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Topic: how to generate a random number following truncated Weibull distribution?
Replies: 1   Last Post: Jul 8, 2006 3:38 PM

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Anon.

Posts: 379
Registered: 6/2/05
Re: how to generate a random number following truncated Weibull distribution?
Posted: Jul 8, 2006 3:38 PM
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john2 wrote:
> Lucy wrote:
>> Hi all,
>>
>> I want to generate random numbers that follow truncated Weibull
>> distribution.
>>
>> In Matlab, there is a "wblcdf" function that can generate Weibull
>> random numbers, but their support spams the half-line (0, +inf). But
>> that's not what I want, I want to derive and generate random numbers
>> that fall in (0, d), where "d" is a positive constant. I call this
>> "Truncated Weibull distribution", but I am not sure if there is such a
>> name existing.
>>

>
>
> Usually what is required is just the part of the pdf on [0,d] but
> rescaled by its cumulative distribution D(x) value at d.
> So W'(x) = W(x)/D(d) 0<x<d
> where W(x) is a standard Weibull on [0, inf]
> If you're generating random numbers, simply use a standard Weibull,
> discard the values which exceed d, and resample, which is the same thing.
>

That might not be efficient, if D(d)<<1. If it's not, then there are
other ideas, such as rejection sampling (which is actually what you're
suggesting, but a more efficient proposal distribution could be used),
or slice sampling
(http://www.cs.toronto.edu/~radford/slice-aos.abstract.html). I'm not
going to claim that any one method is better than another, but at least
the OP has a couple more options.

Bob

--
Bob O'Hara
Department of Mathematics and Statistics
P.O. Box 68 (Gustaf Hällströmin katu 2b)
FIN-00014 University of Helsinki
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