I am trying to prove that adding up all the individual probabilities of a Poisson distribution by multiplying between trials (fundamental counting principle) is equivalent to using a Poisson Cdf function. Conceptually, all disjoint trials should be independent and thus this should work, but again I want to prove it.
A binomial probability of random variable X is given by nCr * p(success)^number of successes * p(failure)^number of failures.
Originally, the binomial PDF was found with an assumption of independence, you multiply all the probabilities of getting a success times failures times the number of ways those could be arranged. The problem with this is that you can only account for 1 success in 1 trial. You now use Poisson dist. to account for any number of successes in a certain time period.
All this now does is allows you to find the probability that n successes in a time period rather than a single trial, which is like adding up infinitely small trial periods that have either 1 success or 0 success. Because this is assumed to be homogenous, the mean can be changed proportional to the time period.
Instead of probability of success in one trial, you have probability of number of successes over a number of trials (time period).
Since Poisson is derived from binomial and therefore must have independent intervals if disjoint, why can’t you do the fundamental counting principle on the poisson distribution, only this time it can’t be expressed as simply as P(success)*number of successes because you are looking for a number of successes in a given period?
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Independence (the reason the above is true)
The binomial distribution assumes that all events are independent of one another, hence the form of it: p(success)^#ofsuccesses * p(failures)^#offailures * arrangements is essentially the fundamental counting method * number of arrangements. The FTM relies on independence of events, and because the poisson distribution is binomial with infinite trials, this distribution must also follow the same independence principle - if x trials are independent of all other trials, then they must also as a group be independent of all other trials. Therefore x and y trials with the same properties of independence must both be independent of one another even when treated as groups of trials.
One other way of seeing it is that the poisson distribution split up into individual results with half the mean is equivalent to splitting up a binomial distribution of x trials into x different single trial distributions, where the probability of each distribution is p/x. The difference is, for the poisson, you have a number of trials approaching infinity, so the expected value gets divided up linearly when splitting up such a homogenous distribution, rather than a probability.
The notation I used here is the following: C(u,m) is the cumulative probability of events with m or less successes. P(u,m) is the individual for m successes. Both with mean u for that given period.
My method of doing so begins with the following real-world problem as I'm not proficient with mathematical terminology. Let's say u cars pass a bridge every minute, and you must find the probability that at most n cars pass over m minutes.
$ = 1 - C(mu,n) = 1- \displaystyle \sum_{i=0}^n P(mu,i)$
Therefore if you could find $ P(mu,i) $ for any parameters $m,u,i$ with the FCP then this should be proved. I see that there would be an outside factor of $me^{-mu}$ on the outside of this, and also that you could create a pattern of all the possible ways to create exactly the parameters given. That would look something like i,0,0,0,0... m-1 times, i,i-1,0,0,0,0,0 m-2 times, i,i-1,i-2,0,0,0,0 m-3 times and so on. However, i should stop when it reaches i-i. I am not sure how to express this idea in a mathematical notation, likely a product notation within a summation?
For example, Let's say the mean number of cars that passes a street per minute is 2. With a Poisson distribution, what is the probability that at most 2 cars are seen over a period of 2 minutes? This is the Cdf of 2 cars with mean 4, because it is over two minutes.
Adding up probabilities method:
Minute 1 * Minute 2
poissonPdf(2,0)*poissonPdf(2,0)+
2*poissonPdf(2,0)*poissonPdf(2,1)+
poissonPdf(2,1)*poissonPdf(2,1)+
2*poissonPdf(2,2)*poissonPdf(2,0).