Choosing an Almost-Sure Argument with Independence

Choosing the Right Almost-Sure Argument

An almost-sure conclusion says that an event has probability one. The challenge is often not the definition of “almost surely,” but choosing a route that actually proves the desired claim.

Three useful tools have different jobs. A summable bound on event probabilities controls how often events can occur, even without independence. The independent-events Borel–Cantelli lemma handles divergent probability sums when the events are independent. A tail zero–one law says that a suitable event has probability either zero or one, but does not by itself tell us which.

The key is to check both the structure of the events and the conclusion sought. “Only finitely often,” “infinitely often,” and “probability is zero or one” are different claims, and their hypotheses are not interchangeable.

The Three Routes

The first Borel–Cantelli lemma: If \((A_n)_{n\in\mathbb N}\) is a sequence of events and \[ \sum_{n=1}^{\infty}\mathbb P(A_n)<\infty, \] then \[ \mathbb P(A_n\text{ infinitely often})=0. \] No independence assumption is required.

Here, “\(A_n\) infinitely often” means that infinitely many of the events \(A_n\) occur. In other words, only finitely many occur almost surely. This is the right tool when the probabilities themselves have a finite sum.

The independent-events Borel–Cantelli lemma: If the events \((A_n)_{n\in\mathbb N}\) are mutually independent and \[ \sum_{n=1}^{\infty}\mathbb P(A_n)=\infty, \] then \[ \mathbb P(A_n\text{ infinitely often})=1. \]

For this converse direction, independence matters. A divergent sum of event probabilities alone does not guarantee infinitely many occurrences. In particular, pairwise independence is not a substitute for mutual independence in the stated theorem.

The third route applies when the event is unchanged by discarding any finite initial segment of an independent sequence. Such an event belongs to the tail sigma-algebra.

Kolmogorov’s zero–one law: Let \((X_n)_{n\in\mathbb N}\) be an independent sequence, and let \[ \mathcal T=\bigcap_{m=1}^{\infty}\sigma(X_n:n>m) \] be its tail sigma-algebra. If \(E\in\mathcal T\), then \(\mathbb P(E)\in\{0,1\}\).

A zero–one law gives a dichotomy, not a verdict. To prove that \(\mathbb P(E)=1\), one must still rule out \(\mathbb P(E)=0\), for example by establishing a positive lower bound. To prove \(\mathbb P(E)=0\), one must rule out probability one.

Why Summable Probabilities Control Infinitely Often Occurrence

The first Borel–Cantelli lemma is useful because a union bound turns the tail of a convergent series into a bound on all later occurrences.

Theorem: If \(\sum_{n=1}^{\infty}\mathbb P(A_n)<\infty\), then \(\mathbb P(A_n\text{ infinitely often})=0\).

Proof. For every \(m\in\mathbb N\), the event that \(A_n\) occurs for some \(n\ge m\) is \(\bigcup_{n=m}^{\infty}A_n\). By countable subadditivity, \[ \mathbb P\left(\bigcup_{n=m}^{\infty}A_n\right) \le \sum_{n=m}^{\infty}\mathbb P(A_n). \] The event that \(A_n\) occurs infinitely often is contained in \(\bigcup_{n=m}^{\infty}A_n\) for every \(m\). Therefore its probability is at most \(\sum_{n=m}^{\infty}\mathbb P(A_n)\) for every \(m\). Since the series converges, these tails tend to zero as \(m\) tends to infinity. The probability of infinitely many occurrences is consequently zero. \(\square\)

Worked Example: Summable Threshold Events

Let \(U_1,U_2,\ldots\) be independent uniform random variables on \([0,1]\), and define \(A_n=\{U_n\le 1/n^2\}\). Uniformity gives \(\mathbb P(A_n)=1/n^2\), so \[ \sum_{n=1}^{\infty}\mathbb P(A_n) =\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty. \] The first Borel–Cantelli lemma implies that \(A_n\) occurs only finitely often almost surely. Independence happens to hold here, but it is not needed for this conclusion: the finite sum alone is sufficient.

Why Independence Helps When the Sum Diverges

When the probability sum diverges, the first lemma gives no conclusion. For mutually independent events, however, the probability of avoiding every event in a finite block is a product. That product becomes small when the sum of the probabilities in the block becomes large.

Theorem: If \((A_n)_{n\in\mathbb N}\) are mutually independent and \(\sum_{n=1}^{\infty}\mathbb P(A_n)=\infty\), then \(\mathbb P(A_n\text{ infinitely often})=1\).

Proof. Write \(p_n=\mathbb P(A_n)\). For fixed \(m\le N\), independence gives \[ \mathbb P\left(\bigcap_{n=m}^{N}A_n^c\right) =\prod_{n=m}^{N}(1-p_n) \le \exp\left(-\sum_{n=m}^{N}p_n\right), \] using \(1-x\le e^{-x}\) for \(x\in[0,1]\). Since \(\sum_n p_n=\infty\), the right-hand side tends to zero as \(N\) tends to infinity. By continuity of probability for decreasing events, \[ \mathbb P\left(\bigcap_{n=m}^{\infty}A_n^c\right)=0. \] Thus, for each fixed \(m\), the probability that no \(A_n\) occurs from index \(m\) onward is zero. The event that only finitely many \(A_n\) occur is the countable union, over \(m\), of these events. Its probability is zero. Hence \(A_n\) occurs infinitely often with probability one. \(\square\)

Worked Example: Independent Events with a Divergent Sum

Let \(A_n\) be mutually independent events with \(\mathbb P(A_n)=1/n\). The first probability is \(1\), which causes no difficulty. The harmonic series diverges: \[ \sum_{n=1}^{\infty}\mathbb P(A_n) =\sum_{n=1}^{\infty}\frac1n=\infty. \] The independent-events Borel–Cantelli lemma therefore gives \(\mathbb P(A_n\text{ infinitely often})=1\). The independence assumption is essential to this application; the divergence calculation by itself is not enough.

Recognizing When a Zero–One Law Applies

For an independent sequence, the tail sigma-algebra contains events whose membership is determined by coordinates arbitrarily far out in the sequence. A practical way to check tail measurability is to ask whether changing finitely many initial coordinates can change whether the event occurs.

For example, convergence of a sequence is unchanged by altering or deleting finitely many terms. More precisely, for real-valued random variables, the event that \((X_n)\) converges can be written as the Cauchy event \[ \bigcap_{r=1}^{\infty}\ \bigcup_{N=1}^{\infty}\ \bigcap_{j,k\ge N} \{|X_j-X_k|<1/r\}. \] For every fixed \(m\), this event is also expressible using only indices greater than \(m\): removing the first \(m\) terms does not change whether the sequence is Cauchy. Thus the convergence event belongs to \(\mathcal T\). If the \(X_n\) are independent, Kolmogorov’s zero–one law says its probability is zero or one.

Worked Example: A Convergence Event for Independent Bernoulli Variables

Let \(X_1,X_2,\ldots\) be independent Bernoulli random variables with \(\mathbb P(X_n=0)=\mathbb P(X_n=1)=1/2\). The event \(C=\{(X_n)\text{ converges}\}\) is a tail event, so the zero–one law gives \(\mathbb P(C)\in\{0,1\}\).

To determine which value holds, apply the independent-events Borel–Cantelli lemma separately to \(A_n=\{X_n=0\}\) and \(B_n=\{X_n=1\}\). Each sequence consists of independent events with probability \(1/2\), and both probability sums diverge. Almost surely, zeros occur infinitely often and ones occur infinitely often. A sequence with infinitely many terms equal to zero and infinitely many terms equal to one cannot converge: if it converged, its terms would eventually be within \(1/3\) of its limit, but no real number is within \(1/3\) of both zero and one. Hence \(\mathbb P(C)=0\).

The zero–one law identifies the available dichotomy; the occurrence argument supplies the value. Here the Borel–Cantelli argument determines the value directly as well, illustrating that a zero–one law can organize the reasoning without replacing the need for evidence about which alternative holds.

A Practical Decision Process

Write the target event precisely.
Decide whether the claim concerns finitely often, infinitely often, convergence, or merely a zero–one probability.

Check for a summable bound.
If \(\sum_n\mathbb P(A_n)<\infty\), the first Borel–Cantelli lemma proves only finitely many occurrences almost surely; no independence is needed.

If the sum diverges, check independence.
For mutually independent events, the second Borel–Cantelli lemma proves infinitely many occurrences almost surely. Without its independence hypothesis, this conclusion does not follow from divergence alone.

Check whether the target is a tail event.
If it is measurable with respect to the tail sigma-algebra of an independent sequence, a zero–one law restricts its probability to zero or one.

Supply the missing direction.
A zero–one conclusion does not choose between zero and one. Establish a positive-probability lower bound for the value one, or rule out probability one to obtain zero.

Common Pitfalls

The most frequent error is treating a divergent sum as if it automatically implied infinitely many occurrences. The independent-events Borel–Cantelli lemma needs mutual independence; without it, events can overlap so heavily that a divergent sum still fails to force infinitely many distinct occurrences.

A second error is using a zero–one law as though it proved an event has probability one. It proves only that the probability belongs to \(\{0,1\}\). A separate bound or argument must distinguish the alternatives.

Finally, the summable and divergent cases point in different directions. A finite sum gives an almost-sure “finitely often” conclusion without independence. A divergent sum gives an “infinitely often” conclusion only under the independent-events hypothesis. Matching the hypothesis to the intended conclusion is the central decision.

Check Your Understanding

For each question, identify the relevant hypothesis and the exact conclusion it supports.

If \(\sum_n\mathbb P(A_n)<\infty\), which Borel–Cantelli conclusion follows, and is independence required?
If \(\sum_n\mathbb P(A_n)=\infty\), what additional condition allows the independent-events Borel–Cantelli lemma to apply?
What does Kolmogorov’s zero–one law say about a tail event, and what does it leave undecided?
Why is convergence of a sequence unchanged by changing finitely many initial terms?
For independent Bernoulli variables with success probability \(1/2\), which argument proves that both values occur infinitely often almost surely?