= Solution
For any finite list $t_1,\ldots,t_k\in[0,1]$, the <random vector> $(U_{t_1},\ldots,U_{t_k},B_1)$ 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> give
$$
\operatorname{Cov}(U_t,B_1)=\operatorname{Cov}(B_t,B_1)-t\operatorname{Var}(B_1)=t-t=0.
$$
The <independence of uncorrelated jointly normal variables> implies that $(U_{t_1},\ldots,U_{t_k})$ is <independence>[independent] of $B_1$. Extending from finite-coordinate cylinder <events> by <independence extended from generating pi-systems> proves
$$
\boxed{\sigma(U_t:0\leq t\leq1)\ \text{is independent of}\ \sigma(B_1).}
$$
Thus \b[the <Brownian bridge> is <independence>[independent] of its endpoint]. Its <covariance kernel> is $\operatorname{Cov}(U_s,U_t)=\min(s,t)-st$, confirming that $U$ is the standard <Brownian bridge>. This is the <Brownian bridge independence from its endpoint> property.
Back to article page
