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\begin{document}
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\title[单死过程首中时的高阶矩]{单死过程首中时的高阶矩}

\author[张余辉]{\sihao 张余辉}

\date[2018年11月11日]{\wuhao 随机分析与偏微分方程研讨会\\ (中南大学, 2018年11月9日-11日) }

\institute[北京师范大学]{\wuhao 北京师范大学}

%\frame{\titlepage}


 \begin{frame}
  \includegraphics[scale=0.3]{math.png}\\
\maketitle
 \end{frame}

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


%----------------------------------------------------------------------------------------2
\frame{ \frametitle{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

$Q$-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}
}
%----------------------------------------------------------------------------------------3
\frame{ \frametitle{Background}\small
Single death process:
$Q=(q_{ij})$ satisfies
$${\red q_{i,i-1}>0},\ q_{i,i-j}=0,\ i\geqslant 1,j> 1.$$
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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{Background}\small
\begin{itemize}

\item branching processes:

\end{itemize}

$$\aligned
 i&\to i-1\quad\text{at rate}\quad \alpha ip_0\\
  &\to i\qquad\quad \text{at rate}\quad -\alpha i(1-p_1)\\
  &\to i+1\quad\text{at rate}\quad \alpha ip_2\\
  &\to i+2\quad\text{at rate}\quad \alpha ip_3\\
  &\to \cdots
\endaligned$$
}

%----------------------------------------------------------------------------------------5
\frame{ \frametitle{Background}\small

\begin{itemize}
\item 2017.6.29-2017.7.1, 南宁, The 6th IMS-China International Conference on Statistics and Probability, Notes on single death processes.
\item 2017.7.2-2017.7.9, 北师大的随机过程暑期学校, 从单生到单死.
\item 2017.10.25-2017.10.25, 中国农业大学, 非对称随机过程：从单生到单死.
\item 2017.11.25-2017.11.26, 中南大学, 马氏过程与随机分析学学术论坛, 单死过程遍历性和强遍历性.
\item Z. Frontiers of Mathematics in China, 2018, 13(5): 1215-1243.
\item 2018.1.11-2018.1.14, 南方科技大学, 单死过程和树上生灭过程.
\item 2018.4.27-2018.4.28, 天津大学, 单死过程和树上生灭过程.
\item 2018.7.16-2018.7.20, The 14th Workshop on Markov Processes and Related Topics, Moments of the first hitting times for birth-death processes on trees.
\item Z. submitted to Frontiers of Mathematics in China (2018).
\end{itemize}

\qquad 这个报告的工作是与周晓凤合作完成的.
}
\section{Main results}

%----------------------------------------------------------------------------------------6
\frame{\frametitle{Main results}\small

Define
$$\tau_i=\inf\{t\geqslant 0: X(t)=i\}, \ \sigma_i=\inf\{t\geqslant \text{the first jumping time}: X(t)=i\}.$$

\begin{block}{} Define $q_n^{(k)}:=\sum_{j=k}^\infty q_{nj},\  k>n\geqslant 0$.
$$
G_i^{(i)}=1,\qquad  G_n^{(i)}=\frac 1{q_{n,n-1}}
\sum_{k=n+1}^{i} q_n^{(k)} G_k^{(i)},\qquad 1\leqslant n< i.
$$
\end{block}

\begin{block}{Theorem 1.} Assume that the single death $Q$ is irreducible and the corresponding process is recurrent. Given $i_0\in \mathbb Z_+$ and a positive integer $n\geqslant 1$ arbitrarily. Then
$$\mathbb E_i\tau_{i_0}^n=n\sum_{i_0+1\leqslant k\leqslant i}\sum_{\ell\geqslant k}\frac{G_k^{(\ell)}\mathbb E_\ell\tau_{i_0}^{n-1}}{q_{\ell,\ell-1}},\qquad i\geqslant i_0+1.$$
\end{block}

}
%----------------------------------------------------------------------------------------7
\frame{\frametitle{Main results}\small
\begin{block}{Corollary 2.} Assume that the single death $Q$-matrix is regular and the proc. is exp. erg.. Then
$$\delta:=\sup_{i\geqslant 1}\sum_{1\leqslant k\leqslant i}\sum_{\ell\geqslant i}\frac {G_k^{(\ell)}}{q_{\ell,\ell-1}}<\infty.$$
\end{block}

\begin{block}{} $\ell$-ergodicity: $\mathbb E_j\sigma_j^\ell<\infty$ for some $j$. {\red $1$-erg. $=$ posit. recur., $0$-erg. $=$ null recur.}\end{block}

%\begin{block}{}\end{block}

\begin{block}{Corollary 3.} Assume that the single death $Q$-matrix is regular and the process is recurrent. Then the process is $\ell$-ergodic if and only if
$$d_\ell:=\sum_{k\geqslant 1}q_{0}^{(k)}\sum_{j\geqslant k}\frac{G_k^{(j)}}{q_{j,j-1}}\mathbb E_j\tau_0^{\ell-1}<\infty.$$
\end{block}
}
%----------------------------------------------------------------------------------------8
\frame{\frametitle{Main results}
\small

\begin{block}{Proposition 4}
Let the single death $Q$-matrix be regular and irreducible. Assume that
$$%\begin{equation}\label{e4.03}
\sum_{k\geqslant 1} \frac{q_0^{(k)}}{G_1^{(k)}}<\infty.$$%\end{equation}
If
$$q:=\inf_{n\geqslant 0}q_n>0\quad\text{and}\quad M:=\sup_{n\geqslant 1}\Bigg(\sum_{k=1}^{n}\frac 1{G_1^{(k)}}\Bigg)\Bigg(\sum_{j=n}^\infty\frac {G_1^{(j)}}{q_{j,j-1}}\Bigg)<\infty,$$
then the process is exponentially ergodic.
\end{block}

\begin{block}{} $$M\geqslant \delta.$$\end{block}
}

\section{Sketch of proof}
%----------------------------------------------------------------------------------------9
\frame{\frametitle{Sketch of proof}\small
\begin{block}{}
Define
$$
M_{ik}^{(n-1)}=\sum_{i+1\leqslant \ell\leqslant k}\sum_{1\leqslant s\leqslant n-1}\binom {n}{s}\mathbb E_{\ell}\tau_{\ell-1}^{n-s}\mathbb E_{\ell-1}\tau_{i-1}^s,$$
$$%\begin{equation}\label{e2.2-}
M_{i}^{(n-1)}=\mathbb E_i\tau_{i-1}^{n-1}+\frac 1n\sum_{k\geqslant i+1}q_{ik}M_{ik}^{(n-1)},\qquad n\geqslant 1.
$$%\end{equation}
\end{block}

\begin{itemize}
\item Single death property: $\mathbb E_i\tau_{i-1}^n=?$
\end{itemize}

\begin{block}{}
$$x_i=\frac{ q_i^{(i+1)}}{ q_i}x_i+\sum_{k\geqslant i+1}\frac{ q_i^{(k)}}{ q_i}x_k+\frac n{q_{i}}M_i^{(n-1)},\ i\geqslant 1.\qquad\qquad (*)$$
\end{block}

}

%----------------------------------------------------------------------------------------10
\frame{\frametitle{Sketch of proof}

\begin{block}{Proposition 5. Under the conditions of Theorem 1}
$$\mathbb E_i\tau_{i-1}^n=n\sum_{k\geqslant i}\frac{G_i^{(k)}M_k^{(n-1)}}{q_{k,k-1}},\ i\geqslant 1, n\geqslant 1.$$
\end{block}

\vskip0.5cm
Denote the left hand side by $m_i$, the right hand side by $h_i$.

$$h_i=\frac 1{q_{i,i-1}}\Bigg(nM_i^{(n-1)}+\sum_{k\geqslant i+1}q_i^{(k)}h_k\Bigg).$$
}


%----------------------------------------------------------------------------------------11
\frame{\frametitle{Sketch of proof}

\begin{block}{Lemma 6. }
$(h_i)$ is the minimal nonnegative solution to
$$x_i=\frac{ q_i^{(i+1)}}{ q_i}x_i+\sum_{k\geqslant i+1}\frac{ q_i^{(k)}}{ q_i}x_k+\frac n{q_{i}}M_i^{(n-1)},\ i\geqslant 1.\qquad\qquad (*)$$
\end{block}

{\red $$m_i\geqslant h_i.$$}

}

%----------------------------------------------------------------------------------------12
\frame{\frametitle{Sketch of proof}\small

Sketch proof of Lemma 6:

Fix $N\geqslant 2$, define $Q$-matrix $Q^{(N)}=(\tilde q_{ij})$ on $\{0,1,2,\cdots, N\}$:
$$\tilde q_{ij}=\begin{cases} q_{ij} &\text{if}\ i<N, j<N;\\ q_i^{(N)} &\text{if}\ i<N,j=N;\\ (q_N\vee N)(1+nG^{(N)}a_N) &\text{if}\ i=N, j=N-1;\\
-(q_N \vee N)(1+nG^{(N)}a_N) &\text{if}\ i=N, j=N;\\
0, &\text{if}\ i=N, j<N-1,\end{cases}$$
where $G^{(N)}=\max_{1\leqslant i\leqslant N}G_i^{(N)}$ and
$$a_N=\begin{cases} M_{N}^{(n-1)}&\text{if}\ M_{N}^{(n-1)}<\infty; \\ 1 & \text{if}\ M_{N}^{(n-1)}=\infty.\end{cases}$$
Define $\tilde q_n^{(k)}=\sum_{j=k}^N \tilde q_{nj},\ 0\leqslant n< k\leqslant N,$
$$
\widetilde G_i^{(i)}=1,\qquad   \widetilde G_n^{(i)}=\frac 1{\tilde q_{n,n-1}}
\sum_{k=n+1}^{i} \tilde q_n^{(k)} \widetilde G_k^{(i)},\qquad 1\leqslant n< i\leqslant N.
$$
}

%----------------------------------------------------------------------------------------13
\frame{\frametitle{Sketch of proof}\small

Sketch proof of Lemma 6(continued):

$$
h_n^{(N)}:=n\sum_{k=i}^N\frac{\widetilde G_i^{(k)}M_k^{(n-1)}}{\tilde q_{k,k-1}}\quad (1\leqslant i\leqslant N)
$$
is a unique solution (the minimal non-negative solution) to the following equations
$$
x_i=\frac {\tilde q_{i}^{(i+1)}}{\tilde q_i}\cdot x_i+\sum_{k=i+1}^N\frac {\tilde q_{i}^{(k)}}{\tilde q_i}\cdot x_k+\frac n{\tilde q_i}M_i^{(n-1)},\qquad 1\leqslant i\leqslant N.
$$

$$
x_N=\frac {nM_N^{(n-1)}}{(q_N\vee N)(1+nG^{(N)}a_N)},\ x_i=\frac {q_{i}^{(i+1)}}{q_i}\cdot x_i+\sum_{k=i+1}^N\frac {q_i^{(k)}}{q_i}\cdot x_k+\frac n{q_i}M_i^{(n-1)}.
$$

$ (h_n^{(N)})$ is increasing to the minimal non-negative solution of ($*$) as $N\to \infty$.

$$h_n^{(N)}=n\sum_{k=i}^{N-1}\frac{G_i^{(k)}M_k^{(n-1)}}{q_{k,k-1}}+\frac {nG_i^{(N)}M_N^{(n-1)}}{(q_N\vee N)(1+nG^{(N)}a_N)}\to n\sum_{k=i}^{\infty}\frac{G_i^{(k)}M_k^{(n-1)}}{q_{k,k-1}}=h_i.$$
}

%----------------------------------------------------------------------------------------14
\frame{\frametitle{Sketch of proof}\small

\begin{block}{Lemma 7. Under the conditions of Theorem 1}
$(\mathbb E_i\tau_{i_0}^n: i\ge i_0)$ is the minimal nonnegative solution to
$$
x_{i_0}=0,\qquad x_i=\sum_{j\ne i,j> i_0}\frac {q_{ij}}{q_i} x_j+\frac n{q_i}\mathbb E_i\tau_{i_0}^{n-1},\qquad  i> i_0.
$$
\end{block}

$\displaystyle n\sum_{i_0+1\leqslant k\leqslant i}\sum_{\ell\geqslant k}\frac {G_k^{(\ell)}M_{\ell}^{(n-1)}}{q_{\ell,\ell-1}}+M_{i_0+1,i}^{(n-1)}$ satisfies the equation above.

$$\mathbb E_i\tau_{i_0}^n\leqslant n\sum_{i_0+1\leqslant k\leqslant i}\sum_{\ell\geqslant k}\frac {G_k^{(\ell)}M_{\ell}^{(n-1)}}{q_{\ell,\ell-1}}+M_{i_0+1,i}^{(n-1)}\Rightarrow  {\red m_i\leqslant h_i}.$$
}

%----------------------------------------------------------------------------------------15
\frame{\frametitle{Sketch of proof}\small

\begin{block}{Proposition 5. Under the conditions of Theorem 1}
$$\mathbb E_i\tau_{i-1}^n=n\sum_{k\geqslant i}\frac{G_i^{(k)}M_k^{(n-1)}}{q_{k,k-1}},\ i\geqslant 1, n\geqslant 1.$$
\end{block}

\begin{block}{Lemma 8. For all $1\leqslant i\leqslant v<u$,}
$$G_i^{(u)}=\sum_{i\leqslant k\leqslant v}\frac{G_i^{(k)}}{q_{k,k-1}}\sum_{\ell=v+1}^u q_{k}^{(\ell)}G_\ell^{(u)}, \qquad 1\leqslant i\leqslant v<u.$$
\end{block}

\begin{block}{Lemma 9. Under the conditions of Theorem 1} for any nonnegative sequence $\{a_n\}$ and $n\geqslant 1$,
$$\sum_{k\geqslant i}\frac{G_i^{(k)}}{q_{k,k-1}}\sum_{\ell_1\geqslant k+1}q_{k}^{(\ell_1)}\mathbb E_{\ell_1}\tau_{\ell_1-1}^{n}\sum_{\ell_2=k}^{\ell_1-1}a_{\ell_2}
=n\sum_{u\geqslant i+1}\frac{G_i^{(u)}M_u^{(n-1)}}{q_{u,u-1}}\sum_{\ell_2=i}^{u-1}a_{\ell_2}.$$
\end{block}

}
%----------------------------------------------------------------------------------------16
\frame{\frametitle{Sketch of proof}\small
\begin{block}{Lemma 10. Under the conditions of Theorem 1} for any $1\leqslant i\leqslant k< \ell_1$ and $n\geqslant 2$,
$$M_{k,\ell_1-1}^{(n-1)}+\sum_{s=1}^{n-1}\binom{n}{s}\mathbb E_{\ell_1-1}\tau_{k-1}^{n-s}\mathbb E_{k-1}\tau_{i-1}^{s}=\sum_{\ell_2=k}^{\ell_1-1}\sum_{s=1}^{n-1}\binom{n}{s}\mathbb E_{\ell_2}\tau_{\ell_2-1}^{n-s}\mathbb E_{\ell_2-1}\tau_{i-1}^{s}.$$
\end{block}

\begin{block}{Lemma 11. Under the conditions of Theorem 1} for all $1\leqslant s_1< n$,
$$\aligned &\sum_{k\geqslant i}\frac{G_i^{(k)}}{q_{k,k-1}}\sum_{\ell_1\geqslant k+1}q_{k}^{(\ell_1)}\mathbb E_{\ell_1}\tau_{\ell_1-1}^{n-s_1}\Bigg(M_{k,\ell_1-1}^{(s_1-1)}+\sum_{s_2=1}^{s_1-1}\binom{s_1}{s_2}\mathbb E_{\ell_1-1}\tau_{k-1}^{s_1-s_2}\mathbb E_{k-1}\tau_{i-1}^{s_2}\Bigg)\\
&\qquad =(n-s_1)\sum_{u\geqslant i+1}\frac{G_i^{(u)}}{q_{u,u-1}}M_u^{(n-s_1-1)}M_{i,u-1}^{(s_1-1)}.\endaligned$$
\end{block}
}

%----------------------------------------------------------------------------------------17
\frame{\frametitle{Sketch of proof}\small
\begin{block}{Sketch of proof of Theorem 1.}
$$\mathbb E_i\tau_{i-1}^{n}=h_i=n\sum_{k\geqslant i}\frac{G_i^{(k)}\mathbb E_k\tau_{i-1}^{n-1}}{q_{k,k-1}},\qquad i\geqslant 1,n\geqslant 1.$$

$$\mathbb E_i\tau_{k}^{n}=n\sum_{k+1\leqslant j\leqslant i}\sum_{\ell\geqslant j}\frac{G_j^{(\ell)}\mathbb E_\ell\tau_{k}^{n-1}}{q_{\ell,\ell-1}},\qquad 0\leqslant k<i,n\geqslant 1.$$
\end{block}
}
\section{Future work}
%----------------------------------------------------------------------------------------18
\frame{\frametitle{Future work}\small
\begin{block}{} Define a positive matrix $H=(h_{ij})_{i,j\geqslant 1}$:
$$h_{ij}=\frac 1{q_{j,j-1}}\sum_{k=1}^{i\wedge j}G_{k}^{(j)}.$$
\end{block}

$$\mathbb E_k\text{e}^{\alpha\tau_0}=\sum_{n=0}^\infty((\alpha H)^n\mathbbold 1)(k).$$

\begin{block}{}
$$\frac 1{\alpha_{\text{max}}}=\inf_{y\in\mathscr F}\sup_{k\geqslant 1}\frac{(Hy+H\mathbbold 1)(k)}{y_k}<\infty \Leftrightarrow \text{exp. erg.}\Leftrightarrow \lambda_{\text{max}}<\infty.$$
where
$$\mathscr F=\{y: 0<y\upuparrows, \sum_{k\geqslant 1}q_{0k}y_k<\infty\}.$$
\end{block}
}

%----------------------------------------------------------------------------------------19
\frame{\frametitle{Future work }\small

\begin{block}{Theorem [Tweedie(1981)]} Given a regular irreducible $Q$-matrix $Q=(q_{ij})$. Then the process
$P(t)$ is exponentially ergodicity if and only if for some $\alpha>0$ with $\alpha <q_i$ for all $i$,
$$\begin{cases}y_i\geqslant 1,\quad i\in E&\\ \sum_j q_{ij}y_j\leqslant -\alpha y_i,\quad i\notin H&\\
\sum_{i\in H}\sum_{j\ne i}q_{ij}y_j<\infty&
\end{cases}$$
has a finite solution $(y_i)$.
\end{block}

\begin{block}{}
$$H=((-Q)|_{i\geqslant 1})^{-1}=\bigg(\int_{0}^{\infty}p_{ij}^{\text{min}}(t)\text{d}t\bigg).$$
\end{block}
}
%----------------------------------------------------------------------------------------20
\frame{\frametitle{Future work}\small

\begin{block}{Example 12.} Given a constant $b>2$ (for regularity only needed that $b>1$). Let
$$q_{ij}=\frac{b-1}{b^{j-i+2}},\quad j\geqslant i+1;\quad q_{i,i-1}=\frac{b-1}b,\quad q_i=-q_{ii}=\frac {b^2-b+1}{b^2},\quad i\geqslant 1;$$
$$q_{0j}=\frac{b-1}{b^{j+1}},\qquad j\geqslant 1;\qquad q_0=-q_{00}=\frac 1b.$$
\end{block}

Then $q_n^{(k)}=1/b^{k-n+1}(1\leqslant n<k)$,  $q_0^{(k)}=1/b^{k}(k\geqslant 1)$, $q=1/b$ and
$$G_n^{(i)}=\frac1 {b(b-1)^{i-n}},\qquad 1\leqslant n <i.$$
We know that the process is exp. erg. but not strongly ergodic. Now
$$\delta=\frac{b-1}{(b-2)\wedge(b-2)^2}, \quad \sum_{k\geqslant 1} \frac{q_0^{(k)}}{G_1^{(k)}}=\frac{b^2-b+1}b \quad \text{and}\quad M=\frac {b(b-1)}{(b-2)^2}.$$
}


\section{References}
\frame{\frametitle{References}\small

\begin{itemize}
\item Z. Criteria on ergodicity and strong ergodicity of single death processes, Frontiers of Mathematics in China, 2018, 13(5): 1215-1243.
\item Z. \& Yan. 单死过程一些数字特征的概率意义, 北京师范大学学报, 2019, 55(3).
\item Z. 单死过程研究的一些最新进展, 中国科学: 数学, 2019.
\item Z. Moments of the first hitting times for birth-death processes on trees, 2019.
\item Z. \& Zhou. High-order moments of the first hitting times for single death processes, 2019.
\end{itemize}
}

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

\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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