Modeling data with a truncated and inflated Poisson distribution

Abstract

Zero inflated Poisson regression is a model commonly used to analyze data with excessive zeros. Although many models have been developed to fit zero-inflated data, most of them strongly depend on the special features of the individual data. For example, there is a need for new models when dealing with truncated and inflated data. In this paper, we propose a new model that is sufficiently flexible to model inflation and truncation simultaneously, and which is a mixture of a multinomial logistic and a truncated Poisson regression, in which the multinomial logistic component models the occurrence of excessive counts. The truncated Poisson regression models the counts that are assumed to follow a truncated Poisson distribution. The performance of our proposed model is evaluated through simulation studies, and our model is found to have the smallest mean absolute error and best model fit. In the empirical example, the data are truncated with inflated values of zero and fourteen, and the results show that our model has a better fit than the other competing models.

Appendix

Proof of Lemma 1

Since

$$\begin{aligned}&\frac{\partial }{\partial \eta _{i}}\left[ {\left( {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{y_{i}}}{y_{i}!}}\right) \left( {\sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}}\right) ^{-1}}\right] \\&\quad =\left( {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{y_{i} }}{y_{i}!}}\right) \left( {y_{i}\sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i} }}\right) ^{z}}{z!}- \sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}z}{z!}}}}\right) \left( {\sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}}\right) ^{-2}\\&\quad =\left[ {\left( {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{y_{i}}}{y_{i}!}}\right) \left( {\sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}}\right) ^{-1}}\right] \left[ y_{i}- \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}z}{z!}}\right) \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1}\right] , \end{aligned}$$

hence, we have

$$\begin{aligned} \frac{\partial }{\partial \eta _{i}}\left[ {f_{\eta _{i}}(y_{i})}\right]= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}- \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i} }}\right) ^{z}z}{z!}}\right) \left( \sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}- \left( 0+\sum \limits _{z=1}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}z}{z!}}\right) \left( \sum \limits _{z=0}^K {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}- \left( \sum \limits _{z=1}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{(z-1)!}}\right) \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}- \left( \text{ e }^{\eta _{i}}\sum \limits _{z=1}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z-1}}{(z-1)!}}\right) \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}-\text{ e }^{\eta _{i}} \left( \sum \limits _{z=0}^{K-1}{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}-\text{ e }^{\eta _{i}} \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}- \frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{K}}{K!}\right) \left( \sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}\right) ^{-1} \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}-\text{ e }^{\eta _{i}} \left( {1-\left( {\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{K}}{K!}} \right) \left( {\sum \limits _{z=0}^K{\frac{\left( {\text{ e }^{\eta _{i}}}\right) ^{z}}{z!}}}\right) ^{-1}}\right) \right] \\= & {} f_{\eta _{i}}(y_{i}) \left[ y_{i}-\text{ e }^{\eta _{i}} \left( {1-f_{\eta _{i}}(K)}\right) \right] . \end{aligned}$$

This proves the assertion. $\square$