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For $E=C\mu /(\beta +C\mu )<1$ and mean lifetime $\tau =(\beta +C\mu {)}^{-1}$, the normalized restoring matrix is $M={\tau }^{-1}\left(\begin{array}{cc}1& E\\ E& 1\end{array}\right)$ and the diffusion matrix is $D=(\tau \mu {)}^{-1}\left(\begin{array}{cc}2& E\\ E& 2\end{array}\right)$. Solving the normalized stationary fluctuation-dissipation relation for a reaction network gives the displayed covariance. The sum mode remains damped, but the difference mode's restoring rate is $\beta =(1-E)/\tau$. Its noise grows without bound relative to that restoring rate as $E\to 1$, giving strong negative correlations and invalidating small-fluctuation closure near exclusive joint removal.

// Target: mathematical-biology.bigb

For a stable one-species linear noise approximation, the normalized stationary fluctuation-dissipation relation for a reaction network becomesTo see the factors explicitly, the linearized restoring rate is $a=H/\tau =4\beta m$, while the chemical noise intensity is the rate-weighted squared step sizeStationarity of the variance gives $2a\mathrm{Var}x=B$. Using $\lambda =2\beta m^2$ therefore yieldsThis is the stationary fluctuation result within the stated negligible-fluctuation/linear-noise approximation, not an exact identity for the boundary-corrected discrete chemistry.

Let $m_i=⟨x_i⟩$. The mean equations give $m_1={\lambda }_1/{\beta }_1$ and $m_2={\lambda }_2{m}_1/{\beta }_2$. For relative deviations, the drift Jacobian and normalized reaction-noise covariance areThe normalized stationary fluctuation-dissipation relation for a reaction network is $A\eta +\eta A^T+D=0$ with this negative-drift convention. Its entries successively giveThe birth and death channels act on one species at a time, so the diffusion matrix is diagonal even though the stationary covariance is not.

The normalized stationary fluctuation-dissipation relation for a reaction network isHere $M$ is the negative drift Jacobian matrix expressed in normalized population coordinates, and $D$ is the normalized noise matrix. More explicitly, with $S=\mathrm{diag}({\mu }_i)$, drift Jacobian $A$, stoichiometric vectors ${\nu }_r$, and stationary reaction propensity functions $a_r$, these matrices are $M=-S^{-1}AS$ and $D=S^{-1}B{S}^{-1}$, where $B={\sum }_r{a}_r{\nu }_r{\nu }_r^{\mathsf{T}}$. The linear noise approximation gives a covariance evolution $\dot{\eta }=-M\eta -\eta M^{\mathsf{T}}+D$; stationarity is the displayed Continuous Lyapunov equation.

In one dimension, the normalized restoring coefficient is $M=H/\tau =1/(2\tau )$. Squared chemical jumps determine the noise:The factor nine from a birth burst is essential; its three molecules arise in one correlated reaction event. The scalar Continuous Lyapunov equation is $2M\eta =D$, givingEquivalently, the normalized variance is $⟨s⟩/(H\mu )=2/((1/2)\mu )$. The Fano factor is approximately four. This calculation uses the prescribed mean closure and the linear noise approximation, so its small-relative-fluctuation regime is $\mu \gg 1$; it is not an exact assertion for arbitrarily small molecular copy numbers.

For $\beta >0$, symmetry of the unique stationary mean-field state gives $\mu =⟨X_1⟩=⟨X_2⟩$ and the balance $\lambda =\beta \mu +C{\mu }^2$. The fraction of the molecular removal flux using the joint reaction and the mean molecular lifetime are consequentlyIn particular $0\le E<1$, $C\mu =E/\tau$, and $\beta =(1-E)/\tau$.

The total loss flux of species one is $v_1^{-}={\mu }_1(\beta +C{\mu }_2)$, whereas its production flux $v_1^{+}=\lambda$ is constant. The reaction-rate elasticities are therefore $H_{11}=1$ and $H_{12}=C{\mu }_2/(\beta +C{\mu }_2)=E$ at the symmetric state. Species two gives $H_{22}=1$, $H_{21}=E$. All chemical jumps in either component have unit absolute size, so their molecular-flux-weighted reaction event sizes satisfy $⟨r_1⟩=⟨r_2⟩=1$. With $M_{ij}=H_{ij}/{\tau }_i$, this provesThe same matrix follows from $M=-S^{-1}AS$ with $S=\mu I$ and the drift Jacobian matrix in part (b); its positive eigenvalues are the restoring rates.

For the chemical diffusion matrix, each component receives noise from one unit birth, its own unit death, and the joint unit loss. The unnormalized diagonal noise intensity isOnly the joint reaction changes both components in one event. Its stoichiometric vector is $(-1,-1)$, so $B_{12}=B_{21}=(-1)(-1)C{\mu }^2=C{\mu }^2$. The positive sign reflects simultaneous decreases, rather than independent death events. Divide by the product of the means to normalize:Thus the chemical diffusion matrix in the normalized stationary fluctuation-dissipation relation for a reaction network isThe unequal roles of $M$ and $D$ matter: $M$ describes restoration of perturbations, while $D$ records the covariance of the immediate stochastic reaction increments.

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.