Past exam of the probability course of the University of Oxford / 2023 / 3 / b / iv / Solution

Every step changes the population by one, so $X_n$ has the parity of $n$ when $X_0=0$. Hence $\mathbb{P}(X_{2m+1}=0)=0$. A return in two steps has positive probability from every state, and all return times are even, so the period of a state in a Markov chain is exactly two.

Each transition exchanges the even and odd classes. Since $\pi$ is an invariant distribution, their masses are equal: $\pi (\text{even})=\pi (\text{odd})=1/2$. The chain sampled at times $2m$, restricted to the even states, is irreducible and positive recurrent, and it is aperiodic because its diagonal entries are positive. Its stationary distribution is $2{\pi }_k$ on even $k$. The countable-state Markov chain convergence theorem applied to this sampled chain yieldsSince ${\pi }_0>0$, the full sequence has no limit. Its time average nevertheless tends to ${\pi }_0$ by the countable-state Markov chain ergodic theorem.