I have been trying to solve the following problem for quite a while now, but not with much luck.
The Question
Let $P$ be the TPM(Transition Probability Matrix) of a DTMC with state space $\{0,1,2,\cdots, N\}$. Let $\{0,N\}$ be absorbing states and $\{1,\cdots, N-1\}$ be irreducible transient states. Form another TPM $\overline{P}$ with $\overline{P}(0,1) = 1 = \overline{P}(N,1)$ and other elements as in $P$. Then $\overline{P}$ is TPM of an irreducible DTMC with stationary distribution $\pi$. Show that if original DTMC (w/ TPM $P$) starts in state $1$, it's mean time to absorption in $\{0,N\}$ is $$\frac{1}{\pi(0) +\pi(N)} -1$$
My Attempt
The "$-1$" in the result is what was bothering me. I verified the result for the simple 3 state case where $0$ and $2$ are absorbing. I wanted to get the result by understanding the problem as a Geometric random variable with parameter $\pi(0) + \pi(N)$, but I couldn't account for the $-1$ term. Could someone help me on these lines?
Here's one approach: define a further modified TPM $\tilde P$ by combining $0$ and $N$ into a single state $\{0,N\}$, i.e. define $\tilde P(i,\{0,N\}) = P(i,0)+P(i,N)$ for $i=1,\dots,N-1$, and define $\tilde P(\{0,N\},1)=1$. Then $\tilde P$ is the TPM of an irreducible DTMC, and denote its stationary distribution by $\tilde \pi$. Then you can show that $\tilde \pi(i)=\pi(i)$ for $i=1,\dots,N-1$, and $\tilde \pi(\{0,N\})=\pi(0)+\pi(N)$.
Now, beginning in state $\{0,N\}$, according to a standard theorem the mean return time to $\{0,N\}$ under $\tilde P$ is $\frac1{\tilde \pi(\{0,N\})}=\frac1{\pi(0)+\pi(N)}$. Since, beginning in state $\{0,N\}$, with probability 1 the next state is state 1, it follows that beginning in state 1 the mean time to hit $\{0,N\}$ is $\frac1{\pi(0)+\pi(N)}-1$.