131A_1_ds10_2010fall_Sol

131A_1_ds10_2010fall_Sol - X i represent an error for the i...

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

View Full Document Right Arrow Icon
EE 131A Discussion Set 10 Probability Wednesday, December 1, 2010 Instructor: Lara Dolecek Friday, December 3, 2010 Reading: Chapter 7of Probability, Statistics, and Random Processes by A. Leon-Garcia 1. Sum of RVs. Problem 7.2, page 402 of ALG. Let the mean of the random variables written as E ( X ). The mean is given by: E ( S n ) = E ( X i X i ) = n × E ( X ) The variance is given by: V ar ( S n ) = i j E [( X i - E ( X i ))( X j - E ( X j ))] = nE [( X - E ( X )) 2 ] + ∑∑ i 6 = j E [( X i - E ( X i ))( X j - E ( X j ))] The summation ∑∑ i 6 = j only has non-zero values when | i - j | = 1, and there are 2( n - 1) such terms. Therefore, we have: V ar ( S n ) = 2 + 2( n - 1) ρσ 2 2. Sum of 2 exponentials. Problem 7.7, page 403 of ALG. (a) Since X and Y are independent, the characteristic function is given as: Φ Z ( ω ) = Φ X ( ω Y ( ω ) = 2 2 - × 10 10 - = 20 ω 2 - 12 + 20 (b) Following the result from above, we have: Φ Z ( ω ) = 20 ω 2 - 12 + 20 = 2 . 5 2 - - 2 . 5 10 - So we have f Z ( t ) = 2 . 5( e - 2 t - e - 10 t ) , t 0. 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
3. CLT. Problem 7.29, page 405 of ALG. Let the random variable
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: X i represent an error for the i th bit. Then we have E ( X i ) = 0 . 15 and V ar ( X i ) = 0 . 1275. The total number of error bits is given as S 100 = X 1 + + X 100 , with the normalized form: Z 100 = S 100- 10 = S 100-15 3 . 57 Therefore we have the probability of having 20 or less bit errors to be: P ( S 100 20) = P ( Z 100 20-15 3 . 57 ) 1-Q (1 . 4) = 0 . 92 4. Chebyshev bound . Problem 7.17 , page 404 of ALG The mean of a die toss is: = 3 . 5. The variance of a die toss is: 2 = 2 . 92. We can then represent the probability of having the sample mean between 60 / 20 = 3 and 80 / 20 = 4 by Chebyshev inequality as: P ( | M n-3 . 5 | < . 5) 1- 2 n 2 1-2 . 91 20 . 5 2 = 0 . 418 2...
View Full Document

This note was uploaded on 02/03/2011 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

Page1 / 2

131A_1_ds10_2010fall_Sol - X i represent an error for the i...

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