Skip to contents

Evaluate the cumulative distribution function of a Poisson distribution

Usage

# S3 method for Poisson
cdf(d, x, drop = TRUE, elementwise = NULL, ...)

Arguments

d

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

x

A vector of elements whose cumulative 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 ppois. 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.seed(27)

X <- Poisson(2)
X
#> [1] "Poisson distribution (lambda = 2)"

random(X, 10)
#>  [1] 5 0 4 1 1 1 0 0 1 1

pdf(X, 2)
#> [1] 0.2706706
log_pdf(X, 2)
#> [1] -1.306853

cdf(X, 4)
#> [1] 0.947347
quantile(X, 0.7)
#> [1] 3

cdf(X, quantile(X, 0.7))
#> [1] 0.8571235
quantile(X, cdf(X, 7))
#> [1] 7