For any observable $f$, the <Markov jump-process generator> identity gives
$$
\frac d{dt}\langle f\rangle=\langle\mathcal Lf\rangle.
$$
Set $m(t)=\langle x_2\rangle$ and $s(t)=\langle x_2^2\rangle$. With $\alpha_7=0$, reaction 2 changes $x_2$ by $+1$ and reaction 3 changes it by $-1$. Therefore
$$
\boxed{\dot m=\alpha_2\langle x_1\rangle-\alpha_3\langle x_2^2\rangle}
$$
and, using $(x_2+1)^2-x_2^2=2x_2+1$ and $(x_2-1)^2-x_2^2=-2x_2+1$,
$$
\boxed{\dot s
=\alpha_2\left(2\langle x_1x_2\rangle+\langle x_1\rangle\right)
-2\alpha_3\langle x_2^3\rangle
+\alpha_3\langle x_2^2\rangle.}
$$
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 $x_1^*=1$. Put
$$
\mu=\langle x_2^*\rangle,\qquad
q=\frac{\alpha_2}{\alpha_3}.
$$
The stationary first-moment equation gives
$$
\langle(x_2^*)^2\rangle=q.
$$
The stationary second-moment equation then gives
$$
\langle(x_2^*)^3\rangle=q(\mu+1).
$$
Under the prescribed <central-moment closure>,
$$
0=\left\langle(x_2^*-\mu)^3\right\rangle
=\langle(x_2^*)^3\rangle
-3\mu\langle(x_2^*)^2\rangle+2\mu^3.
$$
Substitution yields the requested <polynomial equation>
$$
\boxed{2\mu^3-2\frac{\alpha_2}{\alpha_3}\mu
+\frac{\alpha_2}{\alpha_3}=0.}
$$
This cubic comes from the closure approximation; it is not an exact equation for the stationary mean.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-356/1/d/i/solution