reservr (version 0.0.2)

quantile.Distribution: Quantiles of Distributions

Description

Produces quantiles corresponding to the given probabilities with configurable distribution parameters.

Usage

# S3 method for Distribution
quantile(x, probs = seq(0, 1, 0.25), with_params = list(), ..., .start = 0)

Value

The quantiles of x corresponding to probs with parameters with_params.

Arguments

x

A Distribution.

probs

Quantiles to compute.

with_params

Optional list of distribution parameters. Note that if x$has_capability("quantile") is false, with_params is assumed to contain only one set of parameters.

...

ignored

.start

Starting value if quantiles are computed numerically. Must be within the support of x.

Details

If x$has_capability("quantile") is true, this returns the same as x$quantile(probs, with_params = with_params). In this case, with_params may contain separate sets of parameters for each quantile to be determined.

Otherwise, a numerical estimation of the quantiles is done using the density and probability function. This method assumes with_params to cantain only one set of parameters. The strategy uses two steps:

1. Find the smallest and largest quantiles in probs using a newton method starting from .start.

2. Find the remaining quantiles with bisection using stats::uniroot().

Examples

# With quantiles available
dist <- dist_normal(sd = 1)
qqs <- quantile(dist, probs = rep(0.5, 3), with_params = list(mean = 1:3))
stopifnot(all.equal(qqs, 1:3))

# Without quantiles available
dist <- dist_erlangmix(shapes = list(1, 2, 3), scale = 1.0)
my_probs <- c(0, 0.01, 0.25, 0.5, 0.75, 1)
qqs <- quantile(
  dist, probs = my_probs,
  with_params = list(probs = list(0.5, 0.3, 0.2)), .start = 2
)

all.equal(dist$probability(qqs, with_params = list(probs = list(0.5, 0.3, 0.2))), my_probs)
# Careful: Numerical estimation of extreme quantiles can result in out-of-bounds values.
# The correct 0-quantile would be 0 in this case, but it was estimated < 0.
qqs[1L]