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Unformatted text preview: Solutions of Selected Problems from Probability Essentials, Second Edition Solutions to selected problems of Chapter 2 2.1 Let’s first prove by induction that #(2 Ω n ) = 2 n if Ω = { x 1 ,...,x n } . For n = 1 it is clear that #(2 Ω 1 ) = #( {∅ , { x 1 }} ) = 2. Suppose #(2 Ω n 1 ) = 2 n 1 . Observe that 2 Ω n = {{ x n } ∪ A,A ∈ 2 Ω n 1 } ∪ 2 Ω n 1 } hence #(2 Ω n ) = 2#(2 Ω n 1 ) = 2 n . This proves finiteness. To show that 2 Ω is a σalgebra we check: 1. ∅ ⊂ Ω hence ∅ ∈ 2 Ω . 2. If A ∈ 2 Ω then A ⊂ Ω and A c ⊂ Ω hence A c ∈ 2 Ω . 3. Let ( A n ) n ≥ 1 be a sequence of subsets of Ω. Then S ∞ n =1 A n is also a subset of Ω hence in 2 Ω . Therefore 2 Ω is a σalgebra. 2.2 We check if H = ∩ α ∈ A G α has the three properties of a σalgebra: 1. ∅ ∈ G α ∀ α ∈ A hence ∅ ∈ ∩ α ∈ A G α . 2. If B ∈ ∩ α ∈ A G α then B ∈ G α ∀ α ∈ A . This implies that B c ∈ G α ∀ α ∈ A since each G α is a σalgebra. So B c ∈ ∩ α ∈ A G α . 3. Let ( A n ) n ≥ 1 be a sequence in H . Since each A n ∈ G α , S ∞ n =1 A n is in G α since G α is a σalgebra for each α ∈ A . Hence S ∞ n =1 A n ∈ ∩ α ∈ A G α . Therefore H = ∩ α ∈ A G α is a σalgebra. 2.3 a. Let x ∈ ( ∪ ∞ n =1 A n ) c . Then x ∈ A c n for all n , hence x ∈ ∩ ∞ n =1 A c n . So ( ∪ ∞ n =1 A n ) c ⊂ ∩ ∞ n =1 A c n . Similarly if x ∈ ∩ ∞ n =1 A c n then x ∈ A c n for any n hence x ∈ ( ∪ ∞ n =1 A n ) c . So ( ∪ ∞ n =1 A n ) c = ∩ ∞ n =1 A c n . b. By parta ∩ ∞ n =1 A n = ( ∪ ∞ n =1 A c n ) c , hence ( ∩ ∞ n =1 A n ) c = ∪ ∞ n =1 A c n . 2.4 lim inf n →∞ A n = ∪ ∞ n =1 B n where B n = ∩ m ≥ n A m ∈ A ∀ n since A is closed under taking countable intersections. Therefore lim inf n →∞ A n ∈ A since A is closed under taking countable unions. By De Morgan’s Law it is easy to see that lim sup A n = (lim inf n →∞ A c n ) c , hence lim sup n →∞ A n ∈ A since lim inf n →∞ A c n ∈ A and A is closed under taking complements. Note that x ∈ lim inf n →∞ A n ⇒ ∃ n * s.t x ∈ ∩ m ≥ n * A m ⇒ x ∈ ∩ m ≥ n A m ∀ n ⇒ x ∈ lim sup n →∞ A n . Therefore lim inf n →∞ A n ⊂ lim sup n →∞ A n . 2.8 Let L = { B ⊂ R : f 1 ( B ) ∈ B} . It is easy to check that L is a σalgebra. Since f is continuous f 1 ( B ) is open (hence Borel) if B is open. Therefore L contains the open sets which implies L ⊃ B since B is generated by the open sets of R . This proves that f 1 ( B ) ∈ B if B ∈ B and that A = { A ⊂ R : ∃ B ∈ B with A = f 1 ( B ) ∈ B} ⊂ B . 1 Solutions to selected problems of Chapter 3 3.7 a. Since P ( B ) > P ( .  B ) defines a probability measure on A , therefore by Theorem 2.4 lim n →∞ P ( A n  B ) = P ( A  B )....
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This document was uploaded on 03/28/2011.
 Spring '11
 Probability

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