Entry
Math: Probability: Distribution: Poisson: Can you derive the Poisson distribution L^k . e^(L) / k!?
Feb 16th, 2005 08:20
Knud van Eeden,

 Knud van Eeden  22 December 2004  00:09 am 
Math: Probability: Distribution: Poisson: Can you derive the Poisson
distribution L^k . e^(L) / k!?


The idea is to reduce the amount of information (by letting it cancel
away), when N grows very large.


The Poisson distribution is directly derived from the Bernouilli chance
formula.
The Poisson distribution can be viewed as the limiting case of a
binomial distribution as N approaches infinity and p approaches zero,
but N . p remaining constant.


from Bernouilli:
chance on k out of N, given a chance p is:
N k N  k
( ) . p . (1p) (1)
k

given: L = N . p {=average rate} (2)
(L gives also the expected number of successes, but is also
equal to the average rate (e.g. 16 incoming telephones per day on
average))
The purpose of the work out below is to replace any occurrence of p
(and thus 1p) (as it shows to be inconvenient to work with these
values) by this average rate L.
This L is the average vertical value of the Poisson curve.
As for almost every curve a mean value can be calculated, this
is a quite general value.
So instead of p you replace it to work with L.

Another way of looking at L is:
Imagine an interval of 1 unit of time divided into N tiny
subintervals, each so small that it is not physically possible for
more than one event to occur in a subinterval.
Poisson:
...
0 1 2 3 4 N1 N
> number of intervals
If p is the probability that an event does occur in a subinterval,
then the average number of events in 1 unit of time will be
proportionally larger, and because there N subintervals here,
the number of events will be N times as large.
So you scale with a factor N.
p + p + p + p + p + p + p + p + p + p + p + p + p + p + ... + p
 
+ as much p's as there are subintervals, thus N totally +
or thus N . p
And L = N . p

Now, (1) can be written out, using the definition of N over
k combinations, as:
N.(N1).(N2).(N3)...(Nk+1) k Nk
 . p . (1p)
k.(k1).(k2).(k3)...1
or further as, using the law of the exponents
(you can split the difference of powers in
a quotient)
N.(N1).(N2).(N3)...(Nk+1) k N
 . p . (1p) (3)
k.(k1).(k2).(k3)...1 
k
(1p)
Working towards the goal is to get
average
the e expression in. You know maybe that
x
(another) definition of e is:
x
1 1
(1 + ) = e (4)
x
lim x > +infinity
An analog expression is:
x
1 1
(1  ) = e (5)
x
lim x > +infinity
and similarly:
p x +p
(1 + ) = e (6)
x
lim x > +infinity
and similarly:
p x p
(1  ) = e (7)
x
lim x > +infinity
Now, you can write (3) also as (as the law of the products
says that you can exchange the factors)
k N
N.(N1).(N2).(N3)...(Nk+1) p . (1p)
 .  (8)
(1p).(1p).(1p).(1p)...(1p) k.(k1).(k2).(k3)...1
 
+ k factors +
Now, from (2) follows equivalently: p = L / N, (9)
so you replace p by L and N instead
L k L N
N.(N1).(N2).(N3)...(Nk+1) (  ) . (1  )
N N
 .  (10)
(1L).(1L).(1L).(1L)...(1L) k.(k1).(k2).(k3)...1
      
 N N N N N 
 
+ k factors +
This can be written as:

A.
You have, by using the law of the exponents
(as you can distribute the power of a quotient over the nominator and
the denominator)
k
L k L
(  ) = 
N k
N
1 1
By writing  as  (11)
k N . N . N . N . N . N ...
N  
+ k factors +
and writing this underneath every corresponding factor in the
N . (N1)...
product.
This denominator will cancel against the numerator, when N grows very
large, as every term (Ni) becomes more and more equal,

N
thus growing towards 1 for every factor. And the product of all 1's is
again 1, which is of no influence in a product, so can be cancelled.

B.
We write the form
N
(1  p)
in the equivalent form
L N
(1  )
N
as this is approaches (when N grows larger and larger),
according the definition and canonic form (7) to
L
e (12)

C.
As every factor (1L) converges to 1  0 (in other words 1), when

N
N grows larger and larger (as L grows then to 0),

N
this product (1L) . (1L) . (1L) ...
  
N N N
grows to (10) . (10) . (10) ...
or thus to 1 . 1 . 1 ... = 1 (13)

D.
According to the definition of faculty,
k.(k1).(k2)...1
is of course k!
(I did expand this, but it was not really necessary here) (14)
Now putting (11), (12), (13) and (14) in (15) leaves the much reduced
form (given that N grows larger and larger):
+ k factors +
  k L N
N .(N1).(N2).(N3)...(Nk+1) L . (1  )
N
 .  (15)
N . N . N . N ... N k.(k1).(k2).(k3)...1

(1L).(1L).(1L).(1L)...(1L)
      
 N N N N N 
 
+ k factors +
Or thus:
k L
1 L . e
 . 
1 k!
So this leaves
k L
L . e
= 
k!
(given:
that L is constant, and
N grows very large)
further: k maybe any value between 0 and N

Note that the sample size N as well as the probability p have
completely dropped out of the expression, which has the same functional
form for all values of L.

Or thus in words:
k (average over your interval)
= (average over your interval) . e

k!


Book: see also:

[book: Spiegel, Murray R.  Probability and statistics  McGrawHill
(Schaum series)  p. 129 'The Poisson distribution']


Internet: see also:

Poisson Distribution
http://home.ched.coventry.ac.uk/Volume/Vol4/derive.htm

Poisson Distribution
http://mathworld.wolfram.com/PoissonDistribution.html

Math: Probability: Distribution: Link: Overview: Can you give an
overview of links?
http://www.faqts.com/knowledge_base/view.phtml/aid/32917/fid/815
