# proof of Borel-Cantelli 2

Let $E$ denote the set of samples that are in ${A}_{i}$ infinitely often. We want to show that the complement of $E$ has probability zero.

As in the proof of Borel-Cantelli 1, we know that

$${E}^{c}=\bigcup _{k=1}^{\mathrm{\infty}}\bigcap _{i=k}^{\mathrm{\infty}}{A}_{i}^{c}$$ |

where the superscript ${}^{c}$ means set complement. But for each $k$,

$P({\cap}_{i=k}{A}_{i}^{c})$ | $={\displaystyle \prod _{i=k}^{\mathrm{\infty}}}P({A}_{i}^{c})$ | ||

$={\displaystyle \prod _{i=k}^{\mathrm{\infty}}}(1-P({A}_{i}))$ |

Here we use the assumption^{} that the event ${A}_{i}$’s are independent^{}. The inequality $1-a\le {e}^{-a}$ and the assumption that the sum of $P({A}_{i})$ diverges together imply that

$$P({\cap}_{i=k}{A}_{i}^{c})\le \mathrm{exp}(-\sum _{i=k}^{\mathrm{\infty}}P({A}_{i}))=0$$ |

Therefore ${E}^{c}$ is a union of countable^{} number of events, each of them has probability zero. So $P({E}^{c})=0$.

Title | proof of Borel-Cantelli 2 |
---|---|

Canonical name | ProofOfBorelCantelli2 |

Date of creation | 2013-03-22 14:29:35 |

Last modified on | 2013-03-22 14:29:35 |

Owner | kshum (5987) |

Last modified by | kshum (5987) |

Numerical id | 4 |

Author | kshum (5987) |

Entry type | Proof |

Classification | msc 60A99 |