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131A_1_ds10_2010fall_Sol

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

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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

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3. CLT. Problem 7.29, page 405 of ALG. Let the random variable
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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-μ 10 σ = 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: μ = 3 . 5. The variance of a die toss is: σ 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-σ 2 n × ε 2 ≥ 1-2 . 91 20 × . 5 2 = 0 . 418 2...
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131A_1_ds10_2010fall_Sol - X i represent an error for the i...

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