introduction-probability.pdf

# Finally we define p g 0 1 by p a 1 1 a 1 2 a n 1 a n

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Finally, we define P 0 : G → [0 , 1] by P 0 (( A 1 1 × A 1 2 ) ∪ · · · ∪ ( A n 1 × A n 2 ) ) := n k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) . Proposition 1.2.18 The system G is an algebra. The map P 0 : G → [0 , 1] is correctly defined and satisfies the assumptions of Carath´ eodory’s extension theorem Proposition 1.2.17. Proof . (i) Assume A = ( A 1 1 × A 1 2 ) ∪ · · · ∪ ( A n 1 × A n 2 ) = ( B 1 1 × B 1 2 ) ∪ · · · ∪ ( B m 1 × B m 2 ) where the ( A k 1 × A k 2 ) n k =1 and the ( B l 1 × B l 2 ) m l =1 are pair-wise disjoint, respec- tively. We find partitions C 1 1 , ..., C N 1 1 of Ω 1 and C 1 2 , ..., C N 2 2 of Ω 2 so that A k i and B l i can be represented as disjoint unions of the sets C r i , r = 1 , ..., N i . Hence there is a representation A = ( r,s ) I ( C r 1 × C s 2 ) for some I ⊆ { 1 , ..., N 1 } × { 1 , .... , N 2 } . By drawing a picture and using that P 1 and P 2 are measures one observes that n k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) = ( r,s ) I P 1 ( C r 1 ) P 2 ( C s 2 ) = m l =1 P 1 ( B l 1 ) P 2 ( B l 2 ) .

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1.2. PROBABILITY MEASURES 23 (ii) To check that P 0 is σ -additive on G it is sufficient to prove the following: For A i , A k i ∈ F i with A 1 × A 2 = k =1 ( A k 1 × A k 2 ) and ( A k 1 × A k 2 ) ( A l 1 × A l 2 ) = for k = l one has that P 1 ( A 1 ) P 2 ( A 2 ) = k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) . Since the inequality P 1 ( A 1 ) P 2 ( A 2 ) N k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) can be easily seen for all N = 1 , 2 , ... by an argument like in step (i) we concentrate ourselves on the converse inequality P 1 ( A 1 ) P 2 ( A 2 ) k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) . We let ϕ ( ω 1 ) := n =1 1I { ω 1 A n 1 } P 2 ( A n 2 ) and ϕ N ( ω 1 ) := N n =1 1I { ω 1 A n 1 } P 2 ( A n 2 ) for N 1, so that 0 ϕ N ( ω 1 ) N ϕ ( ω 1 ) = 1I A 1 ( ω 1 ) P 2 ( A 2 ). Let ε (0 , 1) and B N ε := { ω 1 Ω 1 : (1 - ε ) P 2 ( A 2 ) ϕ N ( ω 1 ) } ∈ F 1 . The sets B N ε are non-decreasing in N and N =1 B N ε = A 1 so that (1 - ε ) P 1 ( A 1 ) P 2 ( A 2 ) = lim N (1 - ε ) P 1 ( B N ε ) P 2 ( A 2 ) . Because (1 - ε ) P 2 ( A 2 ) ϕ N ( ω 1 ) for all ω 1 B N ε one gets (after some calcu- lation...) (1 - ε ) P 2 ( A 2 ) P 1 ( B N ε ) N k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) and therefore lim N (1 - ε ) P 1 ( B N ε ) P 2 ( A 2 ) lim N N k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) = k =1 P 1 ( A k 1 ) P 2 ( A k 2 ) . Since ε (0 , 1) was arbitrary, we are done. Definition 1.2.19 [product of probability spaces] The extension of P 0 to F 1 ⊗ F 2 according to Proposition 1.2.17 is called product measure and denoted by P 1 × P 2 . The probability space (Ω 1 × Ω 2 , F 1 ⊗ F 2 , P 1 × P 2 ) is called product probability space .
24 CHAPTER 1. PROBABILITY SPACES One can prove that ( F 1 ⊗ F 2 ) ⊗ F 3 = F 1 ( F 2 ⊗ F 3 ) and ( P 1 × P 2 ) × P 3 = P 1 × ( P 2 × P 3 ) , which we simply denote by 1 × Ω 2 × Ω 3 , F 1 ⊗ F 2 ⊗ F 3 , P 1 × P 2 × P 3 ) . Iterating this procedure, we can define finite products 1 × Ω 2 × · · · × Ω n , F 1 ⊗ F 2 ⊗ · · · ⊗ F n , P 1 × P 2 × · · · × P n ) by iteration. The case of infinite products requires more work. Here the interested reader is referred, for example, to [5] where a special case is con- sidered.

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

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