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There is an old conundrum in queueing theory that goes like this:
  • A passenger arrives at a bus-stop at some arbitrary point in time
  • Buses arrive according to a Poisson process
  • The mean interval between the buses is 10 min.
What is the mean waiting time until the next bus?
busline
Answer: 10 min. This is an example of length-biased sampling. The explanation of the paradox lies therein that the passengers' probability to arrive during a long interarrival interval is greater than during a short interval. ] Here is a neat non-technical explanation (taken from this book). [

Given the interarrival interval, within that interval the arrival instant of the passanger is uniformly distributed and the expected waiting time is one half of the total duration of the interval. The point is that in the selection by the random instant the long intervals are more frequently represented than the short ones (with a weight proportional to the length of the interval).

Consider a long period of time t. The waiting time to the next bus arrival W(τ) as a function of the arrival instant τ of the passenger is represented by:
waittwhere the Xi are the interarrival intervals. The mean waiting time, W_bar, is the average value of this sawtooth curve:waitt01
Note that long interarrival intervals contribute much more than short ones to the average waiting time. As t grows, t/n -> X_bar, hence,
waitt02For exponential distribution (as the Xi are distributed),
waitt03Therefore,
waitt04Altogether,
waitt05
Q.E.D.

Sources:
Advanced Course in Operating Systems (University of Haifa), Lecture 1 & 2
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John (guest) meinte am 17. Jan, 01:25:
Close, but not quite
Hey, I've been puzzling over this for the last couple of days. The spirit of your argument is correct, but your result is a bit off.

You're correct up to your formula for the average wait time (although I'm not quite sure I agree with your reasoning... but you do end up with the correct formula):

W_bar=(1/2) /

but then you say that the variance of a poisson process is ^2 when in fact, it's just . So,

=+^2

and your final value for your average wait time should be:

w_bar=(1/2)*(1+).

I simulated this problem in matlab with the code for an average time between buses of 10 minutes as in your problem. We would expect to get a value of about 5.5 minutes (using my formula) and we do.

Great problem though. Gave me a something to think about.

Matlab Code:

count=1;
wait_time=zeros(10000,1);
while (count<10000)
bus_time=poissrnd(10);
ped_arrival=randi(25);

if ped_arrival<bus_time
wait_time(count)=bus_time-ped_arrival;
count=count+1;
end
end
avg=mean(wait_time);
disp('The average wait time is:')
disp(avg) 
John (guest) meinte am 17. Jan, 01:26:
Sorry
My formulas didn't go through! 
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