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hmwk1sol

# hmwk1sol - ISyE 3232 Stochastic Manufacturing and Service...

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ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 Professors Hayriye Ayhan and Jim Dai Solutions to Homework 1 1. (a) Poisson (b) exponential (c) geometric (d) Bernoulli (e) binomial (f) normal 2. We are given that E [ X ] = 2 and V ar ( X ) = 16. Thus a) E [6 - 2 X ] = 6 - 2 E [ X ] = 2 and V ar (6 - 2 X ) = 4 V ar ( X ) = 64. b) E [( X - 3) / 4] = 1 4 E [ X ] - 3 4 = - 1 / 4 and V ar (( X - 3) / 4) = 1 16 V ar ( X ) = 1. 3. We know that a) 1 = 3 X k =1 P ( X = 2 k - 1) = 3 X k =1 (2 k - 1) c = 9 c so it follows that c = 1/9. b) E [ X ] = 3 i =1 (2 k - 1) P ( X = 2 k - 1) = (1 / 9)(1 + 9 + 25) = 35 / 9 . c) E [ X 2 ] = 3 i =1 (2 k - 1) 2 P ( X = 2 k - 1) = (1 / 9)(1 + 27 + 125) = 17 . d) V ar ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 17 - (35 / 9) 2 = 1 . 8765 . e) Note that ( X - 2) + = 0 with probability 1 / 9 + 3 / 9 2 with probability 5 / 9 Hence, E [( X - 2) + ] = 2 * 5 / 9 = 10 / 9 . 4. a) P ( X = k ) = 1 k e - 1 /k ! b) E ( X ) = 1 c) V ar ( X ) = 1. d) Suppose 0 k 1, k integer. Then P ( Y = k ) = P ( min ( X, 2) = k ) = P ( X = k ) = 1 k e - 1 /k ! P ( Y = 2) = P ( min ( X, 2) = 2) = X k =2 P ( X = k ) = 1 - 1 X k =0 1 k e - 1 /k ! . e) E [ Y ] = 2 X k =0 kP ( Y = k ) = e - 1 + 2 - 2 1 X k =0 1 k e - 1 /k ! . 5. a) We know that 1 = Z 0 ce - 2 s ds = c/ 2 so it follows that c = 2 (thus, Y is an exponential random variable with rate 2).

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hmwk1sol - ISyE 3232 Stochastic Manufacturing and Service...

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