Stats - Problem 3.10 Show that if E F , then P (E) P (F )....

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Problem 3.10 Show that if E F , then P ( E ) P ( F ). Proof. We can represent event F as a union of two mutually exclusive events: F = E [ ( F \ E ) ,whereE \ ( F \ E ) = Then P ( F ) = P ( E ) + P ( F \ E ) P ( E ) 1
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Prove Broke’s inequality: P ˆ n [ i =1 E i ! n X i =1 P ( E i ) Proof . We will use mathematical induction 1 as a method for statement proof. The basis step: showing that the statement holds when n = 2. P ˆ 2 [ i =1 E i ! = P E 1 [ E 2 · = P ( E 1 )+ P ( E 2 ) - P ( E 1 \ E 2 ) 2 X i =1 P ( E i ) The inductive step: showing that if the statement holds for n = k P ˆ k [ i =1 E i ! k X i =1 P ( E i ) then the same statement also holds for n = k + 1. For n = k + 1 we get P ˆ k +1 [ i =1 E i ! = P ˆ k [ i =1 E i [ E k +1 ! P ˆ k [ i =1 E i ! + P ( E k +1 ) k X i =1 P ( E i ) + P ( E k +1 ) = k +1 X i =1 P ( E i ) 1 Induction was a topic of MAT4 course. See e.g. Finn V. Jensen, Sven Skyum, ”Induktion og Rekursion” or
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Stats - Problem 3.10 Show that if E F , then P (E) P (F )....

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