[R] fitting truncated normal distribution

[Sundar Dorai-Raj]

aon.912182281.tmp at aon.at wrote:
> Hello,
> I am a new user of R and found the function dtnorm() in the package msm.
> 
> My problem now is, that it is not possible for me to get the mean and sd out of a sample when I want a left-truncated normal distribution starting at "0".
> 
> fitdistr(x,dtnorm, start=list(mean=0, sd=1))
> 
> returns the error message 
> "Fehler in "[<-"(`*tmp*`, x >= lower & x <= upper, value = numeric(0)) :    nichts zu ersetzen"
> 
> I don't know, where to enter the lower/upper value. Is there a possibility to program the dtnorm function by myself?
> 
> Thank you very much in advance for your help,
> markus
> 
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Hi, Markus,

You should always supply the package name where dtnorm is located. My 
guess is most don't know (as I didn't) it is part of the msm package. 
Also, you should supply a reproducible example so others may understand 
your particular problem. For example, when I ran your code on data 
generated from "rtnorm" (also part of msm) I got warnings related to the 
NaNs generated in pnorm and qnorm, but no error as you reported. Both of 
these suggestions are in the posting guide (see signature above).

So, to answer your problem, here's a quick example.

library(MASS) ## for fitdistr
library(msm) ## for dtnorm

dtnorm0 <- function(x, mean = 0, sd = 1, log = FALSE) {
   dtnorm(x, mean, sd, 0, Inf, log)
}

set.seed(1) ## to others may reproduce my results exactly
x <- rtnorm(100, lower = 0)
fitdistr(x, dtnorm0, start = list(mean = 0, sd = 1))

Note, the help page ?fitdistr suggests additional parameters may be 
passed to the density function (i.e. dtnorm) or optim. However, this 
won't work here because "lower" is an argument for both functions. This 
is the reason for writing dtnorm0 which has neither a lower or an upper 
argument.

HTH,

--sundar