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RE: mean of geometric and negative binomial distributions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg24077] RE: [mg24066] mean of geometric and negative binomial distributions
*From*: Tomas.Garza at smc.vnet.net
*Date*: Fri, 23 Jun 2000 02:26:44 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
It is a matter of definition. Most sources define the geometric distribution as the waiting time for the occurrence of a binomial event, i.e., the number of trials up to and including the first "success". In this case the expected value is 1/p. However, Mathematica uses the definition "number of trials before the first success", and then the expected value is (1 - p)/p. Here the number of trials is one less than in the former case, and then the expected value is equal to 1/p - 1 = (1 - p)/p. A similar reasoning provides the answer for the negative binomial distribution (which is the distribution of a sum of independent and identically distributed geometric random variables).
Tomas Garza
Mexico City (temporarily in Barcelona)
----
Gareth Russell wrote:
Can anyone tell me why Mathematica returns (1-p)/p for Mean[
GeometricDistribution[p]] and n(1-p)/p for Mean[NegativeBinomialDistribution[
p]], when all sources I have to hand (such as CRC Standard Mathematical
Formulae) give these as 1/p and n/p respectively? The variance expressions
agree with CRC, it's just the means that are different.
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