then, we have
$$(Qx)_i=q_{i0}x_0-1,\qquad i\ge 1;\qquad (Qx)_0=-1.$$
Applying Theorem 1 to $c= 0$ and $f_i=q_{i0}(1-\delta_{i0})x_0-1$ ($i\ge 0$),
we obtain the solution to the last equation:
$$x_n=\sum_{k=0}^{n-1}\bigg[F_k^{(0)}\bigg(x_0-\frac{1}{q_{01}}\bigg)-d_k\bigg],\quad n\ge 1.$$
}

%---------------------------------------------------------54
\frame{\small
$$x_0^*
=\frac 1{q_{01}}+\sup_{n\ge 1}\frac{\sum_{k=0}^{n-1} d_k}{\sum_{k=0}^{n-1}F_k^{(0)}}$$
and
$$x_n^*=\sum_{k=0}^{n-1}\left(F_k^{(0)}\sup_{n\ge 1}\frac{\sum_{j=0}^{n-1} d_j}{\sum_{j=0}^{n-1}F_j^{(0)}}
-d_k\right),\quad n\ge 1.$$
}
%---------------------------------------------------------------------------------55
\frame{\frametitle{\small 5. Polynomial moments of hitting time and life time}\small
\begin{block}{Proposition[Z.(2004, 2013); Chen \& Z.(2014)]}  Assume that the single birth $Q$-matrix $Q=(q_{ij})$ is irreducible and the
corresponding process is $(\ell-1)$-ergodic ($\ell\ge 1$), i.e. ${\mathbb E}_i\sigma_{i_0}^{\ell-1}<\infty$ for every $i\ge 0$. When $\ell=1$, assume additionally that
the process is unique. Then we have
$$\mathbb E_n\sigma_{i_0}^\ell=\begin{cases}\ell \sum_{n\le k\le i_0-1} v_k^{(\ell)}+\big[1-\sum_{n\le k\le i_0-1} u_k\big]{\mathbb E}_{i_0}\sigma_{i_0}^\ell\quad & 0\le n\le i_0;\\
-\ell\sum_{i_0\le k\le n-1} v_k^{(\ell)}+\big[1+\sum_{i_0\le k\le n-1}u_k\big]{\mathbb E}_{i_0}\sigma_{i_0}^\ell,\quad & n> i_0;\end{cases}$$
\end{block}
}

%---------------------------------------------------------------------------------56
\frame{\small
\begin{block}{Proposition(contin.)}
\begin{gather}
u_k={\begin{cases}
\sum_{j=i_0-1}^k {q_{j, j+1}}^{-1} {F_k^{(j)}} q_{ji_0}(1-\delta_{ji_0}),\qquad &k\ge i_0,\nonumber\\
1, &k=i_0-1,\\
0,  &0\le k\le i_0-2\end{cases}}\\
v_k^{(\ell)}=\sum_{j=0}^k \frac{F_k^{(j)}}{q_{j, j+1}} {\mathbb E}_j\sigma_{i_0}^{\ell-1},\qquad k\ge 0,\nonumber
\end{gather}
$$
{\mathbb E}_{i_0}\sigma_{i_0}^\ell=\ell\varlimsup_{n\to\infty} \bigg({\sum_{i_0\le k\le n} v_k^{(\ell)}}\bigg)\bigg[{1+\sum_{i_0\le k\le n}u_k}\bigg]^{-1}
=\ell\lim_{n\to\infty} \frac{v_n^{(\ell)}}{u_n}\; \text{\rm if}\; \exists.$$
\end{block}
}
%%---------------------------------------------------------------------------------57
%\frame{\small
%
%Proof.
%$(y_i^*:={\mathbb E}_i\sigma_{i_0}^\ell: i\in E)$ is the the minimal nonnegative solution to the following equation:
%$$y_i=\sum_{k\ne i, i_0}\frac{1}{q_i} q_{ik}y_k+\frac{\ell}{q_i} {\mathbb E}_i \sigma_{i_0}^{\ell-1},\qquad i\in E.$$
%Then, we obtain the Poisson equation:
%$$(Qy)_i= q_{ii_0}(1-\delta_{ii_0})y_{i_0}-\ell\mathbb E_i\sigma_{i_0}^{\ell-1},\qquad i\in E. $$
%Applying Theorem 1 to $c= 0$ and
%$f_i= q_{ii_0}(1-\delta_{ii_0})y_{i_0}-\ell\mathbb E_i\sigma_{i_0}^{\ell-1}$, it follows that
%the solution to the last equation is as follows:
%$$y_n=y_0+\sum_{0\le k\le n-1}\sum_{j=0}^k\frac{F_k^{(j)}f_j}{q_{j,j+1}}
%=y_0+ y_{i_0}\!\!\sum_{0\le k\le n-1}u_k-\ell\!\! \sum_{0\le k\le n-1} v_k^{(\ell)},\quad n\ge 0.$$
%}
%
%%-------------------------------------------------------------------------58
%\frame{\small
%
%$$y_0=\ell \sum_{0\le k\le i_0-1} v_k^{(\ell)}+ y_{i_0} \bigg(1-\sum_{0\le k\le i_0-1} u_k\bigg).$$
%$$
%y_n={\begin{cases}
%-\ell\sum_{i_0\le k\le n-1} v_k^{(\ell)}+y_{i_0} \big[1+\sum_{i_0\le k\le n-1}u_k\big],\ & n\ge i_0+1\\
%\ell \sum_{n\le k\le i_0-1} v_k^{(\ell)}+y_{i_0} \big[1-\sum_{n\le k\le i_0-1} u_k\big], \ & n\le i_0.
%\end{cases}} $$
%$$y_{i_0}^*= \ell\varlimsup_{n\to\infty} \frac{\sum_{i_0\le k\le n} v_k^{(\ell)}}{1+\sum_{i_0\le k\le n}u_k}$$
%}
%-------------------------------------------------------------------------59
\frame{\small


\begin{block}{Proposition[Z.(2013); Chen \& Z.(2014)]}
Let the single birth $Q$-matrix $Q=(q_{ij})$ be irreducible and explosive
(i.e. $\sum_n m_n<\infty$). Assume that
the minimal process has finite $(\ell-1)$-th moments of $\tau_\infty$ for some integer $\ell\ge 1$
(i.e. $E_i\tau_\infty^{\ell-1}<\infty$ for all $i\ge 0$). Then
$$\mathbb E_n\tau_\infty^\ell=\ell\sum_{k\ge n}\overline m_k^{(\ell)},\qquad n\ge 0,$$
where
$$\overline m_n^{(\ell)}=\frac{1}{q_{n,n+1}}\bigg[{\mathbb E}_n\tau_{\infty}^{\ell-1}+\sum_{0\le k\le n-1}q_n^{(k)}{\overline m}_k^{(\ell)}\bigg]
=\sum_{j=0}^n\frac{F_n^{(j)}\mathbb E_j\tau_\infty^{\ell-1}}{q_{j,j+1}},\quad n\ge 0.$$
\end{block}
}
%%--------------------------------------------------------------------60
%
%\frame{\small
%Proof. $(\mathbb E_i\tau_\infty^\ell: i\in E)$
%is the the minimal nonnegative solution $(y_i^*: i\in E)$ to the following equation:
%$$y_i=\sum_{k\ne i}\frac{1}{q_i} q_{ik}y_k+\frac{\ell}{q_i} {\mathbb E}_i \tau_\infty^{\ell-1},\qquad i\in E.$$
%That is,
%$$(Qy)_i= -\ell\mathbb E_i\tau_\infty^{\ell-1},\qquad i\in E. $$
%Applying Theorem 1 to $c= 0$ and $f_i=-\ell\mathbb E_i\tau_\infty^{\ell-1}$ ($i\ge 0$),
%$$y_n=y_0-\ell\sum_{k=0}^{n-1}\overline m_k^{(\ell)},\qquad n\ge 1.$$
%By the nonnegative and minimal properties, it follows that
%$$y_0^*=\sup_{n\ge 1}\Bigg(\ell\sum_{k=0}^{n-1}\overline m_k^{(\ell)}\Bigg)=\ell\sum_{k=0}^\infty\overline m_k^{(\ell)}, \quad y_n^*=\ell\sum_{k=n}^\infty\overline m_k^{(\ell)},\quad n\ge 1.$$
%}
%----------------------------------------------------------------61
\frame{\frametitle{\small 6. Exponential ergodicity and Laplace transform of return time}\small

\begin{block}{Proposition[Z.(2003); Chen \& Z.(2014)]} Let the single birth $Q$-matrix $(q_{ij})$ be irreducible. Assume that its process is ergodic. Define $\big(\widetilde F_k^{(i)}\big)$ and $\big(\tilde d_k\big)$ by setting $c_i\equiv\lambda>0$. Then for small $\lambda$,
$$\mathbb E_0\text{e}^{\lambda\sigma_0}\!=\!\frac {q_{01} (1+\lambda\tilde d)}{q_{01}-\lambda}\!<\!\infty,\
\mathbb E_n\text{e}^{\lambda\sigma_0}\!=\!1+ \lambda \sum_{k=0}^{n-1}\Big(\widetilde F_k^{(0)}\tilde d-\tilde d_k\Big)\!<\!\infty,\ n\ge 1$$
iff
$${\tilde d}:=\varlimsup_{n\to \infty}
\mathbbm{1}_{\big\{{\sum_{k=0}^{n}\widetilde F_k^{(0)}}>0\big\}}
\frac{\sum_{k=0}^{n}\tilde d_k}{\sum_{k=0}^{n}\widetilde F_k^{(0)}}<\infty$$
and
\begin{equation}\label{e4.4-0}
{\tilde d}\, \sum_{k=0}^{n-1}\widetilde F_k^{(0)}>\sum_{k=0}^{n-1} \tilde d_k\quad \text{  whenever } \sum_{k=0}^{n-1}\widetilde F_k^{(0)}\le 0 \quad\text{  for }\ n\ge 2.
\end{equation}
\end{block}
}
%--------------------------------------------------------62
\frame{\small

\begin{block}{Proposition(contin.)}
Furthermore, once ${\widetilde F}_n^{(0)}>0$ for large enough $n$ and $\sum_n {\widetilde F}_n^{(0)}=\infty$, we have $${\tilde d}=\lim_{n\to\infty} \frac{{\tilde d}_n}{\widetilde F_n^{(0)}}\;\;\text{\rm if the limit exists}.$$
Finally, the process is exponentially ergodic iff both $\tilde d<\infty$ and (\ref{e4.4-0}) holds.
\end{block}
}
%%----------------------------------63
%\frame{\small
%Proof. $(\mathbb E_i\text{e}^{\lambda\sigma_0}: i\in E)$ is the minimal solution $(x_i^*)$ of the following equation
%$$x_i=\frac{1}{q_i-\lambda}\sum_{k\notin \{0,i\}}
%  q_{ik}x_k+\frac{q_{i0}(1-\delta_{i0})}{q_i-\lambda}, \qquad x_i\ge1,\qquad i\in E. $$
%Then, we have
%$$
%(Qx)_i+\lambda x_i=q_{i0}(x_0-1),\quad i\ge 1;\quad (Qx)_0+\lambda x_0=0.
%$$
%Applying Theorem 1 to $c_i\equiv\lambda$ and $f_i=q_{i0}(1-\delta_{i0})(x_0-1)$ for all $i\ge 0$,
%$$x_n=x_0\bigg(1-\frac \lambda{q_{01}}\bigg)\sum_{k=0}^{n-1}\widetilde F_k^{(0)}-\sum_{k=0}^{n-1}\big(\widetilde F_k^{(0)}+\lambda\tilde d_k\big)+1,\quad n\ge 1.$$
%}
%----------------------------------------------------64
\frame{\small
\begin{block}{Proposition[Brockwell(1986), Anderson(1991), Chen \& Z.(2014)]}
Define $\big(\widetilde F_k^{(i)}\big)$ and $\big(\tilde d_k\big)$ with $c_i\equiv -\lambda<0$.
Let the single birth process be recurrent. Then
the Laplace transform of $\sigma_0$ is given by
$$\mathbb E_0\text{e}^{-\lambda\sigma_0}=\frac {q_{01} (1-\lambda\tilde d)}{q_{01}+\lambda},\ \mathbb E_n\text{e}^{-\lambda\sigma_0}=1- \lambda \sum_{k=0}^{n-1}\Big(\widetilde F_k^{(0)}\tilde d-\tilde d_k\Big),\ n\ge 1,$$
where
$$\tilde d=\lim_{n\to\infty}\frac{\sum_{k=0}^{n-1}\tilde d_k}{\sum_{k=0}^{n-1}\widetilde F_k^{(0)}}
=\lim_{n\to\infty}\frac{{\tilde d}_n}{ {\widetilde F}_n^{(0)}} \quad \text{\rm if the limit exists}.$$
\end{block}


}
%----------------------------------------------------------------------------------------65
\frame{\small
\begin{block}{Proposition[Chen \& Z.(2014)]} Assume that the single birth $Q$-matrix $Q=(q_{ij})$ is explosive and irreducible. Define $(\widetilde m_k)$ with $c_i\equiv \lambda$.
For the corresponding minimal process,

(i) if $\exists\lambda>0$ such that $\lambda\sum_{k=0}^{n-1}\widetilde m_k<1$ for every $n>1$,
then
$$\mathbb E_n\text{e}^{\lambda\tau_\infty}=1+ \lambda \Bigg[\bar c\Bigg(1-\lambda\sum_{k=0}^{n-1}\widetilde m_k\Bigg)-\sum_{k=0}^{n-1}\widetilde  m_k\Bigg],\quad n\ge 0,$$
$$\bar c=\varlimsup_{n\to\infty }\frac{\sum_{k=0}^{n}\widetilde m_k}{1-\lambda\sum_{k=0}^{n}\widetilde m_k}.$$
Furthermore, the process decays exponentially fast provided $\bar c<\infty$.
%\end{block}
%}
%%----------------------------------------------------------------------------------------66
%\frame{\small
%\begin{block}{Proposition(contin.)}
(ii) For $\lambda>0$, the Laplace transform of $\tau_\infty$ is given by
$$\mathbb E_n\text{e}^{-\lambda\tau_\infty}=\frac{1+\lambda\sum_{0\le k\le n-1}\widetilde m_k}{1+\lambda\sum_{k\ge 0}\widetilde  m_k},\quad n\ge 0.$$
\end{block}
}
%Proof. Define
%$$e_{i\infty}(\lambda)=\int_0^\infty\text{e}^{\lambda t}\mathbb P_i(\tau_\infty>t)\,\text{d} t$$
%with $\lambda<q_i$ for all $i\ge 0$.
%$$\mathbb E_i\text{e}^{\lambda\tau_\infty}=1+\lambda e_{i\infty}(\lambda).$$
%$(e_{i\infty}(\lambda))$ is the minimal solution to the equation
%$$x_i=\frac{q_i}{q_i-\lambda}\sum_{k}\Pi_{ik}x_k+\frac 1{q_i-\lambda}, \quad i\ge 0.$$
%}
%%---------------------------------------------67
%\frame{\small
%$$(Qx)_i+\lambda x_i=-1,\qquad i\ge 0.$$
%Applying Theorem 1 to $c_i\equiv \lambda$ and $f_i\equiv -1$, the solution of the equation has
%the form:
%$$x_n=x_0\Bigg(1-\lambda\sum_{k=0}^{n-1}\widetilde m_k\Bigg)-\sum_{k=0}^{n-1}\widetilde  m_k,\quad n\ge 1.$$
%Note that $\lambda<q_0=q_{01}$ and $\lambda\widetilde m_0<1$. If there exists a positive
%$\lambda$ small enough so that $\lambda\sum_{k=0}^{n-1}\widetilde m_k<1$ for every $n>1$,
%then one gets
%$$
%e_{0\infty}(\lambda)=\sup_{n\ge 1}\frac{\sum_{k=0}^{n-1}\widetilde m_k}{1-\lambda\sum_{k=0}^{n-1}\widetilde m_k}=
%\varlimsup_{n\to\infty}\frac{\sum_{k=0}^{n}\widetilde m_k}{1-\lambda\sum_{k=0}^{n}\widetilde m_k}=:\bar c
%$$
%$$e_{n\infty}(\lambda)=\bar c\Bigg(1-\lambda\sum_{k=0}^{n-1}\widetilde m_k\Bigg)-\sum_{k=0}^{n-1}\widetilde  m_k,\quad n\ge 1.$$
%}
%---------------------------------------------68
\frame{\frametitle{\small 7. Examples}\small
\begin{block}{Example 1[uniform catastrophes]}
Let
$$q_{i,i+1}=b\, i,\qquad i\ge 0;\qquad q_{ij}=a,\qquad j=0,1,\dots, i-1;$$
and $q_{ij}=0$ for other $j>i+1$, where $a$ and $b$ are positive constants. Then the extinction of the process
has an exponential distribution
$${\mathbb E}_n\text{e}^{-\lambda\tau_0}=\frac{a}{a+\lambda},
\qquad \lambda>0,\; n\ge 1.$$
It is surprising that the distribution is independent of $b$ and the starting point $n$. Redefine $q_{01}=1$. Then the irreducible process is indeed
strongly ergodic.
\end{block}

P. J. Brockwell et al(1982)
}
%---------------------------------------------69
\frame{\small
\begin{block}{Example 2[Chen \& Z.(2014)]}
Consider the single birth $Q$-matrix $(q_{ij})$
with
$$q_{i0}>0, \;\; q_{i, i+1}>0,\;\; q_{ij}=0\;\;\text{\rm for all other }j\ne i.$$
Assume that $q_{n0}-c_n\equiv q_{10}-c_1$ for
every $n\ge 1$. Then the process is explosive if
$$\kappa':=\lim_{n\to\infty}
\frac{n(q_{n+1, n+2}-q_{n, n+1}-q_{10})}{q_{n, n+1}+q_{10}}>1$$
($q_{n, n+1}=(n+1)^{\gamma}$ for $\gamma> 1$ for example).
Otherwise, if $\kappa'<1$ ($q_{n, n+1}=(n+1)^{\gamma}$ for some $\gamma\le 1$ for instance), then the process is
unique. If so, the process is indeed strongly ergodic.
\end{block}
}
%---------------------------------------------70
\frame{\frametitle{\small 8. Related topics and applications}\small
\begin{itemize}
\item Exponential ergodicity: $\displaystyle\lim_{t\to\infty}{\red e^{\alpha t}}|p_{ij}(t)-\pi_j|=0$.
\end{itemize}

\begin{block} {Theorem[Mao \&  Z.(2004)]} Given a regular irreducible single
birth $Q$-matrix $Q=(q_{ij})$. If

$$\inf_i q_i>0\ \hbox{and}\ M:=\sup_{i> 0}\sum_{j=0}^{i-1}F_j^{(0)}\sum_{j=i}^\infty \frac 1{q_{j,j+1}F_j^{(0)}}
<\infty,$$
then the $Q$-process is exponential ergodicity.\end{block}

\begin{block}{Example} Let $q_{n,n+1}=1$ for all $n\ge 0$, $q_{10}=1$,
$q_{n,n-2}=1$ for all $n\ge 2$ and $q_{ij}=0$ for other $i\ne j$.
Then the single birth process is exponentially
ergodic and not strongly ergodic.
\end{block}

}
%---------------------------------------------71
\frame{\small
\begin{block} {Theorem([Chen(2000)]} Given a regular birth-death $Q$-matrix $(a_i,b_i)$. Then the process
is exponentially ergodic iff
$$\delta:=\sup_{i> 0}\sum_{j=0}^{i-1}\frac 1{\mu_jb_j}\sum_{j=i}^\infty \mu_j<\infty.$$
\end{block}

$$\frac 1{4\delta}\le\lambda_1\le\frac 1{\delta},\quad \lambda_1>0\Longleftrightarrow \delta<\infty.$$

$$\lambda_1=\ \hbox{exponential convergence rate.}$$

Key: birth-death processes are reversible!
}
%---------------------------------------------72
\frame{\small
\begin{block} {Theorem} Given a regular irreducible $Q$-matrix $Q=(q_{ij})$. Then the process
$P(t)$ is exponentially ergodicity if and only if for some $\lambda>0$ with $\lambda <q_i$ for all $i$,
$$\begin{cases}\sum_j q_{ij}y_j\le -\lambda y_i-1,\quad i\notin H&\\
\sum_{i\in H}\sum_{j\ne i}q_{ij}y_j<\infty&
\end{cases}$$
has a nonnegative finite solution $(y_i)$.
\end{block}

\vskip0.2cm
``$\inf_iq_i>0$'' is necessary essentially for exponential ergodicity.

\vskip0.2cm

Difficulty: two parameters!
}
%---------------------------------------------73
\frame{\small
Mao and Zhang (2004) gives another proof.
\vskip0.2cm
Keys:

1. Construct a test function with double summations for sufficiency.
The construction comes from the study of spectral gap essentially.

2. Necessity. Note that
$$\aligned
\mathbb E_{i}\sigma_0^{n}\ge n\bigg(&\sum_{j=0}^{i-1}\frac1{\mu_jb_j}\sum_{k=i}^\infty\mu_k\bigg)\mathbb E_{i}\sigma_0^{n-1}\\
\ge\cdots&\ge n!\bigg(\sum_{j=0}^{i-1}\frac1{\mu_jb_j}\sum_{k=i}^\infty\mu_k\bigg)^n.\\
\mathbb E_i\textrm{e}^{\lambda\sigma_0}<\infty
&\Rightarrow\sum_{n=1}^\infty\bigg(\lambda\sum_{j=0}^{i-1}\frac 1{\mu_jb_j}\sum_{k=i}^\infty\mu_k\bigg)^n<
\infty\\
&\Rightarrow\lambda\sum_{j=0}^{i-1}\frac 1{\mu_jb_j}\sum_{k=i}^\infty\mu_k< 1\Rightarrow\delta\le \lambda^{-1}<\infty.
\endaligned$$
}
%---------------------------------------------74
\frame{\small

Unfortunately, ``$M<\infty$'' is not necessary for exponential ergodicity.

\begin{block}{Example} Given $q_{01}>0$ arbitrarily. Let $q_{i,i+1}=i^\gamma$, $q_{i0}=i^{\gamma-1}$
for all $i\ge 1$ where the constant $\gamma\in (1,2)$, and $q_{ij}=0$ for other $i\ne j$.
Obviously, the $Q$-matrix is irreducible.
It is easily computed that $F_i^{(0)}=1$ for all $i\ge 0$.
So the single birth process is recurrent and furthermore
regular. Note that $\inf_{i\ge 1}q_{i0}=1>0$. It follows that the single birth
process is strongly ergodic and furthermore exponentially ergodic immediately. But
$M=\infty$.\end{block}

Now, for the single birth $Q$-matrix $(q_{ij})$,
define a conservative birth-death $Q$-matrix $(a_i, b_i)$ as follows.
$$
 a_i=\frac{q_{i,i+1}F_i^{(0)}}{F_{i-1}^{(0)}},\quad i\ge 1;\qquad b_i=q_{i,i+1},\quad i\ge 0.
$$
Then $\mathbb E_i\tau_0^n\le\mathbb E_i\overline\tau_0^n\,(i\ge 1,n\ge 0)$, furthermore
$\mathbb E_i\textrm{e}^{\lambda\tau_0}\le \mathbb E_i\textrm{e}^{\lambda\overline\tau_0}\,(i\ge 1,\lambda>0)$.
}
%---------------------------------------------------------------------- 76
\frame{\small

\begin{itemize}
\item Single birth $Q$-matrix with absorbing boundary: $1\le N:=\max\{i+1:q_{i,i+1}=0\}<\infty$. When $N=\max\emptyset=0$, it is single birth.

M.F. Chen(1999), Single birth processes, Chin. Ann. Math., \textbf{20B}, 77-82. Uniqueness.

\item Single birth $Q$-matrix with immigration: $q_{0j}>0$ for $j\ge 2$.

Y.H. Zhang, Q.Q. Zhao, 2010.

\item Cutoff

Mao \& Z.(2014). Explicit criteria on separation cutoff for birth and death chains, Frontiers of Mathematics in China, 9(4): 881-898.

Mao, C. Zhang \& Z.(2016). Separation cutoff for upward skip-free chains, J. Applied Prob., 53(1): 299-306.

\end{itemize}
}
%---------------------------------------------------------------------- 77
\frame{\small

\begin{itemize}
\item Applications
\end{itemize}

{\blue Yan \& Chen(1986). Multidimensional $Q$-processes. Chinese Ann. Math. 7B: 90-110.}

\vskip0.2cm
To study sufficient conditions for these problems of multidimensional $Q$-processes,
we use the comparison method, i.e., compare them with a single birth $Q$-processes.
The method reduces the multidimensional problems to one-dimensional ones.

\vskip0.2cm
The typical example
is Sch\"ogl's model: a model of chemical reaction with diffusion in a container.


\vskip0.2cm The method is efficient for {\red necessary condition}.

\begin{block}{} The $Q$-process corresponding to Brusselator model is not strongly ergodic.\end{block}

}
%---------------------------------------------------------------------- 78
\frame{\small
Let $E$ be a countable set. $Q=(q(x,y):x,y\in E)$ be a
conservative $Q$-matrix. Suppose that there exists a partition $\{E_k\}$ of $E$
such that $\sum_{k=0}^\infty E_k=E$ and

(i) If $q(x,y)>0$ and $x\in E_k$, then $y\in\sum_{j=0}^{k+1}E_j$ for all
$k\ge 0$.

(ii) $\sum_{y\in E_{k+1}}q(x,y)>0$ for all $x\in E_k$ and all $k\ge 0$.

(iii) $C_k:=\sup\{q(x):x\in E_k\}<\infty$ for all $k\ge 0$.
\vskip0.3cm
Define a conservative single birth $Q$-matrix $Q=(q_{ij})$:
$$q_{ij}=\begin{cases}
\sup\{\sum_{y\in E_j}q(x,y):x\in E_i\},\qquad\text{if}\ j=i+1;\\
\inf\{\sum_{y\in E_j}q(x,y):x\in E_i\},\qquad\,\text{if}\ j<i;\\
0,\qquad\qquad\qquad\qquad\,\text{other cases of}\ j\ne i.
\end{cases}$$
}
%---------------------------------------------------------------------- 79
\frame{\small
(0) {\blue If the $(q_{ij})$-process is unique (i.e. $R=\infty$), then so is the
$(q(x,y))$-process.}

\vskip0.2cm

Now suppose that $E_0=\{\theta\}$ where $\theta\in E$ is a reference point,
and that both $(q(x,y))$ and $(q_{ij})$ are irreducible and $(q_{ij})$ is regular.
\vskip0.2cm
(1) Moreover, assume that $E_k$ is finite for all $k\ge 1$.
If the $(q_{ij})$-process is recurrent (i.e. $\sum_{n=0}^\infty F_n^{(0)}=\infty$), then so is the
$(q(x,y))$-process.

(2) If $\widehat d:=\sup_{k\ge 0}d_k/F_k^{(0)}<\infty$, then both processes are ergodic.

(3) If $\inf_i q_i>0$ and $M=\sup_{i>0}\sum_{j=0}^{i-1}F_j^{(0)}\sum_{j=i}^\infty(q_{j,j+1}F_j^{(0)})^{-1}<\infty$,
then both processes are exponentially ergodic.

(4) If $\sup_{k\ge 0} \sum_{n=0}^k(F^{(0)}_n\widehat d-d_n)<\infty$,
then both processes are strongly ergodic.

\vskip0.2cm Keys: increasing solution of equations.
}
%---------------------------------------------------------------------- 80
\frame{\small

Let $S$ be a finite set and $E=\mathbb Z_+^S$. The model is defined by the
following $Q$-matrix $Q=(q(x,y):x,y\in E)$:
$$q(x,y)=\begin{cases} \lambda_1\binom{x(u)}2+\lambda_4,\qquad&\text{if}
\ y=x+e_u,\\
\lambda_2 \binom{x(u)}3+\lambda_3x(u),&\text{if}
\ y=x-e_u,\\
x(u)p(u,v), &\text{if}\ y=x-e_u+e_v,\\
0,&\text{other}\ y\ne x,
\end{cases}$$
$q(x)=-q(x,x)=\sum_{y\ne x}q(x,y)$,
where $x=(x(u):u\in S)$, $\binom nk$ is the usual combination,
$\lambda_1,\cdots,\lambda_4$ are positive constants. $(p(u,v):u,v\in S)$ is a
transition probability matrix on $S$ and $e_u$ is the element in $E$ having
value $1$ at $u$ and $0$ elsewhere.

\begin{block}{Conclusion}
The $Q$-process corresponding to finite dimensional Schl\"ogl model is strongly ergodic.
\end{block}

}

%----------------------------------------------------------------------------------------38
\frame{\small
第三部分: 单死过程

\begin{enumerate}
\item Motivation
\item Dual approach
\item Zero-entrance
\item Mean hitting time
\end{enumerate}
}

%----------------------------------------------------------------------------------------3
\frame{\frametitle{1. Motivation}\small Single death process:
$Q=(q_{ij})$ satisfies
$$q_{i,i-1}>0,\ q_{i,i-j}=0,\ i\ge 1,j> 1.$$
%\vspace{0.3cm}
\begin{center}
\setlength{\unitlength}{1cm}
\begin{picture}(11,2)
\put(4.9,-0.5){$i$}
\put(3.55,-0.5){$i-1$}
\put(5.55,-0.5){$i+1$}
\put(6.55,0.5){$i+2$}
\put(9.55,0.5){$i+k$}
\put(-0.15,-0.5){$0$}
\put(0.85,-0.5){$1$}
\put(0,0){\line(1,0){11}}
\qbezier(9,0)(6.5,2)(5,0)
\qbezier(10,0)(7.5,2)(5,0)
\qbezier(7,0)(6,2)(5,0)
\qbezier(4,0)(4.5,2)(5,0)
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\vskip0.3cm
$$Q=\left(%
\begin{array}{ccccc}
  - & * & * & * & \cdots \\
  + & - & *& * & \cdots \\
  0 & + & - & * & \cdots\\
  \vdots & \vdots & \vdots & \vdots & \ddots \\
  \end{array}%
\right)$$
}
%----------------------------------------------------------------------------------------4

\frame{ \frametitle{1. Motivation}\small
\begin{itemize}

\item branching processes: reference model

\end{itemize}

$$\aligned
 i&\to i-1\quad\text{at rate}\quad ib_0\\
  &\to i\qquad\quad \text{at rate}\quad ib_1=-i\\
  &\to i+1\quad\text{at rate}\quad ib_2\\
  &\to i+2\quad\text{at rate}\quad ib_3\\
  &\to \cdots
\endaligned$$
$$b_1=-\sum_{k\ne 1}b_k=-1.$$

\begin{itemize}
\item single birth processes
\end{itemize}

{\blue Y.R. Li, A.G. Pakes, J. Li, A.H. Gu(2008). The limit behavior of dual Markov branching processes. J. Appl. Prob. 45, 176-189.}
}
%-----------------------------------------------------------------------------------------14
\frame{\frametitle{2. Dual approach }
\small

Given a $Q$-matrix $Q$, denote by $P(t)=(P_{ij}(t))$ the corresponding minimal $Q$-process. The $Q$-process $P(t)$
is said to be {\blue stochastically monotone} if
$$
\sum_{j\geq k} P_{ij}(t)\leq \sum_{j\geq k} P_{mj}(t), \quad i\le m, k=0,1,2,\cdots.
$$
{\blue$\Rightarrow$}
$$
\sum_{j\geq k} q_{ij}\leq \sum_{j\geq k} q_{mj},\quad i\le m, k\in\{0,\cdots,i\}\cup\{m+1, m+2,\cdots\}.
$$
We also say a $Q$-matrix $Q=(q_{ij})$ is {\blue monotone} if the inequality above holds. If a $Q$-matrix is regular and monotone, then the minimal $Q$-process is also {\blue stochastically monotone}.
}
%-----------------------------------------------------------------------------------------15
\frame{\frametitle{2. Dual approach }
\small

\begin{block}{Propositioin[Siegmund(1976)]}
A transition function $P(t)$ is stochastically monotone if and only if there exists a dual process $\widetilde{P}(t)$ of $P(t)$ such that
$$
\widetilde{P}(t)\Lambda=\Lambda P(t)^{\text{T}},
$$
where
$$
\Lambda(i, j)=1_{\{i\leq j\}}.
$$
\end{block}

\vskip0.2cm{\blue
D. Siegmund(1976): The equivalence of absorbing and reflecting barrier problems for stochastically monotone Markov processes. Ann. Probab. 41, 914-924.}
}

%-----------------------------------------------------------------------------------------16
\frame{%\frametitle{2. Dual approach }
\small

\begin{block}{Theorem[M.F. Chen, P.S. Li \& Z.(2017+)]}
Let $Q=(q_{ij})$ be a regular stochastically monotone single birth $Q$-matrix and  let $\tilde{Q}$ be the dual of $Q$: $\tilde Q=\Lambda Q^{\text{T}}\Lambda^{-1}$. Denote the single birth process by $(X_t)_{t\geq 0}$ and denote the (dual) single death process by $(\widetilde{X}_t)_{t\geq 0}$. Then we have:

(1) The (dual) single death matrix $\widetilde{Q}$ is regular if and only if the single birth process $(X_t)_{t\geq 0}$ is not strongly ergodic, i.e.,
$$\sup_{i\geq 1}\sum^{i-1}_{j=0}(F_j^{(0)}d-d_j)=\infty.$$

\end{block}

\vskip0.3cm
{\blue H.J. Zhang, A.Y. Chen, X. Lin, Z.T. Hou(2001): Strong ergodicity of monotone transition functions, Statist. Probab. Lett. 55: 63-69.}

}
%-----------------------------------------------------------------------------------------17
\frame{\frametitle{2. Dual approach }
\small

\begin{block}{Theorem(continued)}
(2) The single birth process is ergodic if and only if for any initial state $i> 0$, the extinction probability of the (dual) single death process $\mathbf P_i(\tilde{\sigma}_0<\infty)<1$. Indeed the extinction probability is given by
$$\mathbf P_i(\tilde{\sigma}_0<\infty)=1-\sum^{i-1}_{j=0} \pi_j,\quad i=1,2,\cdots.$$
\end{block}
}
%-----------------------------------------------------------------------------------------18
\frame{\frametitle{2. Dual approach }
\small

\begin{block}{Theorem(continued)}
(3) Denote the occupation time of the dual single birth process by $\widetilde{T}_j=\int^\infty_0 1_{\{\widetilde{X}_t=j\}}dt$ $j=1,2,\ldots$. Then
$$
\mathbf E_i(\tilde{T}_j)=\int^\infty_0 \mathbf P_{i}(\tilde{X}_t= j)dt=\sum^{i-1}_{k=0}\Bigg(1_{\{k\leq j-1\}}\frac{F^{(k)}_{j-1}}{q_{k, k+1}}-m_{j-1}\pi_k\Bigg).
$$
Moreover, when the extinction probability of the (dual) single death process is $1$, then
$$
\mathbf E_i\tilde{T}_j=\sum^{i-1}_{k=0}1_{\{k\leq j-1\}}\frac{F^{(k)}_{j-1}}{q_{k, k+1}}=\sum^{i\wedge j-1}_{k=0}\frac{F^{(k)}_{j-1}}{q_{k, k+1}}.
$$
\end{block}
}
%-----------------------------------------------------------------------------------------19
\frame{%\frametitle{2. Dual approach }
\small

\begin{block}{Theorem(continued)}
(4) Define the explosion time of the (dual) single death process $\tilde{\zeta}=\lim_{n\rightarrow\infty} \tilde{\zeta}_n$, where $\tilde{\zeta}_n= \inf\{t\geq0: \widetilde{X}_t\geq n\}$. The conditional expectation of the explosion time and the extinction time are given by
$$
\mathbf E_i(\tilde{\zeta}|\tilde{\sigma}_0=\infty)
=
\sum^\infty_{j=1}\frac{\sum^{j-1}_{k=0} \pi_k}{\sum^{i-1}_{k=0} \pi_k}\sum^{i-1}_{k=0}\Bigg(1_{\{k\leq j-1\}}\frac{F^{(k)}_{j-1}}{q_{k, k+1}}-m_{j-1}\pi_k\Bigg),
$$
$$\mathbf E_i(\tilde{\sigma_0}|\tilde{\sigma}_0< \infty)\!\!= \!\!\sum^\infty_{j=1}\frac{1\!-\!\sum^{j-1}_{k=0} \pi_k}{1\!-\!\sum^{i-1}_{k=0} \pi_k}\sum^{i-1}_{k=0}\Bigg(1_{\{k\leq j-1\}}\frac{F^{(k)}_{j-1}}{q_{k, k+1}}-m_{j-1}\pi_k\Bigg).
$$

%\end{block}
%}
%%-----------------------------------------------------------------------------------------20
%\frame{\frametitle{2. Dual approach }
%\small
%
%\begin{block}{Theorem(continued)}
If the extinction probability of the (dual) single death process is $1$, then the mean extinction time of the (dual) single death process is given by
$$
\mathbf E_i\tilde{\sigma}_0= \sum^\infty_{j=1}\sum^{i\wedge j-1}_{k=0}\frac{F_{j-1}^{(k)}}{q_{k,k+1}},\qquad i=1,2,\cdots.
$$
\end{block}
}

%-----------------------------------------------------------------------------------------21
\frame{\frametitle{3. Zero-entrance }\small

Define $c_i:=q_i-\sum_{j\ne i}q_{ij}$ (non-conservative quantity).

\begin{block}{[L.D. Wang \&Z.(2014)]} Given a totally stable single birth $Q$-matrix $Q=(q_{ij})$. Then the single birth $Q$-matrix is zero-exit(equivalently, $BQ$-process is unique) if and only if
$$R:=\sum_{n=0}^\infty m_n=\infty,$$
$$
m_0=\frac {1+c_0}{q_{01}},\ m_n=\frac 1{q_{n,n+1}}\bigg(1+c_n+\sum_{k=0}^{n-1}(c_n+q_n^{(k)})m_k\bigg),\ n\ge 1.
$$
\end{block}

\begin{block}{}$$\lambda x=Qx,\ 0\le x\le 1.$$\end{block}
}
%-----------------------------------------------------------------------------------------21
\frame{\frametitle{3. Zero-entrance }\small
\begin{block}{Theorem[L.D. Wang \& Z.(2014)]}

Given a totally stable single death $Q$-matrix $Q=(q_{ij})$. Define $\bar q_n^{(k)}=\sum_{j=k}^\infty q_{nj}$ $(k>n)$. Then the single death $Q$-matrix is zero-entrance if and only if
~$\overline R:=\sum_{n=0}^\infty \overline m_n=\infty$, where
$$\overline m_0=0,\ \overline m_n=\frac 1{q_{n,n-1}}\bigg(1+\sum_{k=0}^{n-1}(c_k+\bar q_n^{(k)})\overline m_k\bigg),\ n\ge 1.$$
If the single death $Q$-matrix is non-conservative, then $FQ$-process is unique if and only if $\overline R=\infty$.
\end{block}

\begin{block}{}$$\lambda y=yQ,\  y\ge 0, ||y||_1<\infty.$$\end{block}
}
%-----------------------------------------------------------------------------------------22
\frame{\frametitle{3. Zero-entrance }\small
\begin{block}{Theorem[L.D. Wang \& Z.(2014)]}

Given a totally stable single death $Q$-matrix $Q=(q_{ij})$. Then there exists a unique invariant measure as follows:
$$\mu_0=1,\ \mu_n=\frac 1{q_{n,n-1}}\sum_{k=0}^{n-1}(c_k+\bar q_n^{(k)})\mu_k,\ n\ge 1.$$
\end{block}

The stationary distribution of the single death process is represented as follows:
$\pi_n={\mu_n}/{\mu},\ n\ge 0$, where
$$\mu_0=1,\ \mu_n=\frac 1{q_{n,n-1}}\sum_{k=0}^{n-1}\bar q_n^{(k)}\mu_k,\ n\ge 1;$$
$$\mu=\sum_{n=0}^\infty \mu_n.$$
}
%-----------------------------------------------------------------------------------------23
\frame{\frametitle{3. Zero-entrance }\small
\small

$$\pi Q=\mathbf 0.$$

Transpose the matrix equation $\pi Q=\mathbf 0$ to
$$Q^{\text{T}}\pi^{\text{T}} =\mathbf 0^{\text{T}}.$$
Then
$$Q^{\text{T}}g=:\Omega g=Q^*g+cg=\mathbf 0^{\text{T}},$$
where $Q^*$ is a totally stable and conservative single birth $Q$-matrix and
\begin{block}{}
$$c_i=\sum_{j=0}^{i-1}q_{ji}+q_{i+1,i}-\sum_{j= i+1}^\infty q_{ij}-q_{i,i-1},\ i\ge0; $$
$$f_i\equiv  0.$$\end{block}
}

%----------------------------------------------------------------------------------------24
\frame{\frametitle{3. Zero-entrance }
\small
\begin{block}{Proposition[Z.(2016)]} The stationary distribution of the single death process is represented as follows:
$\pi_n={\mu_n}/{\mu},\ n\ge 0$, where
$$\mu_0=1,\  \mu_n=1-\sum_{k=0}^{n-1}\sum_{j=0}^k\frac{\widetilde F_k^{(j)}c_j}{q_{j+1,j}}, n\ge 1;\ \mu:=\sum_{n=0}^\infty \mu_n;$$
\begin{align}
&\widetilde F_i^{(i)}=1,\quad  \widetilde F_n^{(i)}=\frac 1{q_{n+1,n}}
\sum_{k=i}^{n-1}\tilde q_n^{(k)}\widetilde F_k^{(i)},\quad n>i\ge 0,\nonumber\\
&\tilde q_n^{(k)}=\sum_{j=0}^kq_{jn}-c_n,\quad n>k\ge 0.\nonumber\end{align}
\end{block}
}

%----------------------------------------------------------------------------------------24
\frame{\frametitle{4. Mean hitting time }

\small \begin{itemize}
\item single birth process
\end{itemize}
$$m_n
=\sum_{k=0}^n\frac{F_n^{(k)}}{q_{k,k+1}},\ n\ge
0.$$

$$\mathbb E_n\tau_{n+1}=m_n=\sum_{k=0}^n\frac{F_n^{(k)}}{q_{k,k+1}},\ n\ge
0.$$

\begin{block}{}
$$x_n=\frac 1{q_{n}}\bigg(1+\sum_{k=0}^{n-1}q_n^{(k)}x_k+q_n^{(n-1)}x_n\bigg),\ n\ge 0.$$
\end{block}

\begin{block}{}
$$F_i^{(i)}=1, \ F_n^{(i)}=\frac 1{q_{n,n+1}} \sum_{k=i}^{n-1}q_n^{(k)}F_k^{(i)},\ 0\le i<n.$$
\end{block}
}
%----------------------------------------------------------------------------------------25
\frame{\frametitle{4. Mean hitting time }
\small \begin{itemize}
\item single death process: $\mathbb E_n\tau_{n-1}=?$
\end{itemize}
\begin{block}{}
$$x_n=\frac 1{q_{n}}\bigg(1+\sum_{k=n+1}^{\infty}\bar q_n^{(k)}x_k+\bar q_n^{(n+1)}x_n\bigg),\ n\ge 1.$$
\end{block}

$$\bar q_n^{(k)}:=\sum_{j=k}^\infty q_{nj},\  k>n\ge 0$$

\begin{block}{}
$$
G_i^{(i)}=1,\  G_n^{(i)}=\frac 1{q_{n,n-1}}
\sum_{k=n+1}^{i} q_n^{(k)} G_k^{(i)},\ 0\le n< i.
$$
\end{block}
}
%----------------------------------------------------------------------------------------25
\frame{\frametitle{4. Mean hitting time }\small

\begin{block}{}
$$x_n=\frac 1{q_{n}}\bigg(1+\sum_{k=n+1}^{\infty}\bar q_n^{(k)}x_k+\bar q_n^{(n+1)}x_n\bigg),\ n\ge 1.$$
\end{block}

\begin{block}{}
$$\mathbb E_n\tau_{n-1}=\sum_{k=n}^\infty\frac{G_n^{(k)}}{q_{k,k-1}} (n\ge 1)?$$
\end{block}

\begin{block}{The minimal nonnegative solution}
$$h_n=\sum_{k=n}^\infty\frac{G_n^{(k)}}{q_{k,k-1}} (n\ge 1).$$
\end{block}


\begin{block}{}
$$\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{G_n^{(k)}}{q_{k,k-1}}<\infty \Rightarrow \exists\ \text{a QSD uniquely}.$$
\end{block}

}

%----------------------------------------------------83
\frame{

\centerline{ \emph{\textbf{\color{blue} Thank you for your attention!}}}
\vskip0.5cm
\centerline{Homepage: http://math0.bnu.edu.cn/${}^\sim$zhangyh/}

}

\end{CJK*}

\end{document}
