introduction-probability.pdf

# 2 if a b f then p a b p a p a b 3 if b f then p b c 1

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(2) If A, B ∈ F , then P ( A \ B ) = P ( A ) - P ( A B ) . (3) If B ∈ F , then P ( B c ) = 1 - P ( B ) . (4) If A 1 , A 2 , ... ∈ F then P ( i =1 A i ) i =1 P ( A i ) . (5) Continuity from below : If A 1 , A 2 , ... ∈ F such that A 1 A 2 A 3 ⊆ · · · , then lim n →∞ P ( A n ) = P n =1 A n . (6) Continuity from above : If A 1 , A 2 , ... ∈ F such that A 1 A 2 A 3 ⊇ · · · , then lim n →∞ P ( A n ) = P n =1 A n . Proof . (1) We let A n +1 = A n +2 = · · · = , so that P n i =1 A i = P i =1 A i = i =1 P ( A i ) = n i =1 P ( A i ) , because of P ( ) = 0. (2) Since ( A B ) ( A \ B ) = , we get that P ( A B ) + P ( A \ B ) = P (( A B ) ( A \ B )) = P ( A ) . (3) We apply (2) to A = Ω and observe that Ω \ B = B c by definition and Ω B = B . (4) Put B 1 := A 1 and B i := A c 1 A c 2 ∩· · ·∩ A c i - 1 A i for i = 2 , 3 , . . . Obviously, P ( B i ) P ( A i ) for all i . Since the B i ’s are disjoint and i =1 A i = i =1 B i it follows P i =1 A i = P i =1 B i = i =1 P ( B i ) i =1 P ( A i ) . (5) We define B 1 := A 1 , B 2 := A 2 \ A 1 , B 3 := A 3 \ A 2 , B 4 := A 4 \ A 3 , ... and get that n =1 B n = n =1 A n and B i B j =

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1.2. PROBABILITY MEASURES 17 for i = j . Consequently, P n =1 A n = P n =1 B n = n =1 P ( B n ) = lim N →∞ N n =1 P ( B n ) = lim N →∞ P ( A N ) since N n =1 B n = A N . (6) is an exercise. The sequence a n = ( - 1) n , n = 1 , 2 , . . . does not converge, i.e. the limit of ( a n ) n =1 does not exist. But the limit superior and the limit inferior for a given sequence of real numbers exists always. Definition 1.2.7 [ lim inf n a n and lim sup n a n ] For a 1 , a 2 , ... R we let lim inf n a n := lim n inf k n a k and lim sup n a n := lim n sup k n a k . Remark 1.2.8 (1) The value lim inf n a n is the infimum of all c such that there is a subsequence n 1 < n 2 < n 3 < · · · such that lim k a n k = c . (2) The value lim sup n a n is the supremum of all c such that there is a subsequence n 1 < n 2 < n 3 < · · · such that lim k a n k = c . (3) By definition one has that -∞ ≤ lim inf n a n lim sup n a n ≤ ∞ . Moreover, if lim inf n a n = lim sup n a n = a R , then lim n a n = a . (4) For example, taking a n = ( - 1) n , gives lim inf n a n = - 1 and lim sup n a n = 1 . As we will see below, also for a sequence of sets one can define a limit superior and a limit inferior. Definition 1.2.9 [ lim inf n A n and lim sup n A n ] Let (Ω , F ) be a measurable space and A 1 , A 2 , ... ∈ F . Then lim inf n A n := n =1 k = n A k and lim sup n A n := n =1 k = n A k . The definition above says that ω lim inf n A n if and only if all events A n , except a finite number of them, occur, and that ω lim sup n A n if and only if infinitely many of the events A n occur.
18 CHAPTER 1. PROBABILITY SPACES Proposition 1.2.10 [Lemma of Fatou] 3 Let , F , P ) be a probability space and A 1 , A 2 , ... ∈ F . Then P lim inf n A n lim inf n P ( A n ) lim sup n P ( A n ) P lim sup n A n . The proposition will be deduced from Proposition 3.2.6.

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• Spring '17
• Probability, Probability theory, Probability space, measure, lim P, Probability Spaces

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