From: Leaf Sun <leaflovesun_at_yahoo.ca>

Date: Tue 18 Jul 2006 - 04:18:08 EST

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Jul 18 04:26:07 2006

Date: Tue 18 Jul 2006 - 04:18:08 EST

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:

> 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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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Jul 18 04:26:07 2006

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