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### Moment Generating Functions

```
Date: 03/30/99 at 12:46:17
From: Pritish Nandy
Subject: Moment generating functions

1. What can be established about the sum of negative binomial random
variables? Using the moment generating function gave me a
complicated mess that did not seem to resemble any of the known
MGF of discrete distributions.

2. How can we find the MGF of the negative binomial distribution
and prove that the sum of geometric random variables is a negative
binomial using moment generating functions?

I have been trying hard to figure this out. Any assistance would be
tremendously appreciated.

Thank you.
Pritish
```

```
Date: 03/30/99 at 15:08:55
From: Doctor Anthony
Subject: Re: Moment generating functions

The Negative Binomial or Pascal distribution.

p = probability of success at each trial = constant

R = r if there are m-1 successes in r-1 trials and the mth success
occurs at the rth trial.

The probability of this is  C(r-1,m-1)p^(m-1).q^(r-m).p

= C(r-1,m-1).p^m.q^(r-m)

= C(r-1,m-1).p^m.(1-p)^(r-m)

also      = C(r-1,r-m).p^m.(1-p)^(r-m)

inf.
so  mgf = E[e^(tx)] =  SUM[e^(tr).C(r-1,r-m).p^m.q^(r-m)]
r=m

= e^(tm).C(m-1,0).p^m.q^0 + e^(t(m+1).C(m,1).p^(m+1).q + ....

= e^(tm)[p^m + e^t.C(m,1)p^(m+1).q + e^(2t).C(m+1,2).p^(m+2).q^2 +
...

= e^(tm).p^m[1 + C(m,1)e^t.(pq) + C(m+1,2)e^(2t).(pq)^2 + ......

= (p.e^t)^m[1 + C(m,1).(pq.e^t) + C(m+1,2).(pq.e^t)^2 + ........ (1)

consider the expansion of

(1 - pq.e^t)^(-m)

= 1 + m(pq.e^t) + m(m+1)/2 (pq.e^t)^2 + m(m+1)(m+2)/3! (pq.e^t)^3 +

= 1 + C(m,1)(pq.e^t) + C(m+1,2)(pq.e^t)^2 + C(m+2,3)(pq.e^t)^3 + ...

comparing with equation (1) we see that this represents the series in
(1).  It follows that

(p.e^t)^m          (p.e^t)^m
mgf = ------------   =   --------------
(1 - pq.e^t)^m      [1 - p(1-p).e^t]^m

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

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