Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 3

Fix $0\le r\le t$. Since $M$ is a martingale,Every finite vector consisting of $M_{t+s}-M_t$ and past values $M_{r_1},\dots ,M_{r_n}$ has a multivariate normal distribution. Therefore uncorrelated jointly normal variables are independent, so the increment is independent of every finite vector of past values. A Monotone class theorem then extends this to independence from ${\mathcal{F}}_t=\sigma (M_r:0\le r\le t)$. This is the independent increments of a Gaussian martingale.

Define the deterministic function $f(t)=\mathbb{E}[M_t^2]$. The independent increments from part (i) show that $f$ is increasing and that $M_t^2-f(t)$ is a martingale. Mean-square continuity follows from path continuity and the Gaussian laws, so $f$ is continuous.

The Itô formula also says that $M_t^2-[M{]}_t$ is a local martingale. Their difference $[M{]}_t-f(t)$ is therefore a continuous finite-variation process that is also a local martingale. By the theorem that a continuous finite-variation local martingale is constant, and because the difference starts at zero,for all $t\ge 0$ almost surely.

The assertion is false. Let $Z\sim N(0,1)$ and define $X_t=tZ$. This is a centered continuous Gaussian process. Its natural filtration satisfies $Z=X_s/s\in {\mathcal{F}}_s^X$ for every $s>0$, and hence, for $0<s<t$,with positive probability. Thus $X$ is not a martingale and does not belong to the stated martingale class.

The assertion is true. The Dambis-Dubins-Schwarz theorem, with an independent continuation of the Brownian motion if $f$ is bounded, representsBecause $f$ is deterministic, every finite vector $(M_{t_1},\dots ,M_{t_n})$ is a finite vector of a Brownian motion at deterministic times and therefore has a multivariate normal distribution. Hence $M$ is a Gaussian process. This is the deterministic quadratic variation characterizes a Gaussian continuous local martingale result.

The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r_1<|z|<r_2$. The Itô formula therefore makes $\log |B_{t\wedge \tau }|$ a bounded martingale. The optional sampling theorem for a supermartingale giveswhere $p=\mathbb{P}(|B_{\tau }|=r_1)$. Solving this linear equation yields the planar Brownian annulus hitting probability

No such function exists. Continuity on the compact closed unit disk makes $u$ bounded near the origin. The removable singularity for a bounded harmonic function extends $u$ harmonically across the origin. The extended function is continuous on the closed disk and vanishes on its boundary, so the maximum principle for harmonic functions, applied to both $u$ and $-u$, forces $u=0$ throughout the disk. This contradicts $u(0,0)=1$.