Computing probability of ultimate absorption in B&D processes

Question

Consider a B&D process with infinitely many states, State 0 absorbing. Probability of ultimate absorption when starting in State $n$ (denoted $a_n$) is computed by solving $$a_n=\frac{\lambda_na_{n+1}+\mu_na_{n-1}}{\lambda_n+\mu_n}$$ for $n\ge1$, where $\lambda_n$ and $\mu_n$ are the up and down rates (respectively), all positive, and $a_0=1$. Seeking a solution other than $a_n\equiv1$ one defines $d_n=a_n-a_{n+1}$ which then leads to$$d_n=\frac{\alpha\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}{\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}$$ where $\alpha$ is a constant, and $$a_n=1-\sum_{j=0}^{n-1}d_j$$ whenever the infinite sum converges (otherwise, $a_n=1$ is the correct solution). To establish the value of $\alpha$, there is a 'simple probabilistic argument' (Karlin) that $\lim_{n\to\infty}a_n$ cannot be positive, thus must be equal to zero (implying that $\alpha=1$). My question is: what is that argument?

Answer 1

$\newcommand{\intr}[2]{\overline{#1,#2}}$ 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).