VGAMdata (version 1.1-9)

Oipospois: One-Inflated Positive Poisson Distribution

Description

Density, distribution function, quantile function and random generation for the one-inflated positive Poisson distribution with parameter pstr1.

Usage

doipospois(x, lambda, pstr1 = 0, log = FALSE)
poipospois(q, lambda, pstr1 = 0)
qoipospois(p, lambda, pstr1 = 0)
roipospois(n, lambda, pstr1 = 0)

Value

doipospois gives the density,

poipospois gives the distribution function,

qoipospois gives the quantile function, and

roipospois generates random deviates.

Arguments

x, p, q, n

Same as Pospois.

lambda

Vector of positive means.

pstr1

Probability of a structural one (i.e., ignoring the positive Poisson distribution), called \(\phi\). The default value of \(\phi = 0\) corresponds to the response having a positive Poisson distribution.

log

Logical. Return the logarithm of the answer?

Author

T. W. Yee

Details

The probability function of \(Y\) is 1 with probability \(\phi\), and \(PosPoisson(\lambda)\) with probability \(1-\phi\). Thus $$P(Y=1) =\phi + (1-\phi) P(W=1)$$ where \(W\) is distributed as a positive \(Poisson(\lambda)\) random variate.

See Also

Pospois, oapospoisson, oipospoisson, otpospoisson, pospoisson, dpois, poissonff.

Examples

lambda <- 3; pstr1 <- 0.2; x <- (-1):7
(ii <- doipospois(x, lambda, pstr1 = pstr1))
table(roipospois(100, lambda, pstr1 = pstr1))
round(doipospois(1:10, lambda, pstr1 = pstr1) * 100)  # Similar?

if (FALSE)  x <- 0:10
par(mfrow = c(2, 1))  # One-Inflated Positive Poisson
barplot(rbind(doipospois(x, lambda, pstr1 = pstr1),
                dpospois(x, lambda)),
   beside = TRUE, col = c("blue", "orange"),
   main = paste0("OIPP(", lambda, ", pstr1 = ", pstr1,
                 ") (blue) vs PosPoisson(", lambda, ") (orange)"),
   names.arg = as.character(x))
# 0-deflated Pos Poisson:
deflat.limit <- -lambda / (expm1(lambda) - lambda)
newpstr1 <- round(deflat.limit, 3) + 0.001  # Near the boundary
barplot(rbind(doipospois(x, lambda, pstr1 = newpstr1),
                dpospois(x, lambda)),
   beside = TRUE, col = c("blue","orange"),
   main = paste0("ODPP(", lambda, ", pstr1 = ", newpstr1,
                 ") (blue) vs PosPoisson(", lambda, ") (orange)"),
   names.arg = as.character(x))