Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 27I / c / Solution

The stationary equations simplify by setting $a_i={\pi }_i{\lambda }_i$. They read $2a_i={\sum }_{j\ne i}{a}_j$, so all three $a_i$ are equal. If all rates are positive, this gives the unique equilibriumIf any rate is zero, the common $a_i$ is zero, forcing zero equilibrium mass on every positive-rate state. Every zero-rate state is absorbing. With one such state $z$, the unique equilibrium is ${\delta }_z$; with two, all probability mixtures on those two states are equilibria, giving nonuniqueness. The excluded all-zero case would make every distribution an equilibrium.