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In a population of $N$ playing rock paper scissors, let losers copy winners. With pair-selection events of rate $(N-1)/2$, the proportion process has jumps $(e_1-e_3)/N$, $(e_2-e_1)/N$, $(e_3-e_2)/N$ at rates $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. Its fluid limit obeys ${\dot{x}}_i=x_i(x_{i-1}-x_{i+1})$, with indices interpreted cyclically. The product of numbers $x_1{x}_2{x}_3$ is conserved in this ordinary differential equation, whereas the finite chain eventually reaches a monochromatic absorbing state.

A fluid model replaces discrete counts by continuous quantities whose evolution follows averaged large-scale drifts. A heuristic drift calculation motivates such a model; proving a fluid limit requires a separate convergence argument.

With jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, compensate each jump count by its integrated rate. The drift is $F(z)={\sum }_{\ell }\ell {\beta }_{\ell }(z)$ and the predictable quadratic variation matrix of the martingale term is $N^{-1}{\int }_0^t{\sum }_{\ell }\ell {\ell }^{\mathsf{T}}{\beta }_{\ell }(Z_s^N)ds$. On a compact region with bounded rates this bracket is of order $1/N$. A Lipschitz drift and the Gronwall inequality transfer martingale concentration to the fluid limit.

Use coordinates $1,2,3$ for paper, scissors and stone. Let $X^N$ take values in the lattice simplex $S_N=\{(n_1,n_2,n_3)/N:n_i\ge 0,{\sum }_i{n}_i=N\}$. A uniformly selected unordered pair consists of types $i$ and $j$, $i\ne j$, with probability $2n_i{n}_j/(N(N-1))$. Losing copies the winner, so the possible jumps of the continuous-time Markov chain areTies make no change. For arbitrary event rate $\lambda$, the respective transition rates are $2\lambda N{x}_1{x}_3/(N-1)$, $2\lambda N{x}_2{x}_1/(N-1)$ and $2\lambda N{x}_3{x}_2/(N-1)$. ChooseThen the rates are exactly $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. The infinitesimal generator of this cyclic imitation chain iswhere $\mathcal{J}=\{e_1-e_3,e_2-e_1,e_3-e_2\}$ and the three $\beta$ functions are, in that order, $x_1{x}_3,x_2{x}_1,x_3{x}_2$. Its drift is precisely

Here is a precise fluid limit with its error control. Suppose $X_0^N\to x_0$ in probability, and let $x$ solve $\dot{x}=F(x)$, $x(0)=x_0$. For every finite $T$,Indeed, the martingale decomposition of a density-dependent jump process givesThis follows by compensating each transition count by its integrated rate; the compensated counts are square-integrable martingales, with variance equal to the expected compensator. Since $\Vert \ell {\Vert }^2=2$ and $x_1{x}_2+x_2{x}_3+x_3{x}_1\le 1/3$ on the simplex, the Doob L2 maximal inequality, applied coordinatewise, givesThe polynomial drift is Lipschitz on the compact simplex, with some finite constant $L$. It preserves the sum of the coordinates, and each coordinate solves an equation of the form ${\dot{x}}_i=x_i{h}_i(x)$, so nonnegativity is preserved. Thus the ordinary differential equation stays in the simplex and exists for all time. The Gronwall inequality giveswhich proves the claim by Markov's inequality. This proves the required version directly, rather than invoking an unspecified approximation theorem.

For an interior trajectory, differentiating the logarithm of the product of numbers givesThe same identity for the product itself holds on the boundary, so $x_1(t)x_2(t)x_3(t)$ is conserved. A strictly positive initial product keeps the deterministic trajectory away from the boundary. Except at $(1/3,1/3,1/3)$ these interior trajectories are periodic: the strictly concave function ${\sum }_i\log x_i$ has smooth compact level curves surrounding its unique maximum, and the vector field is nonzero and tangent to each such curve.

The finite Markov chain behaves differently at long times. For $P(x)=x_1{x}_2{x}_3$, the change in $P$ during the first jump isThe other two terms are cyclic permutations. Their first-order contributions cancel, leaving the exact product decay in a cyclic imitation chain identityFor $N\ge 3$, while all three counts are positive, their product is at least $N-2$, so $P(X_t^N)\ge (N-2)/N^3$. A missing type cannot reappear. If $\tau$ is the first loss of a type, thenOnce only two types remain, one always wins, so its count increases until an absorbing monochromatic vertex is reached, almost surely in finite time. The cases of one or two initially present types follow the same argument without the first stage. Every finite chain eventually becomes monochromatic, whereas an interior deterministic solution retains its positive product. The fluid limit holds on each fixed finite interval; it does not justify exchanging the limits $N\to \infty$ and $t\to \infty$.

At the unscaled state $(x,y)$, write the Lévy kernel of a Markov jump process on jump displacements $z$ asThe destination-kernel convention translates each Dirac mass by $(x,y)$. The total holding rate is the mass of this kernel. At either zero population the corresponding death rates vanish, so no negative states are reached. Births in the total population are bounded by a Yule process with per-individual rate $\lambda$, while deaths only decrease it; this also gives nonexplosion.

Put $Z^N=(X/N,Y/N)$. This is a density-dependent Markov jump process, with jumps of size $1/N$ and rates of order $N$. Its drift gives the deterministic approximationUniqueness preserves the diagonal $x=y=z$. For positive $\lambda ,\mu$, the resulting logistic differential equation has solutionIf $\lambda <\mu$, it decreases from one to $\lambda /\mu$; if $\lambda >\mu$, it increases to that same equilibrium; equality gives the constant solution one. In particular positive rates do not give extinction on this exactly symmetric deterministic trajectory. Transverse perturbations behave differently: subtracting the equations gives $(x-y{)}^{\prime }=\lambda (x-y)$, so the positive symmetric equilibrium has an unstable difference direction. The symmetric competition model distinguishes diagonal logistic relaxation from this symmetry-breaking instability.

For an explicit error estimate, use the martingale decomposition of a density-dependent jump process:Fix $t_0,\delta >0$ and choose $R>{\sup }_{t\le t_0}z(t)+\delta$. Stop at the first exit ${\sigma }_R$ from the box $[0,R{]}^2$. On that box, $F$ has Lipschitz constant $L_R=\lambda +2\mu R$ in the supremum norm. Each coordinate of $M^N$ has jumps bounded by $1/N$ and predictable quadratic variation at mostIndeed a coordinate's positive and negative jump rates are $N\lambda x$ and $N\mu xy$, and the squared jump size is $1/N^2$. Independent jump clocks give zero cross-bracket.

The Bernstein bound for bounded-jump martingales gives, for each coordinate,The Gronwall inequality applied up to ${\sigma }_R$ givesAn exit from the box already entails an error larger than $\delta$, by the choice of $R$, so the stopped estimate controls the event of any error on the full interval. Take $\eta =\delta e^{-L_R{t}_0}$ and use the union bound on the two coordinates:For fixed time horizon and tolerance this is exponentially small in $N$, proving a fluid limit. The transverse instability explains why this finite-horizon estimate should not be extrapolated to arbitrarily long times without changing its constants.

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 $N\to \infty$.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.

For the cyclic imitation chain, its infinitesimal generator satisfies ${\mathcal{L}}_N(x_1{x}_2{x}_3)=-3x_1{x}_2{x}_3/N$. Therefore $\mathbb{E}[X_1^N(t)X_2^N(t)X_3^N(t)]=e^{-3t/N}\mathbb{E}[X_1^N(0)X_2^N(0)X_3^N(0)]$. While all three types remain, the product of numbers is at least $(N-2)/N^3$ for $N\ge 3$, so this identity proves that one type is lost almost surely. The remaining pair has a fixed winner, which eventually occupies the whole population. The conserved product in the deterministic fluid limit therefore describes fixed finite horizons, not the chain's infinite-time behavior.

Two populations grow at the same per-capita rate and each loses members at a rate proportional to their product. On the invariant diagonal $x=y=z$, the logistic differential equation gives $z(t)=\lambda /[\mu +(\lambda -\mu )e^{-\lambda t}]$ for $z(0)=1$ and positive rates. The positive symmetric equilibrium $(\lambda /\mu ,\lambda /\mu )$ is stable along that diagonal but unstable transversely: $(x-y{)}^{\prime }=\lambda (x-y)$, while the Jacobian matrix has eigenvalues $-\lambda ,+\lambda$. Thus an exactly symmetric deterministic solution can relax toward coexistence while a small imbalance grows. A corresponding density-dependent Markov jump process has a controlled fluid limit on each fixed finite interval, rather than uniform control for all time.