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# hw1sol - HW1 ST520 Hannig Adam Labadorf September 3 2007 1...

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HW1 ST520 - Hannig Adam Labadorf ST520 - Hannig September 3, 2007 HW1 1. 1.2.2 Definition of a σ -field: (a) Ø ∈ F (b) if A 1 , A 2 , ... ∈ F then i =1 A i ∈ F (c) if A ∈ F then A c ∈ F Proofs: (i) A B : A, B ∈ F by the problem definition, and by definition (c) of σ -fields we know that A c , B c ∈ F and by definition (b) we know that A B ∈ F . By these axioms we state that A c B c ∈ F . By applying Morgan’s Law, we can show that ( A c B c ) c = ( A B ) . Therefore, ( A B ) ∈ F . (ii) A \ B : By definition A \ B = A B c . We know that if B ∈ F then B c ∈ F by (c) and that A B ∈ F by (i). Therefore, A \ B ∈ F . (iii) A Δ B : By definition A Δ B = A B \ ( A B ) = A B ( A B ) c . By De Morgan’s Law, A B ( A B ) c = A B A c B c . A B ∈ F from (b), A c , B c ∈ F from (c), and A B ∈ F from (i). Therefore, A Δ B ∈ F . 1.2.4 We show that G is a σ -field by proving each condition listed in 1.2.2 separately: (a) Ø ∈ G Let some A ∈ F be such that A / B . From the definition of G , the set A B ∈ G and, since they are disjoint, A B = Ø . Therefore, Ø ∈ G . (b) if A 1 , A 2 , ... ∈ G then i =1 A i ∈ G Let some series C 1 , C 2 , C 3 , ... be such that C 1 = A 1 B, C 2 = A 2 B, ... . Therefore, C i ∈ G . Since we know that any A 1 A 2 = A 3 ∈ F , there must be some ( A 1 B ) ( A 2 B ) = ( A 3 B ) ∈ G by the definition of G . It follows that there is some A 4 such that A 3 A 4 = A 5 ∈ F and, therefore, ( A 3 B ) ( A 4 B ) = ( A 5 B ) ∈ G . Thus, i =1 A i ∈ G . (c) if A ∈ G then A c ∈ G Let some C ∈ F . Since F is a σ -field we know that C c ∈ F . Let A ∈ G such that C B = A and C c B = A c with respect to B . Therefore, A, A c ∈ G . 1.3.1 To prove that 1 12 P ( A B ) 1 3 we must first show that A and B are not disjoint. We do this by contradicting the axiom: P i =1 A i = i =1 P ( A i ) if all A i are disjoint. Using the sets A and B , we see that A + B = 3 4 + 1 3 = 13 12 > 1 . We cannot have a probability greater than 1 by definition, so there must be some elements in common between A and B . If we then think of Ω as a set of 12 distinct outcomes (any number will do so long as the proportions of A and B are respected) and A, B Ω

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