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Unformatted text preview: Stat 310A/Math 230A Theory of Probability Homework 1 Solutions Andrea Montanari Due on 9/30/2010 Option 1: Exercises on measure spaces Exercise [1.1.4] 1. A and B \ A are disjoint with B = A ∪ ( B \ A ) so P ( A ) + P ( B \ A ) = P ( B ) and rearranging gives the desired result. 2. Let A n = A n ∩ A , B 1 = A 1 and for n > 1, B n = A n \ ∪ n 1 m =1 A m . Since the B n are disjoint and have union A we have using (a) and B m ⊆ A m P ( A ) = ∞ X m =1 P ( B m ) ≤ ∞ X m =1 P ( A m ) 3. Consider the disjoint sets B n = A n \ A n 1 for which ∪ ∞ m =1 B m = A , and ∪ n m =1 B m = A n . Then, P ( A ) = ∞ X m =1 P ( B m ) = lim n →∞ n X m =1 P ( B m ) = lim n →∞ P ( A n ) 4. A c n ↑ A c , so (c) implies P ( A c n ) ↑ P ( A c ) . Since P ( B c ) = 1 P ( B ) it follows that P ( A n ) ↓ P ( A ). Exercise [1.1.13] (a) Let G = T α F α , with each F α a σalgebra. Since F α a σalgebra, we have that Ω ∈ F α , and as this applies for all α , it follows that Ω ∈ G . Suppose now that A ∈ G . That is, A ∈ F α for all α . Since each F α is a σalgebra, it follows that A c ∈ F α for all α , and hence A c ∈ G . Similarly, let A = S i A i for some countable collection A 1 ,A 2 ,... of elements of G . By definition of G , necessarily A i ∈ F α for all i and every α . Since F α is a σalgebra, we deduce that A ∈ F α , and as this applies for all α , it follows that A ∈ G . (b) Since ∅∩ H = ∅ ∈ H , we have ∅ ∈ H H . Next assume A ∈ H H . Then A c ∩ H = H \ ( A ∩ H ) ∈ H whence A c ∈ H H . Finally for { A i } i ∈ N ⊆ H H , we have ( ∪ ∞ i =1 A i ) ∩ H = ∪ ∞ i =1 ( A i ∩ H ) whence ( ∪ ∞ i =1 A i ) ∈ H H . This proves that H H is a σalgebra....
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This note was uploaded on 08/20/2011 for the course STATS 310A at Stanford.
 '11
 MONTANARI,A.
 Probability

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