Re: Mean of Geometric and Negative Binomial distributions
- To: mathgroup at smc.vnet.net
- Subject: [mg24087] Re: [mg24066] Mean of Geometric and Negative Binomial distributions
- From: "Richard Finley" <rfinley at medicine.umsmed.edu>
- Date: Fri, 23 Jun 2000 02:26:55 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Gareth,
With the geometric distribution it is because there is a BUG...with the negative
binomial, it is a misunderstanding. Your reference formulae are correct. Clearly
the problem with Mathematica's implementation of the Geometric Distribution
is that they have neglected the fact that you cannot have a success if there are
no trials (ie if n=0) but in their case PDF[GeometricDistribution[p],0] = p
instead of DF[GeometricDistribution[p],1] = p. So since it is off by one you can
get the correct mean by calculating
Sum[ n PDF[GeometricDistribution[p], n-1], {n,1,Infinity}] which gives
1/p as it should.
For the negative binomial distribution, there is no bug ...but only appears so in the
sense that they calculate a nonstandard result....that is the mean of the total
number of "failures" to get r successes rather than the number of "trials" to get
r successes. You can see this by calculating the results yourself directly rather
than using Mean....
"mean number of trials to get r successes"....
Sum[ (n+r) PDF[NegativeBinomialDistribution[r,p], n] , {n,0,Infinity}] = r/p
"mean number of failures before getting r successes"....
Sum[ n PDF[NegativeBinomialDistribution[r,p], n], {n, 0, Infinity}] = r (1-p)/p
(Note that I use r for the number of successes and n is the number of failures as opposed to the n in your post which is the number of successes).
regards, RF
>>> "Gareth J. Russell" <russell at cerc.columbia.edu> 06/21/00 11:01PM >>>
Dear Group,
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.
Thanks,
Gareth
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Dr. Gareth J. Russell
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