VGAM (version 1.1-2)

erlang: Erlang Distribution

Description

Estimates the scale parameter of the Erlang distribution by maximum likelihood estimation.

Usage

erlang(shape.arg, lscale = "loglink", imethod = 1, zero = NULL)

Arguments

shape.arg

The shape parameters. The user must specify a positive integer, or integers for multiple responses. They are recycled by.row = TRUE according to matrix.

lscale

Link function applied to the (positive) \(scale\) parameter. See Links for more choices.

imethod, zero

See CommonVGAMffArguments for more details.

Value

An object of class "vglmff" (see vglmff-class). The object is used by modelling functions such as vglm and vgam.

Details

The Erlang distribution is a special case of the gamma distribution with shape that is a positive integer. If shape.arg = 1 then it simplifies to the exponential distribution. As illustrated in the example below, the Erlang distribution is the distribution of the sum of shape.arg independent and identically distributed exponential random variates.

The probability density function of the Erlang distribution is given by $$f(y) = \exp(-y/scale) y^{shape-1} scale^{-shape} / \Gamma(shape)$$ for known positive integer \(shape\), unknown \(scale > 0\) and \(y > 0\). Here, \(\Gamma(shape)\) is the gamma function, as in gamma. The mean of Y is \(\mu=shape \times scale\) and its variance is \(shape \times scale^2\). The linear/additive predictor, by default, is \(\eta=\log(scale)\).

References

Most standard texts on statistical distributions describe this distribution, e.g.,

Forbes, C., Evans, M., Hastings, N. and Peacock, B. (2011) Statistical Distributions, Hoboken, NJ, USA: John Wiley and Sons, Fourth edition.

See Also

gammaR, exponential, simulate.vlm.

Examples

# NOT RUN {
rate <- exp(2); myshape <- 3
edata <- data.frame(y = rep(0, nn <- 1000))
for (ii in 1:myshape)
  edata <- transform(edata, y = y + rexp(nn, rate = rate))
fit <- vglm(y ~ 1, erlang(shape = myshape), data = edata, trace = TRUE)
coef(fit, matrix = TRUE)
Coef(fit)  # Answer = 1/rate
1/rate
summary(fit)
# }