 Solution

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.

Hence ${\pi }_i=(p/q{)}^i{\pi }_0$. Since $p<q$, the geometric series is summable, and normalization givesThese probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.