Return to Answer

Take any decreasing sequence $(a_n)_{n=0}^\infty$ and let $$\lambda_n:=a_{n-1}-a_n, \quad \mu_n:=a_n-a_{n+1} $$ for natural $n$. Then the sequence $(a_n)_{n=0}^\infty$ is a solution of the equation \begin{equation*} a_n=\frac{\lambda_n a_{n+1}+\mu_n a_{n-1}}{\lambda_n+\mu_n} \tag{1} \end{equation*} for natural $n$.

In general, $a_n\not\to0$ as $n\to\infty$. Indeed, let \begin{equation*} q_n:=\frac{\mu_n}{\lambda_n+\mu_n}. \end{equation*} Then, by (1), for all natural $n$ $$a_n\ge q_na_{n-1}\ge q_n q_{n-1}a_{n-2}\ge\cdots\ge q_n q_{n-1}\cdots q_1\ge Q:=\prod_{j=1}^\infty q_j>0 $$ if $\sum_{j=1}^\infty(1-q_j)<\infty$. So, in such a case we indeed have $a_n\not\to0$ as $n\to\infty$.

I guess you misunderstood something in Karlin's book. It could help if you reproduced here all the relevant definitions and exact statements from that book or, at least, gave us an accessible reference to the book.

$\newcommand{\intr}[2]{\overline{#1,#2}}$ The desired result follows immediately from

Theorem

(I) If $a_1=1$, then $a_j=1$ for all $j\in\intr1\infty$.

(II) If $a_1<1$, then $S_\infty<\infty$ and $a_j=(S_\infty-S_j)(1-a_1)$ for all $j\in\intr1\infty$, where \begin{equation*} S_j:=\sum_{i=1}^j s_i,\quad s_j:=r_{j-1}\cdots r_1,\quad r_j:=q_j/p_j, \quad p_j:=\frac{\lambda_j}{\lambda_j+\mu_j},\quad q_j:=1-p_j. \end{equation*}

Proof Consider the embedded discrete-time Markov chain $(Y_t)_{t\in\intr0\infty}$, with state space $\intr0\infty$ and transition probabilities $P(Y_{t+1}=j+1|Y_t=j)=p_j=1-P(Y_{t+1}=j-1|Y_t=j)$ for $j\in\intr1\infty$ and $P(Y_{t+1}=0|Y_t=0)=1$, for all $t\in\intr0\infty$. Then the probabilities of the absorption (at $0$) for the embedded chain are the same $a_j$'s, as for the original birth-and-death process.

The key observation is the following simple one: Fix any $j\in\intr0\infty$. For any natural $N\ge j$, let $a^N_j$ denote the conditional probability that the embedded chain reaches the state $0$ before it reaches the state $N$ given that the chain starts in state $j$. Then, by the continuity of probability theorem (Theorem 10.2),
\begin{equation} \text{$a^N_j\to a_j$.}\tag{0} \end{equation} Everywhere here, the convergence is as $N\to\infty$.

Let us now compute $a^N_j$. We have \begin{equation*} a^N_0=1,\quad a^N_N=0,\quad a^N_j=p_ja^N_{j+1}+q_ja^N_{j-1}\ \forall j\in\intr1{N-1}. \end{equation*} The latter equality can be rewritten as $h^N_{j+1}=r_jh^N_j$, where \begin{equation*} h^N_j:=a^N_{j-1}-a^N_j, \end{equation*} whence \begin{equation*} h^N_j=r_{j-1}\cdots r_1h^N_1=s_jh^N_1\tag{0.5} \end{equation*} and, further, \begin{equation*} a^N_j=\sum_{i=j+1}^Nh^N_i=\sum_{i=j+1}^Ns_ih^N_1=(S_N-S_j)(1-a^N_1) \tag{1} \end{equation*} for all $j\in\intr0N$. In particular, for $j=0$ formula (1) yields \begin{equation*} 1=S_N(1-a^N_1). \end{equation*} So, by (0), \begin{equation} a_1=1\iff S_\infty=\infty. \tag{2} \end{equation}

Now consider the following two cases:

*Case I: $a_1=1$. Then $a^N_1\to1$, $h^N_1\to0$, and hence, by (0.5), $h^N_j\to0$ for all $j\in\intr1\infty$. So, $a^N_j=1-\sum_{i=1}^j h^N_i\to1$ and hence $a_j=1$ for all $j\in\intr1\infty$. This proves part (I) of the theorem.

*Case II: $a_1<1$. Then part (II) of the theorem follows immediately from (2), (1), and (0).

Take any decreasing sequence $(a_n)_{n=0}^\infty$ and let $$\lambda_n:=a_{n-1}-a_n, \quad \mu_n:=a_n-a_{n+1} $$ for natural $n$. Then the sequence $(a_n)_{n=0}^\infty$ is a solution of the equation $$a_n=\frac{\lambda_n a_{n+1}+\mu_n a_{n-1}}{\lambda_n+\mu_n} \tag{1}$$ for natural $n$.

In general, $a_n\not\to0$ as $n\to\infty$. Indeed, the (right-continuous version of the) birth-and-death process can be given by the formula $$X(t)=\sum_{n=1}^\infty\xi_n\,I\Big\{\sum_{j=1}^n\tau_j\le t\Big\} $$ for real $t\ge0$, where $I$ denotes the indicator, $\xi_1,\tau_1,\xi_2,\tau_2,\dots$ are independent random variables, $\tau_n$ has the exponential distribution with mean $1/(\lambda_n+\mu_n)$, and $P(\xi_n=1)=p_n:=\lambda_n/(\lambda_n+\mu_n)=1-P(\xi_n=-1)$, for each natural $n$. Let $q_n:=1-p_n$.

Then $$a_n\ge q_n\cdots q_1\ge Q:=\prod_{j=1}^\infty q_j>0 $$ if $\sum_{j=1}^\infty p_j<\infty$. So, in such a case we indeed have $a_n\not\to0$ as $n\to\infty$.

I guess you misunderstood something in Karlin's book. It could help if you reproduced here all the relevant definitions and exact statements from that book or, at least, gave us an accessible reference to the book.

Take any decreasing sequence $(a_n)_{n=0}^\infty$ and let $$\lambda_n:=a_{n-1}-a_n, \quad \mu_n:=a_n-a_{n+1} $$ for natural $n$. Then the sequence $(a_n)_{n=0}^\infty$ is a solution of the equation \begin{equation*} a_n=\frac{\lambda_n a_{n+1}+\mu_n a_{n-1}}{\lambda_n+\mu_n} \tag{1} \end{equation*} for natural $n$.

In general, $a_n\not\to0$ as $n\to\infty$. Indeed, let \begin{equation*} q_n:=\frac{\mu_n}{\lambda_n+\mu_n}. \end{equation*} Then, by (1), for all natural $n$ $$a_n\ge q_na_{n-1}\ge q_n q_{n-1}a_{n-2}\ge\cdots\ge q_n q_{n-1}\cdots q_1\ge Q:=\prod_{j=1}^\infty q_j>0 $$ if $\sum_{j=1}^\infty(1-q_j)<\infty$. So, in such a case we indeed have $a_n\not\to0$ as $n\to\infty$.

I guess you misunderstood something in Karlin's book. It could help if you reproduced here all the relevant definitions and exact statements from that book or, at least, gave us an accessible reference to the book.

Take any decreasing sequence $(a_n)_{n=0}^\infty$ and let $$\lambda_n:=a_{n-1}-a_n, \quad \mu_n:=a_n-a_{n+1} $$ for natural $n$. Then the sequence $(a_n)_{n=0}^\infty$ is a solution of the equation $$a_n=\frac{\lambda_n a_{n+1}+\mu_n a_{n-1}}{\lambda_n+\mu_n} \tag{1}$$ for natural $n$.

I guess you misunderstood something in Karlin's book. It could help if you reproduced here all the relevant definitions and exact statements from that book or, at least, gave us an accessible reference to the book.

Take any decreasing sequence $(a_n)_{n=0}^\infty$ and let $$\lambda_n:=a_{n-1}-a_n, \quad \mu_n:=a_n-a_{n+1} $$ for natural $n$. Then the sequence $(a_n)_{n=0}^\infty$ is a solution of the equation $$a_n=\frac{\lambda_n a_{n+1}+\mu_n a_{n-1}}{\lambda_n+\mu_n} \tag{1}$$ for natural $n$.

In general, $a_n\not\to0$ as $n\to\infty$. Indeed, the (right-continuous version of the) birth-and-death process can be given by the formula $$X(t)=\sum_{n=1}^\infty\xi_n\,I\Big\{\sum_{j=1}^n\tau_j\le t\Big\} $$ for real $t\ge0$, where $I$ denotes the indicator, $\xi_1,\tau_1,\xi_2,\tau_2,\dots$ are independent random variables, $\tau_n$ has the exponential distribution with mean $1/(\lambda_n+\mu_n)$, and $P(\xi_n=1)=p_n:=\lambda_n/(\lambda_n+\mu_n)=1-P(\xi_n=-1)$, for each natural $n$. Let $q_n:=1-p_n$.

Then $$a_n\ge q_n\cdots q_1\ge Q:=\prod_{j=1}^\infty q_j>0 $$ if $\sum_{j=1}^\infty p_j<\infty$. So, in such a case we indeed have $a_n\not\to0$ as $n\to\infty$.

I guess you misunderstood something in Karlin's book. It could help if you reproduced here all the relevant definitions and exact statements from that book or, at least, gave us an accessible reference to the book.