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Unformatted text preview: EE 178 Handout #24 Probabilistic Systems Analysis Sunday, March 16, 2008 Sample Final Solution 1. a. =. Consider P ( n i =1 A i ) = 1 P ( n i =1 A c i ) = 1 n productdisplay i =1 P( A c i ) , by independence = 1 (1 P( A )) n = 1 2 2 n . b. . For each Z = z , Var( X + Y  Z = z ) = Var( X  Z = z ) + 2Cov( X,Y  Z = z ) + Var( Y  Z = z ) = Var( X  Z = z ) + Var( Y  Z = z ) , by independence . Thus E[Var( X + Y  Z )] = E[Var( X  Z )] + E[Var( Y  Z )]. But since E(Var( X  Z )) Var( X ) and E(Var( Y  Z )) Var( Y ) (from the law of conditional variances), it follows that E [Var( X + Y  Z )] Var( X ) + Var( Y ) . c. . Let Z = X 2 , then E( Z 2  Y ) = E( X 4  Y ) and E( Z  Y ) = E( X 2  Y ). The result follows by the fact that the second moment is larger than or equal to the square of the mean. d. =. Using the moment generating property of the moment generating function, we know that if X N (0 , 2 ), then E( X 8 ) = 7 5 3 8 . Thus E( X 8 ) = 105 2 4 = 1680. e. NONE. Uncorrelation does not necessarily imply independence, which is needed for equality. f. . Using the Markov iequality P { XY < 16 } 1 E( XY ) 16 = 1 1 4 = 3 4 . 2. We find the pdf by first finding the cdf (via the complementary cdf) and then differenti ating it. The complementary cdf can be found as follows P { Z > z } = P { Z > z  B = 0 } P { B = 0 } + P { Z > z  B = 1 } P { B = 1 } = P { min { X 1 ,X 2 } > z } P { B = 0 } + P { max { X 1 ,X 2 } > z } P { B = 1 } = P { X 1 > z,X 2 > z } P { B = 0 } + (1 P { X 1 z,X 2 z } )P { B = 1 } = 1 2 e 1 z e 2 z + 1 2 bracketleftbig 1 ( 1 e 1 z )( 1 e 2 z )bracketrightbig = 1 2 ( e 1 z + e 2 z ) for z . Thus, f Z ( z ) = d dz braceleftbigg 1 1 2 ( e 1 z + e 2 z ) bracerightbigg = 1 2 ( 1 e 1 z + 2 e 2 z ) for z ....
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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.
 Spring '08
 Hewlett

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