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Math Help - De Moivre Jordan Theorem

  1. #1
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    De Moivre Jordan Theorem

    Is anybody here acquainted with the De Moivre-Jordan Theorem?
    Cannot locate anything on the internet, so here I shall reproduce a rather terse account of it from a textbook on Fundamental Probability:


    For events B_1, .......B_n, the probability that exactly m of the B_i occur, m = 0, 1, . . . , n, is given by


    \sum_{i=m}^n (-1)^{i-m} {i\choose m} S_i


    An Example

    Given n=4, m=2, calculate the probability that exactly 2 out of 4 B_i occur is:


    P_{2,4}= {2\choose 2} S_2 - {3\choose 2} S_3 + {4\choose 2} S_4

    <br />
= \sum_{i<j} P(B_i B_j) - 3\sum_{i<j<k} P(B_i B_j B_k) + 6 P(B_1 B_2 B_3 B_4)

    where

    \sum_{i<j} P(B_i B_j) is given by:

    P(B_1 B_2) + P(B_1 B_3) + P(B_1 B_4) + P(B_2 B_3) + P(B_2 B_4) + P(B_3 B_4)

    Looks like its something to do with the double counting like the inclusion-exclusion theory, but cannot figure it out.
    Can anyone link me to the correct wiki page or try to explain to me what is happening here?
    Last edited by chopet; October 10th 2007 at 04:09 AM.
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  2. #2
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    Ok. Think I got it. Its simply stating that there are 4 possible events which we are treating as non-disjointed, that is, they may share similar sample points.

    So what is the probability that exactly any 2 such events of the 4 possible are occuring?

    Using the Venn diagram to depict the 4 events where all 4 intersect each other, we can see the double-counting clearly if we just add up the 6 areas where 2 circles intersect (depicting the probability of exactly 2 events happening). This is very much like the inclusion-exclusion theory.

    So, what should we call this theory? Can anyone send me more info here?
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