Let $\mathcal{T}$ be the set of three-vertex subsets and let $I_A$ indicate that $A$ forms a triangle. ThenIn particular $p\sim (6\alpha {)}^{1/3}/n$. We will use the Stein-Chen method with edge-sharing dependency neighborhoods, rather than wrongly treating triangle indicators as mutually independent.

Here is the needed Poisson approximation with dependency neighborhoods, together with its proof. For indicators $I_A$ with means $p_A$, let $B_A$ contain $A$ and suppose $I_A$ is independent of the entire family of indicators outside $B_A$. With $\mu ={\sum }_A{p}_A$ and $W={\sum }_A{I}_A$, defineThe total variation distance convention is $d_{TV}(P,Q)={\sup }_E|P(E)-Q(E)|$. We claimFor an indicator test function $g$ on the nonnegative integers, solve the Poisson Stein equationwith $\Vert \Delta f{\Vert }_{\infty }\le 1$. This elementary Stein factor can be proved as follows. Take the immigration--death process with immigration rate $\mu$ and per-particle death rate one. Starting at $k$, its time-$t$ distribution iswith independent summands. Its stationary law is that of $Z$. Coupling the initial particles with a stationary Poisson initial population makes the centered expectations decay at least as a constant times $(k+\mu )e^{-t}$, sois finite. Its generator equation is $\mu (u(k+1)-u(k))+k(u(k-1)-u(k))=g(k)-\mathbb{E}g(Z)$, obtained by integrating the semigroup derivative. Set $f(k)=u(k)-u(k-1)$ for $k\ge 1$; choose $f(0)=f(1)$, which does not affect the equation. To bound $\Delta f(k)$, couple the processes starting at $k-1,k,k+1$ by adding two independent initial particles. Both survive to time $t$ with probability $e^{-2t}$, and otherwise their second difference cancels in expectation. Since $|g(j+2)-2g(j+1)+g(j)|\le 2$,This proves the Stein factor for $k\ge 1$, and the chosen $f(0)$ handles $k=0$.

Put $W_A=W-{\sum }_{C\in B_A}{I}_C$. Independence gives $\mathbb{E}[I_{A}f(W_A+1)]=p_A\mathbb{E}f(W_A+1)$. Inserting this identity into the Stein equation yieldsThe first sum in absolute value is at most $b_1$, since the two arguments differ by ${\sum }_{C\in B_A}{I}_C$. In the second sum, on $I_A=1$ the arguments differ by ${\sum }_{C\in B_A\setminus \{A\}}{I}_C$, giving bound $b_2$. Taking the supremum over event indicators $g$ proves the claimed total variation distance bound.

For triangle $A$, choose $B_A$ to consist of $A$ itself and all triangles sharing an edge with $A$. There are exactly $1+3(n-3)$ such triangles. Every outside triangle uses none of the three edges of $A$, even if it shares one vertex, so $I_A$ is independent of their entire indicator family. For distinct neighboring triangles, their union has five edges and $\mathbb{E}(I_A{I}_C)=p^5$. ThereforeThe Stein-Chen method gives $d_{TV}(\mathcal{L}(X),\mathrm{Poisson}({\mu }_n))\to 0$. Coupling Poisson distributions of two means by adding an independent Poisson variable givesThis proves the Poisson limit for triangles in a binomial random graph:Applying the same bound to the event $\{1,2,\dots \}$ givesFor finite $n$ this is an approximation, not an exact formula: the error from $1-e^{-{\mu }_n}$ is at most $b_1+b_2$, and the error from $1-e^{-\alpha }$ is at most $b_1+b_2+|{\mu }_n-\alpha |$.

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 $N\to \infty$.

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.