拟合3参数Weibull分布

Question

我一直在R做一些数据分析,我试图找出如何使我的数据适合3参数Weibull分布.我找到了如何使用2参数Weibull来完成它,但是在找到如何使用3参数进行操作方面做得不够.

以下是我使用包中的fitdistr函数来拟合数据的方法MASS:

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

x[[6]] 是我的数据的子集,y是我存储拟合结果的地方.

Answer 1

首先,您可能想要查看FAdist包.但是,这并不难rweibull3做到rweibull:

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

和同样来自dweibull3于dweibull

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

所以我们有这个

> 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)

编辑:如评论中所述,在尝试以这种方式拟合分布时会出现各种警告

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

对我来说起初它只是NaNs produced,而且这不是我第一次看到它所以我认为它没有那么有意义,因为估计是好的.经过一番搜索,这似乎是一个非常受欢迎的问题,我既找不到原因也找不到解决方案.一种替代方案可能是使用stats4包和mle()功能,但它似乎也有一些问题.但是我可以让你使用danielmedic 修改过的代码版本,我已经检查了几次:

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