Hi William,
Thanks a lot for your response. I checked the package and found that what I
want to solve was the opposite, that is, from mean and sd to parameters shape
and scale. Could anyone give some hints please? Any suggestion would be
appreciated!
Leaf
----- Original Message -----
From: William Asquith, [EMAIL PROTECTED]
Sent: 2006-07-17, 16:18:31
To: Leaf Sun, [EMAIL PROTECTED]
Subject: Re: [R] Weibull distribution
Do not have answer per se, but if you are seeking some
comparisons--
try three parameter Weibull as implemented by the lmomco package.
William
On Jul 17, 2006, at 1:18 PM, Leaf Sun wrote:
> Hi all,
>
> By its definition, the mean and variance of two-par. Weibull
> distribution are:
>
>
>
>
>
> (www.wikipedia.org)
>
>
> I was wondering, if given mean and sd. could we parameterize the
>
> distribution? I tried this in R.
>
> gamma.fun <- function(mu,sd,start=100)
> {
> f.fn <- function(alpha) sd^2-mu^2/(gamma(1+1/alpha))^2*(gamma(1+2/
> alpha)-(gamma(1+1/alpha))^2)
> alpha <- optim(start, f.fn,method='BFGS')
> beta <- mu/gamma(1+1/alpha$par)
> return(list=c(a=alpha$par,b=beta));
> }
>
>
> But the problems come up here:
>
> 1) the return values of a and b are only related to the input
>
> mean, and nothing to do with the sd. For instance, when I apply
> a
> mean mu = 3 whatever I use sd=2, sd=4, the function returned the
>
> same scale and shape values.
>
> > gamma.fun(3,4,10);
> a b
> 5.112554 3.263178
>
> > gamma.fun(3,2,10);
> a b
> 5.112554 3.263178
>
> 2) the start value determines the results: if I apply mean = 3,
> and
> sd=2, with a start of 10, it would return alpha close to 10, if
> I
> use a start = 100, it would return alpha close to 100.
>
> > gamma.fun(3,2,10);
> a b
> 5.112554 3.263178
>
> > gamma.fun(3,2,100);
> a b
> 99.999971 3.017120
>
> Since I am not a statistician, I guess there must be some
> theoretical reasons wrong with this question. So I am looking
> forward to some correction and advice to solve these. Thanks a
> lot
> in advance!
>
> Leaf
>
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>
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