John Westbury wrote: > Ben, > > Say you have some variable (x) ~ exp f(x|lambda,theta) where f(x|lambda, > theta) = lamba*(e^(minus lambda*(x-theta))) > > Lambda is the inverse scale parameter and theta is a location parameter > (I erred in calling theta a shape parameter in the original question). > > John
Presumably P(x)=0 for x<theta? In which case
dexp2 <- function(x,rate=1,loc=0,log=FALSE) {
dexp(x-loc,rate=rate,log=log)
}
rexp2 <- function(n,rate=1,loc=0) {
rexp(n,rate=rate)+loc
}
etc.
It turns out that dexp(x) gives 0/-Inf for x<0 (I thought it would
give a warning, but it doesn't), so this is relatively simple.
Of course the next question is what you actually want to do with this
distribution ...
>
> On Mon, Jan 11, 2010 at 4:23 PM, Ben Bolker <[email protected]
> <mailto:[email protected]>> wrote:
>
> John Westbury <jrwestbury <at> gmail.com <http://gmail.com>> writes:
>
> >
> > Hello,
> >
> > I am new to R and am trying to figure out how to specify an
> exponential
> > distribution with scale and shape parameters. I can specify an
> exponential
> > distribution with an inverse scale but would like to know how to
> specify the
> > distribution with a shape parameter as well.
> >
> > thanks much,
> >
> > John
> >
>
> Do you mean a gamma distribution? I'm unfamiliar with any
> definition of the exponential distribution with more than one
> parameter. Can you point to a reference/example somewhere?
>
> Ben Bolker
>
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--
Ben Bolker
Associate professor, Biology Dep't, Univ. of Florida
[email protected] / people.biology.ufl.edu/bolker
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