introduction-probability.pdf

Now we continue with the fundamental lemma of borel

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Now we continue with the fundamental Lemma of Borel-Cantelli. Proposition 1.2.16 [Lemma of Borel-Cantelli] 5 Let , F , P ) be a probability space and A 1 , A 2 , ... ∈ F . Then one has the following: (1) If n =1 P ( A n ) < , then P (lim sup n →∞ A n ) = 0 . (2) If A 1 , A 2 , ... are assumed to be independent and n =1 P ( A n ) = , then P (lim sup n →∞ A n ) = 1 . Proof . (1) It holds by definition lim sup n →∞ A n = n =1 k = n A k . By k = n +1 A k k = n A k and the continuity of P from above (see Proposition 1.2.6) we get that P lim sup n →∞ A n = P n =1 k = n A k = lim n →∞ P k = n A k lim n →∞ k = n P ( A k ) = 0 , where the last inequality follows again from Proposition 1.2.6. (2) It holds that lim sup n A n c = lim inf n A c n = n =1 k = n A c n . So, we would need to show P n =1 k = n A c n = 0 . Letting B n := k = n A c k we get that B 1 B 2 B 3 ⊆ · · · , so that P n =1 k = n A c n = lim n →∞ P ( B n ) 5 Francesco Paolo Cantelli, 20/12/1875-21/07/1966, Italian mathematician.
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1.2. PROBABILITY MEASURES 21 so that it suffices to show P ( B n ) = P k = n A c k = 0 . Since the independence of A 1 , A 2 , ... implies the independence of A c 1 , A c 2 , ... , we finally get (setting p n := P ( A n )) that P k = n A c k = lim N →∞ ,N n P N k = n A c k = lim N →∞ ,N n N k = n P ( A c k ) = lim N →∞ ,N n N k = n (1 - p k ) lim N →∞ ,N n N k = n e - p k = lim N →∞ ,N n e - P N k = n p k = e - P k = n p k = e -∞ = 0 where we have used that 1 - x e - x . Although the definition of a measure is not difficult, to prove existence and uniqueness of measures may sometimes be difficult. The problem lies in the fact that, in general, the σ -algebras are not constructed explicitly, one only knows their existence. To overcome this difficulty, one usually exploits Proposition 1.2.17 [Carath´eodory’s extension theorem] 6 Let Ω be a non-empty set and G an algebra on Ω such that F := σ ( G ) . Assume that P 0 : G → [0 , ) satisfies: (1) P 0 (Ω) < . (2) If A 1 , A 2 , ... ∈ G , A i A j = for i = j , and i =1 A i ∈ G , then P 0 i =1 A i = i =1 P 0 ( A i ) . 6 Constantin Carath´ eodory, 13/09/1873 (Berlin, Germany) - 02/02/1950 (Munich, Ger- many).
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22 CHAPTER 1. PROBABILITY SPACES Then there exists a unique finite measure P on F such that P ( A ) = P 0 ( A ) for all A ∈ G . Proof . See [3] (Theorem 3.1). As an application we construct (more or less without rigorous proof) the product space 1 × Ω 2 , F 1 ⊗ F 2 , P 1 × P 2 ) of two probability spaces (Ω 1 , F 1 , P 1 ) and (Ω 2 , F 2 , P 2 ). We do this as follows: (1) Ω 1 × Ω 2 := { ( ω 1 , ω 2 ) : ω 1 Ω 1 , ω 2 Ω 2 } . (2) F 1 ⊗ F 2 is the smallest σ -algebra on Ω 1 × Ω 2 which contains all sets of type A 1 × A 2 := { ( ω 1 , ω 2 ) : ω 1 A 1 , ω 2 A 2 } with A 1 ∈ F 1 , A 2 ∈ F 2 . (3) As algebra G we take all sets of type A := ( A 1 1 × A 1 2 ) ∪ · · · ∪ ( A n 1 × A n 2 ) with A k 1 ∈ F 1 , A k 2 ∈ F 2 , and ( A i 1 × A i 2 ) ( A j 1 × A j 2 ) = for i = j .
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