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For discrete-time Markov chains run at time step $1/n$, suppose the displayed conditional moments converge uniformly on compact sets to continuous coefficients, the local conditional large-jump second moments vanish, initial laws converge, and compact containment holds on each finite horizon. If the limiting diffusion martingale problem is well-posed and nonexplosive, the chain laws converge to its solution. Vanishing jumps give a continuous limit; linearly interpolated paths converge in the uniform path topology and step paths in Skorokhod J1 topology. Local moment bounds give tightness after exit-time localization. Taylor expansion identifies the limiting test-function martingales, and uniqueness of the martingale problem makes every subsequential limit the same. The diffusion approximation of a birth-death process is an important example.

For bounded measurable $a,b$, with $a$ symmetric positive semidefinite, an L-diffusion is a continuous adapted process $X$ such that $g(X_t)-g(X_0)-{\int }_0^{t}Lg(X_s)ds$ is a true martingale for every $g\in C_b^2$. The general local martingale problem may instead use compactly supported test functions and require only a local martingale. Specify the test class and the true or local convention when coefficients are unbounded.

For a Markov process with infinitesimal generator $L$ and a suitable test function $f$, the displayed process is a local martingale. With integrability it is a true martingale and yields the corresponding expectation identity at admissible stopping times. For bounded $f,Lf$ in the semigroup generator domain, $P_{t}f-f={\int }_0^t{P}_{s}Lfds$. The Markov property applies this identity conditionally over each interval, proving the martingale increments have zero conditional expectation. Localization gives the local version. For a continuous diffusion, these identities are the martingale problem for $L$.

A martingale describes an observable after its predictable drift has been removed. For a Markov chain the drift is the change produced by its transition operator; for Brownian motion it is one half of the Laplacian. Requiring the resulting processes to be martingales for a sufficiently rich class of observables determines the transition law itself. The same identities, combined with optional stopping, determine hitting probabilities, expected exit times and recurrence or transience. The duplicated Q5 paragraph in the PDF is a repeated printing of this one essay request.

For a finite or countable-state Markov chain with transition matrix $P$, define $(Pf)(x)={\sum }_{y}P(x,y)f(y)$ for a bounded test function $f$. The Markov property says $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=(Pf)(X_n)$. Consequentlyis a martingale, since each increment has conditional expected value zero. Conversely, if these processes are martingales for every bounded $f$, taking the conditional expectation of an increment recovers that identity. With $f=1_{\{y\}}$ it gives $\mathbb{P}(X_{n+1}=y\mid {\mathcal{F}}_n)=P(X_n,y)$ for every state $y$. Thus, given the initial law, these identities characterize the Markov chain, not merely a few of its moments. This is the discrete-time Markov chain martingale characterization.

The drift operator is $P-I$. If $(P-I)h=0$ away from a target set, $h(X_{n\wedge T})$ is a stopped martingale, where $T$ is the first hit of that set. Under boundedness or suitable uniform integrability, the optional stopping theorem expresses $h$ in terms of boundary hitting probabilities. Similarly, if $(P-I)v=-1$ away from the target and $v=0$ on it, then $v(X_{n\wedge T})+(n\wedge T)$ is a martingale, giving $v(x)={\mathbb{E}}_{x}T$ when the terminal limit is justified.

For a concrete calculation, take a simple symmetric random walk on $\{0,1,\dots ,N\}$ stopped at $T$, its first hit of $0$ or $N$. In the interior, $h(i)=i$ has zero drift, while $v(i)=i(N-i)$ satisfiesThe stopped identity gives ${\mathbb{E}}_i(T\wedge n)=v(i)-{\mathbb{E}}_{i}v(X_{T\wedge n})\le v(i)$, proving $T$ finite with finite mean. Bounded convergence in both stopped identities now yields the gambler's ruin answersThis illustrates why the integrability justification is part of using a martingale, rather than an automatic consequence of writing an equation.

For a finite-state continuous-time Markov chain with generator matrix $Q=(q_{ij})$, the corresponding martingale problem for a continuous-time Markov chain usesThese are martingales. Conversely, suppose an adapted càdlàg finite-state process satisfies these identities. Let $P_t=e^{tQ}$, fix $T$, and set $g(s,i)=(P_{T-s}f)(i)$. Then ${\partial }_{s}g+Qg=0$. Apply deterministic integration by parts to the finite collection of indicator martingales to obtain the time-dependent Dynkin formula: $g(s,X_s)-g(0,X_0)-{\int }_0^s({\partial }_{r}g+Qg)(r,X_r)dr$ is a martingale. ThereforeThis recovers the Markov property and the entire semigroup, proving the characterization. On countable spaces one uses nonexplosion and localization to justify the analogous unbounded-rate formulas.

For $d$-dimensional Brownian motion, the Itô formula shows that, for $f\in C_c^{\infty }({\mathbb{R}}^d)$,is a true martingale, since its bounded derivatives make the integral square-integrable on each bounded horizon. This is the martingale problem for Brownian motion with generator $L=\frac{1}{2}\Delta$.

Here is why it also characterizes Brownian motion. Suppose a continuous adapted process $X$, with $X_0=x$, satisfies that martingale problem. Use smooth cutoff test functions equal to $x_i$ and $x_i{x}_j$ inside successively larger balls, and stop before leaving those balls. The coordinate tests show that $X^i-x_i$ are continuous local martingales; the product tests show that $X^i{X}^j-x_i{x}_j-{\delta }_{ij}t$ are local martingales. The Itô product rule consequently givesthe difference between the two proposed compensators is a continuous finite-variation local martingale, hence constant and zero at time zero. For each $u\in {\mathbb{R}}^d$, another application of the Itô formula gives the exponential local martingaleIts modulus is the deterministic quantity $e^{|u{|}^{2}t/2}$, so it is a true martingale on every bounded time interval. Taking its conditional expectation and dividing by its known value at time $s$ givesThe conditional characteristic function identifies a centered multivariate normal distribution of covariance $(t-s)I$, independent of the past. Together with continuity, this is the defining law of Brownian motion. This proves the Lévy characterization of multidimensional Brownian motion in the form needed for the generator characterization.

The analytic use of these martingales parallels the chain case. A harmonic function $h$ satisfies $\Delta h=0$, so it yields a stopped local martingale until the process leaves its domain of harmonicity. Bounded stopped observables permit optional stopping and solve boundary hitting problems. A solution of $\frac{1}{2}\Delta v=-1$ with zero boundary values gives an expected exit time. For example, in the ball of radius $R$, $v(x)=(R^2-|x{|}^2)/d$ gives ${\mathbb{E}}_x{T}_R=(R^2-|x{|}^2)/d$; stopping $|B_t{|}^2-dt$ first at $T_R\wedge n$ proves finiteness and justifies the limiting calculation. The Brownian local time construction in Q4 is another example: the predictable compensators of approximations to $|B_t|$ recover its increasing occupation term.

We now prove the required three-dimensional transience of Brownian motion in dimension at least three, in the strong sense $|B_t|\to \infty$ almost surely. Start first at a point $x$ with $r<|x|<R$, and let $T$ be the first exit from this annulus. The radial Laplacian in dimension three givesThus $|B_{t\wedge T}{|}^{-1}$ is a bounded martingale. The exit time is finite: stopping $|B_t{|}^2-3t$ gives $3\mathbb{E}(t\wedge T)\le R^2-|x{|}^2$. At exit, continuity puts $|B_T|$ at either $r$ or $R$. The optional stopping theorem therefore yieldsand hence the Brownian sphere-hitting probability in dimension threeAs $R\to \infty$, the events $\{{\tau }_r<{\tau }_R\}$ increase to $\{{\tau }_r<\infty \}$: a continuous path is bounded on the finite interval preceding any finite inner hit. This argument assumes no transience. ConsequentlyTo pass from a positive escape probability to almost-sure eventual escape, take Brownian motion started at zero. Let ${\sigma }_R$ be its first hit of the sphere of radius $R$. Stopping the squared-norm martingale proves $\mathbb{E}{\sigma }_R=R^2/3<\infty$. At ${\sigma }_R$, the Strong Markov property and the preceding hitting formula giveThe Strong Markov property here follows from independent Brownian increments by approximating a stopping time from above with dyadic-valued stopping times and using path continuity. Choose $R_m=2^{m}r$. The exit times ${\sigma }_{R_m}$ increase to infinity, since a continuous path is bounded on every compact time interval. The events of a return to the closed ball of radius $r$ after ${\sigma }_{R_m}$ decrease and have probabilities $2^{-m}$. Their intersection has probability zero. Thus almost surely there is some finite exit time after which the path never again enters that ball. Apply this simultaneously to all positive integer $r$. This proves the last visit to a bounded ball for three-dimensional Brownian motion conclusion and therefore

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

Set $u_0=\alpha /\beta$ and use the accelerated grid time $j/n$. At population $k=n{u}_0+\sqrt{n}x$, the fluctuation increment is $h_n=n^{-1/2}$, $-h_n$, or zero. Its first and second conditional moments, multiplied by $n$, areThus $b_n\to b(x)=-\alpha x$ and $a_n\to a=2{\alpha }^2/\beta$, uniformly on every compact set. The centered conditional variance has the same limit, since the squared conditional mean contributes only $b_n(x{)}^2/n$.

Here is the diffusion approximation theorem being applied. For Markov chains sampled with time step $1/n$, suppose the first and second conditional increment moments multiplied by $n$ converge uniformly on compact sets to continuous coefficients $b,a$; the large-jump conditional second moments satisfy the local Lindeberg condition; the initial laws converge; compact containment holds on finite time intervals; and the limiting martingale problem is well-posed and nonexplosive. Then the interpolated chains converge weakly, uniformly on finite time intervals, to that problem's continuous solution. The step versions converge in the Skorokhod J1 topology. Local moment bounds supply the stopped-process tightness, and the limiting test-function identities identify every subsequential limit by the martingale problem.

We verify the global condition rather than assume it. Before the stopping rule is triggered, $p_k+q_k\le 1$, and $k\ge 0$ implies $x\ge -u_0\sqrt{n}$. ThereforeFor the Lyapunov function $V(x)=x^2$, the exact one-step generator is consequentlyAfter stopping, the process is constant. Applying the elementary stopped supermartingale inequality for $V(X_{j/n}^n)-j/n$, and stopping also on first reaching $|X^n|\ge L$, givesThis is compact containment. At zero population we use the natural absorbing convention $p_0=q_0=0$; the same calculation is valid there.

The invalid-probability threshold is at the positive rootBecause $2{\alpha }^2<\beta$, we have $u_{\ast }>u_0$. The corresponding fluctuation level $\sqrt{n}(u_{\ast }-u_0)$ tends to infinity, so the stopping rule changes none of the compact-set moment limits. The initial fluctuation satisfies $|X_0^n|\le n^{-1/2}$ and hence tends to zero. All jumps have magnitude at most $n^{-1/2}$, so the Lindeberg condition is immediate. The limiting equation has globally Lipschitz coefficients and is well-posed and nonexplosive. The theorem therefore givesThis is the Ornstein-Uhlenbeck fluctuation limit at a stable population equilibrium. The limiting Ornstein-Uhlenbeck process has explicit solutionThe PDF samples the linearly interpolated population at $\lfloor nt\rfloor$, so its displayed process is actually the step version. Using $nt$ instead gives the continuous interpolation; their uniform distance is at most $n^{-1/2}$. Both therefore have the asserted continuous diffusion limit, with the step version interpreted in Skorokhod J1 topology.

Finally, if ${\tau }_n\le 1$, the stopped fluctuation reaches a value greater than $\sqrt{n}(u_{\ast }-u_0)$. The preceding Lyapunov function bound gives directlyThis also shows explicitly why the artificial stopping rule is asymptotically irrelevant.

For the diffusion operatora continuous adapted process solves the diffusion martingale problem when, for every $f\in C_b^2({\mathbb{R}}^d)$,Here $C_b^2$ means that the function and its derivatives through order two are bounded and continuous. An equivalent compact-support/local convention is often used for the general martingale problem; the bounded-coefficient setting lets us use the true-martingale convention explicitly. The matrix $a$ is interpreted as a covariance matrix in the diffusion formulation, hence is symmetric nonnegative; the displayed operator in any case uses only its symmetric part.

A well-posed martingale problem has existence and uniqueness in law on continuous path space for each specified starting point, or for each initial law in the initial-law formulation. This is uniqueness of the process law, not an assertion of pathwise uniqueness on a prescribed Brownian probability space.

Let the infinitesimal generator be the second-order operatorwhere $a(x)$ is symmetric positive semidefinite and $b(x)$ is a vector. In the standard bounded-coefficient setting take $a,b$ bounded and measurable. An L-diffusion is a continuous adapted process $X$ for which, for every $g\in C_b^2({\mathbb{R}}^d)$,Here $C_b^2$ includes boundedness of the function and its derivatives through order two. This is the true-martingale version of the diffusion martingale problem. The matrix $a$ is the diffusivity, and $b$ is the drift. An Itô diffusion with coefficient $\sigma$ has $a=\sigma {\sigma }^{\mathsf{T}}$.

A more general local martingale problem may permit unbounded coefficients and require only a local martingale. That convention is weaker and must not silently replace the true martingale property used here and in part (b). The bounded-coefficient convention guarantees all finite-horizon integrability needed below; a sufficient more general replacement is $\mathbb{E}{\int }_0^T({\sum }_i|b_i(X_s)|+{\sum }_i{a}_{ii}(X_s))ds<\infty$ for every $T$.

We prove the extension from spatial test functions to time-dependent test functions for a diffusion martingale problem directly, without assuming a driving Brownian motion. Fix $0\le s<t$ and a deterministic partition $s=t_0<\cdots <t_n=t$. For each $j$, $g_j(x)=f(t_j,x)$ lies in $C_b^2$. The defining martingale problem givesCall the sum $S_n$. Telescoping separates the spatial and temporal changes:The sample path $X$ is uniformly continuous on $[s,t]$ and has compact range. Joint continuity of ${\partial }_{t}f$, the spatial first derivatives, and the spatial second derivatives therefore gives, as the mesh tends to zero,almost surely. For bounded $a,b$ and bounded derivatives of $f$, both integrals are uniformly bounded on this finite horizon, as are the two endpoint terms. The dominated convergence theorem permits passage to the limit in the expectation against $1_A$. Thus $M_t^f$ is integrable and $\mathbb{E}[1_A(M_t^f-M_s^f)]=0$ for every $A\in {\mathcal{F}}_s$, provingAdaptation and continuity follow from the formula, so $M^f$ is a continuous martingale. Under the more general integrability condition in part (a), the same proof uses an integrable bound proportional to $1+{\int }_s^t({\sum }_i|b_i(X_u)|+{\sum }_i{a}_{ii}(X_u))du$; positive semidefiniteness bounds every off-diagonal entry by the diagonal ones. With only local assumptions this reasoning proves a local martingale, and an integrability hypothesis is needed to upgrade that conclusion.

An $L$-diffusion solves the martingale problem forfor every $f\in C_c^2(\mathbb{R})$,is a local martingale. Applying this to cutoff approximations of $x$ and $x^2$ shows thatis a continuous local martingale withPart b gives $M=\int |\sigma (X_s)|d{W}_s$. Changing the sign of $W$ predictably where $\sigma <0$, and filling the zero set with independent Brownian noise, produces a Brownian motion $B$ such that $M=\int \sigma (X_s)d{B}_s$. Thus

Use first the test function $f(x)=x$. The assumed martingale problem says thatis a continuous local martingale. Next use $f(x)=x^2$ to see thatis a local martingale. On the other hand, Itô formula applied to $X=X_0+\int b(X_s)ds+Y$ shows thatis a local martingale. Their difference is both a continuous local martingale and a finite-variation process, so

Part (b), with $H_s=\sigma (X_s{)}^2>0$, supplies a Brownian motion $W$ such thatThereforeso $X$ is a weak solution of a stochastic differential equation to the stated equation.

For an L-diffusion with bounded coefficients and $f\in C_b^{1,2}$, $M^f$ is a continuous martingale. Freeze the time argument along a deterministic partition, apply the spatial martingale problem on each interval, and pass to the limit by the dominated convergence theorem. The same proof works if the absolute drift coefficients and the trace of the diffusivity have integrable time integrals along the path on every finite horizon. Local hypotheses alone give a local martingale conclusion.