Past exam of the options I course of the University of Oxford / 2012 / M / M1 / b / ii / Solution

Within the valid parameter rectangle, suppose $a>0$ and $b<1/3$. The chain is then irreducible by part (i). State $1$ has a two-step return $1\to 2\to 1$ and a three-step return $1\to 2\to 4\to 1$, both of positive probability. Its period of a state in a Markov chain therefore divides both two and three, hence equals one. Irreducibility gives the same period to every state, so the chain is aperiodic.

If $b=1/3$ and $a>0$, every transition goes between the two classes $\{1,3\}$ and $\{2,4\}$. Returns must have even length, while $1\to 2\to 1$ has length two. Thus the irreducible chain has period two.

If $a=0$, the closed communicating class $\{1,3,4\}$ alternates between $\{1,3\}$ and $\{4\}$. Its states have period two, using the positive two-step returns through $4$. Consequently the chain is not aperiodic, regardless of how the transient state $2$ is classified. Combining the cases,