Question

find the moment genariting function for the negative binomial

Answer 1

In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. People will still help, so don't worry.

Please consider rewriting your post.

Answer 2

where are you stuck?

Answer 3

are you using \[{k+r-1\choose k}(1-p)^{r}p^{k}\]
\[k\in\{0,1,2,3,\ldots\}\]

Answer 4

\[M(t)=\sum_{k=0}^{\infty}e^{kt}{k+r-1\choose k}(1-p)^{r}p^{k}\]

\[M(t)=\sum_{k=0}^{\infty}{k+r-1\choose k}(1-p)^{r}(e^tp)^{k}\]

\[=\cdots\]

Answer 5

the proble i have is that i don't know how to remove the factorials

Answer 6

ah...you don't need to :)

Answer 7

\[M(t)=\sum_{k=0}^{\infty}{k+r-1\choose k}(1-p)^{r}(e^tp)^{k}\]

\[M(t)=(1-p)^{r}\sum_{k=0}^{\infty}{k+r-1\choose k}(e^tp)^{k}\]

\[M(t)=\frac{(1-p)^{r}}{(1-e^tp)^r}\sum_{k=0}^{\infty}{k+r-1\choose k}(1-e^tp)^r(e^tp)^{k}\]


\[M(t)=\frac{(1-p)^{r}}{(1-e^tp)^r}\cdot 1=\frac{(1-p)^{r}}{(1-e^tp)^r}\]

for \[e^tp<1\]

Answer 8

notice that

\[\sum_{k=0}^{\infty}{k+r-1\choose k}(1-e^tp)^r(e^tp)^{k}\]

let \[q=e^tp\] then we have

\[\sum_{k=0}^{\infty}{k+r-1\choose k}(1-q)^r(q)^{k}\]

which is the sum of a negative binomial over all of its support...that is equal to 1

Answer 9

P(X=x)=\[\left(\begin{matrix}x-1 \\ r-1\end{matrix}\right)\]\[p^{r}\](1-P)\[^{x-r}\] ,x=r,r+1,...
(a) find the moment generating function of X

Answer 10

in the text book they are saying that the moment generating function is (p/(1-qe^t))^r
now i'm confused

Answer 11

the question is like that

Answer 12

so they have the negative binomial in an alternate form. you can still apply the same technique I used above on this form of the negative binomial

Answer 13

oh thanks let me start again

Answer 14

you can also use a substitution with the solution I gave above...

Answer 15

\[M(t)=\frac{(1-p)^{r}}{(1-e^tp)^r}\]

redefine p to be 1-p then we get

\[M(t)=\frac{(1-(1-p))^{r}}{(1-e^t(1-p))^r}\]
\[=\frac{(p)^{r}}{(1-e^tq)^r}\]

where q=1-p

\[=\left(\frac{p}{1-qe^t}\right)^r\]

Answer 16

I think if we are going to change it a little if we are starting from k=r and not k=0

Answer 17

if you are starting at k=r I believe we should get

\[\left(\frac{pe^t}{1-qe^t}\right)^r\]

Answer 18

\[\sum_{}^{}\left(\begin{matrix}x-1 \\ r-1\end{matrix}\right)p ^{r}e ^{tr}(1-p)^{x-r}\]
this comes before that one

Answer 19

\[M(t)=\sum_{x=0}^{\infty}e^{xt}{x-1\choose r-1}p^{r}(1-p)^{x-r}\]

\[M(t)=\sum_{x=0}^{\infty}e^{xt-r+r}{x-1\choose r-1}p^{r}(1-p)^{x-r}\]

\[M(t)=\sum_{x=0}^{\infty}e^{xt-tr}e^{tr}{x-1\choose r-1}p^{r}(1-p)^{x-r}\]

\[M(t)=\sum_{x=0}^{\infty}e^{(x-r)t}e^{tr}{x-1\choose r-1}p^{r}(1-p)^{x-r}\]

\[M(t)=\sum_{x=0}^{\infty}e^{tr}{x-1\choose r-1}p^{r}((1-p)e^t)^{x-r}\]

\[M(t)=(pe^{t})^r\sum_{x=0}^{\infty}{x-1\choose r-1}((1-p)e^t)^{x-r}\]

\[M(t)=\frac{(pe^{t})^r}{(1-(1-p)e^t)^r}\sum_{x=0}^{\infty}{x-1\choose r-1}(1-(1-p)e^t)^r((1-p)e^t)^{x-r}\]

\[M(t)=\frac{(pe^{t})^r}{(1-(1-p)e^t)^r}\]

\[M(t)=\frac{(pe^{t})^r}{(1-qe^t)^r}\]

\[M(t)=\left(\frac{pe^{t}}{1-qe^t}\right)^r\]

Answer 20

dang...x starts from x=r not x=0...so just make those changes...I fort to

Answer 21

*i forgot to

Answer 22

(Pe\[^{t}\])\[^{r}\]/(1-P)\[^{r}\]\[\sum_{?}^{?}\]\[\left(\begin{matrix}x-1 \\ r-1\end{matrix}\right)\](1-p)\[^{k}\]

Answer 23

\[(pe ^{t})^{r} /(1-p)^{r}\]

Answer 24

thanks