Zero–One Laws for Convergence Events

Why Convergence Is a Tail Question

Changing finitely many terms of a sequence cannot affect whether it converges. A sequence may begin differently after such a change, but its eventual behavior is untouched. This simple observation has a precise probability-theoretic consequence: for a sequence of independent random variables, the event of convergence is governed by Kolmogorov’s zero–one law.

The conclusion is powerful but limited. It says the probability of convergence must be either zero or one. It does not say which one, and therefore does not establish almost-sure convergence on its own.

The Convergence Event Is a Tail Event

Let \(X_1,X_2,\ldots\) be real-valued random variables on a probability space \((\Omega,\mathcal F,\mathbb P)\). We mean convergence to a finite real number. Since \(\mathbb R\) is complete, a sequence of real numbers converges to a finite limit if and only if it is Cauchy. The Cauchy property can be expressed using only terms arbitrarily far out in the sequence.

Definition: The convergence event of \((X_n)\) is \[ C=\{\omega\in\Omega:(X_n(\omega))_{n\geq 1}\text{ converges in }\mathbb R\}. \] For each \(m\in\mathbb N\), let \(\sigma(X_n:n\geq m)\) be the sigma-algebra generated by the variables from index \(m\) onward. The tail sigma-algebra is \[ \mathcal T=\bigcap_{m=1}^{\infty}\sigma(X_n:n\geq m). \] An event in \(\mathcal T\) is called a tail event.

To see that \(C\) is measurable and belongs to every sigma-algebra in this intersection, write the Cauchy condition with the tolerances \(1/j\). For each \(m\), define

C_m= \bigcap_{j=1}^{\infty} \bigcup_{N=m}^{\infty} \bigcap_{r,s\geq N} \left\{|X_r-X_s|<\frac{1}{j}\right\}.

Every event inside this expression involves only variables with indices at least \(m\). Thus \(C_m\in\sigma(X_n:n\geq m)\). The lower bound \(N\geq m\) does not change the Cauchy condition: if a suitable \(N\) is smaller than \(m\), the same inequalities hold with \(N\) replaced by \(m\). Consequently \(C_m=C\).

Proposition: For any sequence of real-valued random variables, its finite convergence event belongs to the tail sigma-algebra: \[ C\in\bigcap_{m=1}^{\infty}\sigma(X_n:n\geq m). \]

Proof: Fix \(m\in\mathbb N\). The sequence \((X_n(\omega))\) converges in \(\mathbb R\) exactly when it is Cauchy. In symbols, this is the event \[ \bigcap_{j=1}^{\infty} \bigcup_{N=m}^{\infty} \bigcap_{r,s\geq N} \left\{|X_r-X_s|<\frac{1}{j}\right\}. \] For each \(N\geq m\), every index \(r,s\geq N\) is at least \(m\), so each event in the innermost intersection belongs to \(\sigma(X_n:n\geq m)\). Sigma-algebras are closed under countable intersections and unions, so the displayed event belongs to \(\sigma(X_n:n\geq m)\). It is exactly \(C\), by the Cauchy criterion and the fact that \(\mathbb R\) is complete. Since \(m\) was arbitrary, \(C\) belongs to their intersection. \(\square\)

Kolmogorov’s Zero–One Law for Convergence

Kolmogorov’s zero–one law states that if \(X_1,X_2,\ldots\) are independent, every event in their tail sigma-algebra has probability zero or one. Apply it to the event just identified.

Theorem: If \(X_1,X_2,\ldots\) are independent real-valued random variables, then \[ \mathbb P\bigl((X_n)_{n\geq 1}\text{ converges in }\mathbb R\bigr)\in\{0,1\}. \]

Proof: By the proposition, the convergence event \(C\) belongs to the tail sigma-algebra of the sequence. Independence and Kolmogorov’s zero–one law therefore give \(\mathbb P(C)\in\{0,1\}\). \(\square\)

No identical-distribution assumption is needed. Independence alone is enough. The proposition also shows that the convergence event is a tail event even when the variables are not independent; independence is used for the zero–one conclusion, not for the tail property.

Worked Examples

Worked Example: Independent Signs Scaled by the Index

Suppose \((\varepsilon_n)_{n\geq1}\) are independent random variables satisfying \(\mathbb P(\varepsilon_n\in\{-1,1\})=1\) for each \(n\), and set \(X_n=\varepsilon_n/n\). Define \[ E=\bigcap_{n=1}^{\infty}\{\varepsilon_n\in\{-1,1\}\}. \] Each event in the intersection has probability one. By countable subadditivity applied to the complements, \(\mathbb P(E)=1\). On \(E\), for every \(n\), \[ |X_n|=\frac{|\varepsilon_n|}{n}=\frac{1}{n}. \] Since \(1/n\to0\), \(X_n\to0\) on \(E\), and hence the sequence converges almost surely. The calculation is asserted on \(E\), not necessarily at every outcome: the assumptions allow \(\varepsilon_n\) to take other values on null sets. The scaled variables remain independent, since each \(X_n\) is a measurable function of \(\varepsilon_n\).

Worked Example: An Independent Sequence That Does Not Converge

Let \(X_n(\omega)=(-1)^n\) for every outcome \(\omega\). These are deterministic random variables and are independent: for any finite collection of indices, each event \(\{X_n\in A_n\}\) has probability zero or one, and the probability of their intersection equals the product of their probabilities. But the numerical sequence alternates between \(-1\) and \(1\), so it does not converge. Its convergence event is empty and has probability zero. This example shows directly that independence does not imply convergence.

Worked Example: Independent Bernoulli Variables

Let \(X_n\) be independent Bernoulli random variables with \(\mathbb P(X_n=1)=p\) and \(\mathbb P(X_n=0)=1-p\), where \(0<p<1\). If a sequence taking only the values zero and one converges, it must eventually be constant. Indeed, if it converges to \(L\), then for all sufficiently large \(n\), \(|X_n-L|<1/3\). Both zero and one cannot satisfy this inequality, since their distance is \(1\), greater than \(2/3\). Thus either all sufficiently late terms are zero or all sufficiently late terms are one. For fixed \(N\), independence gives \[ \mathbb P(X_n=0\text{ for every }n\geq N) =\lim_{k\to\infty}(1-p)^{k+1}=0, \] because \(0<1-p<1\). Similarly, \[ \mathbb P(X_n=1\text{ for every }n\geq N) =\lim_{k\to\infty}p^{k+1}=0. \] Convergence requires one of these two events to occur for some \(N\in\mathbb N\). A countable union of probability-zero events has probability zero, so the sequence converges with probability zero. This agrees with the zero–one theorem, but the calculation—not the theorem alone—identifies which of the two possible probabilities occurs.

What the Zero–One Conclusion Does—and Does Not—Say

For an independent sequence, the convergence probability cannot be, for example, \(0.4\). It must be \(0\) or \(1\). However, both possibilities occur, as the examples show. In the scaled-sign example, convergence holds almost surely; in the alternating and Bernoulli examples, it does not.

This distinction is important when using the theorem in a proof. A zero–one law rules out intermediate probabilities, but it supplies no direct estimate that favors one endpoint. To prove almost-sure convergence, one still needs an additional argument that forces the probability to be one. To prove nonconvergence with positive probability, the zero–one law immediately strengthens that fact to probability one for nonconvergence: if \(\mathbb P(C)<1\), then the theorem implies \(\mathbb P(C)=0\).

The scope of the statement also matters. The theorem here concerns convergence to a finite real limit. Its proof uses the equivalence between finite convergence and the Cauchy property in \(\mathbb R\). If a different notion of convergence is intended, its event must be defined and checked for tail measurability before applying a zero–one law.

Takeaway: Convergence is determined by arbitrarily late terms, so it is a tail event. For independent real-valued variables, its probability is therefore zero or one—but deciding which value holds requires further information.

Check Your Understanding

Use the Cauchy characterization and the zero–one conclusion to answer these questions.

Why does the convergence event belong to \(\sigma(X_n:n\geq m)\) for every \(m\)?
Which part of the zero–one argument requires independence: proving the convergence event is a tail event, or applying Kolmogorov’s zero–one law?
Can an independent sequence have convergence probability strictly between zero and one? Explain.
Why is the identity \(|\varepsilon_n/n|=1/n\) in the scaled-sign example only needed on an event of probability one?
What additional task remains after a zero–one law shows that convergence has probability zero or one?