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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.

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.