45-733 PROBABILITY AND STATISTICS I

The Poisson Distribution

**
æ [(l**^{x})e^{-l}]/x!
ç
f(x) = ç x=0,1,2,3,4,...
ç
è 0 otherwise

**Mean = E(X) = l; and
Variance = VAR(X) = l**

where **Lambda**,
**l**, is the
*average number of occurrences of the
phenomenon in 1 unit of time, space, or volume*.

The classic example of the Poisson distribution is the distribution of deaths
in the Prussian Calvary due to horse-kicks to the head. The table below
shows deaths from horse-kicks 10 Army Corps over a 20 year period for a
total of 200 "Corps-Years".
**
Number Reports
With This
Deaths Many Deaths Proportion
---------------------------------
0 109 .545
1 65 .325
2 22 .110
3 3 .015
4 1 .005
---------------------------------
200 1.000
**

It is reasonable to assume that deaths from horse-kicks should occur
randomly over time. Consequently, the number per Corps-Year should
have a Poisson distribution. To test this, we can treat the above
data as a random sample and compute the sample mean. Using the sample
mean to estimate **l**, we can
compute the Poisson probabilities from this l
and compare them to the actual
frequencies in the above table and see how closely they match.

**_**

**X**_{n} = [å_{i=1,n} X_{i}]/n =
(total deaths)/(200 Corps-Years) =

(0*109 + 1*65 + 2*22 + 3*3 + 4*1)/200 = 122/200 = .61

Using **l = .61** produces the following
probabilities:
**
Number Reports
With This
Deaths Many Deaths Proportion l=.61
----------------------------------------
0 109 .545 .543
1 65 .325 .331
2 22 .110 .101
3 3 .015 .021
4 1 .005 .003
----------------------------------------
200 1.000 .999
**

Note that only .001 of the Poisson probability is above **X=4**.

The Poisson Distribution and its
Relationship to the Exponential Distribution

The Poisson Distribution can also be interpreted in the following form:
**
æ [(lt)**^{x}e^{-lt}]/x!
ç
f(x) = ç x=0,1,2,3,4,...
ç
è 0 otherwise

**Mean = E(X) = lt; and
Variance = VAR(X) = lt**^{2}

where **l** is, as above, the average
number of occurrences in one unit of time, and **t** is the length of
the time period.

Let **T** denote the amount of time until the first occurrence. If
**T > t** then there must have been no occurrences in the first t units
of time. The probability of this event is:

**P(T > t) = P(X = 0)= e**^{-lt}

and by the rule of the complement:

**P(T £ t) =
1 - e**^{-lt}

The distribution function for **T** is therefore:

**
æ 0 t < 0
F(t) = ç
è 1 - e**^{-lt} t ³ 0

Since this is a continuous function, we can take the derivative and obtain
the distribution:
**
æ le**^{-lt} t ³ 0
f(t) = ç
è 0 otherwise

Where
**Mean = E(T) = 1/l; and
Variance = VAR(T) = 1/l**^{2}

This is the Exponential Distribution. It is the
distribution of the time until the first occurrence of an event or the
distribution of the amount of time *between* events where the
occurrences of the phenomenon are governed by a Poisson process.

For example, suppose the arrival of cars at a Turnpike entrance is
governed as a Poisson process with a mean of 3 per minute. Suppose the
ticket taker has to leave the booth for 1 minute. What is the probability
that no cars arrive during the ticket taker's absence?

**P(T > 1) = 1 - F(1) = e**^{-3*1} = .04979

As shown by this simple example the Exponential and Poisson distributions
have many useful applications to what are known as *
Waiting Time Problems*. For example, the arrival of ships
at a port (both number and the time in between the arrivals); the arrival
of freight trains at a classification yard; etc. The examples are legion.