Talk:PEG
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This is an interesting formula which seems to give good results.
It is a very simple and usefull formula to determine for exemple;
If I had to wait 10 minutes to see a falling star how long will I have to wait to be 90% sure of seeing it again.
The formula is ;
p=m/(x+m) or x=m*p/(1-p)
Where x is the time that I would have to wait to be p% sure that the event that I waited m for will occur again.
Entering the values I find ;
x=10*.9/(1-.9) =90 minutes.
This means that after 90 minutes I would be 90% sure of seeing another falling star.
It can be used in all kinds of interesting cases.Scientist can use it to find when they would see another new exotic particle.
Stock brockers could also use it to find out how long it would take for a certain stock to rise say 10%.
A Poisson Exponential Geometric Distribution
This distribution answers the questions ;
1.If I had to wait m to see one event how long should I have to wait (x) to be p% sure of seeing it again.
2.I had to go through m samples to find one with a special property how many more samples (x) do I have to go through to be p% sure of finding this property again.
It can also be used in the opposite way.
3.I waited m for an event to occur. If I only wait x I can be p% sure that the event will not occur.
4.I went through m samples until 1 defective sample was found.If I have only x I will be p% sure that there are no defects.
Note !!! It relates to 2 and only 2 random samples.These 2 have to be sampled each time. You cannot sample the first once and compare it to many samples. It relates to two samples and completely bypasses the means of the source exponential or geometric distributions.
It of course assumes that the mean which we do not really know is the same in both cases.It completely bypasses the mean.
Do not make the mistake of taking a best guess at the mean of the exponential distribution and then calculating your chances.The exponential distribution is 100 times more precise than this distribution at the 99.9% level !!!!!!
The only problem is that with one sample you cannot have an accurate estimation of the mean.
We could compare it to the exponential distribution; p=1-exp(-x/m) ; exponential distribution p=x/(m+x) ; PEG
OK the results are a bit wide, but there is no point in looking for your lost wallet under a street light if you lost it somewhere else !!!!!
I already posted this article over here and it was taken off.
The reason was probably that this distribution is new and totally different from most of the other statistical distributions around.
It also has an infinite mean and standard deviation. This does not mean that it is totally useless. Please go through it again carefully before you take it out. I am sure that although its application is slightly limited there are certain cases where it can really be usefull.
Professor John C. Pezzullo, PhD who has a large statistiacal site at http://Pezzullo.net commented my distribution by saying "That is a very interesting result, and one that is simpler than I would have expected." (...) "I ran several million simulations, and the numbers do seem to be perfectly consistent with your function." (...) "It's hard to believe that no one has addressed this question before, since it seems so fundamental."
Please before you erase it go through it carefully. If you find that it is not a new distribution or that it does not work please let me know why.It might also be a subbranch of some other really complicated distribution.
If it is your policy not to "publish" anything new please let me know as well.
My e-mail address is rselsick@bouygtel.fr
