# Thread: moment generating function for poisson disttribution

1. ## 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.

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!

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

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

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

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

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

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

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

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

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

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

* 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.

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

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

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

[snip]
Let $\mu = \lambda e^t$:

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

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

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

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

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

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

5. ahhhh..... now thats a better way to understand ^-^ thanks again