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 therefore
$$
\frac d{dt}\langle x_3\rangle
=\alpha_4-\alpha_5\langle x_3x_4\rangle,
\qquad
\frac d{dt}\langle x_4\rangle
=\alpha_6\langle x_2\rangle
-\alpha_5\langle x_3x_4\rangle.
$$
At the assumed unique <stationary distribution>, both left-hand sides vanish. Eliminating the common mixed moment gives
$$
\alpha_4=\alpha_6\langle x_2^*\rangle,
\qquad
\boxed{\langle x_2^*\rangle=\frac{\alpha_4}{\alpha_6}.}
$$
No <moment closure> or independence assumption is involved.

Solution

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

 Solution

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