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Unformatted text preview: Solutions to Homework 09 1. a. Since X is uniform in the range [ b, b ), we have F X ( x ) = x < b x + b 2 b b ≤ x < b 1 x ≥ 1 (1) and m = E [ X ] = 0 and σ 2 = V AR [ X ] = b 2 / 3. Now, P [  X m  > c ] = P [ c m > X > m + c ]. Thus we get P [  X  > c ] = ( F X ( c ) F X ( c ) ≤ c ≤ b x > b = ( 1 c b ≤ c ≤ b x > b (2) The Chebychev bound gives P [  X m  > c ] ≤ σ 2 c 2 = b 2 3 c 2 (3) b. For the Laplacian random variable, m = E [ X ] = 0 and σ 2 = V AR [ X ] = 2 /α 2 and its pdf is given by f X ( x ) = α 2 e α  x  ∞ < x < ∞ (4) Then, the exact probability is given by P [  X m  > c ] = Z c + m c m α 2 e α  x  dx = e αc (5) The Chebychev bound gives P [  X m  > c ] ≤ σ 2 c 2 = 2 α 2 c 2 (6) c. For our Gaussian random variable, m = E [ X ] = 0 and σ 2 = V AR [ X ] and its pdf is given by f X ( x ) = 1 √ 2 πσ 2 e x 2 2 σ 2 ∞ < x < ∞ (7) Then, the exact probability is given by P [  X  > c ] = F...
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This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.
 Fall '08
 GELFAND

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