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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 F , with each F a algebra. Since F a algebra, we have that F , and as this applies for all , it follows that G . Suppose now that A G . That is, A F for all . Since each F is a algebra, it follows that A c F for all , 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 for all i and every . Since F is a algebra, we deduce that A F , and as this applies for all , 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 algebra....
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 MONTANARI,A.
 Probability

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