Borel–Cantelli lemma

Template:Short description Template:More footnotes 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.<ref>E. Borel, "Les probabilités dénombrables et leurs applications arithmetiques" Rend. Circ. Mat. Palermo (2) 27 (1909) pp. 247–271.</ref><ref>F.P. Cantelli, "Sulla probabilità come limite della frequenza", Atti Accad. Naz. Lincei 26:1 (1917) pp.39–45.</ref> 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 E₁, E₂, ... be a sequence of events in some probability space. The Borel–Cantelli lemma states:<ref>Template:Cite book</ref><ref name=":0">Template:Cite book</ref>

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, <math display="block">\limsup_{n\to\infty} E_n = \bigcap_{n=1}^\infty \bigcup_{k = n}^\infty E_k.</math>The set lim sup E_n is sometimes denoted {E_n i.o.}, 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/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 ΣPr(X_n = 0) converges to Template:Nowrap 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 <math display="inline">\left\{\bigcup_{n=N}^\infty E_n\right\}^\infty_{N=1}</math> is non-increasing: <math display="block">\bigcup_{n=1}^\infty E_n \supseteq \bigcup_{n=2}^\infty E_n \supseteq \cdots \supseteq \bigcup_{n=N}^\infty E_n \supseteq \bigcup_{n=N+1}^\infty E_n \supseteq \cdots \supseteq \limsup_{n\to\infty} E_n.</math>By continuity from above, <math display="block">\Pr(\limsup_{n \to \infty} E_n) = \lim_{N\to\infty}\Pr\left(\bigcup_{n=N}^\infty E_n\right).</math>By subadditivity, <math display="block">\Pr\left(\bigcup_{n=N}^\infty E_n\right) \leq \sum^\infty_{n=N} \Pr(E_n).</math>By original assumption, <math display="inline">\sum_{n=1}^\infty \Pr(E_n)<\infty.</math> As the series <math display="inline"> \sum_{n=1}^\infty \Pr(E_n)</math> converges, <math display="block">\lim_{N\to\infty} \sum^\infty_{n=N} \Pr(E_n)=0,</math> as required.<ref name="math.ucdavis.edu">Template:Cite web</ref>

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:<ref name=":0" />

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 Rⁿ. Specifically Template:Harv, if E_j is a collection of Lebesgue measurable subsets of a compact set in Rⁿ such that <math display="block">\sum_j \mu(E_j) = \infty,</math> then there is a sequence F_j of translates <math display="block">F_j = E_j + x_j </math> such that <math display="block">\lim\sup F_j = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty F_k = \mathbb{R}^n</math> apart from a set of measure zero.

Proof

Suppose that <math display="inline">\sum_{n = 1}^\infty \Pr(E_n) = \infty</math> and the events <math>(E_n)^\infty_{n = 1}</math> 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 <math display="block"> 1-\Pr(\limsup_{n \to \infty} E_n) = 0. </math>

Noting that: <math display="block">\begin{align} 1 - \Pr(\limsup_{n \to \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 \to \infty}E_n^{c}\right)= \lim_{N \to \infty}\Pr\left(\bigcap_{n=N}^\infty E_n^c \right), \end{align} </math> it is enough to show: <math display="inline">\Pr\left(\bigcap_{n=N}^{\infty} E_n^{c}\right) = 0</math>. Since the <math>(E_n)^{\infty}_{n = 1}</math> are independent: <math display="block">\begin{align} \Pr\left(\bigcap_{n=N}^\infty E_n^c\right) &= \prod^\infty_{n=N} \Pr(E_n^c) \\ &= \prod^\infty_{n=N} (1-\Pr(E_n)). \end{align} </math> The convergence test for infinite products guarantees that the product above is 0, if <math display="inline">\sum_{n = N}^\infty \Pr(E_n)</math> 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 <math>(A_n)</math> is monotone increasing for sufficiently large indices. This Lemma says:

Let <math>(A_n)</math> be such that <math>A_k \subseteq A_{k+1}</math>, and let <math>\bar A</math> denote the complement of <math>A</math>. Then the probability of infinitely many <math>A_k</math> occur (that is, at least one <math>A_k</math> occurs) is one if and only if there exists a strictly increasing sequence of positive integers <math>( t_k)</math> such that <math display="block"> \sum_k \Pr( A_{t_{k+1}} \mid \bar A_{t_k}) = \infty. </math>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 <math>(t_k)</math> usually being the essence.

Kochen–Stone

Let <math>(A_n)</math> be a sequence of events with <math display="inline">\sum\Pr(A_n)=\infty</math> and <math display="inline"> \limsup_{k\to\infty} \frac{\left(\sum_{n=1}^k\Pr(A_n)\right)^2} {\sum_{1\le m,n \le k} \Pr(A_m\cap A_n)} > 0.</math> Then there is a positive probability that <math>A_n</math> occur infinitely often.

Proof

Let <math> S_{m,n} = \sum^n_{i=m} \mathbf{1}_{A_i}</math>. Then, note that <math display="block"> E[S_{m,n}]^2 = \left(\sum^n_{i=m} \Pr(A_i)\right)^2 </math> and <math display="block"> E[S_{m,n}^2] = \sum_{1\le i \le j\le n} \Pr(A_i\cap A_j). </math> Hence, we know that <math> \limsup_{n\to\infty} \frac{\mathbb{E}[S_{1,n}]^2}{\mathbb{E}[S_{1,n}^2]} > 0.</math> We have that <math display="block> \Pr\left(\bigcup^n_{i=m} A_i\right) = \Pr(S_{m,n} > 0). </math> Now, notice that by the Cauchy-Schwarz Inequality, <math display="block"> \mathbb{E}[X]^2 \le \mathbb{E}[X\mathbf{1}_{\{X>0\}}]^2 \le \mathbb{E}[X^2]\Pr(X>0), </math> therefore, <math display="block"> \Pr(S_{m,n} > 0) \ge \frac{\mathbb{E}[S_{m,n}]^2}{\mathbb{E}[S_{m,n}^2]}. </math> We then have <math display="block"> \frac{\mathbb{E}[S_{m,n}]^2}{\mathbb{E}[S_{m,n}^2]} \ge \frac{E[S_{1,n} - S_{1,m-1}]^2}{E[S_{1,n}^2]}. </math> Given <math> m </math>, since <math> \lim_{n\to\infty} \mathbb{E}[S_{1,n}] = \infty </math>, we can find <math> n </math> large enough so that <math display="block"> \biggr|\frac{\mathbb{E}[S_{1,n}]-\mathbb{E}[S_{1,m-1}]}{\mathbb{E}[S_{1,n}]} - 1\biggr| < \epsilon, </math> for any given <math> \epsilon > 0 </math>. Therefore, <math display="block"> \lim_{m\to\infty}\sup_{n\ge m}\Pr\left(\bigcup_{i=m}^n A_i\right) \ge \lim_{m\to\infty}\sup_{n\ge m}\frac{E[S_{1,n}]^2}{E[S_{1,n}^2]} > 0. </math> But the left side is precisely the probability that the <math> A_n </math> occur infinitely often since <math> \{A_k \text{ i.o.}\} = \{\omega\in\Omega : \forall m, \exists n\ge m \text{ s.t. } \omega\in A_n\}. </math> We're done now, since we've shown that <math> P(A_k \text{ i.o.}) > 0.</math>

See also

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

References

• Template:Springer
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• 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