Stata's zioprobit command fits zero-inflated ordered probit (ZIOP) models.
ZIOP models are used for ordered response variables, such as (1) fully ambulatory, (2) ambulatory with restrictions, and (3) partially ambulatory, when the data exhibit a high fraction of observations at the lowest end of the ordering. It's called zero-inflated because the idea started with Poisson regression, and it was the lower-end zeros that were overly prevalent. Given the category values we just used, Stata's zioprobit command could fit 1-inflated models. Or we could have numbered the categories 0, 1, and 2, and fit a 0-inflated model. The results would be the same either way.
Standard ordered probit models cannot account for the preponderance of zero observations when the zeros relate to an extra, distinct source. Consider a study of tobacco use in which the outcome of interest, smoking, is an ordered discrete response with four levels coded as 0, 1, 2, and 3, with 0 meaning "Nonsmoker" and 3 meaning "Daily, 20+ cigarettes/day".
Many of the individuals in the first category will be nonsmokers who have never smoked and will never smoke. The rest of them will be ex-smokers. Think of the standard ordered probit model as fitting the behavior of smokers, including ex-smokers. The zero inflation arises because the first group now includes those who have never smoked.
We have fictional data on the smoking study just described. The outcome variable is called tobacco and contains
|Category Frequency Meaning|
|0 78.1% Nonsmoker|
|1 3.6% Weekly or less|
|2 13.0% Daily, less than 20 cigarettes/day|
|3 5.3% Daily, 20 or more cigarettes/day|
We believe that the 0 is inflated.
We want to fit a model in which smoking by those who have ever smoked is given by
And membership in the never-smoked group is determined by
To fit the model, we type
. zioprobit tobacco income i.female age, inflate(income i.female age i.parent i.religion) Zero-inflated ordered probit regression Number of obs = 14,899 Wald chi2(3) = 751.43 Log likelihood = -10299.787 Prob > chi2 = 0.0000
|tobacco||Coef. Std. Err. z P>|z| [95% Conf. Interval]|
|income||.1503256 .0057582 26.11 0.000 .1390398 .1616113|
|female||-.2726466 .047975 -5.68 0.000 -.3666759 -.1786173|
|age||-.1394573 .011523 -12.10 0.000 -.1620419 -.1168727|
|income||-.0654874 .0087703 -7.47 0.000 -.082677 -.0482979|
|female||-.2166707 .0509783 -4.25 0.000 -.3165863 -.1167552|
|age||.1205886 .0165181 7.30 0.000 .0882136 .1529636|
|smoking||.7219495 .0436831 16.53 0.000 .6363321 .8075669|
|discourages||-.2095319 .0586036 -3.58 0.000 -.3243927 -.094671|
|_cons||-.5335904 .0873953 -6.11 0.000 -.7048821 -.3622987|
|/cut1||.0683114 .0881964 -.1045504 .2411731|
|/cut2||.2977055 .0804097 .1401054 .4553055|
|/cut3||1.402649 .067253 1.270836 1.534463|
The standard ordered probit parameters, coefficients and cutpoints, are displayed in the first and last parts of the output, respectively.
The middle part of the output reports the probit coefficients for the inflation.
Coefficients can be difficult to interpret. For instance, what does a parent smoking coefficient of 0.72 mean? It means that, on average in the data, those whose parents are smokers are about 27% less likely to be never-smokers than those whose parents did not use tobacco. We obtained the 27% by using Stata's margins command:
. margins, predict(pnpar) dydx(parent) Average marginal effects Number of obs = 14,899 Model VCE : OIM Expression : Pr(nonparticipation), predict(pnpar) dy/dx w.r.t. : 1.parent
|dy/dx Std. Err. z P>|z| [95% Conf. Interval]|
|smoking||-.266089 .015175 -17.53 0.000 -.2958314 -.2363467|
|Note: dy/dx for factor levels is the discrete change from the base level.|
The predict(pnpar) option is unique to margins when used after zioprobit. We asked margins to calculate predictions of the probability of nonparticipation, which in this example means the probability of being a never-smoker.
You can also fit Bayesian zero-inflated ordered probit models using the bayes prefix.