Renaud Lancelot <[EMAIL PROTECTED]> writes: > >>>And that's great; but I need the log likelihood. > >>> > >>>Anyone know? > >> > >> The deviance will not suffice? sum(dpois(nfix, fitted(freq.mod), > >> log.p=T)) > >> > >>should do the trick otherwise.
> > Thank you, that did the trick. I should note that the method > > doesn't work when I have a noninteger in Y. I'm doing a "log > > linear" analysis > > of a cross table, and one of the cells of the cross table has a zero > > value. I put a .5 in that cell. With the ".5 analysis" the above > > method doesn't work. So I had to revert to an analysis with a zero > > cell to make it work. Which has some problems. So I'm still left > > looking for the log likelihood of the ".5 analysis". I've had success > > with Stata using this method, but not R. > > Thanks again. > > ?logLik Right. Forgot about that. It's the same sum(dpois(....,log=T)) calculation, though (after a detour around the AIC), so that too will generate -Inf if noninteger values are plugged in. Fair enough: The likelihood is by definition the probability of obtaining the data, and the Poisson distribution has probability zero of generating fractional values. However, the deviance is well-defined in such cases. I'd suspect that what Stata outputs is not the true log-likelihood, but a number obtained by "analytic continuation" (wrong term, I know) of the Poisson density to nonintegers, e.g. > p <- function(x,lambda) lambda^x*exp(-lambda)/gamma(x+1) > p(1,1) [1] 0.3678794 > dpois(1,1) [1] 0.3678794 > dpois(.5,1) [1] 0 Warning message: non-integer x = 0.500000 > p(.5,1) [1] 0.4151075 -- O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
