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proof of Borel-Cantelli 2
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(Proof)
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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 $$ E^c = \bigcup_{k=1}^\infty \bigcap_{i=k}^\infty A_i^c $$ where the superscript $^c$ means set complement. But for each $k$ ,
Here we use the assumption that the event $A_i$ 's are independent. The inequality $1-a \leq e^{-a}$ and the assumption that the sum of $P(A_i)$ diverges together imply that $$ P(\cap_{i=k} A_i^c) \leq \exp(-\sum_{i=k}^\infty P(A_i)) = 0 $$ Therefore $E^c$ is a union of countable number of events, each of them has probability zero. So $P(E^c)=0$ .
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"proof of Borel-Cantelli 2" is owned by kshum.
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Cross-references: number, countable, union, imply, diverges, sum, inequality, independent, event, superscript, proof of Borel-Cantelli 1, complement, infinitely often
This is version 1 of proof of Borel-Cantelli 2, born on 2004-07-25.
Object id is 6028, canonical name is ProofOfBorelCantelli2.
Accessed 5052 times total.
Classification:
| AMS MSC: | 60A99 (Probability theory and stochastic processes :: Foundations of probability theory :: Miscellaneous) |
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Pending Errata and Addenda
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