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\title[单生(死)~$Q$~矩阵零流出(入)的判别准则]{\bf  单生(死)~$Q$~矩阵零流出(入)的\\ 判别准则}
\author[张余辉]{张余辉}

\date[8.15, 2012]{\wuhao 第八届海峡两岸统计与概率研讨会(8.15-16, 2012. 哈尔滨). \\ 此工作与王玲娣合作完成.}

\institute[北京师范大学数学科学学院]{\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
%----------------------------------------------------------------------------------------1

\frame{\frametitle{内容}
%\wuhao
\begin{enumerate}
\item 研究背景
\item 单生~$Q$~矩阵零流出
\item 带移民单生~$Q$~矩阵零流出
\item 单死~$Q$~矩阵零流入
\end{enumerate}
}
%----------------------------------------------------------------------------------------2
\frame{\frametitle{研究背景}

\begin{itemize}
\item 状态空间~$\mathbb Z_+:=\{0, 1, 2, \cdots\}$~上的马氏链~$(X(t))_{t\geqslant0}$, 转移概率矩阵~$P(t)=(p_{ij}(t), i, j\in\mathbb Z_+)$.
\item 转移速率矩阵 $Q=(q_{ij})$ 即~$P(t)$~在时刻~$0$~的导数:
$$q_{ij}:=\lim_{t\to 0}\frac{p_{ij}(t)-\delta_{ij}}t,\quad i, j\in\mathbb Z_+.$$
\item $$0\leqslant q_{ij}<\infty,\ i\ne j;\ \sum_{j\ne i}q_{ij}\leqslant -q_{ii}=:q_i\leqslant \infty,\ i\in \mathbb Z_+.$$
\item 全稳定: 对所有~$i\in\mathbb Z_+$~都有~$q_i<\infty$.
\item 保守: 对所有~$i\in\mathbb Z_+$~都有~$q_i=\sum_{j\ne i}q_{ij}$ (等价地, $\sum_{j\in\mathbb Z_+}q_{ij}=0$)
\end{itemize}
}
%----------------------------------------------------------------------------------------3
\frame{\frametitle{研究背景}
\begin{block}{假定}
\centerline {~$Q$~矩阵是全稳定的.}
\end{block}

\begin{itemize}
\item 零流出: 若对某个(等价地, 对所有)~$\lambda>0$, 方程
$$
\lambda x_i=\sum_{j=0}^\infty q_{ij}x_j,\qquad 0\le x_i\le 1,\ i\in \mathbb Z_+
$$
只有平凡解(零解).
\item 零流入: 若对某个(等价地, 对所有)~$\lambda>0$, 方程
$$
\lambda y_i=\sum_{j} y_jq_{ji},\qquad y_i\ge 0,\ i\in \mathbb Z_+,\text{ 且 } \sum_{i} y_i<\infty
$$
只有平凡解.
\end{itemize}
}
%----------------------------------------------------------------------------------------4
\frame{\frametitle{研究背景}
唯一性:  给定一个矩阵~$Q$, 是否存在唯一过程~$P(t)$~使得其转移速率矩阵为~$Q$?

\begin{theorem}[Hou(1974)]\rm\kaishu
给定~$Q$~矩阵~$Q=(q_{ij}: i, j\in\mathbb Z_+)$, 过程唯一当且仅当以下条件同时成立:

(1) 对某~$\lambda>0$, $\inf_{i\in E^0}\sum_{j\in\mathbb Z_+} P^{\min}_{ij}(\lambda)>0$, 其中\\(非保守点集)~$E^0:=\{i\in\mathbb Z_+: q_i-\sum_{j\ne i}q_{ij}>0\}$,\\ ~$P^{\min}(\lambda)$~是最小~$Q$~过程~$P^{\min}(t)$~的~Laplace~变换;

(2) $Q$~矩阵零流出;

(3) $Q$~矩阵保守或者~$Q$~矩阵非保守但零流入.
\end{theorem}
}
%----------------------------------------------------------------------------------------5
\frame{\frametitle{研究背景}

\begin{itemize}
\item 单生~$Q$~矩阵: 对所有~$i, j\in\mathbb Z_+$, $j\geqslant i+2$~都有~$q_{ij}=0$, 且
$$\aligned N:&=\sup\{i+1: q_{i,i+1}=0,i\in\mathbb Z_+\}\\
&=\inf\{i:\text{对}\ k\ge i\ \text{都有}\ q_{k,k+1}>0\}<\infty,\endaligned$$
当~$N\geqslant 1$~时, 也称该~$Q$~矩阵为带吸收边界的单生~$Q$~矩阵.

\item 单死~$Q$~矩阵: 若对所有~$i, j\in\mathbb Z_+$, $j\leqslant i-2$~都有~$q_{ij}=0$, $q_{i,i-1}>0$.
\end{itemize}
}
%----------------------------------------------------------------------------------------6

\frame{\frametitle{研究背景}
单生~$Q$~矩阵
\vspace{-1.2cm}
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\vskip0.3cm
$$Q=\left(%
\begin{array}{ccccc}
  - & + & 0 & 0 & \cdots \\
  * & - & + & 0 & \cdots \\
  * & * & - & + & \cdots\\
  \vdots & \vdots & \vdots & \vdots & \ddots \\
  \end{array}%
\right)$$

}
%----------------------------------------------------------------------------------------7
\frame{\frametitle{研究背景}

\begin{itemize}
\item 对无吸收边界的保守单生~$Q$~矩阵, 张建康(1984)用概率方法首先给出了零流出的判别准则. 严士健和陈木法(1986)给出该准则的分析证明.

\item 对带吸收边界的保守单生~$Q$~矩阵, 陈木法(1999)用分析方法得到零流出的判别准则.

\item Chen A-Y~(2005)用分析方法研究了无吸收边界的保守单生~$Q$~矩阵零流出和保守单死~$Q$~矩阵零流入的判别准则, 并且本质上给出了非保守生灭~$Q$~矩阵的零流出和零流入的判别.

\end{itemize}
}
%----------------------------------------------------------------------------------------8

\frame{\frametitle{研究背景}

\vskip0.3cm
\begin{block}{}
我们将单生~$Q$~矩阵零流出和单死~$Q$~矩阵零流入的判别准则推广至非保守情形, 而且将两种情形下的判别相统一. 还统一处理了保守和非保守两种情形下带移民的单生~$Q$~矩阵零流出的判别问题.
\end{block}
\vskip0.3cm
\begin{block}{}
Chen  A-Y, Pollett P, Zhang H-J, Cairns B. (2005). Uniqueness criteria for continuous-time Markov chains with general transition structure. Adv Appl Probab 37(4): 1056-1074.
\end{block}
\begin{block}{}
Chen M-F.(1999). Single birth processes. Chinese Ann Math 20B: 77-82.
\end{block}
}

%---------------------------------------------------------------------------------------9
\frame{\frametitle{单生~$Q$~矩阵零流出}
记~$c_i:=q_i-\sum_{j\ne i}q_{ij}$~为~$Q$~矩阵在~$i$~点的非保守量.

\begin{theorem}\kaishu\rm 给定单生~$Q$~矩阵~$Q=(q_{ij}: i, j\in\mathbb Z_+)$. 定义~$q_n^{(k)}=\sum_{j=0}^k q_{nj}$ $(k<n)$.
则~$Q$~矩阵是零流出的当且仅当 ~$R:=\sum_{n=N}^\infty m_n=\infty$, 其中
$$
m_n=\frac 1{q_{n,n+1}}\bigg(1+c_n+q_n^{(N-1)}+\sum_{k=N}^{n-1}q_n^{(k)} m_k\bigg),\ n\ge N.
$$
\end{theorem}
}

%---------------------------------------------------------------------------------------10
\frame{\frametitle{单生~$Q$~矩阵零流出}

证明思路:
\begin{block}{}
(1) 在扩大的状态空间~$\overline {\mathbb Z}_+:=\{-1, 0, 1, \cdots\}$~上定义新的带吸收边界的保守单生~$Q$~矩阵~$\widehat Q=(\hat q_{ij}: i,j\in \overline {\mathbb Z}_+)$~如下:
$$\hat q_{-1,j}=0,\ j\in  \overline {\mathbb Z}_+; \ \hat q_{i,-1}=c_i,\
\hat q_{ij}=q_{ij},\ i, j\in\mathbb Z_+.$$
\end{block}

\begin{block}{}
(2) 证明矩阵~$Q$~是零流出的当且仅当~$\widehat Q$~是零流出的. 因此, 将状态空间~$\mathbb Z_+$~上非保守单生~$Q$~矩阵零流出的判定转化成了扩大状态空间~$\overline {\mathbb Z}_+$~上带吸收边界的保守单生~$Q$~矩阵零流出的判定.
\end{block}

}
%---------------------------------------------------------------------------------------11
\frame{\frametitle{单生~$Q$~矩阵零流出}

证明思路(续):
\begin{block}{}
(3) 由~Chen(1999), $\widehat Q$~零流出当且仅当~$\sum_{n=N}^\infty \widehat m_n=\infty$.
$$
\widehat m_n=\frac 1{q_{n,n+1}}\bigg(1+c_n+q_n^{(N-1)}+\sum_{k=N}^{n-1}(c_n+q_n^{(k)})\widehat m_k\bigg),\ n\ge N;
$$
\end{block}

\begin{block}{}
(4) 证明两个级数~$\sum_{n=N}^\infty \widehat m_n$~和~$\sum_{n=N}^\infty m_n$~同敛散.
\end{block}
}
%---------------------------------------------------------------------------------------12
\frame{\frametitle{单生~$Q$~矩阵零流出}

\begin{corollary}\kaishu\rm 给定生灭~$Q$~矩阵~$Q=(a_i, b_i, c_i: i\in\mathbb Z_+)$, 其中~$a_i:=q_{i,i-1}, b_i:=q_{i,i+1}, c_i:=q_i-(a_i+b_i)$, 约定~$a_0=0$.
假定
$$b_i>0,\quad  i\ge N;\qquad a_i>0,\quad i\ge N+1,$$
令
$$\mu_N=1,\qquad \mu_n=\frac{b_N b_{N+1}\cdots b_{n-1}}{a_{N+1} a_{N+2}\cdots a_n},\qquad n\ge N+1.$$
\end{corollary}
}

%---------------------------------------------------------------------------------------13
\frame{\frametitle{单生~$Q$~矩阵零流出}

\begin{corollary}[continued]\kaishu\rm 
则该生灭~$Q$~矩阵零流出当且仅当
$$
\sum_{n=N}^\infty \frac 1{\mu_n b_n}\bigg(\sum_{k=N}^n\mu_k(1+c_k)+\mu_Na_N\bigg)=\infty.
$$
特别地, 当~$N=0$~时, 该生灭~$Q$~矩阵零流出当且仅当
$$
\sum_{n=0}^\infty \frac 1{\mu_n b_n}\sum_{k=0}^n\mu_k(1+c_k)=\infty.
$$
\end{corollary}
}

%---------------------------------------------------------------------------------------14
\frame{\frametitle{带移民单生~$Q$~矩阵零流出}

\begin{block}{}
带移民单生~$Q$~矩阵:
$$q_{01}>0 ( q_{0k}\ge 0, k\ge2);$$
$$q_{i,i+1}>0,\ q_{ij}=0,\ i\ge 1,\; j\ge i+2.$$
\end{block}
\begin{block}{}
张余辉, 赵倩倩.(2010). 带移民的单生过程. 数学学报, 53(5): 833-846.
\end{block}
}
%---------------------------------------------------------------------------------------15
\frame{\frametitle{带移民单生~$Q$~矩阵零流出}

\begin{theorem}\kaishu\rm 对带移民单生~$Q$~矩阵~$Q=(q_{ij}: i, j\in\mathbb Z_+)$, 则~$Q$~矩阵是零流出的当且仅当~$R:=\sum_{n=0}^\infty m_n=\infty$, 其中
$$
m_0=\frac1{q_0},\ m_n=\frac 1{q_{n,n+1}}\bigg(1+c_n+\sum_{k=0}^{n-1}q_n^{(k)} m_k\bigg),\ n\ge 1.
$$
\end{theorem}

}

%---------------------------------------------------------------------------------------16
\frame{\frametitle{单死~$Q$~矩阵零流入}

\begin{theorem}\kaishu\rm
给定单死~$Q$~矩阵~$Q=(q_{ij}: i, j\in\mathbb Z_+)$. 定义~$\bar q_k^{(n)}=\sum_{j=n}^\infty q_{kj}$ $(n>k)$. 则~$Q$~矩阵是零流入的当且仅当~$\overline R:=\sum_{n=0}^\infty \overline m_n=\infty$, 其中
$$\overline m_0=0,\ \overline m_n=\frac 1{q_{n,n-1}}\bigg(1+\sum_{k=0}^{n-1}(c_k+\bar q_k^{(n)})\overline m_k\bigg),\ n\ge 1.$$
\end{theorem}
}
%---------------------------------------------------------------------------------------18
\frame{\frametitle{单死~$Q$~矩阵零流入}

\begin{theorem}\kaishu\rm
给定单死~$Q$~矩阵~$Q=(q_{ij}: i, j\in\mathbb Z_+)$.  则该~$Q$~矩阵有唯一不变测度~$(\mu_i)$~如下:
$$
\mu_0=1,\ \mu_n=\frac 1{q_{n,n-1}}\sum_{k=0}^{n-1}(c_k+\bar q_k^{(n)})\mu_k,\ n\ge 1.
$$
进一步, 该~$Q$~矩阵有唯一不变概率测度~$(\pi_i)$~当且仅当~$\mu:=\sum_{n=0}^\infty \mu_n<\infty$, 此时~$\pi_i=\mu_i/\mu$.
\end{theorem}
}

%----------------------------------------------------------------------------------------19
\frame{\frametitle{}

\begin{center}{\sihao\Large\blue 谢谢!}
\end{center}
}

\end{CJK*}

\end{document}
