Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-1/27i/c/solution
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 1 27I c Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
The stationary equations simplify by setting . They read , so all three are equal. If all rates are positive, this gives the unique equilibriumIf any rate is zero, the common is zero, forcing zero equilibrium mass on every positive-rate state. Every zero-rate state is absorbing. With one such state , the unique equilibrium is ; with two, all probability mixtures on those two states are equilibria, giving nonuniqueness. The excluded all-zero case would make every distribution an equilibrium.
New to topics? Read the docs here!