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?

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    $\begingroup$ i don't quite understand what you mean by the "logit part". you might want to look at this review: jstatsoft.org/v27/i08 $\endgroup$ – charles Nov 26 '13 at 17:40
  • $\begingroup$ Zero-inflated models are usually defined as two-component mixture models combining a point mass at zero with a count distribution such as negative binomial. $\endgroup$ – charles Nov 27 '13 at 3:52
  • $\begingroup$ From that review: "the unobserved probability $\pi$ of belonging to the point mass component is modelled by a binomial GLM" (8). I take it @Jin-Dominique has just confused a binomial GLM with logistic regression. Then they're asking, if $\pi = 0$, isn't a ZIM just an ordinary negative binomial regression? In that case, the RHS of eqn 7 collapses to $f_{count}(y; x, \bet)$, so the first-pass answer is "yes." Differences in the way the models are fit in particular implementations might lead to different estimates, perhaps. $\endgroup$ – Dan Hicks Dec 3 '17 at 19:48

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


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