Proof: Borel-Cantelli Lemma 2 (Independence, Divergent Sum)

Q: Can $ P(A_n) $ be very small, say $ 1/n $, and still satisfy the conditions?

Yes, absolutely. If $ A_n $ are independent events with $ P(A_n) = 1/n $, then $ \sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} 1/n = \infty $ (the harmonic series diverges). By BC2, $ P(A_n \text{ i.o.}) = 1 $. This highlights that even individually rare events can happen infinitely often if their cumulative potential is infinite due to independence.

Q: What does "infinitely often" (i.o.) mean mathematically?

An event $ A $ occurs "infinitely often" if it belongs to the set $ \limsup_{n \to \infty} A_n = \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} A_n $. This definition implies that for any arbitrarily large integer $ N $, there exists some $ n \ge N $ such that $ A_n $ occurs. It does not mean $ A_n $ occurs for all $ n $ beyond some point.

Q: What happens if $ P(A_n) $ does not tend to zero?

If $ P(A_n) $ does not tend to zero, then the series $ \sum P(A_n) $ must diverge. In this case, if the events are also independent, the second Borel-Cantelli Lemma still applies, and $ P(A_n \text{ i.o.}) = 1 $. However, the interesting and less intuitive cases for BC2 are when $ P(A_n) \to 0 $ but the sum still diverges (e.g., $ 1/n $).

Master the Borel-Cantelli Lemma 2, a cornerstone of advanced probability. Understand why independence and divergent sums lead to almost certain infinite occurrences.

The Formal Theorem

Let

(\Omega, \mathcal{F}, P)

be a probability space and let

\{A_n\}_{n=1}^{\infty}

be a sequence of independent events in

\mathcal{F}

. If the sum of their probabilities diverges, then the probability that infinitely many of these events occur is 1. That is, if

\sum_{n=1}^{\infty} P(A_n) = \infty

then

P\left(\limsup_{n \to \infty} A_n\right) = 1

Analytical Intuition.

Imagine an endless cosmic casino, where each spin

A_n

represents an independent 'win' event. The first Borel-Cantelli Lemma says if the total 'expected wins' (sum of probabilities) is finite, then eventually, the wins will almost surely stop. This second lemma, however, speaks to a different kind of destiny. If the 'expected wins' diverge (

\sum P(A_n) = \infty

), and each spin is truly independent, then it's not just possible, but *certain*, that you will win infinitely often. It's like playing a game where the odds of winning any specific round

P(A_n)

might be infinitesimally small, but the sheer, relentless number of independent rounds guarantees that victory will repeatedly find you, an endless cosmic jackpot.

CAUTION

Institutional Warning.

Students frequently overlook the *independence* condition, trying to apply Lemma 2 where events are dependent. They might also confuse "infinitely often" with "all the time" or misinterpret the implication of a divergent sum of small probabilities.

Institutional Deep Dive.

Core Logic: The second Borel-Cantelli Lemma, often referred to as the converse Borel-Cantelli for independent events, posits a profoundly powerful truth: if a sequence of independent events

A_n

has a divergent sum of probabilities (i.e.,

\sum_{n=1}^{\infty} P(A_n) = \infty

), then it is almost certain that infinitely many of these events will occur. The 'almost certain' qualifier is crucial, meaning the probability is exactly 1. This isn't about individual events having high probabilities; indeed, they could all be very small, for instance,

P(A_n) = 1/n

. What matters is the cumulative potential for occurrence across an infinite sequence. If events were dependent, this conclusion wouldn't necessarily hold (e.g., if one event's occurrence made subsequent events impossible). But independence is the critical catalyst here. It prevents any single event or finite collection of events from 'using up' the probability space in a way that would preclude future occurrences. The core logic hinges on demonstrating that the probability of *not* occurring infinitely often (i.e., eventually never occurring again) must be zero. This is meticulously done by considering the complement event and ingeniously using the independence assumption to transform products of probabilities into exponential forms, which then demonstrably tend to zero.

Geometric Mechanics: Picture a conceptual timeline stretching infinitely into the future. Each point

n

on this timeline represents an independent trial, and an event

A_n

occurs with probability

P(A_n)

. If

\sum P(A_n) = \infty

, one can conceptualize this as an 'infinite potential' or 'infinite reservoir' for something to happen. The independence condition signifies that the occurrence (or non-occurrence) of any

A_n

has absolutely no bearing on the likelihood of

A_{n+1}

or any subsequent event. To prove that

A_n

happens infinitely often with probability 1, we adopt a common probabilistic strategy: prove the probability of its complement is 0. The complementary scenario is that

A_n

happens only a finite number of times. This implies that there must exist some point

N

on our timeline after which none of the events

A_N, A_{N+1}, \dots

ever occur again. Mathematically, this means that for all

n \ge N

, the complement event

A_n^c

occurs. The probability of this 'eventual quietude' for a finite block of trials from

N

M

P(A_N^c \cap A_{N+1}^c \cap \dots \cap A_M^c)

. Crucially, due to independence, this intersection's probability simplifies to a product of individual probabilities:

\prod_{n=N}^{M} P(A_n^c) = \prod_{n=N}^{M} (1 - P(A_n))

. A fundamental inequality states

1-x \le e^{-x}

for

x \ge 0

. Applying this inequality, the product becomes

\le \prod_{n=N}^{M} e^{-P(A_n)} = e^{-\sum_{n=N}^{M} P(A_n)}

. As

M \to \infty

, our initial condition

\sum_{n=1}^{\infty} P(A_n) = \infty

implies that the tail sum

\sum_{n=N}^{\infty} P(A_n) = \infty

for any chosen

N

. Consequently, the exponent

-\sum_{n=N}^{M} P(A_n)

tends to

-\infty

M \to \infty

, causing

e^{-\sum_{n=N}^{M} P(A_n)}

to tend to 0. This profound result means the probability of *never* seeing another event

A_n

after any given

N

is 0. Since this holds for *any* starting point

N

, the probability of the union of all such 'eventual quietude' scenarios (i.e., the probability of stopping at some finite point) is also 0. Thus, the probability of occurring infinitely often must be 1. It's a delicate probabilistic argument about the 'thinness' of the complement set collapsing to measure zero.

Institutional Pitfalls: A pervasive pitfall for students is to neglect the crucial independence assumption. While the first Borel-Cantelli Lemma holds universally, regardless of event dependencies, the second Borel-Cantelli Lemma *absolutely requires* independence. Without this condition, the factorization of intersection probabilities (i.e.,

P(A_N^c \cap \dots \cap A_M^c) = \prod_{n=N}^{M} P(A_n^c)

) is invalid, rendering the entire proof strategy based on the exponential bound moot. Consider a simple counterexample: if

A_n

is defined as 'a fair coin lands heads on the first toss' for all

n

, then

P(A_n) = 1/2

for all

n

. Here,

\sum P(A_n) = \sum 1/2 = \infty

. However,

P(A_n \text{ i.o.}) = P(A_1) = 1/2

, not 1, because the events

A_n

are not independent (they are all identical). Another common misinterpretation is confusing 'infinitely often' with 'every time'. 'Infinitely often' does not imply that

A_n

must occur for all large

n

, but rather that there is no final

n

after which

A_n

stops occurring. It guarantees an infinite subsequence of occurrences. Finally, students sometimes conflate a divergent sum of probabilities with individual probabilities not tending to zero. While

P(A_n) \to 0

is a necessary condition for the sum to converge (otherwise the series wouldn't pass the term test), it is not sufficient for convergence, and the sum can diverge even if

P(A_n) \to 0

(e.g.,

P(A_n) = 1/n

). The proof's elegance and rigor rely fundamentally on the exponential bound

1-x \le e^{-x}

, a potent tool in advanced probability theory that requires precise application.

Academic Inquiries.

Why is independence crucial for the second Borel-Cantelli Lemma but not the first?

For BC2, independence allows us to factorize probabilities of intersections of complementary events (e.g., $P(A_N^c \cap \dots \cap A_M^c) = \prod_{n=N}^{M} P(A_n^c)$ ), which is vital for the exponential bound proof. BC1 only uses countable subadditivity, which holds generally for any sequence of events.

Can $P(A_n)$ be very small, say $1/n$ , and still satisfy the conditions?

Yes, absolutely. If $A_n$ are independent events with $P(A_n) = 1/n$ , then $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} 1/n = \infty$ (the harmonic series diverges). By BC2, $P(A_n \text{ i.o.}) = 1$ . This highlights that even individually rare events can happen infinitely often if their cumulative potential is infinite due to independence.

What does "infinitely often" (i.o.) mean mathematically?

An event $A$ occurs "infinitely often" if it belongs to the set $\limsup_{n \to \infty} A_n = \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} A_n$ . This definition implies that for any arbitrarily large integer $N$ , there exists some $n \ge N$ such that $A_n$ occurs. It does not mean $A_n$ occurs for all $n$ beyond some point.

What happens if $P(A_n)$ does not tend to zero?

If $P(A_n)$ does not tend to zero, then the series $\sum P(A_n)$ must diverge. In this case, if the events are also independent, the second Borel-Cantelli Lemma still applies, and $P(A_n \text{ i.o.}) = 1$ . However, the interesting and less intuitive cases for BC2 are when $P(A_n) \to 0$ but the sum still diverges (e.g., $1/n$ ).

Standardized References.

Definitive Institutional SourceDurrett, Richard. Probability: Theory and Examples. 5th ed. Cambridge University Press, 2019.

Advanced

Borel-Cantelli

Borel-Cantelli — Advanced Advanced Probability Theory proof with visual geometric intuition and formal theorem statement. Free at NICEFA.

Advanced

Martingale Convergence

Master Martingale Convergence: explore Doob's Theorem, its L1-bounded conditions, and profound implications for random processes in advanced probability.

Advanced

Ergodic Theorem

Master the Ergodic Theorem in Advanced Probability Theory. Understand how time averages converge to space averages in measure-preserving, ergodic systems.

Institutional Citation

Reference this proof in your academic research or publications.

NICEFA Visual Mathematics. (2026). Proof: Borel-Cantelli Lemma 2 (Independence, Divergent Sum): Visual Proof & Intuition. Retrieved from https://www.nicefa.org/library/advanced-probability-theory/borel-cantelli-lemma-2-independence

Dominate the Logic.

"Abstract theory is just a movement we haven't seen yet."

Master the Proof Early Access

The Formal Theorem

Analytical Intuition.

Institutional Warning.

Institutional Deep Dive.

Academic Inquiries.

Why is independence crucial for the second Borel-Cantelli Lemma but not the first?

Can P(An) P(A_n) P(An​) be very small, say 1/n 1/n 1/n, and still satisfy the conditions?

What does "infinitely often" (i.o.) mean mathematically?

What happens if P(An) P(A_n) P(An​) does not tend to zero?

Standardized References.

Related Proofs Cluster.

Borel-Cantelli

Martingale Convergence

Ergodic Theorem

Institutional Citation

Dominate the Logic.

Can $P(A_n)$ be very small, say $1/n$ , and still satisfy the conditions?

What happens if $P(A_n)$ does not tend to zero?