Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 / 6

Define the diffusion generator $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. Fix a horizon $t$ and start the strong solution from the deterministic state $x$. The Itô formula applied to the time-reversed test function givesThe drift vanishes by the Kolmogorov backward equation. This is initially a local martingale; localization on compact state/time sets justifies the stochastic integral without a global derivative bound. Since $u$ itself is bounded, this local martingale is a true martingale on $[0,t]$. Its two endpoint expectations giveThis is the bounded backward-equation stochastic representation. Starting the strong solution at deterministic $x$ is the precise meaning of the conditional notation at $X_0=x$. Also $f=u(0,\cdot )$ is bounded, even though boundedness was not separately imposed on $f$.

Put $q(x)=\sigma (x{)}^{2}p(x)$. The explicit density formula and local integrability assumption giveHence its Fokker-Planck probability current $bp-q^{\prime }/2$ is zero. The stationary adjoint equation is $L^{\ast }p=-(bp{)}^{\prime }+q^{\prime \prime }/2=0$. We prove the required integral identity with cutoffs, avoiding an unstated boundary condition on derivatives of $u$.

Choose smooth $0\le {\chi }_R\le 1$, equal to one on $[-R,R]$, supported in $[-2R,2R]$, with $|{\chi }_R^{\prime }|\le C_1/R$ and $|{\chi }_R^{\prime \prime }|\le C_2/R^2$. Since $pLu=(q{u}_x{)}^{\prime }/2$, integrate the backward equation in time and integrate by parts twice in space:These integrations have compact support, as permitted in the question. Bounded coefficients and normalized $p$ giveThe right side is bounded in absolute value bywhich tends to zero. Dominated convergence on the left therefore provesThis is the cutoff proof of invariance for a zero-flux diffusion density. In particular, if an independent initial state has density $p$, part (a) and Fubini give a constant expectation of this test function at every time. This consequence alone is sufficient for the final coupling argument.

Use the same Brownian motion for both solutions: this is a synchronous coupling. For $D_t=X_t-Y_t$, the Itô formula givesOn any fixed finite horizon the stochastic integral has mean zero. Indeed, boundedness of $\sigma$ and the assumed second-moment bounds make the expectation of its squared integrand integrable in time. With $h(t)=\mathbb{E}{D}_t^2$, the contraction condition yields, for every $s\le t$,The allowed Gronwall inequality gives the mean-square contraction of synchronously coupled diffusionsThe identical estimate can also be obtained by localizing the nonnegative local supermartingale $e^{kt}{D}_t^2$ and using Fatou's lemma, a formulation useful when the drift has linear growth.

There is a genuine compatibility issue in the printed global assumptions. If $|b|\le B$, then for $d=|x-y|$,It cannot be at most $-k{d}^2$ for all $d>4B/k$. Thus a globally bounded drift cannot satisfy the stated strict contraction on all of the real line. The stochastic estimate above is the requested conditional calculation. For a non-vacuous application, global boundedness of the drift must be relaxed while retaining suitable existence and moment hypotheses; it is not silently changed here.

Enlarge the space if needed to choose $Y_0$ with density $p$, independent of the driving Brownian motion, and solve the same equation with the same driver as $X_0=x$. The second moment of $p$ makes the initial difference square integrable. By parts (a) and (b),while part (c) givesWe must not assume that the smooth bounded $f$ is globally Lipschitz. The hypotheses give $C_Y={\sup }_t\mathbb{E}{Y}_t^2<\infty$, hence uniform tightness of $(Y_t)$. For any $R$ and $0<\delta \le 1$, define the modulus of continuity of $f$ on $[-R-1,R+1]$ by ${\omega }_R(\delta )$. Splitting according to $|Y_t|\le R$ and $|X_t-Y_t|\le \delta$, and applying Markov inequality, givesFirst choose $R$ large, then $\delta$ small using uniform continuity on that compact interval, and finally let $t\to \infty$. The right side can be made arbitrarily small. Together with part (a), this provesThis is convergence by synchronous coupling for bounded continuous test functions. It only needs the constant expectation from part (b) and uniform second moments; it does not add an unstated global bound on $f^{\prime }$ or require a separate stationarity theorem. As explained in part (c), the literal bounded-drift plus strict-contraction assumptions have no global example; the calculation records the intended consequence under compatible dissipative-drift hypotheses as well.