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##### Math: Probability: Distribution: Poisson: Can you describe Poisson distribution L^k . e^(-L) / k!?

Feb 16th, 2005 10:55
Knud van Eeden,

```----------------------------------------------------------------------
--- Knud van Eeden --- 25 December 2004 - 08:05 pm -------------------
Math: Probability: Distribution: Poisson: Can you describe Poisson
distribution L^k . e^(-L) / k!?
---
The Poisson distribution is a special case of the Binomial
distribution,
and can be derived from it.
---
---
It is a discrete distribution.
---
---
Both can be seen as variations on N throws, after each other, with a
coin with a certain (larger in the case of the binomial distribution,
or very small in the case of the Poisson distribution) probability.
---
Binomial or Bernouilli distribution:
|----|----|----|----|...|----|----|----|----|
0    1    2    3    4                 N-1   N
-> number of throws
Poisson:
|----|----|----|----|...|----|----|----|----|
0    1    2    3    4                 N-1   N
-> number of intervals
---
---
Each of the N intervals has a probability p of success.
---
---
The difference is that the total amount of throws with a Poisson
distribution is by definition very large, thus N very large, while the
chance that you have success in each interval (e.g. head), thus p, is
very small.
---
---
The difference between these two is further that the Binomial
distribution requires knowledge of the probability that some event will
happen (that is 'p'), and also about the probability that this event
will not happen (that is '1-p' usually).
This probability p is often derived from finding the proportion of
successes in N trials (e.g. throwing a coin N times, and counting
the total amount of heads, and dividing this by the total amount of
throws. This will give p).
As usually if p is known, the probability that this event will not
happen (that is '1-p') is then also known (because the probability
together must equal 1, or 100%, so you can subtract p from 1 to get the
probability '1-p').
Both 'p' and '1-p' are needed in the binomial distribution formula.
Now, there are many situations where we can count the occurrences
of some event, but no figure of non-occurrences is meaningful.
---
For example,
A telephone center receives per agent 16 calls per hour on average.
How many calls do NOT occur?
Obviously, this is not known.
---
For example,
On a road you count 16 cars passing per hour on average.
How many cars did NOT occur?
Obviously, this is not known.
---
To such situations the Binomial distribution formula is not applicable.
Instead you can use the Poisson distribution.
---
The Binomial distribution works thus with the constant and known
probability p.
---
The Poisson distribution works instead with the constant and known
expectation L (this 'L' stands for the Greek character 'L'ambda, which
is often used in this context to indicate the average rate)
---
This expectation of the Poisson distribution can be written as
L = N . p
---
And you see that choosing this expectation L is a smart move, as it
turns out to be equal to the average, something which you can find out
experimentally by measuring via e.g. counting.
And you will see that N and p drop out of the endresult of the
calculations all together, so you do not need them anymore.
---
How can you possibly understand this expression L = N . p?
If you divide your interval in N subintervals, with each subinterval
having a probability of success p, and there are thus totally N of this
subintervals, then you can say that this expression states:
expectation =
(total number of trials) times (probability of success at any trial)
---
as there are in the interval N subintervals, and each subinterval has
a probability p of success, you are saying thus here
(imagine that you, in each of this subintervals, in a serial sequence,
throw a dice with a probability p, then you throw so N times after each
other this dice. You are asking yourself what your expectation is after
N of this throws)
expectation = N . p
---
e.g.
This is similar to doing throws, one after the other, and at each throw
you have a probability of on average 1 times in 6 times that you get a
5.
Thus after you have thrown 600 times, you can expect on average that
you get 600 . 1/6, or thus 100 times the 5
as a result.
So your expectation will be 100 occurrences of the 5.
---
e.g.
If you play 100 times in the lottery, and the probability that
you win something is on average 1 times in 5 times playing, you are
expected on average to win 100 . 1/5, or thus 20 times.
So your expectation will be 20 occurrences of a win.
---
e.g.
If your football team should play 60 matches, and the probability that
they
win is on average 1 times in 2 times playing, then it is expected on
average that they should win 60 . 1/2, or thus 30 times.
So your expectation will be 30 occurrences of a win.
---
It can be shown that the average value of the Binomial
distribution equals to N . p, or thus the average value of the
Binomial distribution equals the expectation
(you calculate this using the mean value theorem in
calculus, applied to this curve)
---
(as the Poisson distribution is just a special case of the
binomial distribution, so it should have the same properties,
in this case the average)
It can be shown that the average value of the Poisson distribution
equals to N . p, or thus the average value of the Poisson distribution
equals the expectation.
(you calculate this using the mean value theorem in
calculus, applied to this curve)
---
Thus N . p, the expectation, equals the average in this special
case.
---
Thus L, the expectation, equals the average value of the Poisson
distribution.
---
It is assumed for the cases where you apply your Poisson distribution
that this mean or average or expectation remains constant.
Thus
L = N . p = constant
Thus if N gets much larger, then p must get much smaller, such that L
remains constant.
---
This L is your average total number of occurrences.
E.g. totally 16 calls (in your time interval of 1 hour, and per agent)
on average.
---
Thus what you then do is to divide your interval (this can in general
be any interval, be it time, space, volume, wire length, area, ...) in
N smaller subintervals of equal length.
---
original interval
[                                         ]
---
original interval divided in smaller equal subintervals.
[ | | | | | | | | | | | | | | | | | ... | ]
---
Events (e.g. incoming telephone calls) should occur at random in your
time, length, volume, ..., subinterval
---
Now if the following conditions are present, you could use the Poisson
distribution:
1. The probability of more than one count in a subinterval is zero
(e.g. in a 1 millisecond interval it is very unlikely that you get
at the same moment 3 telephone calls)
2. The probability in each subinterval is the same for all
(in other words no subinterval is more important than another
subinterval)
3. The probability is proportional to the total length of the
subinterval (if you wait 5 minutes, you should have a 5 times
bigger chance of getting a call)
4. The probability of each subinterval is independent of another
subinterval
(what happens in another subinterval should have no influence
on the probability in the subinterval)
---
---
How to calculate L?
L is a positive real number, equal to the expected total number of
occurrences that occur in the given interval.
---
Thus L follows from the nominator of
(total number of occurrences)
-----------------------------
(the given interval)
---
Note:
So this expression can also be seen as a frequency, if you look at the
units used, that is, a dimension less (integer) value divided by a
dimensioned value, similar to e.g. a clock beats 10 times per minute,
a wave oscillates 60 times per second.
---
For instance, if the events occur on average 1 time every 2 minutes,
thus
( 1 time )
-------------
( 2 minutes )
and you are interested in the number of events occcurring in a 30
minute interval, you would linearly scale it, by multiplying everything
with a constant, as that does not change the quotient (if you multiply
both nominator and denominator in a quotient with the same value, the
end result remains the same, as multiplying with 1 does not change
the endresult).
Thus you can also write:
(some constant) (total number of occurrences)
---------------------------------------------
(some constant) (your given time interval)
---
Another way of saying this is that L, the total number of occurrences
is proportional with the given time interval.
Thus if the time interval is chosen 2 times as large, the total number
of occurrences is going to be 2 times as large.
Thus if the time interval is chosen 3 times as large, the total number
of occurrences is going to be 3 times as large, and so on.
e.g.
15  . ( 1 time )
-----------------
15 . ( 2 minutes )
remains proportionally the same.
---
Given a total occurrences of
15 . 1 times in 15 . 2 minutes,
or thus totally 15 occurrences in a 30 minutes interval.
Thus L would in that case be chosen to be this new total number of
occurrences, or thus 15.
---
For instance, if the events occur on average
24 times every 1 working day
or thus
( 24 times )
----------------
( 1 working day)
and you are interested in the number of events occcurring in a
1 hour interval, or thus you would proportionally or linearly scale it,
by multiplying by 1/8.
or thus
(1/8) . ( 24 times )
------------------------
(1/8) . ( 1 working day)
or thus
(1/8) . ( 24 times )
------------------------
(1/8) . ( 8 hours)
or thus
( 3 times )
------------
( 1 hour )
Thus L, the total number of occurrences would in this case be chosen
equal to 3.
---
---
So a possible algorithm to calculate the Poisson distribution is:
1. Choose a total time interval
e.g. 8 hours
2. Count the total of occurrences in this interval
e.g. 16 calls
3. Determine the interval in which you are interested
e.g. 1 hour
4. Scale
from the proportion:
(new total occurrences)   (old total occurrences)
----------------------- = -----------------------
(new time interval)       (old time interval)
Note: so you are basically saying that
(new frequency) = (old frequency)
follows:
(new time interval) . (old total occurrences)
(new total occurrences) = ---------------------------------------------
(old time interval)
thus you use as the scale factor
(new time interval)
-------------------
(old time interval)
and you multiply the old total occurrences with this number.
e.g.
(1 hour           ) . (16 total occurrences )
(new total occurrences) = ---------------------------------------------
(8 hours          )
= 2 total occurrences
5. Fill in
1. L, or this 'new total occurrences'
2. k, or thus the amount of times you are asking it to occur
3. in the formula
k        -L
L    .   e
------------
k!
e.g. choosing k equal to 3, gives
3       -2
2    .  e
-----------
3!
6. That will give a number between 0 and 1.
e.g. this fraction equals in this example:
0.180447044
7. Interpret this result
You can also use scaling to interpret this result yourself better.
or thus
it will occur 0.18 times in the 1 hour interval.
Usually you can better imagine this result by using whole numbers.
So you scale it up to the whole number 1.
Or thus about, by scaling it also, in this case to whole number, that
is 1, to 5 . 0.18 in 5 . 1 hour interval,
or thus to about 1 time in a 5 hour interval.
So you use the about proportion
( 0.180447044 times )    5 . ( 0.180447044 times )
--------------------- =  -------------------------
( 1 hour interval )      5 . ( 1 hour interval )
in general, you convert the calculated fraction to a whole number 1, by
multiplying by (1 divided by this fraction).
1
(--------)
(new times)     fraction   ( your just calculated times it will occur)
------------- = ------------------------------------------------------
(new interval)      1      ( your used interval )
(--------)
fraction
---
---
As this scaling might occur all the time, you might as well fill it
in directly in the Poisson formula:
from

k        -L
L    .   e
------------
k!
you get thus, using the above terminology

(new total occurrences)   (old total occurrences)
----------------------- = -----------------------
(new time interval)       (old time interval)
which can be rewritten as:

(new time interval) . (old total occurrences)
(new total occurrences) = ---------------------------------------------
(old time interval)
filling in the right side of  in , you get

k   (new interval).(old occurrences)
(new interval).(old occurrences)    (-------------------------------)
(-------------------------------) .         (old time interval)
(old interval)                     e
---------------------------------------------------------------------
k!
---
---
You can ask yourself then questions like:
---
If given that 24 calls per day are coming in on average per agent,
what is the probability that you get 5 calls per 1 hour?
Thus
= ( 24 times )
----------------
( 1 working day)
= 1/8 . ( 24 times )
------------------
1/8 . ( 8 hours )
= ( 3 times )
-----------
( 1 hour )
Thus in this 1 hour interval, you have
L = totally 3 occurrences per time interval on average
or thus here
L = totally 3 calls per hour on average
or thus
L = 3
and
k = 5 occurrences per time interval
or thus here
k = 5 calls per hour
or thus
k = 5
---
Using the Poisson distribution, you get that this probability equals
k        -L
L    .   e
= ------------
k!
5    -3
= (3)  . e
----------
5!
= 0.100818813...
Thus a probability of 1 in 10, or thus this should happen
once every 10 hours, or thus about once in one to two days.
---
---
Another example, using the cumulative Poisson distribution
---
If given that 24 calls per day are coming in on average per agent,
what is the probability that you get 5 or fewer calls per hour?
Thus
= ( 24 times )
----------------
( 1 working day)
= 1/8 . ( 24 times )
------------------
1/8 . ( 8 hours )
= ( 3 times )
-----------
( 1 hour )
Thus in this 1 hour interval, you have:
L = totally 3 occurrences per time interval on average
or thus
L = totally 3 calls per hour on average
or thus
L = 3
and because is asked for 'fewer' than 5, you get:
k = 0, 1, 2, 3, 4 or 5 occurrences per time interval
or thus here
k = 0, 1, 2, 3, 4 or 5 calls per hour
---
Using the Poisson distribution, you get that this probability equals
(similar to the Binomial distribution, where you also have to add
together
all possibilities smaller or equal to in this case)
k        -L
= SUM( k from 0 to 5 ) of ( L    .   e   )
------------
k!
0    -3      1    -3      2    -3
(3)  . e     (3)  . e     (3)  . e
= ---------- + ---------- + ---------- +
0!           1!           2!
3    -3      4    -3      5    -3
(3)  . e     (3)  . e     (3)  . e
+ ---------- + ---------- + ----------
3!           4!           5!
= 0.0497870684 + 0.149361205 + 0.224041808 +
0.224041808 + 0.168031356 + 0.100818813
= 0.9160820584
Thus a probability of about 9 in 10, thus about in 90% of the hours,
you should
have 5 or less calls.
---
---
Note:
History:
Sim<e'>on Denis Poisson (1781-1840) (a student of Simon de Laplace and
Joseph-Louis Lagrange) introduced this Poisson equation for the first
time in 1837 in his work
'Recherches sur la probabilit<e'> des jugements'
---
---
---
[book: Bronshtein / Semendyayev - a guide book to mathematics - ISBN 0-
387-91106-5 - p. 605 'Poisson distribution' (short formula, 1 numeric
example / 1 numeric example of Poisson distribution approximation to
binomial distribution)]
---
[book: Childers, Donald G. - probability and random processes (using
MatLab with applications to continuous and discrete time systems) -
publisher: McGraw-Hill - ISBN 0-256-13361-1 - p. 90 'Expectation
(Averages)' (introducting the mean as the statistical average, or in a
physical way as the 'center of gravity' (also called 'first moment'
here) of the Poisson graph in vertical direction, so putting it
balancing on a cord will leave it in equilibrium) / 'Poisson random
processes' (with good example relation telephone calls arriving
according to a Poisson distribution, and with the telephone call time
length following an exponential distribution, with a formula workout
and numeric example, in a network, M/M/1 queue)]
---
[book: Daintith, John / Nelson, R. David - the penguin dictionary of
mathematics - ISBN 0-14-051119-9 - p. 254 'Poisson distribution' (only
a very short definition)]
---
[book: Gellert, W. / K<u..>stner, H. / Hellwich, M. - kleine
Enzyklop<a..>die Mathematik - ISBN 3-87144-323-9 - p.
640 'Poissonverteilung' (nice description, good example with e.g.
probability of taking out 1 black ball out of a vase (with putting it
back afterwards) with a very large amount of white balls)]
---
[book: Green, David R. / Lewis, John - science with pocket
calculators - ISBN O-085109-560-7 - p. 152 'Poisson distribution']
---
[book: Hastings, Kevin J. - Probability and statistics - ISBN 0-201-
59278-9 - p. 80 'Poisson distribution' with example, p. 82 'Poisson
probability mass function' (definition), p. 83 (comparison values
Binomial and Poisson distribution), p. 83 (example cumulative Poisson
distribution, for users wanting to access a server), p. 84 'Poisson
random processes' (rate parameter, solution of differential equation
gives Poisson formula), p. 215 'Poisson interarrival times', p.
237 'Generating functions' (superposition of Poisson arrival rates,
distribution of arrival times, about telephone calls arriving at a
switch board)]
---
[book: Kramer, Edna Erestine - the nature and growth of modern
mathematics - p. 333 'Realm of random variables' (Shows example of how
to approximate a binomial distribution with a Poisson distribution)]
---
[book: Lipschutz, Seymour - Probability - McGraw-Hill (Schaum) - p.
108 'Poisson distribution' (4 graphs, approximation to binomial
distribution via Taylor expansion)]
---
[book: Montgomery, Douglas, C. / Runger, George C. / Hubele, Norma,
F. - engineering statistics - publisher: John Wiley - ISBN 0-471-17026-
7 - p. 92 'Poisson process' (example with transmission of n bits over
a communication channel, flaws in a copper cable and contamination of
optical storage disks]
---
[book: Spiegel, Murray R. - Probability and statistics - McGraw-Hill
(Schaum series) - p. 123 'Mean equals N . p', p. 129 'The Poisson
distribution', p. 239 'fitting of numerical data (automobile
accidents')]
---
[book: Weisstein, Eric W. - CRC concise encyclopedia of mathematics -
ISBN 0-8493-9640-9 - p. 1388 'Poisson distribution' (1 graph, lot of
---
[book: Zwillinger, Daniel - CRC standard mathematical tables and
formula - p. 582 'Poisson distribution' (only a very short
definition) / 'Poisson arrivals' (only a formula with no further
explanation)]
---
---
---
Poisson Distribution
http://home.ched.coventry.ac.uk/Volume/Vol4/listof.htm
---
Poisson Distribution
http://mathworld.wolfram.com/PoissonDistribution.html
---
Poisson Distribution
http://en.wikipedia.org/wiki/Poisson_distribution
---
Sim<e'>on Denis Poisson
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Poisson.html
---
Math: Probability: Distribution: Link: Overview: Can you give an