Borel–Cantelli Lemma

In probability theory, the Borel–Cantelli lemma is a theorem about sequences of events. In general, it is a result in measure theory. It is named after Émile Borel and Francesco Paolo Cantelli, who gave statement to the lemma in the first decades of the 20th century. A related result, sometimes called the second Borel–Cantelli lemma, is a partial converse of the first Borel–Cantelli lemma. The lemma states that, under certain conditions, an event will have probability of either zero or one. Accordingly, it is the best-known of a class of similar theorems, known as zero-one laws. Other examples include Kolmogorov's zero–one law and the Hewitt–Savage zero–one law.

Statement of lemma for probability spaces

Let ''E''₁, ''E''₂, ... be a sequence of events in some probability space.
The Borel–Cantelli lemma states:

Here, "lim sup" denotes limit supremum of the sequence of events. That is, lim sup ''E''_''n'' is the outcome that infinitely many of the infinite sequence of events (''E''_''n'') actually occur. Explicitly,
$\limsup_ E_n = \bigcap_^\infty \bigcup_^\infty E_k.$The set lim sup ''E''_''n'' is sometimes denoted , where "i.o." stands for "infinitely often". The theorem therefore asserts that if the sum of the probabilities of the events ''E''_''n'' is finite, then the set of all outcomes that contain infinitely many events must have probability zero. Note that no assumption of independence is required.

Example

Suppose (''X''_''n'') is a sequence of random variables with Pr(''X''_''n'' = 0) = 1/''n''² for each ''n''. The probability that ''X''_''n'' = 0 occurs for infinitely many ''n'' is equivalent to the probability of the intersection of infinitely many 'X''_''n'' = 0events. The intersection of infinitely many such events is a set of outcomes common to all of them. However, the sum ΣPr(''X''_''n'' = 0) converges to and so the Borel–Cantelli Lemma states that the set of outcomes that are common to infinitely many such events occurs with probability zero. Hence, the probability of ''X''_''n'' = 0 occurring for infinitely many ''n'' is 0. Almost surely (i.e., with probability 1), ''X''_''n'' is nonzero for all but finitely many ''n''.

Proof

Let (''E''_''n'') be a sequence of events in some probability space.

The sequence of events $\left\^\infty_$ is non-increasing:
$\bigcup_^\infty E_n \supseteq \bigcup_^\infty E_n \supseteq \cdots \supseteq \bigcup_^\infty E_n \supseteq \bigcup_^\infty E_n \supseteq \cdots \supseteq \limsup_ E_n.$By continuity from above,
$\Pr(\limsup_ E_n) = \lim_\Pr\left(\bigcup_^\infty E_n\right).$By subadditivity,
$\Pr\left(\bigcup_^\infty E_n\right) \leq \sum^\infty_ \Pr(E_n).$By original assumption, $\sum_^\infty \Pr(E_n)<\infty.$ As the series $\sum_^\infty \Pr(E_n)$ converges,
$\lim_ \sum^\infty_ \Pr(E_n)=0,$
as required.

General measure spaces

For general measure spaces, the Borel–Cantelli lemma takes the following form:

Converse result

A related result, sometimes called the second Borel–Cantelli lemma, is a partial converse of the first Borel–Cantelli lemma. The lemma states: If the events ''E''_''n'' are independent and the sum of the probabilities of the ''E''_''n'' diverges to infinity, then the probability that infinitely many of them occur is 1. That is:

The assumption of independence can be weakened to pairwise independence, but in that case the proof is more difficult.

The infinite monkey theorem follows from this second lemma.

Example

The lemma can be applied to give a covering theorem in R^''n''. Specifically , if ''E''_''j'' is a collection of Lebesgue measurable subsets of a compact set in R^''n'' such that
$\sum_j \mu(E_j) = \infty,$
then there is a sequence ''F''_''j'' of translates
$F_j = E_j + x_j$
such that
$\lim\sup F_j = \bigcap_^\infty \bigcup_^\infty F_k = \mathbb^n$
apart from a set of measure zero.

Proof

Suppose that $\sum_^\infty \Pr(E_n) = \infty$ and the events $(E_n)^\infty_$ are independent. It is sufficient to show the event that the ''E''_''n'''s did not occur for infinitely many values of ''n'' has probability 0. This is just to say that it is sufficient to show that
$1-\Pr(\limsup_ E_n) = 0.$

Noting that:
$\begin 1 - \Pr(\limsup_ E_n) &= 1 - \Pr\left(\\right) = \Pr\left(\^c \right) \\ & = \Pr\left(\left(\bigcap_^\infty \bigcup_^\infty E_n\right)^c \right) = \Pr\left(\bigcup_^\infty \bigcap_^\infty E_n^c \right)\\ &= \Pr\left(\liminf_E_n^\right)= \lim_\Pr\left(\bigcap_^\infty E_n^c \right), \end$ it is enough to show: $\Pr\left(\bigcap_^ E_n^\right) = 0$. Since the $(E_n)^_$ are independent:
$\begin \Pr\left(\bigcap_^\infty E_n^c\right) &= \prod^\infty_ \Pr(E_n^c) \\ &= \prod^\infty_ (1-\Pr(E_n)). \end$
The convergence test for infinite products guarantees that the product above is 0, if $\sum_^\infty \Pr(E_n)$ diverges. This completes the proof.

Counterpart

Another related result is the so-called counterpart of the Borel–Cantelli lemma. It is a counterpart of the
Lemma in the sense that it gives a necessary and sufficient condition for the limsup to be 1 by replacing the independence assumption by the completely different assumption that $(A_n)$ is monotone increasing for sufficiently large indices. This Lemma says:

Let $(A_n)$ be such that $A_k \subseteq A_$,
and let $\bar A$ denote the complement of $A$. Then the probability of infinitely many $A_k$ occur (that is, at least one $A_k$ occurs) is one if and only if there exists a strictly increasing sequence of positive integers $( t_k)$ such that
$\sum_k \Pr( A_ \mid \bar A_) = \infty.$This simple result can be useful in problems such as for instance those involving hitting probabilities for stochastic process with the choice of the sequence $(t_k)$ usually being the essence.

Kochen–Stone

Let $(A_n)$ be a sequence of events with $\sum\Pr(A_n)=\infty$ and
$\limsup_ \frac > 0.$ Then there is a positive probability that $A_n$ occur infinitely often.

Proof

Let $S_ = \sum^n_ \mathbf_$. Then, note that

E _2 = \left(\sum^n_ \Pr(A_i)\right)^2

and

E _^2= \sum_ \Pr(A_i\cap A_j).

Hence, we know that
$\limsup_ \frac > 0.$
We have that

\mathbb 2 \le \mathbb \mathbf_2 \le \mathbb ^2Pr(X>0),

therefore,
$\Pr(S_ > 0) \ge \frac.$
We then have
$\frac \ge \frac.$
Given $m$, since \lim_ \mathbb _= \infty , we can find $n$ large enough so that
$\biggr, \frac - 1\biggr, < \epsilon,$
for any given $\epsilon > 0$. Therefore,
$\lim_\sup_\Pr\left(\bigcup_^n A_i\right) \ge \lim_\sup_\frac > 0.$
But the left side is precisely the probability that the $A_n$ occur infinitely often since
$\ = \.$
We're done now, since we've shown that $P(A_k \text) > 0.$

See also

* Lévy's zero–one law
* Kuratowski convergence
* Infinite monkey theorem

References

*
* .
* .
* .
* Durrett, Rick. "Probability: Theory and Examples." Duxbury advanced series, Third Edition, Thomson Brooks/Cole, 2005.

External links

Planet Math Proof
Refer for a simple proof of the Borel Cantelli Lemma

{{DEFAULTSORT:Borel-Cantelli lemma
Theorems in measure theory
Theorems in probability theory
Covering lemmas
Lemmas