Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 2

The Strong Markov property says that for every almost surely finite stopping time $T$, the process $(B_{T+t}-B_T{)}_{t\ge 0}$ is a standard Brownian motion independent of ${\mathcal{F}}_T$.

The process $Y_t=B_{T-t}-B_T$ is centered Gaussian and continuous. Its covariance isas follows by expanding Brownian covariances, or by reading its increments backwards. The Gaussian-process characterization of Brownian motion therefore shows that $(Y_t{)}_{0\le t\le T}$ has the same law as $(B_t{)}_{0\le t\le T}$.

The event $\{B_T=I_T\}$ is the event that $B_{T-t}-B_T\ge 0$ for every $t\le T$. By part (ii), its probability equals the probability that Brownian motion started at zero remains nonnegative throughout $(0,T]$. The Brownian reflection principle implies $\mathbb{P}({\min }_{s\le T}{B}_s\ge 0)=0$. Hence $\mathbb{P}(B_T=I_T)=0$.

Continuity gives existence of a minimizer. If there were two, choose a rational $q$ strictly between them. Then the minima on $[0,q]$ and $[q,1]$ would coincide. Conditional on ${\mathcal{F}}_q$, the latter equals $B_q$ plus the minimum of an independent Brownian motion on $[0,1-q]$, whose distribution is continuous by the Brownian reflection principle. Thus equality has conditional probability zero. Taking the countable union over rational $q$ proves almost-sure uniqueness.

Almost surely, Brownian motion has a unique minimizer on every interval with rational endpoints, by rescaling part (iv). Every local minimum is the minimum on some rational interval contained in a witnessing neighbourhood. Each rational interval contributes at most one point, and there are countably many such intervals. The set of local minima is therefore countable.

The function $z\mapsto \log |z|$ is harmonic on the annulus $\epsilon <|z|<R$. Hence $\log |B_{t\wedge {\tau }_{\epsilon }\wedge {\tau }_R}|$ is a bounded martingale by Itô formula. Optional stopping givesSolving yields

For fixed $y\ne x$, part (i), translated by $y$, and then $\epsilon \downarrow 0$ show that planar Brownian motion has probability zero of ever hitting $y$ before leaving any fixed large disk. Letting the disk radius tend to infinity shows ${\mathbb{P}}_x(y\in B([0,1]))=0$. By Tonelli theorem,The nonnegative random area is therefore zero almost surely.