TITLE: Markov chain, being in one state until absorption
QUESTION [0 upvotes]: Assume that $X_t$ is a continuous time-inhomogeneous Markov process on the states $\{1,2,3\}$ where the state $3$ is absorbing, the other two states are transient. Let the random variable $T$ denote the time until absorbtion in state $3$. I want to determine the probability of never entering state $2$ before absorbtion, i.e. $\mathbb{P}(T>t|X_u=1,\quad \forall u\in [0, T])$.
I have tried to look at something called Phase-type distributions, which are distributions governing the time until absorbtion in a Markov chain with a single absorbing state, however it did'nt work out.
Assuming that the transition rates qij(t) are given by the following,
Q(t)=[-exp(t), 1/2exp(t), 1/2exp(t); 1/2exp(t), -exp(t), 1/2exp(t); 0,0,0]
Thanks in advance

REPLY [0 votes]: Let $p_{ij}$ be the probability of moving from state $i$ to state $j$.
You have some values (I assume) for $p_{11}$, $p_{12}$, $p_{13}$, $p_{21}$, $p_{22}$, $p_{23}$. Because state 3 is absorbing we have $p_{31}=0$, $p_{32}=0$, $p_{33}=1$.
You need to divide up your states so that you can distinguish whether or not you have previously been in state 2.
Create a new system with the following five states:
V: You're currently in state 1 and you have never been in state 2.
W: You're currently in state 1 and you have been in state 2 previously.
X: You're currently in state 2.
Y: You're currently in state 3 and you have never been in state 2.
Z: You're currently in state 3 and you have been in state 3 previously.
The transition probabilities are given by:
$p_{VV}=p_{11}$
$p_{VY}=p_{13}$
$p_{VX}=p_{12}$
$p_{XX}=p_{22}$
$p_{XW}=p_{21}$
$p_{XZ}=p_{23}$
$p_{WW}=p_{11}$
$p_{WX}=p_{12}$
$p_{WZ}=p_{13}$
$p_{YY}=p_{33}$
$p_{ZZ}=p_{33}$
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