Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 4 / b / Solution

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.