# Find the MGF of geometric,neg binomial dist.

• ArcanaNoir

## Homework Statement

Find the MGF (Moment generating function) of the
a. geometric distribution
b. negative binomial distribution

## Homework Equations

geometric distribution: $$f(x)=p^x(1-p)^{x-1}$$ where x=1,2,3...

negative binomial distribution: $$f(x)= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r}$$ where x=r, r+1, r+2...

MGF= $$E(e^{tx})$$

## The Attempt at a Solution

a. $$\sum_{x=1}^{\infty}e^{tx}p^x(1-p)^{x-1}$$
let $q=1-p$
$$\sum_{x=1}^{\infty}e^{tx}p^xq^{x-1}$$
$$\sum_{x=0}^{\infty}(pe^t)q^x$$
$$=\frac{pe^t}{1-q}$$
that's as close as I can get to approximating the solution,
but the book says the answer is $$\frac{pe^t}{1-qe^t}$$

b. $$\sum_{x=r}^{\infty}\frac{(x-1)!}{(x-r)!(r-1)!}e^{tx}p^rq^{x-r}$$ where q=1-p

## Homework Statement

Find the MGF (Moment generating function) of the
a. geometric distribution
b. negative binomial distribution

## Homework Equations

geometric distribution: $$f(x)=p^x(1-p)^{x-1}$$ where x=1,2,3...

negative binomial distribution: $$f(x)= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r}$$ where x=r, r+1, r+2...

Those aren't the distributions I'm used to. Are you sure these formula's are correct??

For the geometric distribution, I got

$$f(x)=p^x(1-p)$$

and the negative binomial is

$$f(x)=\binom{x+r-1}{x}(1-p)^rp^x$$

Could you recheck this first?

oh good god. I was looking at point binomial.
so here's what I get now,
$$\sum_{x=1}^{\infty}e^{tx}p(1-p)^{x-1}$$ letting q= (1-p)
$$= \frac{p}{q} \sum_{x=1}^{\infty}e^{tx}q^{x}$$
$$= \frac{p}{q} \sum_{x=1}^{\infty}(e^{t}q)^x$$

oh good god. I was looking at point binomial.
so here's what I get now,
$$\sum_{x=1}^{\infty}e^{tx}p(1-p)^{x-1}$$ letting q= (1-p)
$$= \frac{p}{q} \sum_{x=1}^{\infty}e^{tx}q^{x}$$
$$= \frac{p}{q} \sum_{x=1}^{\infty}(e^{t}q)^x$$

Yes, go on... You have a geometric series right now.

the negative binomial I have as being : $$\binom{x-1}{r-1} p^r(1-p)^{x-r}$$
which I figured equals : $$\frac{(x-1)!}{(x-1-(r-1))!(r-1)!}p^r(1-p)^{x-r}$$
$$= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r}$$

Yes, go on... You have a geometric series right now.

so, $\frac{1}{1-r}$ ?
$$( \frac{p}{q} ) \frac{1}{1-e^{t}q}$$ ?

so, $\frac{1}{1-r}$ ?
$$( \frac{p}{q} ) \frac{1}{1-e^{t}q}$$ ?

That's the sum of the geometrci series when you start summing from 0. But you start summing from 1 here.

but if I sum from zero, won't I have $$\frac{p}{q} \sum_{x=0}^{\infty}(e^{t}q)^{x+1}$$ oh wait. I see. hold on

nevermind. I don't see. am I suppose to have x-1 at x=0 to do the sum?

oop, there it is! found the answer to the geometric. thanks for the support on that.

## Homework Statement

Find the MGF (Moment generating function) of the
a. geometric distribution
b. negative binomial distribution

## Homework Equations

geometric distribution: $$f(x)=p^x(1-p)^{x-1}$$ where x=1,2,3...

negative binomial distribution: $$f(x)= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r}$$ where x=r, r+1, r+2...

MGF= $$E(e^{tx})$$

## The Attempt at a Solution

a. $$\sum_{x=1}^{\infty}e^{tx}p^x(1-p)^{x-1}$$
let $q=1-p$
$$\sum_{x=1}^{\infty}e^{tx}p^xq^{x-1}$$
$$\sum_{x=0}^{\infty}(pe^t)q^x$$
$$=\frac{pe^t}{1-q}$$
that's as close as I can get to approximating the solution,
but the book says the answer is $$\frac{pe^t}{1-qe^t}$$

b. $$\sum_{x=r}^{\infty}\frac{(x-1)!}{(x-r)!(r-1)!}e^{tx}p^rq^{x-r}$$ where q=1-p

The negative binomial with parameters p and r is the distribution of a sum of r independent geometric random variables with parameter p. What do you know about the MGF of a sum of independent random variables?

RGV