moment generating function of geometric distribution

**lllll** · March 28th 2008, 05:01 PM

I seem to be stuck on the moment generating function of a geometric distribution.

so far

$m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{n} e^{ty} pq^{y-1} = p \sum_{y=0}^{n} e^{ty} q^{y-1}$

how do you go from $p \sum_{y=0}^{n} e^{ty} q^{y-1} \ \ \mbox{to} \ \ p \sum_{y=0}^{n} (qe^t)^y$ where those the -1 in $p \sum_{y=0}^{n} e^{ty} q^{y {\color{red}-1}}$ go?

**lllll** · March 28th 2008, 05:24 PM

and is the final solution

1) $\frac{p}{1-qe^t}$ or

2) $\frac{pe^t}{1-qe^t}$ ?

**mr fantastic** · March 28th 2008, 06:26 PM

Originally Posted by lllll

I seem to be stuck on the moment generating function of a geometric distribution.

so far

$m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{n} e^{ty} pq^{y-1} = p \sum_{y=0}^{n} e^{ty} q^{y-1}$

how do you go from $p \sum_{y=0}^{n} e^{ty} q^{y-1} \ \ \mbox{to} \ \ p \sum_{y=0}^{n} (qe^t)^y$ where those the -1 in $p \sum_{y=0}^{n} e^{ty} q^{y {\color{red}-1}}$ go?

$m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{ty} pq^{y-1} = p \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{ty} q^{y-1}$

$= p e^{t} \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{t(y-1)} q^{y-1}$

$= p e^{t} \sum_{y={\color{red}1}}^{{\color{red}\infty}} (q e^{t})^{y-1}$

$= p e^{t} \sum_{y={\color{blue}0}}^{{\color{red}\infty}} (q e^{t})^{y}$

Now note that the for an infinite geometric series with |r| < 1: $\sum_{y={\color{blue}0}}^{{\color{red}\infty}} r^{y} = \frac{1}{1-r}$ . In the present case, $r = q e^t$ (justification for $0 \leq q e^t < 1$ is left to you).

$= p e^{t} \, \frac{1}{1 - q e^t} = \frac{p e^{t}}{1 - q e^t}$ .

**canuckbassist** · March 31st 2008, 11:37 AM

Originally Posted by lllll

and is the final solution

1) $\frac{p}{1-qe^t}$ or

2) $\frac{pe^t}{1-qe^t}$ ?

My prof uses $p(y) = pq^y$ for y = 0, 1, ..., and that gives me the first result rather than the second one.

**Moo** · February 23rd 2009, 10:16 AM

Geometric distributions can be defined over {0,1,2,...} or {1,2,3,...}
It depends on the context...

**morathyl** · December 5th 2009, 09:15 AM

i know that this post is quite old but I can't get in your proof why we have qe^t smaller than one for me e^t is a constant so we should have in some case qe^t bigger than one no ?
thanks

**mr fantastic** · December 5th 2009, 12:21 PM

Originally Posted by morathyl

i know that this post is quite old but I can't get in your proof why we have qe^t smaller than one for me e^t is a constant so we should have in some case qe^t bigger than one no ?
thanks

There is no finite solution for such a case so it is excluded.

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