We have a Markov chain given by the transition matrix
$$M= \begin{pmatrix} 1/3 & 2/3 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 1/4 & 0 & 1/4 & 1/2\\ 0 & 0 & 0 & 1 \end{pmatrix}$$
Classify the states of the Markov chain which is given by the transition matrix $M$.
Can you please tell me if I did it correctly? I'm pretty sure this will be asked in the test I write next week!
For a better illustration, I converted this matrix to a graph:
We see, these are accessible states: $(a \rightarrow b),(a \rightarrow c),(b \rightarrow a),(d \rightarrow c)$
communicative states: $(a \leftrightarrow b)$ and the other states are only self communicative, so we have classes:
$C_1= \left\{a,b\right\}$
$C_2 = \left\{c\right\}$
$C_3 = \left\{d\right\}$
And we can also say that the markov chain is not irreducible since we have more than one class (three we have).
$C_2=\left\{c\right\}$ so state $c$ is alone in a class and you cannot escape from that state if you entered it once, thus it is a transient state.
Furthermore $C_3=\left\{d\right\}$ is alone in a class and we have that $p_{ii}=1$ and thus it is an absorbing state.
$C_1=\left\{a,b\right\}$ is not transient because you can always escape from one state to another and so they are recurrent states.
Your conclusions seem correct, even though the diagram is not. One problem with the diagram is that you show total probability $1/3 + 1/2 + 1/4 > 1$ on arrows leading out of $a$.
$C_1$ is recurrent (persistent) because its states inter-communicate and there is no way out.
$C_3 = \{d\}$ is absorbing, It can be entered from $b$, but there is no way out (1 on main diagonal).
$C_2$ is transient, but not emphemeral: it leads to itself, but exit is possible.
Given that $C_1$ is visited, there are methods to determine the long-distribution within it.
Given that state $c$ is visited, there are methods to determine the number of steps until absorption into $C_1$ or $C_3.$