= Solution
The chain alternates at every step between the two classes $\{1,2\}$ and $\{3,4\}$. It is irreducible and has period two: returns require an even number of steps, and a two-step return has positive probability.
Writing the off-diagonal block as $B=\begin{pmatrix}1/3&2/3\\2/3&1/3\end{pmatrix}$, its square is $B^2=\begin{pmatrix}5/9&4/9\\4/9&5/9\end{pmatrix}$. Hence
$$
\boxed{P^2=\begin{pmatrix}
5/9&4/9&0&0\\4/9&5/9&0&0\\
0&0&5/9&4/9\\0&0&4/9&5/9
\end{pmatrix}.}
$$
Every two steps remain within the initial class, where the sampled chain is irreducible and aperiodic.
Back to article page
