Date: Tue, 14 Apr 1998 18:22:12 +0200 Message-Id: <199804141622.SAA01845@sophie.ethz.ch> From: Martin Maechler <maechler@stat.math.ethz.ch> To: wsimpson@uwinnipeg.ca Subject: Re: R-beta: SEs for one-param MLE in R? >>>>> "Bill" == Bill Simpson <wsimpson@uwinnipeg.ca> writes: Bill> Simple-mindedly I tried getting MLE and SE for one-parameter Bill> model in the same way as for multi-param models. Bill> out<-nlm(fn,p=c(2),hessian=T) Bill> But sqrt(diag(solve(out$hessian))) gives the answer 1. The Bill> Hessian has only one entry, not really a matrix. diag(x) gives 1 Bill> if x is just a single number. Bill> Is this what I should be doing to get SE for MLE? Bill> sqrt(solve(out$hessian)) Bill> Thanks very much for any help! Well, .Internal(nlm(..)) should be fixed to always return a matrix in $ hessian. However, your problem is solved easily by always using p <- length(estimate) SE <- sqrt(if(p==1) 1/out$hessian else diag(solve(out$hessian))) BTW: I am (we are) interested in the functions that you are writing for nlm(.) It certainly is worthwhile to have nlm(.) return a class "nlm" result and provide print.nlm(.) and summary.nlm(.) functions {{ Jim Lindsey already posted something like this, unfortunately using "nls" which we don't want as long as it is not very close to S' nls(.) function }} Martin -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request@stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._