Hi,
Does anyone know how to show, by means of probability generating functions, that for large n and small p, (n →∞, p→0), such that np →λ, that a binomial distribution will converge to a poisson with parameter λ?
Thanks in advance for your help
Hi,
Does anyone know how to show, by means of probability generating functions, that for large n and small p, (n →∞, p→0), such that np →λ, that a binomial distribution will converge to a poisson with parameter λ?
Thanks in advance for your help
I believe so. This is something I had to prove sometime back. It's a cool proof.
Let . Therefore,
Start with the binomial
Now, if we divide one of the x factors, n, in into each factor of the product , and then write
we get:
Now, let while x and lambda stay fixed.
we get:
Now, we get the famous 'e' limit:
Therefore, hence, and heretofore, we get:
and the limiting distribution becomes the Poisson distribution.
Hi,
First of all thanks so much for the proof.
I do have a few qns:
The problem did ask for proof via probability generating functions, and I can't help but think that there is a more simple proof to it than the method you have shown (i do very much appreciate that it is indeed, quite a cool proof). So do you have any idea how this might work?
[ pgf of binomial(n,p) = (pz + 1-p)^n and pgf of poisson(λ) e^λ(z-1) ]
Also about this proof:
I'm guessing the bit about the famous 'e' limit can be shown via taylor series' expansion??
I fully understand the rest of the proof though, its very thorough and helpful^^
cheers
The proof I gave is via a probability generating function. That is what a binomial is.
Don't worry about deriving the 'e' limit. Though you can if you wish. It's derivation can be found in a calc book.
One can also show, via repeated integration by parts, that
This right side is called the incomplete gamma function
It is so cool how all this stuff ties together.
Hi
Correct me if i'm wrong.. you started with a probability mass function which is given by
the probability generating function on the other hand, is given by G(z) = E(z^X) which results in G(z) = (pz + 1- p)^n for the binomial with parameters n,p. G(z) = e^λ(z-1) for a poisson.
I'm thinking the problem requires the equivalence of both functions as n tends to infinity and p tends to zero but I'm not confident of taking both limits (simultaneously?) without making major conceptual errors.
Really hope I've got this right please let me know if I'm mistaken.