Limit of the intersection of events

  1. Jun 15, 2011 #1
    Hi, I keep seeing this come up
    A1 ⊇ A2 ⊇ A3 .... is an infinite decreasing sequence of events. Prove from first principles
    that
    P(intersection of Ai from i=1 to infinity) = Lim P(An) as n--> infinity

    All i can think of is that since each is a subset of the preceding, then A1 ∩ A2...∩An = An
    So clearly P(A1 ∩ A2...∩An) = P(An) and thus the same for limits.
    I think this is too simplistic though, is it or isnt it ?

    Thanks a lot
     
  2. jcsd
  3. Jun 15, 2011 #2

    micromass

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    Hi stukbv! :smile:

    Did you already encounter the dual version of this? That is, if

    [tex]A_0\subseteq A_1\subseteq A_2\subseteq ...[/tex]

    then

    [tex]P\left(\bigcup_{i=0}^{+\infty}{A_i}\right)=\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

    It is the easiest if we prove this first. Now, this statement actually follows from the [itex]\sigma[/itex]-additivity. However, the [itex]\sigma[/itex]-additivity requires the events to be disjoint, which is not the case here. Is there a way to make the events disjoint, though?
     
  4. Jun 15, 2011 #3
    Yeah ive done it that way I think
    Let B1 = A1
    B2 = A2∩A1C
    B3=A3∩A2C
    And so on,
    Then the Bi are disjoint and the union Bi (up to n) = An, but also U Bi up to ∞ = Union of Ai upto ∞ (since preceding ones are subsets!) so P(An) = P(U Bi) (to n) = Sum P(Bi (to n) as all bi are disjoint, then let RHS and LHS n tend to infinity and we get sum to infinity of P(Bi) = P(U Bi) to ∞ = P(U Ai) to ∞.
    according to my lecturer!
     
  5. Jun 15, 2011 #4

    micromass

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    Ah yes, that's very good!! So now for your question. You need to prove that

    [tex]P\left(\bigcap{A_n}\right)=\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

    Now what happens if you take the complements of these events?
     
  6. Jun 15, 2011 #5
    what like B1 = A1
    B2 = A2\A1
    B2 = A3\A2
    and so on ?
     
  7. Jun 15, 2011 #6

    micromass

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    No, take the complements. What is

    [tex]P\left(\left(\bigcap{A_n}\right)^c\right)[/tex]
     
  8. Jun 15, 2011 #7
    P((U An^c)?
     
  9. Jun 15, 2011 #8

    micromass

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    Yes, carry on...
     
  10. Jun 15, 2011 #9
    SO P(U An^c) is equal to the sum since they are disjoint?
    But i dont see how you can relate the 2 like we did before
     
  11. Jun 15, 2011 #10

    micromass

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    No, they are not disjoint. But you have a formula for

    [tex]P\left(\bigcup{A_n^c}\right)[/tex]

    Right? You've proven the formula above...
     
  12. Jun 15, 2011 #11
    ok so the limit = p(An^c)
     
  13. Jun 15, 2011 #12

    micromass

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    And what the probability of a complement?
     
  14. Jun 15, 2011 #13
    1-p(An).
    I see , but then how do i relate it to my initial statement?
     
  15. Jun 15, 2011 #14

    micromass

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    Well, you know already that the statement

    [tex]P\left(\bigcup{A_n^c}\right)=1-\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

    is true. Now try to evaluate the left side.
     
  16. Jun 16, 2011 #15
    surely if they are all subsets of their preceding ones then the union of the complements is just a1^c ?
     
  17. Jun 16, 2011 #16

    micromass

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    We have

    [tex]A_1\supseteq A_2\supseteq ...[/tex]

    and thus

    [tex]A_1^c\subseteq A_2^c\subseteq ...[/tex]

    So that doesn't really work.
     
  18. Jun 16, 2011 #17
    ohh so is the LHS 1 - P( intersection of Ai)
     
  19. Jun 16, 2011 #18
    Then you can get rid of the ones and then take limits ?
     
  20. Jun 16, 2011 #19
    so we can say lhs = 1-P(intersection Ai) ?
     
  21. Jun 16, 2011 #20

    micromass

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    Yes, that is correct! Just eliminate the ones and you have the result you wanted to prove!
     
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