Question

i found that the mean for negative binomial is qr/p but i'm not sure about my varaince i foud something complicated since my professor said that the moment generating funtion is (p^r)/(1-qe^t)^r

Answer 1

plz help

Answer 2

bek.cud u pls repiy.plz

Answer 3

I put your link in chat, Idk how to solve this

Answer 4

basollate plzz help

Answer 5

can i just ask what topic is this? (i.e. algebra, statistics, calculus, etc.)

Answer 6

statistics

Answer 7

hmm our statistics genius isnt online :/ though i think @Callisto can help you :) i hate statistics sorry :p

Answer 8

how can i find him\her

Answer 9

she'll be coming here..i tagged her...though she may be away from keyboard right now

Answer 10

oh my god

Answer 11

eish she said she ddnt learn it

Answer 12

guys can u pls find someone else who can help

Answer 13

hi.can u plz help zarkon

Answer 14

try this...compute \[E[X(X+1)]\]

Answer 15

how

Answer 16

Same way you would compute \[E(X)\]

Answer 17

any clue

Answer 18

\[\sum_{x=r}^{\infty}x(x+1){x-1\choose x-r}(1-p)^rp^{x-r}\]

Answer 19

write \[{x-1\choose x-r}\] using factorials

Answer 20

\[\sum_{x=r}^{\infty}x(x+1){x-1\choose x-r}(1-p)^rp^{x-r}\]

\[\sum_{x=r}^{\infty}x(x+1)\frac{(x-1)!}{(x-1-(x-r))!(x-r)!}(1-p)^rp^{x-r}\]

Answer 21

\[\sum_{x=r}^{\infty}\frac{(x+1)!}{(x-1-x+r)!(x-r)!}(1-p)^rp^{x-r}\]

\[\sum_{x=r}^{\infty}\frac{(x+1)!}{(r-1)!(x-r)!}(1-p)^rp^{x-r}\]

Answer 22

\[r(r+1)\sum_{x=r}^{\infty}\frac{(x+1)!}{(r+1)!(x-r)!}(1-p)^rp^{x-r}\]

\[\frac{r(r+1)}{(1-p)^2}\sum_{x=r}^{\infty}\frac{(x+1)!}{(r+1)!(x-r)!}(1-p)^{r+2}p^{x-r}\]

Answer 23

\[\frac{r(r+1)}{(1-p)^2}\sum_{x=r}^{\infty}\frac{((x+2)-1)!}{((r+2)-1)!((x+2)-1-((r+2)-1))!}(1-p)^{r+2}p^{x+2-(r+2)}\]

\[y=x+2\]

\[\frac{r(r+1)}{(1-p)^2}\sum_{y=r+2}^{\infty}{y-1\choose (r+2)-1}(1-p)^{r+2}p^{y-(r+2)}\]

\[\frac{r(r+1)}{(1-p)^2}\]

Answer 24

\[Var(X)=E(X^2)-[E(X)]^2=E(X^2)+E(X)-E(X)-[E(X)]^2\]

\[=E(X(X+1))-E(X)-[E(X)]^2\]

plug in your known values

Answer 25

but what if i have to use the moment generating function to find the variance

Answer 26

oh...well that is much easier

Answer 27

can u plz show me how

Answer 28

\[\sigma^2=E(x^2)-E(x)^2\]

\[=M''(0)-[M'(0)]^2\]

Answer 29

my MGF is (p/1-qe^t)^r and the mean i found was qr/p

Answer 30

but i'm having trouble finding M''(t)

Answer 31

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

\[M'(t)=p^r(-r)(1-qe^{t})^{-r-1}(-qe^t)\]

ok

Answer 32

\[M'(t)=p^r(-r)(1-qe^{t})^{-r-1}(-qe^t)\]

\[M''(t)=p^rrq\frac{d}{dt}[(1-qe^{t})^{-r-1}e^t]\]

\[M''(t)=p^rrq[(1-qe^{t})^{-r-1}e^t+(-r-1)(1-qe^t)^{-r-2}(-qe^t)e^t]\]

Answer 33

bekketso do u get this

Answer 34

let me try to substitute

Answer 35

do u get the variance