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\title[Unified representation of formulas for SBP]{Unified representation of formulas for single birth processes}

\author[Yu-Hui Zhang]{Yu-Hui Zhang}

\date[March 31, 2014]{\wuhao (March 31, 2014; Fujian Normal University ) \\
$ $\\ This talk is based on the joint work with Mu-Fa Chen.}

\institute[BNU]{\sihao Beijing Normal University}

\frame{\titlepage}

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         背景及问题        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\large
%----------------------------------------------------------------------------------------1

\frame{
 \frametitle{Outline}\small

\begin{enumerate}
\item Background
\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
\end{enumerate}
}

%----------------------------------------------------------------------------------------2
\frame{ \frametitle{1. Background}\small
\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{1. Background }\small
$Q=(q_{ij}:i,j\ge 0)$: $q_{i,i+1}>0,\ q_{i,i+j}=0,\ i\ge 0,j\ge 2$.
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\vskip0.3cm
$$Q=\left(%
\begin{array}{ccccc}
  - & + & 0 & 0 & \cdots \\
  * & - & + & 0 & \cdots \\
  * & * & - & + & \cdots\\
  \vdots & \vdots & \vdots & \vdots & \ddots \\
  \end{array}%
\right)$$
}
%----------------------------------------------------------------------------------------4
\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
\end{enumerate}

}
%----------------------------------------------------------------------------------------5
\frame{\frametitle{2. Poisson equation}\small
Given a function $c$, define
$$\Omega g=Qg+cg, \quad \text{where}\quad (Qg)_i=\sum_{j}q_{ij}(g_j-g_i).$$
Clearly, if $c\le 0$, then $\Omega$ is an operator corresponding to a single birth
process with killing rates $(-c_i)$. Define
$$
\widetilde F_i^{(i)}=1,\qquad  \widetilde F_n^{(i)}=\frac 1{q_{n,n+1}}
\sum_{k=i}^{n-1}\tilde q_n^{(k)}\widetilde F_k^{(i)},\qquad n>i\ge 0,$$
$$\tilde q_n^{(k)}=q_n^{(k)}-c_n:=\sum_{j=0}^kq_{nj}-c_n,\qquad 0\le k<n.$$
}

%----------------------------------------------------------------------------------------6
\frame{\frametitle{2. Poisson equation}\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}
}
%---------------------------------------------------------7
\frame{\frametitle{2. Poisson equation}\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}
$$g_{i+1}-g_i=\frac 1{q_{i,i+1}}\Bigg(\sum_{0\le k \le i-1}\tilde q_{i}^{(k)}(g_{k+1}-g_k)+f_i-c_ig_0\Bigg).$$
}

%--------------------------------------------------------------------8
\frame{\frametitle{2. Poisson equation}\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}
}
%----------------------------------------------------------------------------9
\frame{\frametitle{2. 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}
}
%----------------------------------------------------------------------------------10
\frame{\frametitle{3. Uniqueness}\small
Define
$${\widetilde m}_0=\frac{1}{q_{01}},\quad
{\widetilde m}_n=\frac 1{q_{n,n+1}}
\bigg(1+\sum_{k=0}^{n-1}{\tilde q}_n^{(k)}{\widetilde m}_k\bigg),\quad n\ge 1.$$

$${\widetilde m}_n
=\sum_{k=0}^n\frac{{\widetilde F}_n^{(k)}}{q_{k,k+1}},\qquad n\ge 0.$$


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

Zhang Jiankang(1984), Yan Shijian \& Chen Mufa(1986)
}

%----------------------------------------------------------11
\frame{\frametitle{3. Uniqueness}\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 $\sum_n {m}_n=\infty$.
}
%----------------------------------------------------------12
\frame{\frametitle{4. Recurrence, extinction/return probability}\small

Yan Shijian \& Chen Mufa(1986)

\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.
}

%----------------------------------------------------------13
\frame{\frametitle{4. Recurrence, extinction/return probability}\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 $\sum_{k=0}^\infty F_k^{(0)}=\infty$.
}
%----------------------------------------------------------14
\frame{\frametitle{4. Recurrence, extinction/return probability}\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}
}
%----------------------------------------------------------15
\frame{\frametitle{4. Recurrence, extinction/return probability}\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:
$$ x_n=x_0\sum_{0\le k\le n-1} F_k^{(0)}-\sum_{1\le k\le n-1}F_k^{(0)},\qquad n\ge 0.$$
$$x_0^*=1-\frac 1{\sum_{k=0}^\infty F_k^{(0)}},\qquad x_n^*=1-\frac{\sum_{k=0}^{n-1}F_k^{(0)}}{\sum_{k=0}^\infty F_k^{(0)}},\qquad n\ge 1.$$
}
%----------------------------------------------------------16
\frame{\frametitle{\small 5. Ergodicity, strong ergodicity, and the first moment of return time}\small
Define
$$\tilde d_0=0,\quad \tilde d_n=\frac 1{q_{n,n+1}}\Bigg(1+\sum_{k=0}^{n-1}\tilde q_n^{(k)}\tilde d_k\Bigg),
\quad n\ge 1,$$
$$\tilde d_n=\sum_{1\le j\le n}\frac{\widetilde F_n^{(j)}}{q_{j,j+1}},\quad n\ge 0.$$

\begin{block}{Proposition} 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}
}



%----------------------------------------------------------17
\frame{\frametitle{\small 5. Ergodicity, strong ergodicity, and the first moment of return time}\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}

Yan Shijian \& Chen Mufa(1986), Zhang(2001, 2003)
}

%----------------------------------------------------------18
\frame{\frametitle{\small 5. Ergodicity, strong ergodicity, and the first moment of return time}\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:
$$x_n=\sum_{k=0}^{n-1}\bigg[F_k^{(0)}\bigg(x_0-\frac{1}{q_{01}}\bigg)-d_k\bigg],\quad n\ge 1.$$
}

%---------------------------------------------------------19
\frame{\frametitle{\small 5. Ergodicity, strong ergodicity, and the first moment of return time}\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
$$x_n^*=\sum_{k=0}^{n-1}\left(F_k^{(0)}\sup_{n\ge 1}\frac{\sum_{j=0}^{n-1} d_j}{\sum_{j=0}^{n-1}F_j^{(0)}}
-d_k\right),\quad n\ge 1.$$
}
%---------------------------------------------------------------------------------20
\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\small
\begin{block}{Proposition}  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}
}

%---------------------------------------------------------------------------------21
\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\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}

Zhang(2004, 2013)
}
%---------------------------------------------------------------------------------22
\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\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:
$$y_n=y_0+\sum_{0\le k\le n-1}\sum_{j=0}^k\frac{F_k^{(j)}f_j}{q_{j,j+1}}
=y_0+ y_{i_0}\!\!\sum_{0\le k\le n-1}u_k-\ell\!\! \sum_{0\le k\le n-1} v_k^{(\ell)},\quad n\ge 0.$$
}

%-------------------------------------------------------------------------23
\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\small

$$y_0=\ell \sum_{0\le k\le i_0-1} v_k^{(\ell)}+ y_{i_0} \bigg(1-\sum_{0\le k\le i_0-1} u_k\bigg).$$
$$
y_n={\begin{cases}
-\ell\sum_{i_0\le k\le n-1} v_k^{(\ell)}+y_{i_0} \big[1+\sum_{i_0\le k\le n-1}u_k\big],\ & n\ge i_0+1\\
\ell \sum_{n\le k\le i_0-1} v_k^{(\ell)}+y_{i_0} \big[1-\sum_{n\le k\le i_0-1} u_k\big], \ & n\le i_0.
\end{cases}} $$
$$y_{i_0}^*= \ell\varlimsup_{n\to\infty} \frac{\sum_{i_0\le k\le n} v_k^{(\ell)}}{1+\sum_{i_0\le k\le n}u_k}$$
}
%-------------------------------------------------------------------------24
\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\small

Zhang(2013)

\begin{block}{Proposition}
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}
}
%--------------------------------------------------------------------25

\frame{\frametitle{\small 6. Polynomial moments of hitting time and life time}\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
$$y_0^*=\sup_{n\ge 1}\Bigg(\ell\sum_{k=0}^{n-1}\overline m_k^{(\ell)}\Bigg)=\ell\sum_{k=0}^\infty\overline m_k^{(\ell)}, \quad y_n^*=\ell\sum_{k=n}^\infty\overline m_k^{(\ell)},\quad n\ge 1.$$
}
%----------------------------------------------------------------26
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\small

\begin{block}{Proposition} 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}
}
%--------------------------------------------------------27
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\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}

Zhang(2003)
}
%----------------------------------28
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\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 $i\ge 0$,
$$x_n=x_0\bigg(1-\frac \lambda{q_{01}}\bigg)\sum_{k=0}^{n-1}\widetilde F_k^{(0)}-\sum_{k=0}^{n-1}\big(\widetilde F_k^{(0)}+\lambda\tilde d_k\big)+1,\quad n\ge 1.$$
}
%----------------------------------------------------29
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\small
\begin{block}{Proposition}
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}

P. J. Brockwell(1986), W. J. Anderson(1991) 
}
%----------------------------------------------------------------------------------------30
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\small
\begin{block}{Proposition} 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
$$\mathbb E_n\text{e}^{\lambda\tau_\infty}=1+ \lambda \Bigg[\bar c\Bigg(1-\lambda\sum_{k=0}^{n-1}\widetilde m_k\Bigg)-\sum_{k=0}^{n-1}\widetilde  m_k\Bigg],\quad n\ge 0,$$
$$\bar c=\varlimsup_{n\ge 0}\frac{\sum_{k=0}^{n}\widetilde m_k}{1-\lambda\sum_{k=0}^{n}\widetilde m_k}.$$
Furthermore, the process decays exponentially fast provided $\bar c<\infty$.
\end{block}
}
%----------------------------------------------------------------------------------------31
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\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
$$x_i=\frac{q_i}{q_i-\lambda}\sum_{k}\Pi_{ik}x_k+\frac 1{q_i-\lambda}, \quad i\ge 0.$$
}
%---------------------------------------------32
\frame{\frametitle{\small 7. Exponential ergodicity and Laplace transform of return time}\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
$$
e_{0\infty}(\lambda)=\sup_{n\ge 1}\frac{\sum_{k=0}^{n-1}\widetilde m_k}{1-\lambda\sum_{k=0}^{n-1}\widetilde m_k}=
\varlimsup_{n\to\infty}\frac{\sum_{k=0}^{n}\widetilde m_k}{1-\lambda\sum_{k=0}^{n}\widetilde m_k}=:\bar c
$$
$$e_{n\infty}(\lambda)=\bar c\Bigg(1-\lambda\sum_{k=0}^{n-1}\widetilde m_k\Bigg)-\sum_{k=0}^{n-1}\widetilde  m_k,\quad n\ge 1.$$
}
%---------------------------------------------33
\frame{\frametitle{\small 8. 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)
}
%---------------------------------------------34
\frame{\frametitle{\small 8. Examples}\small
\begin{block}{Example 2}
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}
}
%---------------------------------------------35
\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 $f_i^{(\ell)}=q_{ii_0}(1-\delta_{ii_0})g_{i_0}-\ell\mathbb E_i\sigma_{i_0}^{\ell-1}$.
}
%----------------------------------------------------36
\begin{frame}

\centerline{ \emph{\textbf{\color{blue} Thank you for your attention!}}}

\end{frame}

\end{CJK*}

\end{document}
