Borel-Cantelli lemmas

The Borel-Cantelli lemmas turn a series of event probabilities into a statement about an entire infinite sequence of observations. A summable sequence of probabilities forces only finitely many occurrences almost surely, regardless of independence. A divergent series forces infinitely many occurrences when the events are mutually independent events. The precise event being measured is the limsup of events.

Imagine alarms with probabilities $p_n=1/(n+1{)}^2$ or $p_n=1/(n+1)$, starting at $n=1$. Both individual probabilities tend to zero. Nevertheless, the first alarm sequence eventually stops almost surely, whereas the second keeps returning almost surely if its events are mutually independent events. Reusing one uniform random variable for every alarm instead produces the shared-uniform counterexample to the second Borel-Cantelli lemma. Thus the individual probabilities alone do not determine long-run recurrence.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.

For independent standard normal random variables $X_n$, the displayed limit superior equals two almost surely. The Mills ratio asymptotic gives, for $a>0$,The series converges for $a>2$ and diverges for $0<a\le 2$. The two Borel-Cantelli lemmas, applied at countably many thresholds approaching two, give the result. In particular, the value two is also a threshold exceeded infinitely often, despite being the almost-sure limit superior.

For arbitrary events $A_n$ in one probability space,Equivalently, only finitely many events occur almost surely; no independence is needed.^[ref]

Proof outline: put $B_m={\bigcup }_{n\ge m}{A}_n$. The union bound givesThe tail of a convergent nonnegative series tends to zero.

If the events $A_n$ are mutually independent events and ${\sum }_{n=1}^{\infty }\mathbb{P}(A_n)=\infty$, then $\mathbb{P}({\mathrm{limsup}}_n{A}_n)=1$. For an infinite family, mutual independence means every finite subfamily satisfies the factorization defining independent events.^[ref]

Proof outline: for $M\ge m$, independent events and $1-x\le e^{-x}$ giveContinuity of probability makes each infinite tail union have probability one; their countable intersection has probability one as well.

Choose one random variable $U$ with the continuous uniform distribution on $[0,1]$, and reuse it in every event:Each probability is $1/(n+1)$, so their harmonic series diverges, just as in summable and harmonic event probabilities. But the events are decreasing: $A_{n+1}\subseteq A_n$. For every outcome $U=u>0$, all indices with $n+1>1/u$ fail. Only finitely many events occur on that outcome. The limsup of events is exactly $\{U=0\}$, a zero-probability event. This is a concrete closed-interval version of the shrinking-interval dependence example.^[ref]

The missing assumption is visible already in two events. For $i<j$, their intersection is $A_j$, henceMoreover, the whole tail union is just $A_m$, with probability $1/(m+1)\to 0$. This calculation explains directly why divergence alone fails: overlapping events keep spending their individual probabilities on the same outcomes.

Let $N={\sum }_{n\ge 1}{1}_{A_n}$ be the total number of occurrences, a sum of indicator random variables. Although $N<\infty$ almost surely, Tonelli theorem gives $\mathbb{E}[N]={\sum }_{n\ge 1}1/(n+1)=\infty$. More explicitly, $\mathbb{P}(N\ge k)=1/(k+1)$ for $k\ge 1$: very small positive values of $U$ give very long, but finite, initial runs. Infinite expected value does not imply a positive probability of an infinite count.

For a nonnegative random variable $X$, Tonelli theorem givesConsequently, for each $a>0$, $\mathbb{E}[X]=\infty$ exactly when ${\sum }_{n\ge 1}\mathbb{P}(X>an)=\infty$; compare the integral on successive intervals of length $a$.

For independent rate-one exponential variables $X_n$ and $M_n={\max }_{m\le n}{X}_m$,almost surely. The first follows by applying both Borel--Cantelli lemmas to thresholds $(1\pm \epsilon )\log n$; the second uses
$\mathbb{P}(M_n\le (1-\epsilon )\log n)\le e^{-n^{\epsilon }}$.

For events $B_n\in {\mathcal{F}}_n$, put $p_n=\mathbb{P}(B_n\mid {\mathcal{F}}_{n-1})$. Then ${\sum }_n{1}_{B_n}<\infty$ almost surely on $\{{\sum }_n{p}_n<\infty \}$, while ${\sum }_n{1}_{B_n}=\infty$ almost surely on $\{{\sum }_n{p}_n=\infty \}$. The proof applies the martingale convergence theorem to compensated sums ${\sum }_{j\le n}(1_{B_j}-p_j)$ and uses Kronecker lemma in the divergent case.