# GLMM hurdle model for continuous data -Truncated negative binomial family in glmmTMB?

I am running a hurdle model using the glmmTMB function. My dependent variable is continuous and >= 0. I was looking for a function that would allow me to model the binary response in a logistic regression (i.e the zero inflated model) and the non-zero response in a Gamma (log) regression (i.e the conditional model). However, the glmmTMB function does not allow to use the gamma family.

I have two questions:

1. is there a function that allows to run a GLMM with gamma family? I tried to write my own code to have these two models but I am not sure about how to predict and calculate CIs.

2. Is it appropriate to use a truncated negative binomial family for the conditional (non-zero values) model? I ran the model:

hpm_nb <- glmmTMB(percapita_dia ~ Tipo_residuo + (1|Trip_ID), zi=~Tipo_residuo, all, family=truncated_nbinom2)

The diagnostics plots look good and model predictions are reasonable. Is it still wrong to have a truncated negative binomial family?

I'm not sure why you say that glmmTMB can't handle zero-inflated Gamma responses: the glmmTMB news file says (for version 1.0.0, release 2020-02-03):

new ziGamma family (minor modification of stats::Gamma) allows zero-inflation (i.e., Gamma-hurdle models)

I'd say it's not crazy to use a truncated negative binomial, but I'd be worried as it doesn't make statistical sense (technically, the likelihood of any non-integer value is 0 ...) (If you really had count data, a zero-inflated NB rather than a hurdle would be a reasonable option ...)

Given the distribution/density functions for each distribution, parameterized in terms of the mean $$\mu$$ and a shape/dispersion parameter ($$k$$ for NB, $$a$$ for Gamma):

$$\textrm{NB}: \qquad \frac{(k/(k+\mu))^k}{\Gamma(k)} \cdot \frac{\Gamma(k+x)}{x!} \cdot (\mu/(k+\mu))^x$$

$$\textrm{Gamma}: \qquad \frac{1}{(\mu/a)^a \Gamma(a)} \cdot x^{a-1} \cdot e^{-(x/(\mu/a))}$$

I think you may? be able to show that NB converges approximately to Gamma for large $$x$$ (but someone better/more dedicated than I am will need to do the math ...)

An empirical demonstration (not "proof"!)

hist(rnbinom(100000,mu=100,size=2),freq=FALSE,ylim=c(0,0.008), breaks=100)