Past exam of the options I course of the University of Oxford / 2006 / N / N1 / ii / Solution

A stationary distribution is a nonnegative row vector $\pi$ with $\pi P=\pi$ and ${\sum }_i{\pi }_i=1$. The balance equations giveThe first yields ${\pi }_1={\pi }_4$. Substitution into the middle pair yields ${\pi }_2=3{\pi }_4/4$ and ${\pi }_3={\pi }_4$; the fourth equation is then satisfied. Normalization gives $15{\pi }_4/4=1$, henceAll states communicate, so this is the unique stationary distribution. The self-loop at state 1 additionally makes this finite Markov chain aperiodic, although aperiodicity is not required for the balance calculation.