Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 29 / 6

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.