Find the MGF of geometric,neg binomial dist.

  • Thread starter ArcanaNoir
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  • #1
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Homework Statement



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

Homework Equations



geometric distribution: [tex] f(x)=p^x(1-p)^{x-1} [/tex] where x=1,2,3...

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

MGF= [tex] E(e^{tx}) [/tex]

The Attempt at a Solution



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

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

Answers and Replies

  • #3
22,089
3,296

Homework Statement



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

Homework Equations



geometric distribution: [tex] f(x)=p^x(1-p)^{x-1} [/tex] where x=1,2,3...

negative binomial distribution: [tex] f(x)= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r} [/tex] 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

[tex]f(x)=p^x(1-p)[/tex]

and the negative binomial is

[tex]f(x)=\binom{x+r-1}{x}(1-p)^rp^x[/tex]

Could you recheck this first?
 
  • #4
768
4
oh good god. I was looking at point binomial.
so here's what I get now,
[tex] \sum_{x=1}^{\infty}e^{tx}p(1-p)^{x-1} [/tex] letting q= (1-p)
[tex] = \frac{p}{q} \sum_{x=1}^{\infty}e^{tx}q^{x} [/tex]
[tex] = \frac{p}{q} \sum_{x=1}^{\infty}(e^{t}q)^x [/tex]
 
  • #5
22,089
3,296
oh good god. I was looking at point binomial.
so here's what I get now,
[tex] \sum_{x=1}^{\infty}e^{tx}p(1-p)^{x-1} [/tex] letting q= (1-p)
[tex] = \frac{p}{q} \sum_{x=1}^{\infty}e^{tx}q^{x} [/tex]
[tex] = \frac{p}{q} \sum_{x=1}^{\infty}(e^{t}q)^x [/tex]
Yes, go on... You have a geometric series right now.
 
  • #6
768
4
the negative binomial I have as being : [tex] \binom{x-1}{r-1} p^r(1-p)^{x-r} [/tex]
which I figured equals : [tex] \frac{(x-1)!}{(x-1-(r-1))!(r-1)!}p^r(1-p)^{x-r} [/tex]
[tex] = \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r} [/tex]
 
  • #7
768
4
Yes, go on... You have a geometric series right now.
so, [itex] \frac{1}{1-r} [/itex] ?
[tex] ( \frac{p}{q} ) \frac{1}{1-e^{t}q} [/tex] ?
 
  • #8
22,089
3,296
so, [itex] \frac{1}{1-r} [/itex] ?
[tex] ( \frac{p}{q} ) \frac{1}{1-e^{t}q} [/tex] ?
That's the sum of the geometrci series when you start summing from 0. But you start summing from 1 here.
 
  • #9
768
4
but if I sum from zero, won't I have [tex] \frac{p}{q} \sum_{x=0}^{\infty}(e^{t}q)^{x+1} [/tex] oh wait. I see. hold on
 
  • #10
768
4
nevermind. I don't see. am I suppose to have x-1 at x=0 to do the sum?
 
  • #11
768
4
oop, there it is! found the answer to the geometric. thanks for the support on that.
 
  • #12
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement



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

Homework Equations



geometric distribution: [tex] f(x)=p^x(1-p)^{x-1} [/tex] where x=1,2,3...

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

MGF= [tex] E(e^{tx}) [/tex]

The Attempt at a Solution



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

b. [tex] \sum_{x=r}^{\infty}\frac{(x-1)!}{(x-r)!(r-1)!}e^{tx}p^rq^{x-r} [/tex] 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
 

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