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Borel-Cantelli Lemma

  1. Mar 10, 2008 #1
    Let (A_n)_n>=1 be any event in some probability space { Omega, F, P }, then

    (i) SUM_n (P(A_n)) < oo => P( limsup_n->oo (A_n) ) = 0

    (ii) If in addition the A_n are independent then
    P( limsup_n->oo (A_n) ) <1 => SUM_n (P(A_n)) < oo

    Does that mean if the A_n are independent then P( limsup_n->oo (A_n)) must be either 0 or 1??

    If so, why bother using "<1" in (ii) and not just use "=0" instead?

    If not, then when it is strictly between 0 and 1 we have
    from (ii) that SUM_n (P(A_n)) < oo
    and then from (i) we get P( limsup_n->oo (A_n) ) = 0, a contradiction.
  2. jcsd
  3. Mar 12, 2008 #2
    It must be zero or one in any case.
  4. Mar 17, 2008 #3
    It's a weaker condition, and so could (potentially) be easier to employ. That said, I suspect the specific phrasing is an artifact. I.e., the usual way of phrasing the second theorem is to say that if the sum diverges, then the probability of the limsup is 1. The phrasing you've used is the contrapositive of that.
    Last edited by a moderator: Mar 17, 2008
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