Description Usage Arguments Author(s) References Examples

View source: R/pgfDpolyaaeppli.R

This function calculates value of the pgf's first derivative of the Polya Aeppli distribution.

1 | ```
pgfDpolyaaeppli(s, params)
``` |

`s` |
Value of the parameter of the pgf. It should be from interval [-1,1]. In the opposite pgf diverges. |

`params` |
List of the parameters of the Polya Aeppli distribution, such that params<-c(theta,p), where theta is the positive number, and p is the probability. |

S. Nadarajah, B. V. Popovic, M. M. Ristic

Johnson N, Kotz S, Kemp A (1992) Univariate Discrete Distributions, John Wiley and Sons, New York

http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap3.pdf

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | ```
params<-c(5,.4)
pgfDpolyaaeppli(.5,params)
## The function is currently defined as
pgfDpolyaaeppli <- function(s,params) {
k<-s[abs(s)>1]
if (length(k)>0)
warning("At least one element of the vector s are out of interval [-1,1]")
if (length(params)<2)
stop("At least one value in params is missing")
if (length(params)>2)
stop("The length of params is 2")
theta<-params[1]
p<-params[2]
if (theta<=0)
stop ("Parameter theta must be positive")
if ((p>=1)|(p<=0))
stop ("Parameter p belongs to the interval (0,1)")
theta*(1-p)/(1-p*s)^2*exp(theta/p*((1-p)/(1-p*s)-1))
}
``` |

Compounding documentation built on May 30, 2017, 4:02 a.m.

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