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eppmsadnb |
EPPM Negative
Binomial Saddlepoint |

**DESCRIPTION**- Computes a saddlepoint approximation to the probabilities for an Extended Poisson Process Model.
**USAGE**`eppmsadnb(lambda)`**REQUIRED ARGUMENTS**`lambda`vector of positive birth rates. Missing values (NAs) are allowed but will usually produce an NA result. **OPTIONAL ARGUMENTS**`second`Logical variable. If `second=T`the second term correction to the saddlepoint approximation is included.**VALUE**- Numerical value giving the log-probability that N = n -1 where n = length(lambda).
**DETAILS**- 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. **REFERENCES**- 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.
**SEE ALSO**- eppmsadno, eppmsadzc, S-Plus programs for EPPM by Heather Podlich.

**EXAMPLES**`# 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*