Expected time till extinction in a B&D process

Question

This is computed based on the following recursive formula $$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$ where: $n$ is the inital state, State $0$ is absorbing, $\lambda_n$ and $\mu_n$ are the up and down rates respectively and $$\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}$$diverges (to make extinction certain). To get the recursion started, we need $w_0=0$ and $$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$The derivation of the last formula can be found in S. Karlin's classic book "A first course in stochastic processes". The last step of his proof requires showing that $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$To prove that is, according to Karlin, "more involved but still possible" (but he does not do it). How does one prove that the last limit must equal to $0$?

Answer 1

Clearly, the expected time $w_n$ till extinction from the initial state $n$ is nondecreasing in $n$. So, if $w_1=\infty$, then $w_n=\infty$ for all natural $n$, so that the difference $w_{n+1}-w_n$ makes no sense, and hence the desired conclusion \begin{equation} \lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0 \tag{1} \end{equation} makes no sense either.

It remains to consider the case $w_1<\infty$. Then the equation \begin{equation} w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n} \tag{2} \end{equation} (together with the condition $w_0=0$) implies that $w_n<\infty$ for all natural $n$. For natural $j$, let then $$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i} $$ (so that $\pi_1=1$), and $$u_j:=\pi_j h_j.$$ Then (2) can be rewritten as $$u_{n+1}=u_n-\frac{\pi_n}{\mu_n}, $$ which implies \begin{equation} u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{3} \end{equation}

Also, the equality $$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$ can be rewritten as $$u_1[=h_1=w_1]=\sum_{j=1}^\infty\frac{\pi_j}{\mu_j}. $$ This and (3) imply $$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}. $$ Now the case condition $w_1<\infty$ implies $\pi_n h_n\underset{n\to\infty}\longrightarrow0$, which means that the desired conclusion (1) holds.

Answer 2

Here is a suggestion: $\omega_n - \omega_{n+1}$ is the time to move from state n+1 to n, and it can't be much worse that the return time to state n+1 (proof: every time you return you have a fixed probability of succeeding in getting to state n on next trial, so no worse than a geometric number of return times). It ought to be easy to write down the expected return time explicitly, as it is the reciprocal of the probability under the stationary distribution of being in that state, which is easily got for a B&D process. Whether this is then tractable, I do not know.

Answer 3

A solution for $w_i$ can be built directly by defining $$\delta_i=w_{i+1}-w_i$$ where $\delta_i$ is clearly the expected time to reach State $i$ (for the first time) from State $i+1$.

We then need to solve $$\delta_i=\frac{\mu_i}{\lambda_i}\delta_{i+1}-\frac{1}{\lambda_i}$$

The general solution is$$\delta_i=\sum_{n=i+1}^\infty\frac{1}{\lambda_n}\prod_{j=i+1}^n\frac{\lambda_j}{\mu_j}+c\prod_{j=1}^i\frac{\mu_j}{\lambda_j}$$Realizing that $\delta_i$ cannot be a function of a rate corresponding to a state lower than State $i+1$, $c$ must be equal to $0$. The rest easily follows.