Past exam of the stochastic analysis and partial differential equations course of the University of Oxford / 2025 / 2 / c

Here the diffusion generator is $\mathcal{L}=ax{\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. The Feynman-Kac formula givesTry $F(t,x)=\exp [iu\alpha (t)x+\beta (t)]$, where $\alpha (T)=1$ and $\beta (T)=0$. Substitution and collection of the coefficients of $x$ and the constant term giveThe solutions of these ordinary differential equations, with the terminal conditions imposed, areFor $\Delta =T-t$, define the conditional varianceThen the required conditional characteristic function isIn particular, for $a\ne 0$ this exponent is $iu{e}^{a\Delta }x-\frac{{\sigma }^2{u}^2}{4a}(e^{2a\Delta }-1)$. The variance $v_{\Delta }$ is nonnegative for either sign of $a$, and its limit at $a=0$ is ${\sigma }^2\Delta$. Hence the solution is bounded in absolute value by one and the Feynman-Kac formula identifies it with the required expectation. The value $u=0$ gives one directly, including that case without dividing by $u$.

For an independent check, the integrating factor $e^{-as}$ and the Itô product rule giveThe stochastic integral has a normal distribution because its integrand is deterministic; the Itô isometry gives its variance as $v_{\Delta }$. For $a<0$ this is the Ornstein-Uhlenbeck transition distribution. The formula also covers the unstable case $a>0$, and degenerates to deterministic motion when $\sigma =0$.

For $a<0$, set $v_{\infty }=-{\sigma }^2/(2a)>0$. The candidate invariant probability law of a Markov process isTo prove invariance rather than merely guess a limiting probability distribution, initialize $X_t\sim \pi$ independently of future Brownian increments. Integrating the conditional characteristic function from part (i) against this centered normal distribution givesbecause $v_{\Delta }=v_{\infty }(1-e^{2a\Delta })$. The uniqueness theorem for characteristic functions implies $X_{t+\Delta }\sim \pi$ for every $\Delta \ge 0$. This proves $\pi P_{\Delta }=\pi$, as required. The time-homogeneous Markov property then also makes the process with this Ornstein-Uhlenbeck stationary initialization a strictly stationary process.