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For a positive function $g$ in the domain of a Markov jump-process generator, the generator martingale for $g(X_t)$ and semimartingale integration by parts give the displayed local martingale. Suitable finite-horizon deterministic bounds on $g$ and $Lg/g$ make it a true martingale. For a single marked exponential jump process, take $g(x)=e^{\theta x}$. Then the exponent's compensating process is $r_{\theta }(t\wedge T)$, where $r_{\theta }=(e^{\theta }-1)/\theta -1$ for $\theta \ne 0$ and $r_0=0$; it is continuous adapted and hence predictable.

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.

A population jumps from $n$ to $n+2$ at a constant immigration rate $\lambda$ and from $n$ to $n-1$ at rate $\beta n$. Applying its Markov jump-process generator to $n$ and $n^2$ gives the displayed mean and variance equations. For $\beta >0$ and finite initial second moment, the stationary mean and variance are $2\lambda /\beta$ and $3\lambda /\beta$. Interpreting the pair-arrival rate as proportional to population instead would produce a different branching process.

At a positive state $x$, the future is constant. At zero, the residual holding time is independent rate-one exponential and the next state is an independent uniform mark. Thus the conditional future given the natural filtration depends only on the present state: $X$ is a Markov jump process. For any bounded Borel $g$, its transition semigroup isThe Lévy kernel of a Markov jump process, written in destination-state convention, isIn displacement convention it has the same expression $K(x,dz)=1_{\{x=0\}}{1}_{(0,1)}(z)dz$, since its only nonzero jump rate is from zero. The Markov jump-process generator is $Lg(x)=1_{\{x=0\}}({\int }_0^{1}g(y)dy-g(0))$.

We state the generator martingale theorem used here: for a nonexplosive Markov jump process and a bounded $g$ with bounded $Lg$, the process $g(X_t)-g(X_0)-{\int }_0^{t}Lg(X_{s-})ds$ is a true martingale. In this one-jump model it also follows directly from part (a), first for mark indicators and then by bounded-function approximation.

Take $g_{\theta }(x)=e^{\theta x}$, and putHere $L{g}_{\theta }(x)/g_{\theta }(x)=r_{\theta }{1}_{\{x=0\}}$. Therefore choose the previsible processIt is continuous adapted and thus predictable. To verify the exponential, write $d{g}_{\theta }(X_t)=d{N}_t+L{g}_{\theta }(X_{t-})dt$ with the generator martingale $N$. The semimartingale integration by parts formula, using the continuous finite-variation factor $e^{-Y_t}$, givesThus $Z_t=e^{\theta X_t-Y_t}$ is a local martingale. On each fixed horizon it is bounded by $e^{|\theta |+|r_{\theta }|t}$, so it is a true martingale with $Z_0=1$. This is the exponential martingale from a jump generator; for $\theta =0$ it reduces to the constant process one.

A Markov jump process, considered up to its explosion time, is an adapted càdlàg piecewise-constant process whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. Describe its rates by an off-diagonal kernel $q(x,dy)$ with finite total rate $q(x,E)$. In state $x$ the holding time has the exponential distribution with this rate, followed by a destination distributed as $q(x,dy)/q(x,E)$; a zero-rate state is absorbing. Its Markov jump-process generator acts byThe Markov property must hold relative to the specified filtration, not merely the natural filtration: future-revealing information could change the jump compensator.

Let $\mu$ count jumps, marked by their destination states. For this jump measure of a Markov jump process, the predictable compensator isThe compensated random-measure theorem says that a predictable $H(s,y)$ with $\mathbb{E}{\int }_0^t\int |H|d\nu <\infty$ gives a true martingaleThe same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $1_A{1}_{(s,t]}H$, for $A\in {\mathcal{F}}_s$, shows that each increment has zero conditional expectation. Taking $H(s,y)=f(y)-f(X_{s-})$ gives the generator local martingale $f(X_t)-f(X_0)-{\int }_0^{t}Qf(X_s)ds$. This states the integrability conditions in the compensated jump-measure local martingale result explicitly.

For the birth-process claim first take the customary fixed initial state $X_0=i$. Write $N_t=X_t-i$, ${\Lambda }_t={\int }_0^t\lambda (X_s)ds$ and $a=e^{\theta }-1$. The compensator of $N$ is $\Lambda$, whose integrand may equivalently use $X_{s-}$ because jump times have zero Lebesgue measure. A birth multiplies $e^{\theta X}$ by $e^{\theta }$, so the jump product rule givesFor an explicit localization, stop at ${\tau }_n=\inf \{t:X_t\ge n\}\wedge n$ with $n>i$. Up to that time the statewise rates are bounded by the maximum of the finitely many rates below $n$, the jump count is bounded, and $M$ and its compensated-integral integrand are bounded on the finite stopped horizon. The compensated-measure theorem therefore makes each stopped process a true martingale. The times increase to the explosion lifetime (or infinity if there is no explosion), proving $M$ is a local martingale on $[0,\zeta )$ for every real $\theta$.

Now suppose $\lambda (j)\le C$ uniformly. For $C>0$, propose candidate births at the times of a rate-$C$ Poisson process, accepting each with probability $\lambda (X_{s-})/C$ using fresh independent uniform marks. While the state is fixed, Poisson thinning gives a holding time with the required exponential distribution of rate $\lambda$; accepted births have exactly the prescribed law. Thus there can be no explosion, and the true birth count $N_T$ is stochastically dominated by a Poisson random variable of mean $CT$. In particular,The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,This is one integrable dominating variable for the entire stopped family. Conditional dominated convergence removes the localization and provesso the exponential martingale of a pure birth process is a genuine martingale for uniformly bounded rates. Only uniform integrability on each finite horizon is asserted; it need not be uniformly integrable over infinite time.

For a random initial state, the same proof requires $\mathbb{E}{e}^{\theta X_0}<\infty$. The source does not explicitly specify the initial distribution, so that qualification cannot be dropped: take constant rate one and an independent initial law $\mathbb{P}(X_0=n)=6/({\pi }^2{n}^2)$. For $\theta >0$, already $\mathbb{E}{M}_0=\infty$, precluding even the integrability in the definition of a martingale. The fixed-initial-state interpretation above supplies the intended complete proof; bounded rates alone do not remedy an arbitrary heavy-tailed initial law.

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.

The pair-birth rate here is the constant immigration rate $\lambda$, while deaths occur at total rate $\beta n$. This interpretation is fixed by the requested mean formula. With $P_n(t)=\mathrm{Pr}(N_t=n)$ and $P_j=0$ for $j<0$, the master equation isThe Markov jump-process generator acts as $Lf(n)=\lambda [f(n+2)-f(n)]+\beta n[f(n-1)-f(n)]$. Taking $f(n)=n$ gives $m^{\prime }=2\lambda -\beta m$, henceFor the second moment $s=\mathbb{E}{N}_t^2$, the two jumps give $s^{\prime }=4\lambda m+4\lambda -2\beta s+\beta m$. Differentiating $V=s-m^2$ therefore yields $V^{\prime }=4\lambda +\beta m-2\beta V$. Substitution of the mean and solution of this scalar linear equation giveThus $V(t)\to 3\lambda /\beta$, assuming $\beta >0$ and finite initial second moment. This is pair immigration with linear deaths; births at rate $\lambda n$ would be a different model and would not give the printed mean.

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.