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Unformatted text preview: k ) + P ( A i A j ( k A k ) c ) # Now to get to P 3 we drop terms from this last expression. That is n1 X i =1 n X j = i +1 " n X k =1 P ( A i A j A k ) + P ( A i A j ( k A k ) c ) # n1 X i =1 n X j = i +1 " n X k =1 P ( A i A j A k ) # n2 X i =1 n1 X j = i +1 n X k = j +1 P ( A i A j A k ) = X 1 i<j<k n P ( A i A j A k ) = P 3 . The sequence of bounds is improving because the bounds P 1 ,P 1P 2 + P 3 ,P 1P 2 + P 3P 4 + P 5 ,... , are getting smaller since P i P j if i j and therefore the termsP 2 k + P 2 k +1 0. The lower bounds P 1P 2 ,P 1P 2 + P 3P 4 ,P 1P 2 + P 3P 4 + P 5P 6 ,... , are getting bigger since P i P j if i j and therefore the terms P 2 k +1P 2 k 0....
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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