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/
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