No3(2004)solutions

# No3(2004)solutions - MATH 587 Assignment 3 Solutions...

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MATH 587 Assignment 3 Solutions November 18, 2004 (1) (a) Let µ ( A )= µ ( B )=0 , | λ 1 | ( A c )= | λ 2 | ( B c ) = 0. Then µ ( A B )=0and | λ 1 | ( C )= | λ 2 | ( C )=0 , and so λ 1 ( C )= λ 2 ( C ) = 0 for every C ( A B ) c .Thu s λ 1 + λ 2 ( C ) = 0 for every C ( A B ) c , so | λ 1 + λ 2 | (( A B ) c )=0. (b) Let µ ( A )=0 . I f λ + 1 ( A ) > 0, then λ ( 1 B ) > 0 for some B A , contradicting µ ( B )=0 . So λ + 1 ( A ) = 0 and similarly λ 1 ( A ) = 0, so | λ 1 | ( A )=0. (c) Let µ ( A )=0 , | λ 2 ( A c ) = 0. By (b), | λ 1 | ( A )=0. (d) By (c), λ 1 λ 1 . Hence for some A ∈F , | λ 1 | ( A )= | λ 1 | ( A c )=0. (2) We have Z 0 px p 1 P { X>x } dx = Z 0 px p 1 Z I { X>x } dP dx = ZZ 0 px p 1 I { X>x } dxdP = ZZ X 0 px p 1 dxdP = Z X p dP. (3) Z −∞ F ( x + a ) F ( x ) dx = Z −∞ Z I { x X x + a } dP dx = ZZ −∞ I { x X
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## This note was uploaded on 01/25/2011 for the course STAT 235a at Stanford.

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