Or, to save a little trouble: in the notation of the BUGS manual, z =
(lambda*x)^v has an exponential distribution, so that E(z^w) = Gamma(w+1)
for any w > -1. If w = 1/v, then E(z) = Gamma(1+1/v) = lambda*E(x) so that
E(x) = Gamma(1+1/v)/lambda. log(Gamma(x)) = loggam(x) is a built-in
function in BUGS.
-----Original Message-----
From: W. Jaross [mailto:[log in to unmask]]
Sent: Wednesday, April 18, 2001 8:14 PM
To: [log in to unmask]
Subject: Re: Calculating the mean of a Weibull distribution
Check out www.weibull.com/LifeDataWeb/
Weikko
On Wed, 18 Apr 2001, Stephen Palmer wrote:
> Dear All
>
> I'm currently trying to model survival data, with right hand censoring,
> using a Weibull distribution. The example provided in Volume 1
> (Mice:Weibull regression) provides a useful framework to adapt. For the
> purposes of my analysis, however, I need to calculate mean (rather than
> median) survival times. I would be extremely grateful if anyone can
> provide help with the equations/alterations to the above example
> required to estimate the mean of a Weibull distribution.
>
> Many thanks in advance
>
> Stephen Palmer
> Centre for Health Economics
>
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