midtermsol - EE 178 Thursday, February 12, 2009...

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Unformatted text preview: EE 178 Thursday, February 12, 2009 Probabilistic Systems Analysis Handout #13 Midterm Solution 1. a. (=). Consider 1- P( A c ∪ B ) = P(( A c ∪ B ) c ) = P( A ∩ B c ) by DeMorgan = P( A )- P( A ∩ B ) . b. ( ≥ ). Since P( A | B ) = P( B | A ) , P( A,B ) P( B ) = P( A,B ) P( A ) . Thus, P( A ) = P( B ) . This implies that P( A ∪ B ) = 2P( A )- P( A ∩ B ) . But since P( A ∪ B ) = 1 , P( A ) ≥ 1 / 2 . c. (=). Consider P { X ≤ 5 | X > 2 } = 1- P { X > 5 | X > 2 } = 1- P { X > 3 } by memorylessness of Exp pdf = P { X ≤ 3 } d. ( ≤ ). Let z = g ( y ) , then p Z ( g ( y )) = p Z ( z ) = summationdisplay { y ′ : g ( y ′ )= z } p Y ( y ′ ) ≥ p Y ( y ) . e. ( ≤ ). This is because Var( X ) = E( X 2 )- (E( X )) 2 . Page 2 of 4 EE 178, Winter 2009 2. a. The tree: a120 a1 a1 a1 a1 a1 a1 a1 P( A ) = 1 / 2 a120 a0 a0 a0 a0 P( H 1 | A ) = p 1 a120 a64 a64 a64 a64 ¯ p 1 a120 a65 a65 a65 a65 a65 a65 a65 P( A c ) = 1 / 2 a120 a0 a0 a0 a0 p 2 a120 a64 a64 a64 a64 ¯ p 2 a120...
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This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Spring '09 term at Aarhus Universitet.

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midtermsol - EE 178 Thursday, February 12, 2009...

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