Can someone explain the Borel-Cantelli Lemma?

Question

I’m looking for an informal and intuitive explanation of the Borel-Cantelli Lemma. The symbolic version can be found here.

What is confusing me is what ‘probability of the limit superior equals $ 0 $’ means.

Thanks!

Answer 1

Let $\{E_n\}$ be a sequence of events. Each event $E_i$ is a collection of outcomes. The limit superior of the collection $\{E_n\}$ is the collection of all those outcomes that appear in infinitely many events. The Borel Cantelli Lemma says that if the sum of the probabilities of the $\{E_n\}$ are finite, then the collection of outcomes that occur infinitely often must have probability zero.

To give an example, suppose I randomly pick a real number $x \in [0,1]$ using an arbitrary probability measure $\mu$. I then challenge my (infinitely many) friends to guess a subset $E_n$ of $[0,1]$ containing $x$. If infinitely many friends guess correctly, I have to buy pizza for everyone. To make sure they don't just guess the whole interval, I make them pay for choosing large subsets -- the cost of choosing a set $E$ is $\mu(E)$. I check my bank account at the end of the guessing and find that I have only made a finite amount of money, i.e. $\sum \mu(E_n) < \infty$ (cheap friends!). Is it likely I will have to buy everyone pizza? No! Because the set $E_{\infty}$ of numbers that infinitely many friends agreed on has measure zero, so my random number $x$ has probability zero of being in there.

Answer 2

Here is a proof I wrote up for sci.math, but never posted:

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Borel-Cantelli: Suppose that $\sum_{i=1}^\infty P(A_i)$ is finite, then the probability that infinitely many of the $A_i$ occur is $0$.

Proof: Let $B_k$ be the event $\cup_{i=k}^\infty A_i$ for $k=1,2,\ldots,$. If $x$ is in the event $A_i$'s i.o., then $x\in B_k$ for all $k$. So $x\in \cap_{k=1}^\infty B_k$.

Conversely, if $x\in B_k$ for all $k$, then we can show that $x$ is in $A_i$'s i.o. Indeed, $x\in B_1 = \cup_{i=1}^\infty A_i$ means that $x\in A_{j(1)}$ for some $j(1)$. However $x\in B_{j(1)+1}$ implies that $x\in A_{j(2)}$ for some $j(2)$ that is strictly larger than $j(1)$. Thus we can produce an infinite sequence of integer $j(1)< j(2)< j(3)<\ldots$ such that $x\in A_{j(i)}$ for all $i$.

Let $E$ be the event $\{x:\, x\in A_i \mbox{ i.o.}\}$. We have $$ E = \bigcap_{k=1}^\infty \bigcup_{i=k}^\infty A_i\tag{1} $$ From $E\subseteq B_k$ for all $k$, it follows that $P(E)\leq P(B_k)$ for all $k$. By union bound, we know that $P(B_k)\leq \sum_{i=k}^\infty P(A_i)$. So $P(B_k)\rightarrow 0$, by the hypothesis that $\sum_{i=1}^\infty P(A_i)$ is finite. Therefore, $P(E)=0$.

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For events, $\bigcap\approx\inf$ and $\bigcup\approx\sup$. Since $$ \limsup_{n\to\infty}a_n=\operatorname*{\inf\vphantom{p}}_{n\ge1}\sup_{k\ge n}a_k\tag{2} $$ it makes sense to call $$ \limsup_{n\to\infty}A_n=\bigcap_{n\ge1}\bigcup_{k\ge n} A_k\tag{3} $$

Answer 3

In infinite probability spaces the probability of an event being $0$ does not mean it can't happen. This can be confusing, but such is life. Consider for instance flipping a fair coin infinitely many times. What is the probability that in all these flips you'll get heads? Answer: 0 (the probability of all flips heads must be less than the probability of first $n$ flips being heads, but that is equal precisely to $(1/2)^n$, so the probability of all heads is smaller then $(1/2)^n$ for all $n$, which means it's equal to $0$). It can of course happen, but with probability $0$.

In finite probability spaces an event having probability $0$ can safely be interpreted as "can't happen". But in infinite probability spaces you need to refine your intuition.

If I understand your question correctly, this is really what you are asking about and not so much about Borel-Cantelli. Just to make one more comment then, Borel-Cantelli says that if the total sum of the probabilities of events is finite (i.e., small) then the probability for such events occurring infinitely often is $0$. It can happen, but it's bloody unlikely. For some events to occur infinitely often it must be the case that the total sum of individual probabilities is large (i.e., infinity). The Wiki page you cited gives several examples. I hope this helps.

Answer 4

This is how I intuitively understand the lemma.

In general, if you have a series of events whose probabilities become arbitrarily small, as long as they are not zero, in many cases it may be possible to have a set of sub-events, whose collective probability is >0, that appear within the series of events at infinitely many steps. This is because their occurrences may be "spread out" across the infinite future, even if the probability within each individual event becomes small. As the probability of events become small, the sub-events can't all occur at the same step, but can collectively across different steps.

The Borel-Cantelli lemma describes a situation where the entire summed future probability of the sequence becomes arbitrarily small. In this case, even "spreading the probability" of sub-events infinitely into the future is not possible. So, any collection of sub-events with probability >0 cannot occur an infinite number of times.

Answer 5

Application of First Borel-Cantelli Lemma

let's consider a scenario where we flip a fair six-sided die infinitely many times.

Let's define an event $A_n$ as the event that the die shows up 6 on the $n$-th flip, but only if $n$ is a perfect square (i.e., $n = 1, 4, 9, 16, 25, 36, \ldots$). For all other $n$, we'll say that $A_n$ cannot occur.

Since the die is fair, the probability of getting a 6 on any given flip is $\frac{1}{6}$. Therefore, the probability of each event $A_n$ is $P(A_n) = \frac{1}{6}$ if $n$ is a perfect square, and $P(A_n) = 0$ otherwise.

Now, let's consider the sum of the probabilities of the events $A_n$. Since there are infinitely many natural numbers but only finitely many perfect squares less than or equal to any given number, the sum of the probabilities is a finite number. Specifically, it's $\sum_{i=1}^\infty P(A_{i^2}) = \sum_{i=1}^\infty \frac{1}{6} = \lim_{n \to \infty}n \cdot \frac{1}{6} =$ finite number.

According to the first Borel-Cantelli lemma, if the sum of the probabilities of the events $A_n$ is finite, then the probability that infinitely many of the events $A_n$ occur is 0. In this case, since the sum of the probabilities is finite, we can conclude that the probability that the die shows up 6 on an infinite number of perfect square flips is 0.

So, in this scenario, the first Borel-Cantelli lemma allows us to conclude that, almost surely, only a finite number of the events $A_n$ will occur.

Application of second Borel-Cantelli lemma

In the context of flipping a fair die infinitely many times, the event $A_1$ represents the die showing up 1 on a flip. If we denote by $A_i$ the event that the die shows up 1 on the $i$-th flip, then each $A_i$ is an independent event with probability $\frac{1}{6}$.

The event $\limsup A_1$ represents the event that $A_1$ occurs infinitely often, i.e., the die shows up 1 infinitely many times in the infinite sequence of flips.

According to the first Borel-Cantelli lemma, if the sum of the probabilities of the events $A_i$ is finite, then the probability that infinitely many of the events $A_i$ occur is 0. In this case, since each $A_i$ has probability $\frac{1}{6}$, the sum of the probabilities is $\sum_{i=1}^\infty P(A_i) = \sum_{i=1}^\infty \frac{1}{6} = \infty$. Therefore, the first Borel-Cantelli lemma does not apply here.

However, the second Borel-Cantelli lemma states that if the events $A_i$ are independent and the sum of their probabilities is infinite, then the probability that infinitely many of the events $A_i$ occur is 1. In this case, since the events $A_i$ are independent and the sum of their probabilities is infinite, the second Borel-Cantelli lemma applies, and we can say that the probability that the die shows up 1 infinitely many times is 1, i.e., $P(\limsup A_1) = 1$.

So, if we flip a fair die infinitely many times, the probability that the die shows up 1 infinitely many times is 1, not 0.