shape1, shape2 and scale.dinvburr(x, shape1, shape2, rate = 1, scale = 1/rate,
log = FALSE)
pinvburr(q, shape1, shape2, rate = 1, scale = 1/rate,
lower.tail = TRUE, log.p = FALSE)
qinvburr(p, shape1, shape2, rate = 1, scale = 1/rate,
lower.tail = TRUE, log.p = FALSE)
rinvburr(n, shape1, shape2, rate = 1, scale = 1/rate)
minvburr(order, shape1, shape2, rate = 1, scale = 1/rate)
levinvburr(limit, shape1, shape2, rate = 1, scale = 1/rate,
order = 1)length(n) > 1, the length is
taken to be the number required.TRUE, probabilities/densities
$p$ are returned as $\log(p)$.TRUE (default), probabilities are
$P[X \le x]$, otherwise, $P[X > x]$.dinvburr gives the density,
invburr gives the distribution function,
qinvburr gives the quantile function,
rinvburr generates random deviates,
minvburr gives the $k$th raw moment, and
levinvburr gives the $k$th moment of the limited loss
variable. Invalid arguments will result in return value NaN, with a warning.
shape1 $=
\tau$, shape2 $= \gamma$ and scale
$= \theta$, has density:
$$f(x) = \frac{\tau \gamma (x/\theta)^{\gamma \tau}}{ x [1 + (x/\theta)^\gamma]^{\tau + 1}}$$
for $x > 0$, $\tau > 0$, $\gamma > 0$ and
$\theta > 0$.The Inverse Burr is the distribution of the random variable $$\theta \left(\frac{X}{1 - X}\right)^{1/\gamma},$$ where $X$ has a Beta distribution with parameters $\tau$ and $1$.
The Inverse Burr distribution has the following special cases:
shape1
== 1;shape2 == 1;shape1 == shape2.The $k$th raw moment of the random variable $X$ is $E[X^k]$ and the $k$ limited moment at some limit $d$ is $E[\min(X, d)]$.
exp(dinvburr(2, 3, 4, 5, log = TRUE))
p <- (1:10)/10
pinvburr(qinvburr(p, 2, 3, 1), 2, 3, 1)
minvburr(2, 1, 2, 3) - minvburr(1, 1, 2, 3) ^ 2
levinvburr(10, 1, 2, 3, order = 2)Run the code above in your browser using DataLab