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. A related result, sometimes called the second Borel–Cantelli lemma, is a partial converse of the first Borel–Cantelli lemma.

Statement of lemma for probability spaces

Let (E_n) be a sequence of events in some probability space. The Borel–Cantelli lemma states:

If the sum of the probabilities of the E_n is finite

$\sum_{n=1}^\infty \Pr(E_n)<\infty,$

then the probability that infinitely many of them occur is 0, that is,

$\Pr\left(\limsup_{n\to\infty} E_n\right) = 0.\,$

Here, "lim sup" denotes limit superior of the sequence of events, and each event is a set of outcomes. That is, lim sup E_n is the set of outcomes that occur infinitely many times within the infinite sequence of events (E_n). Explicitly,

$\limsup_{n\to\infty} E_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k.$

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 are "repeated" infinitely many times must occur with 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/n² 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 = 0] events. The intersection of infinitely many such events is a set of outcomes common to all of them. However, the sum ∑Pr(X_n = 0) converges to π²/6 ≈ 1.645 < ∞, 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 and suppose that the sum of the probabilities of the E_n is finite. That is suppose:

$\sum_{n=1}^\infty \Pr(E_n)<\infty.$

Note that the convergence of this sum implies:

$\inf_{N\geq 1} \sum_{n=N}^\infty \Pr(E_n) = 0. \,$

Therefore it follows that:

$\Pr\left(\limsup_{n\to\infty} E_n\right) = \Pr(E_n \text{ i.o.}) = \Pr\left(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n\right) \leq \inf_{N \geq 1} \Pr\left( \bigcup_{n=N}^\infty E_n\right) \leq \inf_{N\geq 1} \sum_{n=N}^\infty \Pr(E_n) = 0$

where the abbreviation "i.o." denotes "infinitely often."

General measure spaces

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

Let μ be a (positive) measure on a set X, with σ-algebra F, and let (A_n) be a sequence in F. If

$\sum_{n=1}^\infty\mu(A_n)<\infty,$

then

$\mu\left(\limsup_{n\to\infty} A_n\right) = 0.\,$

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:

If $\sum^{\infty}_{n = 1} \Pr(E_n) = \infty$ and the events $(E_n)^{\infty}_{n = 1}$ are independent, then $\Pr(\limsup_{n \rightarrow \infty} E_n) = 1.$

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

Example

The infinite monkey theorem is a special case of this lemma.

The lemma can be applied to give a covering theorem in Rⁿ. Specifically (Stein 1993, Lemma X.2.1), if E_j is a collection of Lebesgue measurable subsets of a compact set in Rⁿ 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

$\lim\sup F_j = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty F_k = \mathbb{R}^n$

apart from a set of measure zero.

Proof

Suppose that $\sum_{n = 1}^\infty \Pr(E_n) = \infty$ and the events $(E_n)^\infty_{n = 1}$ 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

$1-\Pr(\limsup_{n \rightarrow \infty} E_n) = 0. \,$

Noting that:

$\begin{align} 1 - \Pr(\limsup_{n \rightarrow \infty} E_n) &= 1 - \Pr\left(\{E_n\text{ i.o.}\}\right) = \Pr\left(\{E_n \text{ i.o.}\}^{c}\right) \\ & = \Pr\left(\left(\bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty}E_n\right)^{c}\right) = \Pr\left(\bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty}E_n^{c}\right)\\ &= \Pr\left(\liminf_{n \rightarrow \infty}E_n^{c}\right)= \lim_{N \rightarrow \infty}\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right) \end{align}$

it is enough to show: $\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right) = 0$. Since the $(E_n)^{\infty}_{n = 1}$ are independent:

$\begin{align} \Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right) &= \prod^{\infty}_{n=N}\Pr\left(E_n^{c}\right) \\ &= \prod^{\infty}_{n=N}\left(1-\Pr\left(E_n\right)\right) \\ &\leq \prod^{\infty}_{n=N}\left(1-\Pr(E_n)+\frac{(\Pr(E_n))^{2}}{2!}-\frac{(\Pr(E_n))^{3}}{3!}+\cdots\right) \\ & = \prod^{\infty}_{n=N}\left(\sum^{\infty}_{m=1}\frac{(-\Pr(E_n))^{m}}{m!}\right) \\ &=\prod^{\infty}_{n=N}\exp\left(-\Pr\left(E_n\right)\right)\\ &=\exp\left(-\sum^{\infty}_{n=N}\Pr(E_n)\right)\\ &= 0. \end{align}$

This completes the proof. Alternatively, we can see $\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right) = 0$ by taking negative the logarithm of both sides to get:

$\begin{align} -\log\left(\Pr\left(\bigcap_{n=N}^{\infty}E_n^{c}\right)\right) &= -\log\left(\prod^{\infty}_{n=N} (1-\Pr(E_n))\right) \\ &= - \sum^{\infty}_{n=N}\log(1-\Pr(E_n)) \end{align}$

Since −log(1 − x) ≥ x for all x > 0, the result similarly follows from our assumption that $\sum^\infty_{n = 1} \Pr(E_n) = \infty.$

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 $A_k \subseteq A_{k+1}$, 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

$\sum_{k} \Pr( A_{t_{k+1}}| \bar A_{t_k}) = \infty.$

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.

References

• Prokhorov, A.V. (2001), "Borel–Cantelli lemma", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1556080104, http://www.encyclopediaofmath.org/index.php?title=B/b017040 
• Feller, William (1961), An Introduction to Probability Theory and Its Application, John Wiley & Sons .
• Stein, Elias (1993), Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals, Princeton University Press .
• Bruss, F. Thomas (1980), "A counterpart of the Borel Cantelli Lemma", J. Appl. Prob. 17: 1094–1101 .
• 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