P Ehlers <[EMAIL PROTECTED]> writes: > I prefer a (consistent) NaN. What happens to our notion of a > Binomial RV as a sequence of Bernoulli RVs if we permit n=0? > I have never seen (nor contemplated, I confess) the definition > of a Bernoulli RV as anything other than some dichotomous-outcome > one-trial random experiment.
What's the problem ?? An n=0 binomial is the sum of an empty set of Bernoulli RV's, and the sum over an empty set is identically 0. > Not n trials, where n might equal zero, > but _one_ trial. I can't see what would be gained by permitting a > zero-trial experiment. If we assign probability 1 to each outcome, > we have a problem with the sum of the probabilities. Consistency is what you gain. E.g. binom(.,n=n1+n2,p) == binom(.,n=n1,p) * binom(.,n=n2,p) where * denotes convolution. This will also hold for n1=0 or n2=0 if the binomial in that case is defined as a one-point distribution at zero. Same thing as any(logical(0)) etc., really. -- O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
