Some new statistical methods for a class of zero-truncated discrete distributions with applications

Abstract

Counting data without zero category often occurs in various fields. A class of zero-truncated discrete distributions such as the zero-truncated Poisson, zero-truncated binomial and zero-truncated negative-binomial distributions are proposed in literature to model such count data. In this paper, three main contributions have been made for better studying the zero-truncated discrete distributions: First, a novel unified expectation–maximization (EM) algorithm is developed for calculating the maximum likelihood estimates (MLEs) of parameters in general zero-truncated discrete distributions and an important feature of the proposed EM algorithm is that the latent variables and the observed variables are independent, which is unusual in general EM-type algorithms; Second, for those who do not understand the principle of latent variables, a unified minorization–maximization algorithm, as an alternative to the EM algorithm, for obtaining the MLEs of parameters in a class of zero-truncated discrete distributions is discussed; Third, a unified method is proposed to derive the distribution of the sum of i.i.d.zero-truncated discrete random variables, which has important applications in the construction of the shortest Clopper–Pearson confidence intervals of parameters of interest and in the calculation of the exact p value of a two-sided test for small sample sizes in one sample problem.

Appendices

Appendix A: A review on other commonly used univariate zero-truncated discrete distributions

1.1 Zero-truncated negative-binomial distribution

The pmf of the zero-truncated negative-binomial distribution is

$$\begin{aligned} {p}_{_{\tiny {\mathrm{T}}}}(y|r, \pi ) = \Pr (Y=y) = \frac{1}{1-(1-\pi )^r} \cdot \frac{\Gamma (y+r)}{y!\Gamma (r)} \pi ^{y}(1-\pi )^{r}, \quad y \geqslant 1. \end{aligned}$$

(A.1)

We denote it by $Y \sim \text{ ZTNB }(r, \pi )$. The mean and variance of Y are given by

$$\begin{aligned} E(Y)= \frac{r \pi }{[1- (1-\pi )^r](1-\pi )}, \end{aligned}$$

and

$$\begin{aligned} \text{ Var }(Y) = \frac{r \pi [1 - (1-\pi )^r - r \pi (1-\pi )^r]}{(1-\pi )^2[1-(1-\pi )^r]^2}. \end{aligned}$$

(A.2)

Rider (1955) derived the MLE of $\pi$ when r is known and When $r>0$ is unknown, Sampford (1955) discussed the maximum likelihood estimates of both r and $\pi$ at the same time and pointed out the slowness of the iteration process, which requires a special aid table. He then suggested an alternative method to obtain the solutions to the MLEs of the parameters by trial and error process. This iterative process starts with an inefficient estimate of r and then the corresponding $\pi$ is obtained. By making use of the result from Bliss and Fisher (1953), Hartley (1958) proposed a method to obtain the MLEs of r and $\pi$. The idea is to start the iteration process by choosing some computationally convenient “pivotal values” of r and calculate the corresponding MLE of $\pi$. The final MLEs of r and $\pi$ are solved by interpolation.

Ahuja (1971) derived the distribution of the sum of several i.i.d. ZTNB random variables by applying one of the results obtained by Patil (1963) for the generalized power series distribution. The explicit form of the distribution of Y was given as a linear combination of the incomplete beta functions.

1.2 Zero-truncated generalized negative-binomial distribution

By introducing an additional parameter $\beta$ to the negative-binomial distribution using Lagrange’s expansion, Jain and Consul (1971) is the first paper to introduce the generalized negative-binomial distribution. Its pmf is given by

$$\begin{aligned} p(x|r, \pi , \beta ) = \frac{r\Gamma {(r+\beta x)}}{x!\Gamma {(r+\beta x-x+1)}}\pi ^x(1-\pi )^{r+\beta x-x}, \end{aligned}$$

(A.3)

for $x=0,1, \ldots , \infty$, where $r>0$, $\pi \in (0, 1)$ and $\beta =0$ or $1\leqslant \beta \leqslant \pi ^{-1}$. We denote it by $X \sim \text{ GNB }(r, \pi , \beta )$. It was shown that the mean and variance of X are given by (Jain and Consul 1971)

$$\begin{aligned} E(X) = \frac{r \pi }{1-\beta \pi } \quad \text{ and } \quad \text{ Var }(X) = \frac{r \pi (1-\pi )}{(1-\beta \pi )^3}, \end{aligned}$$

(A.4)

which exist for $\beta =0$ or $\beta \in [1, \pi ^{-1})$, see Consul and Famoye (2006, p.191). In particular, the $\text{ GNB }(r, \pi , \beta )$ distribution reduces to the $\text{ Binomial }(r, \pi )$ when $\beta =0$ and reduces to the $\text{ NBinomial }(r, \pi )$ when $\beta =1$.

Thus, the pmf of a zero-truncated generalized negative-binomial distribution is

$$\begin{aligned} {p}_{_{\tiny {\mathrm{T}}}}(y|r, \pi , \beta ) = \frac{r\Gamma {(r+\beta y)}}{y!\Gamma {(r+\beta y-y+1)}} \cdot \frac{\pi ^y(1-\pi )^{r+\beta y-y}}{1 - (1-\pi )^r}, \qquad y \geqslant 1. \end{aligned}$$

(A.5)

However, to our best knowledge, we did not find any papers on estimating the parameters in a zero-truncated generalized negative-binomial distribution. Medhi (1975) derived the distribution of the convolution of left-truncated generalized negative-binomial variables by making using of the same property of the pmf.

1.3 Zero-truncated generalized Poisson distribution

By considering a limiting form of the generalized negative binomial distribution, Consul and Jain (1973) proposed the generalized Poisson distribution. A non-negative integer valued random variable X is said to have a generalized Poisson (GP) distribution with parameters $\lambda >0$ and $\theta$, if its pmf is given by (Consul and Jain 1973; Consul 1989, p.4)

$$\begin{aligned} p(x|\lambda , \theta ) = \left\{ \begin{array}{ll} \displaystyle \frac{\lambda (\lambda +\theta x)^{x-1}\text{ e }^{-\lambda -\theta x}}{x!}, &{}\quad x=0, 1, \ldots , \infty , \\ 0, &{}\quad \text{ for } x > q, \text{ when } \theta < 0, \end{array}\right. \end{aligned}$$

(A.6)

where $\max (-1, -\lambda /q) < \theta \leqslant 1$ and $q \,(\geqslant 4)$ is the largest positive integer for which $\lambda + \theta q > 0$ when $\theta < 0$. We denote it by $X \sim \text{ GP }(\lambda , \theta )$. It was shown that the mean and variance of X are given by (Consul 1989, p.14)

$$\begin{aligned} E(X) = \frac{\lambda }{1-\theta } \quad \text{ and } \quad \text{ Var }(X) = \frac{\lambda }{(1-\theta )^3}, \quad \text{ where } \;\, \theta < 1. \end{aligned}$$

(A.7)

The $\text{ GP }(\lambda , \theta )$ distribution reduces to the $\text{ Poisson }(\lambda )$ when $\theta =0$, and it has the twin properties of over-dispersion when $\theta >0$ and under-dispersion when $\theta <0$. The most frequently used version of the GP distribution assumes that $\lambda >0$ and $\theta \in [0, 1)$.

Thus, the pmf of a zero-truncated generalized Poisson distribution is

$$\begin{aligned} {p}_{_{\tiny {\mathrm{T}}}}(y|\lambda , \theta ) = \frac{\lambda (\lambda +\theta y)^{y-1}\text{ e }^{-\lambda -\theta y}}{(1 - \text{ e }^{-\lambda })y!}, \qquad y = 1, 2, \ldots , \infty . \end{aligned}$$

(A.8)

We denote it by $Y \sim \text{ ZTGP }(\lambda , \theta )$. The distribution of the convolution of zero-truncated generalized Poisson variables was firstly derived by Medhi (1975) by considering a general problem and later, Consul (1989, p. 70) provided a simple and independent proof for it.

Appendix B: A general strategy for establishing an EM algorithm for truncated-data problems (Meng 1997)

Let $f(x|\varvec{\theta })$ denote the untruncated pdf, where $x \in {\mathbb {A}}\cup {\bar{{\mathbb {A}}}}$ and ${\mathbb {A}}$ is a region such that value outside ${\mathbb {A}}$ will be truncated (e.g., ${\mathbb {A}}= \{1,2,\ldots \}$ and ${\bar{{\mathbb {A}}}}=\{0\}$ for the ZTP distribution). Therefore, the truncated pdf is $f(x|\varvec{\theta })/\Pr ({\mathbb {A}}|\varvec{\theta })$ with

$$\begin{aligned} \Pr ({\mathbb {A}}|\varvec{\theta }) = \int _{{\mathbb {A}}} f(x|\varvec{\theta }) \; \text{ d }x. \end{aligned}$$

Let $X_1,\ldots ,X_n {\mathop {\sim }\limits ^{\mathrm{iid}}}f(x|\varvec{\theta })/\Pr ({\mathbb {A}}|\varvec{\theta })$, where $x \in {\mathbb {A}}$. The likelihood function of $\varvec{\theta }$ based on the observed data $Y_{\mathrm{obs}} = \{x_1,\ldots ,x_n\}$ is

$$\begin{aligned} L(\varvec{\theta }|Y_{\mathrm{obs}}) = \frac{\prod _{i=1}^n f(x_i|\varvec{\theta })}{[\Pr ({\mathbb {A}}|\varvec{\theta })]^n}. \end{aligned}$$

Augmenting the $Y_{\mathrm{obs}}$ to include the sample values that were truncated out and the number of such values: $Y_{\mathrm{mis}} = \{x_{n+1}, \ldots , x_{n+m}; m\}$ with m being a random variable, we obtain the complete-data likelihood function of $\varvec{\theta }$, which is given by

$$\begin{aligned} L(\varvec{\theta }|Y_{\mathrm{obs}}, Y_{\mathrm{mis}}) \propto f(x_1,\ldots ,x_{n+m}, m|\varvec{\theta }) = \prod _{i=1}^{n+m} f(x_i|\varvec{\theta }). \end{aligned}$$

Note that the joint distribution $(Y_{\mathrm{obs}}, Y_{\mathrm{mis}})$ can be rewritten as

$$\begin{aligned}&f(x_1,\ldots ,x_{n+m}, m|\varvec{\theta }) \\&\quad = f(x_1,\ldots ,x_{n+m}|\varvec{\theta }) \times \Pr (m|x_1,\ldots ,x_{n+m},\varvec{\theta }) \\&\quad = \frac{\prod _{i=1}^{n+m} f(x_i|\varvec{\theta })}{[\Pr ({\mathbb {A}}|\varvec{\theta })]^n[1-\Pr ({\mathbb {A}}|\varvec{\theta })]^m} \times \Pr (m|x_1,\ldots ,x_{n+m},\varvec{\theta }). \end{aligned}$$

Therefore,

$$\begin{aligned} \Pr (m|x_1,\ldots ,x_{n+m},\varvec{\theta }) = [\Pr ({\mathbb {A}}|\varvec{\theta })]^n[1-\Pr ({\mathbb {A}}|\varvec{\theta })]^m \end{aligned}$$

and

$$\begin{aligned} \Pr (m|Y_{\mathrm{obs}},\varvec{\theta }) = \bigg (\begin{array}{c} m+n-1 \\ m \end{array} \bigg ) [\Pr ({\mathbb {A}}|\varvec{\theta })]^n[1-\Pr ({\mathbb {A}}|\varvec{\theta })]^m, \quad m \geqslant 0, \end{aligned}$$

which is a negative binomial distribution. The E-step is to replace m in the expression of the MLE of $\varvec{\theta }$ via the M-step by its conditional expectation

$$\begin{aligned} E(m|Y_{\mathrm{obs}},\varvec{\theta }) = n \frac{1-\Pr ({\mathbb {A}}|\varvec{\theta })}{\Pr ({\mathbb {A}}|\varvec{\theta })}. \end{aligned}$$

(B.1)

Appendix C: Derivation of EM-type iterations for a class of zero-truncated discrete distributions

Example C.1

(Zero-truncated Poisson). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTP }(\lambda )$ and the observed data be $Y_{\mathrm{obs}}=\{y_1, \ldots , y_n\}$. Then, the observed-data likelihood function for $\lambda$ is given by

$$\begin{aligned} L(\lambda |Y_{\mathrm{obs}}) = \prod _{i=1}^n \frac{\lambda ^{y_i}\text{ e }^{-\lambda }}{(1-\text{ e }^{-\lambda })y_i!} \propto \lambda ^{n{\bar{y}}} \text{ e }^{-n\lambda }(1-\text{ e }^{-\lambda })^{-n}, \end{aligned}$$

where ${\bar{y}}= (1/n)\sum _{i=1}^n y_i$ is a sufficient statistic of $\lambda$. In (3.4), let $\varvec{\theta }=\lambda$ and $p(x|\lambda ) = \lambda ^x \text{ e }^{-\lambda }/x!$. Then, the complete-data MLE of $\lambda$ given by

$$\begin{aligned} {\hat{\lambda }}= \frac{\sum _{i=1}^n z_iy_i}{n}. \end{aligned}$$

(C.1)

The E-step is to replace $\{z_i\}_{i=1}^n$ in (C.1) by their conditional expectations:

$$\begin{aligned} E(Z_i|Y_{\mathrm{obs}}, \lambda ) = 1-\text{ e }^{-\lambda }, \quad i=1,\ldots , n. \end{aligned}$$

(C.2)

By combining (C.1) with (C.2), we have the following EM iterations:

$$\begin{aligned} \lambda ^{(t+1)} = {\bar{y}}\Big (1-\text{ e }^{-\lambda ^{(t)}} \Big ), \quad t=0, 1, \ldots , \end{aligned}$$

(C.3)

where $\lambda ^{(0)}$ is the initial value of the MLE ${\hat{\lambda }}$. $\square$

Example C.2

(Zero-truncated binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTB }(m,\pi )$ and the observed data be $Y_{\mathrm{obs}}=\{y_1, \ldots , y_n\}$. Then, the observed-data likelihood function is given by

$$\begin{aligned} L(\pi |Y_{\mathrm{obs}}) = \prod _{i=1}^n \frac{ {m \atopwithdelims ()y_i} \pi ^{y_i}(1-\pi )^{m-y_i}}{1-(1-\pi )^m} \propto [1-(1-\pi )^m]^{-n} \pi ^{n{\bar{y}}}(1-\pi )^{n(m-{\bar{y}})}, \end{aligned}$$

where ${\bar{y}}= (1/n)\sum _{i=1}^n y_i$ is a sufficient statistic of $\pi$. In (3.4), let $\varvec{\theta }=\pi$ and $p(x|\pi ) = {m \atopwithdelims ()x} \pi ^x(1-\pi )^{m-x}$. Then, the complete-data MLE of $\pi$ given by

$$\begin{aligned} {\hat{\pi }} = \frac{\sum _{i=1}^n z_iy_i}{nm}. \end{aligned}$$

(C.4)

The E-step is to replace $\{z_i\}_{i=1}^n$ in (C.4) by their conditional expectations:

$$\begin{aligned} E(Z_i|Y_{\mathrm{obs}}, \pi ) = 1-(1-\pi )^m, \quad i=1,\ldots , n. \end{aligned}$$

(C.5)

By combining (C.4) with (C.5), we have the following EM iterations:

$$\begin{aligned} \pi ^{(t+1)} = \frac{[1-(1-\pi ^{(t)})^m]{\bar{y}}}{m}, \quad t=0, 1, \ldots , \end{aligned}$$

(C.6)

where $\pi ^{(0)}$ is the initial value of the MLE ${\hat{\pi }}$. $\square$

Example C.3

(Zero-truncated negative-binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTNB }(r,\pi )$ and the observed data be $Y_{\mathrm{obs}}=\{y_1, \ldots , y_n\}$, where $r>0$ is assumed to be known. Then, the observed-data likelihood function is

$$\begin{aligned} L(\pi |Y_{\mathrm{obs}}) = \prod _{i=1}^n \left[ \frac{\Gamma (y_i+r)}{y_i!\Gamma (r)} \cdot \frac{\pi ^{y_i}(1-\pi )^r }{1-(1-\pi )^r} \right] \propto [1-(1-\pi )^r]^{-n}\pi ^{n{\bar{y}}}(1-\pi )^{nr}, \end{aligned}$$

where ${\bar{y}}= (1/n)\sum _{i=1}^n y_i$ is a sufficient statistic of $\pi$. In (3.4), let $\varvec{\theta }=\pi$ and $p(x|\pi ) = \frac{\Gamma (x+r)}{x!\Gamma (r)} \pi ^x(1-\pi )^{r}$. Then, the complete-data MLE of $\pi$ given by

$$\begin{aligned} {\hat{\pi }} = \frac{\sum _{i=1}^n z_iy_i}{\sum _{i=1}^n z_iy_i+nr}. \end{aligned}$$

(C.7)

The E-step is to replace $\{z_i\}_{i=1}^n$ in (C.7) by their conditional expectations:

$$\begin{aligned} E(Z_i|Y_{\mathrm{obs}}, \pi ) = 1-(1-\pi )^r, \quad i=1,\ldots , n. \end{aligned}$$

(C.8)

By combining (C.7) with (C.8), we have the following EM iterations:

$$\begin{aligned} \pi ^{(t+1)} = \frac{[1-(1-\pi ^{(t)})^r]{\bar{y}}}{[1-(1-\pi ^{(t)})^r]{\bar{y}}+r}, \quad t=0, 1, \ldots , \end{aligned}$$

(C.9)

where $\pi ^{(0)}$ is the initial value of the MLE ${\hat{\pi }}$. $\square$

Example C.4

(Zero-truncated generalized negative-binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTGNB }(r,\pi , \beta )$ and the observed data be $Y_{\mathrm{obs}}=\{y_1, \ldots , y_n\}$, where $r>0$ is assumed to be known. Then, the observed-data likelihood function of $(\pi , \beta )$ is

$$\begin{aligned} L(\pi , \beta |Y_{\mathrm{obs}})= & {} \prod _{i=1}^n \left[ \frac{r\Gamma {(r+\beta y_i)}}{y_i!\Gamma {(r+\beta y_i-y_i+1)}} \cdot \frac{\pi ^{y_i}(1-\pi )^{r+\beta y_i-y_i}}{1 - (1-\pi )^r} \right] \\\propto & {} \left[ \prod _{i=1}^n\frac{\Gamma {(r+\beta y_i)}}{\Gamma {(r+\beta y_i-y_i+1)}}\right] \cdot [1-(1-\pi )^r]^{-n}\pi ^{n{\bar{y}}}(1-\pi )^{nr+(\beta -1)n{\bar{y}}}, \end{aligned}$$

where ${\bar{y}}= (1/n)\sum _{i=1}^n y_i$. From (3.4), the complete-data log-likelihood is proportional to

$$\begin{aligned}&\sum _{i=1}^n \Big [\log \Gamma {(r+\beta x_i)} - \log \Gamma {(r+\beta x_i-x_i+1)} \Big ] \\&\quad +\,n{\bar{x}}\log (\pi ) + [nr+(\beta -1)n{\bar{x}}]\log (1-\pi ). \end{aligned}$$

Therefore, given $\pi =\pi ^{(t)}$, the complete-data MLE $\beta ^{(t+1)}$ of $\beta$ is the root of the equation

$$\begin{aligned}&h(\beta | \pi ^{(t)}) \nonumber \\&\quad = \sum _{i=1}^n \left[ \frac{\Gamma '{(r+\beta z_iy_i)z_iy_i}}{\Gamma {(r+\beta z_iy_i)}} - \frac{\Gamma '{(r+\beta z_iy_i-z_iy_i+1)z_iy_i}}{\Gamma {(r+\beta z_iy_i-z_iy_i+1)}}\right] \nonumber \\&\qquad +\, \left( \sum _{i=1}^n z_iy_i \right) \log [1- \pi ^{(t)}] = 0, \end{aligned}$$

(C.10)

and given $\beta = \beta ^{(t+1)}$, the complete-data MLE $\pi ^{(t+1)}$ of $\pi$ is

$$\begin{aligned} \pi = \frac{\sum _{i=1}^n z_iy_i}{\beta ^{(t+1)} \sum _{i=1}^n z_iy_i + nr}. \end{aligned}$$

(C.11)

From (3.6) and (3.5), the E-step is to replace $\{z_i\}_{i=1}^n$ and $\{g_i(z_i)\}_{i=1}^n$ in (C.10) and (C.11) by their conditional expectations

$$\begin{aligned} E[Z_i|Y_{\mathrm{obs}}, \pi ^{(t)}, \beta ^{(t)}]= & {} 1-(1 - \pi ^{(t)})^r, \quad i=1,\ldots , n, \quad \text{ and } \quad \nonumber \\ E[g_i(Z_i)|Y_{\mathrm{obs}}, \pi ^{(t)}, \beta ^{(t)}]= & {} g_i(0)(1-\pi ^{(t)})^r + g_i(1)[1-(1 - \pi ^{(t)})^r],\nonumber \\ \end{aligned}$$

(C.12)

respectively, resulting in the following ECM iterations:

• CM-Step 1: Given $\pi =\pi ^{(t)}$, calculate $\beta ^{(t+1)}$, which is the solution to the following equation

  $$\begin{aligned} \sum _{i=1}^n y_i\left[ \frac{\Gamma '{(r+\beta y_i)}}{\Gamma {(r+\beta y_i)}} - \frac{\Gamma '{(r+\beta y_i-y_i+1)}}{\Gamma {(r+\beta y_i- y_i+1)}} \right] \, +\, n{\bar{y}}\log [1-\pi ^{(t)}] = 0. \quad \end{aligned}$$

  (C.13)

• CM-Step 2: Given $\beta = \beta ^{(t+1)}$, calculate

  $$\begin{aligned} \pi ^{(t+1)} = \frac{{\bar{y}}[1-(1-\pi ^{(t)})^r ]}{\beta ^{(t+1)}{\bar{y}}[1-(1-\pi ^{(t)})^r] + r}, \quad t=0, 1, \ldots , \end{aligned}$$

  (C.14)

  where $\pi ^{(0)}$ is the initial value of the MLE of ${\hat{\pi }}$.

$\square$

Example C.5

(Zero-truncated generalized Poisson). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTGP }(\lambda ,\theta )$ and the observed data be $Y_{\mathrm{obs}}=\{y_1, \ldots , y_n\}$. Then, the observed-data likelihood function of $(\lambda , \theta )$ is

$$\begin{aligned} L(\lambda , \theta |Y_{\mathrm{obs}})= & {} \prod _{i=1}^n \frac{\lambda (\lambda +\theta y_i)^{y_i-1}\text{ e }^{-\lambda -\theta y_i}}{(1 - \text{ e }^{-\lambda })y_i!} \propto \lambda ^n\\&\times \left[ \prod _{i=1}^n(\lambda + \theta y_i)^{y_i - 1} \right] \text{ e }^{-n\lambda - \theta n{\bar{y}}} (1 - \text{ e }^{-\lambda } )^{-n}, \end{aligned}$$

where ${\bar{y}}= (1/n)\sum _{i=1}^n y_i$. From (3.4), the complete-data log-likelihood is proportional to

$$\begin{aligned} n \log (\lambda ) - n\lambda - \theta n {\bar{x}}+ \sum _{i=1}^n (x_i-1) \log (\lambda + \theta x_i), \end{aligned}$$

Therefore, given $\lambda =\lambda ^{(t)}$, the complete-data MLE $\theta ^{(t+1)}$ of $\theta$ is the root of the equation

$$\begin{aligned} H(\theta | \lambda ^{(t)}) = \sum _{i=1}^n \frac{(x_i - 1)x_i}{\lambda + \theta x_i} - n {\bar{x}}= \sum _{i=1}^n \frac{z_i^2y_i^2 - z_iy_i}{\lambda ^{(t)} + \theta z_iy_i} - \sum _{i=1}^n z_iy_i = 0, \end{aligned}$$

(C.15)

and given $\theta = \theta ^{(t+1)}$, the complete-data MLE $\lambda ^{(t+1)}$ of $\lambda$ is given by (Consul 1989, p.102)

$$\begin{aligned} \lambda = {\bar{x}}(1-\theta ) = \frac{\sum _{i=1}^n z_iy_i}{n} (1 - \theta ^{(t+1)}), \end{aligned}$$

(C.16)

where ${\bar{x}}= (1/n)\sum _{i=1}^n z_iy_i$. From (3.6) and (3.5), the E-step is to replace $\{z_i\}_{i=1}^n$ and $\{g_i(z_i)^2\}_{i=1}^n$ in (C.15) and (C.16) by their conditional expectations

$$\begin{aligned} E[Z_i|Y_{\mathrm{obs}}, \lambda ^{(t)}, \theta ^{(t)}]= & {} 1-\text{ e }^{-\lambda ^{(t)}}, \quad i=1,\ldots , n, \quad \text{ and } \quad \nonumber \\ E[g_i(Z_i)|Y_{\mathrm{obs}}, \lambda ^{(t)}, \theta ^{(t)}]= & {} g_i(0)\text{ e }^{-\lambda ^{(t)}} + g_i(1)[1-\text{ e }^{-\lambda ^{(t)}}], \end{aligned}$$

(C.17)

respectively, resulting in the following ECM iterations:

• CM-Step 1: Given $\lambda =\lambda ^{(t)}$, calculate $\theta ^{(t+1)}$, which is the solution to the following equation

  $$\begin{aligned} \sum _{i=1}^n \frac{y_i^2 - y_i }{\lambda ^{(t)} + \theta y_i} - n {\bar{y}}= 0. \end{aligned}$$

  (C.18)

• CM-Step 2: Given $\theta = \theta ^{(t+1)}$, calculate

  $$\begin{aligned} \lambda ^{(t+1)} = {\bar{y}}\Big (1-\text{ e }^{-\lambda ^{(t)}} \Big )(1-\theta ^{(t+1)}), \quad t=0, 1, \ldots , \end{aligned}$$

  (C.19)

  where $\lambda ^{(0)}$ is the initial value of the MLE of ${\hat{\lambda }}$.

$\square$

Appendix D: Derivation of MM iterations for a class of zero-truncated discrete distributions

Example D.1

(Zero-truncated Poisson). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTP }(\lambda )$ and the observed data be $Y_{\mathrm{obs}}=\{y_i\}_{i=1}^n$. The corresponding parent random variable $X \sim \text{ Poisson }(\lambda )$ with the pmf $p(x|\lambda ) = \lambda ^x \text{ e }^{-\lambda }/x!$. In (4.4), let $\varvec{\theta }=\lambda$, we have

$$\begin{aligned} Q(\lambda |\lambda ^{(t)})= & {} \sum _{i=1}^n \left[ \log \left( \frac{\lambda ^{y_i} \text{ e }^{-\lambda }}{y_i!} \right) + \frac{\text{ e }^{-\lambda ^{(t)}} }{1-\text{ e }^{-\lambda ^{(t)}}} \log (\text{ e }^{-\lambda }) \right] \\= & {} n{\bar{y}}\log (\lambda ) - \frac{n\lambda }{1-\text{ e }^{-\lambda ^{(t)}}} + c. \end{aligned}$$

Setting $\text{ d }Q(\lambda |\lambda ^{(t)})/\text{ d }\lambda =0$, we obtain an MM iteration, which is identical to (C.3).

$\square$

Example D.2

(Zero-truncated binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTB }(m,\pi )$ and the observed data be $Y_{\mathrm{obs}}=\{y_i\}_{i=1}^n$. The corresponding parent random variable $X \sim \text{ Binomial }(m, \pi )$ with the pmf $p(x|\pi ) = {m \atopwithdelims ()x} \pi ^x (1-\pi )^{m-x}$. In (4.4), let $\varvec{\theta }=\pi$, we have

$$\begin{aligned} Q(\pi |\pi ^{(t)})= & {} \sum _{i=1}^n \left\{ \log \left[ {m \atopwithdelims ()y_i} \pi ^{y_i} (1-\pi )^{m-y_i} \right] + \frac{(1-\pi ^{(t)})^m }{1-(1-\pi ^{(t)})^m } \log [(1-\pi )^m] \right\} \\= & {} n{\bar{y}}\log (\pi ) + n\left[ m-{\bar{y}}+ \frac{m(1-\pi ^{(t)})^m}{1-(1-\pi ^{(t)})^m} \right] \log (1-\pi ) + c. \end{aligned}$$

Setting $\text{ d }Q(\pi |\pi ^{(t)})/\text{ d }\pi =0$, we obtain an MM iteration, which is identical to (C.6). $\square$

Example D.3

(Zero-truncated negative-binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTNB }(r,\pi )$ and the observed data be $Y_{\mathrm{obs}}=\{y_i\}_{i=1}^n$, where $r>0$ is assumed to be known. The corresponding parent random variable $X \sim \text{ NBinomial }(r, \pi )$ with the pmf $p(x|\pi ) = \frac{\Gamma (x+r)}{x!\Gamma (r)} \pi ^x (1-\pi )^r$. In (4.4), let $\varvec{\theta }=\pi$, we have

$$\begin{aligned} Q(\pi |\pi ^{(t)})= & {} \sum _{i=1}^n \left\{ \log \left[ \frac{\Gamma (y_i+r)}{y_i!\Gamma (r)} \pi ^{y_i} (1-\pi )^r \right] + \frac{(1-\pi ^{(t)})^r }{1-(1-\pi ^{(t)})^r } \log [(1-\pi )^r] \right\} \\= & {} n{\bar{y}}\log (\pi ) + \frac{nr}{1-(1-\pi ^{(t)})^r} \log (1-\pi ) + c. \end{aligned}$$

Setting $\text{ d }Q(\pi |\pi ^{(t)})/\text{ d }\pi =0$, we obtain an MM iteration, which is identical to (C.9). $\square$

Example D.4

(Zero-truncated generalized negative-binomial). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTGNB }(r,\pi , \beta )$ and the observed data be $Y_{\mathrm{obs}}=\{y_i\}_{i=1}^n$, where $r>0$ is assumed to be known. The corresponding parent random variable $X \sim \text{ GNB }(r, \pi , \beta )$ with pmf given by (A.3). In (4.4), let $\varvec{\theta }=(\pi , \beta )$, we have

$$\begin{aligned}&Q(\pi , \beta | \pi ^{(t)}, \beta ^{(t)}) \\&\quad = \sum _{i=1}^n \left\{ \log \left[ \frac{r\Gamma (r+ \beta y_i) \cdot \pi ^{y_i} (1-\pi )^{r+\beta y_i-y_i} }{y_i!\Gamma (r + \beta y_i - y_i +1)} \right] + \frac{(1-\pi ^{(t)})^r }{1-(1-\pi ^{(t)})^r } \log [(1-\pi )^r] \right\} \\&\quad = \sum _{i=1}^n \Big \{\log [\Gamma {(r+\beta y_i)}]-\log [\Gamma {(r+\beta y_i-y_i+1)}] \Big \} \nonumber \\&\qquad +\, n \left\{ {\bar{y}}\log (\pi )+ \left[ r+(\beta -1){\bar{y}}+ \frac{r(1-\pi ^{(t)})^r }{1-(1-\pi ^{(t)})^r } \right] \log (1-\pi ) \right\} + c. \end{aligned}$$

By setting

$$\begin{aligned} \left\{ \begin{array}{l} 0 = \displaystyle \frac{\partial Q(\pi , \beta | \pi ^{(t)}, \beta ^{(t)})}{\partial \pi } = \frac{n {\bar{y}}}{\pi } - \frac{n \left[ r+(\beta -1){\bar{y}}+ \displaystyle \frac{r(1-\pi ^{(t)})^r }{1-(1-\pi ^{(t)})^r } \right] }{1-\pi }, \\ 0 = \displaystyle \frac{\partial Q(\pi , \beta | \pi ^{(t)}, \beta ^{(t)})}{\partial \beta } = \sum _{i=1}^n y_i \left[ \frac{\Gamma '{(r+\beta y_i) }}{\Gamma {(r+\beta y_i)}} - \frac{\Gamma '{(r+\beta y_i-y_i+1)}}{\Gamma {(r+\beta y_i- y_i+1)}} \right] + n{\bar{y}}\log (1-\pi ), \end{array} \right. \end{aligned}$$

we have the following MM iterations:

• Step 0: Let $\pi ^{(0)}$ be the initial value of the MLE of ${\hat{\pi }}$ and set $t=0$.

• Step 1: Given $\pi =\pi ^{(t)}$, calculate $\beta ^{(t+1)}$, which is the solution to the following equation

  $$\begin{aligned} \sum _{i=1}^n y_i\left[ \frac{\Gamma '{(r+\beta y_i)}}{\Gamma {(r+\beta y_i)}} - \frac{\Gamma '{(r+\beta y_i-y_i+1)}}{\Gamma {(r+\beta y_i- y_i+1)}} \right] + n{\bar{y}}\log [1-\pi ^{(t)}] = 0. \quad \end{aligned}$$

  (D.1)

• Step 2: Given $\beta = \beta ^{(t+1)}$, calculate

  $$\begin{aligned} \pi ^{(t+1)} = \frac{{\bar{y}}[1-(1-\pi ^{(t)})^r ]}{\beta ^{(t+1)}{\bar{y}}[1-(1-\pi ^{(t)})^r] + r}. \end{aligned}$$

  (D.2)

We can see that (D.1) is identical to (C.13) and (D.2) is identical to (C.14). $\square$

Example D.5

(Zero-truncated generalized Poisson). Let $Y_1, \ldots , Y_n {\mathop {\sim }\limits ^{\mathrm{iid}}}\text{ ZTGP }(\lambda ,\theta )$ and the observed data be $Y_{\mathrm{obs}}=\{y_i\}_{i=1}^n$. The corresponding parent random variable $X \sim \text{ GP }(\lambda , \theta )$ with pmf given by (A.6). In (4.4), let $\varvec{\theta }=(\lambda , \theta )$, we have

$$\begin{aligned}&Q(\lambda , \theta | \lambda ^{(t)}, \theta ^{(t)}) \\&\quad = \sum _{i=1}^n \left\{ \log \left[ \frac{\lambda (\lambda +\theta y_i)^{y_i-1} \text{ e }^{-\lambda -\theta y_i}}{y_i!} \right] + \frac{\text{ e }^{-\lambda ^{(t)}} }{1- \text{ e }^{-\lambda ^{(t)}} } \log (\text{ e }^{-\lambda }) \right\} \\&\quad = \sum _{i=1}^n (y_i - 1)\log (\lambda + \theta y_i) + n\left[ \log (\lambda )- \lambda -\theta {\bar{y}}- \frac{\lambda \text{ e }^{-\lambda ^{(t)}} }{1- \text{ e }^{-\lambda ^{(t)}} } \right] + c. \end{aligned}$$

By setting

$$\begin{aligned} \left\{ \begin{array}{l} 0 = \displaystyle \frac{\partial Q(\lambda , \theta | \lambda ^{(t)}, \theta ^{(t)})}{\partial \lambda } = \sum _{i=1}^n \frac{y_i - 1}{\lambda + \theta y_i} + n \left( \frac{1}{\lambda } - \frac{1}{1- \text{ e }^{-\lambda ^{(t)}}} \right) , \\ 0 = \displaystyle \frac{\partial Q(\lambda , \theta | \lambda ^{(t)}, \theta ^{(t)})}{\partial \theta } = \sum _{i=1}^n \frac{y_i^2 - y_i}{\lambda + \theta y_i} - n {\bar{y}}, \end{array} \right. \end{aligned}$$

we have the following MM iterations:

• Step 0: Let $\lambda ^{(0)}$ be the initial value of the MLE of ${\hat{\lambda }}$ and set $t=0$.

• Step 1: Given $\lambda =\lambda ^{(t)}$, calculate $\theta ^{(t+1)}$, which is the solution to the following equation

  $$\begin{aligned} \sum _{i=1}^n \frac{y_i^2 - y_i}{\lambda ^{(t)} + \theta y_i} - n {\bar{y}}= 0. \quad \end{aligned}$$

  (D.3)

• Step 2: Given $\theta = \theta ^{(t+1)}$, calculate

  $$\begin{aligned} \lambda ^{(t+1)} = {\bar{y}}(1 - \text{ e }^{-\lambda ^{(t)}})(1 - \theta ^{(t+1)}), \end{aligned}$$

  (D.4)

  which can be obtained by first multiplying $\lambda$ and $\theta$ to the above two partial differential equations, respectively, and then adding them together.

We can see that (D.3) is identical to (C.18) and (D.4) is identical to (C.19). $\square$