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

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A moment closure replaces higher moments in a <moment hierarchy> by functions of retained lower moments. It converts an infinite hierarchy into a finite approximate system.

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 Incoming links: Moment closure