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\begin{document}
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  \includegraphics[scale=0.3]{math.png}\\

\title[稳定性]{连续时间向上一致有限程马氏链的稳定性}

\author[张余辉]{\sihao 张余辉}

\date[2024年1月10日]{\wuhao (河南大学邀请报告, 2024年1月10日)}

\institute[北京师范大学]{\wuhao 北京师范大学}

\maketitle
\vskip0.2cm
{ 此工作是与赵奕, 闫艳艳合作完成.}

 \end{frame}



%----------------------------------------------------------------------------------------1
\begin{frame}
  \frametitle{CONTENTS}
  \tableofcontents%[sections]
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section{Background}
%----------------------------------------------------------------------------------------2
\section{Background}
\frame{ \frametitle{Background}\small

Stability:

\begin{itemize}
\item classical problems (uniqueness(cont.-time), recurrence, ergodicity(exp., strong ergodicity)).
\item integral-type functionals: moments and distribution (Laplace transform).
$$\int_0^{\tau_0}V(X(t))\text{d}t.$$
\item qualitative and quantitative results.\end{itemize}
}

\frame{\frametitle{Background}\small

连续时间向上一致有限程马氏链？

Assume that $Q$ is totally stable and conservative.

There exists such a positive integer $m\geqslant 1$ that
$$q_{i, i+m}>0, q_{ij}=0\, (j>i+m),\ \text{\blue upward}.$$
(or
$$q_{i, i-m}>0, q_{ij}=0\, (0<j<i-m).\ \text{\blue downward})$$
Generally we call them $m$-birth and $m$-death respectively.

\vskip0.2cm

When $m=1$, {\blue single birth, single death}.

\begin{itemize}
\item single birth: Yan \& Chen (1986), Z.(2001). [$21$ papers]
\item single death: Z.(2018) [$10$ papers]
%\item single death processes: ergodicity (Z., 2018), uniqueness, recurrence (Mao, Yan, Z., 2021+)
%\item birth-death processes: (Z., 2019), uniqueness, recurrence (Wang, Yan, Z. 2021+)
%\item Uniformly finite rang upward or downward: $\exists m\geqslant 1$,
%$$q_{i, i+m}>0, q_{ij}=0\, (j>i+m), i> 0,\ \text{\blue upward};$$
%or
%$$q_{i, i-m}>0, q_{ij}=0\, (0<j<i-m), i> m.\ \text{\blue downward}$$
\item $m$-birth \& $m$-death: Z.(2023), uniqueness and recurrence.
\item $2$-death: Z. \& Zhao.(2024+), invariant measures \& ergodicity.
\item $2$-birth: Yan, Z.\& Zhao (2024+).
\end{itemize}
}


\section{Ergodicity, strong ergodicity \& Laplace transform}

\frame{\frametitle{以$\mathbb E_i\tau_0$为例}\small

Given a totally stable, conservative and irreducible $Q$. For its minimal process $X(t)$, denote
$$\tau_0=\inf\{t>0: X(t)=0\}.$$

It is well known that $(\mathbb E_i\tau_0:i\geqslant1)$ is the minimal nonnegative solution to the equations
$$x_i=\sum_{j\ne i,0}\frac{q_{ij}}{q_i}\cdot x_j+\frac 1{q_i},\quad i\geqslant 1.$$

{\blue Approximation.} Fix $n$, define a totally stable and conservative $Q^{(n)}$ on $\{0, 1, \cdots, n\}$:
$$ q_{ij}^{(n)}=\begin{cases} q_{ij}, &\text{if}\ 1\leqslant i, j\leqslant n;\\ q_{i0}+\sum_{k>n}q_{ik},&\text{if}\ 1\leqslant i\leqslant n, j=0.
\end{cases}$$

$(\mathbb E_i \tau_0^{(n)}:1\leqslant i\leqslant n)$ is the minimal nonnegative solution to the equations
$$x_i=\sum_{1\leqslant j\leqslant n, j\ne i}\frac{q_{ij}}{q_i}\cdot x_j+\frac 1{q_i},\quad 1\leqslant i\leqslant n.$$
}

\frame{\frametitle{以$\mathbb E_i\tau_0$为例}\small
$$x_i=\sum_{1\leqslant j\leqslant n, j\ne i}\frac{q_{ij}}{q_i}\cdot x_j+\frac 1{q_i},\quad 1\leqslant i\leqslant n.$$
Rewrite the above equation as $x=T^{(n)}x+f^{(n)}$, where the elements of $T^{(n)}$ and $f^{(n)}$ have the form
	$$T_{ij}^{(n)}=\frac{q_{ij}}{q_{i}}\mathbbold{1}_{\{1\leqslant i\neq j\leqslant n\}},\  f_i^{(n)}=\frac{1}{q_{i}}\mathbbold{1}_{\{1\leqslant i\leqslant n\}},\  i,j\in\mathbb{Z}_{+}.
	$$
Then it is obvious that $T^{(n)}$ and $f^{(n)}$ are increasing in $n$ in the element-wise sense.

{\blue $$\mathbb E_i \tau_0^{(n)}\ \uparrow\ \mathbb E_i \tau_0\quad \text{as}\quad n\uparrow +\infty.$$}
}

\frame{\frametitle{以$\mathbb E_i\tau_0$为例}\small
$$x_i=\sum_{1\leqslant j\leqslant n, j\ne i}\frac{q_{ij}}{q_i}\cdot x_j+\frac 1{q_i},\quad 1\leqslant i\leqslant n.$$

The equations can be rewritten as
$$(I-\Pi)\mathbf x=\mathbf r,$$
where $I$ is the $n\times n$ unit matrix, $\mathbf x=(x_1,\cdots, x_{n})^{\text{T}}$ and $\mathbf r=(\frac{1}{q_1},\cdots, \frac{1}{q_{n}})^{\text{T}}$.
By the irreducibility of $Q$,  the local embedding chain is transient,
$$(I-\Pi)^{-1}=\sum_{n=0}^\infty\Pi^n<\infty\Rightarrow\mathbf x=\sum_{n=0}^\infty\Pi^n\mathbf r,$$
there exists exactly a unique solution. The decomposition according to the first time arriving $0$.
}
\frame{\frametitle{以$\mathbb E_i\tau_0$为例}\small
\begin{itemize}
\item Z.F. Wei. Inverse problems for ergodicity of Markov chains. J. Math. Anal. Appl. 505 (2022) 125483.
\end{itemize}
%\centerline{ \includegraphics[scale=0.15]{20240109.pdf}}
\centerline{ \includegraphics[scale=0.08]{20240109.jpg}}
}

\frame{\frametitle{以$2$生, $\mathbb E_i\tau_0$为例}\small
Consider the $2$-birth process. Now let $q_{i}^{(k)}=\sum_{\ell=0}^k q_{i\ell}$ for all $0\leqslant k\leqslant i$ and
$$\aligned &F_n^{(n)}=1,\qquad F_{i}^{(n)}=\frac 1{q_{i,i+2}}\sum_{k=n}^{i-1} q_i^{(k+1)} F_k^{(n)},\qquad i>n\geqslant 0;\\
&d_0=0,\qquad d_{i}=\frac 1{q_{i,i+2}}\bigg(1+\sum_{k=0}^{i-1} q_i^{(k+1)} d_k\bigg),\qquad i\geqslant 1;\\
&c_0=0,\qquad c_{i}=\frac 1{q_{i,i+2}}\bigg(-q_{i1}+\sum_{k=0}^{i-1} q_i^{(k+1)} c_k\bigg),\qquad i\geqslant 1.\endaligned$$
}

\frame{\frametitle{遍历与强遍历准则}\small

\begin{block}{Theorem 1 (Regodicity \& strong ergodicity)} Given a regular and irreducible $2$-birth $Q$-matrix $Q=(q_{ij})$ on $\mathbb Z_+$. Then
$$
					\mathbb{E}_{k}\tau_{0}=\sum\limits_{i=0}^{k-2}c_i\cdot \mathbb{E}_{1}\tau_{0}+\sum\limits_{i=0}^{k-2}F_i^{(0)}\cdot\mathbb{E}_{2}\tau_{0}-\sum\limits_{i=0}^{k-2}d_{i},\ k\geqslant 3;
				$$
and as $n\to+\infty$,
$$\aligned
&\mathbb E_1\tau_0^{(n)}=\frac{F_n^{(0)}\sum_{i=0}^{n-1}d_i-d_n\sum_{i=0}^{n-1}F_i^{(0)}}{F_n^{(0)}\sum_{i=0}^{n-1}c_i-c_n\sum_{i=0}^{n-1}F_i^{(0)}}\quad \bigg\uparrow\quad D_1=:%\sup_{N\geqslant1}\frac{F_n^{(0)}\sum_{i=0}^{n-1}d_i-d_n\sum_{i=0}^{n-1}F_i^{(0)}}{F_n^{(0)}\sum_{i=0}^{n-1}c_i-c_n\sum_{i=0}^{n-1}F_i^{(0)}}=
\mathbb E_1\tau_0,\\
&\mathbb E_2\tau_0^{(n)}=\frac{d_n\sum_{i=0}^{n-1}c_i-c_n\sum_{i=0}^{n-1}d_i}{F_n^{(0)}\sum_{i=0}^{n-1}c_i-c_n\sum_{i=0}^{n-1}F_i^{(0)}}\quad \bigg\uparrow\quad
D_2=:\mathbb E_2\tau_0.
\endaligned$$
Moreover, the process is erg. iff $D_1$ and $D_2$ is finite; str. erg. iff
$$\sup_{n\geqslant 1}\Bigg(\bigg(\sum_{i=0}^{n}c_i\bigg)D_1+\bigg(\sum_{i=0}^{n}F_{i}^{(0)}\bigg)D_2-\sum_{i=0}^{n}d_i\Bigg)<+\infty.$$
\end{block}
}

\begin{frame}{以$2$生, $\mathbb E_i\tau_0$为例}\small
\begin{block}{Remark}
Single birth case:
\begin{itemize}

\item Yan \& Chen (1986), Z. (2003).
$$d:=\sup_{n\geqslant 1}\frac{\sum_{i=0}^{n-1}d_i}{\sum_{i=0}^{n-1}F_i^{(0)}}=\mathbb E_1\tau_0.$$
\item Chen \& Z. (2014).
$$d:=\varlimsup_{n\to+\infty}\frac{\sum_{i=0}^{n-1}d_i}{\sum_{i=0}^{n-1}F_i^{(0)}}.$$
\item
$$\mathbb E_1\tau_0^{(n)}=\frac{\sum_{i=0}^{n-1}d_i}{\sum_{i=0}^{n-1}F_i^{(0)}}\quad \bigg\uparrow\quad \mathbb E_1\tau_0 =d.$$
\end{itemize}
\end{block}
\end{frame}

\frame{\frametitle{以$2$生, $\mathbb E_i\tau_0$为例}\small

\begin{equation*}
					q_{ij}^{(n)}=\left\{
					\begin{array}{ll}
                        q_{ij},& \text{ if}~ 0\leqslant i\leqslant n-2, 0\leqslant j\leqslant n,\\
						q_{ij},& \text{ if}~ i=n-1,n, 1\leqslant j\leqslant n,\\
						q_{i0}+q_{i,n+1}, &\text{ if}~i=n-1, j=0,\\
						q_{i0}+q_{i,n+1}+q_{i,n+2}, &\text{ if}~i=n, j=0.
					\end{array}	
					\right.
				\end{equation*}
{\blue $$x_i=\sum_{1\leqslant j\leqslant n, j\ne i}\frac{q_{ij}}{q_i}\cdot x_j+\frac 1{q_i},\quad 1\leqslant i\leqslant n.$$}

Let $w_i=x_{i+1}-x_{i}$ for all $i\geqslant 1$.
			\begin{equation*}
				w_{i+1}=\sum\limits_{k=1}^{i}\frac{F_i^{(k)}(-1+q_{k}^{(1)}x_2-q_{k1}x_{1})}{q_{k,k+2}},\ 1\leqslant i\leqslant n-2.
			\end{equation*}
}

\frame{\frametitle{以$2$生, $\mathbb E_i\tau_0$为例}\small

			Hence for all $j\geqslant m+1$ we have
		\begin{align*}
			x_j&=\sum\limits_{i=2}^{j-1}w_i+x_2=x_2\bigg(\sum\limits_{i=1}^{j-2}F_i^{(0)}+1\bigg)+x_{1}\sum\limits_{i=1}^{j-2}c_i-
			\sum\limits_{i=0}^{j-2}d_i\\&=\sum\limits_{i=0}^{j-2}c_i\cdot x_{1}+\sum\limits_{i=0}^{j-2} F_i^{(0)}\cdot x_{2}-
			\sum\limits_{i=0}^{j-2}d_i.
		\end{align*}
			\begin{itemize}
				\item  Rewrite the equation as
			\end{itemize}
%\vspace{-1em}
\begin{equation*}
			\left\{
			\begin{aligned}
				&x_i=\sum\limits_{k=1,k\neq i}^{i+2}\frac{q_{ik}}{q_{i}}\cdot x_k+\frac{1}{q_{i}},\qquad 1\leqslant i\leqslant n-2,\\
				&x_{n-1}=\sum\limits_{k=1}^{n-2}\frac{q_{n-1,k}}{q_{n-1}}\cdot x_{k}+\frac{q_{n-1,n}}{q_{n-1}}\cdot x_{n}+\frac{1}{q_{n-1}},\\
				&x_{n}=\sum\limits_{k=1}^{n-1}\frac{q_{nk}}{q_{n}}\cdot x_{k}+\frac{1}{q_{n}}.
			\end{aligned}	
			\right.
		\end{equation*}
}
\frame{\frametitle{以$2$生, $\mathbb E_i\tau_0$为例}\small
\begin{itemize}
			\item Solving the above equations we could derive that
			\begin{align*}
				D^{(n)}\mathbf{x}=\mathbf{h}^{(n)},\quad \text{i.e.}\quad\left\{
				\begin{aligned} d_{11}^{(n)}x_{1}+d_{12}^{(n)}x_{2}=h_1^{(n)},\\d_{21}^{(n)}x_{1}+d_{22}^{(n)}x_{2}=h_2^{(n)},
				\end{aligned}\right.
			\end{align*}
			where $h_1^{(n)}=\sum\limits_{k=0}^{n-1}d_k$ and $h_2^{(n)}=d_{n}$.
			\item The assertion follows immediately by applying Cramer's rule.
		\end{itemize}

\begin{itemize}
			\item For general $i_0\geqslant 1$, the method also works.
				\begin{equation*}
				x_i=\frac{1}{q_i}\sum_{k\notin \{i_0,i\}}q_{ik}x_k+\frac{1}{q_i},\quad i\neq i_0 .
			\end{equation*}
			\item For higher moments $\mathbb{E}_i\tau_0^{\ell},~\ell>1$, it can be expressed by lower moments.
		\end{itemize}
}
\begin{frame}{Laplace transform of $\tau_0$, $m=2$}\small
		\begin{itemize}
			\item Now we turn to focus on $\mathbb{E}_i \mathrm{e}^{-\lambda \tau_0}\,(i\geqslant 0)$ to investigate the distribution of $\tau_0$.
		\item 	Define $\psi_{i, i_0}(\lambda)=1-\mathbb{E}_i \mathrm{e}^{-\lambda \xi_{i_0}}$.
		\end{itemize}
		\begin{block}{Lemma (Z.-T. Hou and Q.-F. Guo, 1978)}
		$\left(\psi_{i, i_0}(\lambda): i \geqslant 0\right)$ is the minimal nonnegative solution to:
		$$
		y_i=\sum_{k \neq i, i_0} \frac{q_{i k}}{\lambda V(i)+q_i} \cdot y_k+\frac{\lambda V(i)}{\lambda V(i)+q_i}, \ i \geqslant 0.
		$$
		\end{block}
		\begin{itemize}
			\item Here $V\equiv 1$, $i_0=0$.
		\end{itemize}
	\end{frame}
\begin{frame}{Laplace transform of $\tau_0$, $m=2$}
		Let $\tilde{q}_{i}^{(j)}:=q_i^{(j)}+\lambda$ for $0\leqslant j< i$ and let $\tilde{q}_i^{(i)}:=q_i^{(i)}$.   Correspondingly, we define the following notations.\vspace{-0.5em}
		$$\begin{aligned}
			&\tilde{F}_n^{(n)}=1,\ \tilde{F}_i^{(n)}=\frac{1}{q_{i,i+2}}\sum\limits_{k=n}^{i-1}\tilde{q}_i^{(k+1)}\tilde{F}_k^{(n)},\ 0\leqslant n < i,\\
			&\tilde{c}_0=0,\ \tilde{c}_i=\frac{1}{q_{i,i+2}}\bigg(-q_{i1}+\sum\limits_{k=0}^{i-1}\tilde{q}_i^{(k+1)}\tilde{c}_k\bigg),\ i\geqslant 1,\\
			&\tilde{d}_0=0,\  \tilde{d}_i=\frac{1}{q_{i,i+2}}\bigg(\lambda +\sum\limits_{k=0}^{i-1}\tilde{q}_i^{(k+1)}\tilde d_k\bigg),\ i\geqslant 1,
		\end{aligned}$$
			\end{frame}

\begin{frame}{Laplace transform of $\tau_0$, $m=2$}\small

\begin{block}{Theorem 2 (Laplace transform)}
		Assume that the $2$-birth $Q$-matrix $Q=(q_{ij})$ is irreducible and the corresponding process is recurrent. Then the following relations holds:
		\begin{equation*}
			\psi_{i0}(\lambda)=\sum\limits_{k=0}^{i-2}\tilde{c}_k\cdot \psi_{10}(\lambda)+\sum\limits_{k=0}^{i-2}\tilde{F}_k^{(0)}\cdot\psi_{20}(\lambda) -\sum\limits_{\ell=0}^{i-2}\tilde{d}_{\ell},\quad i\geqslant 3,
		\end{equation*}
		and the Laplace transform of $\tau_0$ has the representation that
		\begin{align*}
			&\mathbb{E}_1\mathrm{e}^{-\lambda\tau_0}=1-\lim\limits_{n\to\infty}\frac{\tilde F_n^{(0)}\sum_{i=0}^{n-1}\tilde d_i-\tilde d_n\sum_{i=0}^{n-1}\tilde F_i^{(0)}}{\tilde F_n^{(0)}\sum_{i=0}^{n-1}\tilde c_i-\tilde c_n\sum_{i=0}^{n-1}\tilde F_i^{(0)}},\\
&\mathbb{E}_2\mathrm{e}^{-\lambda\tau_0}=1-\lim\limits_{n\to\infty}
\frac{\tilde d_n\sum_{i=0}^{n-1}\tilde c_i-\tilde c_n\sum_{i=0}^{n-1}\tilde d_i}{\tilde F_n^{(0)}\sum_{i=0}^{n-1}\tilde c_i-\tilde c_n\sum_{i=0}^{n-1}\tilde F_i^{(0)}},\\
			&\mathbb{E}_0\mathrm{e}^{-\lambda\sigma_0}=\frac{q_{01}}{q_0+\lambda }\mathbb{E}_1\mathrm{e}^{-\lambda\tau_0}+\frac{q_{02}}{q_0+\lambda }\mathbb{E}_2\mathrm{e}^{-\lambda\tau_0}.
		\end{align*}
	\end{block}
	\end{frame}


\begin{frame}{Example}\small
\vspace{-0.5em}	\begin{block}{Example}
			Given a regular and irreducible single death $Q$-matrix $Q=(q_{ij})$ satisfying:
		$$q_{i,i-1}=a\geqslant0,\quad q_{i,i+1}=b\geqslant 0,\quad q_{i,i+2}=d>0,\quad i\geqslant 1,$$and $q_{ij}=0$  for other $i,j\geqslant 1,~ i\neq
		j$. Then the process is recurrent if and only if $a\geqslant b+2d$ and it is ergodic if and only if $a>b+2d$. Moreover in the recurent case, starting form $j\geqslant 1$, the moment of the hitting time of $0$ has the form:
		$$\mathbb{E}_{j}\tau_0=\frac{j}{a-b-2d},\quad j\geqslant 1.$$Therefore the process could not be strongly ergodic.
	\end{block}
\begin{itemize}
	\item It can be regarded both as $2$-birth processes and single death process.
	\item For single death processes, $\mathbb{E}_j\tau_0=\sum\limits_{k=1}^{j}\sum\limits_{\ell=k}^{\infty}\displaystyle\frac{G_k^{(\ell)}}{q_{\ell,\ell-1}},\quad j\geqslant 1.$
\end{itemize}
\end{frame}

\begin{frame}{Laplace transform of integral-type functionals, $m=2$}\small
\begin{itemize}
			\item For general $V$, the method also works.
				\begin{equation*}
				x_i=\sum_{k \neq i, i_0} \frac{q_{i k}}{\lambda V(i)+q_i} \cdot x_k+\frac{\lambda V(i)}{\lambda V(i)+q_i}, \ i \geqslant 0.
			\end{equation*}
\end{itemize}
\end{frame}

\section{Recurrence \& uniqueness}

\begin{frame}%{Recurrence \& uniqueness}
\small
		\begin{block}{Theorem 3 (Recurrence)}
			Assume that the $2$-birth $Q$-matrix $Q=(q_{ij})$ is irreducible. Then recur. iff
			$$\lim\limits_{n\to\infty}\frac{q_{01}F_n^{(0)}-q_{02}c_n}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}}=0.$$
			Moreover the return/extinction probability can be obtained by
			\begin{align*}				&\mathbb{P}_0(\sigma_0<\infty)=1-\frac{1}{q_{0}}\lim\limits_{n\to\infty}\frac{q_{01}F_n^{(0)}-q_{02}c_n}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}},\\
&\mathbb{P}_1(\sigma_0<\infty)=1-\lim\limits_{n\to\infty}\frac{F_n^{(0)}}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}},\\ &\mathbb{P}_2(\sigma_0<\infty)=1+\lim\limits_{n\to\infty}\frac{c_n}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}},\\
&\mathbb{P}_i(\sigma_0<\infty)=1-\lim\limits_{n\to\infty}\frac{F_n^{(0)}\sum_{k=0}^{i-2}c_k-c_n\sum_{k=0}^{i-2} F_k^{(0)}}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}},\ i\geqslant 3.
			\end{align*}
		\end{block}
	\end{frame}

\begin{frame}{Recurrence \& uniqueness}\small
			Fix $n\in\mathbb{Z}_{+}$, define the first hitting time of the set $\{n+1,n+2,\cdots\}$:
			$$\tau_{n+}=\inf\{t>0:X(t)\geqslant n+1\}.$$
{\blue $$\text{recur.}\Leftrightarrow \lim_{n\to+\infty}\mathbb{P}_{i}\left(\tau_{n+}<\sigma_0\right)=0.$$}

Denote $x_i:=\mathbb{P}_{i}\left(\tau_{n+}<\sigma_0\right),\, i\geqslant 0$. By the strong Markov property and jump as well as $2$-birth property of the process, it is obtained that
			$$
				\left\{
				\begin{aligned}
					&x_0=\frac{q_{01}}{q_{0}}\cdot x_1+\frac{q_{02}}{q_{0}}\cdot x_2,\\
					&x_i=\sum\limits_{j=1}^{i-1}\frac{q_{ij}}{q_{i}}\cdot x_j+\frac{q_{i,i+1}}{q_{i}}\cdot x_{i+1}+\frac{q_{i,i+2}}{q_{i}}\cdot x_{i+2},\qquad 1\leqslant i\leqslant n-2,\\
					&x_{n-1}=\sum\limits_{j=1 }^{n-2}\frac{q_{n-1,j}}{q_{n-1}}\cdot x_j+\frac{q_{n-1,n}}{q_{n-1}}\cdot x_{n}+\frac{q_{n-1,n+1}}{q_{n-1}},\\
					&x_{n}=\sum\limits_{j=1 }^{n-1}\frac{q_{n,j}}{q_{n}}\cdot x_j+\frac{q_{n,n+1}+q_{n,n+2}}{q_{n}}.
				\end{aligned}\right.
			$$
		\end{frame}

\begin{frame}{Recurrence \& uniqueness}\small
%For uniqueness of the $2$-birth processes, we have the following result.
\begin{block}{Theorem 4 (Uniqueness)}
Assume that the $2$-birth $Q$-matrix $\hat{Q}=(\hat{q}_{ij})_{i,j\in\mathbb{Z}_{+}}$ is totally stable and conservative. Define an irreducible, totally stable and conservative $Q$-matrix $Q$ on $\mathbb{Z}_{+}$:
$$q_{01}=q_{02}=1,\ q_{0j}=0,\ j>2;\ q_{i0}=1,\ q_{ij}=\hat{q}_{i-1,j-1},\ i,j\geqslant 1,j\neq i.$$
Then the process determined by $\hat{Q}$ is unique iff
%$$\lim\limits_{n\to\infty}\frac{\det(D_1^{(n)})+\det(D_2^{(n)})}{\det(D^{(n)})}=0.$$
$$\lim\limits_{n\to\infty}\frac{F_n^{(0)}-c_n}{F_n^{(0)}\sum_{k=0}^{n-1}c_k-c_n\sum_{k=0}^{n-1}F_k^{(0)}}=0.$$
\end{block}
\end{frame}

\begin{frame}{Recurrence \& uniqueness}\small
			\begin{itemize}
				\item uniqueness:
				$$u_i=\sum_{j\ne i}\frac{q_{ij}}{1+q_i}u_j, \quad 0\leqslant u_i\leqslant 1, i\geqslant 0,$$
				$$u=\Pi(1)u.$$
			\end{itemize}
			
			We introduce a fictitious state $\Delta$ and
			define on the enlarged state space $E_\Delta=E\cup\{\Delta\}$ a new transition probability
			matrix
			$$\Pi_{ij}^\Delta(1)=\begin{cases}\Pi(1), &i,j\in E;\\
				\frac{1}{1+q_i}, &i\in E, j=\Delta;\\
				p_j, &i=\Delta, j\in E,\end{cases}$$
			where $(p_j, j\in E)$ is a positive probability measure.
			\begin{block}{Theorem}
				The original $Q$-process is unique iff the $\Pi^{\Delta}(1)$-chain is recurrent.
			\end{block}
		\end{frame}
\begin{frame}{Recurrence \& uniqueness}\small
$$\begin{matrix}
       &\Delta& 0    & 1    & 2    &3     &\cdots\\
\Delta &-1    & p_0    & p_1    &p_2     &p_3     &\cdots\\
0      &1     &-1-q_0&q_{01}&q_{02}&q_{03}&\cdots\\
1      &1     &q_{10}&-1-q_1&q_{12}&q_{13}&\cdots\\
2      &1     &q_{20}&q_{21}&-1-q_2&q_{23}&\cdots\\
3      &1     &q_{30}&q_{31}&q_{32}&-1-q_3&\cdots\\
\vdots &\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{matrix}$$
\end{frame}


\section{Further work}
\begin{frame}{Further work}\small
\begin{itemize}
\item Exponential moment \& exponential ergodicity:
$( \mathbb E_i\mathrm{e}^{\lambda\tau_0}-1, i\geqslant 1)$ is the minimal nonnegative solution to
$$x_i=\sum_{j\ne i,0}\frac{q_{ij}}{q_i-\lambda}\cdot x_j+\frac {\lambda}{q_i-\lambda},\ i\geqslant 1,$$
where $0<\lambda<q_i$ for all $i$.

In the single birth case, let $\tilde q_i^{(k)}=q_i^{(k)}-\lambda$, $\tilde d_i=\cdots$, $\tilde F_i^{(k)}=\cdots$. Now the approximating equation is
$$x_i=\sum_{j=1}^{i-1}\frac{q_{ij}}{q_i-\lambda}\cdot x_j+\frac{q_{i,i+1}}{q_i-\lambda}\cdot x_{i+1}+
\frac {\lambda}{q_i-\lambda},\ 1\leqslant i\leqslant n-1,$$
$$x_n=\sum_{j=1}^{n-1}\frac{q_{nj}}{q_n-\lambda}\cdot x_j+\frac {\lambda}{q_n-\lambda}.$$
\end{itemize}
\end{frame}

\begin{frame}{Further work}\small
Then
$$w_i=x_{i+1}-x_i=\tilde F_i^{(0)}x_1-\lambda \tilde d_i,\ 1\leqslant i<n.$$
Further
$$x_i=\sum_{k=0}^{i-1}(\tilde F_k^{(0)}x_1-\lambda \tilde d_k),\ 1\leqslant i\leqslant n.$$
Substituting it into
$$(q_n-\lambda)x_n=\sum_{j=1}^{n-1}q_{nj}x_j+\lambda,$$
we have
$$x_1=\frac{\lambda\sum_{k=0}^{n}\tilde d_k}{\sum_{k=0}^{n}\tilde F_k^{(0)}}=\mathbb E_1\textrm{e}^{\lambda\tau_0^{(n)}}-1.$$

{\blue $$\text{exp. ergoc} \Leftrightarrow \exists 0<\lambda<q_i\,(i\geqslant 0)\ \textrm{s.t.}\ %\frac{\sum_{k=0}^{n}\tilde d_k}{\sum_{k=0}^{n}\tilde F_k^{(0)}}\geqslant 0\,(n\geqslant 1),$$} and {\blue $$
\frac{\sum_{k=0}^{n}\tilde d_k}{\sum_{k=0}^{n}\tilde F_k^{(0)}}\uparrow \tilde d<+\infty.$$}
\end{frame}

\begin{frame}{Further work}\small
\begin{itemize}
\item For $m$-death?
\begin{block}{}
	Given a regular and irreducible single death $Q$-matrix $Q=(q_{ij})$ on $\mathbb{Z}_{+}$. Then it follows that\vspace{-0.8em}
	$$	\mathbb{E}_i\tau_0=\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{i}\sum\limits_{\ell=k}^{n}\frac{G_k^{(\ell)}}{q_{\ell,\ell-1}}-
\sum\limits_{k=1}^{i}G_k^{(n+1)}\cdot\frac{\sum\limits_{k=1}^{n}\sum\limits_{\ell=k}^{n}\dfrac{G_k^{(\ell)}}{q_{\ell,\ell-1}}}
{\sum\limits_{k=1}^{n+1}G_k^{(n+1)}}\right)  ,\  i\geqslant 1.
	$$
\end{block}

[Z., Front. Math. China 2018]
$$	\mathbb{E}_i\tau_0=\sum_{k=1}^{i}\sum_{\ell=k}^{+\infty}\frac{G_k^{(\ell)}}{q_{\ell,\ell-1}},\ i\geqslant 1.$$

For $2$-death?
\end{itemize}
\end{frame}

\begin{frame}{Further work}\small
\begin{itemize}
\item QSD? $2$-death?
\item Y.-Y. Li \&J.-P. Li. Down/up crossing properties of weighted Markov collision processes. Front. Math. China 2021, 16(2): 525–542.
$$q_{ij}=\begin{cases}w_ib_{j-i+2},&i\geqslant 2, j\geqslant i-2,\\ 0,&\text{otherwise}.\end{cases}$$
where $b_j \geqslant 0\,(j \ne  2)$, $0 <-b_2 =\sum_{j\ne 2}b_j <+\infty$, $w_i>0 \,(i \geqslant  2)$.

They consider the $(\mathbb S-2)$-range crossing number of the process until its extinction, i.e., the joint probability distribution
of the $|\mathbb S|$-dimensional random vector ($Y(t)=(Y_i(t), i\in\mathbb S$), where $Y_i(t)$ is the $(i-2)$-step crossing number of the process until $X(t)$'s extinction.
$$\mathbb S=\{0,1\}, \{k\}, \{0, 1, k\}\,(k\geqslant 3),\ (X(t), Y(t))$$
{\blue Dual method? 2-birth?}
\end{itemize}
\end{frame}


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

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