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

  • $\begingroup$ Are you asking for anything beyond explication of the set theory notation? $\endgroup$
    – whuber
    Commented Dec 15, 2012 at 21:32
  • $\begingroup$ 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? $\endgroup$
    – user60887
    Commented Oct 27, 2015 at 5:18

3 Answers 3


$T = \{1,2\}$ 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, $d(3)=gcd\{n:p_{33}^{n}>0\}=1$


The following is the state transition diagram of the given Markov Chain.

enter image description here

Consider state $1$.

As $p_{11}^{(1)}>0$, state $1$ communicates with itself. State 2 is accessible from state $1$, but state $1$ is not accessible from state $2$, and hence, State $1$ itself forms communicating class $\{1\}$, which is non-closed. In a finite state, homogeneous Markov chain, states belonging to a non-closed communicating class are transient states. State $1$ is transient.

Consider state $2$.

As $p_{22}^{(1)}>0$, state $2$ communicates with itself. As $p_{23}^{(1)}>0$, state $3$ is accessible from state $2$. However, if the process moves into the state $3$, it will never come back to state $2$. Like state $1$, state $2$ itself forms a non-closed communicating class $\{2\}$, and hence, is transient.

Consider state $3$.

We note that, it is not possible for the process to move to either state $1$ or state $2$, once the process occupies the state $3$. Such a state is called an absorbing state. State $3$ itself forms a closed communicating class $\{3\}$. State $3$ is a recurrent state.

In the given Markov Chain, the states $\{1,2\}$ are transient and the state $\{3\}$ is recurrent.


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