6.262.PS4.sol

6.262.PS4.sol - 61/562 Discrete Stochastic Processes Spring 2010 Problem Set 4 Solutions due Friday March 5 2010 Problem 1(Exercise 2.22 a Each of

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61/562 Discrete Stochastic Processes, Spring 2010 Problem Set 4 — Solutions due: Friday, March 5, 2010 Problem 1 (Exercise 2.22) a Each of the 1000 voters vote independently for A with probability 1 / 2 each, so P ( n votes for A | 1000 voters) = ± 1000 n ² (1 / 2) n (1 / 2) 1000 - n = ± 1000 n ² (1 / 2) 1000 for n = 0 , 1 ,..., 1000. b Given the condition of 1000 arrivals in the first ten hours, each arrival can be regarded as uniformly and independently distributed over the 10 hour period, and thus each arrives in the first four hours independently with probability 4 / 10. Thus each of the 1000 independently arrives in the first four hours and votes for A with probability (4/10)(1/2)=1/5, so P ( n votes for A in 4 hrs | 1000 voters in 10 hrs) = ± 1000 n ² (1 / 5) n (4 / 5) 1000 - n for n = 0 , 1 1000 . Comment : Alternatively one can do it by conditioning on the number of voters in the first 4 hours. P ( n votes for A in 4 hrs | 1000 voters in 10 hrs) = 1000 X m = n P ( n votes for A in 4 hrs | m voters in 4 hrs , 1000 in 10 hrs) P ( m in 4 hrs | 1000 in 10 hrs) = 1000 X m = n ± m n ² (1 / 2) m ± 1000 m ² (4 / 10) m (6 / 10) 1000 - m = 1000! (1000 - n )! n ! 1000 - n X i =0 (1000 - n )! i !(1000 - n - i )! (1 / 5) n (1 / 5) i (6 / 10) 1000 - n - i (setting i = m - n ) = ± 1000 n ² (1 / 5) n (4 / 5) 1000 - n (using the binomial expansion to simplify the summation) c The voters split into two Poisson processes, one for A and one for B , each of rate 50. Letting T be the arrival epoch of the first voter for A , f T ( t ) = 50 e - 50 t for t 0. d Assume voting goes on indefinitely. The probability there are n B voters before the first A voter is equal to the probability that the first n voters are split as B voters followed by a voter for A and this is equal to (1 / 2) n (1 / 2) = 2 - ( n +1) , n 0. 1
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e Given an A B reversal, the time to the next reversal is the time to the next A voter. This time is exponential with rate 50 and independent of all previous arrivals and thus reversals. Given a B A reversal, the time to the next reversal is again exponential with the same parameter. So, starting with some voter the inter-arrival time between reversals is i.i.d exponentially distributed. However if we look at the time to the first reversal then it includes the time until the first voter plus the time until reversal which is the sum of two exponentials one of rate 100 and the other of rate 50. Hence the distribution of time until the first reversal is not the same as the other inter-arrival times between reversals. Thus, the reversal counting process is not a Poisson process. Instead, we will consider it to be a delayed Poisson process , in the sense that restarting the counting process after the first reversal yields a Poisson process. (Do we have to restart the process immediately after the first reversal for
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This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.

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6.262.PS4.sol - 61/562 Discrete Stochastic Processes Spring 2010 Problem Set 4 Solutions due Friday March 5 2010 Problem 1(Exercise 2.22 a Each of

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