I came across several publications dealing with overdispersed zero-inflated count data that "simply" modelled presence absence in one model and then postive counts in a second model. This led to two models with two different outcomes. The authors stated they were using "hurdle models".

In my opinion hurdle models do not work like that but "integrate" the results of both models in one step in one final output, is this right?

Is the other approach now "wrong" ??

  • 1
    $\begingroup$ I've not seen results from hurdle models 'integrated'; what does that look like? Do you have examples of papers that do that? (It might be helpful to link to the papers you found that have two separate models as well.) $\endgroup$ Commented Aug 13, 2012 at 13:55
  • $\begingroup$ soorry for answering so late. In the paper "Identifying essential summer habitat of the endangered beluga whale Delphinapterus leucas in Cook Inlet, Alaska" by Goetz et al. Endangered Species Res. 16: 135-147 (you can find it on google), they build two single models with glmmPQL() and did not combine the models "statistically"; only the predicted results were multiplied at the end(!!!). Is this okay? For sure not I would say $\endgroup$
    – Jens
    Commented Aug 29, 2012 at 16:08
  • $\begingroup$ Hi @Jens, sorry I don't have time to look at it further right now with the semester starting next week. Hopefully someone else will; that's the great thing about this site! $\endgroup$ Commented Aug 29, 2012 at 16:36
  • $\begingroup$ sure, no problem. I´ll hang on here ;) $\endgroup$
    – Jens
    Commented Aug 29, 2012 at 19:00
  • $\begingroup$ I think that ZIPs have only recently been fully integrated into the world of generalized linear mixed models. So, if Goetz et al wanted random effects in both the binomial and the poisson models they might have been unable to fit an integrated model. The package glmmADMB (glmmadmb.r-forge.r-project.org/glmmADMB.html) can fit mixed model ZIPS. Zuur et al 2012 get at this I think via WinBUGS (highstat.com/book4.htm) $\endgroup$
    – N Brouwer
    Commented Sep 12, 2012 at 1:28

1 Answer 1


I just randomly came across this question. Sorry it's so late. It is fine to model the two parts of a hurdle model separately. This paper addresses hurdle and zero-inflated models https://journal.r-project.org/archive/2017/RJ-2017-066/index.html

Using data from the glmmTMB package here's an example. You could fit the data with either a zero-inflated model (zinb below, where zeros can be from either the negative binomial or the zero-inflation) or with a hurdle model (hnb below). Then we can see that hnb is statistically equivalent to the combination of a binomial model for the zeros and a zero-truncated model for the positive counts.

> zinb = glmmTMB(count~spp * mined + (1|site), zi=~spp * mined, data=Salamanders, family=nbinom2)
> hnb = glmmTMB(count~spp * mined + (1|site), zi=~spp * mined, data=Salamanders, family=truncated_nbinom2)
> zeros = glmmTMB(count<.5 ~spp * mined, data=Salamanders, family=binomial)
> pos = glmmTMB(count ~spp * mined + (1|site), data=subset(Salamanders, count>0), family= truncated_nbinom2)
> logLik(pos)
'log Lik.' -491.5107 (df=16)
> logLik(zeros)
'log Lik.' -315.2394 (df=14)
> logLik(pos)+logLik(zeros)
'log Lik.' -806.7501 (df=16)
> logLik(hnb)
'log Lik.' -806.7501 (df=30)

You can also see the coefficients if you try it.

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    $\begingroup$ Many thanks! It is never too late for a good answer. $\endgroup$
    – Jens
    Commented Dec 19, 2019 at 6:56

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