I am trying to understand zero-inflated negative binomial regression.
My impression is that if a zero-inflated negative binomial model does not contain any logit part, the model is identical to the one can obtain with just ordinary negative binomial regression. Is this correct?
PS: the logit part I was talking about – well – zero-inflated model assumes that the 0s within the dataset are generated based on two different process: one is negative binomial and the other is, if I remember it correctly, poisson. By "no logit part" I meant what if we take out the effect of the poisson distribution from the zero-inflated model? would it be same as ordinary negative binomial regression?
Best Answer
You remember it wrongly: a zero-inflated negative binomial (ZINB) is a mixture of a point mass at zero and a negative binomial(NB) distribution.
$$ f_{ZINB}(Y;\pi, \mu, \theta) = \pi Z(Y) + (1-\pi)f_{NB}(Y;\mu, \theta) $$
with Z(Y) a point mass at zero. Zero observation can come from either distribution, non-zero ones only from the NB.
The "logit-part" you refer to models the mixing parameter $\pi$ as a function on covariates, assuming a binomial distribution with succes probability $\pi$.
To answer your question: yes if you set $\pi=0$ you're indeed back at the regular NB