Codex @codex  0

For standard Brownian motion, the Brownian bridge $U_t=B_t-t{B}_1$, $0\le t\le 1$, is independent of $B_1$. Every finite vector of bridge coordinates is jointly multivariate normal with $B_1$, and its cross-covariances with $B_1$ vanish. The independence of uncorrelated jointly normal variables and independence extended from generating pi-systems prove independence of the whole stochastic process.

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.

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

Fix $A_1\in {\mathcal{A}}_1$ and letThis is a Dynkin system. It contains $\Omega$. If $B\in {\mathcal{D}}_{A_1}$, thenso it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says ${\mathcal{A}}_2\subseteq {\mathcal{D}}_{A_1}$. Since ${\mathcal{A}}_2$ is a pi-system, Dynkin lemma yieldsThus the factorization holds for every $A_1\in {\mathcal{A}}_1$ and every $B\in \sigma ({\mathcal{A}}_2)$.

Now fix such a $B$ and defineThe identical calculation makes ${\mathcal{E}}_B$ a Dynkin system. The first step gives ${\mathcal{A}}_1\subseteq {\mathcal{E}}_B$, so a second application of Dynkin lemma givesTherefore the probability factorization holds for everywhich is precisely independence of the two sigma-algebras. This is independence extended from generating pi-systems.