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Let $\xi =x-⟨x⟩$ and $C_{ij}=⟨{\xi }_i{\xi }_j⟩$. In the linear noise approximation, $A_i=a_{ik}{\xi }_k$ and $B_{ij}=b_{ij}$. At the stationary mean, the preceding moment identity gives ${\dot{C}}_{ij}=a_{ik}{C}_{kj}+a_{jk}{C}_{ik}+b_{ij}$. Hence the stationary covariance matrix solves the Continuous Lyapunov equationwhich is precisely the requested component formula because $C$ is symmetric.

At the stationary mean,The $(1,1)$ entry of the Continuous Lyapunov equation gives $-2\gamma C_{11}+2\lambda =0$, so $C_{11}=\lambda /\gamma$. Its $(1,2)$ entry gives $\gamma C_{11}-(\gamma +\beta )C_{12}-\lambda =0$, hence $C_{12}=0$. Finally $(2,2)$ gives $2\gamma C_{12}-2\beta C_{22}+2\lambda =0$. ThusBoth variances equal their means; the absence of stationary covariance does not mean that the maturation jumps are independent events in the two populations.

Comparison with the multivariate normal density shows that $D$ is its precision matrix. Since the mean is zero,

Multiplying $a+\frac{1}{2}bD=0$ on the right by $C=D^{-1}$ gives $aC=-b/2$. Taking the transpose and using the symmetry of $b$ and $C$ gives $C{a}^T=-b/2$. Hence the stationary covariance obeys the Continuous Lyapunov equationThe same equation follows directly by setting the time derivative of the second-moment equation from part (c) to zero.

For linear drift $A(x)=ax$ and constant symmetric positive-definite diffusion matrix $b$, a zero-current stationary Gaussian $P(x)\propto \exp (-x^{T}Dx/2)$ obeysThis reversible solution exists when $D=-2b^{-1}a$ is symmetric positive definite. Its covariance $C=D^{-1}$ satisfies the Continuous Lyapunov equation