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midtermsol_2003 - MS&E 221 Stochastic Modeling Prof Peter W...

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MS&E 221 Midterm Examination Solutions Stochastic Modeling 21st February 2003 Prof. Peter W. Glynn Page 1 of 4 Midterm Examination Solutions 1. (30 points) Consider a driver for whom the probability of having an accident in any given year depends on whether she had an accident the year before. Specifically, the probability of her having an accident in a particular year given that she did not have an accident the year before is 0 . 1, while the probability of her having an accident given that she had an accident the year before is 0 . 6. (a) (4 points) Assuming our driver did not have an accident in 2002, what is the expected number of years till her next accident? The time until her next accident is a geometric r.v. with success probability 0.1, so its expected value is 10. (b) (5 points) Suppose that our driver is currently 20 years old. What is the probability that our driver will have more than 14 accidents by the time she is 70? Consider the Markov Chain Y = ( Y n : n 0) on state space { A, N } where Y n = A if the driver has an accident in year n and Y n = N otherwise. Its graph representation is A B 0.6 0.9 0.1 0.4 We use a CLT-based approximation for K ( n ) 4 = n - 1 j =0 I ( Y j = A ) , (i.e. for the number of accidents she has had by year n ). Let f = ( 1 0 ) be the corresponding “reward function” and π be the stationary probabilities. Solving π = πP , πe = 1 , π 0 we get π ( A ) = 1 / 5 , π ( N ) = 4 / 5 . Hence, the long term average reward per year (mean number of accidents per year) is α 4 = πf = 1 / 5 . Let ˆ f be the centered reward, ˆ f = 4 / 5 - 1 / 5 · . From the CLT from DTMC we know that K ( n ) - σ n = N (0 , 1) as n -→ ∞ , where σ 2 = π ( A ) E A T A - 1 j =0 ˆ f ( X j ) · 2 and T A 4 = inf { j 1 : Y j = A } .
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