I’m getting NaN from a multinomial distribution, where I expected 1. Is this a bug or am I doing something wrong?
using Distributions
dist = Multinomial(2, [0.999,0.001])
pdf(dist,[2,0]) ## 0.998, ok
dist = Multinomial(2, [1.0,0.0])
pdf(dist,[2,0]) #NaN, not ok
The equivalent limit of the binomial has no problem:
pdf(Binomial(1,0.999),[0,0.5,1]) ## [0.001, 0.0, 0.999]
pdf(Binomial(1,1.0),[0,0.5,1]) ## [0.0, 0.0, 1.0]
You are getting a NaN from a 0^0, which is correct. See the PMF of a multinomial.
EDIT: correcting my mistake — in these contexts it is customary to interpret 0^0=1. See the famous “Two notes on notation” paper by Knuth for example.
The formula has a 0^0, I’m sure this is what’s causing the NaN.
But the limiting distribution puts all the mass on one point, and this limit is what I was hoping the function would return.
This works for the binomial case, where again the simple formula has an indefinite 0^0, but I guess someone has caught the special case.
Maybe file a bug against Distributions.