Past exam of the options I course of the University of Oxford / 2010 / M / M2 / c / iii / Solution

On the class $\{1,2\}$, the two-step transition matrix is symmetric with positive entries, so its stationary distribution is $(1/2,1/2)$ and its distribution converges to that law. ThereforeExplicitly the two eigenvalues are $1$ and $1/9$, giving $P(X_{2n}=2\mid X_0=1)=\frac{1}{2}(1-9^{-n})$. The original chain has probability zero of being in state $2$ at odd times, consistent with its period two.