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Unformatted text preview: EE 511 Solutions to Problem Set 4 1. (i) φ X ( s ) = e s 2 / 2 . (ii) We have P [ X ≥ a ] = e as φ X ( s ) for all s > . This upper bound should be minimized with respect to s to obtain the Chernoff bound. e as φ X ( s ) = e as e s 2 / 2 Setting the derivative with respect to s to 0, we get se as e s 2 / 2 + ( a ) e as e s 2 / 2 = 0 i.e., s = a . The second derivative at s = a can be shown to be positive. Therefore, the Chernoff bound is given by P [ X ≥ a ] ≤ e a 2 / 2 (iii) From the Chebyshev inequality, we get P [  X  ≥ a ] ≤ 1 a 2 . Since f X ( x ) is symmetric, we get P [ X ≥ a ] ≤ 1 2 a 2 . 2. E [ Z ] = E [ X ] + aE [ Y ] = 0. E [ X  Y = y ] = E [ Z  Y = y ] aE [ Y  Y = y ] = E [ Z ] ay = ay. 3. E [ X ] = Z 100 xf X ( x ) dx = Z 100 x 100 dx = 50 . Given that X ≥ 65, X is uniformly distributed in [65 , 100]. Therefore, we have E [ X  X ≥ 65] = Z 100 65 xf X ( x  X ≥ 65) dx = Z 100 65 x 35 dx = 82 . 5 . 4. E [ X ] = ∞ X k =0 ke a a k k ! . We know ∞ X k =0 e a a k k ! = 1 . (1) Differentiating with respect to a , we get ∞ X k =0 ke a a k 1 k ! e a ∞ X k =0 a k k ! = 0 . 1 1 a E [ X ] = e a ∞ X k =0 a k k ! = 1 . Therefore, we have E [ X ] = a . Differentiating (1) twice with respect to a , we get ∞ X k =0 k ( k 1) e a a k 2 k ! ∞ X k =0 ke a a k 1 k ! ∞ X k =0 ke a a k 1 k ! + ∞ X k =0 e a a k k ! = 0 1 a 2 ∞ X k =0 k 2 e a a k k ! 1 a 2 ∞ X k =0 ke a a k k !...
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This note was uploaded on 10/26/2011 for the course ECE 171 taught by Professor Tu during the Spring '11 term at Aarhus Universitet.
 Spring '11
 TU

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