SampleMTsol

SampleMTsol - EE 178 Sunday Probabilistic Systems Analysis Handout#10 Sample Midterm Problem Solutions 1 a(= since P A ∪ B = P A P B P A ∩ B =

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Unformatted text preview: EE 178 Sunday, February 10, 2008 Probabilistic Systems Analysis Handout #10 Sample Midterm Problem Solutions 1. a. (=) since P( A ∪ B ) = P( A ) + P( B )- P( A ∩ B ) = P( A ) + P( B )- P( A )P( B ) = P( A ) + P( B )(1- P( A )) = P( A ) + P( B )P( A c ) , where the second line follows from the independence of A and B . b. ( ≤ ) By DeMorgan’s law, ( A ∩ B ) c = A c ∪ B c . Thus P (( A ∩ B ) c ) = P( A c ∪ B c ) ≤ P( A c ) + P( B c ) . c. ( ≥ ) Since p X ( x ) ≥ p X ( x ) p Y | X ( y | x ) = p X,Y ( x,y ). d. ( ≤ ) Note that p Z ( g ( y )) = summationdisplay y ′ : g ( y ′ )= g ( y ) p Y ( y ′ ) ≥ p Y ( y ) . Thus p Y ( y ) ≤ p Z ( g ( y )). e. ( ≥ ) This follows from the fact that Var( X ) = E ( X 2 )- ( E ( X )) 2 ≤ E ( X 2 ) . 2. a. The tree: a120 a1 a1 a1 a1 a1 a1 a1 P( E 1 ) = 0 . 2 a120 a0 a0 a0 a0 P( E 2 | E 1 ) = 0 . 4 a120 a64 a64 a64 a64 . 6 a120 a65 a65 a65 a65 a65 a65 a65 . 8 a120 a0 a0 a0 a0 . 1 a120 a64 a64 a64 a64 . 9 a120 a8 a8 a8 a8 P( E 3 | E 2 ) = 0 . 4 a72 a72 a72 a72 . 6 a8 a8 a8 a8 . 1 a72 a72 a72 a72 . 9 a8 a8 a8 a8 . 4 a72 a72 a72 a72 . 6 a8 a8 a8 a8 . 1 a72 a72 a72 a72 . 9 a120 E 1 , E 2 , E 3 a120 E 1 , E 2 , E c 3 a120 E 1 , E c 2 , E 3 a120 E 1 , E c 2 , E c 3 a120 E c 1 , E 2 , E 3 a120 E c 1 , E 2 , E c 3 a120 E c 1 , E c 2 , E 3 a120 E c 1 , E c 2 , E c 3 Page 2 of 6 EE 178, Winter 2008 b. The probability of the second bit being in error is: P( E 2 ) = P( E 1 )P( E 2 | E 1 ) + P( E c 1 )P( E 2 | E c 1 ) = 0 . 2 × . 4 + 0 . 8 × . 1 = 0 . 16 ....
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This note was uploaded on 09/29/2008 for the course EE 178 taught by Professor Hewlett during the Spring '08 term at Stanford.

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SampleMTsol - EE 178 Sunday Probabilistic Systems Analysis Handout#10 Sample Midterm Problem Solutions 1 a(= since P A ∪ B = P A P B P A ∩ B =

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