I have a matrix 4-state Markov Chain that looks like this with state space $\{1,2,3,0\}$
\begin{pmatrix} 0.2 & 0.5 & 0.2 & 0.1\\ 0.2 & 0.6 & 0.1& 0.1\\ 0.2 & 0.3 & 0.3 & 0.2\\ 0 & 0 & 0& 0.1\\ \end{pmatrix} My question is: what is the probability that state 2 is never reached.
So my understanding is that this is equal to: probability of being abosrbed into state 0 without ever reaching state 2.
But I don't know how the standard $(I - Q)$ inverse method will work to avoid ever reaching state 2.
Any hint is appreciated!
For $i\in\{1,2,3\}$ let $$ \tau_i = \mathbb P\left(\bigcup_{n=0}^\infty \{X_n=2\mid X_0=i\} \right). $$ Clearly $\tau_2$ is 1, and we have the system of linear equations \begin{align} \tau_1 &= \frac15\tau_1 + \frac12 + \frac 15\tau_3\\ \tau_3 &= \frac15\tau_1 + \frac3{10}+\frac3{10}\tau_3, \end{align} which has solution $(\tau_1,\tau_3) = \left(\frac{41}{52}, \frac{17}{26}\right)$. So the probability that state $2$ is never reached is $$ 1-\tau_i = \begin{cases} \frac{11}{52},& i=1\\ \frac9{26},& i=3. \end{cases} $$