solution09

solution09 - Solutions to Homework 09 1. a. Since X is...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.

Page1 / 3

solution09 - Solutions to Homework 09 1. a. Since X is...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online