Why Tail Events Have Special Probabilities
A tail event is determined by what happens arbitrarily far along a sequence. Changing or deleting finitely many initial terms cannot change whether such an event occurs. For an independent sequence, this makes a tail event independent of every finite initial segment—and, ultimately, independent of itself.
That last conclusion is highly restrictive. If an event is independent of itself, its probability must equal its square, so it can only be zero or one. The main work is making the independence argument rigorous: first for each finite segment, and then for the sigma-algebra generated by the entire sequence.
The Tail Sigma-Algebra
Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, and let \(X_1,X_2,\ldots\) be random elements, where \(X_n\) takes values in a measurable space \((S_n,\mathcal S_n)\). For \(m\in\mathbb N\), the sigma-algebra \(\sigma(X_k:k\ge m)\) records the information contained in the sequence from index \(m\) onward.
Thus, \(A\in\mathcal T\) means that for every \(m\), the event \(A\) can be described using only \(X_m,X_{m+1},\ldots\). In particular, \(A\) belongs to the sigma-algebra generated by the full sequence, since \(\sigma(X_k:k\ge m)\subseteq\sigma(X_k:k\ge1)\).
Throughout, an independent sequence means that every finite subfamily of the random elements is independent. This mutual independence is important: pairwise independence alone does not provide the factorization used below.
Independence from a Finite Initial Segment
Fix \(n\in\mathbb N\). Write \(\mathcal F_n=\sigma(X_1,\ldots,X_n)\) for the information in the first \(n\) random elements, and write \(\mathcal H_n=\sigma(X_k:k\ge n+1)\) for the information in the remaining sequence. We first show that these two sigma-algebras are independent. The proof uses generating classes so that independence of finite collections of random elements extends to the whole infinite remainder.
Proof. For each \(i\), choose \(B_i\in\mathcal S_i\). The sets \(\{X_i\in B_i\}\), for \(1\le i\le n\), generate a pi-system \(\mathcal P_n\) for \(\mathcal F_n\), when finite intersections and \(\Omega\) are included. Likewise, finite intersections of events of the form \(\{X_j\in B_j\}\), with \(j\ge n+1\), together with \(\Omega\), form a pi-system \(\mathcal Q_n\) generating \(\mathcal H_n\). Each member of \(\mathcal Q_n\) involves only finitely many coordinates.
For \(C\in\mathcal P_n\) and \(D\in\mathcal Q_n\), independence of the finite subfamily of random elements involved gives \(\mathbb P(C\cap D)=\mathbb P(C)\mathbb P(D)\). Fix such a \(C\). The collection of all \(D\in\mathcal H_n\) satisfying this identity is a lambda system: it contains \(\Omega\), is closed under complements, and is closed under countable disjoint unions. Since it contains \(\mathcal Q_n\), the pi-lambda theorem shows that it contains \(\mathcal H_n\).
Now fix any \(D\in\mathcal H_n\). The collection of all \(C\in\mathcal F_n\) satisfying the identity is also a lambda system. The preceding step shows that it contains \(\mathcal P_n\). A second application of the pi-lambda theorem gives the identity for every \(C\in\mathcal F_n\). Therefore \(\mathcal F_n\) and \(\mathcal H_n\) are independent. \(\square\)
If \(A\) is a tail event, then \(A\in\mathcal H_n\) for every \(n\), because \(A\in\sigma(X_k:k\ge n+1)\). The lemma therefore implies that \(A\) is independent of \(\mathcal F_n\) for every finite initial segment. That is not yet independence from the sigma-algebra generated by the entire sequence; a further generating-class step supplies it.
Kolmogorov’s Zero–One Law
Proof. Fix \(A\in\mathcal T\), and let \(\mathcal F_\infty=\sigma(X_k:k\ge1)\). Consider the collection $$ \mathcal D_A=\{B\in\mathcal F_\infty:\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)\}. $$ This is a lambda system. It contains \(\Omega\), because \(\mathbb P(A\cap\Omega)=\mathbb P(A)=\mathbb P(A)\mathbb P(\Omega)\). If \(B\in\mathcal D_A\), then \(\mathbb P(A\cap B^c)=\mathbb P(A)-\mathbb P(A\cap B)=\mathbb P(A)(1-\mathbb P(B))=\mathbb P(A)\mathbb P(B^c)\). For pairwise disjoint \(B_1,B_2,\ldots\in\mathcal D_A\), countable additivity gives \(\mathbb P(A\cap\bigcup_j B_j)=\sum_j\mathbb P(A\cap B_j)=\mathbb P(A)\sum_j\mathbb P(B_j)=\mathbb P(A)\mathbb P(\bigcup_j B_j)\).
The sigma-algebras \(\mathcal F_n\) are nested. Consequently, \(\mathcal P=\bigcup_{n\ge1}\mathcal F_n\) is a pi-system: if \(B\in\mathcal F_n\) and \(C\in\mathcal F_m\), then \(B\cap C\in\mathcal F_{\max(n,m)}\). It generates \(\mathcal F_\infty\). Since \(A\) is independent of each \(\mathcal F_n\), the collection \(\mathcal D_A\) contains \(\mathcal P\). The pi-lambda theorem now gives \(\mathcal D_A=\mathcal F_\infty\).
Finally, \(A\in\mathcal F_\infty\), so we may take \(B=A\) in the defining identity for \(\mathcal D_A\). This yields $$ \mathbb P(A)=\mathbb P(A\cap A)=\mathbb P(A)^2. $$ The only real numbers satisfying \(p=p^2\) are \(p=0\) and \(p=1\). Hence \(\mathbb P(A)\in\{0,1\}\). \(\square\)
Worked Examples
Worked Example: Infinitely Many Successes
Let \(X_n\) be independent Bernoulli random variables with \(\mathbb P(X_n=1)=p\), where \(0<p<1\). Consider the event $$ A=\{X_n=1\text{ for infinitely many }n\}. $$ Whether infinitely many successes occur is unaffected by any finite initial segment, so \(A\) is a tail event. The zero–one law says its probability is zero or one. To identify which, for each \(N\) let \(E_N\) be the event that there are no successes at any index \(n\ge N\). For \(m\ge N\), independence gives $$ \mathbb P(X_N=0,\ldots,X_m=0)=(1-p)^{m-N+1}. $$ As \(m\) increases, these events decrease to \(E_N\). Continuity of probability for decreasing events therefore gives $$ \mathbb P(E_N)=\lim_{m\to\infty}(1-p)^{m-N+1}=0, $$ since \(0<1-p<1\). Having only finitely many successes is the event \(\bigcup_{N=1}^{\infty}E_N\), which has probability zero by countable subadditivity. Thus \(\mathbb P(A)=1\).
Worked Example: Convergence of an Independent Bernoulli Sequence
Again suppose \(X_n\) are independent Bernoulli random variables with a common parameter \(p\in(0,1)\). Let \(C\) be the event that the numerical sequence \(X_n\) converges. A convergent sequence taking values only in \(\{0,1\}\) must eventually be constant: if it converges to \(x\), then for all sufficiently large \(n\), \(|X_n-x|<1/2\), and at most one of 0 and 1 can satisfy this inequality. Conversely, an eventually constant sequence converges. Therefore $$ C=\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}\{X_n=0\} \;\cup\; \bigcup_{N=1}^{\infty}\bigcap_{n\ge N}\{X_n=1\}. $$ This expression shows that \(C\) belongs to \(\sigma(X_n:n\ge m)\) for every \(m\): in either union, the condition “for all \(n\ge N\)” is unchanged if the finitely many indices below \(m\) are omitted. Thus \(C\) is a tail event.
For each fixed \(N\), the probability that all terms from \(N\) through \(m\) are zero is \((1-p)^{m-N+1}\), which tends to zero. Continuity for decreasing events shows that the probability of being zero at every index \(n\ge N\) is zero. The corresponding probability for being one at every \(n\ge N\) is \(\lim_{m\to\infty}p^{m-N+1}=0\). Taking the countable unions over \(N\) shows \(\mathbb P(C)=0\). The zero–one law guarantees the probability is extreme; the calculation identifies it as zero.
Worked Example: Infinitely Often When the Probabilities Have Divergent Sum
Let \(A_1,A_2,\ldots\) be independent events with \(\sum_{n=1}^{\infty}\mathbb P(A_n)=\infty\). The event that infinitely many \(A_n\) occur is a tail event: removing any finite number of events does not affect it. For \(m\ge N\), independence gives $$ \mathbb P\left(\bigcap_{n=N}^{m}A_n^c\right) =\prod_{n=N}^{m}\bigl(1-\mathbb P(A_n)\bigr) \le \exp\left(-\sum_{n=N}^{m}\mathbb P(A_n)\right). $$ Here the inequality follows from \(1-x\le e^{-x}\) for \(0\le x\le1\), applied to each factor. The right-hand side tends to zero as \(m\to\infty\). By continuity for decreasing events, the probability that none of \(A_n\) occur for \(n\ge N\) is zero. The event that only finitely many occur is the union of these zero-probability events over \(N\), and so has probability zero. Therefore the probability that infinitely many \(A_n\) occur is one.
What the Law Does—and Does Not—Say
The law gives a sharp restriction on the probability of tail events, but it does not decide between zero and one. In the examples, additional calculations determined which value applied. In other problems, estimates or separate limit arguments are needed to identify the value.
It is also essential that the event be a tail event. For example, \(\{X_1=1\}\) depends on the first coordinate and need not have probability zero or one. Nor does the law apply merely because an event concerns an infinite sequence: one must check that the event is measurable with respect to every tail sigma-algebra. Finally, the proof relies on independence of the sequence, not just pairwise independence.
The central mechanism is useful to remember: a tail event is independent of every finite initial segment; a pi-lambda argument extends this to the full sigma-algebra generated by the sequence; and an event independent of itself must have probability zero or one.
Check Your Understanding
Use the definitions and proof above to answer the following questions.
- Why does a tail event \(A\) belong to \(\sigma(X_k:k\ge n+1)\) for every fixed \(n\)?
- What role does the pi-lambda theorem play in extending independence from finite-coordinate events?
- In the proof of the zero–one law, why is it valid to take \(B=A\) after showing \(A\) is independent of \(\mathcal F_\infty\)?
- For independent Bernoulli variables with parameter \(p\in(0,1)\), why does convergence imply eventual constancy?
- Does Kolmogorov’s zero–one law alone determine whether a given tail event has probability zero or one? Explain.