Exponential Poisson Distribution Family Function
exppoisson.RdEstimates the two parameters of the exponential Poisson distribution by maximum likelihood estimation.
Arguments
- lshape, lrate
Link function for the two positive parameters. See
Linksfor more choices.- ishape, irate
Numeric. Initial values for the
shapeandrateparameters. Currently this function is not intelligent enough to obtain better initial values.- zero
Details
The exponential Poisson distribution has density function
$$f(y; \beta = rate, \lambda = shape) =
\frac{\lambda \beta}{1 - e^{-\lambda}} \,
e^{-\lambda - \beta y + \lambda \exp{(-\beta y)}}$$
where \(y > 0\),
and the parameters shape, \(\lambda\),
and rate, \(\beta\), are positive.
The distribution implies a population facing discrete
hazard rates which are multiples of a base hazard.
This VGAM family function requires the hypergeo package
(to use their genhypergeo function).
The median is returned as the fitted value.
Value
An object of class "vglmff" (see vglmff-class).
The object is used by modelling functions such as vglm
and vgam.
References
Kus, C., (2007). A new lifetime distribution. Computational Statistics and Data Analysis, 51, 4497–4509.
Examples
if (FALSE) { # \dontrun{
shape <- exp(1); rate <- exp(2)
rdata <- data.frame(y = rexppois(n = 1000, rate = rate, shape = shape))
library("hypergeo") # Required!
fit <- vglm(y ~ 1, exppoisson, data = rdata, trace = FALSE, maxit = 1200)
c(with(rdata, median(y)), head(fitted(fit), 1))
coef(fit, matrix = TRUE)
Coef(fit)
summary(fit)
} # }