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Sept 2014 - Q7(vii)
- Thread starter omurice
- Start date
Hello
Good question! We can use a Poisson process when we are just counting the arrivals of something but we need a compound Poisson process when each of those arrivals have an associated value (here the toll).
Cars
Let's consider a slightly simpler question that addresses the same issue:
Why can't we model the total amount of the toll collected from cars over time (measure in minutes) as a Poisson process with the per minute rate being 10, which comes comes from 2 * 5?
Well one reason we can see this doesn't work is the state space. The possible values that the total amount of tolls can take is 0, 2, 4, 6, 8, 10, 12, ... etc. However, the Poisson process can take values 0, 1, 2, 3, 4, 5, ... . So, for example, if we use the Poisson process as a model, the probability of collecting £3 in tolls in any given minute is non-zero, where as it is 0 in reality.
The reason this doesn't work is that the Poisson process takes the £2 collected per car and 'breaks it down' into the individual £1 elements collected by effectively doubling the rate. So using a Poisson process here is almost like counting the number of 'half-cars' that arrive!
All vehicles
When it comes to all vehicles, as in part (vii), we can't make the same argument using the state space because the correct process has state space 0, 1, 2, 3, 4, ... etc just like the Poisson process with parameter 19.5.
However, hopefully you can agree that we have the same problem. The Poisson process with parameter 19.5 is counting 'half-cars' and 'lorry-fifths' at twice and five times the rates of car and lorry arrivals, respectively. This is not the same as modelling £2 of income for each car arrival and £5 of income for each lorry arrival.
For the probability in question, when using the Poisson process, P(X=4) takes account of the following impossible scenarios:
4/5ths of a lorry arrival
2/5ths of a lorry arrival and a car
1/2th of a car arrival and 3 motorbikes
etc
Even though the Poisson process includes these additional scenarios, note that the overall probability is actually lower in this case:
> dpois(4,19.5)
[1] 2.047316e-05
This is because, for example, the scenario leading to £4 being collected from 4 motorcycles has probability:
P(4 motorcycles and 0 'truck fifths' and 0 'car halfs') (1)
for the Poisson process, compared to:
P(4 motorcycles and 0 trucks and 0 cars) (2)
for the compound process and (1) < (2) ('truck fifths' arrive at a much higher rate than trucks so the probability of seeing 0 is lower).
Hope this helps!
Andy
Good question! We can use a Poisson process when we are just counting the arrivals of something but we need a compound Poisson process when each of those arrivals have an associated value (here the toll).
Cars
Let's consider a slightly simpler question that addresses the same issue:
Why can't we model the total amount of the toll collected from cars over time (measure in minutes) as a Poisson process with the per minute rate being 10, which comes comes from 2 * 5?
Well one reason we can see this doesn't work is the state space. The possible values that the total amount of tolls can take is 0, 2, 4, 6, 8, 10, 12, ... etc. However, the Poisson process can take values 0, 1, 2, 3, 4, 5, ... . So, for example, if we use the Poisson process as a model, the probability of collecting £3 in tolls in any given minute is non-zero, where as it is 0 in reality.
The reason this doesn't work is that the Poisson process takes the £2 collected per car and 'breaks it down' into the individual £1 elements collected by effectively doubling the rate. So using a Poisson process here is almost like counting the number of 'half-cars' that arrive!
All vehicles
When it comes to all vehicles, as in part (vii), we can't make the same argument using the state space because the correct process has state space 0, 1, 2, 3, 4, ... etc just like the Poisson process with parameter 19.5.
However, hopefully you can agree that we have the same problem. The Poisson process with parameter 19.5 is counting 'half-cars' and 'lorry-fifths' at twice and five times the rates of car and lorry arrivals, respectively. This is not the same as modelling £2 of income for each car arrival and £5 of income for each lorry arrival.
For the probability in question, when using the Poisson process, P(X=4) takes account of the following impossible scenarios:
4/5ths of a lorry arrival
2/5ths of a lorry arrival and a car
1/2th of a car arrival and 3 motorbikes
etc
Even though the Poisson process includes these additional scenarios, note that the overall probability is actually lower in this case:
> dpois(4,19.5)
[1] 2.047316e-05
This is because, for example, the scenario leading to £4 being collected from 4 motorcycles has probability:
P(4 motorcycles and 0 'truck fifths' and 0 'car halfs') (1)
for the Poisson process, compared to:
P(4 motorcycles and 0 trucks and 0 cars) (2)
for the compound process and (1) < (2) ('truck fifths' arrive at a much higher rate than trucks so the probability of seeing 0 is lower).
Hope this helps!
Andy
