solution_set4 - EE 511 Solutions to Problem Set 4 1. (i) φ...

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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.

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solution_set4 - EE 511 Solutions to Problem Set 4 1. (i) φ...

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