Poisson Distribution: Is adding up individual probabilities between 2 trials equal to doubling the mean?

Question

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?

poissonCdf(a,b,c) is the cumulative probability with mean a and bounds b to c.

poissonPdf(a,b) is the individual probability of mean a and # of cars value b.

I understand one method is poissonCdf(4,0,2) with the mean doubled because it is over a period of 2 minutes. Equivalently, you could do poissonPdf(4,0)+poissonPdf(4,1)+poissonPdf(4,2).

My question is, why is it also equivalent to say (or is it) the answer equals to

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)?

Why is it accurate to say that doubling the mean to account for both minutes at once is equivalent to taking account of all the possibilities of events between those two minutes that are at most 2 car? Are those events independent?

Answer 1

This is really a question about a Poisson process. The assumption should be that the occurrences of a car passing the street form a Poisson process of rate $2$ per minute. This implies that in any time interval of length $L$ minutes, the number of occurrences in that interval is a Poisson random variable with parameter $2L$, and the numbers in disjoint intervals are independent. These are consistent because the sum of independent Poisson random variables of parameters $\lambda$ and $\mu$ has a Poisson distribution with parameter $\lambda + \mu$.

Answer 2

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?

———————

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.