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
[[alternative HTML version deleted]]
______________________________________________
[email protected] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
