Overfitting of Zero-inflated Poisson

Question

I'm trying to fit the data with 0s with Poisson, Negative Binomial, Zero-inflated Poisson and Zero-inflated negative binomial and discuss which model is more desired. When I'm trying to calculate the zero proportion from the model, I found that ZIP models the zero proportion exact the same with the original zero proportion. So my questions are:

1. Is there a reason to explain this situation? I would assume ZIP is overfitting since the AIC is not the smallest but I'm not sure about it.

2. If ZIP is overfitting, is there a way I can prove it? I don't think I can use a test dataset since I only have the intercept.

library(MASS)
library(pscl)
arr = c(rep(0,2387), rep(1,273), rep(2,36), rep(3,3), rep(4,3))
po = glm(arr ~ 1, family = 'poisson')
mu_po = exp(coef(po))
nb = glm.nb(arr ~ 1)
mu_nb = exp(coef(nb))
alpha_nb = 1/nb$theta
zpo = zeroinfl(as.numeric(arr) ~ 1|1, dist = "poisson")
mu_zpo = exp(coef(zpo)[1])
pi_zpo = exp(coef(zpo)[2])/(1+exp(coef(zpo)[2]))
znb = zeroinfl(arr ~ 1|1, dist = "negbin")
mu_znb = exp(coef(znb)[1])
pi_znb = exp(coef(znb)[2])/(1+exp(coef(znb)[2]))
alpha_znb = 1/znb$theta
#zero proportion
data.frame(true = sum(arr==0)/length(arr),
           poisson = exp(-mu_po),
           NB = (1+alpha_nbmu_nb)^(-1/alpha_nb),
           ZIP = pi_zpo+(1-pi_zpo)exp(-mu_zpo) ,
           ZINB = pi_znb+(1-pi_znb)* 
                 (1+alpha_znb*mu_znb)^(-1/alpha_znb))
#AIC
data.frame(poisson = AIC(po), NB = AIC(nb), ZIP = AIC(zpo), 
           ZINB = AIC(znb))

Answer 1

I think it's expected that the zero-inflated Poisson model fits the original zero proportion exactly, because the $\pi$ parameter exists solely to inflate $\mathbb{P}(X=0)$ to the desired value.

The process of fitting a ZIP model can be thought of like this:

1. Ignoring all zero observations, find the value of $\lambda$ which best fits the rest of your data.
2. Choose $\pi$ to solve $\pi + (1-\pi)e^{-\lambda} = \frac{z}{n}$, where $z$ is the number of observed zeros and $n$ is the sample size.

Step 2 will exactly fit the observed zero proportion, which matches what you saw.

This fails if the value of $\lambda$ you got from step 1 is low enough that even if you set $\pi = 0$, the model's $\mathbb{P}(X=0)$ is already greater than $\frac{z}{n}$. But if that were the case, you wouldn't have observed enough zeroes to justify using a zero-inflated Poisson model.

We can check the algebra directly for maximum likelihood estimators. Wikipedia gives equations satisfied by the maximum likelihood estimators for $\pi$ and $\lambda$:

$${\displaystyle {\hat {\pi }}_{ml}=1-{\frac {m}{{\hat {\lambda }}_{ml}}}}$$

$${\displaystyle m(1-e^{-{\hat {\lambda }}_{ml}})={\hat {\lambda }}_{ml}\left(1-{\frac {n_{0}}{n}}\right)}$$

Where $m$ is the sample mean and $\frac{n_0}{n}$ is the observed proportion of zeros.

If you plug these into the PMF $P(X = 0) = \pi + (1-\pi)e^{-\lambda}$, you get $\frac{n_0}{n}$, i.e. the MLE estimators exactly predict the observed proportion of zeros, as expected.

See also An Illustrated Guide to the Zero Inflated Poisson Regression Model.