VaRES (version 1.0.1)

moexp: Marshall-Olkin exponential distribution

Description

Computes the pdf, cdf, value at risk and expected shortfall for the Marshall-Olkin exponential distribution due to Marshall and Olkin (1997) given by $$\begin{array}{ll} &\displaystyle f (x) = \frac {\displaystyle \lambda \exp (\lambda x)} {\displaystyle \left[ \exp (\lambda x) - 1 + a \right]^2}, \\ &\displaystyle F (x) = \frac {\displaystyle \exp (\lambda x) - 2 + a}{\displaystyle \exp (\lambda x) - 1 + a}, \\ &\displaystyle {\rm VaR}_p (X) = \frac {1}{\lambda} \log \frac {2 - a - (1 - a) p}{1 - p}, \\ &\displaystyle {\rm ES}_p (X) = \frac {1}{\lambda} \log \left[ 2 - a - (1 - a) p \right] - \frac {2 - a}{\lambda (1 - a) p} \log \frac {2 - a - (1 - a) p}{2 - a} + \frac {1 - p}{\lambda p} \log (1 - p) \end{array}$$ for \(x > 0\), \(0 < p < 1\), \(a > 0\), the first scale parameter and \(\lambda > 0\), the second scale parameter.

Usage

dmoexp(x, lambda=1, a=1, log=FALSE)
pmoexp(x, lambda=1, a=1, log.p=FALSE, lower.tail=TRUE)
varmoexp(p, lambda=1, a=1, log.p=FALSE, lower.tail=TRUE)
esmoexp(p, lambda=1, a=1)

Arguments

x

scaler or vector of values at which the pdf or cdf needs to be computed

p

scaler or vector of values at which the value at risk or expected shortfall needs to be computed

a

the value of the first scale parameter, must be positive, the default is 1

lambda

the value of the second scale parameter, must be positive, the default is 1

log

if TRUE then log(pdf) are returned

log.p

if TRUE then log(cdf) are returned and quantiles are computed for exp(p)

lower.tail

if FALSE then 1-cdf are returned and quantiles are computed for 1-p

Value

An object of the same length as x, giving the pdf or cdf values computed at x or an object of the same length as p, giving the values at risk or expected shortfall computed at p.

References

S. Nadarajah, S. Chan and E. Afuecheta, An R Package for value at risk and expected shortfall, submitted

Examples

# NOT RUN {
x=runif(10,min=0,max=1)
dmoexp(x)
pmoexp(x)
varmoexp(x)
esmoexp(x)
# }