# Second Borel-Cantelli Lemma

(Redirected from Second Borel-Cantelli lemma)

## 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 $\displaystyle \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$ and the events $\displaystyle \sequence {E_n}^\infty_{n \mathop = 1}$ are independent, then:

$\displaystyle \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 1$

## Proof

Let $\displaystyle \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$.

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:

$\displaystyle 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 0$

Noting that:

 $\displaystyle 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n}$ $=$ $\displaystyle 1 - \map \Pr {\set {E_n \text { i.o.} } }$ $\displaystyle$ $=$ $\displaystyle \map \Pr {\set {E_n \text{ i.o.} }^c}$ $\displaystyle$ $=$ $\displaystyle \map \Pr {\paren {\bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty E_n}^c}$ $\displaystyle$ $=$ $\displaystyle \map \Pr {\bigcup_{N \mathop = 1}^\infty \bigcap_{n \mathop = N}^\infty E_n^c}$ $\displaystyle$ $=$ $\displaystyle \map \Pr {\liminf_{n \mathop \to \infty} E_n^c}$ $\displaystyle$ $=$ $\displaystyle \lim_{N \mathop \to \infty} \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c}$

it is enough to show:

$\displaystyle \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$

Since the $\sequence {E_n}^\infty_{n \mathop = 1}$ are independent:

 $\displaystyle \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c}$ $=$ $\displaystyle \prod^\infty_{n \mathop = N} \map \Pr {E_n^c}$ $\displaystyle$ $=$ $\displaystyle \prod^\infty_{n \mathop = N} \paren {1 - \map \Pr {E_n} }$ $\displaystyle$ $\le$ $\displaystyle \prod_{n \mathop = N}^\infty \map \exp {-\map \Pr {E_n} }$ $\displaystyle$ $=$ $\displaystyle \map \exp {-\sum_{n \mathop = N}^\infty \map \Pr {E_n} }$ $\displaystyle$ $=$ $\displaystyle 0$

This completes the proof.

$\blacksquare$

Alternatively, we can see:

$\displaystyle \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$

by taking negative the logarithm of both sides to get:

 $\displaystyle \map \ln {\map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} }$ $=$ $\displaystyle -\map \ln {\prod_{n \mathop = N}^\infty \paren {1 - \map \Pr {E_n} } }$ $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = N}^\infty \map \ln {1 - \map \Pr {E_n} }$

Since $-\map \ln {1 - x} \ge x$ for all $x > 0$, the result similarly follows from our assumption that $\displaystyle \sum^\infty_{n \mathop = 1} \map \Pr {E_n} = \infty$.

$\blacksquare$

## Source of Name

This entry was named for Émile Borel and Francesco Paolo Cantelli.