Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 1 / d / ii / Solution

Reaction 4 creates $X_3$, reaction 5 removes one $X_3$ and one $X_4$, and reaction 6 creates $X_4$; reaction 7 leaves $X_4$ unchanged. The exact first-moment equations are thereforeAt the assumed unique stationary distribution, both left-hand sides vanish. Eliminating the common mixed moment givesNo moment closure or independence assumption is involved.