Poisson Distribution: Mean & Variance Explained

[dervast]

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

 

[HallsofIvy]

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]<br /> = e^{-\lambda}\Sigma_{n=1}^\infty \frac{n^2\lambda^n}{n!}-\frac{n\lambda^n}{n!}+ \frac{n\lambda^n}{n!}-\lambda^2<br /> = e^{-\lambda}\Sigma_{n=1}^\infty \frac{n(n-1)\lambda^n}{n!}+ \frac{n\lambda^n}{n!}<br /> = \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<br /> Now we can recognize both of those sums as Taylor&#039;s series for e^{\lamba} and so the variance is \lambda^2+ \lambda- \lamba^2= \lambda.

 

[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

 

[mathman]

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