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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 occur with probability zero or with probability one. As such, it is the best-known of a class of similar theorems, known as zero-one laws. Other examples include the Kolmogorov 0-1 law and the Hewitt–Savage zero-one law.

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

$\sum _{n=1}^{\mathrm{\infty }}\mathrm{Pr}(E_n)<\mathrm{\infty },$

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

$\mathrm{Pr}\left({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n\right)=0.$

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,

${\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n=\bigcap _{n=1}^{\mathrm{\infty }}\bigcup _{k=n}^{\mathrm{\infty }}{E}_k.$

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:

$\sum _{n=1}^{\mathrm{\infty }}\mathrm{Pr}(E_n)<\mathrm{\infty }.$

Note that the convergence of this sum implies:

${\inf }_{N\ge 1}\sum _{n=N}^{\mathrm{\infty }}\mathrm{Pr}(E_n)=0.$

Therefore it follows that

$\begin{array}{rl}& \mathrm{Pr}\left({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n\right)=\mathrm{Pr}(E_n\text{ infinitely often})\\ & =\mathrm{Pr}\left(\bigcap _{N=1}^{\mathrm{\infty }}\bigcup _{n=N}^{\mathrm{\infty }}{E}_n\right)\le {\inf }_{N\ge 1}\mathrm{Pr}\left(\bigcup _{n=N}^{\mathrm{\infty }}{E}_n\right)\le {\inf }_{N\ge 1}\sum _{n=N}^{\mathrm{\infty }}\mathrm{Pr}(E_n)=0.\end{array}$

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

$\sum _{n=1}^{\mathrm{\infty }}\mu (A_n)<\mathrm{\infty },$

then

$\mu \left({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{A}_n\right)=0.$

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 $\sum _{n=1}^{\mathrm{\infty }}\mathrm{Pr}(E_n)=\mathrm{\infty }$ and the events $(E_n{)}_{n=1}^{\mathrm{\infty }}$ are independent, then $\mathrm{Pr}({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n)=1.$

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 Template:Harv, if E_j is a collection of Lebesgue measurable subsets of a compact set in Rⁿ such that

$\sum _j\mu (E_j)=\mathrm{\infty },$

then there is a sequence F_j of translates

$F_j=E_j+x_j$

such that

$\lim \sup F_j=\bigcap _{n=1}^{\mathrm{\infty }}\bigcup _{k=n}^{\mathrm{\infty }}{F}_k={\mathbb{ℝ}}^n$

apart from a set of measure zero.

Proof

Suppose that $\sum _{n=1}^{\mathrm{\infty }}\mathrm{Pr}(E_n)=\mathrm{\infty }$ and the events $(E_n{)}_{n=1}^{\mathrm{\infty }}$ 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

$1-\mathrm{Pr}({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n)=0.$

Noting that:

$\begin{array}{rl}1-\mathrm{Pr}({\mathrm{lim\; sup}}_{n\to \mathrm{\infty }}{E}_n)& =1-\mathrm{Pr}(\{E_n\text{ i.o.}\})=\mathrm{Pr}(\{E_n\text{ i.o.}{\}}^c)\\ & =\mathrm{Pr}\left({\left(\bigcap _{N=1}^{\mathrm{\infty }}\bigcup _{n=N}^{\mathrm{\infty }}{E}_n\right)}^c\right)=\mathrm{Pr}\left(\bigcup _{N=1}^{\mathrm{\infty }}\bigcap _{n=N}^{\mathrm{\infty }}{E}_n^c\right)\\ & =\mathrm{Pr}\left({\mathrm{lim\; inf}}_{n\to \mathrm{\infty }}{E}_n^c\right)=\underset{N\to \mathrm{\infty }}{\lim }\mathrm{Pr}\left(\bigcap _{n=N}^{\mathrm{\infty }}{E}_n^c\right)\end{array}$

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

$\begin{array}{rl}\mathrm{Pr}\left(\bigcap _{n=N}^{\mathrm{\infty }}{E}_n^c\right)& =\prod _{n=N}^{\mathrm{\infty }}\mathrm{Pr}\left(E_n^c\right)\\ & =\prod _{n=N}^{\mathrm{\infty }}(1-\mathrm{Pr}\left(E_n\right))\\ & \le \prod _{n=N}^{\mathrm{\infty }}(1-\mathrm{Pr}(E_n)+\frac{(\mathrm{Pr}(E_n){)}^2}{2!}-\frac{(\mathrm{Pr}(E_n){)}^3}{3!}+\cdots )\\ & =\prod _{n=N}^{\mathrm{\infty }}\left(\sum _{m=0}^{\mathrm{\infty }}\frac{(-\mathrm{Pr}(E_n){)}^m}{m!}\right)\\ & =\prod _{n=N}^{\mathrm{\infty }}\exp (-\mathrm{Pr}\left(E_n\right))\\ & =\exp (-\sum _{n=N}^{\mathrm{\infty }}\mathrm{Pr}(E_n))\\ & =0.\end{array}$

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

$\begin{array}{rl}-\log (\mathrm{Pr}\left(\bigcap _{n=N}^{\mathrm{\infty }}{E}_n^c\right))& =-\log (\prod _{n=N}^{\mathrm{\infty }}(1-\mathrm{Pr}(E_n)))\\ & =-\sum _{n=N}^{\mathrm{\infty }}\log (1-\mathrm{Pr}(E_n)).\end{array}$

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

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 $\overline{A}$ 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

$\sum _k\mathrm{Pr}(A_{t_{k+1}}|{\overline{A}}_{t_k})=\mathrm{\infty }.$

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.

See also

• Lévy's zero-one law
• Kuratowski convergence

References

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  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
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  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
  To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010.
• 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