solutions1 - Homework 1: Solutions 1. Claim 1: B1 , B2 , ....

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Homework 1: Solutions 1. Claim 1: B 1 , B 2 , . . . are disjoint. Proof: Let ω B i . If ω 0 B j for j < i then ω 0 A j and hence ω 0 / B i . Therefore, B i B j = . If ω 0 B j for j > i then ω 0 A j and ω 0 / A j - 1 . Since A i A j - 1 then ω 0 / A i . Hence, ω 0 / B i . Therefore, B i B j = . Claim 2: S n 1=1 A i = S n 1=1 B i . Proof: ω S n 1=1 A i = ω A i for some i = . Let i 0 be the smallest i such that ω A i . If i 0 = 1 then ω B 1 . If i 0 > 1 then ω A i 0 but ω / A i for i < i 0 . Hence, ω B i . Hence ω S n i =1 B i . For the reverse direction, suppose that ω S n i =1 B i . Then ω B i for some i . Hence, ω A i and therefore, ω S n 1=1 A i . Claim 3: S 1=1 A i = S 1=1 B i . The proof is the same as for Claim 2. The monotone decreasing case. Let A = lim n A n = T n =1 A n . Then A c = ( T n A n ) c = S n A c n . Since the A n are monotone decreasing, the A c n are monotone
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This note was uploaded on 05/25/2008 for the course STAT 36-625 taught by Professor Larrywasserman during the Fall '02 term at Carnegie Mellon.

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solutions1 - Homework 1: Solutions 1. Claim 1: B1 , B2 , ....

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