prob: Computing Probabilities of Hypercubes

In copula: Multivariate Dependence with Copulas

Computing Probabilities of Hypercubes

Description

Compute probabilities of a d-dimensional random vector U distributed according to a given copula x to fall in a hypercube (l,u], where l and u denote the lower and upper corners of the hypercube, respectively.

volume(<function>, *) computes the d-dimensional integral over the hypercube for “arbitrary” functions.

Usage

prob(x, l, u)

volume(object, ...) # generic function
## S3 method for class 'function'
volume(object, lower, upper, ...)

Arguments

x	copula of dimension d, that is, an object inheriting from Copula.
l, u, lower, upper	d-dimensional, numeric, lower and upper hypercube boundaries, respectively, satisfying 0 \le l_i \le u_i \le 1, for i\in{1,\dots,d}. No checks are done for volume() except for being of the same length.
object	a function to compute the volume of wrt to the specified hypercube.
...	additional arguments passed to the function object.

Value

prob() returns a numeric in [0,1] which is the probability P(l_i< U_i \le u_i).

volume() is the workhorse underlying prob() and its output is the d-volume, i.e., d-dimensional integral over the given hypercube, of the provided function.

See Also

pCopula(.).

Examples

## Construct a three-dimensional nested Joe copula with parameters
## chosen such that the Kendall's tau of the respective bivariate margins
## are 0.2 and 0.5.
theta0 <- copJoe@iTau(.2)
theta1 <- copJoe@iTau(.5)
C3 <- onacopula("J", C(theta0, 1, C(theta1, c(2,3))))

## Compute the probability of a random vector distributed according to
## this copula to fall inside the cube with lower point l and upper
## point u.
l <- c(.7,.8,.6)
u <- c(1,1,1)
prob(C3, l, u)

## ditto for a bivariate normal copula with rho = 0.8 :
Cn <- normalCopula(0.8)
(prob(Cn, c(.2,.4), c(.3,.6)) -> pr) # 0.0222...
## prob() just using volume(), internally:
pr == volume(function(z) pCopula(z, Cn), c(.2,.4), c(.3,.6))

copula documentation built on Feb. 20, 2026, 5:07 p.m.