Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 36 (Stochastic Calculus and Its Applications)

Take a complete, right-continuous filtration $({\mathcal{F}}_t)$, an increasing family of sigma-algebras. Adapted means that $X_t$ is ${\mathcal{F}}_t$-measurable at each $t$; càdlàg means right-continuous with left limits. A local martingale is an adapted càdlàg process $X$, with integrable initial value, for which there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $X_t^{{\tau }_n}=X_{t\wedge {\tau }_n}$ is a true martingale. Such a sequence is a localizing sequence. A true martingale is integrable at each time and satisfies $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_s$ for $s\le t$.

The necessary and sufficient condition is that $X$ is a Class DL process:A family $\mathcal{Z}$ is uniformly integrable if its members are integrable and ${\sup }_{Z\in \mathcal{Z}}\mathbb{E}(|Z|1_{\{|Z|>K\}})\to 0$ as $K\to \infty$. A stopping time $\tau$ satisfies $\{\tau \le t\}\in {\mathcal{F}}_t$ for all $t$.

For necessity, on $[0,T]$ the optional stopping theorem gives $X_{\tau }=\mathbb{E}(X_T\mid {\mathcal{F}}_{\tau })$; conditional expectations of one integrable variable form a uniformly integrable family. For sufficiency, $X_{t\wedge {\tau }_n}\to X_t$ almost surely, and Class DL upgrades this convergence to $L^1$ on each finite horizon. Pass to the limit in $\mathbb{E}(X_{t\wedge {\tau }_n}\mid {\mathcal{F}}_s)=X_{s\wedge {\tau }_n}$ to obtain the martingale identity. Requiring uniform integrability of all times on the entire half-line would be stronger than necessary; Brownian motion is already a counterexample to that stronger requirement.

The Lévy characterization of multidimensional Brownian motion says that a continuous adapted ${\mathbb{R}}^d$-valued process $Z$, with $Z_0=0$, is standard Brownian motion relative to the given filtration if and only if its coordinates are local martingales andFor Brownian motion, independent centered Gaussian increments give the martingale property of each coordinate and of $Z_t^i{Z}_t^j-{\delta }_{ij}t$. The defining uniqueness of quadratic covariation consequently gives the bracket condition.

Conversely, suppose the local martingale and bracket conditions hold. For $\theta \in {\mathbb{R}}^d$, the Itô formula givessince the bracket term cancels the time derivative. On $[0,T]$, $|E_t|\le e^{|\theta {|}^{2}T/2}$, so its real and imaginary parts are bounded local martingales and hence true martingales. Thus, for $s\le t$,This is the characteristic function of $N(0,(t-s)I_d)$ and is independent of ${\mathcal{F}}_s$. Uniqueness of characteristic functions shows that $Z_t-Z_s$ has that Gaussian distribution and is independent of the past. Iterating over ordered times gives independent Gaussian increments; together with continuous paths and $Z_0=0$, these are precisely the defining properties of standard Brownian motion.

First justify that the logarithm is defined at all times from one onward. For planar Brownian motion started at $z\ne 0$, stop on exiting the annulus $\epsilon <|x|<R$, where $\epsilon <|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t{|}^2-2t$ gives $\mathbb{E}(t\wedge \rho )\le R^2/2$. The function $f(x)=\log |x|$ is harmonic on the annulus, so its stopped process is a bounded martingale. Optional stopping givesHitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that planar Brownian motion avoids a fixed point. Since $B_1$ has a density and is nonzero almost surely, condition on ${\mathcal{F}}_1$ to apply this result to the future path.

For $x\ne 0$, $\nabla f(x)=x/|x{|}^2$ and $\Delta f(x)=0$ in two dimensions. The Itô formula on annuli therefore givesTo localize explicitly, let ${\tau }_n=\inf \{t\ge 1:|B_t|\notin (1/n,n)\}$. These stopping times increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to ${\tau }_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log |B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{{\tau }_n}$ is a true martingale from time one and $X$ is a local martingale. This also gives its quadratic variation from time one as ${\int }_1^t|B_s{|}^{-2}ds$.

Brownian scaling gives $|B_t|\stackrel{d}{=}\sqrt{t}R$, where $R=|B_1|$ has Rayleigh distribution with density $r{e}^{-r^2/2}$, $r>0$. For every $p>0$,Near zero this is bounded by ${\int }_0^1|\log r{|}^{p}rdr={\int }_0^{\infty }{u}^p{e}^{-2u}du$, which is finite; near infinity the Gaussian tail dominates any logarithmic power. The inequality $|a+b{|}^p\le C_p(|a{|}^p+|b{|}^p)$, valid also for $0<p<1$, now yieldsLet $c=\mathbb{E}\log R$, a finite constant. The same scaling identity givesAn integrable martingale has constant expectation, so the logarithm of planar Brownian radius is a strict local martingale on $[1,\infty )$. Fixed-time integrability, even of every positive power, does not imply the stopping-time uniform integrability required in part a.

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by left-continuous adapted processes. Equivalently it is generated by $A\times \{0\}$ with $A\in {\mathcal{F}}_0$ and $A\times (s,t]$ with $A\in {\mathcal{F}}_s$, $0\le s<t$. A previsible process is a process measurable for this sigma-algebra; predictable is a synonym.

Use the zero-starting convention for ${\mathcal{M}}_c^2$: continuous martingales with $M_0=0$ and ${\sup }_t\mathbb{E}{M}_t^2<\infty$. For such an L2-bounded continuous martingale, the square-integrable stochastic integrand space isIdentify processes that agree almost everywhere for the quadratic-variation measure ${\nu }_M(d\omega ,ds)=\mathbb{P}(d\omega )d[M{]}_s$. This makes the displayed norm an actual Hilbert-space norm on equivalence classes. On a fixed horizon $[0,T]$, replace infinity by $T$; this is the same construction for continuous martingales square-integrable on that horizon. Nonzero initial values may be subtracted because the integral depends only on increments.

For an elementary previsible process $H={\sum }_{j=0}^{r-1}{H}_j{1}_{(t_j,t_{j+1}]}$, with bounded ${\mathcal{F}}_{t_j}$-measurable coefficients, defineThe Itô isometry isMore generally the squared $L^2$ distance between the integrals of $H$ and $K$ is the integral of $(H-K{)}^2$ on the right. The integral is a zero-starting continuous square-integrable martingale.

Elementary predictable processes are dense in $L^2(M)$. Given $H$ in this space, choose elementary $H_n$ converging in its norm and define the Itô integral by the $L^2$ limit of $H_n\cdot M$. The isometry proves that this is independent of the approximating sequence and remains valid for the limit. The Doob L2 maximal inequality bounds the expected squared path supremum of the difference by four times its terminal second moment; hence the convergence also gives a continuous martingale version, with uniform convergence on compacts in probability.

For a continuous local martingale $M$ and locally bounded process $H$ that is previsible, choose increasing stopping times ${\tau }_n\uparrow \infty$ such that $M^{{\tau }_n}$ is L2-bounded, $|H|\le K_n$ up to ${\tau }_n$, and ${\tau }_n\le n$. This can be done by combining localizers with first exits of $M$ from bounded intervals and the local bounds of $H$. Then $H{1}_{[0,{\tau }_n]}$ is in the integrand space for $M^{{\tau }_n}$. Define the integral on each stopped interval by the preceding construction. The stopping property of elementary integrals, extended by the isometry, makes these definitions agree on overlaps. Pasting givesThis is a localization construction; global square integrability of the resulting process is not asserted.

For every zero-starting continuous local martingale $M$, there is a unique, up to indistinguishability, continuous adapted increasing process $[M]$, with $[M{]}_0=0$, such thatThis is its quadratic variation. It also satisfies, for deterministic partitions whose mesh tends to zero,with convergence uniformly on compact time intervals in probability. The theorem includes existence and finiteness on each compact interval; monotonicity is understood pathwise outside one null set.

Uniqueness follows because the difference of two candidate increasing continuous processes is both a finite-variation process and a continuous local martingale, hence is identically zero from its zero initial value. For a continuous semimartingale $X=X_0+M+A$, its quadratic variation is $[X]=[M]$, since continuous finite-variation terms have zero quadratic variation. Polarization defines the quadratic covariation by $[M,N]=([M+N]-[M-N])/4$.

The correct quadratic variation of a stochastic integral identity isThe brackets around the final integrator are missing in the PDF. Without them the assertion is false: for $H=1$ and $M$ Brownian motion, the left side is $t$ while the printed right side is $B_t$.

For elementary predictable $H$, the integral is obtained by scaling each martingale increment by its known coefficient. Quadratic variation on each interval is therefore scaled by the square of that coefficient, giving the displayed formula. To pass to a general $L^2(M)$ integrand on $[0,T]$, take elementary $H_n\to H$ in the integrand norm and put $N_n=H_n\cdot M$, $N=H\cdot M$. The Itô isometry gives $\mathbb{E}(N_n-N{)}_T^2\to 0$, hence $\mathbb{E}[N_n-N{]}_T\to 0$. The bracket Cauchy-Schwarz boundand polarization show that $[N_n]\to [N]$ uniformly in probability. On the other side, Cauchy-Schwarz for the quadratic-variation measure givesTaking limits proves the formula in the square-integrable case. Stopping and pasting as in part b proves it for locally bounded predictable integrands and continuous local martingales.

For an elementary previsible integrand, the integral is the increment sum ${\sum }_j{H}_j(X_{t\wedge t_{j+1}}-X_{t\wedge t_j})$. This expression uses only $H$ and $X$, so is identical under the two measures, irrespective of how the local martingale and finite-variation parts change.

We may use the stochastic dominated convergence theorem: if bounded predictable $H_n$ converge pointwise to $H$ under a common deterministic bound, their integrals against a semimartingale converge uniformly on compact intervals in probability. Also convergence in $\mathbb{P}$-probability implies convergence in $\mathbb{Q}$-probability when $\mathbb{Q}\ll \mathbb{P}$. To see the latter directly, let $D=d\mathbb{Q}/d\mathbb{P}$. For any event $E$,and let first $\mathbb{P}(E)\to 0$, then $K\to \infty$. Apply this to events involving the supremum on a finite horizon for the uniform-in-probability statement.

Let $\mathcal{C}$ be the bounded predictable integrands for which the two integral processes agree up to $\mathbb{Q}$-indistinguishability. It is a vector space containing elementary rectangle indicators generating the predictable sigma-algebra. It is closed under uniformly bounded pointwise limits: stochastic dominated convergence theorem gives convergence of the integrals under each measure, and the preceding transfer makes the $\mathbb{P}$ limit also a $\mathbb{Q}$ limit. Uniqueness of limits in probability makes the two limiting integrals equal. The functional monotone-class theorem therefore puts every bounded predictable integrand in $\mathcal{C}$.

Finally, for locally bounded $H$, stop on intervals where it has a deterministic bound. Such stopping times tend to infinity under $\mathbb{P}$ and hence also under $\mathbb{Q}$. Apply the bounded result to the stopped integrands and use the stopping property of the integral. This proves the stochastic integral under an absolutely continuous measure change identityThe decompositions $H\cdot M+H\cdot A$ and $H\cdot N+H\cdot B$ thus represent the same process under $\mathbb{Q}$, although their individual components need not agree. With only absolute continuity, equality is asserted outside a $\mathbb{Q}$-null set, not necessarily outside a $\mathbb{P}$-null set for a chosen $\mathbb{Q}$ version.

For càdlàg semimartingales, the semimartingale integration by parts formula isFor continuous processes the left limits can be replaced by the values. The bracket includes the jump products in the general formula.

For a continuous ${\mathbb{R}}^d$-valued semimartingale $X$ and $f\in C^2({\mathbb{R}}^d)$, the Itô formula isThe formula is localized when the derivatives are unbounded; continuity of $X$ makes them bounded after stopping on compact sets.

To prove the one-dimensional formula for polynomials without assuming that formula, use induction. It is immediate for constants and $X$. SupposeWrite $X=X_0+M+A$ with $M$ a continuous local martingale and $A$ finite variation. The induction hypothesis identifies the martingale part of $X^k$ as $k{X}^{k-1}\cdot M$. The stochastic-integral covariation identity, obtained by polarization from the quadratic-variation identity, givesApply semimartingale integration by parts to $X^{k}X$. Its first-order terms combine to $(k+1)X^{k}dX$, and its bracket terms combine toThis proves the induction step. Taking linear combinations proves the polynomial Itô formula from integration by parts for every polynomial $f$.

Fix a finite horizon $T$ and define ${\tau }_n=\inf \{s\ge 0:|B_s|\ge n\}$. These stopping times increase to infinity almost surely because continuous paths are bounded on each finite interval. Applying the Itô formula to $B_{t\wedge {\tau }_n}^k$, for $k\ge 2$, givesThe first integrand is bounded by $k{n}^{k-1}$, so its integral is square-integrable and has expectation zero. ThusTo justify both limits explicitly, let $S_T={\sup }_{s\le T}|B_s|$ and choose an integer $q>\max \{k,1\}$. The Gaussian density gives $\mathbb{E}|B_T{|}^q<\infty$ by its exponential tail. The Doob Lp maximal inequality statesTherefore $S_T^k$ and $1+S_T^{k-2}$ are integrable. The stopped terminal powers converge almost surely to $B_t^k$ and are dominated in absolute value by $S_T^k$. The integrands in the time integral converge almost surely and are bounded by $1+S_T^{k-2}$, so their integrals are dominated by $T(1+S_T^{k-2})$. Dominated convergence theorem on both sides, and then Fubini justified by the same bound, give the Brownian moment recursionThis uses only finiteness of Gaussian moments, not the closed moment formula requested in the next part.

The initial values are ${\beta }_0(t)=1$ and ${\beta }_1(t)=0$. Repeatedly applying the Brownian moment recursion to odd indices reduces them to ${\beta }_1$, so every odd moment vanishes. For even indices, if the formula holds at $2k-2$, thenThe base case $k=0$ is ${\beta }_0=1$. ThusEquivalently the even moments are $(2k-1)!!t^k$, the familiar Gaussian moment formula with variance $t$.

Interpret the otherwise undefined $b_k(t)$ here as the moment ${\beta }_k(t)$ defined above. For $k=1$, the process is $B_t$, and for $k=2$ it is $B_t^2-t$; both are martingales.

For $k\ge 3$, the Itô formula and the derivative of the Brownian moment recursion yieldThe stochastic integral is a true martingale on finite horizons, since the required $2k-2$ moment is integrable in time. The continuous finite-variation term is not identically zero. If it were, its continuous derivative would vanish at every time, whereas at a fixed $t>0$, the nondegenerate Gaussian $B_t$ does not almost surely satisfy the polynomial equation $B_t^{k-2}={\beta }_{k-2}(t)$. Uniqueness of continuous semimartingale decomposition therefore rules out even the local martingale property. HenceFor example $\mathbb{E}(B_t^3\mid {\mathcal{F}}_s)=B_s^3+3(t-s)B_s$, so merely subtracting its zero expectation cannot make the cubic power a martingale. The centered Brownian powers need not be martingales distinction is resolved by subtracting the random drift compensator $k(k-1){\int }_0^t{B}_s^{k-2}ds/2$ instead of the deterministic mean.

A weak solution of a stochastic differential equation consists of a choice of probability space, filtration, driving Brownian motion and adapted solution with the specified initial law. The driving noise and the space are part of what may be chosen.

A strong solution of a stochastic differential equation is constructed for a prescribed Brownian motion and initial value, as a nonanticipating measurable functional of that data; for deterministic initial value it is adapted to the usual augmentation of the Brownian filtration. No further independent randomness is required.

Uniqueness in law means that any two weak solutions of a stochastic differential equation with the same initial distribution have the same law as solution processes. Pathwise uniqueness means that two solutions on the same filtered space, driven by the same Brownian motion and with the same initial value almost surely, are indistinguishable: $\mathbb{P}(X_t={\tilde{X}}_t\text{ for every }t)=1$. The latter concerns equality of paths on a common space, whereas uniqueness in law concerns equality of probability distributions.

Put $S=\Vert \sigma {\Vert }_{\infty }$, $K=\Vert b{\Vert }_{\infty }$, and write $X_t-x_0=M_t+A_t$, where $M_t={\int }_0^t\sigma (X_s)d{B}_s$ and $A_t={\int }_0^{t}b(X_s)ds$. The Itô isometry and boundedness giveContinuity and boundedness of $b$ then imply $\mathbb{E}b(X_s)\to b(x_0)$. Since $M$ has mean zero,For the variance, $\mathbb{E}{M}_t^2={\int }_0^t\mathbb{E}\sigma (X_s{)}^{2}ds$. The remaining terms satisfyContinuity of $\sigma$ and the initial $L^2$ convergence give $\mathbb{E}\sigma (X_s{)}^2\to \sigma (x_0{)}^2$. Expanding the variance and dividing by $t$ proves the initial mean and variance derivatives of an Itô diffusion:The derivatives at the initial time are right derivatives because the process is indexed by $t\ge 0$.

Apply the Itô formula separately to cosine and sine, using $[B{]}_t=t$:In vector standard form $d{Z}_t=\sigma (Z_t)d{B}_t+b(Z_t)dt$, with one-dimensional driving noise and two-dimensional state, the coefficient functions areThe original trigonometric process starts at $(1,0)$, but these coefficient functions define a stochastic differential equation on all of ${\mathbb{R}}^2$.

Both coefficient functions are linear, hence globally Lipschitz. Their norms are bounded by a constant times $|(x,y)|$, so they satisfy the linear growth condition for an SDE. The existence and pathwise uniqueness theorem for a stochastic differential equation gives a unique global strong solution of a stochastic differential equation for every deterministic starting point $(x_0,y_0)\in {\mathbb{R}}^2$. Linear growth prevents finite-time explosion.

The theorem gives pathwise uniqueness, and strong existence with pathwise uniqueness gives uniqueness in law. Thus every initial value is permitted, and both forms of uniqueness hold. The coefficients need not be bounded for this theorem; their global Lipschitz and linear-growth properties are sufficient.

For any initial point, rotate its vector through the Brownian angle:This is a nonanticipating function of the prescribed Brownian path and has the required initial value. Equivalently, $Z_t=X_t+i{Y}_t=(x_0+i{y}_0)e^{i{B}_t}$. The Itô formula gives $d{Z}_t=i{Z}_{t}d{B}_t-Z_{t}dt/2$, whose real and imaginary parts are exactly the equations in part c. Therefore the displayed formula is the strong solution of a stochastic differential equation, and uniqueness from part d makes it the only one.

The Brownian rotation in the plane preserves radius: $X_t^2+Y_t^2=x_0^2+y_0^2$. Nonzero starting points move around their initial circle, and the origin remains fixed.

Use $M_0=0$, as implicit in the specified initial value of $Y$. First let $[M{]}_t=t$. The Lévy characterization of Brownian motion makes $M$ Brownian motion relative to the given filtration. SetOn this event at time one, the drift has absolute size at most $C$, while $|Y_0|<1$, so $|Y_1|\ge |M_1|-|Y_0|-C>1$. Continuity forces exit by time one. ThereforeThe same estimate applies conditionally at each integer time without assuming anything about future drift independence. On $\{T>n\}$, $|Y_n|<1$; if $|M_{n+1}-M_n|>C+2$, the drift bound forces exit by $n+1$. That Brownian increment is independent of ${\mathcal{F}}_n$. HenceIn particular $T<\infty$ almost surely.

For general $Q$, write $K_t=[M{]}_t$ and $D_t={\int }_0^t{A}_{s}ds$. Continuity of $Q$ and monotonicity of its integral imply $Q_s\ge 0$ at all times outside one null set. The drift condition givesLet ${\tau }_u=\inf \{t:K_t>u\}$. The divergent clock makes ${\tau }_u$ finite for every $u$, and its continuity gives $K_{{\tau }_u}=u$. By the Dambis-Dubins-Schwarz theorem, $W_u=M_{{\tau }_u}$ is Brownian motion for ${\mathcal{G}}_u={\mathcal{F}}_{{\tau }_u}$. On intervals where $K$ is constant, both $M$ and $D$ are constant. Thus $d_u=D_{{\tau }_u}$ is a continuous adapted function of clock time satisfying $|d_v-d_u|\le C(v-u)$. The time-changed process isThe unit-clock argument uses only this increment bound on the drift, so its first exit time $S$ obeys $\mathbb{P}(S>n)\le (1-\epsilon {)}^n$. It is finite almost surely, and a corresponding finite ordinary time has $|Y|\ge 1$. This proves the exit with drift controlled by quadratic variation resultThis proof needs only a nonnegative measurable bracket density, a continuous divergent integral clock, and the drift increment bound; pointwise continuity of the density is unnecessary. This observation permits the application below with measurable SDE coefficients.

For the multidimensional SDE, put $a_{ij}(x)={\sum }_{k=1}^m{\sigma }_k^i(x){\sigma }_k^j(x)$. The generator of an SDE diffusion isFor every $f\in C_b^2({\mathbb{R}}^d)$, the Itô formula givesBounded derivatives and bounded noise fields make the right side square-integrable on each finite horizon, hence a true martingale. This is precisely the defining test-function property of an L-diffusion, in the martingale-problem sense.

Now fix $x$ and $\xi \ne 0$. Its scalar projection has martingale part $M_t^{\xi }={\sum }_k{\int }_0^t⟨\xi ,{\sigma }_k(X_s^x)⟩d{B}_s^k$ and drift $⟨\xi ,b(X_s^x)⟩$. The uniform ellipticity assumption givesLet $B_{\ast }={\sup }_z|b(z)|$ and take $R>|⟨\xi ,x⟩|$. Rescale the projection by $R$. Its bracket density is at least $\lambda |\xi {|}^2/R^2$, its absolute drift is at most $B_{\ast }|\xi |/R$, and hence that drift is at most $C_R$ times its bracket density, with $C_R=B_{\ast }R/(\lambda |\xi |)$. Its clock diverges, so the exit lemma forces it out of $(-1,1)$ in finite time. Equivalently it leaves the slab $|⟨\xi ,X_t^x⟩|<R$. Taking a countable sequence of integer $R$ tending to infinity proves the projection unboundedness of a uniformly elliptic diffusion:For the final uniqueness result, let ${\tau }_D=\inf \{t:X_t^x\notin D\}$, with $x\in D$. Since $D$ is bounded, it lies in a finite slab; the preceding finite slab-exit time bounds ${\tau }_D$ from above. Thus ${\tau }_D<\infty$ almost surely. Continuity gives $X_{{\tau }_D}^x\in \partial D$.

Set $h=u-v$. It is bounded, has bounded derivatives, is zero on the boundary, and satisfies $Lh=0$ in $D$. The stopped test-function identity says $h(X_{t\wedge {\tau }_D}^x)$ is a martingale; equivalently it is a bounded local martingale, which suffices. Taking expectations and then using bounded convergence as $t\to \infty$ givesTherefore the bounded-domain harmonic uniqueness for a diffusion conclusion isIn particular the argument supplies the representation $u(x)={\mathbb{E}}_{x}u(X_{{\tau }_D})$ and establishes uniqueness without selecting a unique diffusion 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 $N\to \infty$.