Poisson Calculations

[ptcane]

hi all,
i would like to know whether i can / how to calculate the mean value
of occurrences from a given percentage probability for a given number
of occurrences.

for example, if know that there is a 40% chance there will be 2 or
less occurrences, how do i calculate the mean number of occurrences?

(i will then use this value as the mean part of various poisson
equations to work out the probability of various numbers of
occurrences) .

thanks for your help.

 

[Jerry W. Lewis]

Probability of 2 or less occurances is =POISSON(2,mean,TRUE)
Given that, there are various ways to answer your question:
- You can plot this by various values of mean and get a rough visual
estimate of what mean gives a value of 0.4.
- You can use Solver to numerically find the mean that makes this equal to
0.4
- You can download Ian Smith's probabilty library
http://members.aol.com/iandjmsmith/examples.xls
and use it to calculate =ucb_poisson(2,0.4)
- A less accurate implementation of the theoretically exact calculation for
a poisson upper confidence bound would be to use the native Excel functions
=GAMMAINV(1-0.4,x+1,1) or =CHIINV(0.4,2*(2+1))/2, where x=2 for your question.

The Poisson mean that has exactly 40% probability of 2 or less is 3.105379...

Jerry

 

[Dana DeLouis]

=GAMMAINV(1-0.4,x+1,1)

Hi Jerry. Very nice.
For fun, here's my vba attempt at trying to beat Excel's 7-9 digit accuracy.
If the op's " # of occurrences" is an integer, than the Newton method
appears to have terms that cancel. Hence the following idea:

' Check for occurrences of 2 & 10

Sub Testit()
Debug.Print FindMean(2, 0.4)
Debug.Print FindMean(10, 0.4)
End Sub

Function FindMean(val, perc)
Dim p
Dim k
Dim c
Dim m
Dim j As Long
Dim cnt As Long

m = val ' Best Guess for m
p = perc

With WorksheetFunction
For cnt = 1 To 15
k = 0
p = val
For j = 0 To val
c = .Permut(val, j)
k = k + c * m ^ p
p = p - 1
Next j

m = m - (c * Exp(m) * perc - k) / (m ^ val)
Next cnt
End With

FindMean = m
End Function

Returns:
3.10537859726335
11.5153304496102

If I check it with a math program:

InverseGammaRegularized[1+2,Infinity,-0.4]

3.10537859726335

InverseGammaRegularized[1+10,Infinity,-0.4]

11.51533044961022

They appear to match pretty well.

Excel worksheet functions had the following which we know to be caused by
GAMMAINV not being very accurate.
3.10537859748574
11.51533046679880

Anyway, very interesting.

 

[ptcane]

=GAMMAINV(1-0.4,x+1,1)

sorry for the delay, many thanks for your help. just what i needed.

 

[Jerry W. Lewis]

Better late than never; thanks for the feedback.

Jerry