Let $X$ be a markov chain on state space $\{1,2,3,4\}$ with transition matrix given by $$P=\begin{pmatrix}0&1/2&1/2&0\\1/2&0&1/2&0\\2/3&0&0&1/3\\0&0&0&1\end{pmatrix}$$
I want to determine the following:
(i) Is $X$ irreducible?
(ii) Which states are recurrent and which are transient?
(iii) Find the periods of each state
(iv) What is the probability, given $X_0=1$, that the chain will ever visit state $3$?
My solutions:
(i) $X$ is not irreducible; drawing the state space diagram we see that $4$ is in its own communicating class, and so $X$ is not one single communicating class which violates the definition of an irreducible chain.
(ii) $4$ is the only reucrrent state - starting at any other state we could reach state $4$ which is absorbing, and so we would never return to the initial state.
(iii) The period of each state is the $gcd$ of the number of steps it has to take to return to that state. State $4$ is absorbing and hence aperiodic (period $1$). I think states $1,2,3$ have undefined periods as they could reach $4$ and hence never return.
(iv) The probability we require is $f_{13}^*=p_{13}+p_{11}f_{13}^*+p_{12}f_{23}^*+p_{14}f_{14}^*$. This gives $$f_{13}^*=\frac{1}{2}+\frac{1}{2}f_{23}^*$$
Doing the same for $f_{23}^*$ we get $f_{23}^*=p_{23}+p_{21}f_{13}^*+p_{22}f_{23}^*+p_{24}f_{43}^*=\frac{1}{2}+\frac{1}{2}f_{13}^*$
Subbing into or equation for $f_{23}^*$ gives $f_{13}^*=\frac{1}{2}+\frac{1}{2}(\frac{1}{2}+\frac{1}{2}f_{13}^*)$ and so $f_{13}^*=1$
Is this correct?
Answers to (i) and (ii) look good.
For (iii), I think you can still make sense of the period for states 1, 2, 3; at least using the definition I'm familiar with, it's not important whether a chain actually returns to a state, but rather whether it could, and there's a well-defined answer to that question for states 1, 2, 3 (as well as state 4, whose period you correctly stated already).
For (iv), your solution is good. If you want some intuitive verification that the probability should be one, note that (a) the walk will almost surely end up in the absorbing state 4, and (b) the only way to get there is via state 3.