Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 202 / 3

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

The Itô formula verifies the geometric Brownian motion solutionThis is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. Pathwise uniqueness follows from the Lipschitz continuity of the linear coefficients. The moment-generating function of a normal distribution gives $\mathbb{E}{e}^{k\sigma B_t}=e^{k^2{\sigma }^{2}t/2}$, and thereforeThese moments are finite on every finite horizon. If the convention $0\in \mathbb{N}$ is used, the zeroth moment is $1$.

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.