On the robustness of truncated negative binomial regression model: application to field epidemiology

Abstract

Truncated count data are often obtained from field investigations conducted for individuals with some health-related symptoms to discover the possible causes of food-borne outbreaks quickly and accurately. This study shows two robust properties of the truncated negative binomial (TNB) model. First, by characterizing the whole set of models leading to the same likelihood function as the TNB model, we find a practical meaning that the TNB model gives reliable inference for the regression coefficients even zero inflation is allowed, but a careful interpretation of the regression coefficients is needed. Second, the TNB model can be derived from the Poisson distribution with the random intercept following a gamma distribution, however, it is difficult to justify the distribution assumption for the random intercept. We find that the TNB model presents robust inference for the slope parameters against a misspecified random effect distribution. With some analytic justifications, our numerical study shows that the empirical coverage based on the TNB model is close to its nominal level, even when the random effect distribution is misspecified. The TNB model is applied to analyze truncated count data from the food-borne outbreak that occurred in South Korea.

1. Introduction

In August 2021, a food poisoning incident occurred in Gyeonggi-do, South Korea. Epidemiological investigators were dispatched urgently to investigate the situation. To expedite the investigation, the investigators first examined people with symptoms related to the exposure in question. The outcome variable is the number of diarrhea cases, and data are obtained from individuals with at least one case of diarrhea. This is a truncated random sample due to the investigation characteristic of field epidemiology. Therefore, instead of the standard count data analysis, an analysis of truncated count data should be performed. In addition to the above-mentioned application in field epidemiology, truncated count data have appeared in various contexts such as capture-recapture studies [1] and medical studies on the number of cholera cases [6].

The truncated negative binomial (TNB) model is one of the most popular parametric models for truncated count data analysis [10]. The truncated Poisson (TPO) model is a special case of the TNB model, such that the latter is more flexible than the former at the cost of estimating an additional dispersion parameter. It is important to recognize how the relationship between Poisson and negative binomial (NB) distributions relies on the presence of truncation. At first, suppose that there is no truncation, and the true data-generating model is an NB regression model. It is well known that the use of Poisson regression guarantees the consistency of regression coefficients in the mean parameter as long as the mean model is correctly specified. However, in the presence of truncation, this property no longer holds because the functional form of the mean from the TNB model differs from that from the TPO model [10], which implies that the robustness of the Poisson regression to distributional assumption can be vulnerable to how the samples are obtained.

In addition, several streams of the literature [4,9,10] involving modeling with truncated count data often start with parametric distributions defined on the whole non-negative integers. For example, the TNB and TPO distributions are obtained from the NB and Poisson distributions, respectively. However, starting with a parametric distribution defined on both truncated and observed parts leaves researchers with a serious concern in modeling the truncated count data because it is difficult to justify the truncated and unobserved part. For example, in the food poisoning incident in Gyeonggi-do, the number of purchasers of food related to the incident is approximately 8000, but the number of subjects that are epidemiologically investigated is 339. This motivating example shows that there may be many people who have had the food but have not contracted food poisoning, which corresponds to the existence of zero inflation in the truncated part. A natural question entails whether the analysis of truncated count data using the TNB model provides reliable results, even when the truncated part is not generated from the NB distribution. We investigate the robustness of the TNB model, and we fully characterize the class of distributions that lead to the TNB that can be identified using observed data. In addition, we show that a careful interpretation of the regression coefficients of the TNB model is required because the interpretation can depend on the assumption of the truncated part.

The Poisson distribution assumes that the variance is equal to the mean, which is often considered an inflexible part when analyzing overdispersed count data. The NB distribution is derived from the Poisson regression model with the random intercept following a gamma distribution. Therefore, as a more flexible method for handling overdispersion, the NB distribution has been widely used in various areas [10]. However, the use of a specific random effect distribution in deriving the NB distribution leaves researchers with a serious concern because the random effect distribution can be different from the gamma distribution in practice. Because it is difficult to know which random effect distribution is correct in advance, a natural question entails whether the analysis of truncated count data using the TNB model provides reliable results, even when the distribution of the random intercept differs from the gamma distribution. Previous studies, such as Neuhaus et al. [14,15] and McCulloch and Neuhaus [12], investigated the impact of misspecified random effect distributions in generalized linear mixed models. Their key findings were that the estimates of slope parameters associated with the covariates, which do not involve random slope, have no or little bias, while the intercept parameter can be biased. However, it remains unclear whether these findings generalize to the type of non-standard data structure involving outcome truncation. Therefore, we check how a misspecified distribution for the random intercept affects the biases and empirical coverage results of confidence intervals of the TNB estimates. Herein, we show some analytic results for the robustness of the TNB model against a misspecified random effect distribution. For other non-tractable cases, we perform a numerical study to check its robustness to various misspecified situations.

The remainder of this paper is organized as follows: Section 2 introduces parametric models for truncated count data. The first part of Section 3 explains robustness against the model assumption for the truncated part by characterizing the class of probability mass functions leading to the same truncated distribution. In addition, we explain why it is necessary to be cautious when interpreting the regression coefficients in the mean model. The second part of Section 3 explains the robustness against the misspecified random effect distribution. Some analytic results are obtained. Section 4 includes a numerical study designed to confirm the findings in Section 3. Section 5 analyzes the actual truncated count data from the food-borne outbreak, while Section 6 provides the conclusion.

2. Parametric models for truncated count data

Suppose that Y is an outcome variable of a non-negative integer type and $\mathit{X}$ is a p-dimensional vector of covariates, and $(Y_i,{\mathit{X}}_{\mathit{i}})$ $i=1,\dots ,n$ are independently and identically distributed. Let $f(y\mid \mathit{x},\mathit{\theta })$ be a probability mass function for Y = y given $\mathit{X}=\mathit{x}$ parametrized by $\mathit{\theta }$. $\mathcal{F}$ denotes a family of parametric models for count data $\mathcal{F}=\left\{f\right(y\mid \mathit{x},\mathit{\theta })\mid \mathit{\theta }\in \mathbf{\Theta },y\in \{0,1,2\dots ,\left\}\right\}$ where $\mathbf{\Theta }$ is the whole parametric space. The k-truncated density function from $f(y\mid \mathit{x},\mathit{\theta })$, which corresponds to the distribution of Y conditioned on $Y\ge k+1$, is represented as

$$f^{\left(k\right)}(y\mid \mathit{x},\mathit{\theta })=f(y\mid \mathit{x},\mathit{\theta })/\{1-\sum _{j=0}^{k}f(j\mid \mathit{x},\mathit{\theta }\left)\right\},$$

and the corresponding log-likelihood function for the whole data is

$${\ell }^{\left(k\right)}\left(\mathit{\theta }\right)=\sum _{i=1}^n[\log f(y_i\mid {\mathit{x}}_i,\mathit{\theta })-\log \{1-\sum _{j=0}^{k}f(j\mid {\mathit{x}}_i,\mathit{\theta }\left)\right\}].$$

The superscript (k) of $f(y\mid \mathit{x},\mathit{\theta })$ and $\ell \left(\mathit{\theta }\right)$ denotes truncation of the data up to k. For estimating $\mathit{\theta }$, note that ${\ell }^{\left(k\right)}\left(\mathit{\theta }\right)$ relies on $f(j\mid \mathit{x},\mathit{\theta })$ for $j=0,1,\dots ,k$, and no data belonging to this truncated part is available.

Two widely-used parametric models for truncated count data are TPO and TNB [7,8]. Here, we use the canonical link function for the conditional mean as $\log \mathbb{E}(Y\mid \mathit{X})=\log \mu ={\mathit{X}}^{\mathrm{T}}\mathrm{\text{β}}$ where T denotes the transpose. Using the probability mass function of TPO, the corresponding log-likelihood function for the whole data is

$${\ell }_{\mathrm{PO}}^{\left(k\right)}\left(\mathrm{\text{β}}\right)=\sum _{i=1}^n[y_i\log {\mu }_i-{\mu }_i-\log \{1-\sum _{j=0}^k({\mu }_i^j\exp (-{\mu }_i)/j!\left)\right\}].$$ (1)

The probability mass function of the TNB model is

$$f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )=\frac{\mathrm{\Gamma }(\frac{1}{\varphi }+y)}{y!\mathrm{\Gamma }\left(\frac{1}{\varphi }\right)}\frac{(\mathrm{\varphi \mu }{)}^y}{(1+\mathrm{\varphi \mu }{)}^{y+1/\varphi }}/\{1-\sum _{j=0}^k\frac{\mathrm{\Gamma }(\frac{1}{\varphi }+j)}{j!\mathrm{\Gamma }\left(\frac{1}{\varphi }\right)}\frac{(\mathrm{\varphi \mu }{)}^j}{(1+\mathrm{\varphi \mu }{)}^{j+1/\varphi }}\}$$

and the corresponding log-likelihood function is

$$\begin{array}{rl}{\ell }_{\mathrm{NB}}^{\left(k\right)}(\mathrm{\text{β}},\varphi )& =\sum _{i=1}^n[\log \frac{\mathrm{\Gamma }(\frac{1}{\varphi }+j)}{\mathrm{\Gamma }\left(\frac{1}{\varphi }\right)}+\log \frac{(\varphi {\mu }_i{)}^{y_i}}{(1+\varphi {\mu }_i{)}^{y_i+1/\varphi }}\\ & -\log \{1-\sum _{j=0}^k\frac{\mathrm{\Gamma }(\frac{1}{\varphi }+j)}{j!\mathrm{\Gamma }\left(\frac{1}{\varphi }\right)}\frac{(\varphi {\mu }_i{)}^j}{(1+\varphi {\mu }_i{)}^{j+1/\varphi }}\}],\end{array}$$ (2)

where ϕ is the dispersion parameter. It is well known that (2) converges to (1) as ϕ goes to zero. Therefore, the TPO model is a special case of the TNB model so that the latter is more robust to the overdispersion of count data. In addition, unlike the relationship between the Poisson and NB models when there is no truncation, the maximum likelihood estimators of $\mathrm{\text{β}}$ from (1) are generally inconsistent for $\mathrm{\text{β}}$ in (2) [4,7]. Considering these points, in the following discussion, the TNB model is our basic model for analyzing truncated count data.

3. Robust properties of truncated negative binomial model

We investigate two robustness properties of the TNB model. The first property is about the robustness against the truncated part, which is the unobserved part. The second property is about the robustness against the distribution assumption of the random intercept in the Poisson distribution.

3.1. Robustness against the model assumption for truncated part

To figure out the first property, we characterize the set of probability mass functions leading to the common truncated distribution (2). Once this characterization is provided, we can understand robustness against the model assumption for the truncated part.

Suppose that $f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ is known or consistently estimated from the truncated data. Define

$$\begin{array}{rl}{\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}& =\{h\mid \mathrm{the}\mathrm{probability}\mathrm{vector}\mathrm{is}\mathrm{equal}\mathrm{to}(h(0\mid \mathit{x}),\dots ,h(k\mid \mathit{x}),\dots )\\ & \mathrm{satisfying}{h}^{\left(k\right)}(y\mid \mathit{x})=f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )I_{(y>k)}\},\end{array}$$

where $I_{(\cdot )}$ is an indicator function that has a value of 1 when the condition in parentheses is true, and 0 otherwise. The set ${\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$ is the whole collection of probability mass functions h leading to $f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ as its probability mass function for the truncated data up to k. Define

$$\begin{array}{rl}\mathcal{H}& =\{h\mid \mathrm{the}\mathrm{probability}\mathrm{vector}\mathrm{is}\mathrm{equal}\mathrm{to}(h(0\mid \mathit{x}),\dots ,h(k\mid \mathit{x}),\dots )\\ & \mathrm{where}h(y\mid \mathit{x})=\alpha \left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )\mathrm{for}y>k\},\end{array}$$

where $\sum _{y=0}^{k}h(y\mid \mathit{x})=1-\alpha \left(\mathit{x}\right)<1$ and $h(y\mid \mathit{x})\ge 0$ for $y=0,\dots ,k$. This representation says that $\mathcal{H}$ consists of the probability mass function h where the observed part (y>k) is proportional to $f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$, and the remaining probability from the observed part is arbitrarily distributed on the truncated part ( $0,1,\dots ,k$). The following show the equivalent relationship between these two representations.

Proposition 3.1

${\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}=\mathcal{H}$ and it is convex.

For the proof, first we show $\mathcal{H}\subset {\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$. Let $h(y\mid \mathit{x})$ be any probability mass function in $\mathcal{H}$ and consider its k-truncated probability mass function $h^{\left(k\right)}(y\mid \mathit{x})$, then

$$h^{\left(k\right)}(y\mid \mathit{x})=\frac{h(y\mid \mathit{x})}{1-\sum _{j=0}^{k}h(j\mid \mathit{x})}=\frac{\alpha \left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )}{1-(1-\alpha (\mathit{x}\left)\right)}=f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi ),$$

which shows that every element of $\mathcal{H}$ becomes an element of ${\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$.

To show $\mathcal{H}\supset {\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$, let $h(y\mid \mathit{x})$ be any element of ${\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$, then

$$\begin{array}{rl}{h}^{\left(k\right)}(y\mid \mathit{x})& =\frac{h(y\mid \mathit{x})}{1-\sum _{j=0}^{k}h(j\mid \mathit{x})}=f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi ),\mathrm{and}\\ h(y\mid \mathit{x})& =(1-\sum _{j=0}^{k}h(j\mid \mathit{x}\left)\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )=\alpha \left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )\end{array}$$

Thus, $h(y\mid \mathit{x})\in \mathcal{H}$.

The remainder of the proof is to show that $\mathcal{H}$ is convex, which requires that $w_1{h}_1+w_2{h}_2$ belongs to $\mathcal{H}$ for any $h_1,h_2\in \mathcal{H}$ where $w_1\ge 0$ and $w_2\ge 0$ and $w_1+w_2=1$.

Suppose that $h_1(y\mid \mathit{x})$ and $h_2(y\mid \mathit{x})$ belong to $\mathcal{H}$. Then,

$$h(y\mid \mathit{x})=w_1{h}_1(y\mid \mathit{x})+w_2{h}_2(y\mid \mathit{x})=\left(w_1{\alpha }_1\right(\mathit{x})+w_2{\alpha }_2(\mathit{x}\left)\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi ),$$

where $h_i(y\mid \mathit{x})={\alpha }_i\left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ for $i=1\mathrm{or}2$. And

$$\begin{array}{rl}\sum _{j=0}^{k}h(j\mid \mathit{x})& =1-\sum _{j=k+1}^{\mathrm{\infty }}h(j\mid \mathit{x})=1-\left(w_1{\alpha }_1\right(\mathit{x})+w_2{\alpha }_2(\mathit{x}\left)\right)\sum _{y\ge k+1}^{\mathrm{\infty }}{f}_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )\\ & =1-\left(w_1{\alpha }_1\right(\mathit{x})+w_2{\alpha }_2(\mathit{x}\left)\right)<1.\end{array}$$

Thus, $h(y\mid \mathit{x})$ satisfies all conditions to be an element of the set $\mathcal{H}$ so that $\mathcal{H}$ is a convex set.

This result shows that any probability mass function leading to $f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )I_{(y>k)}$ for the truncated data can be represented as an element in $\mathcal{H}$, and the representation in $\mathcal{H}$ is useful to examine some geometric property of ${\mathrm{\Omega }}_{f_{\mathrm{NB}}^{\left(k\right)}}$. Examining the proof, this result can be extended beyond the TNB model.

It is interesting to investigate some specific elements in $\mathcal{H}$. First, suppose that we consider a mixture of Y and $Y^k$, $Y^{\ast }=\mathrm{ZY}+(1-Z)Y^k$, where Y follows the NB distribution, $Y^k$ is drawn from the probability mass function $g(y\mid \mathit{x})$ which has an integer support ranging from 0 to k, and $Z\mid \mathit{X}=\mathit{x}$ follows the Bernoulli distribution with probability $\omega \left(\mathit{x}\right)$. The probability mass function of $Y^{\ast }$ has the form of,

$$h_{Y^{\ast }}(y^{\ast }\mid \mathit{x})=\omega \left(\mathit{x}\right)f_{\mathrm{NB}}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )+(1-\omega (\mathit{x}\left)\right)g(y\mid \mathit{x}),$$

which is included in the set $\mathcal{H}$ as the truncated density function of $h_{Y^{\ast }}(y^{\ast }\mid \mathit{x})$ has the form of $\alpha \left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ because

$$\frac{\omega \left(\mathit{x}\right)f_{\mathrm{NB}}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )}{1-\sum _{j=0}^k\left\{\omega \right(\mathit{x}\left)f_{\mathrm{NB}}\right(j\mid \mathit{x},\mathrm{\text{β}},\varphi )+(1-\omega \left(\mathit{x}\right)\left)g\right(j\mid \mathit{x}\left)\right\}}=f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi ),$$

and $\sum _{j=0}^{k}g(j\mid \mathit{x})=1$. This case corresponds to the zero-inflated negative binomial(ZINB) model [4] when k=0, which leads to the following conditional mean function

$$\mathbb{E}(Y\mid \mathit{x})=\omega \left(\mathit{x}\right)\mu +(1-\omega (\mathit{x}\left)\right)\sum _{y=0}^k\{y\cdot g(y\mid \mathit{x}\left)\right\}.$$ (3)

Another example in $\mathcal{H}$ is

$$(1-\omega (\mathit{x}\left)\right)g(y\mid \mathit{x})+\omega \left(\mathit{x}\right)f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi ),$$ (4)

which belongs to the hurdle model [5,13]. Note that the support of $g(y\mid \mathit{x})$ and that of $f_{\mathrm{NB}}^{\left(k\right)}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ are separated, and the conditional mean is

$$\mathbb{E}(Y\mid \mathit{x})=\frac{\omega \left(\mathit{x}\right)(\mu -\sum _{y=0}^{k}y{f}_{\mathrm{NB}}(y\mid \mathit{x},\mathrm{\text{β}},\varphi \left)\right)}{1-\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid \mathit{x},\mathrm{\text{β}},\varphi )}+(1-\omega (\mathit{x}\left)\right)\sum _{y=0}^k\{y\cdot g(y\mid \mathit{x}\left)\right\}.$$ (5)

The above-stated observations imply that the maximum likelihood estimates of the TNB model are consistent under various mixture models leading to the same TNB likelihood function. We note that this property requires a care in interpreting the regression coefficient $\mathrm{\text{β}}$ in the TNB model because the interpretation depends on the model for the truncated part, as shown in (3) and (5). To explain this more specifically, first, suppose that the data generating model is $f_{\mathrm{NB}}(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$, which lies in $\mathcal{H}$. Then, the parameter $\mathrm{\text{β}}$ can be interpreted in a standard way because $\mathrm{\text{β}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathit{x}}\log \mathbb{E}(Y\mid \mathit{x})$. Second, suppose that the data-generating model is the ZINB model. Note that (3) implies $\mathrm{\text{β}}\ne \frac{\mathrm{\partial }}{\mathrm{\partial }\mathit{x}}\log \mathbb{E}(Y\mid \mathit{x})$. For interpreting $\mathrm{\text{β}}$ in the ZINB model, it is useful to consider two underlying groups, namely, the structural group producing the zero inflation and responsive group that includes random zeroes, and $\mathrm{\text{β}}$ is interpreted as the differentiation of the log of the responsive group mean with respect to $\mathit{x}$. For example, Sheu et al. [19] analyzed the effect of cigarette price to the number of smoking per day by treating the nonsmokers as structural zeros and the smokers as the responsive group. Third, suppose that the data generating model is the hurdle model, which also implies $\mathrm{\text{β}}\ne \frac{\mathrm{\partial }}{\mathrm{\partial }\mathit{x}}\log \mathbb{E}(Y\mid \mathit{x})$. The hurdle model assumes a two-stage sampling mechanism, where the first stage determines whether each unit crosses the hurdle, and the second stage observes the counts for the units that cross the hurdle. For this model, $\mathrm{\text{β}}$ lies in the mean model for the group that crosses the hurdle, but its interpretation is not straightforward because of the complex functional form of (5).

Because the correctly specified full model cannot be identified using truncated data only, interpreting the regression parameter should be based on untestable assumptions for the truncated part. Therefore, researchers should keep in mind that their interpretation partially relies on their domain knowledge about the truncated part. Even without considering truncation, Rose et al. [18] points out that the choice between the ZINB and hurdle models requires an understanding of study design and research purpose, as the two models are nearly indistinguishable in practice.

3.2. Robustness against misspecified random effect distribution

Suppose that given the random intercept ν with mean 1 and variance ${\sigma }_{\nu }^2$, Y follows the Poisson distribution with mean $\mathrm{\mu v}$. The random intercept ν is often considered for reflecting individual heterogeneity or omitted covariates [8,11]. The marginal probability mass function of Y can be written as

$$f_G(y\mid \mathit{x},\mathrm{\text{β}},\varphi )={\int }_{\nu }\frac{(\mathrm{\mu \nu }{)}^y\exp (-\mathrm{\mu \nu })}{y!}\mathrm{d}{G}_{\varphi }\left(\nu \right),$$ (6)

where $\mu =\exp \left({\mathit{x}}^{\mathrm{T}}\mathrm{\text{β}}\right)$, ϕ is a dispersion parameter, and $G_{\varphi }$ is the distribution of ν. When ν follows a gamma distribution, $f_G(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ becomes an NB distribution. In practice, researchers are uncertain about whether ν follows a gamma distribution. This raises concerns that the truncated distribution from $f_G(y\mid \mathit{x},\mathrm{\text{β}},\varphi )$ is not the same as the TNB model. Due to the challenge of determining whether the random effect distribution is gamma based on observed data, it becomes necessary to explore whether the TNB model experiences biases and low empirical coverage of confidence intervals for the slope parameters in β when the distribution for the random intercept is misspecified. First, we analytically check whether the robustness of the TNB model holds against a misspecified random effect distribution by extending Neuhaus et al. [14,15]'s approach.

Let ${\mathrm{\text{β}}}_T=({\mathrm{\text{β}}}_{T0},{\mathrm{\text{β}}}_{\mathrm{Ts}}{)}^{\mathrm{T}}$ where ${\mathrm{\text{β}}}_{T0}$ and ${\mathrm{\text{β}}}_{\mathrm{Ts}}$ are the intercept and the slope parameters in the true model $f_G$, respectively, and ${\varphi }_T$ denotes the true dispersion parameter in $f_G$. Considering that the TNB is a working model, the maximum likelihood estimator $\stackrel{ˆ}{{\mathrm{\text{β}}}^{\ast }}=(\stackrel{ˆ}{{\mathrm{\text{β}}}_0^{\ast }},\stackrel{ˆ}{{\mathrm{\text{β}}}_s^{\ast }}{)}^{\mathrm{T}}$ converges to the value $({\mathrm{\text{β}}}_0^{\ast },{\mathrm{\text{β}}}_s^{\ast }{)}^{\mathrm{T}}$ corresponding to the minimizer of the Kullback-Leibler divergence $D(b_0,b_s)$ between the true truncated model and TNB [16],

$$D(b_0,b_s)={\mathbb{E}}_{X\mid Y>k}{\mathbb{E}}_{Y\mid X,Y>k}[\sum _{i=1}^n\log \left\{f_G^{\left(k\right)}\right(Y_i\mid {\mathit{X}}_i,\mathrm{\text{β}},\varphi \left)/f_{\mathrm{NB}}^{\left(k\right)}\right(Y_i\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast }\left)\right\}],$$ (7)

where $b_0$ and $b_s$ correspond to the regression coefficients associated with the intercept and covariates, respectively, and both expectations are taken under the true probability mass function on (6).

We assume that $\mathbb{E}(\mathit{X}\mid Y>k)=0$ and that ν is independent of $\mathit{X}$. To satisfy the first assumption, we can center all observed covariates $\mathit{X}$. Our analytical result is derived when ${\mathrm{\text{β}}}_{\mathrm{Ts}}=0$. Differentiation of (7) with respect to $b_0$ and $b_s$ is

$$\begin{array}{rl}\frac{\mathrm{\partial }D(b_0,b_s)}{\mathrm{\partial }{b}_0}& ={\mathbb{E}}_{\mathit{X}\mid Y>k}{\mathbb{E}}_{Y\mid \mathit{X},Y>k}[\sum _{i=1}^n\left\{r\right(Y_i\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })\\ & +\frac{\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })r(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}{1-\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}\}]\end{array}$$ (8)

and

$$\begin{array}{rl}\frac{\mathrm{\partial }D(b_0,b_s)}{\mathrm{\partial }{b}_s}& ={\mathbb{E}}_{\mathit{X}\mid Y>k}{\mathbb{E}}_{Y\mid \mathit{X},Y>k}[\sum _{i=1}^n\left\{r\right(Y_i\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })\\ & +\frac{\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })r(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}{1-\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}\}{\mathit{X}}_i],\end{array}$$ (9)

where $r(Y_i\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })=\{Y_i-\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right\}/\{1+{\varphi }^{\ast }\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right\}$.

To investigate the property of the solution for $\mathit{b}$, consider the expectation with respect to $Y\mid \mathit{X},Y>k$ in (9). Then,

$$\begin{array}{rl}& {\mathbb{E}}_{Y\mid \mathit{X},Y>k}\left[r\right(Y\mid \mathit{X},\mathit{b},{\varphi }^{\ast }\left)\right]\\ & =(\frac{{\mu }_T-\sum _{j=0}^{k}j\cdot f_G(j\mid \mathit{X},{\mathrm{\text{β}}}_T,\varphi )}{1-\sum _{j=0}^k{f}_G(j\mid \mathit{X},{\mathrm{\text{β}}}_T,\varphi )}-\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right)/(1+{\varphi }^{\ast }\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right),\end{array}$$ (10)

where ${\mu }_T=\exp \left({\mathrm{\text{β}}}_{T0}\right)$. Using this, Equation (9) becomes

$$\begin{array}{rl}& {\mathbb{E}}_{\mathit{X}\mid Y>k}[\sum _{i=1}^n\{(\frac{{\mu }_T-\sum _{j=0}^{k}j\cdot f_G(j\mid {\mathit{X}}_i,{\mathrm{\text{β}}}_T,\varphi )}{1-\sum _{j=0}^k{f}_G(j\mid {\mathit{X}}_i,{\mathrm{\text{β}}}_T,\varphi )}-\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right)/(1+{\varphi }^{\ast }\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right)\\ & +\frac{\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })r(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}{1-\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}\}{\mathit{X}}_i].\end{array}$$ (11)

When $b_s=0$, the inner terms in the curl bracket in (11) are free of $\mathit{X}$ regardless of $b_0$ and ${\varphi }^{\ast }$, and (11) becomes 0 after taking the expectation with respect to $\mathit{X}\mid Y>k$. Thus, $\stackrel{ˆ}{{\mathrm{\text{β}}}_s^{\ast }}$ is a consistent estimator for ${\mathrm{\text{β}}}_s^{\ast }={\mathrm{\text{β}}}_{\mathrm{Ts}}$. This result does not generally hold when ν depends on $\mathit{X}$ as $\log \mathbb{E}(Y\mid \mathit{X})$ is generally not linear in $\mathit{X}$.

However, for the intercept and dispersion parameters, Equation (8), by using (10) and replacing $b_0$ with ${\mathrm{\text{β}}}_{T0}$, becomes

$$\begin{array}{rl}& {\mathbb{E}}_{\mathit{X}\mid Y>k}[\sum _{i=1}^n\{(\frac{{\mu }_T-\sum _{j=0}^{k}j\cdot f_G(j\mid {\mathit{X}}_i,{\mathrm{\text{β}}}_T,\varphi )}{1-\sum _{j=0}^k{f}_G(j\mid {\mathit{X}}_i,{\mathrm{\text{β}}}_T,\varphi )}-\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right)/(1+{\varphi }^{\ast }\exp ({\mathit{X}}_i^{\mathrm{T}}\mathit{b}\left)\right)\\ & +\frac{\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })r(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}{1-\sum _{j=0}^k{f}_{\mathrm{NB}}(j\mid {\mathit{X}}_i,\mathit{b},{\varphi }^{\ast })}\}],\end{array}$$

and this is generally nonzero when $f_G^{\left(k\right)}(\cdot )\ne f_{\mathrm{NB}}^{\left(k\right)}(\cdot )$. To investigate the amount of bias and the intractable case with ${\mathrm{\text{β}}}_{\mathrm{Ts}}\ne 0$, a numerical study will be considered in Section 4.

4. Numerical studies

The robustness discussed in Section 3.1 arises from the fact that various mixture models result in the same TNB likelihood function, ensuring that their likelihood inference is identical. Therefore, a numerical study is not necessary for checking the first robustness property. In this section, we perform a numerical study to check the robustness discussed in Section 3.2 under various simulation settings. For $i=1,\dots ,n$, the outcome variable is generated from $Y_i\mid x_i,{\nu }_i\sim \mathrm{Poisson}\left({\mu }_i{\nu }_i\right)$ where ${\mu }_i=\exp \left(x_i^{\mathrm{T}}\mathrm{\text{β}}\right)$, ${\nu }_i\sim G$ with $\mathbb{E}\left({\nu }_i\right)=1$ and $\mathrm{Var}\left({\nu }_i\right)={\sigma }_{\nu }^2$. We consider the following five distributions for ${\nu }_i$ (1) gamma, (2) uniform, (3) discrete support which contains only two elements, (4) log-normal, and (5) a mixture of two log-normal distributions. Note that the gamma distribution represents the case of correct specification, and serves as a reference for evaluating performance under other misspecification scenarios. For brevity, these are called ‘gamma’, ‘uniform’, ‘two points’, ‘log-normal’, and ‘mixture’, respectively. For the covariate vector ${\mathit{x}}_i=(x_{i1},\dots ,x_{i6}{)}^{\mathrm{T}}$, the first three covariates are drawn from normal distribution with mean 0 and variance 0.5, while the other three covariates are from the Bernoulli trial with probability 0.5. The value of ${\mathrm{\text{β}}}_0$ and ${\mathrm{\text{β}}}_s=({\mathrm{\text{β}}}_1,{\mathrm{\text{β}}}_2,{\mathrm{\text{β}}}_3,{\mathrm{\text{β}}}_4,{\mathrm{\text{β}}}_5,{\mathrm{\text{β}}}_6{)}^{\mathrm{T}}$ are specified according to the four scenarios (S1,S2,S3,S4). S1 and S2 are designed to check the cases with the null effect of covariates. We use $\{{\mathrm{\text{β}}}_0=1.2,{\mathrm{\text{β}}}_s=(0,0,0,0,0,0{)}^{\mathrm{T}}\}$ and $\{{\mathrm{\text{β}}}_0=1.5,{\mathrm{\text{β}}}_s=(0,0,0,0,0,0{)}^{\mathrm{T}}\}$ for S1 and S2, respectively. S3 and S4 are designed to check the cases with the non-null effect of covariates. We use $\{{\mathrm{\text{β}}}_0=0,{\mathrm{\text{β}}}_s=(0.2,0.5,0.8,0.2,0.5,0.8{)}^{\mathrm{T}}\}$ and $\{{\mathrm{\text{β}}}_0=0.5,{\mathrm{\text{β}}}_s=(0.2,0.5,0.8,0.2,0.5,0.8{)}^{\mathrm{T}}\}$ for S3 and S4, respectively. For each scenario, five different distributions for the random intercept, three different values of ${\sigma }_{\nu }^2=(0,0.25,0.5)$, and three different truncation points $k=(0,1,2)$ are considered. In total, 132 different combinations of $\mathrm{\text{β}}$, k, ${\sigma }_{\nu }^2$, and G are studied, and each setting is repeated 1000 times. We set the initial sample sizes as n = 300 for all scenarios, but the number of samples monotonically decreases as k increases and varies for each iteration. Table 1 presents the average number of samples and empirical truncation probabilities depending on $\mathrm{\text{β}}$, k and ${\sigma }_{\nu }^2$. The results with a larger sample size, n = 500, are presented in the supplementary materials because the overall results are similar to those with n = 300.

Table 1.

Average sample sizes and empirical truncation probabilities (in parentheses) after truncation depending on β, k and ${\sigma }_{\nu }^2$ when the initial sample size is 300.

		Truncation
		k = 0			k = 1			k = 2
		${\sigma }_{\nu }^2$			${\sigma }_{\nu }^2$			${\sigma }_{\nu }^2$
Scenario	${\mathrm{\text{β}}}_0$	0	0.25	0.5	0	0.25	0.5	0	0.25	0.5
S1	1.2	289	274	262	253	224	206	193	167	152
		(0.96)	(0.91)	(0.87)	(0.84)	(0.75)	(0.69)	(0.64)	(0.56)	(0.51)
S2	1.5	297	286	277	281	254	235	247	210	189
		(0.99)	(0.95)	(0.92)	(0.94)	(0.85)	(0.78)	(0.82)	(0.70)	(0.63)
S3	0	188	235	224	98	166	156	47	115	108
		(0.63)	(0.78)	(0.75)	(0.33)	(0.55)	(0.52)	(0.16)	(0.38)	(0.36)
S4	0.5	230	265	255	153	216	203	95	171	158
		(0.77)	(0.88)	(0.85)	(0.51)	(0.72)	(0.68)	(0.32)	(0.57)	(0.53)

4.1. Bias results of intercept and slope parameter estimates

To assess the biases of the TNB estimates, we examine the intercept ${\mathrm{\text{β}}}_0$ and a regression coefficient ${\mathrm{\text{β}}}_3$. Here, ${\mathrm{\text{β}}}_3$ is selected because it is the largest coefficient and is expected to show the bias pattern most clearly. For the rest of the coefficients, see the supplementary materials. Figure 1 presents the bias of ${\mathrm{\text{β}}}_0$ from S1–S4. When ${\sigma }_{\nu }^2=0$, that is zero heterogeneity of random intercept, there is no bias for ${\mathrm{\text{β}}}_0$ for all scenarios. However when k=2 and large ${\sigma }_{\nu }^2$, the biases are substantial when the random effect distribution is mixture, which warns of serious bias of ${\mathrm{\text{β}}}_0$ estimates by misspecified G. The interquartile ranges increase as k increases because the sample size monotonically decreases.

Figure 1.

Box plots of bias of the intercept estimate. Each row represents scenarios S1, S2, S3, and S4, respectively, and each column represents a different truncation point k. In each cell, there are 11 box plots corresponding to different distributions of the random effect.

However, there is little bias in ${\mathrm{\text{β}}}_3$ estimates, as shown in Figure 2. For S1 and S2, these observations are consistent with the analytic results presented in Section 3.2. In addition, although the non-null effect is beyond the scope of our analytic result, the analytic results seem to be extended to the covariates with non-null effect for S3 and S4. Unlike ${\mathrm{\text{β}}}_0$, the estimate of ${\mathrm{\text{β}}}_s=({\mathrm{\text{β}}}_1,\dots ,{\mathrm{\text{β}}}_6{)}^{\mathrm{T}}$ in the TNB model has been found to have little bias regardless of the misspecified G.

Figure 2.

Box plots of bias of the third slope estimate. Each row represents scenarios S1, S2, S3, and S4, respectively, and each column represents a different truncation point k. In each cell, there are 11 box plots corresponding to different distributions of the random effect.

4.2. Coverage results of confidence intervals for intercept and slope parameter

We assess the coverage rate of 95% confidence intervals, and the empirical coverage rate is calculated as the number of the confidence intervals containing the true parameter divided by 1000. The 95% confidence intervals are constructed using $\widehat{\mathrm{\text{β}}}\pm 1.96\times$ the model-based standard error estimates. Figures 3 and 4 present empirical coverage rates for ${\mathrm{\text{β}}}_0$ and ${\mathrm{\text{β}}}_3$ under S1–S4, respectively. The red horizontal line corresponds to 0.95. Each line represents how empirical coverage rates change according to ${\sigma }_{\nu }^2$. For the rest of the coefficients, see the supplementary materials.

Figure 3.

Empirical coverage rates of 95% confidence interval for the intercept. Each row represents scenarios S1, S2, S3, and S4, respectively, and each column represents a different truncation point k. Each line corresponds to each random effect distribution and the red horizontal line represents 0.95.

Figure 4.

Empirical coverage rates of 95% confidence interval for the third slope. Each row represents scenarios S1, S2, S3, and S4, respectively, and each column represents a different truncation point k. Each line corresponds to each random effect distribution and the red horizontal line represents 0.95.

On the one hand, in Figure 3, the empirical coverage rates for ${\mathrm{\text{β}}}_0$ are always less than 0.75. The coverage rates in the first two rows in Figure 3 decline with k and ${\sigma }_{\nu }^2$, while the coverage rates in the last two rows in Figure 3 are stable and low. In Figure 4, on the other hand, all the empirical coverage rates are approximately 0.95. The slope estimates keep their empirical coverages close to their nominal levels regardless of the misspecified random effect distribution.

4.3. Comparison of an existing R package

The R-package ‘gamlss’ [17,20] has been widely used in many studies because of its flexibility [10], and it can also implement the TNB model. The TNB of gamlss employs the Newton-Rhapson method which requires the first and second derivatives of the log-likelihood function, but their second derivative does not consider the adjustment term by the truncation. This leads to a lower empirical coverage of 95% confidence interval than the corresponding nominal level. To handle this issue, we make a new R function TruncCount for implementing TNB with fully addressed second derivatives, and it is freely available at https://sites.google.com/view/lwj221. All the algorithmic details of TruncCount are described in the Appendix. We perform a numerical study to compare its performance with the existing one from version 5.4-100.

Figure 5 presents biases and empirical coverage rates for ${\mathrm{\text{β}}}_3$ estimates under S3. The empirical coverage rates of gamlss are lower than 95% because its model-based standard errors are underestimated, while the empirical coverage rates of TruncCount are closer to 95%. However, when k = 2 and ${\sigma }_{\nu }^2$ is large, we observe that its performance can deteriorate because of small sample sizes.

Figure 5.

Performance comparison of the two R-packages, ‘TruncCount’ and ‘gamlss’, in terms of bias and coverage rate. The upper panel plots the biases of the ${\mathrm{\text{β}}}_3$ estimate, and each row and column represents different variances of the random effect ${\sigma }_{\nu }^2$ and truncation point k, respectively. The lower panel plots empirical coverage rates of 95% confidence intervals for ${\mathrm{\text{β}}}_3$. Each row and column represent different software packages and the truncation point k, respectively, and the red horizontal lines represent 0.95.

5. Real data analysis

In this section, we analyzed data from the food-borne outbreak that occurred in Gyeonggi province, South Korea, in 2021. The data are obtained by a field investigation to discover which food ingredient is a cause of the food borne outbreak. Field epidemiologists conducted an in-depth investigation for 339 individuals with a positive number of diarrhea incidents, which is a severity measure of symptoms due to salmonellosis. In addition, they also collect their receipt records that can identify what kind of food ingredients and how much each individual consumes. Therefore, considering the sample selection by truncating the outcome variable, the TNB model is applicable in this situation.

To provide important clues about possible causes of salmonellosis to the field epidemiologists, we investigate the dose-response relationship between the number of diarrhea incidents and the amount of each ingredient. Often, the presence of the dose-response relationship is good evidence for establishing the causality [10]. In this investigation, as likely candidates, the four ingredients (i.e. egg, crab stick, ham, and fish cake) and age are considered to be covariates for the dose-response relationship.

Table 2 presents the analysis results. The regression coefficients are similar between the TNB and TPO models, but the standard errors of the former are nearly twice as those of the latter. This difference can be explained by the uncertainty of the dispersion parameter estimate in the standard error of the TNB model. The estimated dispersion parameter of the TNB models without age adjustment is 0.323 (95% CI: 0.268-0.390), and 0.336 (95% CI: 0.278-0.406) with age adjustment. Given the necessity of a non-zero overdispersion parameter, we conclude that the TNB model is more appropriate than the TPO model. For model diagnostic purposes, we calculated the Pearson residuals and provided normal Q-Q plots for each model. As shown in the Figure in Section 2 of the Supplementary Materials, the TNB model provides a better fit to the data than the TPO model.

Table 2.

Results of analyzing the dose-response relationship between the number of diarrhea and four ingredients based on the truncated samples. Four different models (TNB : TNB with 4 covariates, TNB_age : TNB with 4 covariates and age, TPO : TPO with 4 covariates, TPO_age : TPO with 4 covariates and age) are considered. * denotes the P value is less than 0.05. SE: standard error of the estimate.

	Model
Covariate	Estimate	SE	P value	Estimate	SE	P value
	TNB			TPO
egg	0.005	0.002	0.019*	0.006	0.001	<0.001*
crab stick	0.032	0.013	0.018*	0.031	0.006	<0.001*
fish cake	−0.264	0.188	0.162	−0.254	0.088	0.004*
ham	−0.087	0.206	0.672	−0.065	0.095	0.496
	TNB_age			TPO_age
egg	0.005	0.002	0.028*	0.005	0.001	<0.001*
crab stick	0.022	0.015	0.140	0.022	0.007	0.002*
fish cake	−0.141	0.206	0.491	−0.145	0.096	0.133
ham	−0.054	0.206	0.795	−0.045	0.097	0.641
age	0.004	0.002	0.070	0.004	0.001	<0.001*

Without adjustment for age in the TNB model, egg and crab stick are statistically significant. Only egg is statistically significant when age is adjusted in the TNB model, which presents some statistical reasons to consider eggs first in subsequent examinations. In fact, as many previous research works have reported that eggs are the main cause of the salmonellosis [3], this field investigation also concludes that the cause is egg after checking through various experiments. Compared to the longer list of candidates from the TPO model, the TNB model is useful in providing a narrower list for follow-up investigations.

Considering that a lot of individuals report no diarrhea, it is safe to allow the possibility of zero inflation in our application. First, as shown in Section 3.1, our TNB model results are robust against the existence of the zero inflation. However, caution is needed when interpreting ${\mathrm{\text{β}}}_s$. Considering the possible zero inflation, we may assume that there are two underlying groups, namely, vulnerable and invulnerable, to salmonellosis. Then, ${\mathrm{\text{β}}}_s$ is limited in interpretation to the population vulnerable to salmonellosis rather than the whole population. Specifically, based on the TNB model with adjustment for age, the expected number of diarrhea incidents may increase by a factor of $e^{0.005}\approx 1.005$ for each 1-gram increase in egg intake but only among the vulnerable groups.

6. Conclusion

It is important to perform field investigation quickly and accurately so that truncated count data can be obtained from sampling specialized for field epidemiology. Our study alleviates some concerns about modeling the truncated part and possible misspecification for the random effect distribution by investigating two robust properties of the TNB model. First, the TNB model can be regarded as a partially identified model in the sense that various models lead to the same truncated likelihood function. A practical meaning is that the TNB model gives reliable inference for the regression coefficients $\mathrm{\text{β}}$ and even zero inflation is allowed. Second, the TNB model gives robust inference for the slope parameters against the misspecified random effect distribution. With some analytic justification, our numerical study shows that the empirical coverage based on the TNB model is close to its nominal level, even when the random effect distribution is misspecified. Regarding the intercept, the estimate from the TNB model exhibits bias when the random effect distribution is misspecified. In particular, as shown in the simulation study in Section 4.1, the bias can be substantial when the distribution is a mixture of two log-normal distributions.

Our study has a limitation that it does not include an analytical proof for the non-null slope parameter case. This difficulty has also been noted in previous studies [12,14,15], which provided analytical results only for the case of a null slope. Addressing this limitation is left for future research. In addition, our study focuses on misspecification of the random intercept distribution. Investigating the robustness of more complex models that incorporate random slopes would also be an important direction for future research.

The real situations in field investigation often show more complex points than the truncation. For example, in our application, a knife used for the egg can be contaminated and used for cutting other ingredients such as crap stick or fish cake. Therefore, other ingredients may appear to be a cause; hence the domain knowledge of how the food is made should be exploited in finding the causal relationship. In addition, although we have the receipt records that can identify food ingredients, they can be different from the actual intake by each individual. Therefore, some measurement error problems can be an issue in data analysis. In some other food poisoning incidents, the case under investigation is defined using a complex form of multiple health-related symptoms. For example, the case under investigation can be defined as an individual whose sum of diarrhea and vomiting is more than equal to 2. For this, we need to develop a bivariate TNB regression model. This will be an interesting research topic to develop a sophisticated bivariate regression model that can address both contaminated ingredients and measurement errors. Recently, Böhning and Van Der Heijden [2] considered zero-truncated count data under one-inflation, and showed that generally, the zero-truncated one-inflated likelihood is equivalent to the zero-one truncated likelihood, which implies that the truncation provides interesting views in data analysis. It is also interesting to establish other equivalent relationships between models for truncated count data.

Appendix.

Algorithm of TruncCount

We use Newton-Raphson algorithm to fit the TNB model. To implement this, we need the first and second derivatives of the log-likelihood function of TNB. Their analytic forms are provided below.

Using $\varphi =e^{\psi }$, the probability mass function of negative binomial is given as

$$\begin{array}{rl}f(y\mid \mathit{x},\mathrm{\text{β}},\psi )& =\frac{\mathrm{\Gamma }(e^{-\psi }+y)}{y!\mathrm{\Gamma }\left(e^{-\psi }\right)}\left(e^{\psi +{\mathit{x}}^{\mathrm{T}}\mathrm{\text{β}}}{)}^y\right(1+e^{\psi +{\mathit{x}}^{\mathrm{T}}\mathrm{\text{β}}}{)}^{-(y+e^{-\psi })}\\ & =\frac{{\mathrm{\Pi }}_{m=0}^{y-1}(1+e^{\psi }m)}{y!}\left(e^{{\mathit{x}}^{\mathrm{T}}\mathrm{\text{β}}}{)}^y\right(1+e^{\psi +{\mathit{x}}^{\mathrm{T}}\mathrm{\text{β}}}{)}^{-(y+e^{-\psi })}.\end{array}$$

The log-likelihood for the $i^{\mathrm{th}}$observation $({\mathit{x}}_i,y_i)$ is

$${\ell }_i(\mathrm{\text{β}},\psi ;y_i)=\sum _{m=0}^{y_i-1}\{\log (1+e^{\psi }m\left)\right\}+y_i{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}-(y_i+e^{-\psi })\log (1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}),$$ (A1)

and the log-likelihood function taking into account the truncation is

$$\begin{array}{rl}{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)& =\sum _{m=0}^{y_i-1}\{\log (1+e^{\psi }m\left)\right\}+y_i{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}-(y_i+e^{-\psi })\log (1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}})\\ & -\log \{1-\sum _{j=0}^k\frac{{\mathrm{\Pi }}_{m=0}^{j-1}(1+e^{\psi }m)}{j!}\frac{(e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^j}{(1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^{j+e^{-\psi }}}\}.\end{array}$$ (A2)

Then, based on (A2), the score function ${\mathrm{S}}^{\left(k\right)}(\mathrm{\text{β}},\psi )$ and Fisher information matrix ${\mathrm{I}}^{\left(k\right)}(\mathrm{\text{β}},\psi )$ are

$${\mathrm{S}}^{\left(k\right)}(\mathrm{\text{β}},\psi )={\left(\sum _{i=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{\text{β}}}{\ell }_i^{\left(k\right)}\right(\mathrm{\text{β}},\psi ;y_i{)}^{\mathrm{T}},\sum _{i=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }\psi }{\ell }^{(k{)}_i}(\mathrm{\text{β}},\psi ;y_i))}^{\mathrm{T}}$$ (A3)

and

$${\mathrm{I}}^{\left(k\right)}(\mathrm{\text{β}},\psi )=-\left(\begin{array}{cc}{\displaystyle \sum _{i=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}& {\displaystyle \sum _{i=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }\psi }{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}\\ {\displaystyle \sum _{i=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }\psi \mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}& {\displaystyle \sum _{i=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\psi }^2}{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}\end{array}\right).$$ (A4)

The following results are used to compute (A3) and (A4).

Let $p_i\left(k\right)=\sum _{j=0}^{k}f(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi )$. Then, we have

$$\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{\text{β}}}{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{\text{β}}}{\ell }_i(\mathrm{\text{β}},\psi ;y_i)+\frac{\sum _{j=0}^k\left\{f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{\text{β}}}{\ell }_i\right(\mathrm{\text{β}},\psi ;j\left)\right\}}{1-p_i\left(k\right)}$$

and

$$\frac{\mathrm{\partial }}{\mathrm{\partial }\psi }{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)=\frac{\mathrm{\partial }}{\mathrm{\partial }\psi }{\ell }_i(\mathrm{\text{β}},\psi ;y_i)+\frac{\sum _{j=0}^k\left\{f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }}{\mathrm{\partial }\psi }{\ell }_i\right(\mathrm{\text{β}},\psi ;j\left)\right\}}{1-p_i\left(k\right)},$$

where

$$\begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{\text{β}}}{\ell }_i(\mathrm{\text{β}},\psi ;y_i)& =\frac{(y_i-e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}})}{1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}{\mathit{x}}_i,\mathrm{and}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }\psi }{\ell }_i(\mathrm{\text{β}},\psi ;y_i)& =\sum _{m=0}^{y_i-1}\frac{e^{\psi }m}{1+e^{\psi }m}+e^{-\psi }\log (1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}})-(y_i+e^{-\psi })\frac{e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}{1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}.\end{array}$$

Next, we have

$$\begin{array}{rl}& \frac{{\mathrm{\partial }}^2{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}\\ & =\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}+\frac{\sum _{j=0}^{k}f(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi )\cdot \{\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}+\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}}\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}\}}{1-p_i\left(k\right)}\\ & +\left(\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}}\right)\left(\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}\right)/\\ & (1-p_i(k){)}^2,\\ \frac{{\mathrm{\partial }}^2{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }{\psi }^2}& =\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }{\psi }^2}+\frac{\sum _{j=0}^{k}f(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi )\cdot \{\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }{\psi }^2}+{\left(\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\psi }\right)}^2\}}{1-p_i\left(k\right)}\\ & +{\left(\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\psi }\right)}^2/(1-p_i(k){)}^2,\end{array}$$

and

$$\begin{array}{rl}& \frac{{\mathrm{\partial }}^2{\ell }_i^{\left(k\right)}(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }\psi }\\ & =\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }\psi }\\ & +\left\{\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)(\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }\psi }+\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}}\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\psi })\right\}/(1-p_i(k\left)\right)\\ & +\left\{\left(\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\psi }\right)\left(\sum _{j=0}^{k}f\right(j\mid {\mathit{x}}_i,\mathrm{\text{β}},\psi \left)\frac{\mathrm{\partial }{\ell }_i(\mathrm{\text{β}},\psi ;j)}{\mathrm{\partial }\mathrm{\text{β}}}\right)\right\}/(1-p_i(k){)}^2,\end{array}$$

where

$$\begin{array}{rl}\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}& =-\frac{e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}(1+y_i{e}^{\psi })}{(1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^2}{\mathit{x}}_i{\mathit{x}}_i^{\mathrm{T}}\\ \frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }{\psi }^2}& =\sum _{m=0}^{y_i-1}\frac{e^{\psi }m}{(1+e^{\psi }m{)}^2}-e^{-\psi }\log (1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}})+\frac{2e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}{1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}-\frac{(y_i+e^{-\psi })e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}{(1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^2}\\ \frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\mathrm{\text{β}}\mathrm{\partial }\psi }& ={\left(\frac{{\mathrm{\partial }}^2{\ell }_i(\mathrm{\text{β}},\psi ;y_i)}{\mathrm{\partial }\psi \mathrm{\partial }{\mathrm{\text{β}}}^{\mathrm{T}}}\right)}^{\mathrm{T}}=-\frac{(y_i-e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}})e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}}{(1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^2}{\mathit{x}}_i.\end{array}$$

The R-package ‘gamlss’ uses the same score function, but uses the Fisher information of NB instead of that of TNB. In other words, $\log [1-\sum _{j=0}^k\{\left({\mathrm{\Pi }}_{m=0}^{j-1}\right(1+e^{\psi }m)/j!)\left(\right(e^{{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^j/(1+e^{\psi +{\mathit{x}}_i^{\mathrm{T}}\mathrm{\text{β}}}{)}^{j+e^{-\psi }})\left\}\right]$ in (A2) is the omitted term in their Fisher information.

Funding Statement

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (no. 2021R1A2C1014409).

Disclosure statement

No potential conflict of interest was reported by the authors.

Supplemental Material

Supplemental data for this article can be accessed online at http://dx.doi.org/10.1080/02664763.2025.2545890.