Hi I am trying to recover the hessian of a problem optimised with box-constraints. The problem is that in some cases, my estimates are very close to the boundary, which will make optim(..., hessian=TRUE) or optimHessian() fail, as they do not follow the box-constraints, and hence estimate the function in the unfeasible parameter space.
As a simple example (my problem is more complex though, simple param transformations do not apply ashere), imagine estimating mu and sigma (restricted to be >0) of a simple normally distributed data, where however the true sigma is very close to zero: LLNorm <- function(para, dat=rn) return(-sum(dnorm(dat, para[1], para[2], log=TRUE))) rn2 <- c(rep(10.3, 2000), 10.31) >optim(c(10,1), fn=LLNorm, method="L-BFGS-B", lower=c(-Inf, 0.0000001), dat=rn2,hessian=TRUE) Error in optim(c(10, 1), fn = LLNorm, method = "L-BFGS-B", lower = c(-Inf, : non-finite finite-difference value [2] The only solution/workaround I found is to do a two steps procedure: use optim() without hessian, then use optimHess, specifying the length of the numerical variations (arg ndeps) as smaller as the distance between the parameter and the bound, i.e.: op<-optim(c(10,1), fn=LLNorm, method="L-BFGS-B", lower=c(-Inf, 0.0000001), dat=rn2,hessian=FALSE) H<-optimHess(op$par, LLNorm, dat=rn2, control=list(ndeps=rep((op$par[2]-0)/1000,2))) While this solution "works", it is surely not elegant, and I have following questions: 1) is this solution to use smaller steps good at all? 2) I am not sure either I understand the reasons why the hessian in a constrained optim problem is evaluated without the constraints, or at least the problem transformed into a warning? Furthermore, I realised that using the numDeriv package, the function hessian() does not offer either constraints for the parameters, yet in the special case of the example above, it does not fail (although would have problems with parameter even closer to the bound), unlike the optimHessian() function? See: library(numDeriv) hessian(LLNorm, op$par, dat=rn2) This brings me to the final question: is there a technical reason not to allow a "constrained" hessian, which seems to indicate the fact that the two implementations in R do not do it? I found a very interesting answer by Spencer Grave for a similar question: http://tolstoy.newcastle.edu.au/R/e4/help/08/06/15061.html although this regards more the statistical implications than the numerical issues. Thanks a lot! Matthieu [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
