Poisson distibution

1. Apr 14, 2006

dervast

Hi do u know if the poisson distribution has always the same value for EX(mean value) and variance?

2. Apr 14, 2006

HallsofIvy

Staff Emeritus
Do you mean "how do you know that the mean and variance of a Poisson distribution are the same"? Do the math!

For given parameter, $\lambda$, the Poisson Distribution is
$$P_\lambda(n)= \lambda^n \frac{e^{-\lambda}}{n!}$$
where n can be any positive integer.
The mean is given by
$$\Sigma_{n=1}^\infty \lambda^n \frac{e^{-\lambda}}{(n-1)!}$$
$$= \lambda e^{-\lambda}\Sigma_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}$$
and taking j= n-1,
$$= \lambda e^{-\lambda}\Sigma_{j= 0}^\infty \frac{\lambda^j}{j!}$$
It is easy to recognise that sum as Taylor's series for $e^\lambda$ so the sum is just $\lambda$.

The variance is given by
$$\Sigma_{n=1}^\infty e^{-\lambda}n^2 \frac{\lambda^n}{n!}- \lambda^2[\tex] [tex]= e^{-\lambda}\Sigma_{n=1}^\infty \frac{n^2\lambda^n}{n!}-\frac{n\lambda^n}{n!}+ \frac{n\lambda^n}{n!}-\lambda^2$$
$$= e^{-\lambda}\Sigma_{n=1}^\infty \frac{n(n-1)\lambda^n}{n!}+ \frac{n\lambda^n}{n!}$$
$$= \lambda^2 e^{-\lambda}\Sigma_{n=2}^\infty \frac{\lambda^{n-2}}{(n-2)!}+ \lamba e^{-\lambda}\Sigma_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}- \lambda^2$$
Now we can recognize both of those sums as Taylor's series for $e^{\lamba}$ and so the variance is $\lambda^2+ \lambda- \lamba^2= \lambda$.

Last edited: Apr 14, 2006
3. Apr 18, 2006

dervast

Thx a lot really .. u seem to be really pro :)
I have read somewhere that sometimes we can assume that a poisson distribution is the same as the gaussian one

4. Apr 18, 2006

mathman

Poisson dist. can be approximated by Gaussian when the mean is large (compared to 1). However, Poisson is discrete, while Gaussian is continuous.