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With zero initial messenger RNA and protein concentration, solving the two linear ordinary differential equations in sequence givesThe last expression applies after the short messenger RNA transient when ${\alpha }_R\gg {\alpha }_P$. All three mean curves have this shape. At a fixed steady state, a larger translational burst size requires fewer transcription events: ${\beta }_R={\alpha }_P{X}_{P,\ast }/b$. Thus $b=400$ gives infrequent large upward jumps followed by protein degradation, whereas $b=3$ gives many smaller jumps and a smoother trace.

Noise depends on molecule counts. If the fixed cell volume is $V$, write $\bar{N}=V{X}_{P,\ast }$. In the instantaneous-burst limit, independent Poisson process arrivals of geometrically distributed bursts, with count arrival rate $\lambda =V{\beta }_R$, and unit protein degradation giveIndeed, the moment equations are $d\mathbb{E}N/dt=\lambda b-{\alpha }_P\mathbb{E}N$ and $d\mathrm{var}N/dt=\lambda b(1+2b)+{\alpha }_P\mathbb{E}N-2{\alpha }_P\mathrm{var}N$. With zero initial counts the variance becomes $\bar{N}[(1+b)-e^{-{\alpha }_{P}t}-b{e}^{-2{\alpha }_{P}t}]$. These equations explain both the approach to the steady state and the scatter about it.

For illustration only, $V=1\mu {\mathrm{m}}^3$ gives $\bar{N}=10^4$ at the higher concentration: the two values of the coefficient of variation are $0.020$ and $0.200$. At the lower concentration, $\bar{N}=100$ and $b=3$ gives $0.200$. Decreasing the concentration by a factor of 100 therefore increases relative fluctuations tenfold, despite keeping small bursts. Absolute fluctuation sizes cannot be inferred without a cell volume. The original sketches below stipulate that volume; they are model realizations, not source data.^[ref]

Multiply the Fokker-Planck equation by $B^n$ and integrate over $B>0$ and all $\sigma$, assuming the required boundary terms vanish. Integration by parts givesThe mixed moment is not determined by $\mathbb{E}[B^n]$: the current strain and the accumulated magnetic field are correlated. Zero mean strain does not imply $\mathbb{E}[\sigma B^n]=0$, so this is an unclosed moment equation.

Keep the strain dependence by defining $P_n(\sigma ,t)={\int }_0^{\infty }{B}^{n}P(B,\sigma ,t)dB$. The magnetic drift term satisfiesThe strain derivatives commute with the $B$ integral. HenceThis weighted density obeys a tilted Ornstein-Uhlenbeck Fokker-Planck equation. Its integral over $\sigma$ is the desired moment, while retaining $\sigma$ makes the evolution closed.

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.

In the birth-death master equation, The upward jump has rate $\lambda$ and the downward jump rate $\beta x$, so the first two jump moments per unit time are $A(x)=\lambda -\beta x$ and $B(x)=\lambda +\beta x$. The second-order Kramers-Moyal expansion gives the diffusion approximation of a birth-death processFor the associated Fokker-Planck equation, integrating by parts with vanishing moment boundary terms gives $d⟨f⟩/dt=⟨A{f}^{\prime }+\frac{1}{2}B{f}^{\prime \prime }⟩$. Taking $f=x,x^2$ yieldsFor example, if $m_0=⟨x(0)⟩$, then $m(t)=\lambda /\beta +(m_0-\lambda /\beta )e^{-\beta t}$, while the variance satisfies $v^{\prime }=-2\beta v+\lambda +\beta m$. The two boxed moment equations also follow exactly from the discrete Markov jump-process generator, since quadratic functions have no higher jump terms. A reflecting diffusion confined to $x\ge 0$ can have extra boundary contributions; the formal approximation near extinction is not an exact replacement for the discrete process.