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midtermsol

# midtermsol - EE 178 Probabilistic Systems Analysis Thursday...

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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 .

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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 a8 a8 a8 a8 P( H 2 | H 1 ,A ) = p 1 a72 a72 a72 a72 ¯ p 1 a8 a8 a8 a8 p 2 a72 a72 a72 a72 ¯ p 2 a8 a8 a8 a8 p 2 a72 a72 a72 a72 ¯ p 2 a8 a8 a8 a8 p 1 a72 a72 a72 a72 ¯ p 1 a120 A,H 1 ,H 2 a120 A,H 1 ,H c 2 a120 A,H c 1 ,H 2 a120 A,H c 1 ,H c 2 a120 A c ,H 1 ,H 2 a120 A c ,H 1 ,H c 2 a120 A c ,H c 1 ,H 2 a120 A c ,H c 1 ,H c 2 b. Consider
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