Second Borel-Cantelli Lemma
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Statement
Let the events $E_n$ be independent.
Let 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 $\ds \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$ and the events $\ds \sequence {E_n}^\infty_{n \mathop = 1}$ are independent, then:
- $\ds \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 1$
Proof
Let $\ds \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$.
Let $\sequence {E_n}^\infty_{n \mathop = 1}$ be 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:
- $\ds 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 0$
Noting that:
| \(\ds 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n}\) | \(=\) | \(\ds 1 - \map \Pr {\set {E_n \text { i.o.} } }\) | where i.o. stands for infinitely often | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Pr {\set {E_n \text{ i.o.} }^c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Pr {\paren {\bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty E_n}^c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Pr {\bigcup_{N \mathop = 1}^\infty \bigcap_{n \mathop = N}^\infty E_n^c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Pr {\liminf_{n \mathop \to \infty} E_n^c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{N \mathop \to \infty} \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c}\) |
it is enough to show:
- $\ds \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$
Since the $\sequence {E_n}^\infty_{n \mathop = 1}$ are independent:
| \(\ds \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c}\) | \(=\) | \(\ds \prod^\infty_{n \mathop = N} \map \Pr {E_n^c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \prod^\infty_{n \mathop = N} \paren {1 - \map \Pr {E_n} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \prod_{n \mathop = N}^\infty \map \exp {-\map \Pr {E_n} }\) | Exponential Function Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \exp {-\sum_{n \mathop = N}^\infty \map \Pr {E_n} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
This completes the proof.
$\blacksquare$
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 | This article, or a section of it, needs explaining. In particular: What is the motivation of the 'alternative' explanation below? It is exactly the same as the above proof, just wrapped by a logarithm, using the trivial relation $\map \ln {\exp x} = x$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Alternatively, we can see:
- $\ds \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$
by taking negative the logarithm of both sides to get:
| \(\ds \map \ln {\map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} }\) | \(=\) | \(\ds -\map \ln {\prod_{n \mathop = N}^\infty \paren {1 - \map \Pr {E_n} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = N}^\infty \map \ln {1 - \map \Pr {E_n} }\) |
Since $-\map \ln {1 - x} \ge x$ for all $x > 0$, the result similarly follows from our assumption that $\ds \sum^\infty_{n \mathop = 1} \map \Pr {E_n} = \infty$.
$\blacksquare$
Also see
Source of Name
This entry was named for Émile Borel and Francesco Paolo Cantelli.