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Let $\mu =⟨X⟩$. The Markov jump-process generator applied to $X$ gives the exact moment equation $\dot{\mu }=3\lambda -\beta ⟨\sqrt{X}⟩$. With the prescribed moment closure, $⟨\sqrt{X}⟩\approx \sqrt{\mu }$, stationarity givesThe production flux counts molecules, not reaction events: it is $v^{+}=3\lambda$, while the removal flux is $v^{-}=\beta \sqrt{\mu }=3\lambda$. The reaction-rate elasticity of removal relative to constant production isThe mean molecular lifetime is population divided by its stationary molecular removal flux. ThereforeThe noise-relevant average chemical event size is the molecular-flux-weighted reaction event size. For the stoichiometric vectors $+3$ and $-1$ it isThis weights each absolute jump by the molecular turnover it contributes. Equivalently, the balanced production and removal fluxes give equal weight to their sizes $3$ and $1$, yielding $(3+1)/2$. An event-count-weighted average would be $3/2$, since three quarters of firing events are deaths at stationarity; that different average does not determine the diffusion term in the linear noise approximation.

Let ${\mu }_i=⟨X_i⟩$. Applying the Markov jump-process generator to either coordinate gives the exact moment equationsIn each equation the joint reaction removes one molecule, at reaction propensity function $C{X}_1{X}_2$. Because the two reactants are different chemical species, the propensity uses their product; no same-species factorial correction is needed. The exact identity $⟨X_1{X}_2⟩={\mu }_1{\mu }_2+\mathrm{Cov}(X_1,X_2)$ shows explicitly where the equations fail to close. Neglecting this covariance relative to the product of means is the small-fluctuation moment closure. It gives the mean-field joint removal of two chemical species:The difference equation $d({\mu }_1-{\mu }_2)/dt=-\beta ({\mu }_1-{\mu }_2)$ actually holds exactly as well as after closure, because both the constant production and the joint removal terms cancel. The nonlinear moment closure will be used for the subsequent phase plane and stationary linear noise approximation calculations, as requested.

By symmetry the normalized covariance matrix has the form $\eta =\left(\begin{array}{cc}V& W\\ W& V\end{array}\right)$. Substituting the matrices from part (c) into the normalized stationary fluctuation-dissipation relation for a reaction network, $M\eta +\eta M^{\mathsf{T}}=D$, givesHence $V+EW=1/\mu$ and $W+EV=E/(2\mu )$. Solving these two equations gives the linear noise of two species with joint removal:For $E=0$ the normalized variance is $1/\mu$ and the covariance is zero, as expected for independent immigration–death processes. For $E>0$, $V=1/\mu +E^2/(2\mu (1-E^2))$ exceeds this independent value and $W<0$. The correlation coefficient isThis negative stationary correlation is compatible with positive instantaneous joint-removal noise: the slow difference mode amplifies fluctuations of opposite signs in the two populations.

The symmetric and antisymmetric eigenvectors diagonalize all three matrices. Their normalized covariance eigenvalues areThe corresponding restoring rates are $(1+E)/\tau =\beta +2C\mu$ and $(1-E)/\tau =\beta$. Thus the sum mode retains finite fluctuations, with ${\eta }_{+}\to 3/(4\mu )$, whereas the difference mode becomes arbitrarily noisy. In particular, at a fixed nonzero mean, or along $\beta \to 0$ at fixed $\lambda ,C>0$ so $\mu \to \sqrt{\lambda /C}$,The Fano factor of either population is $\mu V=(2-E^2)/(2(1-E^2))$, which diverges as $E\to 1$ regardless of how the mean is varied. Unnormalized variances are ${\mu }^{2}V$ and covariance ${\mu }^{2}W$, so their absolute magnitudes also depend on the parameter path through $\mu$.

The phase plane in part (b) explains the singularity. At $\beta =0$, the mean-field system has a whole hyperbola of equilibria, rather than an isolated restoring state. A perturbation in the species difference selects another point on that curve. In the actual stochastic reaction network, joint complex formation still leaves $X_1-X_2$ unchanged, but the two constant-rate births change it by $+1$ and $-1$, each at rate $\lambda$. Therefore exclusive joint removal has a diffusing difference mode:Its continuous-time symmetric simple random walk on the integers has no normalizable stationary distribution, so there is no finite stationary covariance at $E=1$. The divergence as $E\uparrow 1$ signals both the loss of difference-mode restoration and eventual failure of the small-fluctuation moment closure; near the limit, the computed normalized variances need not remain small even when the individual mean populations are large.

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.