On Jan 29, 2012, at 12:17 , Christopher Kelvin wrote:
> Hello,
> If i write a function as below using log of weibull distribution i do not get
> the required
>
> results in estimating the parameters what do i do, please
Presumably find and fix the error in your likelihood function!
> z2 <- f
Hello,
If i write a function as below using log of weibull distribution i do not get
the required
results in estimating the parameters what do i do, please
a/b * (t/b)^a-1 * exp(-t/b)^a
n=500
x<-rweibull(n,2,2)
z<-function(p) {(-n*log(p[1])+n*log(p[2])-
(p[1]-1)*sum(log(x))+(p[1]-1)*log(p[2])
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