Negative Binomial Deviance Calculation in H2O

Question

I've been looking at the deviance calculation for negative binomial model in H2O ( code line 580/959) and I'm struggling to reason why it is 0 when yr or ym is/are 0.

(yr==0||ym==0)?0:2*((_invTheta+yr)*Math.log((1+_theta*ym)/(1+_theta*yr))+yr*Math.log(yr/ym))

The formula for the deviance calculation is as below (from H2O Documentation ):

Going with maths, I don't see the deviance is 0 unless both yr and ym are 0.

Does anyone happen to know if there is a special case where deviance for negative binomial needs to be set to 0 when either of the yr and ym is/are 0?

Thanks!

Answer 1

I'm not sure, but it seems to me they maybe just chose the lazy way out of a numerical difficulty.

mu=0 ( ym ) is a degenerate case where p=0 and so y=0 always. It's not interesting, and not really part of any useful analysis. I'm not sure it can even come out with the linear-predictor. With using the natural parameter = linear predictor, you need the linear predictor to be equal to minus infinity...

However, y can be equal to 0 for other mu 's. And what you do in this case, is take the limit of the deviance as y->0 , which is completely defined for Negative-Binomial, and isn't equal to 0. They could have implemented it, but chose not too, so this is why I call it "lazy".