Weibull cumulative distribution function starting

问题:

I've used fitdistr function from R MASS package to adjust a Weibull 2 parameters probability density function (pdf).

This is my code:

require(MASS)

h = c(31.194, 31.424, 31.253, 25.349, 24.535, 25.562, 29.486, 25.680, 26.079, 30.556,      30.552, 30.412, 29.344, 26.072, 28.777, 30.204, 29.677, 29.853, 29.718, 27.860, 28.919, 30.226, 25.937, 30.594, 30.614, 29.106, 15.208, 30.993, 32.075, 31.097, 32.073, 29.600, 29.031, 31.033, 30.412, 30.839, 31.121, 24.802, 29.181, 30.136, 25.464, 28.302, 26.018, 26.263, 25.603, 30.857, 25.693, 31.504, 30.378, 31.403, 28.684, 30.655,  5.933, 31.099, 29.417, 29.444, 19.785, 29.416, 5.682, 28.707, 28.450, 28.961, 26.694, 26.625, 30.568, 28.910, 25.170, 25.816, 25.820)

weib = fitdistr(na.omit(h),densfun=dweibull,start=list(scale=1,shape=5))

hist(h, prob=TRUE, main = "", xlab = "x", ylab = "y", xlim = c(0,40), breaks = seq(0,40,5))
curve(dweibull(x, scale=weib$estimate[1], shape=weib$estimate[2]),from=0, to=40, add=TRUE)

Now, I would like to create the Weibull cumulative distribution function (cdf) and plot it as a graph:

, where x > 0, b = scale , a = shape

I tried to apply scale and shape parameters for h using the formula above, but it was not this way.

回答1:

Here's a stab at a cumulative density function. You just have to remember to include an adjustment for the spacing of the sampling points (note: it works for sampling points with uniform spacing less than or equal to 1):

cdweibull <- function(x, shape, scale, log = FALSE){
  dd <- dweibull(x, shape= shape, scale = scale, log = log)
  dd <- cumsum(dd) * c(0, diff(x))
  return(dd)
}

The discussion above about scale differences notwithstanding, you can plot it over your graph the same as dweibull:

require(MASS)

h = c(31.194, 31.424, 31.253, 25.349, 24.535, 25.562, 29.486, 25.680,
      26.079, 30.556, 30.552, 30.412, 29.344, 26.072, 28.777, 30.204, 
      29.677, 29.853, 29.718, 27.860, 28.919, 30.226, 25.937, 30.594, 
      30.614, 29.106, 15.208, 30.993, 32.075, 31.097, 32.073, 29.600, 
      29.031, 31.033, 30.412, 30.839, 31.121, 24.802, 29.181, 30.136, 
      25.464, 28.302, 26.018, 26.263, 25.603, 30.857, 25.693, 31.504, 
      30.378, 31.403, 28.684, 30.655,  5.933, 31.099, 29.417, 29.444, 
      19.785, 29.416, 5.682, 28.707, 28.450,  28.961, 26.694, 26.625, 
      30.568, 28.910, 25.170, 25.816, 25.820)

weib = fitdistr(na.omit(h),densfun=dweibull,start=list(scale=1,shape=5))

hist(h, prob=TRUE, main = "", xlab = "x", 
     ylab = "y", xlim = c(0,40), breaks = seq(0,40,5), ylim = c(0,1))

curve(cdweibull(x, scale=weib$estimate[1], shape=weib$estimate[2]),
  from=0, to=40, add=TRUE)

回答2:

This works for my data but yours may differ. It uses rweibull3 function from FAdist package.

>h=rweibull3(1000,2,2,2)

>#this gives some warnings...that I ignore.
>weib = fitdistr(h,densfun=dweibull3,start=list(scale=1,shape=5,thres=0.5))

There were 19 warnings (use warnings() to see them)    

The thing to be aware of is that the start values effect the way the fit proceeds. So if the start values are close to the true values, you will get fewer warnings.

>curve(dweibull3(   x, 
            scale=weib$estimate[1], 
            shape=weib$estimate[2], 
            thres=weib$estimate[3]),
            add=TRUE)