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Samplefinalsol

Samplefinalsol - EE 178 Probabilistic Systems Analysis...

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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 , σ 2 ), then E( X 8 ) = 7 × 5 × 3 × σ 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 λ 1 z e λ 2 z + 1 2 bracketleftbig 1 - ( 1 - e λ 1 z ) ( 1 - e λ 2 z )bracketrightbig = 1 2 ( e λ 1 z + e λ 2 z ) for z 0 .
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Thus, f Z ( z ) = d dz braceleftbigg 1 - 1 2 ( e λ 1 z + e λ 2 z ) bracerightbigg = 1 2 ( λ 1 e λ 1 z + λ 2 e λ 2 z ) for z 0 . 3. Note that U and V are independent. a. We first note that, when either x < 0 or u < 0, F X,U ( x, u ) = 0, and when both x > 1 and u > 1, F X,U ( x, u ) = 1. When 0 u x 1, F X,U ( x, u ) = P { X x, U u } = P { U u } = u. When 0 x u 1, F X,U ( x, u ) = P { X x, U u } = P { UV x, U u } = integraldisplay x 0 P { V 1 } du + integraldisplay u x P braceleftBig V x u bracerightBig du = x - integraldisplay u x x u du = x - x ln( x/u ) .
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