Error in fitdist with gamma distribution

Below are my codes:

library(fitdistrplus)

s <- c(11, 4, 2, 9, 3, 1, 2, 2, 3, 2, 2, 5, 8,3, 15, 3, 9, 22, 0, 4, 10, 1, 9, 10, 11,
 2, 8, 2, 6, 0, 15, 0 , 2, 11, 0, 6, 3, 5, 0, 7, 6, 0, 7, 1, 0, 6, 4, 1, 3, 5,
 2, 6, 0, 10, 6, 4, 1, 17, 0, 1, 0, 6, 6, 1, 5, 4, 8, 0, 1, 1, 5, 15, 14, 8, 1,
 3, 2, 9, 4, 4, 1, 2, 18, 0, 0, 10, 5, 0, 5, 0, 1, 2, 0, 5, 1, 1, 2, 3, 7)

o <- fitdist(s, "gamma", method = "mle")
summary(o)
plot(o)

and the error says:

Error in fitdist(s, “gamma”, method = “mle”) : the function mle
failed to estimate the parameters,
with the error code 100

Enquirer: Chuan

||

Solution #1:

The Gamma distribution doesn’t allow zero values (the likelihood will evaluate to zero, and the log-likelihood will be infinite, for a response of 0) unless the shape parameter is exactly 1.0 (i.e., an exponential distribution – see below) … that’s a statistical/mathematical problem, not a programming problem. You’re going to have to find something sensible to do about the zero values. Depending on what makes sense for your application, you could (for example)

  • choose a different distribution to test
  • exclude the zero values (fitdist(s[s>0], ...))
  • set the zero values to some sensible non-zero value (fitdist(replace(s,which(s==0),0.1),...)

which (if any) of these is best depends on your application.


@Sandipan Dey’s first answer (leaving the zeros in the data set) appears to make sense, but in fact it gets stuck at the shape parameter equal to 1.

o <- fitdist(s, "exp", method = "mle")

gives the same answer as @Sandipan’s code (except that it estimates rate=0.2161572, the inverse of the scale parameter=4.626262 that’s estimated for the Gamma distribution – this is just a change in parameterization). If you choose to fit an exponential instead of a Gamma, that’s fine – but you should do it on purpose, not by accident …

To illustrate that the zeros-included fit may not be working as expected, I’ll construct my own negative log-likelihood function and display the likelihood surface for each case.

mfun <- function(sh,sc,dd=s) {
  -sum(dgamma(dd,shape=sh,scale=sc,log=TRUE))
}

library(emdbook)  ## for curve3d() helper function

Zeros-included surface:

cc1 <- curve3d(mfun(x,y),
      ## set up "shape" limits" so we evaluate
      ## exactly shape=1.000 ...
        xlim=c(0.55,3.55),
        n=c(41,41),
        ylim=c(2,5),
        sys3d="none")


png("gammazero1.png")
with(cc1,image(x,y,z))
dev.off()

enter image description here

In this case the surface is only defined at shape=1 (i.e. an exponential distribution); the white regions represent infinite log-likelihoods. It’s not that shape=1 is the best fit, it’s that it’s the only fit …

Zeros-excluded surface:

cc2 <- curve3d(mfun(x,y,dd=s[s>0]),
               ## set up "shape" limits" so we evaluate
               ## exactly shape=1.000 ...
               xlim=c(0.55,3.55),
               n=c(41,41),
               ylim=c(2,5),
               sys3d="none")


png("gammazero2.png")
with(cc2,image(x,y,z))
with(cc2,contour(x,y,z,add=TRUE))
abline(v=1.0,lwd=2,lty=2)
dev.off()

enter image description here

Respondent: Ben Bolker

Solution #2:

Just provide the initial values for the gamma distribution parameters (scale, shape) to be computed with mle using optim and also the lower bounds for the parameters, it should work.

o <- fitdist(s, "gamma", lower=c(0,0), start=list(scale=1,shape=1))
summary(o)

#Fitting of the distribution ' gamma ' by maximum likelihood 
#Parameters : 
#      estimate Std. Error
#scale 4.626262         NA
#shape 1.000000         NA
#Loglikelihood:  -250.6432   AIC:  505.2864   BIC:  510.4766 

enter image description here

As per the comments by @Ben Bolker, we may want to exclude the zero points first:

o <- fitdist(s[s!=0], "gamma", method = "mle", lower=c(0,0), start=list(scale=1,shape=1))
summary(o)
#Fitting of the distribution ' gamma ' by maximum likelihood 
#Parameters : 
#      estimate Std. Error
#scale 3.401208         NA
#shape 1.622378         NA
#Loglikelihood:  -219.6761   AIC:  443.3523   BIC:  448.19

enter image description here

Respondent: Sandipan Dey

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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