Independent sigma-algebras

Two sub-sigma-algebras ${\mathcal{F}}_1,{\mathcal{F}}_2$ are independent whenfor every $A\in {\mathcal{F}}_1$ and $B\in {\mathcal{F}}_2$.

Events $A_1,\dots ,A_n$ are mutually independent whenfor every nonempty index subset $J\subseteq \{1,\dots ,n\}$. For two events and $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning whether $B$ occurred does not change the chance of $A$.

For independent uniform dice, the events that the first result is an even integer and that the second result is an even integer each have probability $1/2$; their intersection has probability $9/36=1/4$. In contrast, "first result is an even integer" and "first result is an odd integer" are disjoint events of positive probability, so they cannot be independent events. Learning one occurred completely rules out the other.

Pairwise independence is weaker than mutual independence. Take two independent fair bits $X,Y$. The three events $X=0$, $Y=0$, and $X=Y$ each have probability $1/2$ and every pair has intersection probability $1/4$. Their triple intersection is the single outcome $(0,0)$, of probability $1/4$ rather than $1/8$. Checking pairs alone misses the joint constraint.

For three pairwise independent events of common probability $a$, write $x=\mathbb{P}(A_1\cap A_2\cap A_3)$. Inclusion-exclusion principle gives avoidance probability $1-3a+3a^2-x$, and $x\le a^2$ gives the displayed lower bound. It is strictly positive for $0<a<1/2$. For $1/2\le a<1$, a symmetric atomic law with no zero-success atom has mass $(1-a{)}^2$ on each singleton success pattern, $(1-a)(2a-1)$ on each doubleton, and $1-3a+3a^2$ on the tripleton. These masses sum to one, have marginal $a$, and have pair intersections $a^2$.

If two pi-systems satisfy the probability factorization pairwise, their generated sigma-algebras are independent. For one event in the first family, the events in the second sigma-algebra satisfying the factorization form a Dynkin system; apply Dynkin lemma, then repeat with the roles reversed.

For a sequence of random variables $(X_n{)}_{n\ge 1}$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, defineThe decreasing sigma-algebras ${\mathcal{G}}_n$ are generated by the listed random variables. Their intersection consists of events expressible after discarding every finite prefix. No independence is required.^[ref]

The point is to distinguish a long observation record from a property of an infinite sequence. The event of seeing infinitely many heads and the event that sample means converge survive any finite change, while an exact first-head time does not. Kolmogorov's zero–one law adds the striking conclusion that under independence each tail event has probability zero or one; the tail sigma-algebra itself can still contain many nonempty sets.

On the full Cartesian product $\Omega ={\mathbb{R}}^{\mathbb{N}}$ with its product sigma-algebra, take the coordinate random variables $X_j(x)=x_j$. A product-measurable set $A$ is a tail event if and only if membership is unchanged whenever two sequences differ in finitely many coordinates. The measurability condition is essential: invariance alone is not a license to assign a probability to an arbitrary subset.

For the forward implication, fix $m$. An event in $\sigma (X_{m+1},X_{m+2},\dots )$ cannot distinguish points with the same remaining coordinates. Since a tail event belongs to this sigma-algebra for every $m$, every finite change preserves membership.

For the converse, freeze the first $m$ coordinates to zero. The insertion mapis a measurable function for the product sigma-algebras. Thus $B_m=i_m^{-1}(A)$ is a measurable set. Invariance gives $A={\pi }_m^{-1}(B_m)$, where the coordinate projection ${\pi }_m$ discards the first $m$ entries. Hence $A\in \sigma (X_{m+1},X_{m+2},\dots )$ for every $m$. The same argument works on $\{0,1{\}}^{\mathbb{N}}$. On a general probability space, coordinates need not admit arbitrary replacements, so the defining intersection of sigma-algebras remains the reliable formulation.

Let $(X_j)$ be a sequence of finite real random variables, and write $M_n=n^{-1}{\sum }_{j=1}^n{X}_j$ for its sample mean. For fixed $m$ and $n>m$, put $M_n^{(m)}=n^{-1}{\sum }_{j=m+1}^n{X}_j$. Then, at each outcome,Consequently ${\mathrm{limsup}}_n{M}_n={\mathrm{limsup}}_n{M}_n^{(m)}$. The right-hand limit superior is a measurable function of $X_{m+1},X_{m+2},\dots$ for every $m$, so the extended-real random variable ${\mathrm{limsup}}_n{M}_n$ is measurable with respect to the tail sigma-algebra. The same holds for the limit inferior. Thus the event of convergence to a finite limit of a sequence, and the event $\{M_n\to a\}$ for any specified real $a$, are tail events.

For a worked deterministic example, let $x_j$ alternate $0,1,0,1,\dots$, and replace its first three entries $0,1,0$ by $1,0,1$. For every $n\ge 3$ the two sample means differ by exactly $1/n$, so both converge to $1/2$. Finite edits can change small-sample answers while preserving the infinite limit of a sequence.

Under independence, Kolmogorov's zero–one law makes each specified convergence event have probability zero or one. It does not select which value occurs. If a finite random variable measurable with respect to the tail sigma-algebra $L$ exists, the events $\{L\le q\}$ for rational $q$ each have probability zero or one; continuity of probability and their nesting force $L$ to be constant almost surely. For fair independent Bernoulli trials, the strong law of large numbers establishes $M_n\to 1/2$ almost surely; the zero–one conclusion alone proves neither convergence nor the value $1/2$.

A tail event is an event $A\in \mathcal{T}$, where $\mathcal{T}$ is the tail sigma-algebra of the chosen sequence of random variables. It concerns information available arbitrarily far along the sequence. In the measurable coordinate models described in finite-prefix invariance of tail events, any finite edit preserves whether it occurs.

This is stronger than merely needing a long waiting time. On $\{0,1{\}}^{\mathbb{N}}$, with $1$ denoting heads, let $K=\inf \{j\ge 1:X_j=1\}$, taking $K=\infty$ if there is no head. The eventis not a tail event: changing only $X_1$ from zero to one changes membership. Under fair independent Bernoulli trials its probability is $1/8$. Even the event $\{K<\infty \}$, of probability one in this model, is not a tail event as a literal set: the all-zero sequence and the sequence with just one initial head differ in one coordinate. A probability of zero or one does not characterize tail events.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

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.

Let $Z$ have the fair Bernoulli distribution and set $X_j=Z$ for every $j$. Each coordinate has the same fair probability distribution as in the independent model, but the coordinates are dependent: $\mathbb{P}(X_1=1,X_2=1)=1/2$, whereas the product of the marginal probabilities is $1/4$. Every future sigma-algebra equals $\sigma (Z)$, so the tail sigma-algebra also equals $\sigma (Z)$.

The tail event of infinitely many heads is now exactly $\{Z=1\}$, with probability $1/2$. For the finite-window events $B_{m,r}$ in infinitely many heads as a tail event,The sample space can even be the same full Cartesian product $\{0,1{\}}^{\mathbb{N}}$ in both models: the dependent probability measure assigns mass $1/2$ to each of the all-zero and all-one sequences. Then $A$ is the same literal tail event in both models. Only the probability measure and its independence properties change.