6.262.PS3.sol

6.262.PS3.sol - 6.262 Discrete Stochastic Processes, Spring...

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Unformatted text preview: 6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 3 Solutions due: Friday, February 26, 2010 Problem 1 (Exercise 2.12) Suppose that buses arrive as a Poisson process with rate with interarrival times denoted by X 1 ,X 2 ,... . Suppose also that commuters arrive as a Poisson process with rate and interarrival times denoted by Y 1 ,Y 2 ,... . The two processes are independent, so they can be viewed as a splitting of a single Poisson process with rate + . a Let N m be the number of passengers on the m th bus. Let t be the time of the previous bus arrival and note that the event { N m = n } = { n passenger arrivals before the next bus arrival } . Due to the memoryless property of the Poisson process, the time t at which we start counting is irrelevant and were simply interested in the probability that out of the first n +1 arrivals of the combined process, the first n arrivals were passengers. Thus, the desired pmf is p N m ( n ) = + n + for n Z + . The mean of N m is + - 1 = (can you see how to obtain this quickly by comparing the above pmf with that of a geometric random variable?). Intuitively, since is the rate of the passenger arrivals and 1 / the average time till the next bus arrival, / ought to give an average, in some sense (not necessarily in the sense of expectation), of the number of passengers on the next bus. We will develop this idea further in Chapter 3 for arbitrary renewal processes. b Given that it takes x units of time for the next bus to arrive, and given that bus arrivals and customer arrivals are independent, the number of customers entering that bus is simply the number of customers arriving during some period [ t ,t + x ]. The desired pmf then becomes p N m | X m ( n | x ) = ( x ) n e- x n ! for n Z + The mean of N m conditional on X m is x , which (rather intuitively again) happens to be the product of the length of the time interval and the rate of customer arrivals. c The argument in part a) applies equally well letting t be 10:30 and letting it be 3 am three thousand years ago. Thus, letting m refer to the first bus arriving after 10:30, we have p N m ( n ) = + n + for n Z + . The mean of N m is again . d Now were told that the last bus arrived at 10:30am and that there were no other busses arriving before 11am. Let N A denote the number of commuters arriving between 10:30am 1 and 11am, and let N B denote the number of commuters arriving between 11am and the arrival of the bus. Since the customer arrival process is independent of the bus arrival process and since the customer arrival process has independent increments, it follows that N A and N B are independent. Assuming arrivals per hour, from part b) we have p N A ( n ) = ( / 2) n e- / 2 n !...
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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.PS3.sol - 6.262 Discrete Stochastic Processes, Spring...

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