Zero-inflated data indicates that the data set contains an excessive number of zeros. The word zero-inflation is used to emphasize that the probability mass at the point zero exceeds than the one allowed under a standard parametric family of discrete distributions. Gupta et al.,¹ Murat & Szynal,² Patil & Shirke³ have contributed to estimation and testing of the parameters involved in Zero Inflated Power Series Distributions. If the data set under study does not contain observations after some known point in the support, we have to modify Zero Inflated Power Series Distribution (ZIPSD) accordingly in order to get better inferential properties. Zero Inflated Truncated Power Series Distribution (ZITPSD) is one of the better options. In the present work we address problem of estimation for ZITPSD with more emphasis on statistical tests. We provide three asymptotic tests for testing the parameter of ZITPSD, using an unconditional (standard) likelihood approach, a conditional likelihood approach and the sample mean, respectively. The performance of first two tests has been studied for Zero Inflated Truncated Poisson Distribution (ZITPD). Asymptotic Confidence Intervals for the parameter are also provided. The model has been applied to a real life data.

Keywords: zero inflation, zero inflated power series distribution, zero inflated truncated power series distribution, zero inflated truncated poisson distribution

In certain applications involving discrete data, we come across data having frequency of an observation ‘zero’ significantly higher than the one predicted by the assumed model. The problem of high proportion of zeros has been an interest in data analysis and modeling. There are many situations in the medical field, engineering applications, manufacturing, economics, public health, road safety epidemiology and in other areas leading to similar situations. In highly automated production process, occurrence of defects is assumed to be Poisson. However, we get no defectives in many samples. This leads to excess number of zeros. Models having more number of zeros significantly are known as zero-inflated models.

In the literature, numbers of researchers have worked on family of zero-inflated power series distributions. Gupta et al.¹ have studied the structural properties and point estimation of parameters of Zero-Inflated Modified Power Series distributions and in particular for zero-inflated Poisson distribution. Murat & Szynal² have studied the class of inflated modified power series distributions where inflation occurs at any of the support points. Moments, factorial moments, central moments, the maximum likelihood estimators and variance-covariance matrix of the estimators are obtained. Murat & Szynal² extended the results of Gupta et al.¹ to the discrete distributions inflated at any point $s$ .

Zero Inflated Truncated Power Series Distribution contains two parameters. The first parameter indicates inflation ( $\pi$ ) of zero and the other parameter ( $\theta$ ) is that of power series distribution. Literature survey reveals that many researchers devoted to the inflation parameter of the model. In the present study, we focus on the referential aspect of the basic parameter of the model. In this article, we provide maximum likelihood parameters, Fisher information and asymptotic tests for testing the parameter of the Zero Inflated Truncated Power Series Distribution. Additionally, asymptotic confidence interval for the parameter is provided.

In section 2.1 we report estimation of both the parameters of ZITPSD and corresponding asymptotic variances using full likelihood approach, conditional likelihood approach and method of moments. In section 2.2, we provide three asymptotic tests for testing the parameter of ZITPSD. Section 2.3 is devoted to asymptotic confidence intervals for the parameters of ZITPSD. In section 3.1 we report estimation of parameters involved in Zero Inflated Truncated Poisson Distribution (ZITPD) and inference related to the model. Section 3.2 is devoted to three asymptotic tests for testing the parameter of ZITPD and in section 3.3 we provide asymptotic confidence intervals for the parameters of ZITPD. Simulation study is carried out in section 4, to study the performance of the tests. Illustrative example is provided in section 5.

Before we define truncated ZIPSD, we first consider the Truncated Power Series Distribution (TPSD) truncated at the support point $\text{'}t\text{'}$  onwards, where $\text{'}t\text{'}$  is known. Then the probability mass function of TPSD is given by

$P(X=x)=\frac{b_x{\theta }^x}{f(\theta )(1-P(X>t)},forx=0,1,2,\mathrm{...},t$

  $=\frac{b_x{\theta }^x}{\left({\displaystyle \sum _{y=0}^t{b}_y{\theta }^y}\right)},$ $=\frac{b_x{\theta }^x}{G(\theta )},$   where $G(\theta )={\displaystyle \sum _{y=0}^t{b}_y{\theta }^y}$          

It is clear that the truncated distribution is also Power series distribution. Based on the same, we define ZITPSD as follows:

Let the probability mass function of a random variable X is given by

$P(X=x)=\{\begin{array}{l}1-\pi +\pi \frac{b_0}{G(\theta )} for x=0\\ \frac{\pi b_x{\theta }^x}{G(\theta )} for x=1,2,3,\mathrm{...},t\end{array}$ …(2.1)

where $G(\theta )={\displaystyle \sum _{y=0}^t{b}_y{\theta }^y}$

Estimation of $\pi$ and $\theta$

Estimation of the parameters using full likelihood function: Suppose a random sample $X_1,X_2,\mathrm{...},X_n$  of size n from ZITPSD is available. Then the likelihood function is given b

$L(\theta ,\pi ;\underset{\_}{x})={{\displaystyle \prod _{i=1}^n\left(1-\pi +\frac{\pi b_0}{G(\theta )}\right)}}^{1-a_i}{\left(\frac{\pi b{x}_i{\theta }^{x_i}}{G(\theta )}\right)}^{a_i}\theta ,\pi >0$   

where $a_i=0$  if $x_i=0$ and $a_i=1$  if $x_i=$ 1,2,3,….t. …(2.2)

then, $\log L(\theta ,\pi ;\underset{\_}{x})=$   $=n_0\log \left(1-\pi +\frac{\pi b_0}{G(\theta )}\right)+{\displaystyle \sum _{i=1}^n{a}_i\log \pi +}{\displaystyle \sum _{i=1}^n{a}_i}\log b{x}_i+{\displaystyle \sum _{i=1}^n{a}_i{x}_i}\log (\theta )-{\displaystyle \sum _{i=1}^n{a}_i\log G(\theta )}$ …(2.3)

Maximum likelihood estimators of $\theta$  and $\pi$ are obtained by solving the following two equations

$\widehat{\pi }=\frac{(n-n_0)G(\widehat{\theta })}{n(G(\widehat{\theta })-b_0)}$                                                       …(2.4)

$\frac{{\displaystyle \sum _{i=1}^n{a}_i{x}_i}}{\widehat{\theta }}=\frac{n_0\widehat{\pi }{b}_0{G}^{\prime }(\widehat{\theta })}{G{(\widehat{\theta })}^2\left(1-\widehat{\pi }+\widehat{\pi }\frac{b_0}{G(\widehat{\theta })}\right)}+\frac{{\displaystyle \sum _{i=1}^n{a}_i{G}^{\prime }(\widehat{\theta })}}{G(\widehat{\theta })}$ , …(2.5)

Substituting $\widehat{\pi }=\frac{(n-n_0)G(\widehat{\theta })}{n(G(\widehat{\theta })-b_0)}$  in eq. (2.5) we get

$\overline{x}=\frac{\widehat{\theta }{G}^{\prime }(\widehat{\theta })}{(G(\widehat{\theta })-b_0)}$ ,                     …(2.6)

which is non-linear equation in $\theta$ , Using Newton-Raphson method first we find $\widehat{\theta }$ , substituting this value of $\widehat{\theta }$  in Eq. (2.4) we find $\widehat{\pi }$ . The Fisher information matrix of $\underset{\_}{\delta }=(\pi ,\theta {)}^{\prime }$  is given by

$I(\underset{\_}{\delta })=\left(\begin{array}{cc}-E\left(\frac{{\partial }^2\log L}{\partial {\pi }^2}\right)& -E\left(\frac{{\partial }^2\log L}{\partial \pi \partial \theta }\right)\\ -E\left(\frac{{\partial }^2\log L}{\partial \pi \partial \theta }\right)& -E\left(\frac{{\partial }^2\log L}{\partial {\theta }^2}\right)\end{array}\right)=\left(\begin{array}{cc}{I}_{11}& I_{12}\\ I_{21}& I_{22}\end{array}\right)$

Where

  $I_{11}=\left(\frac{n\left(G(\theta )-b_0\right)}{\pi \left(G(\theta )-\pi G(\theta )+\pi b_0\right)}\right)$ ,                                           …(2.7)

$I_{12}=\left(\frac{n{b}_0{G}^{\prime }(\theta )}{G(\theta )\left(G(\theta )-\pi G(\theta )+\pi b_0\right)}\right)$                                          …(2.8)

and

$I_{22}=(n\pi b_0\left(\left(\frac{G{(\theta )}^2{G}^{″}(\theta )-2G^{\prime }{(\theta )}^{2}G(\theta )}{G{(\theta )}^4}\right)+\frac{\pi b_0{G}^{\prime }{(\theta )}^2}{G{(\theta )}^4\left(1-\pi +\pi \frac{b_0}{G(\theta )}\right)}\right)$                                  

$+n\pi \left(\frac{G^{\prime }(\theta )}{\theta G(\theta )}\right)+\frac{n\pi \left(1-\frac{b_0}{G(\theta )}\right)\left(G(\theta )G^{″}(\theta )-G^{\prime }{(\theta )}^2\right)}{G{(\theta )}^2}$  …(2.9)

 Assuming that conditions required for asymptotic normality for maximum likelihood estimators are satisfied, we have following theorem:

Theorem 2.1: Let $X_1,X_2,\mathrm{...}{X}_n$  be a random sample from ZITPSD with parameters $\pi$  and $\theta$ . Then the maximum likelihood estimator obtained by solving eq. (2.4) and eq. (2.6), have asymptotic bivariate normal distribution with mean vector $(\pi ,\theta )\text{'}$  and dispersion matrix $I^{-1}(\underset{\_}{\delta })$  for $n$  sufficiently large.

That is as $n\to \infty$ , $\left(\sqrt{n}(\widehat{\pi }-\pi ),\sqrt{n}(\widehat{\theta }-\theta )\right)\to N_2(0,I^{-1}(\underset{\_}{\delta }))$ .

In the following we present conditional likelihood approach and obtain MLEs for $\theta$ .

Conditional likelihood function approach: We observe that the conditional density of $X_i$  given $A_i=a_i$  is independent of inflation parameter $\pi$ , since

$P(X_i=x_i|A_i=a_i)$ $={\left(\frac{b_{x_i}{(\theta )}^{x_i}}{G(\theta )-b_0}\right)}^{a_i}$  …(2.10)

Now the conditional log likelihood function is given by

$\log L^{\ast }(\theta ;\underset{\_}{x})=$ ${\displaystyle \sum _{i=1}^{n-n_0}\log b_{x_i}+{\displaystyle \sum _{i=1}^{n-n_0}{x}_i\log (\theta )}-{\displaystyle \sum _{i=1}^{n-n_0}\log \left(G(\theta )-b_0\right)}}$  …(2.11)

The mle $\tilde{\theta }$  of $\theta$  is the solution to an equation

  $\overline{x}=\frac{\tilde{\theta }{G}^{\prime }(\tilde{\theta })}{(G(\tilde{\theta })-b_0)}$  ,                                     …(2.12)

where $\overline{x}=\frac{{\displaystyle \sum _{i=1}^{n-n_o}{x}_i}}{n-n_0}$  is the mean of the positive observations only. We note that mle of $\theta$  based on full likelihood (eq. 2.6) and based on conditional likelihood (Eq. 2.12) are the same and

$A{V}_{\tilde{\theta }}(\theta )=$ ${\left((n-n_0)\left\{\left(\frac{G^{\prime }(\theta )}{\theta (G(\theta )-b_0)}+\frac{G^{″}(\theta )(G(\theta )-b_0)-G^{\prime }{(\theta )}^2}{{(G(\theta )-b_0)}^2}\right)\right\}\right)}^{-1}$  …(2.13)

Assuming that Cramer-Huzurbazar conditions required for asymptotic normality for MLEs are satisfied, we have following theorem:

Theorem 2.2: Let $X_1,X_2,\mathrm{...}{X}_n$  be a random sample from ZITPSD with parameters $\pi$  and $\theta$ . Then the mle of $\theta$  is solution to the eq. (2.12) and has asymptotic normal distribution with mean $\theta$  and variance $A{V}_{\tilde{\theta }}(\theta )$  for $n$  sufficiently large. That is as $n\to \infty$ , $\left(\sqrt{n}(\tilde{\theta }-\theta )\right)\to N_1(0,A{V}_{\tilde{\theta }}(\theta ))$ .

In the following we present moment estimator of ZITPSD.

Moment estimator of ZITPSD: We have,

$E(X)=$ $\frac{\pi \theta G^{\prime }(\theta )}{G(\theta )}=\pi \mu (\theta )$  

$E(X^2)=$ $\frac{\theta \pi }{G(\theta )}\left(\theta G^{″}(\theta )+G^{\prime }(\theta \right)$  and

$Var(X)=\frac{\theta \pi }{G(\theta )}\left(\theta G^{″}(\theta )+G^{\prime }(\theta )-\frac{\pi \theta {\left(G^{\prime }(\theta \right)}^2}{G(\theta )}\right)$ ,

$={\sigma }^2(\pi ,\theta )$  say.

Let,

  $\overline{X}=\pi \mu (\theta )$  …(2.14)

$\frac{{\displaystyle \sum _{i=1}^n{x}_i{}^2}}{n}=\frac{\pi \theta }{G(\theta )}\left(\theta G^{″}(\theta )+G^{\prime }(\theta \right)$ , …(2.15)

Solving eq. (2.14) and eq. (2.15) we get moment estimators of $\pi$  and $\pi$ .

Theorem 2.3: Let $X_1,X_2,\mathrm{...}{X}_n$  be a random sample from ZITPSD with parameters $\pi$  and $\theta$ . Then the moment estimator of $\pi$  and $\theta$  are obtained by solving in the eq. (2.14) and eq. (2.15). The moment estimator of $\theta$  has asymptotic normal distribution with mean $\pi \mu (\theta )$  and variance $\frac{{\sigma }^2(\pi ,\theta )}{n}$ , for $n$  sufficiently large. That is as $n\to \infty$ , $\left(\sqrt{n}(\overline{\theta }-\theta )\right)\to N_1\left(0,\frac{{\sigma }^2(\pi ,\theta )}{n}\right)$ .

Tests for the parameter $\theta$  of ZITPS distribution

Test based on $\widehat{\theta }$ : Suppose we wish to test $H_0:\theta ={\theta }_0$  vs $H_1:\theta \ne {\theta }_0$ . Let us assume that $\pi$  is known. Therefore, under $H_0$ , from Theorem (2.1) we have

$\left(\widehat{\theta }-{\theta }_0\right)$ ~ $AN\left(0,A{V}_{\widehat{\theta }}(\pi ,{\theta }_0)\right)$ .                                 …(2.16)

Define a test statistic to be $Z_1=\frac{\widehat{\theta }-{\theta }_0}{\sqrt{A{V}_{\widehat{\theta }}(\pi ,{\theta }_0)}}$ . Based on $Z_1$ we define the test ${\psi }_1$  which rejects $H_0$  at α level of significance, if $\left|Z_1\right|>z_{1-\alpha /2}$ , where $z_{1-\alpha /2}$  is the upper $100(\alpha /2)$ ^th percentile of SNV.

Let $\Phi (.)$ be the cumulative distribution function of SNV. Then the power of the test ${\psi }_1$  is given by

${\beta }_{{\psi }_1}(\pi ,\theta )=$ $1-\Phi (B)+\Phi (A)$ ,                                 

where    

  $A=\frac{{\theta }_0-\theta -z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\pi ,{\theta }_0})}{\sqrt{A{V}_{\widehat{\theta }}(\pi ,\theta })}$  and

$B=\frac{{\theta }_0-\theta +z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\pi ,{\theta }_0})}{\sqrt{A{V}_{\widehat{\theta }}(\pi ,\theta })}$ .

However, in practice $\pi$  is unknown. Hence we modify the test statistic by replacing $\pi$  by its maximum likelihood estimator ( ${\widehat{\pi }}_0$ ), when $H_0$  is true. By doing so, we define test ${Z^{\prime }}_1=\frac{\widehat{\theta }-{\theta }_0}{\sqrt{A{V}_{\widehat{\theta }}({\widehat{\pi }}_0,{\theta }_0})}$ , where ${\widehat{\pi }}_0=\frac{(n-n_0)f({\theta }_0)}{n(f({\theta }_0)-b_0)}$

Based on ${Z^{\prime }}_1$ , we propose a test ${{\psi }^{\prime }}_1$  rejects $H_0$  at $\alpha$  level of significance, if $\left|{Z^{\prime }}_1\right|>Z_{1-\alpha /2}$ .

The power of this test is given by

${\beta }_{{{\psi }^{\prime }}_1}(\pi ,\theta )={\displaystyle \sum _{k=0}^n\left(1-\Phi ({\widehat{B}}_k)+\Phi ({\widehat{A}}_k)\right)P(n_0=k)}$ , ...(2.17)

where

${\widehat{B}}_k=\frac{{\theta }_0-\theta +z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}({\widehat{\pi }}_0,{\theta }_0})}{\sqrt{A{V}_{\widehat{\theta }}(\pi ,\theta })}$  , ${\widehat{A}}_k=\frac{{\theta }_0-\theta -z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}({\widehat{\pi }}_0,{\theta }_0})}{\sqrt{A{V}_{\widehat{\theta }}(\pi ,\theta })}$ ,

$P(n_0=k)=\left({}_k{}^n\right)P_0{}^k{(1-P_0)}^{n-k}$ ,with $P_0=\left(1-\pi +\pi \frac{b_0}{f(\theta )}\right)$ . …(2.18)

Below we develop test based on $\tilde{\theta }$ , estimator based on conditional likelihood approach.

Test based on $\tilde{\theta }$ : Theorem (2.5) gives

  $\left(\tilde{\theta }-{\theta }_0\right)$ ~ $AN\left(0,A{V}_{\tilde{\theta }}({\theta }_0)\right)$ . …(2.19)

Hence, we define test statistic $Z_2=\frac{\tilde{\theta }-{\theta }_0}{\sqrt{A{V}_{\tilde{\theta }}({\theta }_0)}}$ . A test based on $Z_2$  which rejects $H_0$  α level of significance, if $\left|Z_2\right|>z_{1-\alpha /2}$ .

The power of the test ${\psi }_2$  is given by

  ${\beta }_{{\psi }_2}(\pi ,\theta )=1-\Phi (B)+\Phi (A)$  , …(2.20)

where, $A=\frac{{\theta }_0-\theta -z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}({\theta }_0)}}{\sqrt{A{V}_{\widehat{\theta }}(\theta })}$ , $B=\frac{{\theta }_0-\theta +z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}({\theta }_0)}}{\sqrt{A{V}_{\widehat{\theta }}(\theta })}$ .      

Test based on the moment estimator $\overline{\theta }$ of $\theta$ : It is clear that the problem of testing $H_0:\theta ={\theta }_0$  vs $H_1:\theta \ne {\theta }_0$  is equivalent to testing $H_0:\mu (\theta )=\mu ({\theta }_0)$  vs $H_1:\mu (\theta )\ne \mu ({\theta }_0)$ , where $\mu (\theta )=\frac{\theta G^{\prime }(\theta )}{G(\theta )}$ . We have from Theorem (2.3), sample mean is consistent and asymptotically normal for the population mean.

That is $\overline{X}$ ~ $AN\left(\pi \mu (\theta ),\frac{{\sigma }^2(\pi ,\theta )}{n}\right)$ .

Therefore, under $H_0$ , we have

$\sqrt{n}\left(\frac{\overline{X}}{\pi }-\mu ({\theta }_0)\right)$ ~ $AN\left(0,\frac{{\sigma }^2(\pi ,{\theta }_0)}{{\pi }^2}\right)$ .

Define test statistic

$Z_3=\frac{\sqrt{n}\left(\frac{\overline{X}}{\pi }-\mu ({\theta }_0)\right)}{\sqrt{\frac{{\sigma }^2(\pi ,{\theta }_0)}{{\pi }^2}}}$ ~ $N\left(0,1\right)$ , when $\pi$  is known.

The test ${\psi }_3$  rejects $H_0$  at α level of significance if $\left|Z_3\right|>z_{1-\alpha /2}$ .

That is, reject $H_0$ if $\left(\frac{\sqrt{n}\left|\frac{\overline{X}}{\pi }-\mu ({\theta }_0)\right|}{\sqrt{\frac{{\sigma }^2(\pi ,{\theta }_0)}{{\pi }^2}}}\right)>z_{1-\alpha /2}$ .

The power of the test ${\psi }_3$ is given by

${\beta }_{{\psi }_3}(\pi ,\theta )=$ $1-\Phi (B^{\prime })+\Phi (A^{\prime })$ , …(2.21)

where $A^{\prime }=\frac{\pi \left(\mu ({\theta }_0)-z_{1-\alpha /2}\sqrt{\frac{{\sigma }^2(\pi ,{\theta }_0)}{n{\pi }^2}}\right)-\pi \mu (\theta )}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{n}}}$  and

$B^{\prime }=\frac{\pi \left(\mu ({\theta }_0)+z_{1-\alpha /2}\sqrt{\frac{{\sigma }^2(\pi ,{\theta }_0)}{n{\pi }^2}}\right)-\pi \mu (\theta )}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{n}}},$  

If $\pi$  is unknown, we modify the test statistic by replacing $\pi$  by its estimate $({\widehat{\pi }}_0)$  under $H_0$ . By doing so, we define test statistic

${Z^{\prime }}_3=\frac{\sqrt{n}\left(\frac{\overline{X}}{{\widehat{\pi }}_0}-\mu ({\theta }_0)\right)}{\sqrt{\frac{{\sigma }^2({\widehat{\pi }}_0,{\theta }_0)}{{\widehat{\pi }}_0{}^2}}}$ , …(2.22)

where ${\widehat{\pi }}_0$  is given by ${\widehat{\pi }}_0=\frac{\overline{X}}{\mu ({\theta }_0)}$ .

Based on ${Z^{\prime }}_3$  we propose a test ${{\psi }^{\prime }}_3$  which rejects $H_0$  at α level of significance if $\left|{Z^{\prime }}_3\right|>z_{1-\alpha /2}$ .

The power of the test is given by

${\beta }_{{{\psi }^{\prime }}_3}(\pi ,\theta )=$ ${\displaystyle \sum _{k=0}^n\left(1-\Phi ({B^{″}}_k)+\Phi ({A^{″}}_k)\right)P(n_0=k)}$ , …(2.23)

where

${A^{″}}_k=\frac{{\widehat{\pi }}_0\left(\mu ({\theta }_0)-z_{1-\alpha /2}\sqrt{\frac{{\sigma }^2({\widehat{\pi }}_0,{\theta }_0)}{n{\pi }^2}}\right)-\pi \mu (\theta )}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{n}}}$  

${B^{″}}_k=\frac{{\widehat{\pi }}_0\left(\mu ({\theta }_0)+z_{1-\alpha /2}\sqrt{\frac{{\sigma }^2({\widehat{\pi }}_0,{\theta }_0)}{n{\pi }^2}}\right)-\pi \mu (\theta )}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{n}}}$

and $P(n_0=k)=\left({}_k{}^n\right)P_0{}^k{(1-P_0)}^{n-k}$ , with $P_0=\left(1-\pi +\pi \frac{a_0}{f(\theta )}\right)$ .

Using the tests developed above, we can define two sided asymptotic confidence intervals for $\theta$ , by inverting acceptance regions of the tests appropriately. Below we report the same.

Asymptotic confidence interval for the parameter $\theta$

Asymptotic confidence interval for $\theta$  based on the test ${\psi }_1$ is given by

$\left(\widehat{\theta }-z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })},\widehat{\theta }+z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })}\right)$  …(2.24)

where, $A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })$  is an estimate of asymptotic variance of $\widehat{\theta }$  and asymptotic confidence interval for θ based on the test ${\psi }_2$  is given by

  $\left(\tilde{\theta }-z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}(\tilde{\theta })},\tilde{\theta }+z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}(\tilde{\theta })}\right)$  …(2.25)

where $A{V}_{\tilde{\theta }}(\tilde{\theta })$  is an estimate of the asymptotic variance of $\tilde{\theta }$  as given in the eq. (2.13) .

Asymptotic confidence interval for $\theta$  based on the test ${\psi }_3$  is given by

$\left(\frac{\overline{X}}{\widehat{\pi }}-z_{1-\alpha /2}\sqrt{A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })},\frac{\overline{X}}{\widehat{\pi }}+z_{1-\alpha /2}\sqrt{A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })}\right)$ , …(2.26)

where $A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })$  = $\frac{\sqrt{n}\left(\frac{\overline{X}}{\pi }-\mu ({\theta }_{})\right)}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{{\pi }^2}}}$ .

In the following we study inference for zero-inflated truncated poisson distribution using results reported in the earlier.

Truncated samples from discrete distributions arise in numerous situations where counts of zero are not observed. As an example, consider the distribution of the number of children per family in developing nations, where records are maintained only if there is at least a child in the family. The number of childless families remains unknown. The resulting sample is thus truncated with zero class missing. In continuous distribution, a sample of this type would be described as singly left truncated. In other situations, sample from discrete distributions might be censored on the right.

In this section, we consider zero-inflated truncated Poisson distribution truncated at right at the support point $\text{'}t\text{'}$  onwards, where $\text{'}t\text{'}$  is known. Moments, maximum likelihood estimators, Fisher information matrix for full and conditional likelihood are provided. We provide three tests for testing the parameter of the ZITPD.

Consider the probability mass function of truncated Poisson distribution (TPD) truncated at the support point $\text{'}t\text{'}$  onwards. The probability mass function of TPD is given by

$P(X=x)=\frac{e^{-\theta }{\theta }^x}{x!(1-P(X>t)},forx=0,1,2,\mathrm{...},t$

$=\frac{e^{-\theta }{\theta }^x}{x!\left({\displaystyle \sum _{y=0}^t\frac{e^{-\theta }{\theta }^y}{y!}}\right)},$                               

$=\frac{{\theta }^x}{x!A(\theta )},$  where $A(\theta )=\left({\displaystyle \sum _{y=0}^t\frac{{\theta }^y}{y!}}\right)$  

Using this truncated distribution, we define the zero-inflated truncated Poisson distribution truncated at $\text{'}t\text{'}$  onwards.

The probability mass function of ZITP distribution is given by

$P(X=x)=\{\begin{array}{l}(1-\pi )+\frac{\pi }{A(\theta )} for x=0\\ \frac{\pi {\theta }^x}{x!A(\theta )} for x=1,2,3,\mathrm{...},t\end{array}$ and $\theta >0,0<\pi <1$  …(3.1)

Estimation of the parameters $\pi$  and $\theta$  

Estimation of the parameters using full likelihood function

Let $X_1,X_2,\mathrm{...},X_n$  be a random sample observed from zero-inflated truncated Poisson distribution truncated at $\text{'}t$  onwards, where $\text{'}t$  is the point in the support defined in the above probability mass function. Then the likelihood function is given by

$L(\theta ,\pi ;\underset{\_}{x})={{\displaystyle \prod _{i=1}^n\left(1-\pi +\frac{\pi }{A(\theta )}\right)}}^{1-a_i}{\left(\frac{\pi {\theta }^x}{x!A(\theta )}\right)}^{a_i}\theta ,\pi >0$  

The corresponding log likelihood function is given by

$\log$

$+$  …(3.2)

To find MLEs of $\theta$  and $\pi$ , we differentiate the eq. (3.2) with respective $\pi$  and $\theta$ , and then equating to zero we get

$\widehat{\pi }=\frac{(n-n_0)A(\widehat{\theta })}{n\left(A(\widehat{\theta })-1\right)}$                                                      …(3.3)

and $\frac{{\displaystyle \sum _{i=1}^n{a}_i{x}_i}}{\widehat{\theta }}=\frac{\pi n_0\left(A^{\prime }(\widehat{\theta })\right)}{\left(1-\pi +\frac{\pi }{A(\widehat{\theta })}\right){\left(A(\widehat{\theta })\right)}^2}+\frac{{\displaystyle \sum _{i=1}^n{a}_i}{A}^{\prime }(\widehat{\theta })}{A(\widehat{\theta })}$

$=\frac{A^{\prime }(\widehat{\theta })}{A(\widehat{\theta })}\left((n-n_0)+\frac{\widehat{\pi }{n}_0}{\left(1-\pi +\frac{\widehat{\pi }}{A(\widehat{\theta })}\right)A(\widehat{\theta })}\right)$

$\frac{{\displaystyle \sum _{i=1}^n{a}_i{x}_i}}{\theta }=\frac{A^{\prime }(\theta )}{A(\theta )}\left(\frac{(n-n_0)(1-\pi )A(\theta )+n\pi }{(1-\pi )A(\theta )+\pi }\right)$  …(3.4)

Substituting $\widehat{\pi }=\frac{(n-n_0)A(\widehat{\theta })}{n\left(A(\widehat{\theta })-1\right)}$  in the above equation we have

$\frac{{\displaystyle \sum _{i=1}^n{a}_i{x}_i}}{\widehat{\theta }}=\frac{(n-n_0)A^{\prime }(\widehat{\theta })}{\left(A(\widehat{\theta })-1\right)}$ ,

${\displaystyle \sum _{i=1}^n{a}_i{x}_i\left(A(\widehat{\theta })-1\right)-(n-n_0)\widehat{\theta }{A}^{\prime }(\widehat{\theta })=0}$ , …(3.5)

which is non-linear equation in $\widehat{\theta }$ . Therefore, we use a numerical technique to solve it. Let

$h(\widehat{\theta })={\displaystyle \sum _{i=1}^n{a}_i{x}_i\left(A(\widehat{\theta })-1\right)-(n-n_0)\widehat{\theta }{A}^{\prime }(\widehat{\theta })}$  and

$h^{\prime }(\widehat{\theta })={\displaystyle \sum _{i=1}^n{a}_i{x}_i{A}^{\prime }(\widehat{\theta })-(n-n_0)(\widehat{\theta }{A^{\prime }}^{\prime }(\widehat{\theta })}+A^{\prime }(\widehat{\theta })$ .

Using Newton-Raphson iterative formula ${\widehat{\theta }}_{i+1}={\widehat{\theta }}_i-\frac{h(\widehat{\theta })}{h^{\prime }(\widehat{\theta })},i=0,1,2,\mathrm{...}$ with suitable initial value of ${\theta }_0$  we get $\widehat{\theta }$ . Substituting this value of $\widehat{\theta }$  in eq. (3.3), we get the value of $\widehat{\pi }$ .

In the following we find the elements of Fisher information matrix

Here we have

$\frac{\partial \log L}{\partial \pi }=\frac{n_0\left(-1+\frac{1}{A(\theta )}\right)}{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}+\frac{{\displaystyle \sum _{i=1}^n{a}_i}}{\pi }$ ,

$\frac{{\partial }^2\log L}{\partial {\pi }^2}=-\frac{n_0{\left(-1+\frac{1}{A(\theta )}\right)}^2}{{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}^2}-\frac{{\displaystyle \sum _{i=1}^n{a}_i}}{{\pi }^2}$ ,

$-E\left(\frac{{\partial }^2\log L}{\partial {\pi }^2}\right)=\frac{E(n_0){\left(-1+\frac{1}{A(\theta )}\right)}^2}{{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}^2}+\frac{E\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}{{\pi }^2}$ ,

$=\frac{n{\left(-1+\frac{1}{A(\theta )}\right)}^2}{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}+\frac{n\left(1-\frac{1}{A(\theta )}\right)}{\pi }$ ,

$-E\left(\frac{{\partial }^2\log L}{\partial {\pi }^2}\right)=\frac{n{\left(-1+\frac{1}{A(\theta )}\right)}^2}{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}-\frac{n\left(-1+\frac{1}{A(\theta )}\right)}{\pi }$ ,                                   

$I_{11}=\frac{n\left(A(\theta )-1\right)}{\pi \left(A(\theta )-\pi A(\theta )+\pi \right)}$ . …(3.6)

Now

$\frac{\partial \log L}{\partial \pi }=\frac{n_0\left(-1+\frac{1}{A(\theta )}\right)}{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}+\frac{{\displaystyle \sum _{i=1}^n{a}_i}}{\pi }$ ,

$\frac{{\partial }^2\log L}{\partial \pi \partial \theta }=\frac{-n_0\frac{A^{\prime }(\theta )}{{\left(A(\theta )\right)}^2}\left(\left(1-\pi +\frac{\pi }{A(\theta )}\right)-\pi \left(-1+\frac{1}{A(\theta )}\right)\right)}{{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}^2}$ ,

$-E\left(\frac{{\partial }^2\log L}{\partial \pi \partial \theta }\right)=\frac{E(n_0)\left(\frac{A^{\prime }(\theta )}{A{(\theta )}^2}\right)\left(\left(1-\pi +\frac{\pi }{A(\theta )}\right)-\pi \left(-1+\frac{1}{A(\theta )}\right)\right)}{{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}^2}$ ,

$-E\left(\frac{{\partial }^2\log L}{\partial \pi \partial \theta }\right)=I_{12}=\frac{n{A}^{\prime }(\theta )}{A(\theta )\left(A(\theta )-\pi A(\theta )+\pi \right)}$  …(3.7)

Further differentiating eq. (3.2) twice with respect to $\theta$ , we get

$\frac{{\partial }^2\log L}{\partial {\theta }^2}=-\frac{n_0\pi \left\{\left(1-\pi +\frac{\pi }{A(\theta )}\right)\left(\frac{A{(\theta )}^2{A}^{″}(\theta )-2A^{\prime }{(\theta )}^{2}A(\theta )}{f{(\theta )}^4}\right)-\frac{A^{\prime }{(\theta )}^2}{A{(\theta )}^2}\left(-\frac{\pi }{A{(\theta )}^2}\right)\right\}}{{\left(1-\pi +\frac{\pi }{A(\theta )}\right)}^2}$

$-\frac{{\displaystyle \sum _{i=1}^n{a}_i{x}_i}}{{\theta }^2}-\frac{{\displaystyle \sum _{i=1}^n{a}_i\left(A(\theta )A^{″}(\theta )-A^{\prime }{(\theta )}^2\right)}}{A{(\theta )}^2}$ .

Therefore,

$-E\left(\frac{{\partial }^2\log L}{\partial {\theta }^2}\right)=$ $(n\pi \left(\left(\frac{A{(\theta )}^2{A}^{″}(\theta )-2A^{\prime }{(\theta )}^{2}A(\theta )}{A{(\theta )}^4}\right)+\frac{\pi A^{\prime }{(\theta )}^2}{A{(\theta )}^4\left(1-\pi +\frac{\pi }{A(\theta )}\right)}\right)$

$+\frac{n\pi A^{\prime }(\theta )}{\theta A(\theta )}+\frac{n\pi \left(1-\frac{1}{A(\theta )}\right)\left(A(\theta )A^{″}(\theta )-A^{\prime }{(\theta )}^2\right)}{A{(\theta )}^2})$ .

Hence,

                $I_{22}=$ $(n\pi \left(\left(\frac{A{(\theta )}^2{A}^{″}(\theta )-2A^{\prime }{(\theta )}^{2}A(\theta )}{A{(\theta )}^4}\right)+\frac{\pi A^{\prime }{(\theta )}^2}{A{(\theta )}^4\left(1-\pi +\frac{\pi }{A(\theta )}\right)}\right)$

$+\frac{n\pi A^{\prime }(\theta )}{\theta A(\theta )}+\frac{n\pi \left(1-\frac{1}{A(\theta )}\right)\left(A(\theta )A^{″}(\theta )-A^{\prime }{(\theta )}^2\right)}{A{(\theta )}^2})$ .

The asymptotic variance of $\widehat{\pi }$  and $\widehat{\theta }$  are

$A{V}_{\widehat{\pi }}(\pi ,\theta )=I^{11}={\left(I_{11}-\frac{I_{12}}{I_{22}}\right)}^{-1}$ .

$A{V}_{\widehat{\theta }}(\pi ,\theta )=I^{22}={\left(I_{22}-\frac{I_{12}}{I_{11}}\right)}^{-1}$  .                                                … (3.8)

1. Conditional likelihood function approach

The conditional likelihood function is given by

$L^{\ast }(\theta ;\underset{\_}{x})={\displaystyle \prod _{i=1}^n{\left(\frac{{\theta }^{x_i}}{x_i!(A(\theta )-1)}\right)}^{a_i},}\theta >0$                   …(3.9)

The corresponding log likelihood function is given by

$\log$  …(3.10)

The corresponding mle $\tilde{\theta }$  is the solution to an equation

$\overline{x}=\frac{\tilde{\theta }A(\tilde{\theta })}{A(\tilde{\theta })-1}$  …(3.11)

Now consider,

$\frac{{\partial }^2\log L^{*}}{\partial {\theta }^2}=-{\displaystyle \sum _{i=1}^{n-n_0}\frac{x_i}{{\theta }^2}}-{\displaystyle \sum _{i=1}^{n-n_0}\frac{\left(A(\theta )-1\right)A^{″}(\theta )-A^{\prime }{(\theta )}^2}{{\left(A(\theta )-1\right)}^2}}$

$-E\left(\frac{{\partial }^2\log L^{*}}{\partial {\theta }^2}\right)=E\left({\displaystyle \sum _{i=1}^{n-n_0}\frac{x_i}{{\theta }^2}}\right)+E\left({\displaystyle \sum _{i=1}^{n-n_0}\frac{\left(A(\theta )-1\right)A^{″}(\theta )-A^{\prime }{(\theta )}^2}{{\left(A(\theta )-1\right)}^2}}\right)$

$=\frac{(n-n_0)\theta A^{\prime }(\theta )}{{\theta }^2(A(\theta )-1)}+\frac{(n-n_0)\left\{(A(\theta )-1)A^{″}(\theta )-A^{\prime }{(\theta )}^2\right\}}{{\left(A(\theta )-1\right)}^2}$

$=\frac{(n-n_0)}{(A(\theta )-1)}\left(\frac{A^{\prime }(\theta )}{\theta }+\frac{A^{″}(\theta )(A(\theta )-1)-A^{\prime }{(\theta )}^2}{(A(\theta )-1)}\right)$ . … (3.12)

Therefore, asymptotic variance of $\tilde{\theta }$  is different than the asymptotic variance of estimate of $\theta$  based on the standard likelihood approach. The same is given by

$A{V}_{\tilde{\theta }}(\theta )={\left(\frac{(n-n_0)}{\left(A(\theta )-1\right)}\left(\frac{A^{\prime }(\theta )}{\theta }+\frac{(A(\theta )-1))A^{″}(\theta )-{(A^{\prime }(\theta ))}^2}{\left(A(\theta )-1\right)}\right)\right)}^{-1}$  … (3.13)

2. Moment estimator of ZITP distribution

 Mean = $E(X)=$ $\frac{\pi \theta A^{\prime }(\theta )}{A(\theta )}$                                  …(3.14)

$E(X^2)=$ $\frac{\theta \pi }{A(\theta )}\left(\theta A^{″}(\theta )+A^{\prime }(\theta )\right)$

$Var(X)=\frac{\theta \pi }{A(\theta )}\left(\theta A^{″}(\theta )+A^{\prime }(\theta )-\frac{\pi \theta A^{\prime }{(\theta )}^2}{A(\theta )}\right)={\sigma }^2(\pi ,\theta )$  say       …(3.15)

$\overline{x}=\frac{\pi \theta A^{\prime }(\theta )}{A(\theta )}$  …(3.16)

$\frac{{\displaystyle \sum _{i=1}^n{x}_i{}^2}}{n}=\frac{\theta \pi }{A(\theta )}\left(\theta A^{″}(\theta )+A^{\prime }(\theta )\right)$  …(3.17)

Solving eq. (3.16) and eq. (3.17), we get moment estimators of $\pi$  and $\theta$ .

Tests for the parameter $\theta$  of ZITP distribution

Suppose we want to test $H_0:\theta ={\theta }_0$  vs $H_1:\theta \ne {\theta }_0$ , (assuming $\pi$  is unknown) [4]

1. Test based on $\widehat{\theta }$

$Z_4=\frac{\widehat{\theta }-{\theta }_0}{\sqrt{A{V}_{\widehat{\theta }}({\widehat{\pi }}_0,{\theta }_0})}$                                   …(3.18)

where $A{V}_{\widehat{\theta }}({\widehat{\pi }}_0,{\theta }_0)$ is defined in eq. (3.8).The test ${\psi }_4$  rejects $H_0$ , if $\left|Z_4\right|>z_{1-\alpha /2}$ .

2. Test based on $\widehat{\theta }$

The test statistic here is $Z_5=\frac{\tilde{\theta }-{\theta }_0}{\sqrt{A{V}_{\tilde{\theta }}({\theta }_0)}}$ , …(3.19)

Where, $A{V}_{\tilde{\theta }}({\theta }_0)$ is as defined in eq. (3.13). The test ${\psi }_5$ rejects $H_0$ if $\left|Z_5\right|>z_{1-\alpha /2}$ .

3. Test based on sample mean

The test statistic

$Z_6=\frac{\sqrt{n}\left(\frac{\overline{X}}{{\widehat{\pi }}_0}-{\theta }_0\right)}{\sqrt{{\widehat{\pi }}_0{}^{-2}A{V}_{\overline{X}}({\widehat{\pi }}_0,{\theta }_0)}}$  , …(3.20)

where ${\widehat{\pi }}_0=\overline{X}\left(\frac{A({\theta }_0)}{{\theta }_0{A}^{\prime }({\theta }_0)}\right)$

Power of the test is given by

${\beta }_{{\psi }_6}(\widehat{\pi },\theta )$   $={\displaystyle \sum _{k=0}^n\left(1-\Phi ({\widehat{B}}_k)+\Phi ({\widehat{A}}_k)\right)P(n_0=k)}$

where , ${\widehat{B}}_k=\frac{{\widehat{\pi }}_0\left({\theta }_0+z_{1-\alpha /2}\sqrt{{\widehat{\pi }}_0{}^{-2}A{V}_{\overline{X}}({\widehat{\pi }}_0,{\theta }_0)}\right)-\pi \theta }{\sqrt{A{V}_{\overline{X}}(\pi ,{\theta }_0)}}$ ,

${\widehat{A}}_k=\frac{{\widehat{\pi }}_0\left({\theta }_0-z_{1-\alpha /2}\sqrt{{\widehat{\pi }}_0{}^{-2}A{V}_{\overline{X}}(\widehat{\pi },{\theta }_0)}\right)-\pi \theta }{\sqrt{A{V}_{\overline{X}}(\pi ,{\theta }_0)}}$  and

$P(n_0=k)=\left({}_k{}^n\right)P_0{}^k{(1-P_0)}^{n-k}$ , with $P_0=1-\pi +\frac{\pi }{A(\theta )}$

Asymptotic confidence interval for the parameter $\theta$

Asymptotic confidence interval for $\theta$  based on the test ${\psi }_4$  is given by

$\left(\widehat{\theta }-z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })},\widehat{\theta }+z_{1-\alpha /2}\sqrt{A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })}\right)$  …(3.21)

where, $A{V}_{\widehat{\theta }}(\widehat{\pi },\widehat{\theta })$  is an estimate of asymptotic variance of $\widehat{\theta }$  and asymptotic confidence interval for q based on the test ${\psi }_5$  is given by

$\left(\tilde{\theta }-z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}(\tilde{\theta })},\tilde{\theta }+z_{1-\alpha /2}\sqrt{A{V}_{\tilde{\theta }}(\tilde{\theta })}\right)$  …(3.22)

where $A{V}_{\tilde{\theta }}(\tilde{\theta })$  is an estimate of the asymptotic variance of $\tilde{\theta }$  as given in the eq. (3.13) .

Asymptotic confidence interval for $\theta$  based on the test ${\psi }_6$  is given by

$\left(\frac{\overline{X}}{\widehat{\pi }}-z_{1-\alpha /2}\sqrt{A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })},\frac{\overline{X}}{\widehat{\pi }}+z_{1-\alpha /2}\sqrt{A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })}\right)$ , …(3.23)

where $A{V}_{\overline{\theta }}(\widehat{\pi },\overline{\theta })$  = $\frac{\sqrt{n}\left(\frac{\overline{X}}{\pi }-\mu ({\theta }_{})\right)}{\sqrt{\frac{{\sigma }^2(\pi ,\theta )}{{\pi }^2}}}$ .