# moment generating function for poisson disttribution

• Mar 29th 2008, 11:19 PM
moment generating function for poisson disttribution
Hi guys i am new. I have a problem in finding moment generating functions. The forum helped me in solving binomial, geometric and other few random equations but i still couldn't understand how the moment generating function for a poisson distribution is derived out from its probability density function.

• Mar 29th 2008, 11:36 PM
mr fantastic
Quote:

Hi guys i am new. I have a problem in finding moment generating functions. The forum helped me in solving binomial, geometric and other few random equations but i still couldn't understand how the moment generating function for a poisson distribution is derived out from its probability density function.

Well, I guess this is about the last one in the list!

$\displaystyle m_X(t) = \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y e^{-\lambda}}{y!} = e^{-\lambda} \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y}{y!}$

$\displaystyle = e^{-\lambda} \sum_{y=0}^{\infty} \frac{(\lambda e^t)^y}{y!}$

using the standard series $\displaystyle \sum_{y=0}^{\infty} \frac{(x)^y}{y!} = e^x$ and substituting $\displaystyle x = \lambda e^t$ *

$\displaystyle = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}$.

* If you don't like doing this there is another clever way of doing it.
• Mar 29th 2008, 11:44 PM
Quote:

Originally Posted by mr fantastic
Well, I guess this is about the last one in the list!

$\displaystyle m_X(t) = \sum_{n=0}^{\infty} e^{nt} \frac{\lambda^n e^{-\lambda}}{n!} = e^{-\lambda} \sum_{n=0}^{\infty} e^{nt} \frac{\lambda^n}{n!}$

$\displaystyle = e^{-\lambda} \sum_{n=0}^{\infty} \frac{(\lambda e^t)^n}{n!}$

using the standard series $\displaystyle \sum_{n=0}^{\infty} \frac{(y)^n}{n!} = e^y$ and substituting $\displaystyle y = \lambda e^t$ *

$\displaystyle = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}$.

* If you don't like doing this there is another clever way of doing it.

ahhh... ic i didnt know there was that standard series thingy xD thanks for the quick reply ^-^ btw what is the other method O.o? no harm in knowing more.
• Mar 30th 2008, 12:41 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
Well, I guess this is about the last one in the list!

$\displaystyle m_X(t) = \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y e^{-\lambda}}{y!} = e^{-\lambda} \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y}{y!}$

$\displaystyle = e^{-\lambda} \sum_{y=0}^{\infty} \frac{(\lambda e^t)^y}{y!}$

[snip]

Let $\displaystyle \mu = \lambda e^t$:

$\displaystyle = e^{-\lambda} \sum_{y=0}^{\infty} \frac{\mu^y}{y!}$

$\displaystyle = e^{-\lambda} \, e^{\mu} \sum_{y=0}^{\infty} \frac{e^{-\mu}\, \mu^y}{y!}$

$\displaystyle \frac{e^{-\mu}\, \mu^y}{y!}$ is recognised as the pdf of a random variable following a Poisson distribution with mean $\displaystyle \mu$. Therefore $\displaystyle \sum_{y=0}^{\infty} \frac{e^{-\mu}\, \mu^y}{y!} = 1$:

$\displaystyle = e^{-\lambda} \, e^{\mu} (1)$

Substitute back that $\displaystyle \mu = \lambda e^t$:

$\displaystyle = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}$.
• Mar 30th 2008, 12:44 AM