Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 35 / 4

Let $X,Y$ be two solutions with the same Brownian driver and the same initial value. Write $D=X-Y$, and take a common Lipschitz constant $L$ for both coefficients. Stop when $|X|+|Y|$ reaches $n$. Before this stopping time ${\tau }_n$, all difference integrands below are bounded. The Itô formula givesThe stopped stochastic integral is a true martingale on every finite horizon. Lipschitz continuity therefore yieldsSince $D_0=0$, the Gronwall inequality makes the left side zero. Let $n\to \infty$; the continuous global solutions are bounded on each compact time interval, so their stopping times exhaust it. Equality first at every rational time and then by continuity gives indistinguishability. This proves pathwise uniqueness for globally Lipschitz drift and diffusion coefficients. The argument does not require the initial value to have a global second moment: after stopping, the difference is bounded and initially zero.

The Itô product rule with a deterministic integrating factor givesIntegrating from zero provesBoth coefficients are globally Lipschitz, so part (a) proves that this is the pathwise unique solution. Under a measure where $B$ is a Brownian motion, the solution is centered Gaussian with variance $e^{2t}{\int }_0^t{e}^{-2s}ds=(e^{2t}-1)/2$. The positive drift sign gives an unstable linear diffusion rather than a mean-reverting one.

Use $\tilde{\mathbb{P}}$ for the reference measure under which $X$ is a Brownian motion. Pathwise, $B_t=X_t-{\int }_0^t{X}_{s}ds$. Put $h_s=X_s{1}_{\{s\le T\}}$ and defineThe stopped integrand has absolute value at most one. The Novikov condition holds on every finite horizon, so $Z$ is a density martingale. The Itô formula also gives the more useful identityThus $Z$ is uniformly integrable over the entire time axis, not just a true martingale on each finite horizon. Under the reference measure, $\tilde{\mathbb{E}}(t\wedge T)=\tilde{\mathbb{E}}{X}_{t\wedge T}^2\le 1$, so $T<\infty$ almost surely. Hence $Z_t\to Z_T>0$ and $\tilde{\mathbb{E}}{Z}_T=1$.

Define the probability measure on the original sigma-algebra byIts restriction to ${\mathcal{F}}_t$ has density $Z_t$. The Girsanov theorem states that subtracting the integrated density integrand from the reference Brownian motion gives a Brownian motion under the new measure. Thereforeis a Brownian motion under $\mathbb{P}$, and $B_{t\wedge T}=W_{t\wedge T}^{\mathbb{P}}$. This is exactly the requested Brownian property until $T$. The bounded Girsanov density for exit of an unstable linear diffusion also proves global absolute continuity, so a mere collection of unspecified finite-horizon measures is unnecessary.

At exit, continuity gives $X_T^2=1$. The preceding density identity becomesSince $|X_s|\le 1$ before exit, ${\int }_0^T{X}_s^{2}ds\le T$. On $\{T\le t\}$ it follows that $Z_T\ge e^{1/2-T}\ge e^{1/2-t}$. The definition of the changed measure now yieldsThe positive sign of the stochastic integral in the density is what introduces the positive drift $X_{t}dt$; reversing that sign would give a different measure and would not prove this inequality.