What is the proper interpretation of the coefficients for the positive count part of a hurdle model (truncated Poisson or Negative Binomial)? I have read that the interpretation of the coefficients for a nontruncated Poisson (one unit increase in x results in an e^β factor increase in the expected count of y, controlling for other predictors) is not appropriate since the expected value is not equal to lambda in the truncated Poisson distribution:

Interpretation of zero-truncated Poisson regression coefficients

This is supported by information on the glmmAdaptive package in R:

"...it has been noted that the fixed-effects coefficients for the positive counts part relate to the mean μ of Poisson distribution that includes the zeros, not the mean for those who experience the event. The mean for those who experience more than one events is μ/(1−e^−μ). Hence, a β=0.1 cannot be interpreted as reflecting a e^β= 10% increase in the mean of the subjects who experience the events."

Link: https://drizopoulos.github.io/GLMMadaptive/articles/ZeroInflated_and_TwoPart_Models.html

However, these sources appear to be contradicted by an article on RPubs related to the pscl package:

"(Positive count model) Given the response is positive (among those who have positive counts), the average count is 3.3. This is increased by a unit increase in hospital stay by 1.24 among those who have positive counts. Poor health increases it by 1.37 times, whereas excellent health decreases it by 0.72, both among those who have positive counts."


What is the appropriate interpretation? Or does it depend on the package used (e.g., glmmAdaptive, pscl, or glmmTMB)?

  • $\begingroup$ 0. Welcome to CV.SE. 1. Nice question (+1) it is easy not torealise the subtly of it. 2. Your intuition is correct, it is fully appropriate to do so but it is not outright false either; please see my answer below when I expand on this a bit more. $\endgroup$
    – usεr11852
    Jul 28, 2021 at 10:17

1 Answer 1


The interpretation does not dependent on the package used; it does dependent on the underlying model and more specifically how close we are to the truncation point of $0$.

Background: The Poisson probability function is $f(y_i;\mu_i| y_i \geq 0) = \frac{\mu^{y_i} e^{-\mu_i}}{y_i!}$ while we can transform it to a truncated Poisson by dividing its probability function by dividing by ($1$ minus the probability that $y_i = 0$). i.e. we have $f(y_i;\mu_i| y_i > 0) = \frac{\mu^{y_i} e^{-\mu_i}}{y_i! (1- e^{-\mu_i})}$. As we can see as $\mu_i$ gets larger, the term $e^{-\mu_i}$ tends to 0 leading us quickly back to be (approximately) Poisson. We could say that for $\hat{\mu} >4.2$, the results is pretty indistinguishable to that of a standard Poisson (as $e^{−4.2} \approx 0.01$).

Specifically for the count model in RPubs linked, the intercept is already ~3.3 so the interpretation them as reflecting a percentage increase while not "great", it is not miles off. After putting the caveat about the "small counts" behaviour, we are fine to use the "standard" interpretation. If we are very particular, we can create a marginal effect plot for our model to visualise this change exactly but that would be a project-specific decision.

  • $\begingroup$ Thank you, this is very helpful. I looked through several published studies where the authors used a hurdle model to see how they interpreted the results. In those papers, the authors are not precise about how a one unit change in predictor x affects the response variable. For example: "λ is reduced by 1 − exp(−0.132) = 12.4%...so the expected number [in the non-zero count] m(λ) = λ/{1 − exp(−λ)} is reduced". I suppose this is "safe" interpretation of the model output? $\endgroup$
    – Andy
    Jul 29, 2021 at 0:20
  • $\begingroup$ I am glad I could help. Almost "safe"; I would probably add the caveat of $\lambda$ not being close to the truncation point $0$, or to quote Rizopoulos' page "for those who experience more than one events". $\endgroup$
    – usεr11852
    Jul 29, 2021 at 1:55
  • $\begingroup$ For the quote you cite, he's talking about the mean of the truncated distribution: "The mean for those who experience more than one events is μ/(1−e−^μ)." He means at least one event, right? So the caveat being if the observation has a high enough lambda (or mu as you put it), away from the truncation point, you can interpret the model like an untruncated Poisson model. What about the truncated negative binomial model? $\endgroup$
    – Andy
    Jul 29, 2021 at 2:44
  • $\begingroup$ Same caveats would apply. :) $\endgroup$
    – usεr11852
    Jul 29, 2021 at 3:27

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