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For a Markov jump process with off-diagonal rate kernel $q(x,dy)$, the jump measure records the jump time and the destination state. Relative to a filtration for which the process has the Markov jump property, its predictable compensator is the displayed rate measure, up to explosion. The corresponding generator is $Qf(x)=\int [f(y)-f(x)]q(x,dy)$. An enlarged filtration revealing future jumps need not preserve this compensator, so the filtration condition is essential.

Work with the filtration relative to which the given Brownian motion or Markov jump process has its stated properties. All martingale assertions below concern every finite time interval; no infinite-horizon uniform integrability is claimed.

First put $C={\sup }_x|\sigma (x)|$. The solution has the representation $X_t={\int }_0^t\sigma (X_s)d{B}_s$, and the Itô isometry givesThus $X$ is a square-integrable martingale, not just a local martingale. Applying the Itô formula to its square cancels the finite-variation term and givesThe expected squared integrand integral is bounded byTherefore both $X$ and $Z$ are true square-integrable martingales. This is the bounded diffusion coefficient gives square martingales criterion. Only bounded measurability of the coefficient and existence of the stipulated solution have been used.

For the jump process, the Lévy kernel of a Markov jump process uses displacements: a mark $y$ sends the state $x$ to $x+y$. Let $\Lambda ={\sup }_x\lambda (x)$ and $V={\sup }_{x}v(x)$. The bounded total rate prevents explosion; the number of jumps on a finite interval is dominated by a rate-$\Lambda$ Poisson process. Let $N(ds,dy)$ record the jump displacements. Its predictable compensator of a jump measure isThe Cauchy-Schwarz inequality gives $\int |y|K(x,dy)\le \sqrt{\lambda (x)v(x)}$, so the zero first moment is absolutely meaningful. Since that moment vanishes,The compensated integral is a square-integrable martingale: its isometry follows first for bounded marks from conditional compensation and then by truncation in the squared-integrand norm.

The jump version of the square Itô formula is $X_t^2=2{\int }_0^t{X}_{s-}d{X}_s+{\sum }_{s\le t}(\Delta X_s{)}^2$. HenceThe first term is square integrable becauseThe second is an integrable martingale: the expected absolute uncompensated jump-square sum is at most $Vt$, and the compensator identity, multiplied by any bounded event known at an earlier time, makes its increments conditionally centered. Values $X_s$ and $X_{s-}$ agree for Lebesgue-almost every time, so the displayed compensation is exactly the one defining $Z$. Thus $X$ and $Z$ are true martingales in the jump case as well. A fourth moment of the jumps is unnecessary; in particular we have not asserted that this $Z$ must be square integrable. This is centered finite-rate jump kernel gives square martingales.

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.

Let $(A_t,B_t)$ be a Markov jump process on integer populations, with rateswhen $a,b>0$, where $\kappa >0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor \alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.

Use the embedded jump chain $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\le 2N$ steps; hold it constant thereafter. PutBefore absorption its two increments are $-2a+1$ and $2b-1$, soAfter absorption set ${\mu }_j=0$. Consequentlyis a discrete-time martingale. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\le 4N$. The total drift up to step $2N$ is at most $2N$. This is the squared-population near-invariant for stochastic attrition.

The two-sided Azuma-Hoeffding inequality states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb{P}(|M_m|\ge z)\le 2\exp (-z^2/(2m{C}^2))$. Its exponential estimate follows by applying the conditional Hoeffding lemma to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives

Let $c=\sqrt{1-{\alpha }^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N{)}^2$. Hence $|X^N-c|>\delta$ impliesIf the first population is extinguished, $X^N=0$ and $D_{2N}\le 0$. This can satisfy the indicated deviation event only when $\delta <c$, and then $|D_{2N}-c^2{N}^2|\ge c^2{N}^2>c\delta N^2$. Thus the same implication holds in both cases.

The initial rounding error obeys $|D_0-c^2{N}^2|=|{\alpha }^2{N}^2-b_N^2|\le 2N$. Combining it with the drift bound shows that the deviation event forcesfor all sufficiently large $N$. Set $\eta =c\delta /2$ in the preceding concentration inequality. We concludefor sufficiently large $N$, and thereforeThe symmetric stochastic attrition process thus has exponential concentration of terminal stochastic attrition survivors. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.

An ancestral recombination graph, or ARG, is the directed ancestry of a sample of chromosomal segments when ancestral lineages can both coalesce and undergo genetic recombination. It records which ancestral material each lineage carries, the times of events and the recombination breakpoints. At any fixed genomic position, following only the ancestral material at that position gives a marginal phylogenetic tree. Different positions can have different trees, while sharing substantial parts of their history. The graph is acyclic in time, although its underlying undirected graph can have loops because a lineage splits backwards and its ancestors can later rejoin.

For the standard neutral, constant-size, randomly mating diploid model, measure time in $2N_e$ generations and set $\rho =4N_{e}r$, where $r$ is the recombination rate across the whole segment. The backward process is a continuous-time Markov jump process on collections of lineages carrying ancestral material:

For a state with $k$ lineages, independent exponential event clocks give total rateThe next waiting time has exponential distribution with rate $R$; choose its event type in proportion to the corresponding rate. A coalescing pair is uniform among unordered pairs; a recombining lineage is selected with weight $b_i$ and its breakpoint is uniform over its active span. With a nonuniform genetic map, use the corresponding integrated intensity and map-weighted breakpoint distribution. Initially all sample lineages span the whole segment, so the initial rate is $\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n\rho /2$. Ancestral material no longer needed after its sample copies have found their local common ancestor can be pruned.

Mutations are placed independently on the resulting ancestral branches, with local rate $\theta /2$ in coalescent units for a locus having scaled mutation parameter $\theta$. Without recombination, the graph reduces to a single Kingman's coalescent tree. With recombination, each individual locus still has that marginal coalescent distribution, but trees at different loci are dependent. The ARG combines backward pairwise mergers with backward recombination splits; local trees are its position-specific projections. The neutral two-locus construction is developed in Hudson's two-locus model.

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.

A Markov jump process relative to $({\mathcal{F}}_t)$ is an adapted càdlàg process with piecewise-constant paths and finitely many jumps on each bounded time interval, whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. For a time-homogeneous process on a countable state space, write $q(x,y)\ge 0$ for its transition rates and $q(x)={\sum }_{y\ne x}q(x,y)<\infty$. In state $x$, the holding time is exponential of rate $q(x)$ and the next state is $y$ with probability $q(x,y)/q(x)$; a zero rate makes the state absorbing. The generator is $\mathcal{A}f(x)={\sum }_{y\ne x}q(x,y)(f(y)-f(x))$. Nonexplosion is included here so that the process is defined at every finite time.

Here is a finite-jump, bounded-rate form of the Kurtz fluid limit theorem. For each $N$, let $Z^N$ be a chain on a countable state space $E_N\subseteq {\mathbb{R}}^d$ with jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, with $\ell$ in a fixed finite subset $\mathcal{J}\subset {\mathbb{R}}^d$. Assume the nonnegative ${\beta }_{\ell }$ are uniformly bounded, admissible jumps remain in the state space, andis globally Lipschitz with constant $L$. Suppose $Z_0^N\to z_0$ in probability, and let $z$ solve $\dot{z}=F(z)$, $z(0)=z_0$. Then for every finite $T$,Bounded total jump rates ensure nonexplosion; the globally Lipschitz drift gives a unique global ODE solution. These explicit hypotheses suffice for the population example, without needing a more general local version of the theorem.

To prove the result, compensate the jumps of this density-dependent Markov jump process. The drift and the coordinate martingales satisfyHere $⟨M^{N,i}⟩$ is predictable quadratic variation, not the sum of realized squared jumps. The compensation identity follows because each possible increment $\ell /N$ has intensity $N{\beta }_{\ell }$, and its squared coordinate increment contributes ${\ell }_i^2{\beta }_{\ell }/N$ per unit time. Bounded rates and jump sizes make these martingales square-integrable.

We use the Doob L2 maximal inequality in the explicit form $\mathbb{E}{\sup }_{s\le T}|M_s{|}^2\le 4\mathbb{E}|M_T{|}^2$ for a scalar square-integrable martingale starting at zero. Applying it coordinatewise and setting $K={\sup }_z{\sum }_{\ell }|\ell {|}^2{\beta }_{\ell }(z)<\infty$ givesSubtract the ODE integral equation. Lipschitz continuity and Gronwall's inequality yieldThe first term tends to zero in probability by hypothesis, and the second by its maximal second-moment estimate. More quantitatively, for $\delta >0$,This proves the stated uniform finite-horizon fluid limit.

For the cells, division replaces one cell by two and increases population by one. At population $k\le N$, the total rate is the sum of the $k$ identical cell rates, namelyThe state $N$ is absorbing. Set $Z_t^N={\xi }_t/N$; its jump is $1/N$ with rate $N\beta (z)$, where $\beta (z)=z(1-z)$ on $[0,1]$. Extending this function by zero outside $[0,1]$ makes it bounded, nonnegative and globally Lipschitz, with Lipschitz constant one. Starting from $Z_0^N=p$, or from $\lfloor pN\rfloor /N$ when integer rounding is necessary, the limiting ODE is the logistic differential equationSeparation of variables, or differentiating $z/(1-z)$, gives its solution. The logistic cell-division process therefore satisfiesHere $K\le 1/4$, so the maximal martingale second moment is at most $T/N$. With exact initial density $p$, the preceding pathwise bound even gives $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le T{e}^{2T}/N$. With integer rounding, the bound $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le 2e^{2T}(T/N+1/N^2)$ follows from $|Z_0^N-p|\le 1/N$ and $(a+b{)}^2\le 2a^2+2b^2$. Thus the deterministic logistic curve approximates the density with fluctuations of order $N^{-1/2}$ on fixed time intervals. The convergence statement is on finite horizons; it does not interchange the limits $t\to \infty$ and $N\to \infty$.

The predictable random measure $\nu$ matching the expected integrals of the jump measure $N$ against every nonnegative predictable integrand. For displacement marks of a Markov jump process, $\nu (ds,dy)=K(X_{s-},dy)ds$. Multiplying the integrand by an event known before an interval proves that the compensated integral has conditionally centered increments; expected absolute integrability makes it a true martingale. Squared-integrand integrability gives the square-integrable compensation isometry.

Let $T$ be a rate-one exponential random variable independent of a uniform random variable $U$ on $(0,1)$. In its natural filtration, this Markov jump process jumps once from zero to its mark and then stays there. Its destination rate kernel is $q(x,dy)=1_{\{x=0\}}{1}_{(0,1)}(y)dy$. For a Borel mark set $B$, the martingale $1_{\{T\le t,U\in B\}}-|B|(t\wedge T)$ follows directly from the memoryless residual lifetime. Information about the future mark or lifetime must not be inserted into the initial filtration.

This finite-state Markov jump process models two opposing populations, each suffering losses at a rate proportional to the surviving opposing population. Both coordinate axes are absorbing. Its embedded jump chain removes one individual at each step, with probabilities $b/(a+b)$ and $a/(a+b)$. The corresponding deterministic equations $\dot{a}=-\kappa b$, $\dot{b}=-\kappa a$ conserve $a^2-b^2$; finite stochastic populations satisfy a near-invariant instead.