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Re: Mean of Geometric and Negative Binomial distributions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg24092] Re: [mg24066] Mean of Geometric and Negative Binomial distributions
*From*: BobHanlon at aol.com
*Date*: Fri, 23 Jun 2000 02:27:00 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
In a message dated 6/22/2000 1:18:29 AM, russell at cerc.columbia.edu writes:
>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.
I don't have the CRC tables; however, Mathematica's results agree with
Abramowitz and Stegun
Needs["Statistics`DiscreteDistributions`"];
distr = GeometricDistribution[p];
PDF[distr, x]
(1 - p)^n*p
As stated in the on-line documentation, the geometric distribution
GeometricDistribution[p] is the distribution of the total number of trials
BEFORE the first success occurs,where the probability of success in each
trial is p. Note that there are (n+1) trials: n failures BEFORE a success and
the success. Consequently, the Domain of the distribution includes zero.
Domain[distr]
Range[0, Infinity]
Verifying validity of the distribution
Sum[PDF[distr, n], {n, 0, Infinity}] == 1
True
Verifying the Mean
Sum[n*PDF[distr, n], {n, 0, Infinity}] == Mean[distr] // Simplify
True
If, however, we look at the TOTAL number of trials, that is the failures plus
the one success, then the distribution is p*(1-p)^(n-1) and
Sum[p*(1 - p)^(n - 1), {n, 1, Infinity}] == 1
True
Sum[n*p*(1 - p)^(n - 1), {n, 1, Infinity}] // Simplify
1/p
I suspect that you are working with this second definition.
As stated in the on-line documentation, the negative binomial distribution
NegativeBinomialDistribution[n, p] is the distribution of the number of
failures that occur in a sequence of trials before n successes have
occurred,where the probability of success in each trial is p.
distr = NegativeBinomialDistribution[n, p];
PDF[distr, m]
(1 - p)^m*p^n*Binomial[m + n - 1, n - 1]
Verifying validity of the distribution
Sum[PDF[distr, x], {x, 0, Infinity}] == 1
True
Verifying the Mean
Sum[x*PDF[distr, x], {x, 0, Infinity}] == Mean[distr] // Simplify
True
Again, if the distribution you are dealing with is describing the TOTAL
number of trials then
Sum[(1 - p)^(m - n) * p^n * Binomial[m - 1, n - 1], {m, n, Infinity}] == 1
True
Sum[m * (1 - p)^(m - n) * p^n * Binomial[m - 1, n - 1], {m, n, Infinity}]
n/p
In general, you need to verify that you are working with the same
definitions.
Bob Hanlon
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