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Independence of events question

  1. Feb 7, 2016 #1
    Hello,

    I am studying independence of events and I came across a formula that I don't understand. It is rather technical, not very interesting, but I feel stuck and it stays in my mind. Could you explain the following :

    If ## (A_i)_{i\in I}## is a sequence of independent events on ##(\Omega,{\cal A}, P)##, then, given the partition ## I_1 \cup I_2## of ##I##, the sequence of events ##(B_i)_{i\in I}## defined by ## B_i = A_i ## if ##i\in I_1## and ##B_i = A_i^c## if ##i\in I_2## is also independent.

    To prove this, one must show that for any finite set ##J\subset I## : ##P(\bigcap_{j\in J} B_j) = \prod_{j\in J} P(B_j)##

    Since ## J = J_1 \cup J_2##, where ## J_i = I_i \cap J ##, we have ## \bigcap_{j\in J} B_j = (\bigcap_{j\in J_1} A_j) \bigcap (\bigcap_{j\in J_2} A_j^c) =(\bigcap_{j\in J_1} A_j) - (\bigcup_{j\in J_2} A_j) ##

    so that
    ##P(\bigcap_{j\in J} B_j) = P(\bigcap_{j\in J_1} A_j) - P(\bigcup_{j\in J_2} A_j) ##

    With the inclusion-exclusion principle
    ## P(\bigcup_{j\in J_2} A_j) = \sum_{ K\subset J_2 \ |K| = 1} P(\bigcap_{j\in K} A_j ) - \sum_{ K\subset J_2 \ |K| = 2} P(\bigcap_{j\in K} A_j ) + ... + (-1)^{|J_2|-1} P(\bigcap_{j\in J_2} A_j) ##

    But I don't see how to finish this once ##P(\bigcap_j A_j ) ## is developped into a product !
     
    Last edited: Feb 7, 2016
  2. jcsd
  3. Feb 7, 2016 #2
    Oh I've made a mistake I get it now :

    ##\bigcap_{j\in J} B_j = A\cap B^c = A - A\cap B##

    with ##A = \bigcap_{j\in J_1} A_j## and ##B = \bigcup_{j\in J_2} A_j ##

    we have ## A \cap B = \bigcup_{j\in J_2} \bigcap_{k\in J_1} (A_k \cap A_j) ##

    And with the principle of inclusion exclusion and the independence of ##(A_i)_{i\in I}## we find ## P( A \cap B) = P(A) P(B) ##
     
    Last edited: Feb 7, 2016
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