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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 mutually independent random variables, every tail event has $\mathbb{P}(A)\in \{0,1\}$. Identical probability distributions are unnecessary. The tail sigma-algebra is trivial in probability: its events have probability zero or one, though it can contain nonempty proper sets.^[ref]

For a proof, set ${\mathcal{F}}_m=\sigma (X_1,\dots ,X_m)$ and ${\mathcal{F}}_X=\sigma (X_1,X_2,\dots )$. Independence extended from generating pi-systems makes ${\mathcal{F}}_m$ independent of $\sigma (X_{m+1},X_{m+2},\dots )$, hence of each tail event $A$. The increasing union ${\bigcup }_m{\mathcal{F}}_m$ is a generating pi-system for ${\mathcal{F}}_X$. For fixed $A$, the events $B$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ form a Dynkin system. The pi-lambda theorem extends this equality to all $B\in {\mathcal{F}}_X$. Taking $B=A$ givesThus $A$ is independent of itself, forcing the two stated values.^[ref]

The useful feature is the absence of intermediate probabilities for infinite properties that ignore every finite prefix. It reduces some calculations to showing that a tail event has positive probability, or that its complement does. The theorem supplies no finite observation deadline, and an extra argument such as the calculation in infinitely many heads as a tail event is needed to decide which value occurs.

Independence of the full sigma-algebras immediately gives independence of their generating pi-systems. For the converse, fix $H\in {\mathcal{H}}_0$. On $\mathcal{G}$, define finite measuresThey have the same total mass $\mathbb{P}(H)$ and agree on ${\mathcal{G}}_0$ by assumption. The supplied measure-uniqueness theorem makes them equal on $\mathcal{G}$. Thus the independence formula holds for every $G\in \mathcal{G}$ and every $H\in {\mathcal{H}}_0$.

Now fix such an arbitrary $G$. The measures $H\mapsto \mathbb{P}(G\cap H)$ and $H\mapsto \mathbb{P}(G)\mathbb{P}(H)$ on $\mathcal{H}$ have equal total mass and agree on ${\mathcal{H}}_0$. A second application gives equality on all of $\mathcal{H}$. This proves independence extended from generating pi-systems.

Equip $S$ with a sigma-algebra $\mathcal{S}$. A Poisson random measure with intensity measure of a point process $\mu$ is a random counting measure $\Pi$ such that, for every measurable $A$ with $\mu (A)<\infty$,and the counts on every finite collection of pairwise disjoint measurable sets are independent. Thus $\mathbb{E}\Pi (A)=\mu (A)$. For a sigma-finite measure $\mu$, sets of infinite intensity contain infinitely many points almost surely: exhaust such a set by increasing finite-intensity sets and use the Poisson count tails as their means tend to infinity. The count is a measurable random variable for every measurable $A$, which is part of saying that $\Pi$ is a random measure.

Under the usual point representation write $\Pi ={\sum }_j{\delta }_{Z_j}$, counting multiplicities. If $\mu$ is a non-atomic measure, this is a simple Poisson point process; if $\mu$ has atoms, several point occurrences can have the same location, and each occurrence is counted separately. In a topological setting, local finiteness requires that the intensity be finite on compact sets.

Sufficient existence conditions are that $S$ is a standard Borel space and $\mu$ is a sigma-finite measure. This gives existence of a Poisson random measure on a standard Borel space. If a simple locally finite configuration is desired, additionally take a non-atomic, locally finite intensity on a locally compact second-countable Hausdorff space. These conditions are sufficient, not claimed necessary. In particular sigma-finiteness suffices for a countable family of occurrences that can each be independently coloured.

Let $R$ and $G$ be the red and green counting measures. The marking assumption means that, conditional on $\Pi$, the colours of all point occurrences are independent, with the same red probability $r$. For a measurable set $A$ of finite intensity $m=\mu (A)$, condition on its total count. To obtain $a$ red and $b$ green points the total must be $a+b$, so direct multiplication gives the joint red and green Poisson count formulaThus the two counts are independent Poisson variables with respective means $rm$ and $(1-r)m$. This computation includes zero intensity with the usual convention for the probability of zero counts.

For disjoint finite-intensity sets $A_1,\dots ,A_k$, the original counts are independent. Conditional marking in different sets uses disjoint collections of independent colours. Repeating the preceding calculation therefore givesThis proves the disjoint-count definition for each coloured Poisson point process and independence of their two count vectors.

To obtain independence of the whole processes, take any finite list of red test sets and any finite list of green test sets, initially all of finite intensity. Refine their union into the disjoint cells given by their membership patterns. Every test-set count is the sum of the counts of its cells. The entire red cell vector is independent of the entire green cell vector by the product formula, hence so are the corresponding test-set count vectors. For arbitrary measurable test sets, intersect first with a finite-intensity exhaustion of $S$ and pass to the increasing count limits. The count cylinder events generate the two evaluation sigma-algebras, and independence extended from generating pi-systems gives independence of those sigma-algebras. ConsequentlyThis is a direct count proof of the Poisson thinning theorem; no general marking or thinning result has been used to deduce it.

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.

Fix $A\in {\mathcal{A}}_1$ and defineThis is a Dynkin system: it contains $\Omega$; taking complements preserves the equality because $\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$; and countable additivity preserves it under disjoint unions. The hypothesis says that ${\mathcal{D}}_A$ contains the pi-system ${\mathcal{A}}_2$. Hence Dynkin lemma gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and putThe same complement and disjoint-union calculations make this a Dynkin system. The first step shows that it contains ${\mathcal{A}}_1$, so Dynkin lemma gives ${\mathcal{E}}_B={\mathcal{G}}_1$. ThusThis is precisely the definition of independent sigma-algebras, and proves independence extended from generating pi-systems.