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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.

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) 

so  mgf = E[e^(tx)] =  SUM[e^(tr).C(r-1,r-m).p^m.q^(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   
Associated Topics:
College Probability
High School Probability

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