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Omk
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Posts: 30
Joined: August 14th, 2006, 11:57 pm

Poisson Process Paradox

September 18th, 2006, 12:44 pm

You are waiting for a bus.They arrive with a rate lambda (poisson process rate) acording to a poisson processes.If we want an average time of 10min during each buses , lambda = 1/(10min)If we arrive at a random time, logickly our average time of wait must be 5min (because you arrive between two buses).However, Poisson processes are independent of the past, so if you take a random time, your average wait is going to be 1/lambda = 10 min..... knowing that the time between two bus is still 10 min.So if you arrive at a random time, are you going to wait 10 or 5 min ?This is a well known paradox ... that is not one ....
 
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karakfa
Posts: 139
Joined: May 25th, 2002, 5:05 pm

Poisson Process Paradox

September 18th, 2006, 2:33 pm

the funny consequence is even after you wait 10 mins, the expected time is still unchanged: it's 10 more minutes.It helps to think in terms of lambda, the rate of arrival doesn't change with time so is the expected time.
 
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amit7ul
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Joined: December 7th, 2004, 8:36 am

Poisson Process Paradox

September 18th, 2006, 5:01 pm

Omk, the case you describe, that if one comes to bus-stop at random time then average time shud be 5 mins would actually be uniform distribution not poisson.
 
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Traden4Alpha
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Joined: September 20th, 2002, 8:30 pm

Poisson Process Paradox

September 18th, 2006, 8:48 pm

This isn't so much a paradox as it is a counterintuitive example -- we would think that if we've waited 10 minutes for a bus already, then we shouldn't have to wait another ten minutes. The problem isn't the Poisson process, its the idea that buses follow a Poisson process. As amit7ul suggests the real distribution isn't Poisson. Instead the frequency of the arrivals is fixed at 10 minutes each cycle and the phase of the cycle is drawn once from a uniform distribution on a 0 to 10 minute interval. (One could argue that the bus is 100% non-stochasitic, especially in Switzerland, and that our arrival at the bus stop is drawn from a uniform distribution on the 10-minute interval). With constant arrival frequency and uniformly distributed phase, the average wait is 5 minutes. In such as model, if we've waited nearly 10 minutes, our expected wait time would be nearly 0.You could also add a second distribution function for the slight scatter of arrival times to account for minor random delays en-route. This complicates the conditional analysis somewhat (the expected wait conditional on having already waited). If you want to be even more realistic, you could add in a non-zero probability of a breakdown. Breakdowns would mean that if you've waited 10 minutes and not seen the bus, then you have fairly high probability evidence of a breakdown and that the actual wait will be 10 more minutes (assuming the next bus isn't delayed due to handling double the average passenger volume to make up for the broken bus). Actual public transportation system can have both positive and negative feedback loops, so a full model can be quite complex.
 
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gentinex
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Joined: June 8th, 2006, 1:16 pm

Poisson Process Paradox

September 19th, 2006, 1:04 am

amit7ul and Traden4Alpha, both of you changed the problem to suit how you "think" that buses should arrive somewhere, by saying that if a bus has just arrived, then it will take the next bus uniform(0,10) time to arrive. To say that the buses arrive according to a Poisson process, as omk originally did, is a well-defined problem (whether or not it approximates reality), and you should do that problem on its merits, rather than say that omk's example is "wrong."(To formulate the problem equivalently, without getting caught up in one's own preconceptions about buses, say that you're away from home, and phone calls are made to your home as some Poisson process with a rate lambda, and you're trying to figure out how long it will take on average, for the phone to ring after you return home.)Without getting into all the gory details (which one can find in, e.g., the first chapter of the second volume of Feller's probability book), the answer to omk's problem is 10. Intuitively, if you think of all the intervals between bus arrivals, you're more likely to come to the bus stop at a long interval rather than a short interval, and thus the average length of the interval in which you come is not 10, but rather 20, which is the number you have to halve to get the correct answer of 10.On a separate note, computer simulations make me believe that the solution to the problem posed by amit7ul and Traden4Alpha (buses come at uniform(0,10) times; how long do you have to wait after you come?) is not 5. Because of length-biasing of intervals as with the Poisson process above, I'd have expected it to be more than 5, but surprisingly, I keep getting answers around 10/3. I don't know why this is, though.EDIT: the mean of a uniform(0,10) is 5, not 10. So the naive answer would be to halve 5, and get 5/2 as the mean waiting time. But again, because of length-biasing of intervals as in the Poisson case, the expected length of the interval should be longer than 5/2, and indeed 10/3 makes sense given this.
Last edited by gentinex on September 18th, 2006, 10:00 pm, edited 1 time in total.
 
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Omk
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Joined: August 14th, 2006, 11:57 pm

Poisson Process Paradox

September 19th, 2006, 11:25 pm

gentinex you are perfectly correct ! THe trick is to understand than when you pick a time at random between two buses, you are more likely to take a long one. It changes everything ! And BTW, people modelise this kind of process by a poisson process even if it seems weird. The problem i pick is a real problem in life.If you would want to estimate an average life time of anything (like light buble) by taking light buble at random time (instead of random light buble) ou will increase its life time a lot !
 
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ge
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Joined: August 1st, 2006, 12:38 pm

Poisson Process Paradox

September 22nd, 2006, 7:53 am

for the uniform distribution 0 to 10, i get the answer 10/3 also.consider the lengths of times between buses, t, this has distribution p(t) = 1/10.now, the probability that you pick an interval of length t, Ppick(t) = t p(t) /<t> = t/50.the mean picked interval is thus (10^3)/(3*50) =20/3 ( from integrating Ppick(t)*t ).But, when you pick an interval, the time you wait until the end of it is on average half of it's length, which gives the answer 10/3 which is what gentinex's simulations found.
 
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timeds
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Joined: October 6th, 2005, 10:50 am

Poisson Process Paradox

September 29th, 2006, 3:36 pm

I find life expectancy to be a good way to explain this. The analogy with busses is hopefully obvious.Suppose the life expectancy of an average person is 70 years. This doesn't mean that the average 69 year old has one year to live, their life expectancy will be greater because it is impossible for them to die in childhood.When you meet an average person, they are unlikely to be a newborn. Consequently the average life expectancy of a random person is greater than 70.It should be easy to see why this means that on picking a random person, on average you will have to wait more than 35 years for them to die, and why on average you have to wait longer than 5 minutes for the bus.