# Identifying the communicating classes and stating whether they are period, aperiodic, recurrent or transient

I don't understand what the answers say

The transition matrix is

$$\begin{pmatrix} 0.5 & 0.5 & 0 \\ 0 & 0.5 & 0.5\\ 0 & 0 & 1 \end{pmatrix}$$

In the answers it says:

$C_1 = \{3\}, T = \{1,2\}$, state 3 is aperiodic because $p_{33} > 0$ and recurrent. States 1 and 2 are aperiodic and transient.

I understand the bits about the individual states, but what does $C_1 = \{3\}, T = \{1,2\}$ mean? The lecturer hasn't explained this or written this in the notes. Is this saying the closed state is 3 and the other communicating classes are 1 and 2?

EDIT: C is the set of irreducible closed classes and T is the set of transient classes

-
Are you asking for anything beyond explication of the set theory notation? – whuber Dec 15 '12 at 21:32
This markov chain is not ergodic since it is reducible? This is because state 1 communicates with state 2 but 2 does not communicate back to 1? – user60887 Oct 27 '15 at 5:18

## 2 Answers

$T = \{1,2\}$ almost certainly means that the set of transient states has as elements the states $1$ and $2$.

$C_1 = \{3\}$ probably means that one of irreducible closed sets has as its only element elements the state $3$. If there were more than one irreducible closed set (not this example) then you might see $C_2$ or others.

-

The states that communicate are said to be in the same class. Periodicity is a class property. So the states in the same class will share this property, or the states will have the same period. Here neither of the states is communicating. So, you will need to check all the 3 states for their periodicity separately. That is also another explanation why the Markov Chain is not reducible.

As Henry pointed out, $T$ is the set of transient states only. $C$ contains only one state and it is a closed set. The subscript 1 comes here, because there is only one such closed set.

$p_{33}>0$ means $p_{33}^{1}>0$ (which is the 1st step transition probability). In fact as $p_{33}=1$, so the last row always remains as it is. Thus, $p_{33}^{n}>0$ always for all $n$. So the period, $d(3)=gcd\{n:p_{33}^{n}>0\}=1$

-