Question

I have been doing some data analysis in R and I am trying to figure out how to fit my data to a 3 parameter Weibull distribution. I found how to do it with a 2 parameter Weibull but have come up short in finding how to do it with a 3 parameter.

Here is how I fit the data using the fitdistr function from the MASS package:

y <- fitdistr(x[[6]], 'weibull')

x[[6]] is a subset of my data and y is where I am storing the result of the fitting.

Answer 1

First, you might want to look at FAdist package. However, that is not so hard to go from rweibull3 to rweibull:

> rweibull3
function (n, shape, scale = 1, thres = 0) 
thres + rweibull(n, shape, scale)
<environment: namespace:FAdist>

and similarly from dweibull3 to dweibull

> dweibull3
function (x, shape, scale = 1, thres = 0, log = FALSE) 
dweibull(x - thres, shape, scale, log)
<environment: namespace:FAdist>

so we have this

> x <- rweibull3(200, shape = 3, scale = 1, thres = 100)
> fitdistr(x, function(x, shape, scale, thres) 
       dweibull(x-thres, shape, scale), list(shape = 0.1, scale = 1, thres = 0))
      shape          scale          thres    
    2.42498383     0.85074556   100.12372297 
 (  0.26380861) (  0.07235804) (  0.06020083)

Edit: As mentioned in the comment, there appears various warnings when trying to fit the distribution in this way

Error in optim(x = c(60.7075705026659, 60.6300379017397, 60.7669410153573,  : 
  non-finite finite-difference value [3]
There were 20 warnings (use warnings() to see them)
Error in optim(x = c(60.7075705026659, 60.6300379017397, 60.7669410153573,  : 
  L-BFGS-B needs finite values of 'fn'
In dweibull(x, shape, scale, log) : NaNs produced

For me at first it was only NaNs produced, and that is not the first time when I see it so I thought that it isn’t so meaningful since estimates were good. After some searching it seemed to be quite popular problem and I couldn’t find neither cause nor solution. One alternative could be using stats4 package and mle() function, but it seemed to have some problems too. But I can offer you to use a modified version of code by danielmedic which I have checked a few times:

thres <- 60
x <- rweibull(200, 3, 1) + thres

EPS = sqrt(.Machine$double.eps) # "epsilon" for very small numbers

llik.weibull <- function(shape, scale, thres, x)
{ 
  sum(dweibull(x - thres, shape, scale, log=T))
}

thetahat.weibull <- function(x)
{ 
  if(any(x <= 0)) stop("x values must be positive")

  toptim <- function(theta) -llik.weibull(theta[1], theta[2], theta[3], x)

  mu = mean(log(x))
  sigma2 = var(log(x))
  shape.guess = 1.2 / sqrt(sigma2)
  scale.guess = exp(mu + (0.572 / shape.guess))
  thres.guess = 1

  res = nlminb(c(shape.guess, scale.guess, thres.guess), toptim, lower=EPS)

  c(shape=res$par[1], scale=res$par[2], thres=res$par[3])
}

thetahat.weibull(x)
    shape     scale     thres 
 3.325556  1.021171 59.975470