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 $\displaystyle np={\lambda}$. Therefore, $\displaystyle p=\frac{\lambda}{n}$
Start with the binomial $\displaystyle b(x;n,p)=\binom{n}{x}\left(\frac{\lambda}{n}\right )^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}$
$\displaystyle =\frac{n(n-1)(n-2)......(n-x+1)}{x!}\left(\frac{\lambda}{n}\right)^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}$
Now, if we divide one of the x factors, n, in $\displaystyle \left(\frac{\lambda}{n}\right)^{x}$ into each factor of the product $\displaystyle n(n-1)(n-2)....(n-x+1)$, and then write
$\displaystyle \left(1-\frac{\lambda}{n}\right)^{n-x}=\left[\left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\right]^{-\lambda}\left(1-\frac{\lambda}{n}\right)^{-x}$
we get:
$\displaystyle \frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)....\left(1-\frac{x-1}{n}\right)}{x!}\cdot {\lambda}^{x}\left[\left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\right]^{-\lambda}\left(1-\frac{\lambda}{n}\right)^{-x}$
Now, let $\displaystyle n\to {\infty}$ while x and lambda stay fixed.
we get:
$\displaystyle \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)....\left(1-\frac{x-1}{n}\right)\to 1$
$\displaystyle \left(1-\frac{\lambda}{n}\right)^{-x}\to 1$
Now, we get the famous 'e' limit:
$\displaystyle \left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\to e$
Therefore, hence, and heretofore, we get:
$\displaystyle p(x;{\lambda})=\frac{{\lambda}^{x}e^{-\lambda}}{x!}$
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 $\displaystyle \sum_{x=0}^{n}\frac{{\lambda}^{x}e^{-\lambda}}{x!}=\frac{1}{n!}\int_{\lambda}^{\infty}t ^{n}e^{-t}dt$
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
$\displaystyle
b(x;n,p)=\binom{n}{x}\left(\frac{\lambda}{n}\right )^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}
$
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.