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For jumps $x\mapsto x+{\nu }_r$ of rates $a_r(x)$, the forward operator acts on a probability mass function byIt is adjoint to the Markov jump-process generator $Lf(x)={\sum }_r{a}_r(x)[f(x+{\nu }_r)-f(x)]$.

For a Markov jump-process generator $L$, differentiating an integrable observable gives $d\mathbb{E}[f(X_t)]/dt=\mathbb{E}[Lf(X_t)]$, when the generator identity and interchange are justified. Choosing powers gives equations for moments. These need not form a closed finite system.

A moment hierarchy is the coupled family of equations obtained by applying a Markov jump-process generator to powers of the state. Nonlinear reaction propensity functions make a moment of one order depend on higher-order moments, so the hierarchy generally does not close after finitely many equations.

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.

For ${\lambda }_n=\alpha n+\beta$,so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function $f(n)=n$ asConsequently $m(t)=\mathbb{E}N(t)$ satisfiesSolving this linear ordinary differential equation gives the mean of the linear birth process with immigration:

Differentiate $⟨f(n)⟩={\sum }_{n\ge 0}{f}_n{p}_n$ and substitute the three master equations. Reindexing every gain term by its source state gives one contribution per possible transition:This is the generator identity for a continuous-time Markov chain applied to the test function $f$.

Applying the reduced Markov jump-process generator to $f(x)=x$ gives the exact moment equationUnder the stated moment closure this becomes the same expression evaluated at $X=m$. Set $m=V\bar{x}$ and let $V\to \infty$ to obtainIts positive stable equilibrium is

With population $n$, the transition diagram has the two outgoing arrowsThus this is a batch-birth linear-death process. If $P_n(t)=\mathbb{P}(N_t=n)$ and $P_j=0$ for $j<0$, probability enters state $n$ from $n-3$ by a birth and from $n+1$ by a death. Hence the master equation is

The Markov jump-process generator isFor $f(n)=n$, this gives $Ln=3\lambda -\beta n$. Therefore the expected value $m(t)=\mathbb{E}[N_t]$ satisfiesso

For $f(n)=n^2$,Writing $s=\mathbb{E}[N_t^2]$ and using the generator identity givesThe variance $v=s-m^2$ consequently obeysIf $v(0)=v_0<\infty$, substitution of the formula for $m$ yieldsAll transient terms vanish, and the moments of a batch-birth linear-death process therefore give

A simple birth process with parameter $\lambda$ starting from one individual is the Yule process: when its population is $n$, it waits an exponential time of rate $n\lambda$ and then moves to $n+1$.

For $f(n)=n$, the Markov jump-process generator satisfiesTherefore $\mu (t)=\mathbb{E}{X}_t$ obeysIt follows, as recorded by the mean population of a Yule process, that

For a function $h$ on the copy-number state space, define the shift operatorWith the propensities from part (a), the forward operator of a Markov jump process isEquivalently, the chemical master equation has the gain-minus-loss formwhere terms outside the state space vanish.

The adjoint Markov jump-process generator acts on observables:Indeed, shifting the summation index in the gain terms gives $⟨f,{\mathcal{L}}^{\ast }p⟩=⟨\mathcal{L}f,p⟩$.

For any observable $f$, the Markov jump-process generator identity givesSet $m(t)=⟨x_2⟩$ and $s(t)=⟨x_2^2⟩$. With ${\alpha }_7=0$, reaction 2 changes $x_2$ by $+1$ and reaction 3 changes it by $-1$. Thereforeand, using $(x_2+1{)}^2-x_2^2=2x_2+1$ and $(x_2-1{)}^2-x_2^2=-2x_2+1$,The equation for the second moment contains the third, whose equation contains the fourth, and so on. This is an infinite moment hierarchy.

Reaction 1 eventually leaves the odd initial copy number at $x_1^{\ast }=1$. PutThe stationary first-moment equation givesThe stationary second-moment equation then givesUnder the prescribed central-moment closure,Substitution yields the requested polynomial equationThis cubic comes from the closure approximation; it is not an exact equation for the stationary mean.

A jump process with event types $k$, increments ${\nu }_k$, and rates $a_k(X)$ can be represented as $X(t)=X(0)+{\sum }_k{\nu }_k{P}_k({\int }_0^t{a}_k(X(s-))ds)$, with independent unit-rate Poisson processes. This represents the Markov jump-process generator through time-changed counts.