Poisson Distribution -- Why No Need To Measure Standard Deviation?

Question

Elementary Statistics (Larson 4th ed) gives this example of Poisson, typical in my opinion. The mean number of accidents per month at a certain intersection is three. What is the probability that in any given month four accidents will occur at this intersection?

That's all that's given, and somehow it's sufficient to find the probability. But it seems common sense that some intersections (with an average of three per month) are more consistently three per month and therefore rarely see four in a month. While other intersections (with an average of three per month) see two, three or four with equal frequency, and often see four in a month. Poisson claims that this is not true: both of these intersections have the same probability of four accidents in the next month, because they both average three per month. How is it possible that this question does not arise, we need not measure the variance, and we just push on?

I'm asking for an explanation that is (as nearly as possible) in English sentences rather than computations.

Answer 1

The mean and the variance of the Poisson distribution cannot be chosen independently. If the mean is $3$ then the variance is also $3$. (If the mean is $\lambda$ then the variance is also $\lambda.$) And the probability that $k$ axidents occur is

$$\frac{\lambda^k}{k!}e^{\lambda}.$$

Even the formula shows that there is only one parameter.

So, if the Poisson approach is good enough to describe the statistics of an intersection from the point of view of axidents then there are no different intersections of the same mean.

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The statement that the Poisson distribution is not good enough is not a mathematical statement.

You have to go out to find two different intersections and do measurements. If you find that the average number of accidents is different then go and find another intersection. If finally you find two "equimean" intersections then calculate the relative frequencies of the number of accidents.

If the relative freqeanties of the same number of accidents are significantly different then the Poisson distribution is not appropriate in the case of your road crossings.

There are phenomena in the case of which the Poisson distribution works quite well. Such a case is the number of accidents at road crossings.

Answer 2

The Poisson distribution is based on a fundamental assumption:

For a "short" period of time $\Delta t$, the average number of events $\Delta N$ that occur is proportional to $\Delta t$: $\Delta N = \lambda \Delta t$. Moreover, this proportionality $\lambda$ is independent of time and of the number of events that have already occurred.

To see how this constrains the distribution of events in a given interval, let's look at a different distribution with the same mean: the intersection always has three accidents per month. No more, no less. In other words, the variance is exactly zero. Now suppose, on April 1st, it so happens that an accident occurs. If it is to be the case that there are exactly three accidents in that month, then the chance that there will be on an accident on any of the remaining 29 days in the month must now be lower than average: it would be something like $\frac{2}{29}$ rather than the $\frac{3}{30}$ that one would expect. In other words, the probability for "an accident will occur on the 2nd given that one occurred on the 1st" is different than the probability that "an accident will occur on the 2nd." This violates the postulate that the probability of an event during a "short" period of time (in this case, a day) is independent of any previous events that have occurred.

One could similarly construct an example where no accidents occur for the first 29 days of April. (This is improbable, but possible.) In that case, there must be three accidents on the 30th; the probability of an accident (in fact, three accidents!) has now shot up to 1. Again, this violates the principle that the probability of an event during a "short" period of time is independent of previous events.

This isn't a completely satisfying argument, since all it does is prove that the variance of a Poisson process cannot be zero. But hopefully it shows that the principle underlying the Poisson distribution does actually constrain its variance: the mean and the variance cannot be specified independently of each other.