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eppmsadnb EPPM Negative Binomial Saddlepoint
Computes a saddlepoint approximation to the probabilities for an Extended Poisson Process Model.
lambda vector of positive birth rates. Missing values (NAs) are allowed but will usually produce an NA result.
second Logical variable. If second=T the second term correction to the saddlepoint approximation is included.
Numerical value giving the log-probability that N = n -1 where n = length(lambda).
The function computes the log-probability mass for the count distribution resulting from a pure birth process at unit time. The waiting time until the next birth is exponential with mean lambda[n], where n is the number of births so far. Let N be the number of births at unit time. The probability that N = n depends on lambda[0:n]. The function takes the input vector to be lambda[0:n] and computes log P(N=n).

The  computation uses a saddlepoint approximation based on the negative binomial distribution. The probabilities are exact whenever the lambda can be sorted to form an arithmetic increasing sequence. Amongst other things, this means that eppmsadnb will compute binomial, negative binomial or Poisson probabilities exactly including the case where the sizes are not integers. In the worst cases, the probabilities are accurate to 2 significant figures.

The computation of probabilities for the pure birth process is central to extended Poisson process models for modelling count data.
Smyth, G. K., and Podlich, H. M. (2002). An improved saddlepoint approximation based on the negative binomial distribution for the general birth process. Computational Statistics 17, 17-28. [PDF]
Podlich, H. M., Faddy, M. J., and Smyth, G. K. (1999). Semi-parametric extended Poisson process models.
eppmsadno, eppmsadzc, S-Plus programs for EPPM by Heather Podlich.
# Probability that N=3 for Poisson with mean 5
# Same as dpois(3,mean=5)
> exp(eppmsadnb(c(5,5,5,5)))
[1] 0.1403739

# Probability that N=3 for negative binomial
# with size=2 successes and success probablity exp(-1)
# Same as dnbinom(3,size=2,prob=exp(-1))
> exp(eppmsadnb(c(2,3,4,5)))
[1] 0.1367322

# Probability that N=3 for binomial distribution
# with 5 trials and success probability 1-exp(-1)
# Same as dbinom(3,size=5,prob=1-exp(-1))
> exp(eppmsadnb(c(5,4,3,2)))
[1] 0.3418305

# Worst case - the lambda form two clusters at very different values
# Exact value is actually 0.368
> exp(eppmsadnb(c(100,1,100,1)))
[1] 0.3736466
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Gordon Smyth. Copyright © 1996-2016. Last modified: 10 February 2004