04q1sol

04q1sol - URBAN OR QUIZ 1 SOLUTIONS The key to this problem...

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U RBAN OR Q UIZ 1 S OLUTIONS The key to this problem is to understand random incidence. The best way to approach the problem is to view the interval h in which Mendel arrives as two pieces. These are: 1. The length of time(t) from the last bus till Mendel arrives 2. The length of time(u) from when Mendel arrives to the next bus. Thus E [ u ] + E [ t ] = E [ h ] Clearly the sum of these two expected values is the expected value of time between busses for a randomly chosen interval. Let’s answer the second one first. Since arrivals are Poisson and the Poisson process is memoryless the time from when Mendel arrives to the next bus is clearly a Poisson random variable with parameter . 1 Thus E [ u ] = . Now in order to evaluate quantity one, we have to realize that a Poisson process is reversible in time. Thus we can view the variable u alternatively as the time between when Mendel arrives till the previous bus arrived. Hence we can look forward and backwards in time. Thus looking at t this way it is clear that u is also a Poisson variable with parameter . 1 Thus E [ t ] = . Combining these two we can find the expected value of h. 1 1 2 E [ h ] = E [ t ] + E [ u ] = + = Alternately, we can use our formula for the mean time between Mendel’s arrival and the end of his interval, based on the mean and variance of the distribution of intervals. Then, knowing that on average he arrives in the middle of the interval, we double the calculated result to get the mean total interval. Once again, the answer is 2/ . 2. We let:

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This note was uploaded on 11/08/2011 for the course AERO 16.72 taught by Professor Hansman during the Fall '06 term at MIT.

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04q1sol - URBAN OR QUIZ 1 SOLUTIONS The key to this problem...

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