# Issues with fitting a non-standard distribution to data and initial values

I'm trying to find a distribution that fits my data in order that I can then predict a 5th percentile but none of the standard distributions seem to fit.

I'll explain my approach so far with examples below. I have then been trying to fit a Burr distribution in fitdistrplus but cannot find any suitable initial values. But I am unsure it this is because this is an incorrect distribution for the data or if its me. I've also experimented with prefit function but am getting a similar problem of not being able to choose feasible starting values. Maybe there is a more appropriate distribution I haven't tried?

My data are:

0.0001900 0.0002100 0.0002200 0.0003000 0.0007800 0.0008400 0.0011000 0.0011300 0.0012000 0.0016000 0.0016000 0.0020000 0.0020000 0.0031000 0.0056500 0.0059000 0.0082449 0.0130000 0.0180000 0.0191000 0.0510000

The Cullen and Frey graph using the following code is as follows:

descdist(Data, boot = 500)


from this I thought the Beta distribution may work best but it isn't quite right. I can only post 2 images so have only included the QQ plot here:

fitln <- fitdist(SSD2$NOEC,"lnorm") fitW <- fitdist(SSD2$NOEC, "weibull")
fitg <- fitdist(SSD2$NOEC, "gamma") fitn <- fitdist(SSD2$NOEC, "norm")
fitexp <- fitdist(SSD2$NOEC,"exp") fitB <- fitdist(SSD2$NOEC,"beta")
fitP <- fitdist(SSD2$NOEC,"pareto") cdfcomp(list(fitW, fitg, fitln, fitn, fitexp, fitB, fitP), legendtext=c("Weibull", "gamma", "lognormal", "norm", "exp", "Beta", "Pareto")) denscomp(list(fitW, fitg, fitln, fitn, fitexp, fitB, fitP), legendtext=c("Weibull", "gamma", "lognormal", "norm", "exp", "Beta", "Pareto")) qqcomp(list(fitW, fitg, fitln, fitn, fitexp, fitB, fitP), legendtext=c("Weibull", "gamma", "lognormal", "norm", "exp", "Beta", "Pareto"))  ppcomp(list(fitW, fitg, fitln, fitn, fitexp, fitB, fitP), legendtext=c("Weibull", "gamma", "lognormal", "norm", "exp", "Beta", "Pareto")) gofstat(list(fitW, fitg, fitln, fitn, fitexp, fitB, fitP)) Goodness-of-fit statistics 1-mle-weibull 2-mle-gamma 3-mle-lnorm 4-mle-norm 5-mle-exp 6-mle-beta Kolmogorov-Smirnov statistic 0.17899327 0.2201135 0.13002018 0.2888904 0.3552864 0.2208923 Cramer-von Mises statistic 0.08976409 0.1567659 0.04560304 0.5880214 0.6158410 0.1589649 Anderson-Darling statistic 0.53379880 0.8341498 0.30557002 3.1553792 3.5184364 0.8457680 7-mle-pareto Kolmogorov-Smirnov statistic 0.12410532 Cramer-von Mises statistic 0.04296412 Anderson-Darling statistic 0.29429372 Goodness-of-fit criteria 1-mle-weibull 2-mle-gamma 3-mle-lnorm 4-mle-norm 5-mle-exp 6-mle-beta Akaike's Information Criterion -173.8742 -171.8001 -177.3758 -124.3444 -167.3059 -171.7056 Bayesian Information Criterion -171.7851 -169.7111 -175.2868 -122.2554 -166.2614 -169.6166 7-mle-pareto Akaike's Information Criterion -176.0548 Bayesian Information Criterion -173.9658  The code I have used so far to try burr distribution is as follows but am struggling with the initial values - I have tried a few variations of shape 1 and 2: fitBurr <- fitdist(SSD2$NOEC,"burr", start = list(shape1 = 1, shape2 = 3, rate = 1))
prefit(SSD2$NOEC,"burr", method = "mle", start = list(shape1 = 1, shape2 = 3))  Any help in getting a suitable distribution for this data would be greatly appreciated. • What's the parameterization of this Burr? Are "shape1" and "shape2" the$c$and$k$(respectively) in the Wikipedia page on the Burr type Xii? Is the rate the inverse of the scale ($\lambda$) there? Looking at it, I'd expect that a best fitting$c$would be less than 1 and$k\$ a bit greater than one, if anything. Commented Aug 8, 2017 at 3:10
• It may be that most "standard" type distributions won't fit the lower tail all that well; the bulk of the distribution is close to lognormal but the left tail is a bit shorter than it would be for a lognormal or even a shifted lognormal; this would suggest a gamma probably wouldn't suit either (since its lower tail will tend to be longer, not shorter, than a lognormal), this would then rule out the less general exponential as well. On similar grounds you could rule out a number of the others. Commented Aug 8, 2017 at 3:11
• Yes - I agree that its the lower tail that is is the cause of the deviation from the distributions. I believe that the shapes 1 and 2 are c and k respectively as you say but im still failing to find feasible starting values. Commented Aug 9, 2017 at 10:13