Question

The truncated normal is given by:

dtnorm<- function(x, mean, sd, a, b) {
dnorm(x, mean, sd)/(pnorm(b, mean, sd)-pnorm(a, mean, sd))
}
ptnorm <- function(x, mean, sd, a, b) {
(pnorm(x,mean,sd) - pnorm(a,mean,sd)) / 
  (pnorm(b,mean,sd) - pnorm(a,mean,sd))
}

The fit is given by:

fitdist( data, tnorm, method="mle",
                    start=list(mean=mapply("[[", results[1], 1),
                               sd=mapply("[[", results[1], 2)),
                    fix.arg=list(a=minLoose,b=maxLoose))

Where results[i] is a matrix with the mle results of fitdist using normal instead of tnormal.

I get the following results for tnorm:

mean=-0.00844725266454969, sd=0.012540928272073

whereas with norm:

mean=0.00748402597402597, sd=0.00614293813955003

The data is all larger than 0 and smaller than 0.04 so the mle obtained for tnorm does not seem right…. Any advise?

Thanks!

Answer 1

The fact that your data is all above normal (er, rather above 0) has little bearing on whether the “mean” of best fit to a truncated distribution does or doesn’t exceed 0. You are fitting a right tail of a Normal distribution to your data. The estimated location parameter for the truncated is not really a mean, but rather where the mean would be in an uncensored dataset with a right tail of the same density “shape” as your data. (This is really a stats question rather than an R question.)

You can find the formula to calculate the expected value of a doubly truncated Normal at the moments section of the Wikipedia article:
http://en.wikipedia.org/wiki/Truncated_normal_distribution It is readily translatable into calls to pnorm and qnorm.

A further thought: Check out the facilities for working with truncated distributions in packages: ‘gamlss’ and ‘gamlss.tr’.