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### 131A_1_ds3_2011fall_sol

Course: EE 132A, Fall 2011
School: UCLA
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Word Count: 456

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131A Probability Instructor: EE Lara Dolecek Discussion Set 3 Wednesday, October 12, 2011 and Friday, October 14, 2011 Reading: Chapters 3 and 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia Concepts covered: What is a random variable ? What is PMF ? What is mean and variance ? Examples of important discrete random variables. Problems: 1. Coin tossing example and RV. Problem 3.9, page...

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131A Probability Instructor: EE Lara Dolecek Discussion Set 3 Wednesday, October 12, 2011 and Friday, October 14, 2011 Reading: Chapters 3 and 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia Concepts covered: What is a random variable ? What is PMF ? What is mean and variance ? Examples of important discrete random variables. Problems: 1. Coin tossing example and RV. Problem 3.9, page 131 of ALG Solution:(a) Let the random variable X be the number of heads, then SX = {0, 1, ..., n}. Then the random variable Y , which is dierence between the number of heads and the number of tails in the n tosses of a coin, is given by Y = X (n X ) = 2X n. Therefore the sample space SY is given by SY = {n, n + 2, ...n 2, n} n (b) P [Y = 0] = P [2X n = 0] = P [X = n/2] = n/2 pn/2 (1 p)n/2 for n is even (c) P [Y = k ] = P [2X n = k ] = (n+n )/2 p(n+k)/2 (1 p)(nk)/2 for n+k is even k Problem 3.26, page 133 of ALG Solution: Note Y = 2X n and X is the number of heads in n tosses of a coin, which is the binomial random variable. In the class, we already proved E (X ) = np and V ar(X ) = np(1 p). Therefore E (Y ) and V ar(Y ) are given by E (Y ) = E (2X n) = 2E (X ) n = 2np n V ar(Y ) = V ar(2X n) = 4V ar(X ) = 4np(1 p) In a large number repetitions of of this random experiment, we expect the dierence between the number of heads and the number of tails to be E (Y ). 1 2. Memoryless property of Geometric RV. Problem 3.54, page 136 of ALG Proof : Let M be a geometric random variable. Then P [M = k ] = (1 p)k1 p k = 1, 2, ... The probability that M k can be written in closed form: k (1 p)j 1 p = (1 p)k P [M > k ] = 1 P [M k ] = 1 j =1 P [M k + j, M > j ] P [M k + j ] = for k 1 P [M > j ] P [M > j ] (1 p)k+j 1 = (1 p)j = (1 p)k1 = P [M k ] P [M k + j |M > j ] = Therefore, we show geometric random variable satises the memoryless property. 3. Poisson RV and examples. Problem 3.61, page 136 of ALG Solution: the mean and variance of a Poisson random variable. k k1 k k e = E [X ] = e = e = k! (k 1)! k! k=0 k=1 k=0 2 k E [X ] = k=0 2 k k! e k=1 = [ k=2 k = k=1 (k 1) = [ k 1 e (k 1)! k 1 k 1 e + e] (k 1)! (k 1)! k=1 k2 k1 e+ e] (k 2)! (k 1)! k=1 = ( + 1) 2 X = E [X 2 ] E [X ]2 = Problem 3.59, page 136 of ALG Solution: =6000 reqs/min=100 reqs/s and = 10 = 10 reqs/s. Then the probabilities can be found as follows: (a) P [N = 0] = e = e10 = 4.54 105 2 (b) 10 P [5 N 10] = k=5 3 k 10 e = 0.554 k!
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