Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 4

A process $(X_t{)}_{t\ge 0}$ is Brownian motion in ${\mathbb{R}}^d$ when $X_0=0$, its paths are almost surely continuous, and for $0\le t_0<\cdots <t_k$ the increments $X_{t_j}-X_{t_{j-1}}$ are independent centered Gaussian vectors with covariance $(t_j-t_{j-1})I_d$.

The paths of $U{X}_t$ are continuous and start at zero. Its increments are independent because they are deterministic functions of the independent increments of $X$. They are centered Gaussian, and orthogonality givesThus $UX$ is Brownian motion. This is the orthogonal invariance of Brownian motion.

Apply the orthogonal transformationThe processes $D$ and $C$ are independent one-dimensional Brownian motions, with $D_0=\sqrt{2}a$ and $C_0=0$. The meeting time is the first time $D$ hits zero, which is almost surely finite by one-dimensional Brownian recurrence.

The Brownian reflection principle gives the first-passage density from $x>0$ to zero asSubstituting $x=\sqrt{2}a$ gives the meeting time of two independent Brownian motions density

At the meeting time,The process $C$ is independent of $T$, so conditional on $T=u$ the meeting position is $N(0,u/2)$. Independently, $B_t-B_u$ is $N(0,t-u)$. ThereforeFor $0<s\le t$, integrate this conditional Gaussian distribution against the density from part c: