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 <- function(p)-sum(dweibull(x,p[1],p[2],log=TRUE))
> z2(c(2,1))
[1] 1359.169> z2(c(2,2))
[1] 634.8413> z(c(2,1))
[1] 736251.1> z(c(2,2))
[1] 184012> optim(c(.5,.5),z2)
$par
[1] 1.971611 1.938388
$value
[1] 633.9709
$counts
function gradient
79 NA
$convergence
[1] 0
$message
NULL
>
> 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])+(sum(x/p[2])^(p[1])) )}
> zz<-optim(c(0.5,0.5),z)
> zz
> [[alternative HTML version deleted]]
>
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
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