Return to 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$$d_n=\frac{\alpha\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}{\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}$$ 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 fix the value of $\alpha$, there is a 'simple probabilistic argument' (Karlin) that $\alpha$ cannot be positive and less than 1; therefore, it must equal to 1, implying that $\lim_{n\to\infty}a_n=0$. My question is: what is that 'simple probabilistic argument'?

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?

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 can show that the corresponding $\lim_{n\to\infty}a_n$ is not allowed to be positive (thus, can only be equal to 0) 'by a simple probabilistic argument' (claims Karlin, without spelling it out). This then adds the second 'boundary' condition on the desired solution, thus making it unique. What exactly is that 'simple probabilistic argument'?

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$$d_n=\frac{\alpha\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}{\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}}$$ 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 fix the value of $\alpha$, there is a 'simple probabilistic argument' (Karlin) that $\alpha$ cannot be positive and less than 1; therefore, it must equal to 1, implying that $\lim_{n\to\infty}a_n=0$. My question is: what is that 'simple probabilistic argument'?