I am studying from Lefebvre book on Applied Stochastic processes. I am now on the chapter about stationary (time-homogeneous) discrete-time and discrete-state space processes, and recently I have come across the following proposition about irreducibility:
“If $i \rightarrow j$ OR $j \rightarrow i$ for all pairs of states $i$ and $j$ of Markov Chain $\{X_n | n=1,2,3 ... \}$, then then the chain is irreducible.”
This feels a little wrong to me.
Suppose the following example: Let MC be defined on the state space $\{1,2,3\}$ and have the following one-step transition matrix, where $*$ denotes strictly positive number less than one: $$ \begin{matrix} * & * & * \\ 0 & * & * \\ 0 & * & * \\ \end{matrix} $$ Then clearly the matrix is not irreducible, because $\{2,3\}$ form a closed set, since there is no way getting back to 1, once you enter the set. However the proposition would mean that the MC is irreducible, because for all pairs of states, at least in one direction the states are accessible, i.e., for example, 1 is not accessible from 2, but 2 is certainly accessible from 1.
Where did my logic go wrong in this one?
I agree with you. Consider a Markov chain with states $\{1,2\}$. Suppose $$\mathbb{P}(X_{0}=2\mid X_{1}=1)=\mathbb{P}(X_{0}=2\mid X_{1}=2)=1.$$ This satisfies the stated requirement since $1\rightarrow2$. However, the transition matrix is clearly reducible: $$ \begin{pmatrix}0 & 1\\ 0 & 1 \end{pmatrix} $$
The issue in your text is that the OR should be an AND: