The Poisson Distribution

The probability that there are x occurences of an event (where x is a non-negative integer) is given by

λ is often a rate per unit time, distance, or area. This is a discrete distribution based on counting events. Note as you tend towards large λ (i.e. long times) this distribution becomes like a normal distribution.

While this may be a scary looking math equation, once again you can use a simple poisson calculator and just insert the relevant numbers and there are only two variables; the expected rate λ (per unit time, per unit length, etc) and the number of events, x.

Example:

If the events occur on average every 4 minutes, and you are interested in the number of events occurring in a 10 minute interval, you woulduse a Poisson distribution with λ = 2.5. (e.g. 10/4 total time window divided by average arrival rate). So say in one sample you detected 6 events in that 10 minute window, so x=6. Plug those numbers directly in to the calculator.

The calculator returns a probablity of 2.7% that there would be 6 events in this 10 minute window indicating that to be an unlikely occurrence (note you can ignore the other returned probablities).

In a Poisson distribution:

• the expected value is equal to λ
• standard deviation is square root of λ
• Median is approximately λ - 1/6
• mode is approximately λ - 1/2.

Ignoring the multi day calculus proof of this:

This distribution arises as a consequence of the summation of many rare events.

Examples of environmentally related events governed by this distribution:

• The number of cars that pass through a certain point on a road during a given period of time.

• The number of spelling mistakes a secretary makes while typing a single page.

• The number of phone calls at a call center per minute Poisson statistics govern the behavior of queues!

• The number of roadkill found per unit length of road

• The number of pine trees per unit area of mixed forest.

• The number of soldiers killed by horse-kicks each year in each corps in the Prussian cavalry.

• The number of Cat 5 Hurricanes that pass by a given latitude and longtitude

Some important assumptions

• arrivals/events occur one at a time (hence discrete)

• probability of an arrival/event in a given time interval depends only on the length of the time interval

• the number of arrivals/events in non-overlapping times intervals are independent - that is the arrival of the next event is not correlated with the fact that a previous event has occurred. This is important. If this is not the case, then you can not use this statistical approach.

Example:

Lollipops arrive to the lollipop tester on average every 10 minutes.

The arrival process is Poissonian.

What is the probability of 2 arrivals in the next 15 minutes?

So, x =2 (that's the number of events we are counting)

λ = 1.5 (we expect one per arrival every 10 minutes; total time = 15 minutes; &lambda = 15/10 = 1.5)

P(x) = .251

What about 4 events?

P(x) = .047

The following animation shows the probability distribution of events as a function of λ (shown in upper right hand corner) you can see as λ increases the probability distribution evolves from being skewed to being normal. this means for large x (x>20 or so) you can just use normal distributions and standard deviations to determine probabilty of occurence).

Another Example:

Determining the probability of a rare event


Assume that the probability of dying from leukemia is described by a Poisson distribution with a mean of 2.3 deaths per 10000 children λ = 2.3

You now have been commissioned to study an apparent increase in leukemia incidences associated with children that live in ToxicWasteVille. Your boss, The Federal Health Authority, has determined that compensation will be based on a 1% criteria - that is, if the probability of the increase in occurrence is less than 1% then they will be compensated.

Your qualitative bias is to be on the side of the children.

You then go to ToxicWasteVille and measure a local death rate of 6 deaths per 10000 children. That is you observed 6 events (X = 6) when the expected event rate was 2.3. Important, you can not observe fractional events!

Will this death rate qualify for compensation?

So, What is the probability of observing a local death rate of 6 deaths per 10000?

P(6) = e^-2.3 2.3⁶ / 6!

P(6) = 0.1003 * 148.04 / 720

P(6) = 0.021 or 2.1% does not meet the compensation criteria

Note: the Poisson distribution would not be used with an infectious disease because the events (people with the disease) would not be independent of each other, i.e. if your partner has an infectious disease the probability of you getting it is higher than some one not in contact with the disease.




Last example (I have used in a climate change paper) was the 2005 Hurricane season in which there were 3 Cat 5 Hurricanes (Katrina, Rita and Wilma) the occured within a 6 week period in the Gulf of Mexico:


• We don't really care about the 6 week period, we can basically ask the question, during a hurricane season, what is the average number of cat 5 storms that enter the Gulf of Mexico.
  
  
• Look at the historical data to find that about once every 4 years there is 1 Cat 5 storm in that location.
  
  
• λ = 0.25 (1 storm every 4 years).
  
  
• x = 3 (3 storms in that one season)
  
  
• P(x) (can't be calculated by the calculator because is so low) but it turns out to be 1 in a million that this could occur if the conditions for producing Cat 5 hurricanes were the same in that season as they were throughout history. This strongly indicates that the physiscal condition in the Gulf of Mexico were not average for that Season (and they were not but that is material outside the scope of this class).