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ProbEssSolutions - Solutions of Selected Problems from...

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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 σ -algebra 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 n =1 A n is also a subset of Ω hence in 2 Ω . Therefore 2 Ω is a σ -algebra. 2.2 We check if H = α A G α has the three properties of a σ -algebra: 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 σ -algebra. So B c ∈ ∩ α A G α . 3. Let ( A n ) n 1 be a sequence in H . Since each A n ∈ G α , n =1 A n is in G α since G α is a σ -algebra for each α A . Hence n =1 A n ∈ ∩ α A G α . Therefore H = α A G α is a σ -algebra. 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 part-a 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 σ -algebra. 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
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Solutions to selected problems of Chapter 3 3.7 a. Since P ( B ) > 0 P ( . | B ) defines a probability measure on A , therefore by Theorem 2.4 lim n →∞ P ( A n | B ) = P ( A | B ). b. We have that A B n A B since 1 A B n ( w ) = 1 A ( w ) 1 B n ( w ) 1 A ( w ) 1 B ( w ). Hence P ( A B n ) P ( A B ). Also P ( B n ) P ( B ). Hence P ( A | B n ) = P ( A B n ) P ( B n ) P ( A B ) P ( B ) = P ( A | B ) .
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