Find the MGF of geometric,neg binomial dist.

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In summary, the MGF of the geometric distribution is given by (pe^t)/(1-qe^t), while the MGF of the negative binomial distribution with parameters p and r is the distribution of a sum of r independent geometric random variables with parameter p.
  • #1
ArcanaNoir
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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
 
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  • #3
ArcanaNoir said:

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
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
ArcanaNoir said:
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
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
micromass said:
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
ArcanaNoir said:
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
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
nevermind. I don't see. am I suppose to have x-1 at x=0 to do the sum?
 
  • #11
oop, there it is! found the answer to the geometric. thanks for the support on that.
 
  • #12
ArcanaNoir said:

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
 

1. What is the formula for finding the MGF of a geometric distribution?

The moment generating function (MGF) of a geometric distribution is given by M(t) = pet / (1-qet), where p is the probability of success and q = 1-p is the probability of failure.

2. How is the MGF of a geometric distribution different from that of a negative binomial distribution?

While both distributions involve a sequence of independent Bernoulli trials, the geometric distribution counts the number of trials until the first success, while the negative binomial distribution counts the number of trials needed to achieve a certain number of successes. As a result, the MGF for a negative binomial distribution involves an additional parameter for the number of successes.

3. Can the MGF of a geometric distribution be used to find the mean and variance?

Yes, the mean and variance of a geometric distribution can be found by taking the first and second derivatives of the MGF, respectively.

4. Is there a specific range of values for which the MGF of a geometric distribution is defined?

Yes, the MGF is only defined for values of t within the interval (-ln q, ln p). Outside of this interval, the MGF does not exist.

5. How can the MGF of a geometric distribution be used in practical applications?

The MGF can be used to calculate the probability of a specific number of successes within a given number of trials, as well as to determine the expected value and variance of the distribution. This can be useful in fields such as finance, where the geometric distribution is often used to model the number of trials until an event occurs.

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