lmoms
. The parameter $\psi$ is the right truncation of the distribution, and $\alpha$ is a scale parameter, letting $\beta = 1/\alpha$ to match nomenclature of Vogel and others (2008), and recalling the L-moments in terms of the parameters and letting $\eta = \mathrm{exp}(-\beta\psi)$,$$\lambda_1 = \frac{1 - \eta + \eta\log(\eta)}{\beta(1-\eta)}\mbox{,}$$ $$\lambda_2 = \frac{1 + 2\eta\log(\eta) - \eta^2}{2\beta(1-\eta)^2}\mbox{, and}$$ $$\tau_2 = \frac{\lambda_2}{\lambda_1} = \frac{1 + 2\eta\log(\eta) - \eta^2}{2(1-\eta)[1-\eta+\eta\log(\eta)]}\mbox{,}$$
and $\tau_2$ is a monotonic function of $\eta$ is decreasing from $\tau_2 = 1/2$ at $\eta = 0$ to $\tau_2 = 1/3$ at $\eta = 1$ the parameters are readily solved given $\tau_2 = [1/3, 1/2]$, the uniroot
function can be used to solve for $\eta$ with a starting interval of $(0, 1)$, then the parameters in terms of the parameters are
$$\alpha = \frac{1 - \eta + \eta\log(\eta)}{(1 - \eta)\lambda_1}\mbox{\ and}$$ $$\psi = -\log(\eta)/\alpha\mbox{.}$$
If the $\eta$ is rooted as equaling zero, then it is assumed that $\hat\tau_2 \equiv \tau_2$ and the exponential distribution triggered, or if the $\eta$ is rooted as equaling unity, then it is assumed that $\hat\tau_2 \equiv \tau_2$ and the uniform distribution triggered (see below).
The distribution is restricted to a narrow range of L-CV ($\tau_2 = \lambda_2/\lambda_1$). If $\tau_2 = 1/3$, the process represented is a stationary Poisson for which the probability density function is simply the uniform distribution and $f(x) = 1/\psi$. If $\tau_2 = 1/2$, then the distribution is represented as the usual exponential distribution with a location parameter of zero and a rate parameter $\beta$. Both of these limiting conditions are supported.
If the distribution shows to be uniform ($\tau_2 = 1/3$), then the third element in the returned parameter vector is used as the $\psi$ parameter for the uniform distribution, and the first and second elements are NA
of the returned parameter vector.
If the distribution shows to be exponential ($\tau_2 = 1/2$), then the second element in the returned parameter vector is the inverse of the rate parameter for the exponential distribution, and the first element is NA
and the third element is 0
(a numeric FALSE
) of the returned parameter vector.
partexp(lmom, checklmom=TRUE)
lmom
be checked for validity using the are.lmom.valid
function. Normally this should be left as the default and it is very unlikely that the L-moments will not be viable (particularly in the $\tau_4$ and $\tau_3$ inequlist
is returned.texp
.NA
if $\tau_2$ tests as being outside the 1/3 and 1/2 limits.lmomtexp
, cdftexp
, pdftexp
, quatexp
# truncated exponential is a nonstationary poisson process
A <- partexp(vec2lmom(c(100, 1/2), lscale=FALSE)) # pure exponential
B <- partexp(vec2lmom(c(100, 0.499), lscale=FALSE)) # almost exponential
BB <- partexp(vec2lmom(c(100, 0.45), lscale=FALSE)) # truncated exponential
C <- partexp(vec2lmom(c(100, 1/3), lscale=FALSE)) # stationary poisson process
D <- partexp(vec2lmom(c(100, 40))) # truncated exponential
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