pdf.ZTPoisson {distributions3}

Evaluate the probability mass function of a zero-truncated Poisson distribution

Description

Evaluate the probability mass function of a zero-truncated Poisson distribution

Usage

## S3 method for class 'ZTPoisson'
pdf(d, x, drop = TRUE, elementwise = NULL, ...)

## S3 method for class 'ZTPoisson'
log_pdf(d, x, drop = TRUE, elementwise = NULL, ...)

Arguments

d

A ZTPoisson object created by a call to ZTPoisson().

x

A vector of elements whose probabilities you would like to determine given the distribution d.

drop

logical. Should the result be simplified to a vector if possible?

elementwise

logical. Should each distribution in d be evaluated at all elements of x (elementwise = FALSE, yielding a matrix)? Or, if d and x have the same length, should the evaluation be done element by element (elementwise = TRUE, yielding a vector)? The default of NULL means that elementwise = TRUE is used if the lengths match and otherwise elementwise = FALSE is used.

...

Arguments to be passed to dztpois. Unevaluated arguments will generate a warning to catch mispellings or other possible errors.

Value

In case of a single distribution object, either a numeric vector of length probs (if drop = TRUE, default) or a matrix with length(x) columns (if drop = FALSE). In case of a vectorized distribution object, a matrix with length(x) columns containing all possible combinations.

Examples

## set up a zero-truncated Poisson distribution
X <- ZTPoisson(lambda = 2.5)
X

## standard functions
pdf(X, 0:8)
cdf(X, 0:8)
quantile(X, seq(0, 1, by = 0.25))

## cdf() and quantile() are inverses for each other
quantile(X, cdf(X, 3))

## density visualization
plot(0:8, pdf(X, 0:8), type = "h", lwd = 2)

## corresponding sample with histogram of empirical frequencies
set.seed(0)
x <- random(X, 500)
hist(x, breaks = -1:max(x) + 0.5)