TITLE: Markov chain capture times
QUESTION [7 upvotes]: Given the Markov process represented by the above diagram,
where the transition probabilites
$q$ and $p$ satisfy $p+q=1$.
What is the probability $P_r$ of ending up in the sink state $r$, supposing that the initial
state is $i$?
The answer is
\begin{aligned}
P_r &= q+p^2q+p^2q^3+p^2q^5+\dots
=q\left[1+p^2\left(1+q^2+q^4+\dots\right)\right]
=q\left[1+p^2\sum_{n=0}^\infty q^{2n}\right]\\
&=q\left[1+\frac{p^2}{1-q^2}\right]=\frac{2q}{1+q},
\end{aligned}
likewise, what is the probability $P_t$ of ending up in the sink state
$t$?
The answer is
$$
P_t =p^2\left(1+q^2+q^4+\dots\right)=\frac{p^2}{1-q^2}=\frac{1-q}{1+q}
$$
Assuming that the transition time between states "1" and "2", in either way, is constant and is
given by $\tau$, the transition time from state $i$ to state "2" is also $\tau$,
while the other transitions have zero time spans.
Hence, the average passage time from state $i$ to state $r$ is given
by
\begin{aligned}
\bar T_r&= \left(2\tau p^2q+4\tau p^2 q^3+6\tau p^2q^5+\dots\right)/P_r
=\frac{2p^2q\tau}{P_r}\sum_{n=0}^\infty(n+1)q^{2n}
=p^2(1+q)\tau\left(\sum_{n=0}^\infty nq^{2n}+\sum_{n=0}^\infty q^{2n}\right)\\
&=(1-q)(1-q^2)\tau\left(\frac{q^2}{(1-q^2)^2}+\frac{1}{1-q^2}\right)
=(1-q)\tau\left(\frac{q^2}{1-q^2}+1\right)
= \frac{\tau}{1+q},\\
\end{aligned}
while the average passage time to state $t$ is given by
$$
\bar T_t =\frac{p^2\tau}{P_t}\left(1+3q^2+5q^4+\dots\right)=(1-q^2)\tau\sum_{n=0}^\infty
(2n+1)q^{2n}=\frac{(1+q^2)\tau}{1-q^2},
$$
where we have used the following results
$$
\sum_{n=0}^\infty q^{2n}=\frac{1}{1-q^2};\qquad
\sum_{n=0}^\infty nq^{2n}=\frac{q^2}{\left(1-q^2\right)^2}.
$$
The problem with these calculations is that when $q\rightarrow 1$, the average time of capture to state $r$, $\bar T_r$, should go to zero and not to $\tau/2$. I don't know where my mistake(s) is/are.

REPLY [3 votes]: I did not check your actual calculation details, but the intuition for why the expected time to hit $r$ (counting only jumps between $1$ and $2$) does not go to zero as $q \to 1^-$ is as follows.
It is true that the probability that the process going directly to $r$ goes to $1$. But there is still a residual probability $p$ that it doesn't do that. If this happens, the process now spends Geo($p$) time bouncing around in $\{ 1,2 \}$ (which gets longer as $p$ goes to zero), before finally going to either $t$ or $r$.
This means the average time to just absorb at all, counting only jumps between $1$ and $2$ and the jump to $2$, is $q \cdot 0 + p \cdot (1+(p^{-1}-1))\tau=\tau$.
Interestingly this is a case of a failure to obtain convergence in mean from convergence in probability (since the absorption time converges in probability to just zero).