# 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.