Hi I've got this question that I've been struggling with for hours.
$f(x)=\binom{r+x-1}{x}p^{r}(1-p)^{x}$
Show that:
$m_{x}(u)=E[e^{uX}]=(\frac{p}{1-(1-p)e^{u}})^{r}$
I've tried this so far:
$m_{x}(u)=E[e^{uX}]=\sum_{x=0}^{r+x-1}\binom{r+x-1}{x}e^{uX}p^{r}(1-p)^{x}$ $=\sum_{x=0}^{n}\binom{r+x-1}{x}p^{r}(e^{u}(1-p))^{x}$
But I get stuck here. Am I on the right track?
Any help much appreciated.
Firstly, The distribution is not binomial but negative binomial with parameter $p$ and $r$.
$$E(e^{uX}) = \sum\limits_{x=0}^{\infty} e^{ux} \binom{r+x-1}{x}p^r(1-p)^x = \sum\limits_{x=0}^{\infty} \binom{r+x-1}{x}p^r(e^u(1-p))^x$$
Multiply and divided by the constant $(1-(e^{u}(1-p))^r$ and take $p^r$ out.
$$ E(e^{uX})= \frac{p^r}{(1-(e^{u}(1-p)))^r}\sum\limits_{x=0}^{\infty} \binom{r+x-1}{x}(1-(e^{u}(1-p)))^r(e^{u}(1-p))^x$$
The summation equals to $1$ since it is the sum of the probability of each point in the sample space of the negative binomial distribution with parameter $(1-(e^{u}(1-p))$ and $r$.
$$E(e^{uX}) = \left(\frac{p}{(1-(e^{u}(1-p)))}\right)^r$$