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

For any observable $f$, the Markov jump-process generator identity givesSet $m(t)=⟨x_2⟩$ and $s(t)=⟨x_2^2⟩$. With ${\alpha }_7=0$, reaction 2 changes $x_2$ by $+1$ and reaction 3 changes it by $-1$. Thereforeand, using $(x_2+1{)}^2-x_2^2=2x_2+1$ and $(x_2-1{)}^2-x_2^2=-2x_2+1$,The equation for the second moment contains the third, whose equation contains the fourth, and so on. This is an infinite moment hierarchy.

Reaction 1 eventually leaves the odd initial copy number at $x_1^{\ast }=1$. PutThe stationary first-moment equation givesThe stationary second-moment equation then givesUnder the prescribed central-moment closure,Substitution yields the requested polynomial equationThis cubic comes from the closure approximation; it is not an exact equation for the stationary mean.

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.